Analytical Chem istry - DePauw University

(a)(b)Phase 2Phase 12.95 3.00 3.05 3.10 3.15 3.20 3.25Mass of Pennies (g)2.0Analytical Chemistry1.0moles metal = moles ligandpH0.8absorbance0.60.4metal in excessligand in excessIn –indicatoris color of In –0.20.00.0 0.2 0.4 0.6 0.8 1.0X LpH = pK a,HInindicator’scolor transitionrangeHInindicatoris color of HIn

Brief Table of Contents1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.Preface.......................................xviiIntroduction to Analytical Chemistry.. . . . . . . . . . . . . . . 1Basic Tools of Analytical Chemistry................ 13The Vocabulary of Analytical Chemistry.. . . . . . . . . . . . 41Evaluating Analytical Data....................... 63Standardizing Analytical Methods................. 153Equilibrium Chemistry .. . . . . . . . . . . . . . . . . . . . . . . . 209Collecting and Preparing Samples................. 285Gravimetric Methods.. . . . . . . . . . . . . . . . . . . . . . . . . . 355Titrimetric Methods............................ 411Spectroscopic Methods......................... 543Electrochemical Methods........................ 667Chromatographic and Electrophoretic Methods..... 783Kinetic Methods............................... 881Developing a Standard Method.. . . . . . . . . . . . . . . . . . 941Quality Assurance............................. 991Additional Resources.. . . . . . . . . . . . . . . . . . . . . . . . . 1017Appendix................................... 1067

Detailed Table of ContentsPreface.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xviiA Organization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xviiiB Role of Equilibrium Chemistry ............................. xviiiC Computational Software . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xixD How to Use The Electronic Textbook’s Features. . . . . . . . . . . . . . . . . . xixE Acknowledgments ........................................ xxi1. Introduction to Analytical Chemistry..............11A What is Analytical Chemistry? ................................21B The Analytical Perspective ...................................51C Common Analytical Problems ................................81D Key Terms ...............................................81E Chapter Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .91F Problems ................................................91G Solutions to Practice Exercises ...............................102. Basic Tools of Analytical Chemistry.. . . . . . . . . . . . . 132A Measurements in Analytical Chemistry ........................142A.1 Units of Measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142A.2 Uncertainty in Measurements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162B Concentration ...........................................182B.1 Molarity and Formality ...........................................192B.2 Normality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202B.3 Molality .......................................................202B.4 Weight, Volume, and Weight-to-Volume Ratios .........................202B.5 Parts Per Million and Parts Per Billion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202B.6 Converting Between Concentration Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212B.7 p-Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222C Stoichiometric Calculations .................................232D Basic Equipment .........................................262D.1 Equipment for Measuring Mass .....................................262D.2 Equipment for Measuring Volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272D.3 Equipment for Drying Samples .....................................292E Preparing Solutions .......................................302E.1 Preparing Stock Solutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302E.2 Preparing Solutions by Dilution .....................................322F Spreadsheets and Computational Software ......................332G The Laboratory Notebook ..................................34

2H Key Terms ..............................................342I Chapter Summary. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .352J Problems ................................................352K Solutions to Practice Exercises ...............................383. The Vocabulary of Analytical Chemistry.. . . . . . . . . 413A Analysis, Determination and Measurement . . . . . . . . . . . . . . . . . . . . .423B Techniques, Methods, Procedures, and Protocols .................433C Classifying Analytical Techniques ............................443D Selecting an Analytical Method ..............................453D.1 Accuracy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453D.2 Precision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463D.3 Sensitivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463D.4 Specificity and Selectivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 473D.5 Robustness and Ruggedness. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503D.6 Scale of Operation ...............................................503D.7 Equipment, Time, and Cost . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 523D.8 Making the Final Choice ..........................................523E Developing the Procedure ..................................533E.1 Compensating for Interferences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533E.2 Calibration .....................................................543E.3 Sampling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553E.4 Validation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553F Protocols ...............................................553G The Importance of Analytical Methodology ....................573H Key Terms ..............................................573I Chapter Summary. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .583J Problems ................................................583K Solutions to Practice Exercises ...............................614. Evaluating Analytical Data .. . . . . . . . . . . . . . . . . . . . 634A Characterizing Measurements and Results ......................644A.1 Measures of Central Tendency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 644A.2 Measures of Spread . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 664B Characterizing Experimental Errors ...........................684B.1 Errors Affecting Accuracy ..........................................684B.2 Errors Affecting Precision ..........................................734B.3 Error and Uncertainty . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 754C Propagation of Uncertainty .................................764C.1 A Few Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 774C.2 Uncertainty When Adding or Subtracting .............................774C.3 Uncertainty When Multiplying or Dividing . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

4C.4 Uncertainty for Mixed Operations ...................................794C.5 Uncertainty for Other Mathematical Functions .........................804C.6 Is Calculating Uncertainty Actually Useful? ............................814D The Distribution of Measurements and Results ..................834D.1 Populations and Samples ..........................................844D.2 Probability Distributions for Populations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 844D.3 Confidence Intervals for Populations .................................904D.4 Probability Distributions for Samples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 924D.5 Confidence Intervals for Samples ....................................964D.6 A Cautionary Statement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 984E Statistical Analysis of Data ..................................984E.1 Significance Testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 994E.2 Constructing a Significance Test .....................................994E.3 One-Tailed and Two-Tailed Significance Tests . . . . . . . . . . . . . . . . . . . . . . . . . 1004E.4 Errors in Significance Testing ......................................1024F Statistical Methods for Normal Distributions ...................1024F.1 Comparing X to μ ......................................................1024F.2 Comparing s 2 to σ 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1054F.3 Comparing Two Sample Variances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1064F.4 Comparing Two Sample Means . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1074F.5 Outliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1144G Detection Limits ........................................1174H Using Excel and R to Analyze Data ..........................1204H.1 Excel ........................................................1204H.2 R ...........................................................1244I Key Terms ..............................................1324J Chapter Summary .......................................1324K Problems ..............................................1334L Solutions to Practice Exercises ..............................1435. Standardizing Analytical Methods .. . . . . . . . . . . . . 1535A Analytical Standards. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1545A.1 Primary and Secondary Standards . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1545A.2 Other Reagents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1545A.3 Preparing Standard Solutions ......................................1555B Calibrating the Signal (S total ) ...............................1565C Determining the Sensitivity (k A ) ............................1565C.1 Single-Point versus Multiple-Point Standardizations . . . . . . . . . . . . . . . . . . . . 1575C.2 External Standards ..............................................1585C.3 Standard Additions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1615C.4 Internal Standards . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1675D Linear Regression and Calibration Curves .....................1705D.1 Linear Regression of Straight Line Calibration Curves ...................172

5D.2 Unweighted Linear Regression with Errors in y ........................1725D.3 Weighted Linear Regression with Errors in y ..........................1825D.4 Weighted Linear Regression with Errors in Both x and y .................1855D.5 Curvilinear and Multivariate Regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1855E Blank Corrections .......................................1865F Using Excel and R for a Regression Analysis ....................1885F.1 Excel .........................................................1885F.2 R ............................................................1925G Key Terms .............................................1975H Chapter Summary .......................................1975I Problems ...............................................1985J Solutions to Practice Exercises ...............................2026. Equilibrium Chemistry .. . . . . . . . . . . . . . . . . . . . . . 2096A Reversible Reactions and Chemical Equilibria . . . . . . . . . . . . . . . . . .2106B Thermodynamics and Equilibrium Chemistry ..................2116C Manipulating Equilibrium Constants ........................2136D Equilibrium Constants for Chemical Reactions .................2146D.1 Precipitation Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2146D.2 Acid–Base Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2156D.3 Complexation Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2206D.4 Oxidation–Reduction (Redox) Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2226E Le Châtelier’s Principle ....................................2266F Ladder Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2286F.1 Ladder Diagrams for Acid–Base Equilibria . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2286F.2 Ladder Diagrams for Complexation Equilibria . . . . . . . . . . . . . . . . . . . . . . . . . 2326F.3 Ladder Diagram for Oxidation/Reduction Equilibria . . . . . . . . . . . . . . . . . . . . 2346G Solving Equilibrium Problems ..............................2356G.1 A Simple Problem—Solubility of Pb(IO 3 ) 2 .....................................2366G.2 A More Complex Problem—The Common Ion Effect . . . . . . . . . . . . . . . . . . 2376G.3 A Systematic Approach to Solving Equilibrium Problems . . . . . . . . . . . . . . . . 2396G.4 pH of a Monoprotic Weak Acid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2416G.5 pH of a Polyprotic Acid or Base ....................................2436G.6 Effect of Complexation on Solubility ................................2466H Buffer Solutions ........................................2496H.1 Systematic Solution to Buffer Problems ..............................2496H.2 Representing Buffer Solutions with Ladder Diagrams . . . . . . . . . . . . . . . . . . . 2526H.3 Preparing Buffers ...............................................2536I Activity Effects ..........................................2546J Using Excel and R to Solve Equilibrium Problems ...............2606J.1 Excel ..........................................................2606J.2 R ............................................................263

8B.1 Theory and Practice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3588B.2 Quantitative Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3728B.2 Qualitative Applications ..........................................3798B.3 Evaluating Precipitation Gravimetry. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3798C Volatilization Gravimetry .................................3818C.1 Theory and Practice .............................................381Thermogravimetry ....................................................3818C.2 Quantitative Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3868C.3 Evaluating Volatilization Gravimetry ................................3908D Particulate Gravimetry ...................................3908D.1 Theory and Practice .............................................3908D.2 Quantitative Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3928D.3 Evaluating Particulate Gravimetry ..................................3948E Key Terms .............................................3948F Chapter Summary ........................................3958G Problems ..............................................3958H Solutions to Practice Exercises ..............................4059. Titrimetric Methods.. . . . . . . . . . . . . . . . . . . . . . . . . 4119A Overview of Titrimetry ...................................4129A.1 Equivalence Points and End points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4129A.2 Volume as a Signal ..............................................4129A.3 Titration Curves ................................................4149A.4 The Buret .....................................................4169B Acid–Base Titrations .....................................4179B.1 Acid–Base Titration Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4189B.2 Selecting and Evaluating the End point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4269B.3 Titrations in Nonaqueous Solvents ..................................4349B.5 Qualitative Applications ..........................................4509B.6 Characterization Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4509B.7 Evaluation of Acid–Base Titrimetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4549C Complexation Titrations ..................................4599C.1 Chemistry and Properties of EDTA .................................4599C.2 Complexometric EDTA Titration Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4629C.3 Selecting and Evaluating the End point ..............................4689C.4 Quantitative Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4739C.5 Evaluation of Complexation Titrimetry ..............................4789D Redox Titrations ........................................4789D.1 Redox Titration Curves ..........................................4799D.2 Selecting and Evaluating the End point ..............................4859D.3 Quantitative Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4919D.4 Evaluation of Redox Titrimetry ....................................5029E Precipitation Titrations ...................................502

12A.4 Electrophoretic Separations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78812B General Theory of Column Chromatography .................78912B.1 Chromatographic Resolution ....................................79112B.2 Solute Retention Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79212B.3 Selectivity ....................................................79512B.4 Column Efficiency .............................................79512B.5 Peak Capacity .................................................79712B.6 Asymmetric Peaks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79812C Optimizing Chromatographic Separations. . . . . . . . . . . . . . . . . . . .79912C.1 Using the Retention factor to Optimize Resolution ....................80012C.2 Using Selectivity to Optimize Resolution ............................80112C.3 Using Column Efficiency to Optimize Resolution .....................80312D Gas Chromatography ...................................80712D.1 Mobile Phase .................................................80812D.2 Chromatographic Columns ......................................80812D.3 Sample Introduction ...........................................81112D.4 Temperature Control ...........................................81512D.5 Detectors for Gas Chromatography ................................81512D.6 Quantitative Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81812D.7 Qualitative Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82212D.8 Evaluation ...................................................82612E High-Performance Liquid Chromatography ..................82712E.1 HPLC Columns ...............................................82712E.2 Mobile Phases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83012E.3 HPLC Plumbing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83412E.4 Detectors for HPLC ............................................83612E.5 Quantitative Applications. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83812E.6 Evaluation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84212F Other Forms of Liquid Chromatography .....................84212F.1 Liquid-Solid Adsorption Chromatography . . . . . . . . . . . . . . . . . . . . . . . . . . . 84212F.2 Ion-Exchange Chromatography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84212F.3 Size-Exclusion Chromatography ...................................84612F.4 Supercritical Fluid Chromatography ................................84712G Electrophoresis ........................................84912G.1 Theory of Capillary Electrophoresis ................................85012G.2 Instrumentation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85512G.3 Capillary Electrophoresis Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85912G.4 Evaluation ...................................................86312H Key Terms ............................................86312I Summary .............................................86412J Problems ..............................................86612K Solutions to Practice Exercises .............................877

13. Kinetic Methods .. . . . . . . . . . . . . . . . . . . . . . . . . . . 88113A Kinetic Methods Versus Equilibrium Methods . . . . . . . . . . . . . . . .88213B Chemical Kinetics ......................................88313B.1 Theory and Practice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88313B.2 Classifying Chemical Kinetic Methods ..............................88613B.3 Making Kinetic Measurements ....................................89613B.4 Quantitative Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89813B.5 Characterization Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90113B.6 Evaluation of Chemical Kinetic Methods ............................904Accuracy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90513C Radiochemistry ........................................90813C.1 Theory and Practice ............................................90913C.2 Instrumentation ...............................................90913C.3 Quantitative Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91013C.4 Characterization Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91413C.5 Evaluation ...................................................91513D Flow Injection Analysis ..................................91613D.1 Theory and Practice ............................................91613D.2 Instrumentation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92013D.3 Quantitative Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92413D.4 Evaluation ...................................................92813E Key Terms ............................................92913F Summary .............................................92913G Problems .............................................93013H Solutions to Practice Exercises .............................93914. Developing a Standard Method................94114A Optimizing the Experimental Procedure .....................94214A.1 Response Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94214A.2 Searching Algorithms for Response Surfaces ..........................94314A.3 Mathematical Models of Response Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . 95014B Verifying the Method ....................................96014B.1 Single Operator Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96014B.2 Blind Analysis of Standard Samples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96114B.3 Ruggedness Testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96114B.4 Equivalency Testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96414C Validating the Method as a Standard Method .................96514C.1 Two-Sample Collaborative Testing .................................96514C.2 Collaborative Testing and Analysis of Variance ........................97014C.3 What is a Reasonable Result for a Collaborative Study? .................97614D Using Excel and R for an Analysis of Variance .................97714D.1 Excel .......................................................977

14D.2 R ..........................................................97814E Key Terms ............................................98014G Problems .............................................98114H Solutions to Practice Exercises .............................99015. Quality Assurance...........................99115A The Analytical Perspective—Revisited .......................99215B Quality Control ........................................99315C Quality Assessment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .99515C.1 Internal Methods of Quality Assessment . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99515C.2 External Methods of Quality Assessment ...........................100015D Evaluating Quality Assurance Data ........................100015D.1 Prescriptive Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100015D.2 Performance-Based Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100315E Key Terms ...........................................101015G Problems ............................................101115H Solutions to Practice Exercises ............................1014Additional Resources .. . . . . . . . . . . . . . . . . . . . . . . . . 1017Chapter 1 .................................................1018Chapter 2 .................................................1019Chapter 3 .................................................1020Chapter 4 .................................................1021Chapter 5 .................................................1025Chapter 6 .................................................1029Chapter 7 .................................................1032Chapter 8 .................................................1036Chapter 9 .................................................1037Chapter 10 ................................................1040Chapter 11 ................................................1048Chapter 12 ................................................1053Chapter 13 ................................................1060Chapter 14 ................................................1063Chapter 15 ................................................1065Appendix.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1067Appendix 1: Normality ......................................1068Appendix 2: Propagation of Uncertainty and Uncertainty Budgets ......1069Appendix 3: Single-SidedNormal Distribution .....................1070Appendix 4: Critical Values for t-Test ............................1072

Appendix 5: Critical Values for the F-Test ........................1073Appendix 6: Critical Values for Dixon’s Q-Test .....................1075Appendix 7: Critical Values for Grubb’s Test .......................1076Appendix 8: Recommended Primary Standards ....................1077Appendix 9: Correcting Mass for the Buoyancy of Air . . . . . . . . . . . . . . .1079Appendix 10: Solubility Products ...............................1081Appendix 11: Acid Dissociation Contants ........................1085Appendix 12: Formation Contants ..............................1093Appendix 13: Standard Reduction Potentials . . . . . . . . . . . . . . . . . . . . . .1098Appendix 14: Random Number Table ...........................1099Appendix 15: PolarographicHalf-Wave Potentials ...................1100Appendix 16: Countercurrent Separations . . . . . . . . . . . . . . . . . . . . . . . .1101Appendix 17: Review of Chemical Kinetics .......................1102

DRAFTPrefaceOverviewABCDEPrefaceOrganizationRole of Equilibrium ChemistryComputational SoftwareHow to Use The Electronic Textbook’s FeaturesAcknowledgmentsAs currently taught in the United States, introductory courses in analytical chemistryemphasize quantitative (and sometimes qualitative) methods of analysis along with a heavydose of equilibrium chemistry. Analytical chemistry, however, is much more than a collection ofanalytical methods and an understanding of equilibrium chemistry; it is an approach to solvingchemical problems. Although equilibrium chemistry and analytical methods are important, theircoverage should not come at the expense of other equally important topics. The introductorycourse in analytical chemistry is the ideal place in the undergraduate chemistry curriculum forexploring topics such as experimental design, sampling, calibration strategies, standardization,optimization, statistics, and the validation of experimental results. Analytical methods comeand go, but best practices for designing and validating analytical methods are universal. Becausechemistry is an experimental science it is essential that all chemistry students understand theimportance of making good measurements.My goal in preparing this textbook is to find a more appropriate balance between theoryand practice, between “classical” and “modern” analytical methods, between analyzing samplesand collecting samples and preparing them for analysis, and between analytical methods anddata analysis. There is more material here than anyone can cover in one semester; it is myhope that the diversity of topics will meet the needs of different instructors, while, perhaps,suggesting some new topics to cover.Copyright: David Harvey, 2009xvii

xviii Analytical Chemistry 2.0AOrganizationThis textbook is organized into four parts. Chapters 1–3 serve as a generalintroduction, providing: an overview of analytical chemistry (Chapter 1); areview of the basic equipment and mathematical tools of analytical chemistry,including significant figures, units, and stoichiometry (Chapter 2);and an introduction to the terminology of analytical chemistry (Chapter3). Familiarity with this material is assumed throughout the remainder ofthe textbook.Chapters 4–7 cover a number of topics that are important in understandinghow analytical methods work. Later chapters are mostly independentof these chapters, allowing an instructor to choose those topicsthat support his or her goals. Chapter 4 provides a thorough introductionto the statistical analysis of data. Methods for calibrating equipment andstandardizing methods are covered in Chapter 5, along with a discussionof linear regression. Chapter 6 provides an introduction to equilibriumchemistry, stressing both the rigorous solution to equilibrium problemsand the use of semi-quantitative approaches, such as ladder diagrams. Theimportance of collecting the right sample, and methods for separating analytesand interferents are the subjects of Chapter 7.Chapters 8–13 cover the major areas of analysis, including gravimetry(Chapter 8), titrimetry (Chapter 9), spectroscopy (Chapter 10), electrochemistry(Chapter 11), chromatography and electrophoresis (Chapter 12),and kinetic methods (Chapter 13). Related techniques, such as acid–basetitrimetry and redox titrimetry, are intentionally gathered together in singlechapters. Combining related techniques in this way encourages students tosee the similarity between methods, rather than focusing on their differences.The first technique in each chapter is generally that which is mostcommonly covered in an introductory course.Finally, the textbook concludes with two chapters discussing the designand maintenance of analytical methods, two topics of importance to all experimentalchemists. Chapter 14 considers the development of an analyticalmethod, including its optimization, its verification, and its validation.Quality control and quality assessment are discussed in Chapter 15.BRole of Equilibrium ChemistryEquilibrium chemistry often receives a significant emphasis in the introductoryanalytical chemistry course. Although it is an important topic, anoveremphasis on the computational aspects of equilibrium chemistry maylead students to confuse analytical chemistry with equilibrium chemistry.Solving equilibrium problems is important—it is equally important, however,for students to recognize when such calculations are impractical, orto recognize when a simpler, more intuitive approach is all they need toanswer a question. For example, in discussing the gravimetric analysis ofAg + by precipitating AgCl, there is little point in calculating the equilib-

Prefacexixrium solubility of AgCl because the equilibrium concentration of Cl – israrely known. It is important, however, for students to understand that alarge excess of Cl – increases the solubility of AgCl due to the formation ofsoluble silver–chloro complexes. To balance the presentation of a rigorousapproach to solving equilibrium problems, this textbook also introducesladder diagrams as a means for rapidly evaluating the effect of solutionconditions on an analysis. Students are encouraged to use the approach bestsuited to the problem at hand.CComputational SoftwareMany of the topics in this textbook benefit from the availability of appropriatecomputational software. There are many software packages availableto instructors and students, including spreadsheets (e.g. Excel), numericalcomputing environments (e.g. Mathematica, Mathcad, Matlab, R), statisticalpackages (e.g. SPSS, Minitab), and data analysis/graphing packages (e.g.Origin). Because of my familiarity with Excel and R, examples of their usein solving problems are incorporated into this textbook. Instructors interestedin incorporating other software packages into future editions of thistextbook are encouraged to contact me at harvey@depauw.edu.DHow to Use The Electronic Textbook’s FeaturesAs with any format, an electronic textbook has advantages and disadvantages.Perhaps the biggest disadvantage to an electronic textbook is thatyou cannot hold it in your hands (and I, for one, like the feel of book inmy hands when reading) and leaf through it. More specifically, you cannotmark your place with a finger, flip back several pages to look at a table or figure,and then return to your reading when you are done. To overcome thislimitation, an electronic textbook can make extensive use of hyperlinks.Whenever the text refers to an object that is not on the current page—a figure, a table, an equation, an appendix, a worked example, a practiceexercise—the text is displayed in blue and is underlined. Clicking on thehyperlink transports you to the relevant object. There are two methods forreturning to your original location within the textbook. When there is noambiguity about where to return to, such as when reviewing an answer toa practice exercise, a second text hyperlink is included.Most objects have links from multiple places within the text, whichmeans that a single return hyperlink is not possible. To return to youroriginal place select View: Go To: Previous View from the menu bar. Ifyou do not like to use pull down menus, you can configure the toolbar toinclude buttons for “Previous View” and for “Next View.” To do this, clickon View: Toolbars: More Tools..., scroll down to the category for PageNavigation Tools and check the boxes for Previous View and for NextView. Your toolbar now includes buttons that you can use to return to yourAlthough you can use any PDF readerto read this electronic textbook, the useof Adobe’s Acrobat Reader is stronglyencouraged. Several features of this electronictextbook—notably the commentingtools and the inclusion of video—areavailable only when using Acrobat Reader8.0 or later. If you do not have AcrobatReader installed on your computer, youcan obtain it from Adobe’s website.Click this button to return to youroriginal position within the text

xx Analytical Chemistry 2.0original position within the text. These buttons will appear each time youopen the electronic textbook.There are several additional hyperlinks to help you navigate within theelectronic textbook. The Table of Contents—both the brief form and theexpanded form—provide hyperlinks to chapters, to main sections, and tosubsections. The “Chapter Overview” on the first page of each chapter provideshyperlinks to the chapter’s main sections. The textbook’s title, whichappears at the top of each even numbered page, is a hyperlink to the brieftable of contents, and the chapter title, which appears at the top of mostodd numbered pages, is a hyperlink to the chapter’s first page. The collectionof key terms at the end of each chapter are hyperlinks to where theterms were first introduced, and the key terms in the text are hyperlinksthat return you to the collection of key terms. Taken together, these hyperlinksprovide for a easy navigation.If you are using Adobe Reader, you can configure the program to rememberwhere you were when you last closed the electronic textbook. SelectAdobe Reader: Preferences... from the menu bar. Select the category“Documents” and check the box for “Restore last view settings when reopeningdocuments.” This is a global change that will affect all documentsthat you open using Adobe Reader.There are two additional features of the electronic textbook that youmay find useful. The first feature is the incorporation of QuickTime andFlash movies. Some movies are configured to run when the page is displayed,and other movies include start, pause, and stopbuttons. If you find that you cannot play a movie checkyour permissions. You can do this by selecting AdobeReader: Preferences... from the menu bar. Choose theoption for “Multimedia Trust” and check the option for“Allow multimedia operations.” If the permission forQuickTime or Flash is set to “Never” you can changeit to “Prompt,” which will ask you to authorize the useof multimedia when you first try to play the movie, orchange it to “Always” if you do not wish to prevent anyfiles from playing movies.A second useful feature is the availability of Adobe’scommenting tools, which allow you to highlight text,to create text boxes in which you can type notes, to addsticky notes, and to attach files. You can access the commentingtools by selecting Tools: Commenting & Markup from the menubar. If you do not like to use the menu bar, you can add tools to the toolbar by selecting View: Toolbars: Commenting & Markup from the menubar. To control what appears on the toolbar, select View: Toolbars: MoreTools... and check the boxes for the tools you find most useful. Details onsome of the tools are shown here. You can alter any tool’s properties, such

Prefacexxias color, by right-clicking on the tool and selecting Tool Default Properties...from the pop-up menu.EAcknowledgmentsThese tools allow you to highlight, to underline,and to cross out text. Select the tool andthe click and drag over the relevant text.The text box tool allows you to provide shortannotations to the textbook. Select the tooland then click and drag to create a text box.Click in the textbook and type your note.The sticky note tool is useful when you have alonger annotation or when you want to makeyour annotation easier to see. Click in thetextbook at the spot where you wish to addthe sticky note. Type your entry and the clickthe close button to collapse the note into itsicon. Hovering the cursor over the icon displaysthe note and clicking on the icon allowsyou to edit your note.The pencil tool allows you to use your mouseto write notes on the text. The eraser tool allowsyou to erase notes.The paperclip tool allows you to attach a fileto the textbook. Select the tool and click inthe textbook at the spot where you wish toattach the file. Use the browser to find the fileand select OK. The resulting icon is a link tothe file. You can open the file by clicking onthe icon.This textbook began as a print version published in 2000 by McGraw-Hill, and the support of the then editorial staff (Jim Smith, Publisher forChemistry; Kent Peterson, Sponsoring Editor for Chemistry; Shirley Oberbroeckling,Developmental Editor for Chemistry; and Jayne Klein, ProjectManager) is gratefully acknowledged. The assistance of Thomas Timp(Sponsoring Editor for Chemistry) and Tamara Hodge (Senior SponsoringEditor for Chemistry) in returning the copyright to me also is acknowledged.The following individuals kindly reviewed portions of the print versionas it worked its way through the publication process:David Ballantine, Northern Illinois UniversityJohn Bauer, Illinois State UniversityAli Bazzi, University of Michigan–Dearborn

xxii Analytical Chemistry 2.0Steven D. Brown, University of DelawareWendy Clevenger, University of Tennessee–ChattanoogaCathy Cobb, Augusta State UniversityPaul Flowers, University of North Carolina–PembrokeGeorge Foy, York College of PennsylvaniaNancy Gordon, University of Southern MaineVirginia M. Indivero, Swarthmore CollegeMichael Janusa, Nicholls State UniversityJ. David Jenkins, Georgia Southern UniversityDavid Karpovich, Saginaw Valley State UniversityGary Kinsel, University of Texas at ArlingtonJohn McBride, Hoftsra UniversityRichard S. Mitchell, Arkansas State UniversityGeorge A. Pearse, Jr., LeMoyne CollegeGary Rayson, New Mexico State UniversityDavid Redfield, NW Nazarene UniversityVincent Remcho, West Virginia UniversityJeanette K. Rice, Georgia Southern UniversityMartin W. Rowe, Texas A&M UniversityAlexander Scheeline, University of IllinoisJames D. Stuart, University of ConnecticutThomas J. Wenzel, Bates CollegeDavid Zax, Cornell UniversityI am particularly grateful for their detailed comments and suggestions.Much of what is good in the print edition is the result of their interest andideas.Without the support of DePauw University and its Faculty DevelopmentCommittee, work on this project would not have been possible. Thisproject began in 1992 with a summer course development grant, and receivedfurther summer support from the Presidential Discretionary Fund.Portions of the first draft were written during a sabbatical leave in fall 1993.The second draft was completed with the support of a Fisher Fellowshipin fall 1995. Converting the print version into this electronic version wascompleted as part of a year-long sabbatical during the 2008/09 academicyear.Ho w To Co n t a c t t h e Au t h o rWorking on this textbook continues to be an interesting challenge and arewarding endeavour. One interesting aspect of electronic publishing is thata book is not static—corrections, new examples and problems, and newtopics are easy to add, and new editions released as needed. I welcome yourinput and encourage you to contact me with suggestions for improving thiselectronic textbook. You can reach me by e-mail at harvey@depauw.edu.

DRAFTChapter 1Introduction to Analytical ChemistryChapter Overview1A What is Analytical Chemistry?1B The Analytical Perspective1C Common Analytical Problems1D Key Terms1E Chapter Summary1F Problems1G Solutions to Practice ExercisesChemistry is the study of matter, including its composition and structure, its physicalproperties, and its reactivity. There are many ways to study chemistry, but, we traditionally divideit into five fields: organic chemistry, inorganic chemistry, biochemistry, physical chemistry, andanalytical chemistry. Although this division is historical and, perhaps, arbitrary—as witnessedby current interest in interdisciplinary areas such as bioanalytical chemistry and organometallicchemistry—these five fields remain the simplest division spanning the discipline of chemistry.Training in each of these fields provides a unique perspective to the study of chemistry.Undergraduate chemistry courses and textbooks are more than a collection of facts; they are akind of apprenticeship. In keeping with this spirit, this chapter introduces the field of analyticalchemistry and highlights the unique perspectives that analytical chemists bring to the study ofchemistry.Copyright: David Harvey, 20091

2 Analytical Chemistry 2.0This quote is attributed to C. N. Reilly(1925-1981) on receipt of the 1965 FisherAward in Analytical Chemistry. Reilly,who was a professor of chemistry at theUniversity of North Carolina at ChapelHill, was one of the most influentialanalytical chemists of the last half of thetwentieth century.You might, for example, have determinedthe amount of acetic acid in vinegar usingan acid–base titration, or used a qualscheme to identify which of several metalions are in an aqueous sample.Seven Stages of an Analytical Method1. Conception of analytical method(birth).2. Successful demonstration that theanalytical method works.3. Establishment of the analytical method’scapabilities.4. Widespread acceptance of the analyticalmethod.5. Continued development of the analyticalmethod leads to significant improvements.6. New cycle through steps 3–5.7. Analytical method can no longer competewith newer analytical methods(death).Steps 1–3 and 5 are the province of analyticalchemistry; step 4 is the realm ofchemical analysis.The seven stages of an analytical methodgiven here are modified from Fassel, V.A. Fresenius’ Z. Anal. Chem. 1986, 324,511–518 and Hieftje, G. M. J. Chem.Educ. 2000, 77, 577–583.1AWhat is Analytical Chemistry?“Analytical chemistry is what analytical chemists do.”Let’s begin with a deceptively simple question. What is analytical chemistry?Like all fields of chemistry, analytical chemistry is too broad and tooactive a discipline for us to define completely. In this chapter, therefore, wewill try to say a little about what analytical chemistry is, as well as a littleabout what analytical chemistry is not.Analytical chemistry is often described as the area of chemistry responsiblefor characterizing the composition of matter, both qualitatively (Isthere any lead in this sample?) and quantitatively (How much lead is in thissample?). As we shall see, this description is misleading.Most chemists routinely make qualitative and quantitative measurements.For this reason, some scientists suggest that analytical chemistry isnot a separate branch of chemistry, but simply the application of chemicalknowledge. 1 In fact, you probably have preformed quantitative and qualitativeanalyses in other chemistry courses.Defining analytical chemistry as the application of chemical knowledgeignores the unique perspective that analytical chemists bring to the study ofchemistry. The craft of analytical chemistry is not in performing a routineanalysis on a routine sample, which more appropriately is called chemicalanalysis, but in improving established analytical methods, in extending existinganalytical methods to new types of samples, and in developing newanalytical methods for measuring chemical phenomena. 2Here is one example of this distinction between analytical chemistryand chemical analysis. Mining engineers evaluate the value of an ore bycomparing the cost of removing the ore with the value of its contents. Toestimate its value they analyze a sample of the ore. The challenge of developingand validating an appropriate quantitative analytical method is theanalytical chemist’s responsibility. After its development, the routine, dailyapplication of the analytical method is the job of the chemical analyst.Another distinction between analytical chemistry and chemical analysisis that analytical chemists work to improve and extend established analyticalmethods. For example, several factors complicate the quantitativeanalysis of nickel in ores, including nickel’s unequal distribution withinthe ore, the ore’s complex matrix of silicates and oxides, and the presenceof other metals that may interfere with the analysis. Figure 1.1 shows aschematic outline of one standard analytical method in use during the latenineteenth century. 3 The need for many reactions, digestions, and filtrationsmakes this analytical method both time-consuming and difficult toperform accurately.1 Ravey, M. Spectroscopy, 1990, 5(7), 11.2 de Haseth, J. Spectroscopy, 1990, 5(7), 11.3 Fresenius. C. R. A System of Instruction in Quantitative Chemical Analysis; John Wiley and Sons:New York, 1881.

Chapter 1 Introduction to Analytical Chemistry3SolidsSolutionskeyPbSO 4; SandOriginal Sample1:3 H 2SO 4/HNO 3, 100°C for 8-10 hrsdilute w/H 2O, digest for 2-4 hrCu 2+ , Fe 3+ , Co 2+ , Ni 2+Start14 hoursdilute; bubble H 2S(g)Fe 3+ , Co 2+ , Ni 2+CuS16 hourscool, add NH 3digest 50 o -70 o for 30 minFe(OH) 3Co 2+ , Ni 2+HClFe 3+ neutralize w/NH 3add Na 2CO 3, CH 3COOHbasic ferric acetatemass AWasteCo 2+ , Ni 2+Co(OH) 2, Ni(OH) 2heatWasteheat; H 2(g)*slightly acidify w/HClheat, bubble H 2S(g)CoS, NiSadd aqua regia and heatadd HCl until strongly acidicbubble H 2S(g)add Na 2CO 3until alkalineadd NaOHCo, Niadd HNO 3, K 2CO 3, KNO 3,and CH 3COOH and digest for 24 hoursCuS, PbSApproximate Elapsed Time17 hours20 hours22 hours23 hours26 hoursNi 2+ K 3Co(NO 3) 5add dilute HCl51 hoursmass BCo 2+Cofollow procedurefrom point * aboveWaste%Ni =mass A - mass Bmass sample x 100Figure 1.1 Fresenius’ analytical scheme for the gravimetric analysis of Ni in ores. Note thatthe mass of nickel is not determined directly. Instead, Co and Ni are isolated and weighed(mass A), and then Co is isolated and weighed (mass B). The timeline shows that afterdigesting a sample, it takes approximately 44 hours to complete an analysis. This schemeis an example of a gravimetric analysis in which mass is the important measurement. SeeChapter 8 for more information about gravimetric procedures.54 hours58 hours

4 Analytical Chemistry 2.0CH 3NOHHONCH 3dimethylglyoximeThe development, in 1905, of dimethylglyoxime (dmg), a reagentthat selectively precipitates Ni 2+ and Pd 2+ , led to an improved analyticalmethod for the quantitative analysis of nickel. 4 The resulting analysis, asshown in Figure 1.2, requires fewer manipulations and less time after completingthe sample’s dissolution. By the 1970s, flame atomic absorptionspectrometry replaced gravimetry as the standard method for analyzingnickel in ores, 5 resulting in an even more rapid analysis. Today, the standardanalytical method utilizes an inductively coupled plasma optical emissionspectrometer.A more appropriate description of analytical chemistry is “the scienceof inventing and applying the concepts, principles, and…strategies formeasuring the characteristics of chemical systems.” 6 Analytical chemists4 Kolthoff, I. M.; Sandell, E. B. Textbook of Quantitative Inorganic Analysis, 3rd Ed., The MacmillanCompany: New York, 1952.5 Van Loon, J. C. Analytical Atomic Absorption Spectroscopy, Academic Press: New York, 1980.6 Murray, R. W. Anal. Chem. 1991, 63, 271A.Residue%Ni =SolidsSolutionskeyOriginal Samplemass A x 0.2031mass sample x 100HNO 3, HCl, heatmass AyesSolutionIssolidpresent?noNi(dmg) 220% NH 4Cl10% tartaric acidmake alkaline w/ 1:1 NH 3make acidic w/ HCl1% alcoholic dmgmake alkaline w/ 1:1 NH 3Approximate Elapsed TimeStart14 hours18 hoursFigure 1.2 Gravimetric analysis for Ni in ores by precipitating Ni(dmg) 2. The timeline shows thatit takes approximately four hours to complete an analysis after digesting the sample, which is 10xshorter than for the method in Figure 1.1. The factor of 0.2301 in the equation for %Ni accounts forthe difference in the formula weights for Ni and Ni(dmg) 2; see Chapter 8 for further details.

Chapter 1 Introduction to Analytical Chemistry5typically operate at the extreme edges of analysis, extending and improvingthe ability of all chemists to make meaningful measurements on smallersamples, on more complex samples, on shorter time scales, and on speciespresent at lower concentrations. Throughout its history, analytical chemistryhas provided many of the tools and methods necessary for researchin the other traditional areas of chemistry, as well as fostering multidisciplinaryresearch in, to name a few, medicinal chemistry, clinical chemistry,toxicology, forensic chemistry, materials science, geochemistry, and environmentalchemistry.You will come across numerous examples of analytical methods in thistextbook, most of which are routine examples of chemical analysis. It isimportant to remember, however, that nonroutine problems promptedanalytical chemists to develop these methods.The next time you are in the library, look through a recent issue of ananalytically oriented journal, such as Analytical Chemistry. Focus on thetitles and abstracts of the research articles. Although you may not recognizeall the terms and analytical methods, you will begin to answer for yourselfthe question “What is analytical chemistry?”1BThe Analytical PerspectiveHaving noted that each field of chemistry brings a unique perspective tothe study of chemistry, we now ask a second deceptively simple question.What is the analytical perspective? Many analytical chemists describe thisperspective as an analytical approach to solving problems. 7 Although thereare probably as many descriptions of the analytical approach as there areanalytical chemists, it is convenient for our purpose to define it as the fivestepprocess shown in Figure 1.3.Three general features of this approach deserve our attention. First, insteps 1 and 5 analytical chemists may collaborate with individuals outsidethe realm of analytical chemistry. In fact, many problems on which analyticalchemists work originate in other fields. Second, the analytical approachincludes a feedback loop (steps 2, 3, and 4) in which the result of one stepmay require reevaluating the other steps. Finally, the solution to one problemoften suggests a new problem.Analytical chemistry begins with a problem, examples of which includeevaluating the amount of dust and soil ingested by children as an indicatorof environmental exposure to particulate based pollutants, resolvingcontradictory evidence regarding the toxicity of perfluoro polymers duringcombustion, and developing rapid and sensitive detectors for chemical andbiological weapons. At this point the analytical approach may involve acollaboration between the analytical chemist and the individual or agencyA recent editorial in Analytical Chemistryentitled “Some Words about Categoriesof Manuscripts” nicely highlights whatmakes a research endeavour relevant tomodern analytical chemistry. The full citationis Murray, R. W. Anal. Chem. 2008,80, 4775.These examples are taken from a seriesof articles, entitled the “Analytical Approach,”which was a regular feature of thejournal Analytical Chemistry, a bimonthlypublication of the American ChemicalSociety. The first issue of each monthcontinues to publish a variety of engagingarticles highlighting current trends inanalytical chemistry.7 For several different viewpoints see (a) Beilby, A. L. J. Chem. Educ. 1970, 47, 237-238; (b)Lucchesi, C. A. Am. Lab. 1980, October, 112-119; (c) Atkinson, G. F. J. Chem. Educ. 1982,59, 201-202; (d) Pardue, H. L.; Woo, J. J. Chem. Educ. 1984, 61, 409-412; (e) Guarnieri, M. J.Chem. Educ. 1988, 65, 201-203, (f) Strobel, H. A. Am. Lab. 1990, October, 17-24.

Chapter 1 Introduction to Analytical Chemistry7and to validate the procedure. Results are compared to the original designcriteria and the experimental design is reconsidered, additional trials arerun, or a solution to the problem is proposed. When a solution is proposed,the results are subject to an external evaluation that may result in a newproblem and the beginning of a new cycle.As noted earlier some scientists question whether the analytical approachis unique to analytical chemistry. 1 Here, again, it helps to distinguishbetween a chemical analysis and analytical chemistry. For other analyticallyoriented scientists, such as a physical organic chemist or a public healthofficer, the primary emphasis is how the analysis supports larger researchgoals involving fundamental studies of chemical or physical processes, orimproving access to medical care. The essence of analytical chemistry, however,is in developing new tools for solving problems, and in defining thetype and quality of information available to other scientists.Chapter 4 introduces the statistical analysisof data.Practice Exercise 1.1As an exercise, let’s adapt our model of the analytical approach to thedevelopment of a simple, inexpensive, portable device for completingbioassays in the field. Before continuing, locate and read the article“Simple Telemedicine for Developing Regions: Camera Phones and Paper-BasedMicrofluidic Devices for Real-Time, Off-Site Diagnosis”by Andres W. Martinez, Scott T. Phillips, Emanuel Carriho, Samuel W.Thomas III, Hayat Sindi, and George M. Whitesides. You will find it onpages 3699-3707 in Volume 80 of the journal Analytical Chemistry, whichwas published in 2008. As you read the article, pay particular attention tohow it emulates the analytical approach. It might be helpful to considerthe following questions:What is the analytical problem and why is it important?What criteria did the authors consider in designing their experiments?What is the basic experimental procedure?What interferences were considered and how did they overcome them?How did the authors calibrate the assay?How did the authors validate their experimental method?Is there evidence of repeating steps 2, 3, and 4?Was there a successful conclusion to the problem?This exercise provides you with an opportunityto think about the analyticalapproach in the context of a real analyticalproblem. Boxed exercises such as thisprovide you with a variety of challengesranging from simple review problems tomore open-ended exercises. You will findanswers to exercises at the end of eachchapter.Use this link to access the article’s abstractfrom the journal’s web site. If your institutionhas an on-line subscription you alsowill be able to download a PDF versionof the article.Don’t let the technical details in the paper overwhelm you. If you skimover these you will find that the paper is well-written and accessible.Click here to review your answers to these questions.

8 Analytical Chemistry 2.0Current research in the areas of quantitativeanalysis, qualitative analysis, andcharacterization analysis are reviewedbiennially (odd-numbered years) in AnalyticalChemistry’s series of “ApplicationReviews.” The 2007 issue, for example,reviews forensic science, water, pharmaceuticals,geochemistry, and protemics, toname a few.Current research in the area of fundamentalanalysis is reviewed biennially (evennumberedyears) in Analytical Chemistry’sseries of “Fundamental Reviews.”The 2008 issue, for example, reviewsfiber-optic chemical sensors and biosensors,thermal analysis, chiral separations,chemometrics, and solid state NMR, toname a few.As you review this chapter, try to define akey term in your own words. Check youranswer by clicking on the key term, whichwill take you to the page where it was firstintroduced. Clicking on the key termthere, will bring you back to this page sothat you can continue with another keyterm.1CCommon Analytical ProblemsMany problems in analytical chemistry begin with the need to identifywhat is present in a sample. This is the scope of a qualitative analysis,examples of which include identifying the products of a chemical reaction,screening an athlete’s urine for the presence of a performance-enhancingdrug, or determining the spatial distribution of Pb on the surface of an airborneparticulate. Much of the early work in analytical chemistry involvedthe development of simple chemical tests to identify inorganic ions andorganic functional groups. The classical laboratory courses in inorganic andorganic qualitative analysis, still taught at some schools, are based on thiswork. 8 Currently, most qualitative analyses use methods such as infrared(IR) spectroscopy and nuclear magnetic resonance (NMR) spectroscopy.These qualitative applications are covered adequately elsewhere in the undergraduatecurriculum and, so, will receive no further consideration inthis text.Perhaps the most common analytical problem is a quantitativeanalysis. Examples of typical quantitative analyses include the elementalanalysis of a newly synthesized compound, measuring the concentrationof glucose in blood, or determining the difference between the bulk andsurface concentrations of Cr in steel. Much of the analytical work in clinical,pharmaceutical, environmental, and industrial labs involves developingnew quantitative methods for trace amounts of chemical species in complexsamples. Most of the examples in this text are quantitative analyses.Another important area of analytical chemistry, which receives someattention in this text, is the development of new methods for characterizingphysical and chemical properties. Determinations of chemical structure,equilibrium constants, particle size, and surface structure are examples of acharacterization analysis.The purpose of a qualitative, quantitative, or characterization analysisis to solve a problem associated with a particular sample. The purpose of afundamental analysis, on the other hand, is to improve our understandingof the theory behind an analytical method. Extending and improvingthe theory on which an analytical method is based, studying an analyticalmethod’s limitations, and designing and modifying existing analyticalmethod are examples of fundamental studies in analytical chemistry.1DKey Termscharacterization analysis fundamental analysis qualitative analysisquantitative analysis8 See, for example, the following laboratory texts: (a) Sorum, C. H.; Lagowski, J. J. Introductionto Semimicro Qualitative Analysis, 5th Ed.; Prenctice-Hall: Englewood, NJ, 1977; (b) Shriner,R. L.; Fuson, R. C.; Curtin, D. Y. The Systematic Identification of Organic Compounds, 5th Ed.;John Wiley and Sons: New York, 1964.

Chapter 1 Introduction to Analytical Chemistry91EChapter SummaryAnalytical chemists work to improve the ability of all chemists to makemeaningful measurements. Chemists working in the other traditional areasof chemistry, as well as in interdisciplinary fields such as medicinal chemistry,clinical chemistry, and environmental chemistry, need better toolsfor analyzing materials. The need to work with smaller samples, with morecomplex materials, with processes occurring on shorter time scales, andwith species present at lower concentrations challenges analytical chemiststo improve existing analytical methods and to develop new ones.Typical problems on which analytical chemists work include qualitativeanalyses (What is present?), quantitative analyses (Who much is present?),characterization analyses (What are the sample’s chemical and physicalproperties?), and fundamental analyses (How does this method work andhow can it be improved?).1FProblems1. For each of the following problems indicate whether its solution requiresa qualitative analysis, a quantitative analysis, a characterizationanalysis, or a fundamental analysis. More than one type of analysis maybe appropriate for some problems.(a) A hazardous-waste disposal site is believed to be leaking contaminantsinto the local groundwater.(b) An art museum is concerned that a recent acquisition is a forgery.(c) Airport security needs a more reliable method for detecting thepresence of explosive materials in luggage.(d) The structure of a newly discovered virus needs to be determined.(e) A new visual indicator is needed for an acid–base titration.(f) A new law requires a method for evaluating whether automobilesare emitting too much carbon monoxide.2. Read the article “When Machine Tastes Coffee: Instrumental Approachto Predict the Sensory Profile of Espresso Coffee,” by several scientistsworking at the Nestlé Research Center in Lausanne, Switzerland. Youwill find the article on pages 1574-1581 in Volume 80 of AnalyticalChemistry, published in 2008. Write an essay summarizing the natureof the problem and how it was solved. As a guide, refer to Figure 1.3for a model of the analytical approach to solving problems.Answers, but not worked solutions, tomost end-of-chapter problems are availablehere.Use this link to access the article’s abstractfrom the journal’s web site. If your institutionhas an on-line subscription you alsowill be able to download a PDF versionof the article.

10 Analytical Chemistry 2.01GSolutions to Practice ExercisesThis is an example of a colorimetric methodof analysis. Colorimetric methods arecovered in Chapter 10.Literature Exercise 1.1What is the analytical problem and why is it important?A medical diagnoses often relies on the results of a clinical analysis. Whenvisiting a doctor, he or she may ask the nurse to draw a sample of your bloodand send it to the lab for analysis. In some cases the result of the analysis isavailable in 10-15 minutes. What is possible in a developed country, suchas the United States, may not be feasible in a country with fewer resourcesbecause lab equipment is expensive, and because there may be a shortage oftrained personnel to run the tests and to interpret the results. The problemaddressed in this paper, therefore, is the development of a reliable devicefor rapidly and quantitatively performing clinical assays in less than idealcircumstances.What criteria did the authors consider in designing their experiments?In considering solutions to this problem, the authors identify seven importantcriteria for the device: it must be inexpensive; it must operate withoutthe need for much electricity, so that it can be taken to remote sites; it mustbe adaptable to many types of assays; its operation must not require a highlyskilled technician; it must be quantitative; it must be accurate; and it mustproduce results rapidly.What is the basic experimental procedure?The authors describe the development of a paper-based microfluidic devicethat allows anyone to run an analysis by dipping the device into a sample(synthetic urine, in this case). The sample moves by capillary action into testzones containing reagents that react with specific species (glucose and protein,for this prototype device). The reagents react to produce a color whoseintensity is proportional to the species’ concentration. Digital pictures ofthe microfluidic device are taken with a cell phone camera and sent to anoff-site physician who analyzes the picture using image editing software andinterprets the assay’s result.What interferences were considered and how did they overcome them?In developing this analytical method the authors considered several chemicalor physical interferences. One concern was the possibility of non-specificinteractions between the paper and the glucose or protein, which couldlead to non-uniform image in the test zones. A careful analysis of thedistribution of glucose and protein in the text zones showed that this wasnot a problem. A second concern was the possible presence in samplesof particulate materials that might interfere with the analyses. Paper is anatural filter for particulate materials and the authors found that samplescontaining dust, sawdust, and pollen did not interfere with the analysis forglucose. Pollen, however, is an interferent for the protein analysis, presumablybecause it, too, contains protein.

Chapter 1 Introduction to Analytical Chemistry11How did the author’s calibrate the assay?To calibrate the device the authors analyzed a series of standard solutionscontaining known concentrations of glucose and protein. Because an image’sintensity depends upon the available light, a standard sample is runwith the test samples, which allows a single calibration curve to be used forsamples collected under different lighting conditions.How did the author’s validate their experimental method?The test device contains two test zones for each analyte, allowing for duplicateanalyses and providing one level of experimental validation. To furthervalidate the device, the authors completed 12 analyses at each of threeknown concentrations of glucose and protein, obtaining acceptable accuracyand precision in all cases.Is there any evidence of repeating steps 2, 3, and 4?Developing this analytical method required several repetitive paths throughsteps 2, 3, and 4 of the analytical approach. Examples of this feedback loopinclude optimizing the shape of the test zones, and evaluating the importanceof sample size.In summary, the authors report the successful development of an inexpensive,portable, and easy-to-use device for running clinical samples indeveloping countries.Click here to return to the chapter.

12 Analytical Chemistry 2.0

DRAFTChapter 2Basic Tools ofAnalytical ChemistryChapter Overview2A Measurements in Analytical Chemistry2B Concentration2C Stoichiometric Calculations2D Basic Equipment2E Preparing Solutions2F Spreadsheets and Computational Software2G The Laboratory Notebook2H Key Terms2I Chapter Summary2J Problems2K Solutions to Practice ExercisesIn the chapters that follow we will explore many aspects of analytical chemistry. In the processwe will consider important questions such as “How do we treat experimental data?”, “Howdo we ensure that our results are accurate?”, “How do we obtain a representative sample?”,and “How do we select an appropriate analytical technique?” Before we look more closely atthese and other questions, we will first review some basic tools of importance to analyticalchemists.Copyright: David Harvey, 200913

14 Analytical Chemistry 2.0Some measurements, such as absorbance,do not have units. Because the meaning ofa unitless number may be unclear, someauthors include an artificial unit. It is notunusual to see the abbreviation AU, whichis short for absorbance unit, following anabsorbance value. Including the AU clarifiesthat the measurement is an absorbancevalue.It is important for scientists to agree upona common set of units. In 1999 NASAlost a Mar’s Orbiter spacecraft becauseone engineering team used English unitsand another engineering team used metricunits. As a result, the spacecraft cameto close to the planet’s surface, causing itspropulsion system to overheat and fail.2AMeasurements in Analytical ChemistryAnalytical chemistry is a quantitative science. Whether determining theconcentration of a species, evaluating an equilibrium constant, measuring areaction rate, or drawing a correlation between a compound’s structure andits reactivity, analytical chemists engage in “measuring important chemicalthings.” 1 In this section we briefly review the use of units and significantfigures in analytical chemistry.2A.1 Units of MeasurementA measurement usually consists of a unit and a number expressing thequantity of that unit. We may express the same physical measurement withdifferent units, which can create confusion. For example, the mass of asample weighing 1.5 g also may be written as 0.0033 lb or 0.053 oz. Toensure consistency, and to avoid problems, scientists use a common set offundamental units, several of which are listed in Table 2.1. These units arecalled SI units after the Système International d’Unités.We define other measurements using these fundamental SI units. Forexample, we measure the quantity of heat produced during a chemical reactionin joules, (J), where1 Murray, R. W. Anal. Chem. 2007, 79, 1765.1J 1 mkg 2=2sTable 2.1 Fundamental SI Units of Importance to Analytical ChemistryMeasurement Unit Symbol Definition (1 unit is...)mass kilogram kg...the mass of the international prototype, a Pt-Ir objecthoused at the Bureau International de Poids and Measuresat Sèvres, France. †distance meter m ...the distance light travels in (299 792 458) -1 seconds.temperature Kelvin K...equal to (273.16) –1 , where 273.16 K is the triple pointof water (where its solid, liquid, and gaseous forms are inequilibrium).time second s...the time it takes for 9 192 631 770 periods of radiationcorresponding to a specific transition of the 133 Cs atom.current ampere A...the current producing a force of 2 × 10 -7 N/m whenmaintained in two straight parallel conductors of infinitelength separated by one meter (in a vacuum).amount of substance mole mol...the amount of a substance containing as many particlesas there are atoms in exactly 0.012 kilogram of 12 C.† The mass of the international prototype changes at a rate of approximately 1 mg per year due to reversible surface contamination. The referencemass, therefore, is determined immediately after its cleaning by a specified procedure.

Chapter 2 Basic Tools of Analytical Chemistry15Table 2.2 Derived SI Units and Non-SI Units of Importance to Analytical ChemistryMeasurement Unit Symbol Equivalent SI Unitslength angstrom (non-SI) Å 1 Å = 1 × 10 –10 mvolume liter (non-SI) L 1 L = 10 –3 m 3force newton (SI) N 1 N = 1 m·kg/s 2pressureenergy, work, heatpascal (SI)atmosphere (non-SI)joule (SI)calorie (non-SI)electron volt (non-SI)PaatmJcaleV1 Pa = 1 N/m 2 = 1 kg/(m·s 2 )1 atm = 101,325 Pa1 J = N·m = 1 m 2·kg/s 21 cal = 4.184 J1 eV = 1.602 177 33 × 10 –19 Jpower watt (SI) W 1 W =1 J/s = 1 m 2·kg/s 3charge coulomb (SI) C 1 C = 1 A·spotential volt (SI) V 1 V = 1 W/A = 1 m2·kg/(s3·A)frequency hertz (SI) Hz 1 Hz = s –1temperatureCelsius (non-SI)o C o C = K – 273.15Table 2.2 provides a list of some important derived SI units, as well as a fewcommon non-SI units.Chemists frequently work with measurements that are very large or verysmall. A mole contains 602 213 670 000 000 000 000 000 particles andsome analytical techniques can detect as little as 0.000 000 000 000 001 gof a compound. For simplicity, we express these measurements using scientificnotation; thus, a mole contains 6.022 136 7 × 10 23 particles, andthe detected mass is 1 × 10 –15 g. Sometimes it is preferable to express measurementswithout the exponential term, replacing it with a prefix (Table2.3). A mass of 1×10 –15 g, for example, is the same as 1 fg, or femtogram.Writing a lengthy number with spacesinstead of commas may strike you as unusual.For numbers containing more thanfour digits on either side of the decimalpoint, however, the currently acceptedpractice is to use a thin space instead ofa comma.Table 2.3 Common Prefixes for Exponential NotationPrefix Symbol Factor Prefix Symbol Factor Prefix Symbol Factoryotta Y 10 24 kilo k 10 3 micro m 10 –6zetta Z 10 21 hecto h 10 2 nano n 10 –9eta E 10 18 deka da 10 1 pico p 10 –12peta P 10 15 - - 10 0 femto f 10 –15tera T 10 12 deci d 10 –1 atto a 10 –18giga G 10 9 centi c 10 –2 zepto z 10 –21mega M 10 6 milli m 10 –3 yocto y 10 –24

16 Analytical Chemistry 2.02A.2 Uncertainty in MeasurementsFigure 2.1 When weighing anobject on a balance, the measurementfluctuates in the final decimalplace. We record this cylinder’smass as 1.2637 g ± 0.0001 g.In the measurement 0.0990 g, the zero ingreen is a significant digit and the zeros inred are not significant digits.A measurement provides information about its magnitude and its uncertainty.Consider, for example, the balance in Figure 2.1, which is recordingthe mass of a cylinder. Assuming that the balance is properly calibrated, wecan be certain that the cylinder’s mass is more than 1.263 g and less than1.264 g. We are uncertain, however, about the cylinder’s mass in the lastdecimal place since its value fluctuates between 6, 7, and 8. The best we cando is to report the cylinder’s mass as 1.2637 g ± 0.0001 g, indicating bothits magnitude and its absolute uncertainty.Significant Fi g u r e sSignificant figures are a reflection of a measurement’s magnitude anduncertainty. The number of significant figures in a measurement is thenumber of digits known exactly plus one digit whose value is uncertain.The mass shown in Figure 2.1, for example, has five significant figures, fourwhich we know exactly and one, the last, which is uncertain.Suppose we weigh a second cylinder, using the same balance, obtaining amass of 0.0990 g. Does this measurement have 3, 4, or 5 significant figures?The zero in the last decimal place is the one uncertain digit and is significant.The other two zero, however, serve to show us the decimal point’s location.Writing the measurement in scientific notation (9.90 × 10 –2 ) clarifies thatthere are but three significant figures in 0.0990.Example 2.1How many significant figures are in each of the following measurements?Convert each measurement to its equivalent scientific notation or decimalform.(a) 0.0120 mol HCl(b) 605.3 mg CaCO 3(c) 1.043 × 10 –4 mol Ag +(d) 9.3 × 10 4 mg NaOHSo l u t i o n(a) Three significant figures; 1.20 × 10 –2 mol HCl.(b) Four significant figures; 6.053 × 10 2 mg CaCO 3 .(c) Four significant figures; 0.000 104 3 mol Ag + .(d) Two significant figures; 93 000 mg NaOH.The log of 2.8 × 10 2 is 2.45. The log of2.8 is 0.45 and the log of 10 2 is 2. The 2in 2.45, therefore, only indicates the powerof 10 and is not a significant digit.There are two special cases when determining the number of significantfigures. For a measurement given as a logarithm, such as pH, the number ofsignificant figures is equal to the number of digits to the right of the decimalpoint. Digits to the left of the decimal point are not significant figures sincethey only indicate the power of 10. A pH of 2.45, therefore, contains twosignificant figures.

Chapter 2 Basic Tools of Analytical Chemistry17An exact number has an infinite number of significant figures. Stoichiometriccoefficients are one example of an exact number. A mole ofCaCl 2 , for example, contains exactly two moles of chloride and one mole ofcalcium. Another example of an exact number is the relationship betweensome units. There are, for example, exactly 1000 mL in 1 L. Both the 1 andthe 1000 have an infinite number of significant figures.Using the correct number of significant figures is important because ittells other scientists about the uncertainty of your measurements. Supposeyou weigh a sample on a balance that measures mass to the nearest ±0.1mg. Reporting the sample’s mass as 1.762 g instead of 1.7623 g is incorrectbecause it does not properly convey the measurement’s uncertainty. Reportingthe sample’s mass as 1.76231 g also is incorrect because it falsely suggestan uncertainty of ±0.01 mg.Significant Fi g u r e s in Ca l c u l at i o n sSignificant figures are also important because they guide us when reportingthe result of an analysis. In calculating a result, the answer can never bemore certain than the least certain measurement in the analysis. Roundinganswers to the correct number of significant figures is important.For addition and subtraction round the answer to the last decimal placethat is significant for each measurement in the calculation. The exact sumof 135.621, 97.33, and 21.2163 is 254.1673. Since the last digit that issignificant for all three numbers is in the hundredth’s place135.62197.3321.2163157.1673The last common decimal place shared by135.621, 97.33, and 21.2163 is shown inred.we round the result to 254.17. When working with scientific notation,convert each measurement to a common exponent before determining thenumber of significant figures. For example, the sum of 4.3 × 10 5 , 6.17 × 10 7 ,and 3.23 × 10 4 is 622 × 10 5 , or 6.22 × 10 7 .617.× 104.3 × 100.323×10621.623×105555The last common decimal place shared by4.3 × 10 5 , 6.17 × 10 7 , and 3.23 × 10 4 isshown in red.For multiplication and division round the answer to the same numberof significant figures as the measurement with the fewest significant figures.For example, dividing the product of 22.91 and 0.152 by 16.302 gives ananswer of 0.214 because 0.152 has the fewest significant figures.22. 91×0.152= 0. 2131 = 0.21416.302

18 Analytical Chemistry 2.0It is important to recognize that the rulesfor working with significant figures aregeneralizations. What is conserved in acalculation is uncertainty, not the numberof significant figures. For example,the following calculation is correct eventhough it violates the general rules outlinedearlier.10199= 102 .Since the relative uncertainty in each measurementis approximately 1% (101 ± 1,99 ± 1), the relative uncertainty in thefinal answer also must be approximately1%. Reporting the answer as 1.0 (two significantfigures), as required by the generalrules, implies a relative uncertainty of10%, which is too large. The correct answer,with three significant figures, yieldsthe expected relative uncertainty. Chapter4 presents a more thorough treatment ofuncertainty and its importance in reportingthe results of an analysis.There is no need to convert measurements in scientific notation to a commonexponent when multiplying or dividing.Finally, to avoid “round-off” errors it is a good idea to retain at least oneextra significant figure throughout any calculation. Better yet, invest in agood scientific calculator that allows you to perform lengthy calculationswithout recording intermediate values. When your calculation is complete,round the answer to the correct number of significant figures using thefollowing simple rules.1. Retain the least significant figure if it and the digits that follow are lessthan half way to the next higher digit. For example, rounding 12.442to the nearest tenth gives 12.4 since 0.442 is less than half way between0.400 and 0.500.2. Increase the least significant figure by 1 if it and the digits that followare more than half way to the next higher digit. For example, rounding12.476 to the nearest tenth gives 12.5 since 0.476 is more than half waybetween 0.400 and 0.500.3. If the least significant figure and the digits that follow are exactly halfwayto the next higher digit, then round the least significant figure tothe nearest even number. For example, rounding 12.450 to the nearesttenth gives 12.4, while rounding 12.550 to the nearest tenth gives 12.6.Rounding in this manner ensures that we round up as often as we rounddown.Practice Exercise 2.1For a problem involving both addition and/or subtraction, and multiplicationand/or division, be sure to account for significant figures ateach step of the calculation. With this in mind, to the correct number ofsignificant figures, what is the result of this calculation?2B−3 −20. 250× ( 993 . × 10 ) − 0. 100× ( 1. 927×10 )=−3 −2993 . × 10 + 1.927×10Click here to review your answer to this exercise.ConcentrationConcentration is a general measurement unit stating the amount of solutepresent in a known amount of solutionamount of soluteConcentration =amount of solutionAlthough we associate the terms “solute” and “solution” with liquid samples,we can extend their use to gas-phase and solid-phase samples as well. Table2.4 lists the most common units of concentration.2.1

Chapter 2 Basic Tools of Analytical Chemistry192B.1 Molarity and FormalityBoth molarity and formality express concentration as moles of soluteper liter of solution. There is, however, a subtle difference between molarityand formality. Molarity is the concentration of a particular chemicalspecies. Formality, on the other hand, is a substance’s total concentrationwithout regard to its specific chemical form. There is no difference betweena compound’s molarity and formality if it dissolves without dissociatinginto ions. The formal concentration of a solution of glucose, for example,is the same as its molarity.For a compound that ionize in solution, such as NaCl, molarity andformality are different. Dissolving 0.1 moles of CaCl 2 in 1 L of water givesa solution containing 0.1 moles of Ca 2+ and 0.2 moles of Cl – . The molarityof NaCl, therefore, is zero since there is essentially no undissociated NaCl.The solution, instead, is 0.1 M in Ca 2+ and 0.2 M in Cl – . The formalityA solution that is 0.0259 M in glucose is0.0259 F in glucose as well.Table 2.4 Common Units for Reporting ConcentrationName Units Symbolmolaritymolessoluteliters solutionMformalitymolessoluteliters solutionFnormalityequivalentssoluteliters solutionNmolalitymolessolutekilogramssolventmweight percentgramssolute100 gramssolution% w/wAn alternative expression for weight percentisvolume percentweight-to-volume percentmL solute100 mL solutiongramssolute100 mL solution% v/v% w/vgramssolutegramssolution ×100You can use similar alternative expressionsfor volume percent and for weight-tovolumepercent.parts per millionparts per billiongramssolute610 gramssolutionppmgramssolute910 gramssolutionppb

20 Analytical Chemistry 2.0One handbook that still uses normality isStandard Methods for the Examination ofWater and Wastewater, a joint publicationof the American Public Health Association,the American Water Works Association,and the Water Environment Federation.This handbook is one of the primaryresources for the environmental analysis ofwater and wastewater.of NaCl, however, is 0.1 F since it represents the total amount of NaCl insolution. The rigorous definition of molarity, for better or worse, is largelyignored in the current literature, as it is in this textbook. When we state thata solution is 0.1 M NaCl we understand it to consist of Na + and Cl – ions.The unit of formality is used only when it provides a clearer description ofsolution chemistry.Molarity is used so frequently that we use a symbolic notation to simplifyits expression in equations and in writing. Square brackets around aspecies indicate that we are referring to that species’ molarity. Thus, [Na + ]is read as “the molarity of sodium ions.”2B.2 NormalityNormality is a concentration unit that is no longer in common use. Becauseyou may encounter normality in older handbooks of analytical methods, itcan be helpful to understand its meaning. Normality defines concentrationin terms of an equivalent, which is the amount of one chemical speciesreacting stoichiometrically with another chemical species. Note thatthis definition makes an equivalent, and thus normality, a function of thechemical reaction in which the species participates. Although a solution ofH 2 SO 4 has a fixed molarity, its normality depends on how it reacts. You willfind a more detailed treatment of normality in Appendix 1.2B.3 MolalityMolality is used in thermodynamic calculations where a temperature independentunit of concentration is needed. Molarity is based on the volumeof solution containing the solute. Since density is a temperature dependentproperty a solution’s volume, and thus its molar concentration, changeswith temperature. By using the solvent’s mass in place of the solution’s volume,the resulting concentration becomes independent of temperature.2B.4 Weight, Volume, and Weight-to-Volume RatiosWeight percent (% w/w), volume percent (% v/v) and weight-to-volumepercent (% w/v) express concentration as the units of solute presentin 100 units of solution. A solution of 1.5% w/v NH 4 NO 3 , for example,contains 1.5 gram of NH 4 NO 3 in 100 mL of solution.2B.5 Parts Per Million and Parts Per BillionParts per million (ppm) and parts per billion (ppb) are ratios givingthe grams of solute to, respectively, one million or one billion gramsof sample. For example, a steel that is 450 ppm in Mn contains 450 µg ofMn for every gram of steel. If we approximate the density of an aqueoussolution as 1.00 g/mL, then solution concentrations can be express in ppmor ppb using the following relationships.

Chapter 2 Basic Tools of Analytical Chemistry21ppm = mg gL= µ gppb = µ g ngL= mLFor gases a part per million usually is a volume ratio. Thus, a helium concentrationof 6.3 ppm means that one liter of air contains 6.3 µL of He.2B.6 Converting Between Concentration UnitsThe most common ways to express concentration in analytical chemistryare molarity, weight percent, volume percent, weight-to-volume percent,parts per million and parts per billion. By recognizing the general definitionof concentration given in equation 2.1, it is easy to convert betweenconcentration units.Example 2.2A concentrated solution of ammonia is 28.0% w/w NH 3 and has a densityof 0.899 g/mL. What is the molar concentration of NH 3 in this solution?You should be careful when using partsper million and parts per billion to expressthe concentration of an aqueous solute.The difference between a solute’s concentrationin mg/L and mg/g, for example, issignificant if the solution’s density is not1.00 g/mL. For this reason many organizationsadvise against using the abbreviationppm and ppb (see www.nist.gov). Ifin doubt, include the exact units, such as0.53 mg Pb 2+ /L for the concentration oflead in a sample of seawater.So l u t i o n28. 0 gNH 30.899 gsolution 1 molNH31000 mL×× × = 14.8 M100 gsolution mL solution 17.04 gNH L3Example 2.3The maximum permissible concentration of chloride in a municipal drinkingwater supply is 2.50 × 10 2 ppm Cl – . When the supply of water exceedsthis limit it often has a distinctive salty taste. What is the equivalent molarconcentration of Cl – ?So l u t i o n2 −−250 . × 10 mg Cl 1g1 molCl× ×L 1000 mg 35.453 gCl−= ×−705 . 10 3MPractice Exercise 2.2Which solution—0.50 M NaCl or 0.25 M SrCl 2 —has the larger concentrationwhen expressed in μg/mL?Click here to review your answer to this exercise.

22 Analytical Chemistry 2.0Figure 2.2 Graph showing theprogress for the titration of 50.0mL of 0.10 M HCl with 0.10 MNaOH. The [H + ] is shown on theleft y-axis and the pH on the righty-axis.0.10.08141210[H] (M)0.060.0486pH0.024200 20 40 60 80Volume NaOH (mL)0Acid–base titrations, as well as severalother types of titrations, are covered inChapter 9.A more appropriate equation for pH ispH = –log(a H +)where a H + is the activity of the hydrogenion. See Chapter 6 for more details. Fornow the approximate equationis sufficient.pH = –log[H + ]2B.7 p-FunctionsSometimes it is inconvenient to use the concentration units in Table 2.4.For example, during a reaction a species’ concentration may change bymany orders of magnitude. If we want to display the reaction’s progressgraphically we might plot the reactant’s concentration as a function of timeor as a function of the volume of a reagent being added to the reaction. Suchis the case in Figure 2.2 for the titration of HCl with NaOH. The y-axison the left-side of the figure displays the [H + ] as a function of the volumeof NaOH. The initial [H + ] is 0.10 M and its concentration after adding80 mL of NaOH is 4.3 × 10 -13 M. We can easily follow the change in [H + ]for the first 14 additions of NaOH. For the remaining additions of NaOH,however, the change in [H + ] is too small to see.When working with concentrations spanning many orders of magnitude,it is often more convenient to express concentration using a p-function.The p-function of X is written as pX and is defined aspX=−log( X)The pH of a solution that is 0.10 M H + isand the pH of 4.3 × 10 -13 M H + is+pH =− log[ H ] =− log( 010 . ) = 100 .

Chapter 2 Basic Tools of Analytical Chemistry23+ −13pH =− log[ H ] =− log( 43 . × 10 ) = 12.37Figure 2.2 shows that plotting pH as a function of the volume of NaOHprovides more detail about how the concentration of H + changes duringthe titration.Example 2.4What is pNa for a solution of 1.76 × 10 -3 M Na 3 PO 4 ?So l u t i o nSince each mole of Na 3 PO 4 contains three moles of Na + , the concentrationof Na + is[Naand pNa is+3 3 molNa] = ( 176 . × 10 M)× = 52 . 8× 10 −3MmolNaPO+ −3 4+ −3pNa=− log[ Na ] =− log( 528 . × 10 ) = 2.277Example 2.5What is the [H + ] in a solution that has a pH of 5.16?So l u t i o nThe concentration of H + is+pH =− log[ H ] = 516 .log[ H + ] =−516 ..[ H + ] antilog( . ) − 516= − 516 = 10 = 6.9×10− 6MIf X = 10 a , then log(X) = a.Practice Exercise 2.3What is pK and pSO 4 for a solution containing 1.5 g K 2 SO 4 in a totalvolume of 500.0 mL?Click here to review your answer to this exercise.2CStoichiometric CalculationsA balanced reaction, which gives the stoichiometric relationship betweenthe moles of reactants and the moles of products, provides the basis formany analytical calculations. Consider, for example, an analysis for oxalicacid, H 2 C 2 O 4 , in which Fe 3+ oxidizes oxalic acid to CO 2 .

24 Analytical Chemistry 2.0HOOoxalic acidOOHOxalic acid, in sufficient amounts, is toxic.At lower physiological concentrations itleads to the formation of kidney stones.The leaves of the rhubarb plant containrelatively high concentrations of oxalicacid. The stalk, which many individualsenjoy eating, contains much smaller concentrationsof oxalic acid.Note that we retain an extra significantfigure throughout the calculation, roundingto the correct number of significantfigures at the end. We will follow this conventionin any problem involving morethan one step.If we forget that we are retaining an extrasignificant figure, we might report the finalanswer with one too many significantfigures. In this chapter we will mark theextra digit in red for emphasis. Be surethat you pick a system for keeping trackof significant figures.3+2Fe ( aq) + HCO ( aq) + 2H O()l →2 2 422++2Fe( aq ) + 2CO( g) + 2H O ( aq )The balanced reaction indicates that one mole of oxalic acid reacts with twomoles of Fe 3+ . As shown in Example 2.6, we can use this balanced reactionto determine the amount of oxalic acid in a sample of rhubarb.Example 2.6The amount of oxalic acid in a sample of rhubarb was determined by reactingwith Fe 3+ . After extracting a 10.62 g of rhubarb with a solvent, oxidationof the oxalic acid required 36.44 mL of 0.0130 M Fe 3+ . What is theweight percent of oxalic acid in the sample of rhubarb?So l u t i o nWe begin by calculating the moles of Fe 3+ used in the reaction0.0130 molFeL3+4× 0. 03644 L= 473 . 7× 10− mol Fe 3+The moles of oxalic acid reacting with the Fe 3+ , therefore, is41473 . 7× 10−3+molH CO2 2 44molFe × = 236 . 8× 10− molH CO3+2mol FeConverting the moles of oxalic acid to grams of oxalic acid490.03 gHCO236 . 8× 10− molC HO ×2 2 4molH C O2 2 42 2 42= 213 . 2× 10−322 2 4gHCO2 2 4and calculating the weight percent gives the concentration of oxalic acidin the sample of rhubarb as2213 . 2×10− gHCO2 2 4× 100 = 020 . 1% w/wHCO10.62 grhubarb2 2 4Practice Exercise 2.4The analyte in Example 2.6, oxalic acid, is in a chemically useful formbecause there is a reagent, Fe 3+ , that reacts with it quantitatively. In manyanalytical methods, we must convert the analyte into a more accessibleYou can dissolve a precipitate of AgBr by reacting it with Na 2 S 2 O 3 , as shown here.AgBr(s) + 2Na 2 S 2 O 3 (aq)→ Ag(S 2 O 3 ) 2 3– (aq) + Br – (aq) + 4Na + (aq)How many mL of 0.0138 M Na 2 S 2 O 3 do you need to dissolve 0.250 g of AgBr?Click here to review your answer to this question

Chapter 2 Basic Tools of Analytical Chemistry25form before we can complete the analysis. For example, one method for thequantitative analysis of disulfiram, C 10 H 20 N 2 S 4 —the active ingedient inthe drug Anatbuse—requires that we convert the sulfur to H 2 SO 4 by firstoxidizing it to SO 2 by combustion, and then oxidizing the SO 2 to H 2 SO 4by bubbling it through a solution of H 2 O 2 . When the conversion is complete,the amount of H 2 SO 4 is determined by titrating with NaOH.To convert the moles of NaOH used in the titration to the moles of disulfiramin the sample, we need to know the stoichiometry of the reactions.Writing a balanced reaction for H 2 SO 4 and NaOH is straightforwardHSO ( aq) + 2NaOH( aq) → 2H O() l + NaSO ( aq )2 4 2 2 4but the balanced reactions for the oxidations of C 10 H 20 N 2 S 4 to SO 2 , andof SO 2 to H 2 SO 4 are not as immediately obvious. Although we can balancethese redox reactions, it is often easier to deduce the overall stoichiometryby using a little chemical logic.Example 2.7An analysis for disulfiram, C 10 H 20 N 2 S 4 , in Antabuse is carried out byoxidizing the sulfur to H 2 SO 4 and titrating the H 2 SO 4 with NaOH. Ifa 0.4613-g sample of Antabuse is taken through this procdure, requiring34.85 mL of 0.02500 M NaOH to titrate the H 2 SO 4 , what is the %w/wdisulfiram in the sample?So l u t i o nCalculating the moles of H 2 SO 4 is easy—first, we calculate the moles ofNaOH used in the titration4( 0. 02500 M) × ( 0. 03485 L) = 8.7125× 10− molNaOHand then we use the balanced reaction to calcualte the correspondingmoles of H 2 SO 4 .48. 7125× 10− molNaOH×1mol HSO2 4= × − 44. 356210 mol HSO2mol NaOH2 4We do not need balanced reactions to convert the moles of H 2 SO 4 tothe corresponding moles of C 10 H 20 N 2 S 4 . Instead, we take advantage of aconservation of mass—all the sulfur in C 10 H 20 N 2 S 4 must end up in theH 2 SO 4 ; thus41 molS4. 3562× 10− molH SO × ×2 4molH SO1mol C H NS10 20 2 44 molS2 4= 1. 0890× 10−4 molC H NS10 20N SS NSdisulfram2 4Here is where we use a little chemical logic!A conservation of mass is the essence ofstoichiometry.S

26 Analytical Chemistry 2.0or41.0890× 10− molC H N S ×10 20 2 4296.54 gC H N S10 20 2 4molC H N S10 20 2 4= 0.032293gC H N S10 20 2 40.032293gC H N S10 20 2 4× 100 = 7.000% w/wC H N S0.4613 gsample10 20 2 42DBasic EquipmentThe array of equipment for making analytical measurements is impressive,ranging from the simple and inexpensive, to the complex and expensive.With three exceptions—measuring mass, measuring volume, and dryingmaterials—we will postpone the discussion of equipment to later chapterswhere its application to specific analytical methods is relevant.Although we tend to use interchangeably,the terms “weight” and “mass,” there isan important distinction between them.Mass is the absolute amount of matter inan object, measured in grams. Weight is ameasure of the gravitational force actingon the object:weight = mass × gravitational accelerationAn object has a fixed mass but its weightdepends upon the local acceleration dueto gravity, which varies subtly from location-to-location.A balance measures an object’s weight, notits mass. Because weight and mass are proportionalto each other, we can calibratea balance using a standard weight whosemass is traceable to the standard prototypefor the kilogram. A properly calibratedbalance will give an accurate value for anobject’s mass.2D.1 Equipment for Measuring MassAn object’s mass is measured using a digital electronic analytical balance(Figure 2.3). 2 An electromagnet levitates the sample pan above a permanentcylindrical magnet. The amount of light reaching a photodetector indicatesthe sample pan’s position. Without an object on the balance, the amountof light reaching the detector is the balance’s null point. Placing an objecton the balance displaces the sample pan downward by a force equal to theproduct of the sample’s mass and its acceleration due to gravity. The balancedetects this downward movement and generates a counterbalancingforce by increasing the current to the electromagnet. The current returningthe balance to its null point is proportional to the object’s mass. A typical2 For a review of other types of electronic balances, see Schoonover, R. M. Anal. Chem. 1982, 54,973A-980A.Figure 2.3 The photo shows a typical digital electronic balancecapable of determining mass to the nearest ±0.1 mg. Thesticker inside the balance’s wind shield is its annual calibrationcertification.

Chapter 2 Basic Tools of Analytical Chemistry27electronic balance has a capacity of 100-200 grams, and can measure massto the nearest ±0.01 mg to ±1 mg.If the sample is not moisture sensitive, a clean and dry container isplaced on the balance. The container’s mass is called the tare. Most balancesallow you to set the container’s tare to a mass of zero. The sample is transferredto the container, the new mass is measured and the sample’s massdetermined by subtracting the tare. Samples that absorb moisture fromthe air are treated differently. The sample is placed in a covered weighingbottle and their combined mass is determined. A portion of the sample isremoved and the weighing bottle and remaining sample are reweighed. Thedifference between the two masses gives the sample’s mass.Several important precautions help to minimize errors when measuringan object’s mass. A balance should be placed on a stable surface to minimizethe effect of vibrations in the surrounding environment, and shouldbe maintained in a level position. The sensitivity of an analytical balance issuch that it can measure the mass of a fingerprint. For this reason materialsbeing weighed should normally be handled using tongs or laboratorytissues. Volatile liquid samples must be weighed in a covered container toavoid the loss of sample by evaporation. Air currents can significantly affecta sample’s mass. To avoid air currents the balance’s glass doors shouldbe closed, or the balance’s wind shield should be in place. A sample thatis cooler or warmer than the surrounding air will create a convective aircurrents that affects the measurement of its mass. For this reason, warm orcool your sample to room temperature before determining its mass. Finally,samples dried in an oven should be stored in a desiccator to prevent themfrom reabsorbing moisture from the atmosphere.2D.2 Equipment for Measuring VolumeAnalytical chemists use a variety of glassware to measure a liquid’s volume.The choice of what type of glassware to use depends on how accurately weneed to know the liquid’s volume and whether we are interested in containingor delivering the liquid.A graduated cylinder is the simplest device for delivering a known volumeof a liquid reagent (Figure 2.4). The graduated scale allows you todeliver any volume up to the cylinder’s maximum. Typical accuracy is±1% of the maximum volume. A 100-mL graduated cylinder, for example,is accurate to ±1 mL.A Volumetric pipet provides a more accurate method for deliveringa known volume of solution. Several different styles of pipets are available,two of which are shown in Figure 2.5. Transfer pipets provide the most accuratemeans for delivering a known volume of solution. A transfer pipetdelivering less than 100 mL generally is accurate to the hundredth of a mL.Larger transfer pipets are accurate to the tenth of a mL. For example, the10-mL transfer pipet in Figure 2.5 will deliver 10.00 mL with an accuracyof ±0.02 mL.Figure 2.4 A 500-mL graduatedcylinder. Source: Hannes Grobe(commons.wikimedia.org).

28 Analytical Chemistry 2.0calibration markNever use your mouth to suck a solutioninto a pipet!Figure 2.6 Digital micropipets.From the left, the pipets deliver volumesof 50 μL–200 μL; 0.5 μL–10μL; 100 μL–1000 μL; 1 μL–20 μL.Source: Retama (commons.wikimedia.org).Figure 2.5 Two examples of 10-mL volumetric pipets. The pipet on the top is atransfer pipet and the pipet on the bottom is a Mohr measuring pipet. The transferpipet delivers a single volume of 10.00 mL when filled to its calibration mark. TheMohr pipet has a mark every 0.1 mL, allowing for the delivery of variable volumes.It also has additional graduations at 11 mL, 12 mL, and 12.5 mL.To fill a transfer pipet suction use a rubber bulb to pull the liquid uppast the calibration mark (see Figure 2.5). After replacing the bulb withyour finger, adjust the liquid’s level to the calibration mark and dry theoutside of the pipet with a laboratory tissue. Allow the pipet’s contents todrain into the receiving container with the pipet’s tip touching the innerwall of the container. A small portion of the liquid will remain in the pipet’stip and should not be blown out. With some measuring pipets any solutionremaining in the tip must be blown out.Delivering microliter volumes of liquids is not possible using transferor measuring pipets. Digital micropipets (Figure 2.6), which come in avariety of volume ranges, provide for the routine measurement of microlitervolumes.Graduated cylinders and pipets deliver a known volume of solution. Avolumetric flask, on the other hand, contains a specific volume of solution(Figure 2.7). When filled to its calibration mark a volumetric flaskcontaining less than 100 mL is generally accurate to the hundredth of amL, whereas larger volumetric flasks are accurate to the tenth of a mL. Forexample, a 10-mL volumetric flask contains 10.00 mL ± 0.02 mL and a250-mL volumetric flask contains 250.0 mL ± 0.12 mL.A recent report describes a nanopipetcapable of dispensing extremely smallvolumes. Scientists at the BrookhavenNational Laboratory used a germaniumnanowire to make a pipet delivering a 35zeptoliter (10 –21 L) drop of a liquid goldgermaniumalloy. You can read about thiswork in the April 21, 2007 issue of ScienceNews.Figure 2.7 A 500-mL volumetric flask. Source:Hannes Grobe (commons.wikimedia.org).

Chapter 2 Basic Tools of Analytical Chemistry29Because a volumetric flask contains a solution, it is useful for preparinga solution with an accurately known concentration. Transfer the reagent tothe volumetric flask and add enough solvent to bring the reagent into solution.Continuing adding solvent in several portions, mixing thoroughlyafter each addition. Adjust the volume to the flask’s calibration mark usinga dropper. Finally, complete the mixing process by inverting and shakingthe flask at least 10 times.If you look closely at a volumetric pipet or volumetric flask you will seemarkings similar to those shown in Figure 2.8. The text of the markings,which reads10 mL T. D. at 20 o C ±0.02 mLindicates that the pipet is calibrated to deliver (T. D.) 10 mL of solutionwith an uncertainty of ±0.02 mL at a temperature of 20 o C. The temperatureis important because glass expands and contracts with changes intemperatures. At higher or lower temperatures, the pipet’s accuracy is lessthan ±0.02 mL. For more accurate results you can calibrate your volumetricglassware at the temperature you are working. You can accomplish thisby weighing the amount of water contained or delivered and calculate thevolume using its temperature dependent density.You should take three additional precautions when working with pipetsand volumetric flasks. First, the volume delivered by a pipet or containedby a volumetric flask assumes that the glassware is clean. Dirt and greaseon the inner surface prevents liquids from draining evenly, leaving dropletsof the liquid on the container’s walls. For a pipet this means that the deliveredvolume is less than the calibrated volume, while drops of liquid abovethe calibration mark mean that a volumetric flask contains more than itscalibrated volume. Commercially available cleaning solutions can be usedto clean pipets and volumetric flasks.Second, when filling a pipet or volumetric flask the liquid’s level mustbe set exactly at the calibration mark. The liquid’s top surface is curvedinto a meniscus, the bottom of which should be exactly even with theglassware’s calibration mark (Figure 2.9). When adjusting the meniscuskeep your eye in line with the calibration mark to avoid parallax errors. Ifyour eye level is above the calibration mark you will overfill the pipet orvolumetric flask and you will underfill them if your eye level is below thecalibration mark.Finally, before using a pipet or volumetric flask rinse it with several smallportions of the solution whose volume you are measuring. This ensures theremoval of any residual liquid remaining in the pipet or volumetric flask.Figure 2.8 Close-up of the 10-mLtransfer pipet from Figure 2.5.A volumetric flask has similar markings,but uses the abbreviation T. C. for “to contain”in place of T. D.calibration markmeniscusFigure 2.9 Proper position of thesolution’s meniscus relative to thevolumetric flask’s calibration mark.2D.3 Equipment for Drying SamplesMany materials need to be dried prior to analysis to remove residual moisture.Depending on the material, heating to a temperature between 110

30 Analytical Chemistry 2.0Figure 2.10 Example of a mufflefurnace.Figure 2.11 Example of a desiccator.The solid in the bottom of thedesiccator is the desiccant, whichin this case is silica gel. Source:Hannes Grobe (commons.wikimedia.org).o C and 140 o C is usually sufficient. Other materials need much highertemperatures to initiate thermal decomposition.Conventional drying ovens provide maximum temperatures of 160 o Cto 325 o C (depending on the model). Some ovens include the ability tocirculate heated air, allowing for a more efficient removal of moisture andshorter drying times. Other ovens provide a tight seal for the door, allowingthe oven to be evacuated. In some situations a microwave oven can replace aconventional laboratory oven. Higher temperatures, up to 1700 o C, requirea muffle furnace (Figure 2.10).After drying or decomposing a sample, it should be cooled to roomtemperature in a desiccator to prevent the readsorption of moisture. Adesiccator (Figure 2.11) is a closed container that isolates the samplefrom the atmosphere. A drying agent, called a desiccant, is placed in thebottom of the container. Typical desiccants include calcium chloride andsilica gel. A perforated plate sits above the desiccant, providing a shelf forstoring samples. Some desiccators include a stopcock that allows them tobe evacuated.2EPreparing SolutionsPreparing a solution of known concentration is perhaps the most commonactivity in any analytical lab. The method for measuring out the soluteand solvent depend on the desired concentration unit and how exact thesolution’s concentration needs to be known. Pipets and volumetric flasksare used when a solution’s concentration must be exact; graduated cylinders,beakers and reagent bottles suffice when concentrations need only beapproximate. Two methods for preparing solutions are described in thissection.2E.1 Preparing Stock SolutionsA stock solution is prepared by weighing out an appropriate portion ofa pure solid or by measuring out an appropriate volume of a pure liquidand diluting to a known volume. Exactly how this is done depends on therequired concentration unit. For example, to prepare a solution with a desiredmolarity you weigh out an appropriate mass of the reagent, dissolveit in a portion of solvent, and bring to the desired volume. To prepare asolution where the solute’s concentration is a volume percent, you measureout an appropriate volume of solute and add sufficient solvent to obtainthe desired total volume.Example 2.8Describe how to prepare the following three solutions: (a) 500 mL of approximately0.20 M NaOH using solid NaOH; (b) 1 L of 150.0 ppm Cu 2+using Cu metal; and (c) 2 L of 4% v/v acetic acid using concentrated glacialacetic acid (99.8% w/w acetic acid).

Chapter 2 Basic Tools of Analytical Chemistry31So l u t i o n(a) Since the concentration is known to two significant figures the mass ofNaOH and the volume of solution do not need to be measured exactly.The desired mass of NaOH is0.20 molNaOHL40.0 gNaOH× × 050 . L = 40 . gmolNaOHTo prepare the solution, place 4.0 grams of NaOH, weighed to thenearest tenth of a gram, in a bottle or beaker and add approximately500 mL of water.(b) Since the concentration of Cu 2+ has four significant figures, the mass ofCu metal and the final solution volume must be measured exactly. Thedesired mass of Cu metal is150.0 mg Cu× 1. 000 L = 150.0 mgCu=0.1500 gCuLTo prepare the solution we measure out exactly 0.1500 g of Cu into asmall beaker and dissolve using small portion of concentrated HNO 3 .The resulting solution is transferred into a 1-L volumetric flask. Rinsethe beaker several times with small portions of water, adding each rinseto the volumetric flask. This process, which is called a quantitativetransfer, ensures that the complete transfer of Cu 2+ to the volumetricflask. Finally, additional water is added to the volumetric flask’s calibrationmark.(c) The concentration of this solution is only approximate so it is notnecessary to measure the volumes exactly, nor is it necessary to accountfor the fact that glacial acetic acid is slightly less than 100% w/w aceticacid (it is approximately 99.8% w/w). The necessary volume of glacialacetic acid is4 mL CH COOH3× 2000 mL = 80 mL CH COOH3100 mLTo prepare the solution, use a graduated cylinder to transfer 80 mL ofglacial acetic acid to a container that holds approximately 2 L and addsufficient water to bring the solution to the desired volume.Practice Exercise 2.5Provide instructions for preparing 500 mL of 0.1250 M KBrO 3 .Click here to review your answer to this exercise.

32 Analytical Chemistry 2.02E.2 Preparing Solutions by DilutionSolutions are often prepared by diluting a more concentrated stock solution.A known volume of the stock solution is transferred to a new containerPractice Exercise 2.6To prepare a standard solution of Zn 2+ you dissolve a 1.004 g sample ofZn wire in a minimal amount of HCl and dilute to volume in a 500-mLvolumetric flask. If you dilute 2.000 mL of this stock solution to 250.0mL, what is the concentration of Zn 2+ , in μg/mL, in your standard solution?Click here to review your answer to this exercise.Equation 2.2 applies only when the concentrationare written in terms of volume,as is the case with molarity. Using thisequation with a mass-based concentrationunit, such as % w/w, leads to an error. SeeRodríquez-López, M.; Carrasquillo, A. J.Chem. Educ. 2005, 82, 1327-1328 forfurther discussion.and brought to a new volume. Since the total amount of solute is the samebefore and after dilution, we know thatC × V = C × Vo o d d2.2where C o is the stock solution’s concentration, V o is the volume of stocksolution being diluted, C d is the dilute solution’s concentration, and V dis the volume of the dilute solution. Again, the type of glassware used tomeasure V o and V d depends on how exact the solution’s concentration mustbe known.Example 2.9A laboratory procedure calls for 250 mL of an approximately 0.10 M solutionof NH 3 . Describe how you would prepare this solution using a stocksolution of concentrated NH 3 (14.8 M).So l u t i o nSubstituting known volumes in equation 2.214. 8 M× V = 010 . M×025 . Loand solving for V o gives 1.69 × 10 -3 liters, or 1.7 mL. Since we are makinga solution that is approximately 0.10 M NH 3 we can use a graduatedcylinder to measure the 1.7 mL of concentrated NH 3 , transfer the NH 3 toa beaker, and add sufficient water to give a total volume of approximately250 mL.As shown in the following example, we can use equation 2.2 to calculatea solution’s original concentration using its known concentration afterdilution.

Chapter 2 Basic Tools of Analytical Chemistry33Example 2.10A sample of an ore was analyzed for Cu 2+ as follows. A 1.25 gram sampleof the ore was dissolved in acid and diluted to volume in a 250-mL volumetricflask. A 20 mL portion of the resulting solution was transferred bypipet to a 50-mL volumetric flask and diluted to volume. An analysis ofthis solution gave the concentration of Cu 2+ as 4.62 μg/L. What is theweight percent of Cu in the original ore?So l u t i o nSubstituting known volumes (with significant figures appropriate for pipetsand volumetric flasks) into equation 2.22+2+( µ g/LCu ) × 20. 00 mL= 462 . µ g/L Cu × 5000mL .oand solving for (μg/L Cu 2+ ) o gives the original solution concentration as11.55 μg/L Cu 2+ . To calculate the grams of Cu 2+ we multiply this concentrationby the total volume11.55µ gCumL2+The weight percent Cu is1g× 250. 0 mL× = 288 . 8× 10−3 gCu 2+610 µ g3288 . 8×10− gCu125 . gsample2+× 100 = 0. 231%w/w Cu 2+2FSpreadsheets and Computational SoftwareAnalytical chemistry is an inherently quantitative discipline. Whether youare completing a statistical analysis, trying to optimize experimental conditions,or exploring how a change in pH affects a compound’s solubility, theability to work with complex mathematical equations is essential. Spreadsheets,such as Microsoft Excel can be an important tool for analyzing yourdata and for preparing graphs of your results. Scattered throughout the textyou will find instructions for using spreadsheets.Although spreadsheets are useful, they are not always well suited forworking with scientific data. If you plan to pursue a career in chemistry youmay wish to familiarize yourself with a more sophisticated computationalsoftware package, such as the freely available open-source program that goesby the name R, or commercial programs such as Mathematica and Matlab.You will find instructions for using R scattered throughout the text.Despite the power of spreadsheets and computational programs, don’tforget that the most important software is that behind your eyes and betweenyour ears. The ability to think intuitively about chemistry is a critically importantskill. In many cases you will find it possible to determine if an ana-If you do not have access to Microsoft Excelor another commercial spreadsheet package,you might considering using Calc, afreely available open-source spreadsheetthat is part of the OpenOffice.org softwarepackage at www.openoffice.org.You can download the current version ofR from www.r-project.org. Click on thelink for Download: CRAN and find a localmirror site. Click on the link for themirror site and then use the link for Linux,MacOS X, or Windows under the heading“Download and Install R.”

34 Analytical Chemistry 2.0lytical method is feasible, or to approximate the optimum conditions for ananalytical method without resorting to complex calculations. Why spendtime developing a complex spreadsheet or writing software code when a“back-of-the-envelope” estimate will do the trick? Once you know the generalsolution to your problem, you can spend time using spreadsheets andcomputational programs to work out the specifics. Throughout the text wewill introduce tools for developing your ability to think intuitively.2GThe Laboratory NotebookFinally, we can not end a chapter on the basic tools of analytical chemistrywithout mentioning the laboratory notebook. Your laboratory notebook isyour most important tool when working in the lab. If kept properly, youshould be able to look back at your laboratory notebook several years fromnow and reconstruct the experiments on which you worked.Your instructor will provide you with detailed instructions on how heor she wants you to maintain your notebook. Of course, you should expectto bring your notebook to the lab. Everything you do, measure, or observewhile working in the lab should be recorded in your notebook as it takesplace. Preparing data tables to organize your data will help ensure that yourecord the data you need, and that you can find the data when it is time tocalculate and analyze your results. Writing a narrative to accompany yourdata will help you remember what you did, why you did it, and why youthought it was significant. Reserve space for your calculations, for analyzingyour data, and for interpreting your results. Take your notebook with youwhen you do research in the library.Maintaining a laboratory notebook may seem like a great deal of effort,but if you do it well you will have a permanent record of your work.Scientists working in academic, industrial and governmental research labsrely on their notebooks to provide a written record of their work. Questionsabout research carried out at some time in the past can be answeredby finding the appropriate pages in the laboratory notebook. A laboratorynotebook is also a legal document that helps establish patent rights andproof of discovery.As you review this chapter, try to define akey term in your own words. Check youranswer by clicking on the key term, whichwill take you to the page where it was firstintroduced. Clicking on the key termthere, will bring you back to this page sothat you can continue with another keyterm.2H Key Termsanalytical balance concentration desiccantdesiccator dilution formalitymeniscus molality molaritynormality parts per million parts per billionp-function quantitative transfer scientific notationsignificant figures SI units stock solutionvolume percent volumetric flask volumetric pipetweight percentweight-to-volume percent

Chapter 2 Basic Tools of Analytical Chemistry352IChapter SummaryThere are a few basic numerical and experimental tools with which youmust be familiar. Fundamental measurements in analytical chemistry, suchas mass and volume, use base SI units, such as the kilogram. Other units,such as energy, are defined in terms of these base units. When reportingmeasurements, we must be careful to include only those digits that aresignificant, and to maintain the uncertainty implied by these significantfigures when transforming measurements into results.The relative amount of a constituent in a sample is expressed as a concentration.There are many ways to express concentration, the most commonof which are molarity, weight percent, volume percent, weight-to-volumepercent, parts per million and parts per billion. Concentrations also can beexpressed using p-functions.Stoichiometric relationships and calculations are important in manyquantitative analyses. The stoichiometry between the reactants and productsof a chemical reaction are given by the coefficients of a balanced chemicalreaction.Balances, volumetric flasks, pipets, and ovens are standard pieces ofequipment that you will routinely use in the analytical lab. You should befamiliar with the proper way to use this equipment. You also should befamiliar with how to prepare a stock solution of known concentration, andhow to prepare a dilute solution from a stock solution.2JProblems1. Indicate how many significant figures are in each of the following numbers.a. 903 b. 0.903 c. 1.0903d. 0.0903 e. 0.09030 f. 9.03x10 22. Round each of the following to three significant figures.a. 0.89377 b. 0.89328 c. 0.89350d. 0.8997 e. 0.08907Answers, but not worked solutions, tomost end-of-chapter problems are availablehere.3. Round each to the stated number of significant figures.a. the atomic weight of carbon to 4 significant figuresb. the atomic weight of oxygen to 3 significant figuresc. Avogadro’s number to 4 significant figuresd. Faraday’s constant to 3 significant figures4. Report results for the following calculations to the correct number ofsignificant figures.a. 4.591 + 0.2309 + 67.1 =

36 Analytical Chemistry 2.0b. 313 – 273.15 =c. 712 × 8.6 =d. 1.43/0.026 =e. (8.314 × 298)/96485 =f. log(6.53 × 10 –5 ) =g. 10 –7.14 =h. (6.51 × 10 –5 ) × (8.14 × 10 –9 ) =5. A 12.1374 g sample of an ore containing Ni and Co was carried throughFresenius’ analytical scheme shown in Figure 1.1. At point A the combinedmass of Ni and Co was found to be 0.2306 g, while at point Bthe mass of Co was found to be 0.0813 g. Report the weight percentNi in the ore to the correct number of significant figures.6. Figure 1.2 shows an analytical method for the analysis of Ni in oresbased on the precipitation of Ni 2+ using dimethylglyoxime. The formulafor the precipitate is Ni(C 4 H 14 N 4 O 4 ) 2 . Calculate the precipitate’sformula weight to the correct number of significant figures.7. An analyst wishes to add 256 mg of Cl – to a reaction mixture. Howmany mL of 0.217 M BaCl 2 is this?8. The concentration of lead in an industrial waste stream is 0.28 ppm.What is its molar concentration?9. Commercially available concentrated hydrochloric acid is 37.0% w/wHCl. Its density is 1.18 g/mL. Using this information calculate (a) themolarity of concentrated HCl, and (b) the mass and volume (in mL) ofsolution containing 0.315 moles of HCl.10. The density of concentrated ammonia, which is 28.0% w/w NH 3 , is0.899 g/mL. What volume of this reagent should be diluted to 1.0 × 10 3mL to make a solution that is 0.036 M in NH 3 ?11. A 250.0 mL aqueous solution contains 45.1 µg of a pesticide. Expressthe pesticide’s concentration in weight percent, in parts per million,and in parts per billion.12. A city’s water supply is fluoridated by adding NaF. The desired concentrationof F – is 1.6 ppm. How many mg of NaF should be added pergallon of treated water if the water supply already is 0.2 ppm in F – ?13. What is the pH of a solution for which the concentration of H + is6.92 × 10 –6 M? What is the [H + ] in a solution whose pH is 8.923?

Chapter 2 Basic Tools of Analytical Chemistry3714. When using a graduate cylinder, the absolute accuracy with which youcan deliver a given volume is ±1% of the cylinder’s maximum volume.What are the absolute and relative uncertainties if you deliver 15 mLof a reagent using a 25 mL graduated cylinder? Repeat for a 50 mLgraduated cylinder.15. Calculate the molarity of a potassium dichromate solution prepared byplacing 9.67 grams of K 2 Cr 2 O 7 in a 100-mL volumetric flask, dissolving,and diluting to the calibration mark.16. For each of the following explain how you would prepare 1.0 L of asolution that is 0.10 M in K + . Repeat for concentrations of 1.0 × 10 2ppm K + and 1.0% w/v K + .a. KCl b. K 2 SO 4 c. K 3 Fe(CN) 617. A series of dilute NaCl solutions are prepared starting with an initialstock solution of 0.100 M NaCl. Solution A is prepared by pipeting10 mL of the stock solution into a 250-mL volumetric flask and dilutingto volume. Solution B is prepared by pipeting 25 mL of solution Ainto a 100-mL volumetric flask and diluting to volume. Solution C isprepared by pipeting 20 mL of solution B into a 500-mL volumetricflask and diluting to volume. What is the molar concentration of NaClin solutions A, B and C?This is an example of a serial dilution,which is a useful method for preparingvery dilute solutions of reagents.18. Calculate the molar concentration of NaCl, to the correct number ofsignificant figures, if 1.917 g of NaCl is placed in a beaker and dissolvedin 50 mL of water measured with a graduated cylinder. If this solution isquantitatively transferred to a 250-mL volumetric flask and diluted tovolume, what is its concentration to the correct number of significantfigures?19. What is the molar concentration of NO 3–in a solution prepared bymixing 50.0 mL of 0.050 M KNO 3 with 40.0 mL of 0.075 M NaNO 3 ?What is pNO 3 for the mixture?20. What is the molar concentration of Cl – in a solution prepared by mixing25.0 mL of 0.025 M NaCl with 35.0 mL of 0.050 M BaCl 2 ? Whatis pCl for the mixture?21. To determine the concentration of ethanol in cognac a 5.00 mL sampleof cognac is diluted to 0.500 L. Analysis of the diluted cognac gives anethanol concentration of 0.0844 M. What is the molar concentrationof ethanol in the undiluted cognac?

38 Analytical Chemistry 2.02KSolutions to Practice ExercisesPractice Exercise 2.1The correct answer to this exercise is 1.9×10 –2 . To see why this is correct,let’s work through the problem in a series of steps. Here is the originalproblem−3 −20. 250× ( 993 . × 10 ) − 0. 100× ( 1. 927×10 )=−3 −2993 . × 10 + 1.927×10Following the correct order of operations we first complete the two multiplicationsin the numerator. In each case the answer has three significantfigures, although we retain an extra digit, highlight in red, to avoid roundofferrors.−248 . 2× 10 − 1.927×103 −3−993 . × 10 + 1.927×103 −2=Completing the subtraction in the numerator leaves us with two significantfigures since the last significant digit for each value is in the hundredthsplace.055 . 5×10−3−993 . × 10 + 1.927×103 −2=The two values in the denominator have different exponents. Because weare adding together these values, we first rewrite them using a commonexponent.055 . 5×10−3−0. 993× 10 + 1.927×102 −2=The sum in the denominator has four significant figures since each valuehas three decimal places.055 . 5×102.920×10−3−2=Finally, we complete the division, which leaves us with a result having twosignificant figures.055 . 5×102.920×10−3−2= 19 . × 10−2Click here to return to the chapter.Practice Exercise 2.2The concentrations of the two solutions are

Chapter 2 Basic Tools of Analytical Chemistry39050 . mol NaCl 58.43 gNaCl 10 6 µ g 1 L× × × = 29 . × 10 3 µg/mLL molNaCl g 1000 mL025 . mol SrCl 158.5 gSrCl 10 62 2µg 1 L× × × = 40 . × 10 3 µ g/mLL molNaCl g 1000 mLThe solution of SrCl 2 has the larger concentration when expressed in μg/mL instead of in mol/L.Click here to return to the chapter.Practice Exercise 2.3The concentrations of K + and SO 2– 4 are+15 . gK SO 1 molK SO2 4 2 42 molK× × = 34 . 4× 10−0. 500 L 174.3 gK SO molK SO2 42 42MK+2−15 . gK SO 1 molK SO 1mol SO2 4 2 44× × = 17 . 2×100. 500 L 174.3 gK SO molK SO2 42 4The pK and pSO 4 values arepK = –log(3.44× 10 –2 ) = 1.46pSO 4 = –log(1.72 × 10 -2 ) = 1.76Click here to return to the chapter.Practice Exercise 2.4First, we find the moles of AgBr1mol AgBr0.250 gAgBr × = 133 . 1× 10 −3187.8 gAgBrand then the moles and volume of Na 2 S 2 O 332mol Na SO133 . 1× 10 −2 2 3molAgBr × = 2.66210molAgBr31 L266 . 2× 10− molNaSO ×2 2 30.0138 molNaSO2 2 3Click here to return to the chapter.Practice Exercise 2.5Preparing 500 mL of 0.1250 M KBrO 3 requiresMSO−2 2−4× −31000 mL× = 193 mLL

40 Analytical Chemistry 2.00. 1250 molKBrO 167.00 gKBrO0.500 L× ×Lmol KBrO3 33= 10.44gKBrO3Because the concentration has four significant figures, we must prepare thesolution using volumetric glassware. Place a 10.44 g sample of KBrO 3 in a500-mL volumetric flask and fill part way with water. Swirl to dissolve theKBrO 3 and then dilute with water to the flask’s calibration mark.Click here to return to the chapter.Practice Exercise 2.6The first solution is a stock solution, which we then dilute to prepare thestandard solution. The concentration of Zn 2+ in the stock solution is61. 004 gZn 10 µ g× = 2008 µ g/mL500 mL gTo find the concentration of the standard solution we use equation 2.32008 µ gZnmL2+× 2. 000 mL = C × 250.0 mLwhere C new is the standard solution’s concentration. Solving gives a concentrationof 16.06 μg Zn 2+ /mL.Click here to return to the chapter.new

DRAFTChapter 3The Vocabularyof Analytical ChemistryChapter Overview3A Analysis, Determination, and Measurement3B Techniques, Methods, Procedures, and Protocols3C Classifying Analytical Techniques3D Selecting an Analytical Method3E Developing the Procedure3F Protocols3G The Importance of Analytical Methodology3H Key Terms3I Chapter Summary3J Problems3K Solutions to Practice ExercisesIf you leaf through an issue of the journal Analytical Chemistry, you will soon discover that theauthors and readers share a common vocabulary of analytical terms. You are probably familiarwith some of these terms, such as accuracy and precision, but other terms, such as analyteand matrix may be less familiar to you. In order to participate in the community of analyticalchemists, you must first understand its vocabulary. The goal of this chapter, therefore, is tointroduce you to some important analytical terms. Becoming comfortable with these terms willmake the material in the chapters that follow easier to read and understand.Copyright: David Harvey, 200941

42 Analytical Chemistry 2.0Fecal coliform counts provide a generalmeasure of the presence of pathogenic organismsin a water supply. For drinkingwater, the current maximum contaminantlevel (MCL) for total coliforms, includingfecal coliforms is less than 1 colony/100mL. Municipal water departments mustregularly test the water supply and musttake action if more than 5% of the samplesin any month test positive for coliformbacteria.3AAnalysis, Determination and MeasurementThe first important distinction we will make is among the terms analysis,determination, and measurement. An analysis provides chemical or physicalinformation about a sample. The component of interest in the sampleis called the analyte, and the remainder of the sample is the matrix. In ananalysis we determine the identity, concentration, or properties of an analyte.To make this determination we measure one or more of the analyte’schemical or physical properties.An example helps clarify the difference between an analysis, a determinationand a measurement. In 1974 the federal government enactedthe Safe Drinking Water Act to ensure the safety of public drinking watersupplies. To comply with this act, municipalities regularly monitor theirdrinking water supply for potentially harmful substances. One such substanceis fecal coliform bacteria. Municipal water departments collect andanalyze samples from their water supply. They determine the concentrationof fecal coliform bacteria by passing a portion of water through a membranefilter, placing the filter in a dish containing a nutrient broth, and incubatingfor 22–24 hr at 44.5 o C ± 0.2 o C. At the end of the incubation period theycount the number of bacterial colonies in the dish and report the result asthe number of colonies per 100 mL (Figure 3.1). Thus, municipal waterdepartments analyze samples of water to determine the concentration offecal coliform bacteria by measuring the number of bacterial colonies thatform during a carefully defined incubation period.Figure 3.1 Colonies of fecal coliform bacteria from a water supply. Source: SusanBoyer (www.ars.usda.gov).

Chapter 3 The Vocabulary of Analytical Chemistry433BTechniques, Methods, Procedures, and ProtocolsSuppose you are asked to develop an analytical method to determine theconcentration of lead in drinking water. How would you approach thisproblem? To provide a structure for answering this question let’s draw adistinction among four levels of analytical methodology: techniques, methods,procedures, and protocols. 1A technique is any chemical or physical principle we can use to studyan analyte. There are many techniques for determining the concentration oflead in drinking water. 2 In graphite furnace atomic absorption spectroscopy(GFAAS), for example, we first convert aqueous lead ions into a free atomstate—a process we call atomization. We then measure the amount of lightabsorbed by the free atoms. Thus, GFAAS uses both a chemical principle(atomization) and a physical principle (absorption of light).A method is the application of a technique for a specific analyte in aspecific matrix. As shown in Figure 3.2, the GFAAS method for determininglead in water is different from that for lead in soil or blood.A procedure is a set of written directions telling us how to apply amethod to a particular sample, including information on obtaining samples,handling interferents, and validating results. A method may have severalprocedures as each analyst or agency adapts it to a specific need. As shownin Figure 3.2, the American Public Health Agency and the American Societyfor Testing Materials publish separate procedures for determining theconcentration of lead in water.See Chapter 10 for a discussion of graphitefurnace atomic absorption spectroscopy.Chapters 8–13 provide coverage for arange of important analytical techniques.1 Taylor, J. K. Anal. Chem. 1983, 55, 600A–608A.2 Fitch, A.; Wang, Y.; Mellican, S.; Macha, S. Anal. Chem. 1996, 68, 727A–731A.TechniquesGraphite Furnace Atomic Absorption Spectroscopy(GFAAS)MethodsPb in SoilPb in WaterPb in BloodProceduresAPHAASTMFigure 3.2 Chart showing the hierarchicalrelationship among a technique,methods using that technique, and proceduresand protocols for one method.ProtocolsEPAThe abbreviations are APHA: AmericanPublic Health Association, ASTM:American Society for Testing Materials,EPA: Environmental ProtectionAgency.

44 Analytical Chemistry 2.01 2Figure 3.3 Graduated cylinderscontaining 0.10 M Cu(NO 3 ) 2 .Although the cylinders containthe same concentration of Cu 2+ ,the cylinder on the left contains1.0 × 10 -4 mol Cu 2+ and the cylinderon the right contains 2.0 × 10 -4mol Cu 2+ .Historically, most early analytical methodsused a total analysis technique. Forthis reason, total analysis techniques areoften called “classical” techniques.Finally, a protocol is a set of stringent guidelines specifying a procedurethat must be followed if an agency is to accept the results. Protocolsare common when the result of an analysis supports or defines public policy.When determining the concentration of lead in water under the SafeDrinking Water Act, for example, labs must use a protocol specified by theEnvironmental Protection Agency.There is an obvious order to these four levels of analytical methodology.Ideally, a protocol uses a previously validated procedure. Before developingand validating a procedure, a method of analysis must be selected. Thisrequires, in turn, an initial screening of available techniques to determinethose that have the potential for monitoring the analyte.3CClassifying Analytical TechniquesAnalyzing a sample generates a chemical or physical signal that is proportionalto the amount of analyte in the sample. This signal may be anythingwe can measure, such as mass or absorbance. It is convenient to divideanalytical techniques into two general classes depending on whether thesignal is proportional to the mass or moles of analyte, or to the analyte’sconcentration.Consider the two graduated cylinders in Figure 3.3, each containing asolution of 0.010 M Cu(NO 3 ) 2 . Cylinder 1 contains 10 mL, or 1.0 × 10 -4moles of Cu 2+ , and cylinder 2 contains 20 mL, or 2.0 × 10 -4 moles of Cu 2+ .If a technique responds to the absolute amount of analyte in the sample,then the signal due to the analyte, S A , isS= k n3.1A A Awhere n A is the moles or grams of analyte in the sample, and k A is a proportionalityconstant. Since cylinder 2 contains twice as many moles of Cu 2+ ascylinder 1, analyzing the contents of cylinder 2 gives a signal that is twicethat of cylinder 1.A second class of analytical techniques are those that respond to theanalyte’s concentration, C AS = k CA A A3.2Since the solutions in both cylinders have the same concentration of Cu 2+ ,their analysis yields identical signals.A technique responding to the absolute amount of analyte is a totalanalysis technique. Mass and volume are the most common signals for atotal analysis technique, and the corresponding techniques are gravimetry(Chapter 8) and titrimetry (Chapter 9). With a few exceptions, the signalfor a total analysis technique is the result of one or more chemical reactionsinvolving the analyte. These reactions may involve any combination of precipitation,acid–base, complexation, or redox chemistry. The stoichiometryof the reactions determines the value of k A in equation 3.1.

Chapter 3 The Vocabulary of Analytical Chemistry45Spectroscopy (Chapter 10) and electrochemistry (Chapter 11), inwhich an optical or electrical signal is proportional to the relative amountof analyte in a sample, are examples of concentration techniques. Therelationship between the signal and the analyte’s concentration is a theoreticalfunction that depends on experimental conditions and the instrumentationused to measure the signal. For this reason the value of k A in equation3.2 must be determined experimentally.Since most concentration techniques relyon measuring an optical or electrical signal,they also are known as “instrumental”techniques.3DSelecting an Analytical MethodA method is the application of a technique to a specific analyte in a specificmatrix. We can develop an analytical method for determining the concentrationof lead in drinking water using any of the techniques mentionedin the previous section. A gravimetric method, for example, might precipitatethe lead as PbSO 4 or PbCrO 4 , and use the precipitate’s mass as theanalytical signal. Lead forms several soluble complexes, which we can useto design a complexation titrimetric method. As shown in Figure 3.2, wecan use graphite furnace atomic absorption spectroscopy to determine theconcentration of lead in drinking water. Finally, the availability of multipleoxidation states (Pb 0 , Pb 2+ , Pb 4+ ) makes electrochemical methods feasible.The requirements of the analysis determine the best method. In choosinga method, consideration is given to some or all the following designcriteria: accuracy, precision, sensitivity, selectivity, robustness, ruggedness,scale of operation, analysis time, availability of equipment, and cost.3D.1 AccuracyAccuracy is how closely the result of an experiment agrees with the “true”or expected result. We can express accuracy as an absolute error, eor as a percentage relative error, %e re = obtained result - expected resultobtained result − expected result% e =×100rexpected resultA method’s accuracy depends on many things, including the signal’s source,the value of k A in equation 3.1 or equation 3.2, and the ease of handlingsamples without loss or contamination. In general, methods relying on totalanalysis techniques, such as gravimetry and titrimetry, produce results ofhigher accuracy because we can measure mass and volume with high accuracy,and because the value of k A is known exactly through stoichiometry.Since it is unlikely that we know the trueresult, we use an expected or acceptedresult when evaluating accuracy. For example,we might use a reference standard,which has an accepted value, to establishan analytical method’s accuracy.You will find a more detailed treatment ofaccuracy in Chapter 4, including a discussionof sources of errors.

46 Analytical Chemistry 2.0(a)Figure 3.4 Two determinations ofthe concentration of K + in serum,showing the effect of precision onthe distribution of individual results.The data in (a) are less scatteredand, therefore, more precisethan the data in (b).5.8 5.9 6.0 6.1 6.2ppm K(b)5.8 5.9 6.0 6.1 6.2ppm KConfusing accuracy and precision is acommon mistake. See Ryder, J.; Clark, A.U. Chem. Ed. 2002, 6, 1–3, and Tomlinson,J.; Dyson, P. J.; Garratt, J. U. Chem.Ed. 2001, 5, 16–23 for discussions of thisand other common misconceptions aboutthe meaning of error.You will find a more detailed treatment ofprecision in Chapter 4, including a discussionof sources of errors.Confidence, as we will see in Chapter 4,is a statistical concept that builds on theidea of a population of results. For thisreason, we will postpone our discussion ofdetection limits to Chapter 4. For now,the definition of a detection limit givenhere is sufficient.3D.2 PrecisionWhen a sample is analyzed several times, the individual results are rarely thesame. Instead, the results are randomly scattered. Precision is a measure ofthis variability. The closer the agreement between individual analyses, themore precise the results. For example, in determining the concentrationof K + in serum the results shown in Figure 3.4(a) are more precise thanthose in Figure 3.4(b). It is important to understand that precision doesnot imply accuracy. That the data in Figure 3.4(a) are more precise doesnot mean that the first set of results is more accurate. In fact, neither set ofresults may be accurate.A method’s precision depends on several factors, including the uncertaintyin measuring the signal and the ease of handling samples reproducibly.In most cases we can measure the signal for a total analysis methodwith a higher precision than the corresponding signal for a concentrationmethod. Precision is covered in more detail in Chapter 4.3D.3 SensitivityThe ability to demonstrate that two samples have different amounts of analyteis an essential part of many analyses. A method’s sensitivity is a measureof its ability to establish that such differences are significant. Sensitivityis often confused with a method’s detection limit, which is the smallestamount of analyte that we can determine with confidence.Sensitivity is equivalent to the proportionality constant, k A , in equation3.1 and equation 3.2. 3 If DS A is the smallest difference that we can measurebetween two signals, then the smallest detectable difference in the absoluteamount or relative amount of analyte is3 IUPAC Compendium of Chemical Terminology, Electronic version, http://goldbook.iupac.org/S05606.html.

Chapter 3 The Vocabulary of Analytical Chemistry47∆nA∆S∆SA= or ∆C=AkkASuppose, for example, that our analytical signal is a measurement of massusing a balance whose smallest detectable increment is ±0.0001 g. If ourmethod’s sensitivity is 0.200, then our method can conceivably detect adifference in mass of as little as∆n A= ± 0.0001 g=± 0.0005 g0.200For two methods with the same DS A , the method with the greater sensitivity—thelarger k A —is better able to discriminate between smaller amountsof analyte.3D.4 Specificity and SelectivityAn analytical method is specific if its signal depends only on the analyte. 4Although specificity is the ideal, few analytical methods are completelyfree from the influence of interfering species. When an interferent contributesto the signal, we expand equation 3.1 and equation 3.2 to includeits contribution to the sample’s signal, S sampS = S + S = k n + knsamp A I A A I I3.3S = S + S = k C + kCsamp A I A A I I3.4AAwhere S I is the interferent’s contribution to the signal, k I is the interferent’ssensitivity, and n I and C I are the moles (or grams) and concentration of theinterferent in the sample.Selectivity is a measure of a method’s freedom from interferences. 5The selectivity of a method for the interferent relative to the analyte is definedby a selectivity coefficient, K A,IKA,IkI= 3.5kwhich may be positive or negative depending on the sign of k I and k A . Theselectivity coefficient is greater than +1 or less than –1 when the method ismore selective for the interferent than for the analyte.Determining the selectivity coefficient’s value is easy if we already knowthe values for k A and k I . As shown by Example 3.1, we also can determineK A,I by measuring S samp in the presence of and in the absence of theinterferent.4 (a) Persson, B-A; Vessman, J. Trends Anal. Chem. 1998, 17, 117–119; (b) Persson, B-A; Vessman,J. Trends Anal. Chem. 2001, 20, 526–532.5 Valcárcel, M.; Gomez-Hens, A.; Rubio, S. Trends Anal. Chem. 2001, 20, 386–393.AAlthough k A and k I are usually positive,they also may be negative. For example,some analytical methods work by measuringthe concentration of a species that reactswith the analyte. As the analyte’s concentrationincreases, the concentration ofthe species producing the signal decreases,and the signal becomes smaller. If the signalin the absence of analyte is assigned avalue of zero, then the subsequent signalsare negative.

48 Analytical Chemistry 2.0If you are unsure why the signal in thepresence of zinc is 100.5, note that thepercentage relative error for this problemis given byobtained result − 100100× 100 = + 05 .%Solving gives an obtained result of 100.5.Example 3.1A method for the analysis of Ca 2+ in water suffers from an interference inthe presence of Zn 2+ . When the concentration of Ca 2+ is 100 times greaterthan that of Zn 2+ the analysis for Ca 2+ gives a relative error of +0.5%. Whatis the selectivity coefficient for this method?So l u t i o nSince only relative concentrations are reported, we can arbitrarily assign absoluteconcentrations. To make the calculations easy, we will let C Ca = 100(arbitrary units) and C Zn = 1. A relative error of +0.5% means that thesignal in the presence of Zn 2+ is 0.5% greater than the signal in the absenceof zinc. Again, we can assign values to make the calculation easier. If thesignal in the absence of zinc is 100 (arbitrary units), then the signal in thepresence of zinc is 100.5.The value of k Ca is determined using equation 3.2kCaSCa100= = = 1C 100In the presence of zinc the signal is given by equation 3.4; thusCaS = 100. 5= k C + k C = ( 1× 100)+ k × 1samp Ca Ca Zn Zn ZnSolving for k Zn gives a value of 0.5. The selectivity coefficient iskZn05 .K = = = 05 .Ca,Znk 1CaPractice Exercise 3.1Wang and colleagues describe a fluorescence method for the analysis ofAg + in water. When analyzing a solution containing 1.0 × 10 -9 M Ag +and 1.1× 10 -7 M Ni 2+ the fluorescence intensity (the signal) was +4.9%greater than that obtained for a sample of 1.0 × 10 -9 M Ag + . What isK Ag,Ni for this analytical method? The full citation for the data in thisexercise is Wang, L.; Liang, A. N.; Chen, H.; Liu, Y.; Qian, B.; Fu, J.Anal. Chim. Acta 2008, 616, 170-176.Click here to review your answer to this exercise.The selectivity coefficient provides us with a useful way to evaluate aninterferent’s potential effect on an analysis. Solving equation 3.5 for k Ik = K × kI A,I A3.6substituting in equation 3.3 and equation 3.4, and simplifying gives

50 Analytical Chemistry 2.0Solving this inequality for the ratio C I /C A and substituting in thevalue for K A,I from part (a) givesCCIA0. 0050 0.0050≤ = = 0.033K 015 .A,ITherefore, the concentration of 6-methoxycodeine can not exceed3.3% of codeine’s concentration.Look back at Figure 1.1, which shows Fresenius’analytical method for the determinationof nickel in ores. The reason thereare so many steps in this procedure is thatprecipitation reactions generally are notvery selective. The method in Figure 1.2includes fewer steps because dimethylglyoximeis a more selective reagent. Even so,if an ore contains palladium, additionalsteps will be needed to prevent the palladiumfrom interfering.Practice Exercise 3.2Mercury (II) also is an interferent in the fluorescence method for Ag +developed by Wang and colleagues (see Practice Exercise 3.1 for the citation).The selectivity coefficient, K Ag,Hg has a value of -1.0 × 10 -3 .(a) What is the significance of the selectivity coefficient’s negative sign?(b) Suppose you plan to use this method to analyze solutions with concentrationsof Ag + that are no smaller than 1.0 nM . What is themaximum concentration of Hg 2+ you can tolerate to ensure that yourpercentage relative errors are less than ±1.0%?Click here to review your answers to this exercise.When a method’s signal is the result of a chemical reaction—for example,when the signal is the mass of a precipitate—there is a good chance thatthe method is not very selective and that it is susceptible to interferences.Problems with selectivity also are more likely when the analyte is present ata very low concentration. 73D.5 Robustness and RuggednessFor a method to be useful it must provide reliable results. Unfortunately,methods are subject to a variety of chemical and physical interferencesthat contribute uncertainty to the analysis. When a method is relativelyfree from chemical interferences, we can use it on many analytes in a widevariety of sample matrices. Such methods are considered robust.Random variations in experimental conditions also introduces uncertainty.If a method’s sensitivity, k, is too dependent on experimental conditions,such as temperature, acidity, or reaction time, then a slight change inany of these conditions may give a significantly different result. A ruggedmethod is relatively insensitive to changes in experimental conditions.3D.6 Scale of OperationAnother way to narrow the choice of methods is to consider three potentiallimitations: the amount of sample available for the analysis, the expectedconcentration of analyte in the samples, and the minimum amount of ana-7 Rodgers, L. B. J. Chem. Educ. 1986, 63, 3–6.

Chapter 3 The Vocabulary of Analytical Chemistry51lyte that produces a measurable signal. Collectively, these limitations definethe analytical method’s scale of operations.We can display the scale of operations graphically (Figure 3.5) by plottingthe sample’s size on the x-axis and the analyte’s concentration on they-axis. 8 For convenience, we divide samples into macro (>0.1 g), meso (10mg–100 mg), micro (0.1 mg–10 mg), and ultramicro (1% w/w), minor (0.01% w/w–1% w/w),trace (10 -7 % w/w–0.01% w/w), and ultratrace (

52 Analytical Chemistry 2.0It should not surprise you to learn thattotal analysis methods typically requiremacro or meso samples containing majoranalytes. Concentration methods areparticularly useful for minor, trace, andultratrace analytes in macro, meso, andmicro samples.We can use Figure 3.5 to establish limits for analytical methods. If amethod’s minimum detectable signal is equivalent to 10 mg of analyte, thenit is best suited to a major analyte in a macro or meso sample. Extending themethod to an analyte with a concentration of 0.1% w/w requires a sampleof 10 g, which is rarely practical due to the complications of carrying sucha large amount of material through the analysis. On the other hand, smallsamples containing trace amounts of analyte place significant restrictionson an analysis. For example, 1-mg sample with an analyte present at 10 –4 %w/w contains just 1 ng of analyte. If we can isolate the analyte in 1 mL ofsolution, then we need an analytical method that can reliably detect it at aconcentration of 1 ng/mL.3D.7 Equipment, Time, and CostFinally, we can compare analytical methods with respect to equipment needs,the time to complete an analysis, and the cost per sample. Methods relyingon instrumentation are equipment-intensive and may require significantoperator training. For example, the graphite furnace atomic absorptionspectroscopic method for determining lead in water requires a significantcapital investment in the instrument and an experienced operator to obtainreliable results. Other methods, such as titrimetry, require less expensiveequipment and less training.The time to complete an analysis for one sample is often fairly similarfrom method to method. This is somewhat misleading, however, becausemuch of this time is spent preparing solutions and gathering togetherequipment. Once the solutions and equipment are in place, the samplingrate may differ substantially from method to method. Additionally, somemethods are more easily automated. This is a significant factor in selectinga method for a laboratory that handles a high volume of samples.The cost of an analysis depends on many factors, including the cost ofequipment and reagents, the cost of hiring analysts, and the number ofsamples that can be processed per hour. In general, methods relying oninstruments cost more per sample then other methods.3D.8 Making the Final ChoiceUnfortunately, the design criteria discussed in this section are not mutuallyindependent. 9 Working with smaller samples or improving selectivity oftencomes at the expense of precision. Minimizing cost and analysis time maydecrease accuracy. Selecting a method requires carefully balancing the designcriteria. Usually, the most important design criterion is accuracy, andthe best method is the one giving the most accurate result. When the needfor results is urgent, as is often the case in clinical labs, analysis time maybecome the critical factor.9 Valcárcel, M.; Ríos, A. Anal. Chem. 1993, 65, 781A–787A.

Chapter 3 The Vocabulary of Analytical Chemistry53In some cases it is the sample’s properties that determine the best method.A sample with a complex matrix, for example, may require a method withexcellent selectivity to avoid interferences. Samples in which the analyte ispresent at a trace or ultratrace concentration usually require a concentrationmethod. If the quantity of sample is limited, then the method must notrequire a large amount of sample.Determining the concentration of lead in drinking water requires amethod that can detect lead at the parts per billion concentration level.Selectivity is important because other metal ions are present at significantlyhigher concentrations. A method using graphite furnace atomic absorptionspectroscopy is a common choice for determining lead in drinking waterbecause it meets these specifications. The same method is also useful fordetermining lead in blood where its ability to detect low concentrations oflead using a few microliters of sample are important considerations.3EDeveloping the ProcedureAfter selecting a method the next step is to develop a procedure that will accomplishthe goals of our analysis. In developing the procedure attention isgiven to compensating for interferences, to selecting and calibrating equipment,to acquiring a representative sample, and to validating the method.3E.1 Compensating for InterferencesA method’s accuracy depends on its selectivity for the analyte. Even the bestmethod, however, may not be free from interferents that contribute to themeasured signal. Potential interferents may be present in the sample itselfor in the reagents used during the analysis.When the sample is free of interferents, the total signal, S total , is a sumof the signal due to the analyte, S A , and the signal due to interferents inthe reagents, S reag ,S = S + S = kn + Stotal A reag A reag3.9S = S + S = kC + Stotal A reag A reag3.10Without an independent determination of S reag we cannot solve equation3.9 or 3.10 for the moles or concentration of analyte.To determine the contribution of S reag in equations 3.9 and 3.10 wemeasure the signal for a method blank, a solution that does not containthe sample. Consider, for example, a procedure in which we dissolve a 0.1-gsample in a portion of solvent, add several reagents, and dilute to 100 mLwith additional solvent. To prepare the method blank we omit the sampleand dilute the reagents to 100 mL using the solvent. Because the analyte isabsent, S total for the method blank is equal to S reag . Knowing the value forS reag makes it is easy to correct S total for the reagent’s contribution to thetotal signal; thusA method blank also is known as a reagentblank.When the sample is a liquid, or is in solution,we use an equivalent volume ofan inert solvent as a substitute for thesample.

54 Analytical Chemistry 2.0( S −S )= S = kntotal reag A A( S −S )= S = kCtotal reag A AMethods for effecting this separation arediscussed in Chapter 7.By itself, a method blank cannot compensate for an interferent that ispart of the sample’s matrix. If we happen to know the interferent’s identityand concentration, then we can be add it to the method blank; however,this is not a common circumstance. A more common approach is to finda method for separating the analyte and interferent by removing one fromthe sample. Once the separation is complete, we can proceed with theanalysis using equation 3.9 or equation 3.10.3E.2 CalibrationA simple definition of a quantitative analytical method is that it is a mechanismfor converting a measurement, the signal, into the amount of analytein a sample. Assuming that we can correct for the method blank and thatwe can compensate for interferents, a quantitative analysis is nothing morethan solving equation 3.1 or equation 3.2 for n A or C A .To solve these equations we need the value of k A . For a total analysismethod we usually know the value of k A because it is defined by the stoichiometryof the chemical reactions generating the signal. For a concentrationmethod, however, the value of k A usually is a complex function of experimentalconditions. A Calibration is the process of experimentally determiningthe value of k A by measuring the signal for one or more standardsamples, each containing a known concentration of analyte. With a singlestandard we can calculate the value of k A using equation 3.1 or equation3.2. When using several standards with different concentrations of analyte,the result is best viewed visually by plotting S A versus the concentration ofanalyte in the standards. Such a plot is known as a calibration curve, anexample of which is shown in Figure 3.6.Figure 3.6 Example of a calibrationcurve. The filled circles (• )are the individual results for thestandard samples and the line isthe best fit to the data determinedby a linear regression analysis. SeeChapter 5 for a further discussionof calibration curves and an explanationof linear regression.S A10.80.60.40.200.00 0.20 0.40 0.60 0.80 1.00Concentration of Analyte (µg/mL)

Chapter 3 The Vocabulary of Analytical Chemistry553E.3 SamplingSelecting an appropriate method and executing it properly helps us ensurethat our analyses are accurate. If we analyze the wrong samples, however,then the accuracy of our work is of little consequence.A proper sampling strategy ensures that our samples are representativeof the material from which they are taken. Biased or nonrepresentativesampling, and contaminating samples during or after their collection aresampling errors that can lead to a significant error in accuracy. It is importantto realize that sampling errors are independent of errors in the analyticalmethod. As a result, sampling errors can not be corrected by evaluatinga reagent blank.3E.4 ValidationIf we are to have confidence in our procedure we must demonstrate that itcan provide acceptable results, a process we call validation. Perhaps themost important part of validating a procedure is establishing that its precisionand accuracy are appropriate for the problem under investigation. Wealso ensure that the written procedure has sufficient detail so that differentanalysts or laboratories will obtain comparable results. Ideally, validationuses a standard sample whose composition closely matches the samples thatwill be analyzed. In the absence of appropriate standards, we can evaluateaccuracy by comparing results to those obtained using a method of knownaccuracy.Chapter 7 provides a more detailed discussionof sampling, including strategies forobtaining representative samples.You will find more details about validatinganalytical methods in Chapter 14.3FProtocolsEarlier we defined a protocol as a set of stringent written guidelines specifyingan exact procedure that must be followed before an agency will acceptthe results of an analysis. In addition to the considerations taken into accountwhen designing a procedure, a protocol also contains explicit instructionsregarding internal and external quality assurance and quality control(QA/QC) procedures. 10 The goal of internal QA/QC is to ensure that a laboratory’swork is both accurate and precise. External QA/QC is a process inwhich an external agency certifies a laboratory.As an example, let’s outline a portion of the Environmental ProtectionAgency’s protocol for determining trace metals in water by graphite furnaceatomic absorption spectroscopy as part of its Contract Laboratory Program(CLP) . The CLP protocol (see Figure 3.7) calls for an initial calibrationusing a method blank and three standards, one of which is at the detectionlimit. The resulting calibration curve is verified by analyzing initialcalibration verification (ICV) and initial calibration blank (ICB) samples.The lab’s result for the ICV sample must fall within ±10% of its expected10 (a) Amore, F. Anal. Chem. 1979, 51, 1105A–1110A; (b) Taylor, J. K. Anal. Chem. 1981, 53,1588A–1593A.

56 Analytical Chemistry 2.0StartFigure 3.7 Schematic diagram showing a portion of the EPA’sprotocol for determining trace metals in water using graphitefurnace atomic absorption spectrometry.Initial CalibrationThe abbreviations are ICV: initial calibration verification;ICB: initial calibration blank; CCV: continuing calibrationverification; CCB: continuing calibration blank.ICV, ICBOK?NoIdentify andcorrect problemYesCCV, CCBOK?NoYesRun 10 SamplesCCV, CCBOK?NoDiscard results forlast set of samplesYesYesMoreSamples?Noconcentration. If the result is outside this limit the analysis is stopped, andthe problem identified and corrected before continuing.After a successful analysis of the ICV and ICB samples, the lab reverifiesthe calibration by analyzing a continuing calibration verification (CCV)sample and a continuing calibration blank (CCB). Results for the CCV alsomust be within ±10% of its expected concentration. Again, if the lab’s resultfor the CCV is outside the established limits, the analysis is stopped, theproblem identified and corrected, and the system recalibrated as describedabove. Additional CCV and the CCB samples are analyzed before the firstsample and after the last sample, and between every set of ten samples. Ifthe result for any CCV or CCB sample is unacceptable, the results for thelast set of samples are discarded, the system is recalibrated, and the samplesreanalyzed. By following this protocol, every result is bound by successfulchecks on the calibration. Although not shown in Figure 3.7, the protocolalso contains instructions for analyzing duplicate or split samples, and forusing of spike tests to verify accuracy.End

Chapter 3 The Vocabulary of Analytical Chemistry573GThe Importance of Analytical MethodologyThe importance of analytical methodology is evident if we examine environmentalmonitoring programs. The purpose of a monitoring program isto determine the present status of an environmental system, and to assesslong term trends in the system’s health. These are broad and poorly definedgoals. In many cases, an environmental monitoring program begins beforethe essential questions are known. This is not surprising since it is difficultto formulate questions in the absence of any results. Without careful planning,however, a poor experimental design may result in data that has littlevalue.These concerns are illustrated by the Chesapeake Bay Monitoring Program.This research program, designed to study nutrients and toxic pollutantsin the Chesapeake Bay, was initiated in 1984 as a cooperative venturebetween the federal government, the state governments of Maryland,Virginia, and Pennsylvania, and the District of Columbia. A 1989 reviewof the program highlights the problems common to many monitoringprograms. 11At the beginning of the Chesapeake Bay monitoring program, little attentionwas given to selecting analytical methods, in large part because theeventual use of the monitoring data had yet to be specified. The analyticalmethods initially chosen were standard methods already approved by theEnvironmental Protection Agency (EPA). In many cases these methodswere not useful because they were designed to detect pollutants at theirlegally mandated maximum allowed concentrations. In unpolluted waters,however, the concentrations of these contaminants are often well belowthe detection limit of the EPA methods. For example, the detection limitfor the EPA approved standard method for phosphate was 7.5 ppb. Sinceactual phosphate concentrations in Chesapeake Bay were below the EPAmethod’s detection limit, it provided no useful information. On the otherhand, the detection limit for a non-approved variant of the EPA method,a method routinely used by chemical oceanographers, was 0.06 ppb. Inother cases, such as the elemental analysis for particulate forms of carbon,nitrogen and phosphorous, EPA approved procedures provided poorer reproducibilitythan nonapproved methods.3HKey Termsaccuracy analysis analytecalibration calibration curve concentration techniquesdetection limit determination interferentmatrix measurement methodmethod blank precision procedureprotocol QA/QC robustAs you review this chapter, try to define akey term in your own words. Check youranswer by clicking on the key term, whichwill take you to the page where it was firstintroduced. Clicking on the key termthere, will bring you back to this page sothat you can continue with another keyterm.11 D’Elia, C. F.; Sanders, J. G.; Capone, D. G. Envrion. Sci. Technol. 1989, 23, 768–774.

58 Analytical Chemistry 2.0rugged selectivity selectivity coefficientsensitivity signal specificitytechnique total analysis techniques validation3IChapter SummaryEvery discipline has its own vocabulary. Your success in studying analyticalchemistry will improve if you master this vocabulary. Be sure that youunderstand the difference between an analyte and its matrix, a techniqueand a method, a procedure and a protocol, and a total analysis techniqueand a concentration technique.In selecting an analytical method we consider criteria such as accuracy,precision, sensitivity, selectivity, robustness, ruggedness, the amountof available sample, the amount of analyte in the sample, time, cost, andthe availability of equipment. These criteria are not mutually independent,and it often is necessary to find an acceptable balance between them.In developing a procedure or protocol, we give consideration to compensatingfor interferences, calibrating the method, obtaining an appropriatesample, and validating the analysis. Poorly designed procedures andprotocols produce results that are insufficient to meet the needs of theanalysis.Answers, but not worked solutions, tomost end-of-chapter problems are availablehere.3JProblems1. When working with a solid sample, it often is necessary to bring theanalyte into solution by digesting the sample with a suitable solvent.Any remaining solid impurities are removed by filtration before continuingwith the analysis. In a typical total analysis method, the proceduremight readAfter digesting the sample in a beaker, remove any solid impuritiesby passing the solution containing the analyte through filterpaper, collecting the filtrate in a clean Erlenmeyer flask. Rinse thebeaker with several small portions of solvent, passing these rinsingsthrough the filter paper and collecting them in the same Erlenmeyerflask. Finally, rinse the filter paper with several portions ofsolvent, collecting the rinsings in the same Erlenmeyer flask.For a typical concentration method, however, the procedure mightstateAfter digesting the sample in a beaker, remove any solid impuritiesby filtering a portion of the solution containing the analyte. Collectand discard the first several mL of filtrate before collecting a sampleof approximately 5 mL for further analysis.

Chapter 3 The Vocabulary of Analytical Chemistry59Explain why these two procedures are different.2. A certain concentration method works best when the analyte’s concentrationis approximately 10 ppb.(a) If the method requires a sample of 0.5 mL, about what mass ofanalyte is being measured?(b) If the analyte is present at 10% w/v, how would you prepare thesample for analysis?(c) Repeat for the case where the analyte is present at 10% w/w.(d) Based on your answers to parts (a)–(c), comment on the method’ssuitability for the determination of a major analyte.3. An analyst needs to evaluate the potential effect of an interferent, I, onthe quantitative analysis for an analyte, A. She begins by measuring thesignal for a sample in which the interferent is absent and the analyte ispresent with a concentration of 15 ppm, obtaining an average signal of23.3 (arbitrary units). When analyzing a sample in which the analyteis absent and the interferent is present with a concentration of 25 ppm,she obtains an average signal of 13.7.(a) What is the sensitivity for the analyte?(b) What is the sensitivity for the interferent?(c) What is the value of the selectivity coefficient?(d) Is the method more selective for the analyte or the interferent?(e) What is the maximum concentration of interferent relative to that ofthe analyte (i.e. [interferent]/[analyte]), if the error in the analysisis to be less than 1%?4. A sample was analyzed to determine the concentration of an analyte.Under the conditions of the analysis the sensitivity is 17.2 ppm -1 . Whatis the analyte’s concentration if S total is 35.2 and S reag is 0.6?5. A method for the analysis of Ca 2+ in water suffers from an interferencein the presence of Zn 2+ . When the concentration of Ca 2+ is 50 timesgreater than that of Zn 2+ , an analysis for Ca 2+ gives a relative error of–2.0%. What is the value of the selectivity coefficient for this method?6. The quantitative analysis for reduced glutathione in blood is complicatedby the presence of many potential interferents. In one study, whenanalyzing a solution of 10 ppb glutathione and 1.5 ppb ascorbic acid,the signal was 5.43 times greater than that obtained for the analysis of

60 Analytical Chemistry 2.010 ppb glutathione. 12 What is the selectivity coefficient for this analysis?The same study found that when analyzing a solution of 350 ppb methionineand 10 ppb glutathione the signal was 0.906 times less than thatobtained for the analysis of 10 ppb glutathione. What is the selectivitycoefficient for this analysis? In what way do these interferents behavedifferently?7. Oungpipat and Alexander described a method for determining the concentrationof glycolic acid (GA) in a variety of samples, including physiologicalfluids such as urine. 13 In the presence of only GA, the signal isgiven asS samp,1 = k GA C GAand in the presence of both glycolic acid and ascorbic acid (AA), thesignal isS samp,2 = k GA C GA + k AA C AAWhen the concentration of glycolic acid is 1.0 × 10 –4 M and the concentrationof ascorbic acid is 1.0 × 10 –5 M, the ratio of the two signalswas found to beSSsamp,2samp,1= 144 .(a) Using the ratio of the two signals, determine the value of the selectivityratio K GA,AA .(b) Is the method more selective toward glycolic acid or ascorbic acid?(c) If the concentration of ascorbic acid is 1.0 × 10 –5 M, what is thesmallest concentration of glycolic acid that can be determined suchthat the error introduced by failing to account for the signal fromascorbic acid is less than 1%?8. Ibrahim and co-workers developed a new method for the quantitativeanalysis of hypoxanthine, a natural compound of some nucleic acids. 14As part of their study they evaluated the method’s selectivity for hypoxanthinein the presence of several possible interferents, includingascorbic acid.12 Jiménez-Prieto, R.; Velasco, A.; Silva, M; Pérez-Bendito, D. Anal. Chem. Acta 1992, 269, 273–279.13 Oungpipat, W.; Alexander, P. W. Anal. Chim. Acta 1994, 295, 36–46.14 Ibrahim, M. S.; Ahmad, M. E.; Temerk, Y. M.; Kaucke, A. M. Anal. Chim. Acta 1996, 328,47–52.

Chapter 3 The Vocabulary of Analytical Chemistry61(a) When analyzing a solution of 1.12 × 10 –6 M hypoxanthine the authorsobtained a signal of 7.45 × 10 –5 amps. What is the sensitivityfor hypoxanthine? You may assume that the signal has been correctedfor the method blank.(b) When a solution containing 1.12 × 10 –6 M hypoxanthine and6.5 × 10 –5 M ascorbic acid was analyzed a signal of 4.04 × 10 –5 ampswas obtained. What is the selectivity coefficient for this method?(c) Is the method more selective for hypoxanthine or for ascorbicacid?(d) What is the largest concentration of ascorbic acid that may be presentif a concentration of 1.12 × 10 –6 M hypoxanthine is to be determinedwithin ±1%?9. Examine a procedure from Standard Methods for the Analysis of Watersand Wastewaters (or another manual of standard analytical methods)and identify the steps taken to compensate for interferences, to calibrateequipment and instruments, to standardize the method and toacquire a representative sample.3KSolutions to Practice ExercisesPractice Exercise 3.1Since the signal for Ag + in the presence of Ni 2+ is given as a relative error,the fact that are not given absolute signals is of no consequence. Instead, wewill assign a value of 100 as the signal for 1 × 10 –9 M Ag + . With a relativeerror of +4.9%, the signal for the solution of 1 × 10 –9 M Ag + and 1.1 × 10 –7M Ni 2+ is 104.9. The sensitivity for Ag+ is determined using the solutionthat does not contain Ni 2+ .kAgSAg100= = = 10 . × 10C ×−91 10 MAgSubstituting into equation 3.4 values for k Ag , S samp for the solution containingAg + and Ni 2+ , and the concentrations of Ag + and Ni 2+Ssamp11= = × M− 1 −9104. 9 ( 10 . 10 ) × (. 1 0× 10 M) + k × (. 11× 10 −7M)and solving gives k Ni as 4.5 × 10 7 M –1 . The selectivity coefficient isKAg,NiClick here to return to the chapter.kNi45 . × 10= =k 10 . × 10AgM7 −1M11 −111MNi−1= 45 . × 10 −4

62 Analytical Chemistry 2.0Practice Exercise 3.2(a) A negative value for K Ag,Hg means that the presence of Hg 2+ decreasesthe signal from Ag + .(b) In this case we need to consider an error of –1%, since the effect of Hg 2+is to decrease the signal from Ag + . To achieve this error, the term K × CA,I Iin equation 3.8 must be less than -1% of C A ; thusK × C =− 001 . × CAg,Hg Hg AgSubstituting in known values for K Ag,Hg and C Ag , we find that the maximumconcentration of Hg 2+ is 1.0 × 10 -8 M.Click here to return to the chapter.

DRAFTChapter 4Evaluating Analytical DataChapter Overview4A Characterizing Measurements and Results4B Characterizing Experimental Errors4C Propagation of Uncertainty4D The Distribution of Measurements and Results4E Statistical Analysis of Data4F Statistical Methods for Normal Distributions4G Detection Limits4H Using Excel and R to Analyze Data4I Key Terms4J Chapter Summary4K Problems4L Solutions to Practice ExercisesWhen using an analytical method we make three separate evaluations of experimental error.First, before beginning an analysis we evaluate potential sources of errors to ensure that they willnot adversely effect our results. Second, during the analysis we monitor our measurements toensure that errors remain acceptable. Finally, at the end of the analysis we evaluate the qualityof the measurements and results, comparing them to our original design criteria. This chapterprovides an introduction to sources of error, to evaluating errors in analytical measurements,and to the statistical analysis of data.Copyright: David Harvey, 200963

64 Analytical Chemistry 2.0Figure 4.1 An uncirculated 2005Lincoln head penny. The “D” belowthe date indicates that thispenny was produced at the UnitedStates Mint at Denver, Colorado.Pennies produced at the PhiladelphiaMint do not have a letter belowthe date. Source: United StatesMint image (www.usmint.gov).4ACharacterizing Measurements and ResultsLet’s begin by choosing a simple quantitative problem requiring a singlemeasurement—What is the mass of a penny? As you consider this question,you probably recognize that it is too broad. Are we interested in the mass ofa United States penny or of a Canadian penny, or is the difference relevant?Because a penny’s composition and size may differ from country to country,let’s limit our problem to pennies from the United States.There are other concerns we might consider. For example, the UnitedStates Mint currently produces pennies at two locations (Figure 4.1). Becauseit seems unlikely that a penny’s mass depends upon where it is minted,we will ignore this concern. Another concern is whether the mass of a newlyminted penny is different from the mass of a circulating penny. Because theanswer this time is not obvious, let’s narrow our question to—What is themass of a circulating United States Penny?A good way to begin our analysis is to examine some preliminary data.Table 4.1 shows masses for seven pennies from my change jar. In examiningthis data it is immediately apparent that our question does not have asimple answer. That is, we can not use the mass of a single penny to draw aspecific conclusion about the mass of any other penny (although we mightconclude that all pennies weigh at least 3 g). We can, however, characterizethis data by reporting the spread of individual measurements around acentral value.4A.1 Measures of Central TendencyOne way to characterize the data in Table 4.1 is to assume that the massesare randomly scattered around a central value that provides the best estimateof a penny’s expected, or “true” mass. There are two common ways toestimate central tendency: the mean and the median.Me a nThe mean, X , is the numerical average for a data set. We calculate the meanby dividing the sum of the individual values by the size of the data setTable 4.1Masses of Seven Circulating U. S. PenniesPennyMass (g)1 3.0802 3.0943 3.1074 3.0565 3.1126 3.1747 3.198

Chapter 4 Evaluating Analytical Data65X=∑inXiwhere X i is the i th measurement, and n is the size of the data set.Example 4.1What is the mean for the data in Table 4.1?So l u t i o nTo calculate the mean we add together the results for all measurements3.080 + 3.094 + 3.107 + 3.056 + 3.112 + 3.174 + 3.198 = 21.821 gand divide by the number of measurements21.821 gX = = 3.117 g7The mean is the most common estimator of central tendency. It is nota robust estimator, however, because an extreme value—one much largeror much smaller than the remainder of the data—strongly influences themean’s value. 1 For example, if we mistakenly record the third penny’s massas 31.07 g instead of 3.107 g, the mean changes from 3.117 g to 7.112 g!An estimator is robust if its value is notaffected too much by an unusually largeor unusually small measurement.Me d i a nThe median, X , is the middle value when we order our data from thesmallest to the largest value. When the data set includes an odd number ofentries, the median is the middle value. For an even number of entries, themedian is the average of the n/2 and the (n/2) + 1 values, where n is thesize of the data set.Example 4.2What is the median for the data in Table 4.1?When n = 5, the median is the third valuein the ordered data set; for n = 6, the medianis the average of the third and fourthmembers of the ordered data set.So l u t i o nTo determine the median we order the measurements from the smallest tothe largest value3.056 3.080 3.094 3.107 3.112 3.174 3.198Because there are seven measurements, the median is the fourth value inthe ordered data set; thus, the median is 3.107 g.As shown by Examples 4.1 and 4.2, the mean and the median providesimilar estimates of central tendency when all measurements are compara-1 Rousseeuw, P. J. J. Chemom. 1991, 5, 1–20.

66 Analytical Chemistry 2.0Problem 12 at the end of the chapter asksyou to show that this is true.ble in magnitude. The median, however, provides a more robust estimate ofcentral tendency because it is less sensitive to measurements with extremevalues. For example, introducing the transcription error discussed earlier forthe mean changes the median’s value from 3.107 g to 3.112 g.4A.2 Measures of SpreadIf the mean or median provides an estimate of a penny’s expected mass,then the spread of individual measurements provides an estimate of thedifference in mass among pennies or of the uncertainty in measuring masswith a balance. Although we often define spread relative to a specific measureof central tendency, its magnitude is independent of the central value.Changing all measurements in the same direction, by adding or subtractinga constant value, changes the mean or median, but does not change thespread. There are three common measures of spread: the range, the standarddeviation, and the variance.Ra n g eThe range, w, is the difference between a data set’s largest and smallestvalues.w = X largest – X smallestThe range provides information about the total variability in the data set,but does not provide any information about the distribution of individualvalues. The range for the data in Table 4.1 isSt a n d a r d Deviationw = 3.198 g – 3.056 g = 0.142 gThe standard deviation, s, describes the spread of a data set’s individualvalues about its mean, and is given ass =∑( X − X) 2iin −1where X i is one of n individual values in the data set, and X is the data set’smean value. Frequently, the relative standard deviation, s r , is reported.sr =The percent relative standard deviation, %s r , is s r × 100.Example 4.3What are the standard deviation, the relative standard deviation and thepercent relative standard deviation for the data in Table 4.1?sX4.1

Chapter 4 Evaluating Analytical Data67So l u t i o nTo calculate the standard deviation we first calculate the difference betweeneach measurement and the mean value (3.117), square the resulting differences,and add them together to give the numerator of equation 4.1.2 2( 3. 080 − 3. 117) = ( − 0. 037) = 0.0013692 2( 3. 094−31. 17) =− ( 0. 023) = 0.00052922( 3. 107 − 3. 117) = ( −0.010) = 0.0001002 2( 3. 056 − 3. 117) = ( − 0. 061) = 00 . 037212 2( 3. 112− 3. 117) = ( − 0. 005) = 0.0000252 2( 31 . 74 − 3. 117) = ( + 0. 057) = 0.003249( 3. 198−3. 117) 2 =+ ( 0. 081) 2 = 0.0065610.015554For obvious reasons, the numerator ofequation 4.1 is called a sum of squares.Next, we divide this sum of the squares by n – 1, where n is the number ofmeasurements, and take the square root.s =0.015554= 0. 051 g7−1Finally, the relative standard deviation and percent relative standard deviationares r= 0.051 g3.117 g=0. 016 %s r = (0.016) × 100% = 1.6%It is much easier to determine the standard deviation using a scientificcalculator with built in statistical functions.Many scientific calculators include twokeys for calculating the standard deviation.One key calculates the standard deviationfor a data set of n samples drawn froma larger collection of possible samples,which corresponds to equation 4.1. Theother key calculates the standard deviationfor all possible samples. The later is knownas the population’s standard deviation,which we will cover later in this chapter.Your calculator’s manual will help you determinethe appropriate key for each.Va r i a n c eAnother common measure of spread is the square of the standard deviation,or the variance. We usually report a data set’s standard deviation, ratherthan its variance, because the mean value and the standard deviation havethe same unit. As we will see shortly, the variance is a useful measure ofspread because its values are additive.Example 4.4What is the variance for the data in Table 4.1?So l u t i o nThe variance is the square of the absolute standard deviation. Using thestandard deviation from Example 4.3 gives the variance ass 2 = (0.051) 2 = 0.0026

68 Analytical Chemistry 2.0The convention for representing statisticalparameters is to use a Roman letter for avalue calculated from experimental data,and a Greek letter for the correspondingexpected value. For example, the experimentallydetermined mean is X , and itsunderlying expected value is μ. Likewise,the standard deviation by experiment is s,and the underlying expected value is s.It is possible, although unlikely, that thepositive and negative determinate errorswill offset each other, producing a resultwith no net error in accuracy.Practice Exercise 4.1The following data were collected as part of a quality control study forthe analysis of sodium in serum; results are concentrations of Na + inmmol/L.4B140 143 141 137 132 157 143 149 118 145Report the mean, the median, the range, the standard deviation, and thevariance for this data. This data is a portion of a larger data set from Andrew,D. F.; Herzberg, A. M. Data: A Collection of Problems for the Studentand Research Worker, Springer-Verlag:New York, 1985, pp. 151–155.Click here to review your answer to this exercise.Characterizing Experimental ErrorsCharacterizing the mass of a penny using the data in Table 4.1 suggeststwo questions. First, does our measure of central tendency agree with thepenny’s expected mass? Second, why is there so much variability in theindividual results? The first of these questions addresses the accuracy of ourmeasurements, and the second asks about their precision. In this section weconsider the types of experimental errors affecting accuracy and precision.4B.1 Errors Affecting AccuracyAccuracy is a measure of how close a measure of central tendency is to theexpected value, μ. We can express accuracy as either an absolute error, eor as a percent relative error, %e r .e= X −µ 4.2X%e = − µ × 100r4.3µAlthough equations 4.2 and 4.3 use the mean as the measure of centraltendency, we also can use the median.We call errors affecting the accuracy of an analysis determinate. Althoughthere may be several different sources of determinate error, eachsource has a specific magnitude and sign. Some sources of determinate errorare positive and others are negative, and some are larger in magnitude andothers are smaller. The cumulative effect of these determinate errors is a netpositive or negative error in accuracy.We assign determinate errors into four categories—sampling errors,method errors, measurement errors, and personal errors—each of whichwe consider in this section.Sa m p l i n g Er r o r sA determinate sampling error occurs when our sampling strategy doesnot provide a representative sample. For example, if you monitor the envi-

Chapter 4 Evaluating Analytical Data69ronmental quality of a lake by sampling a single location near a point sourceof pollution, such as an outlet for industrial effluent, then your results willbe misleading. In determining the mass of a U. S. penny, our strategy forselecting pennies must ensure that we do not include pennies from othercountries.An awareness of potential sampling errorsis especially important when workingwith heterogeneous materials. Strategiesfor obtaining representative samples arecovered in Chapter 5.Me t h o d Er r o r sIn any analysis the relationship between the signal and the absolute amountof analyte, n A , or the analyte’s concentration, C A , isS = k n + Stotal A A mb4.4S = k C + Stotal A A mb4.5where k A is the method’s sensitivity for the analyte and S mb is the signalfrom the method blank. A determinate method error exists when ourvalue for k A or S mb is invalid. For example, a method in which S total isthe mass of a precipitate assumes that k is defined by a pure precipitateof known stoichiometry. If this assumption is not true, then the resultingdetermination of n A or C A is inaccurate. We can minimize a determinateerror in k A by calibrating the method. A method error due to an interferentin the reagents is minimized by using a proper method blank.Me a s u r e m e n t Er r o r sThe manufacturers of analytical instruments and equipment, such as glasswareand balances, usually provide a statement of the item’s maximummeasurement error, or tolerance. For example, a 10-mL volumetricpipet (Figure 4.2) has a tolerance of ±0.02 mL, which means that the pipetdelivers an actual volume within the range 9.98–10.02 mL at a temperatureof 20 o C. Although we express this tolerance as a range, the error isdeterminate; thus, the pipet’s expected volume is a fixed value within thestated range.Volumetric glassware is categorized into classes depending on its accuracy.Class A glassware is manufactured to comply with tolerances specifiedby agencies such as the National Institute of Standards and Technologyor the American Society for Testing and Materials. The tolerance level forClass A glassware is small enough that we normally can use it without calibration.The tolerance levels for Class B glassware are usually twice thosefor Class A glassware. Other types of volumetric glassware, such as beakersand graduated cylinders, are unsuitable for accurately measuring volumes.Table 4.2 provides a summary of typical measurement errors for Class Avolumetric glassware. Tolerances for digital pipets and for balances are listedin Table 4.3 and Table 4.4.We can minimize determinate measurement errors by calibrating ourequipment. Balances are calibrated using a reference weight whose mass canFigure 4.2 Close-up of a 10-mLvolumetric pipet showing that ithas a tolerance of ±0.02 mL at20 o C.

70 Analytical Chemistry 2.0be traced back to the SI standard kilogram. Volumetric glassware and digitalpipets can be calibrated by determining the mass of water that it deliversor contains and using the density of water to calculate the actual volume. Itis never safe to assume that a calibration will remain unchanged during ananalysis or over time. One study, for example, found that repeatedly exposingvolumetric glassware to higher temperatures during machine washingand oven drying, leads to small, but significant changes in the glassware’scalibration. 2 Many instruments drift out of calibration over time and mayrequire frequent recalibration during an analysis.2 Castanheira, I.; Batista, E.; Valente, A.; Dias, G.; Mora, M.; Pinto, L.; Costa, H. S. Food Control2006, 17, 719–726.Table 4.2 Measurement Errors for Type A Volumetric Glassware †Transfer Pipets Volumetric Flasks BuretsCapacity(mL)Tolerance(mL)Capacity(mL)Tolerance(mL)Capacity(mL)Tolerance(mL)1 ±0.006 5 ±0.02 10 ±0.022 ±0.006 10 ±0.02 25 ±0.035 ±0.01 25 ±0.03 50 ±0.0510 ±0.02 50 ±0.0520 ±0.03 100 ±0.0825 ±0.03 250 ±0.1250 ±0.05 500 ±0.20100 ±0.08 1000 ±0.302000 ±0.50† Tolerance values are from the ASTM E288, E542, and E694 standards.Table 4.3 Measurement Errors for Digital Pipets †Pipet Range Volume (mL or mL) ‡ Percent Measurement Error10–100 mL 10 ±3.0%50 ±1.0%100 ±0.8%100–1000 mL 100 ±3.0%500 ±1.0%1000 ±0.6%1–10 mL 1 ±3.0%5 ±0.8%10 ±0.6%† Values are from www.eppendorf.com.‡ Units for volume match the units for the pipet’s range.

Chapter 4 Evaluating Analytical Data71Table 4.4 Measurement Errors for Selected BalancesBalance Capacity (g) Measurement ErrorPrecisa 160M 160 ±1 mgA & D ER 120M 120 ±0.1 mgMetler H54 160 ±0.01 mgPe r s o n a l Er r o r sFinally, analytical work is always subject to personal error, including theability to see a change in the color of an indicator signaling the endpoint ofa titration; biases, such as consistently overestimating or underestimatingthe value on an instrument’s readout scale; failing to calibrate instrumentation;and misinterpreting procedural directions. You can minimize personalerrors by taking proper care.Identifying De t e r m i n a t e Er r o r sDeterminate errors can be difficult to detect. Without knowing the expectedvalue for an analysis, the usual situation in any analysis that matters,there is nothing to which we can compare our experimental result. Nevertheless,there are strategies we can use to detect determinate errors.The magnitude of a constant determinate error is the same for allsamples and is more significant when analyzing smaller samples. Analyzingsamples of different sizes, therefore, allows us to detect a constant determinateerror. For example, consider a quantitative analysis in which weseparate the analyte from its matrix and determine its mass. Let’s assumethat the sample is 50.0% w/w analyte. As shown in Table 4.5, the expectedamount of analyte in a 0.100 g sample is 0.050 g. If the analysis has apositive constant determinate error of 0.010 g, then analyzing the samplegives 0.060 g of analyte, or a concentration of 60.0% w/w. As we increasethe size of the sample the obtained results become closer to the expectedresult. An upward or downward trend in a graph of the analyte’s obtainedTable 4.5 Effect of a Constant Determinate Error on the Analysis of a SampleContaining 50% w/w AnalyteExpected MassObtained Mass Obtained ConcentrationMass SampleConstant Errorof Analyteof Analyteof Analyte(g)(g)(g)(g)(%w/w)0.100 0.050 0.010 0.060 60.00.200 0.100 0.010 0.110 55.00.400 0.200 0.010 0.210 52.50.800 0.400 0.010 0.410 51.21.600 0.800 0.010 0.810 50.6

72 Analytical Chemistry 2.0Figure 4.3 Effect of a constantdeterminate error on the determinationof an analyte in samples ofvarying size.Obtained Concentration of Analyte (% w/w)Mass of Sample (g)concentration versus the sample’s mass (Figure 4.3) is evidence of a constantdeterminate error.A proportional determinate error, in which the error’s magnitudedepends on the amount of sample, is more difficult to detect because theresult of the analysis is independent of the amount of sample. Table 4.6outlines an example showing the effect of a positive proportional error of1.0% on the analysis of a sample that is 50.0% w/w in analyte. Regardless ofthe sample’s size, each analysis gives the same result of 50.5% w/w analyte.One approach for detecting a proportional determinate error is to analyzea standard containing a known amount of analyte in a matrix similarto the samples. Standards are available from a variety of sources, such asthe National Institute of Standards and Technology (where they are calledStandard Reference Materials) or the American Society for Testing andMaterials. Table 4.7, for example, lists certified values for several analytes ina standard sample of Gingko bilboa leaves. Another approach is to compareyour analysis to an analysis carried out using an independent analyticalmethod known to give accurate results. If the two methods give significantlydifferent results, then a determinate error is the likely cause.Table 4.6 Effect of a Proportional Determinate Error on the Analysis of a SampleContaining 50% w/w AnalyteExpected Mass Proportional Obtained Mass Obtained ConcentrationMass Sampleof Analyte Errorof Analyteof Analyte(g)(g)(%)(g)(%w/w)0.100 0.050 1.00 0.0505 50.50.200 0.100 1.00 0.101 50.50.400 0.200 1.00 0.202 50.50.800 0.400 1.00 0.404 50.51.600 0.800 1.00 0.808 50.5

Chapter 4 Evaluating Analytical Data73Table 4.7 Certified Concentrations for SRM 3246: Ginkgo biloba (Leaves) †Class of Analyte Analyte Mass Fraction (mg/g or ng/g)Flavonoids/Ginkgolide B Quercetin 2.69 ± 0.31(mass fractions in mg/g) Kaempferol 3.02 ± 0.41Constant and proportional determinate errors have distinctly differentsources, which we can define in terms of the relationship between the signaland the moles or concentration of analyte (equation 4.4 and equation 4.5).An invalid method blank, S mb , is a constant determinate error as it adds orsubtracts a constant value to the signal. A poorly calibrated method, whichyields an invalid sensitivity for the analyte, k A , will result in a proportionaldeterminate error.4B.2 Errors Affecting PrecisionIsorhamnetin 0.517 ± 0.099Total Aglycones 6.22 ± 0.77Selected Terpenes Ginkgolide A 0.57 ± 0.28(mass fractions in mg/g) Ginkgolide B 0.470 ± 0.090Ginkgolide C 0.59 ± 0.22Ginkgolide J 0.18 ± 0.10Biloabalide 1.52 ± 0.40Total Terpene Lactones 3.3 ± 1.1Selected Toxic Elements Cadmium 20.8 ± 1.0(mass fractions in ng/g) Lead 995 ± 30Mercury 23.08 ± 0.17† The primary purpose of this Standard Reference Material is to validate analytical methods for determining flavonoids,terpene lactones, and toxic elements in Ginkgo biloba or other materials with a similar matrix. Values are from theofficial Certificate of Analysis available at www.nist.gov.Precision is a measure of the spread of individual measurements or resultsabout a central value, which we express as a range, a standard deviation, ora variance. We make a distinction between two types of precision: repeatabilityand reproducibility. Repeatability is the precision when a singleanalyst completes the analysis in a single session using the same solutions,equipment, and instrumentation. Reproducibility, on the other hand, isthe precision under any other set of conditions, including between analysts,or between laboratory sessions for a single analyst. Since reproducibilityincludes additional sources of variability, the reproducibility of an analysiscannot be better than its repeatability.Errors affecting precision are indeterminate and are characterized byrandom variations in their magnitude and their direction. Because theyare random, positive and negative indeterminate errors tend to cancel,

74 Analytical Chemistry 2.0provided that enough measurements are made. In such situations the meanor median is largely unaffected by the precision of the analysis.So u r c e s o f In d e t e r m i n a t e Er r o r3031Figure 4.4 Close-up of a buretshowing the difficulty in estimatingvolume. With scale divisionsevery 0.1 mL it is difficult to readthe actual volume to better than±0.01–0.03 mL.We can assign indeterminate errors to several sources, including collectingsamples, manipulating samples during the analysis, and making measurements.When collecting a sample, for instance, only a small portion ofthe available material is taken, increasing the chance that small-scale inhomogeneitiesin the sample will affect repeatability. Individual pennies, forexample, may show variations from several sources, including the manufacturingprocess, and the loss of small amounts of metal or the additionof dirt during circulation. These variations are sources of indeterminatesampling errors.During an analysis there are many opportunities for introducing indeterminatemethod errors. If our method for determining the mass of apenny includes directions for cleaning them of dirt, then we must be carefulto treat each penny in the same way. Cleaning some pennies more vigorouslythan others introduces an indeterminate method error.Finally, any measuring device is subject to an indeterminate measurementerror due to limitations in reading its scale. For example, a buretwith scale divisions every 0.1 mL has an inherent indeterminate error of±0.01–0.03 mL when we estimate the volume to the hundredth of a milliliter(Figure 4.4).Ev a l u a t i n g In d e t e r m i n a t e Er r o rAn indeterminate error due to analytical equipment or instrumentationis generally easy to estimate by measuring the standard deviation for severalreplicate measurements, or by monitoring the signal’s fluctuations overtime in the absence of analyte (Figure 4.5) and calculating the standarddeviation. Other sources of indeterminate error, such as treating samplesinconsistently, are more difficult to estimate.Signal (arbitrary units)Figure 4.5 Background noise inan instrument showing the randomfluctuations in the signal. Time (s)

Chapter 4 Evaluating Analytical Data75Table 4.8 Replicate Determinations of the Mass of aSingle Circulating U. S. PennyReplicate Mass (g) Replicate Mass (g)1 3.025 6 3.0232 3.024 7 3.0223 3.028 8 3.0214 3.027 9 3.0265 3.028 10 3.024To evaluate the effect of indeterminate measurement error on our analysisof the mass of a circulating United States penny, we might make severaldeterminations for the mass of a single penny (Table 4.8). The standarddeviation for our original experiment (see Table 4.1) is 0.051 g, and it is0.0024 g for the data in Table 4.8. The significantly better precision whendetermining the mass of a single penny suggests that the precision of ouranalysis is not limited by the balance. A more likely source of indeterminateerror is a significant variability in the masses of individual pennies.4B.3 Error and UncertaintyAnalytical chemists make a distinction between error and uncertainty. 3 Erroris the difference between a single measurement or result and its expectedvalue. In other words, error is a measure of bias. As discussed earlier,we can divide error into determinate and indeterminate sources. Althoughwe can correct for determinate errors, the indeterminate portion of the errorremains. With statistical significance testing, which is discussed later inthis chapter, we can determine if our results show evidence of bias.Uncertainty expresses the range of possible values for a measurementor result. Note that this definition of uncertainty is not the same as ourdefinition of precision. We calculate precision from our experimental data,providing an estimate of indeterminate errors. Uncertainty accounts forall errors—both determinate and indeterminate—that might reasonablyaffect a measurement or result. Although we always try to correct determinateerrors before beginning an analysis, the correction itself is subject touncertainty.Here is an example to help illustrate the difference between precisionand uncertainty. Suppose you purchase a 10-mL Class A pipet from a laboratorysupply company and use it without any additional calibration. Thepipet’s tolerance of ±0.02 mL is its uncertainty because your best estimateof its expected volume is 10.00 mL ± 0.02 mL. This uncertainty is primarilydeterminate error. If you use the pipet to dispense several replicateportions of solution, the resulting standard deviation is the pipet’s precision.Table 4.9 shows results for ten such trials, with a mean of 9.992 mL and astandard deviation of ±0.006 mL. This standard deviation is the precision3 Ellison, S.; Wegscheider, W.; Williams, A. Anal. Chem. 1997, 69, 607A–613A.In Section 4E we will discuss a statisticalmethod—the F-test—that you can use toshow that this difference is significant.See Table 4.2 for the tolerance of a 10-mLclass A transfer pipet.

76 Analytical Chemistry 2.0Although we will not derive or furtherjustify these rules here, you may consultthe additional resources at the end of thischapter for references that discuss thepropagation of uncertainty in more detail.Table 4.9 Experimental Results for Volume Delivered by a10-mL Class A Transfer PipetNumber Volume (mL) Number Volume (mL)1 10.002 6 9.9832 9.993 7 9.9913 9.984 8 9.9904 9.996 9 9.9885 9.989 10 9.999with which we expect to deliver a solution using a Class A 10-mL pipet. Inthis case the published uncertainty for the pipet (±0.02 mL) is worse thanits experimentally determined precision (±0.006 ml). Interestingly, thedata in Table 4.9 allows us to calibrate this specific pipet’s delivery volumeas 9.992 mL. If we use this volume as a better estimate of this pipet’s expectedvolume, then its uncertainty is ±0.006 mL. As expected, calibratingthe pipet allows us to decrease its uncertainty. 44CPropagation of UncertaintySuppose you dispense 20 mL of a reagent using the Class A 10-mL pipetwhose calibration information is given in Table 4.9. If the volume and uncertaintyfor one use of the pipet is 9.992 ± 0.006 mL, what is the volumeand uncertainty when we use the pipet twice?As a first guess, we might simply add together the volume and themaximum uncertainty for each delivery; thus(9.992 mL + 9.992 mL) ± (0.006 mL + 0.006 mL) = 19.984 ± 0.012 mLIt is easy to appreciate that combining uncertainties in this way overestimatesthe total uncertainty. Adding the uncertainty for the first delivery tothat of the second delivery assumes that with each use the indeterminateerror is in the same direction and is as large as possible. At the other extreme,we might assume that the uncertainty for one delivery is positiveand the other is negative. If we subtract the maximum uncertainties foreach delivery,(9.992 mL + 9.992 mL) ± (0.006 mL - 0.006 mL) = 19.984 ± 0.000 mLwe clearly underestimate the total uncertainty.So what is the total uncertainty? From the previous discussion we knowthat the total uncertainty is greater than ±0.000 mL and less than ±0.012mL. To estimate the cumulative effect of multiple uncertainties we usea mathematical technique known as the propagation of uncertainty. Ourtreatment of the propagation of uncertainty is based on a few simple rules.4 Kadis, R. Talanta 2004, 64, 167–173.

Chapter 4 Evaluating Analytical Data774C.1 A Few SymbolsA propagation of uncertainty allows us to estimate the uncertainty ina result from the uncertainties in the measurements used to calculate theresult. For the equations in this section we represent the result with thesymbol R, and the measurements with the symbols A, B, and C. The correspondinguncertainties are u R , u A , u B , and u C . We can define the uncertaintiesfor A, B, and C using standard deviations, ranges, or tolerances (orany other measure of uncertainty), as long as we use the same form for allmeasurements.The requirement that we express each uncertaintyin the same way is a critically importantpoint. Suppose you have a rangefor one measurement, such as a pipet’stolerance, and standard deviations for theother measurements. All is not lost. Thereare ways to convert a range to an estimateof the standard deviation. See Appendix 2for more details.4C.2 Uncertainty When Adding or SubtractingWhen adding or subtracting measurements we use their absolute uncertaintiesfor a propagation of uncertainty. For example, if the result is given bythe equationthen the absolute uncertainty in R isExample 4.5R = A + B - Cu = u + u + u2 2 24.6R A B CWhen dispensing 20 mL using a 10-mL Class A pipet, what is the total volumedispensed and what is the uncertainty in this volume? First, completethe calculation using the manufacturer’s tolerance of 10.00 mL ± 0.02 mL,and then using the calibration data from Table 4.9.So l u t i o nTo calculate the total volume we simply add the volumes for each use of thepipet. When using the manufacturer’s values, the total volume isV = 10. 00 mL + 10. 00 mL = 20.00 mLand when using the calibration data, the total volume isV = 9. 992 mL + 9. 992 mL = 19.984 mLUsing the pipet’s tolerance value as an estimate of its uncertainty gives theuncertainty in the total volume as2 2u R= ( 002 . ) + ( 0. 02) = 0.028 mLand using the standard deviation for the data in Table 4.9 gives an uncertaintyof2 2u R= ( 0. 006) + ( 0. 006) = 0.0085 mL

78 Analytical Chemistry 2.0Rounding the volumes to four significant figures gives 20.00 mL ± 0.03mL when using the tolerance values, and 19.98 ± 0.01 mL when usingthe calibration data.4C.3 Uncertainty When Multiplying or DividingWhen multiplying or dividing measurements we use their relative uncertaintiesfor a propagation of uncertainty. For example, if the result is givenby the equationthen the relative uncertainty in R isA BR = ×CuRu u u= ⎛ 2 2⎞⎝⎜A ⎠⎟ +⎛ ⎞⎝⎜B ⎠⎟ +⎛ ⎞⎝⎜C ⎠⎟R A B C24.7Example 4.6The quantity of charge, Q, in coulombs passing through an electrical circuitisQ = I×twhere I is the current in amperes and t is the time in seconds. When a currentof 0.15 A ± 0.01 A passes through the circuit for 120 s ± 1 s, what isthe total charge passing through the circuit and its uncertainty?So l u t i o nThe total charge isQ = ( 015 . A) × ( 120 s)= 18 CSince charge is the product of current and time, the relative uncertaintyin the charge isuRR=2 2⎛ 001 . ⎞⎝⎜⎠⎟ +⎛ 1 ⎞015 . ⎝⎜120⎠⎟= 0.0672The absolute uncertainty in the charge isu = R × 0 . 0672 = ( 18 C)× ( 0 . 0672 ) = 12 . CRThus, we report the total charge as 18 C ± 1 C.

Chapter 4 Evaluating Analytical Data794C.4 Uncertainty for Mixed OperationsMany chemical calculations involve a combination of adding and subtracting,and multiply and dividing. As shown in the following example, we cancalculate uncertainty by treating each operation separately using equation4.6 and equation 4.7 as needed.Example 4.7For a concentration technique the relationship between the signal and thean analyte’s concentration isS = k C + Stotal A A mbWhat is the analyte’s concentration, C A , and its uncertainty if S total is24.37 ± 0.02, S mb is 0.96 ± 0.02, and k A is 0.186 ± 0.003 ppm –1 .So l u t i o nRearranging the equation and solving for C ACAS=total−SkAmb24. 37 −096.== 125. 9 ppm−10.186 ppmgives the analyte’s concentration as 126 ppm. To estimate the uncertaintyin C A , we first determine the uncertainty for the numerator using equation4.6.2 2u R= ( 002 . ) + ( 0. 02) = 0.028The numerator, therefore, is 23.41 ± 0.028. To complete the calculationwe estimate the relative uncertainty in C A using equation 4.7.uRR=2⎛ 0.028⎞⎝⎜⎠⎟ +⎛ 0.003⎞23.41 ⎜⎝0.186⎠⎟ =20.0162The absolute uncertainty in the analyte’s concentration isu R= ( 125. 9 ppm) × ( 0. 0162) = 20 . ppmThus, we report the analyte’s concentration as 126 ppm ± 2 ppm.Practice Exercise 4.2To prepare a standard solution of Cu 2+ you obtain a piece of copper from a spool of wire. The spool’s initialweight is 74.2991 g and its final weight is 73.3216 g. You place the sample of wire in a 500 mL volumetricflask, dissolve it in 10 mL of HNO 3 , and dilute to volume. Next, you pipet a 1 mL portion to a 250-mLvolumetric flask and dilute to volume. What is the final concentration of Cu 2+ in mg/L, and its uncertainty?Assume that the uncertainty in the balance is ±0.1 mg and that you are using Class A glassware.Click here when to review your answer to this exercise.

80 Analytical Chemistry 2.0Table 4.10 Propagation of Uncertainty for SelectedMathematical Functions †Functionu RR = kAu = kuRAR = A+ Bu = u + u2 2R A BR = A−Bu = u + u2 2R A BR = A×BRA=BuRuR2 2u u= ⎛ ⎞⎝⎜A ⎠⎟ +⎛ ⎞⎝⎜B ⎠⎟R A B2 2u u= ⎛ ⎞⎝⎜A ⎠⎟ +⎛ ⎞⎝⎜B ⎠⎟R A BR = ln( A) uR = log( A) uRRuA=A= 0.4343×uAARA= euRR= uARA= 10uRR= 2.303×uA4C.5 Uncertainty for Other Mathematical FunctionsMany other mathematical operations are common in analytical chemistry,including powers, roots, and logarithms. Table 4.10 provides equations forpropagating uncertainty for some of these function.Example 4.8R =A kIf the pH of a solution is 3.72 with an absolute uncertainty of ±0.03, whatis the [H + ] and its uncertainty?uRRuA= k ×A† Assumes that the measurements A and B are independent; k is a constant whose value has nouncertainty.

Chapter 4 Evaluating Analytical Data81So l u t i o nThe concentration of H + ispH.[ H ] = 10 = 10 = 191 . × 10+ − −372 −4or 1.9 × 10 –4 M to two significant figures. From Table 4.10 the relativeuncertainty in [H + ] isuRRM= 2. 303× u = 2. 303× 003 . = 0.069AThe uncertainty in the concentration, therefore, is−4 −5(. 191× 10 M) × ( 0. 069) = 13 . × 10 MWe report the [H + ] as 1.9 (±0.1) × 10 –4 M.Writing this result as1.9 (±0.1) × 10 –4 Mis equivalent to1.9 × 10 –4 M ± 0.1 × 10 –4 MPractice Exercise 4.3A solution of copper ions is blue because it absorbs yellow and orangelight. Absorbance, A, is defined asPA =− ⎛ ⎞log⎝P ⎜ ⎠⎟where P o is the power of radiation from the light source and P is thepower after it passes through the solution. What is the absorbance if P ois 3.80×10 2 and P is 1.50×10 2 ? If the uncertainty in measuring P o and Pis 15, what is the uncertainty in the absorbance?Click here to review your answer to this exercise.4C.6 Is Calculating Uncertainty Actually Useful?Given the effort it takes to calculate uncertainty, it is worth asking whethersuch calculations are useful. The short answer is, yes. Let’s consider threeexamples of how we can use a propagation of uncertainty to help guide thedevelopment of an analytical method.One reason for completing a propagation of uncertainty is that we cancompare our estimate of the uncertainty to that obtained experimentally.For example, to determine the mass of a penny we measure mass twice—once to tare the balance at 0.000 g, and once to measure the penny’s mass.If the uncertainty for measuring mass is ±0.001 g, then we estimate theuncertainty in measuring mass as2 2u mass= ( 0. 001) + ( 0. 001) = 0.0014 go

82 Analytical Chemistry 2.02 ppm× 100 = 16 .%126 ppmIf we measure a penny’s mass several times and obtain a standard deviationof ±0.050 g, then we have evidence that our measurement process is out ofcontrol. Knowing this, we can identify and correct the problem.We also can use propagation of uncertainty to help us decide how toimprove an analytical method’s uncertainty. In Example 4.7, for instance,we calculated an analyte’s concentration as 126 ppm ± 2 ppm, which is apercent uncertainty of 1.6%. Suppose we want to decrease the percent uncertaintyto no more than 0.8%. How might we accomplish this? Lookingback at the calculation, we see that the concentration’s relative uncertaintyis determined by the relative uncertainty in the measured signal (correctedfor the reagent blank)0.028= 0.0012 or 0.12%23.41and the relative uncertainty in the method’s sensitivity, k A ,0.003 ppm0.186 ppm−1= − 10.016 or 1.6%Of these terms, the uncertainty in the method’s sensitivity dominates theoverall uncertainty. Improving the signal’s uncertainty will not improvethe overall uncertainty of the analysis. To achieve an overall uncertainty of0.8% we must improve the uncertainty in k A to ±0.0015 ppm –1 .Practice Exercise 4.4Verify that an uncertainty of ±0.0015 ppm –1 for k A is the correct result.Click here to review your answer to this exercise.Finally, we can use a propagation of uncertainty to determine which ofseveral procedures provides the smallest uncertainty. When diluting a stocksolution there are usually several different combinations of volumetricglassware that will give the same final concentration. For instance, we candilute a stock solution by a factor of 10 using a 10-mL pipet and a 100-mLvolumetric flask, or by using a 25-mL pipet and a 250-mL volumetric flask.We also can accomplish the same dilution in two steps using a 50-mL pipetand 100-mL volumetric flask for the first dilution, and a 10-mL pipet anda 50-mL volumetric flask for the second dilution. The overall uncertainty inthe final concentration—and, therefore, the best option for the dilution—depends on the uncertainty of the transfer pipets and volumetric flasks. Asshown below, we can use the tolerance values for volumetric glassware todetermine the optimum dilution strategy. 55 Lam, R. B.; Isenhour, T. L. Anal. Chem. 1980, 52, 1158–1161.

Chapter 4 Evaluating Analytical Data83Example 4.9Which of the following methods for preparing a 0.0010 M solution froma 1.0 M stock solution provides the smallest overall uncertainty?(a) A one-step dilution using a 1-mL pipet and a 1000-mL volumetricflask.(b) A two-step dilution using a 20-mL pipet and a 1000-mL volumetricflask for the first dilution, and a 25-mL pipet and a 500-mL volumetricflask for the second dilution.So l u t i o nThe dilution calculations for case (a) and case (b) are1.000 mLcase (a): 1.0 M× = 0.0010 M1000.0 mL20.00 mL 25.00 mLcase(b): 1.0 M× × = 0.0010 M1000.0 mL 500.0 mLUsing tolerance values from Table 4.2, the relative uncertainty for case (a)isuRR=2⎛ 0.006⎞⎝⎜⎠⎟ +⎛ 03 . ⎞1.000 ⎝⎜1000.0⎠⎟ =20.006and for case (b) the relative uncertainty isuRR=22 2⎛ 003 . ⎞⎝⎜⎠⎟ +⎛ 03 . ⎞20.00 ⎝⎜1000. 0⎠⎟ +⎛ 003 . ⎞⎝⎜⎠⎟ +⎛ 02 . ⎞25.00 ⎜⎝500.0⎠⎟ =20.002Since the relative uncertainty for case (b) is less than that for case (a), thetwo-step dilution provides the smallest overall uncertainty.4DThe Distribution of Measurements and ResultsEarlier we reported results for a determination of the mass of a circulatingUnited States penny, obtaining a mean of 3.117 g and a standard deviationof 0.051 g. Table 4.11 shows results for a second, independent determinationof a penny’s mass, as well as the data from the first experiment.Although the means and standard deviations for the two experiments aresimilar, they are not identical. The difference between the experimentsraises some interesting questions. Are the results for one experiment betterthan those for the other experiment? Do the two experiments provideequivalent estimates for the mean and the standard deviation? What is ourbest estimate of a penny’s expected mass? To answers these questions weneed to understand how to predict the properties of all pennies by analyz-

84 Analytical Chemistry 2.0Table 4.11 Results for Two Determinations of the Mass ofa Circulating United States PennyFirst ExperimentSecond ExperimentPenny Mass (g) Penny Mass (g)1 3.080 1 3.0522 3.094 2 3.1413 3.107 3 3.0834 3.056 4 3.0835 3.112 5 3.0486 3.1747 3.198X 3.117 3.081s 0.051 0.037ing a small sample of pennies. We begin by making a distinction betweenpopulations and samples.4D.1 Populations and SamplesA population is the set of all objects in the system we are investigating. Forour experiment, the population is all United States pennies in circulation.This population is so large that we cannot analyze every member of thepopulation. Instead, we select and analyze a limited subset, or sample of thepopulation. The data in Table 4.11, for example, are results for two samplesdrawn from the larger population of all circulating United States pennies.4D.2 Probability Distributions for PopulationsTable 4.11 provides the mean and standard deviation for two samples ofcirculating United States pennies. What do these samples tell us about thepopulation of pennies? What is the largest possible mass for a penny? Whatis the smallest possible mass? Are all masses equally probable, or are somemasses more common?To answer these questions we need to know something about how themasses of individual pennies are distributed around the population’s averagemass. We represent the distribution of a population by plotting theprobability or frequency of obtaining an specific result as a function of thepossible results. Such plots are called probability distributions.There are many possible probability distributions. In fact, the probabilitydistribution can take any shape depending on the nature of the population.Fortunately many chemical systems display one of several commonprobability distributions. Two of these distributions, the binomial distributionand the normal distribution, are discussed in this section.

Chapter 4 Evaluating Analytical Data85Binomial DistributionThe binomial distribution describes a population in which the result isthe number of times a particular event occurs during a fixed number oftrials. Mathematically, the binomial distribution isN !PXN ( , ) =× p × ( 1−p)X !( N − X )!X N − Xwhere P(X , N) is the probability that an event occurs X times during N trials,and p is the event’s probability for a single trial. If you flip a coin five times,P(2,5) is the probability that the coin will turn up “heads” exactly twice.A binomial distribution has well-defined measures of central tendencyand spread. The expected mean value isµ=NpThe term N! reads as N-factorial and is theproduct N × (N-1) × (N-2) ×…× 1. Forexample, 4! is 4 × 3 × 2 × 1 = 24. Yourcalculator probably has a key for calculatingfactorials.and the expected spread is given by the varianceor the standard deviation.σ 2 = Np( 1−p)σ= Np( 1−p)The binomial distribution describes a population whose members cantake on only specific, discrete values. When you roll a die, for example,the possible values are 1, 2, 3, 4, 5, or 6. A roll of 3.45 is not possible. Asshown in Example 4.10, one example of a chemical system obeying thebinomial distribution is the probability of finding a particular isotope in amolecule.Example 4.10Carbon has two stable, non-radioactive isotopes, 12 C and 13 C, with relativeisotopic abundances of, respectively, 98.89% and 1.11%.(a) What are the mean and the standard deviation for the number of 13 Catoms in a molecule of cholesterol (C 27 H 44 O)?(b) What is the probability that a molecule of cholesterol has no atomsof 13 C?So l u t i o nThe probability of finding an atom of 13 C in a molecule of cholesterolfollows a binomial distribution, where X is the number of 13 C atoms, Nis the number of carbon atoms in a molecule of cholesterol, and p is theprobability that any single atom of carbon in 13 C.(a) The mean number of 13 C atoms in a molecule of cholesterol is

86 Analytical Chemistry 2.0µ= Np = 27× 0. 0111=0.300with a standard deviation ofσ= Np( 1− p) = 27× 0. 0111× ( 1− 0. 0111) = 0.172(b) The probability of finding a molecule of cholesterol without an atomof 13 C is27!0 27P( 027 , ) =( 0. 0111) ( 1 0. 0111)0!( 27− 0)!× × − − 0 = 0.740There is a 74.0% probability that a molecule of cholesterol will nothave an atom of 13 C, a result consistent with the observation thatthe mean number of 13 C atoms per molecule of cholesterol, 0.300,is less than one.A portion of the binomial distribution for atoms of 13 C in cholesterol isshown in Figure 4.6. Note in particular that there is little probability offinding more than two atoms of 13 C in any molecule of cholesterol.No r m a l DistributionA binomial distribution describes a population whose members have onlycertain, discrete values. This is the case with the number of 13 C atoms incholesterol. A molecule of cholesterol, for example, can have two 13 C atoms,but it can not have 2.5 atoms of 13 C. A population is continuous if its membersmay take on any value. The efficiency of extracting cholesterol froma sample, for example, can take on any value between 0% (no cholesterolextracted) and 100% (all cholesterol extracted).The most common continuous distribution is the Gaussian, or normaldistribution, the equation for which is1.00.8ProbabilityFigure 4.6 Portion of the binomialdistribution for the number0.4of naturally occurring 13 C atomsin a molecule of cholesterol. Only3.6% of cholesterol moleculescontain more than one atom of13 C, and only 0.33% containmore than two atoms of 13 C. 0 1 2 3 4 50.60.0 0.2Number of 13 C Atoms in a Molecule of Cholesterol

Chapter 4 Evaluating Analytical Data87f ( X)=12πσ2e( X )−− 2µ22σwhere m is the expected mean for a population with n members∑ Xiiµ=nand s 2 is the population’s variance.2∑( X −µ)i2 iσ =4.8nExamples of normal distributions, each with an expected mean of 0 andwith variances of 25, 100, or 400, are shown in Figure 4.7. Two featuresof these normal distribution curves deserve attention. First, note that eachnormal distribution has a single maximum corresponding to m, and that thedistribution is symmetrical about this value. Second, increasing the population’svariance increases the distribution’s spread and decreases its height;the area under the curve, however, is the same for all three distribution.The area under a normal distribution curve is an important and usefulproperty as it is equal to the probability of finding a member of the populationwith a particular range of values. In Figure 4.7, for example, 99.99%of the population shown in curve (a) have values of X between -20 and20. For curve (c), 68.26% of the population’s members have values of Xbetween -20 and 20.Because a normal distribution depends solely on m and s 2 , the probabilityof finding a member of the population between any two limits isthe same for all normally distributed populations. Figure 4.8, for example,shows that 68.26% of the members of a normal distribution have a valueFigure 4.7 Normal distributioncurves for:(a) m = 0; s 2 = 25(b) m = 0; s 2 = 100(c) m = 0; s 2 =400

88 Analytical Chemistry 2.0Figure 4.8 Normal distributioncurve showing the area under thecurve for several different rangesof values of X. As shown here,68.26% of the members of a normallydistributed population havevalues within ±1σ of the population’sexpected mean, and 13.59%have values between μ–1σ andu–2σ. The area under the curvebetween any two limits can befound using the probability tablein Appendix 3.-3σ -2σ -1σ μ +1σ +2σ +3σwithin the range m ± 1s, and that 95.44% of population’s members havevalues within the range m ± 2s. Only 0.17% members of a population havevalues exceeding the expected mean by more than ± 3s. Additional rangesand probabilities are gathered together in a probability table that you willfind in Appendix 3. As shown in Example 4.11, if we know the mean andstandard deviation for a normally distributed population, then we can determinethe percentage of the population between any defined limits.Example 4.1134.13% 34.13%13.59 %13.59 %2.14 % 2.14 %Value of XThe amount of aspirin in the analgesic tablets from a particular manufactureris known to follow a normal distribution with m = 250 mg ands 2 = 25. In a random sampling of tablets from the production line, whatpercentage are expected to contain between 243 and 262 mg of aspirin?So l u t i o nWe do not determine directly the percentage of tablets between 243 mgand 262 mg of aspirin. Instead, we first find the percentage of tablets withless than 243 mg of aspirin and the percentage of tablets having more than262 mg of aspirin. Subtracting these results from 100%, gives the percentageof tablets containing between 243 mg and 262 mg of aspirin.To find the percentage of tablets with less than 243 mg of aspirin or morethan 262 mg of aspirin we calculate the deviation, z, of each limit from min terms of the population’s standard deviation, szX= −µσwhere X is the limit in question. The deviation for the lower limit is

Chapter 4 Evaluating Analytical Data89243−250z lower= =−14.5and the deviation for the upper limit is262−250z upper= =+ 24 .5Using the table in Appendix 3, we find that the percentage of tablets withless than 243 mg of aspirin is 8.08%, and the percentage of tablets withmore than 262 mg of aspirin is 0.82%. Therefore, the percentage of tabletscontaining between 243 and 262 mg of aspirin is100.00% - 8.08% - 0.82 % = 91.10%Figure 4.9 shows the distribution of aspiring in the tablets, with the areain blue showing the percentage of tablets containing between 243 mg and262 mg of aspirin.Practice Exercise 4.5What percentage of aspirin tablets will contain between 240 mg and 245mg of aspirin if the population’s mean is 250 mg and the population’sstandard deviation is 5 mg.Click here to review your answer to this exercise.8.08%0.82%230 240 250 260 270Aspirin (mg)91.10% Figure 4.9 Normal distributionfor the population of aspirin tabletsin Example 4.11. The population’smean and standard deviationare 250 mg and 5 mg, respectively.The shaded area shows the percentageof tablets containing between243 mg and 262 mg of aspirin.

90 Analytical Chemistry 2.04D.3 Confidence Intervals for PopulationsIf we randomly select a single member from a population, what is its mostlikely value? This is an important question, and, in one form or another, itis at the heart of any analysis in which we wish to extrapolate from a sampleto the sample’s parent population. One of the most important features ofa population’s probability distribution is that it provides a way to answerthis question.Figure 4.8 shows that for a normal distribution, 68.26% of the population’smembers are found within the range of m ± 1s. Stating this anotherway, there is a 68.26% probability that the result for a single sample drawnfrom a normally distributed population is in the interval m ± 1s. In general,if we select a single sample we expect its value, X i to be in the rangeWhen z = 1, we call this the 68.26% confidenceinterval.X = µ ± z σ 4.9iwhere the value of z is how confident we are in assigning this range. Valuesreported in this fashion are called confidence intervals. Equation 4.9,for example, is the confidence interval for a single member of a population.Table 4.12 gives the confidence intervals for several values of z. For reasonswe will discuss later in the chapter, a 95% confidence level is a commonchoice in analytical chemistry.Example 4.12What is the 95% confidence interval for the amount of aspirin in a singleanalgesic tablet drawn from a population for which μ is 250 mg and σ 2is 25?So l u t i o nUsing Table 4.12, we find that z is 1.96 for a 95% confidence interval.Substituting this into equation 4.9, gives the confidence interval for asingle tablet asX i = μ ± 1.96σ = 250 mg ± (1.96 × 5) = 250 mg ± 10 mgTable 4.12 Confidence Intervals for aNormal Distribution (μ ± zσ)z Confidence Interval (%)0.50 38.301.00 68.261.50 86.641.96 95.002.00 95.442.50 98.763.00 99.733.50 99.95

Chapter 4 Evaluating Analytical Data91A confidence interval of 250 mg ± 10 mg means that 95% of the tablets inthe population contain between 240 and 260 mg of aspirin.Alternatively, we can express a confidence interval for the expectedmean in terms of the population’s standard deviation and the value of asingle member drawn from the population.Example 4.13µ = X ± zσi4.10The population standard deviation for the amount of aspirin in a batch ofanalgesic tablets is known to be 7 mg of aspirin. If you randomly select andanalyze a single tablet and find that it contains 245 mg of aspirin, what isthe 95% confidence interval for the population’s mean?So l u t i o nThe 95% confidence interval for the population mean is given asµ = X ± zσ= 245 mg ± (. 196× 7)mg = 245 mg ± 14 mgiTherefore, there is 95% probability that the population’s mean, μ, lieswithin the range of 231 mg to 259 mg of aspirin.It is unusual to predict the population’s expected mean from the analysisof a single sample. We can extend confidence intervals to include themean of n samples drawn from a population of known σ. The standarddeviation of the mean, σ X, which also is known as the standard errorof the mean, isσX=σnProblem 8 at the end of the chapter asksyou to derive this equation using a propagationof uncertainty.The confidence interval for the population’s mean, therefore, isµ= X ±zσn4.11Example 4.14What is the 95% confidence interval for the analgesic tablets describedin Example 4.13, if an analysis of five tablets yields a mean of 245 mg ofaspirin?So l u t i o nIn this case the confidence interval is

92 Analytical Chemistry 2.0196 . × 7µ= 245 mg ± mg = 245 mg ± 6 mg5Thus, there is a 95% probability that the population’s mean is between 239to 251 mg of aspirin. As expected, the confidence interval when using themean of five samples is smaller than that for a single sample.Practice Exercise 4.6An analysis of seven aspirin tablets from a population known to havea standard deviation of 5, gives the following results in mg aspirin pertablet:246 249 255 251 251 247 250What is the 95% confidence interval for the population’s expectedmean?Click here when you are ready to review your answer.4D.4 Probability Distributions for SamplesIn working example 4.11–4.14 we assumed that the amount of aspirin inanalgesic tablets is normally distributed. Without analyzing every memberof the population, how can we justify this assumption? In situations wherewe can not study the whole population, or when we can not predict themathematical form of a population’s probability distribution, we must deducethe distribution from a limited sampling of its members.Sa m p l e Distributions a n d t h e Ce n t r a l Limit Th e o r e mLet’s return to the problem of determining a penny’s mass to explore furtherthe relationship between a population’s distribution and the distribution ofa sample drawn from that population. The two sets of data in Table 4.11are too small to provide a useful picture of a sample’s distribution. To gaina better picture of the distribution of pennies we need a larger sample, suchas that shown in Table 4.13. The mean and the standard deviation for thissample of 100 pennies are 3.095 g and 0.0346 g, respectively.A histogram (Figure 4.10) is a useful way to examine the data in Table4.13. To create the histogram, we divide the sample into mass intervalsand determine the percentage of pennies within each interval (Table 4.14).Note that the sample’s mean is the midpoint of the histogram.Figure 4.10 also includes a normal distribution curve for the populationof pennies, assuming that the mean and variance for the sample provide appropriateestimates for the mean and variance of the population. Althoughthe histogram is not perfectly symmetric, it provides a good approximationof the normal distribution curve, suggesting that the sample of 100 pennies

Chapter 4 Evaluating Analytical Data93Table 4.13 Masses for a Sample of 100 Circulating U. S. PenniesPenny Mass (g) Penny Mass (g) Penny Mass (g) Penny Mass (g)1 3.126 26 3.073 51 3.101 76 3.0862 3.140 27 3.084 52 3.049 77 3.1233 3.092 28 3.148 53 3.082 78 3.1154 3.095 29 3.047 54 3.142 79 3.0555 3.080 30 3.121 55 3.082 80 3.0576 3.065 31 3.116 56 3.066 81 3.0977 3.117 32 3.005 57 3.128 82 3.0668 3.034 33 3.115 58 3.112 83 3.1139 3.126 34 3.103 59 3.085 84 3.10210 3.057 35 3.086 60 3.086 85 3.03311 3.053 36 3.103 61 3.084 86 3.11212 3.099 37 3.049 62 3.104 87 3.10313 3.065 38 2.998 63 3.107 88 3.19814 3.059 39 3.063 64 3.093 89 3.10315 3.068 40 3.055 65 3.126 90 3.12616 3.060 41 3.181 66 3.138 91 3.11117 3.078 42 3.108 67 3.131 92 3.12618 3.125 43 3.114 68 3.120 93 3.05219 3.090 44 3.121 69 3.100 94 3.11320 3.100 45 3.105 70 3.099 95 3.08521 3.055 46 3.078 71 3.097 96 3.11722 3.105 47 3.147 72 3.091 97 3.14223 3.063 48 3.104 73 3.077 98 3.03124 3.083 49 3.146 74 3.178 99 3.08325 3.065 50 3.095 75 3.054 100 3.104Table 4.14 Frequency Distribution for the Data in Table 4.13Mass Interval Frequency (as %) Mass Interval Frequency (as %)2.991–3.009 2 3.104–3.123 193.010–3.028 0 3.124–3.142 123.029–3.047 4 3.143–3.161 33.048–3.066 19 3.162–3.180 13.067–3.085 15 3.181–3.199 23.086–3.104 23

94 Analytical Chemistry 2.0Figure 4.10 The blue bars showa histogram for the data in Table4.13. The height of a bar correspondsto the percentage of pennieswithin the mass intervals shown inTable 4.14. Superimposed on thehistogram is a normal distributioncurve assuming that m and s 2 forthe population are equivalent toX and s 2 for the sample. The totalarea of the histogram’s bars and thearea under the normal distributioncurve are equal.You might reasonably ask whether thisaspect of the central limit theorem is importantas it is unlikely that we will complete10 000 analyses, each of which isthe average of 10 individual trials. This isdeceiving. When we acquire a sample foranalysis—a sample of soil, for example—it consists of many individual particles,each of which is an individual sample ofthe soil. Our analysis of the gross sample,therefore, is the mean for this large numberof individual soil particles. Because ofthis, the central limit theorem is relevant.2.95 3.00 3.05 3.10 3.15 3.20 3.25Mass of Pennies (g)is normally distributed. It is easy to imagine that the histogram will moreclosely approximate a normal distribution if we include additional penniesin our sample.We will not offer a formal proof that the sample of pennies in Table 4.13and the population of all circulating U. S. pennies are normally distributed.The evidence we have seen, however, strongly suggests that this is true. Althoughwe can not claim that the results for all analytical experiments arenormally distributed, in most cases the data we collect in the laboratoryare, in fact, drawn from a normally distributed population. According tothe central limit theorem, when a system is subject to a variety of indeterminateerrors, the results approximate a normal distribution. 6 As thenumber of sources of indeterminate error increases, the results more closelyapproximate a normal distribution. The central limit theorem holds trueeven if the individual sources of indeterminate error are not normally distributed.The chief limitation to the central limit theorem is that the sourcesof indeterminate error must be independent and of similar magnitude sothat no one source of error dominates the final distribution.An additional feature of the central limit theorem is that a distributionof means for samples drawn from a population with any distributionwill closely approximate a normal distribution if the size of the samples islarge enough. Figure 4.11 shows the distribution for two samples of 10 000drawn from a uniform distribution in which every value between 0 and 1occurs with an equal frequency. For samples of size n = 1, the resulting distributionclosely approximates the population’s uniform distribution. Thedistribution of the means for samples of size n = 10, however, closely approximatesa normal distribution.6 Mark, H.; Workman, J. Spectroscopy 1988, 3, 44–48.

Chapter 4 Evaluating Analytical Data95(a)500400(b)20001500dFrequency300200100Frequency100050000.0 0.2 0.4 0.6 0.8 1.00.2 0.3 _ 0.4 0.5 0.6 0.7 0.8Value of X for Samples of Size n = 1 Value of X for Samples of Size n = 10Figure 4.11 Histograms for (a) 10000 samples of size n = 1 drawn from a uniform distribution with a minimum valueof 0 and a maximum value of 1, and (b) the means for 10 000 samples of size n = 10 drawn from the same uniformdistribution. For (a) the mean of the 10000 samples is 0.5042, and for (b) the mean of the 10000 samples is 0.5006.Note that for (a) the distribution closely approximates a uniform distribution in which every possible result is equallylikely, and that for (b) the distribution closely approximates a normal distribution.0De g r e e s o f Fr e e d o mIn reading to this point, did you notice the differences between the equationsfor the standard deviation or variance of a population and the standarddeviation or variance of a sample? If not, here are the two equations:σs22∑( X −µ)ii=n∑( X − X)ii=n −1Both equations measure the variance around the mean, using m for a populationand X for a sample. Although the equations use different measuresfor the mean, the intention is the same for both the sample and the population.A more interesting difference is between the denominators of the twoequations. In calculating the population’s variance we divide the numeratorby the population’s size, n. For the sample’s variance we divide by n – 1,where n is the size of the sample. Why do we make this distinction?A variance is the average squared deviation of individual results fromthe mean. In calculating an average we divide by the number of independentmeasurements, also known as the degrees of freedom, contributingto the calculation. For the population’s variance, the degrees of freedom isequal to the total number of members, n, in the population. In measuring22

96 Analytical Chemistry 2.0Here is another way to think about degreesof freedom. We analyze samples tomake predictions about the underlyingpopulation. When our sample consists ofn measurements we cannot make morethan n independent predictions aboutthe population. Each time we estimate aparameter, such as the population’s mean,we lose a degree of freedom. If there aren degrees of freedom for calculating thesample’s mean, then there are n – 1 degreesof freedom remaining for calculating thesample’s variance.every member of the population we have complete information about thepopulation.When calculating the sample’s variance, however, we first replace m withX , which we also calculate from the same data. If there are n members inthe sample, we can deduce the value of the n th member from the remainingn – 1 members. For example, if n = 5 and we know that the first foursamples are 1, 2, 3 and 4, and that the mean is 3, then the fifth member ofthe sample must beX = ( X × n) −X −X −X − X = ( 3× 5)−1−2−3− 4=55 1 2 3 4Using n – 1 in place of n when calculating the sample’s variance ensures thats 2 is an unbiased estimator of s 2 .4D.5 Confidence Intervals for SamplesEarlier we introduced the confidence interval as a way to report the mostprobable value for a population’s mean, m,µ= X ±zσn4.11where X is the mean for a sample of size n, and s is the population’s standarddeviation. For most analyses we do not know the population’s standarddeviation. We can still calculate a confidence interval, however, if we maketwo modifications to equation 4.11.The first modification is straightforward—we replace the population’sstandard deviation, s, with the sample’s standard deviation, s. The secondmodification is less obvious. The values of z in Table 4.12 are for a normaldistribution, which is a function of s 2 , not s 2 . Although the sample’s variance,s 2 , provides an unbiased estimate for the population’s variance, s 2 ,the value of s 2 for any sample may differ significantly from s 2 . To accountfor the uncertainty in estimating s 2 , we replace the variable z in equation4.11 with the variable t, where t is defined such that t ≥ z at all confidencelevels.µ= X ±tsn4.12Values for t at the 95% confidence level are shown in Table 4.15. Note thatt becomes smaller as the number of degrees of freedom increases, approachingz as n approaches infinity. The larger the sample, the more closely itsconfidence interval approaches the confidence interval given by equation4.11. Appendix 4 provides additional values of t for other confidence levels.

Chapter 4 Evaluating Analytical Data97Table 4.15 Values of t for a 95% Confidence IntervalDegrees ofDegrees ofFreedomtFreedomt1 12.706 12 2.1792 4.303 14 2.1453 3.181 16 2.1204 2.776 18 2.1015 2.571 20 2.0866 2.447 30 2.0427 2.365 40 2.0218 2.306 60 2.0009 2.262 100 1.98410 2.228 ∞ 1.960Example 4.15What are the 95% confidence intervals for the two samples of pennies inTable 4.11?So l u t i o nThe mean and standard deviation for first experiment are, respectively,3.117 g and 0.051 g. Because the sample consists of seven measurements,there are six degrees of freedom. The value of t from Table 4.15, is 2.447.Substituting into equation 4.12 gives2. 447×0.051 gµ= 3.117 g ±= 3. 117 g±0.047 g7For the second experiment the mean and standard deviation are 3.081 gand 0.073 g, respectively, with four degrees of freedom. The 95% confidenceinterval is2. 776×0.037 gµ= 3.081 g ±= 3. 081 g±0.046 g5Based on the first experiment, there is a 95% probability that the population’smean is between 3.070 to 3.164 g. For the second experiment,the 95% confidence interval spans 3.035 g–3.127 g. The two confidenceintervals are not identical, but the mean for each experiment is containedwithin the other experiment’s confidence interval. There also is an appreciableoverlap of the two confidence intervals. Both of these observationsare consistent with samples drawn from the same population.Our comparison of these two confidenceintervals is rather vague and unsatisfying.We will return to this point in the nextsection, when we consider a statistical approachto comparing the results of experiments.

98 Analytical Chemistry 2.0We will return to the topic of detectionlimits near the end of the chapter.Practice Exercise 4.7What is the 95% confidence interval for the sample of 100 pennies inTable 4.13? The mean and the standard deviation for this sample are3.095 g and 0.0346 g, respectively. Compare your result to the confidenceintervals for the samples of pennies in Table 4.11.Click here when to review your answer to this exercise.4D.6 A Cautionary StatementThere is a temptation when analyzing data to plug numbers into an equation,carry out the calculation, and report the result. This is never a goodidea, and you should develop the habit of constantly reviewing and evaluatingyour data. For example, if an analysis on five samples gives an analyte’smean concentration as 0.67 ppm with a standard deviation of 0.64 ppm,then the 95% confidence interval is2. 776×064 . ppmµ= 067 . ppm ±= 067 . ppm ± 0.79 ppm5This confidence interval suggests that the analyte’s true concentration lieswithin the range of –0.12 ppm to 1.46 ppm. Including a negative concentrationwithin the confidence interval should lead you to reevaluate yourdata or conclusions. A closer examination of your data may convince youthat the standard deviation is larger than expected, making the confidenceinterval too broad, or you may conclude that the analyte’s concentration istoo small to detect accurately.Here is a second example of why you should closely examine your data.The results for samples drawn from a normally distributed population mustbe random. If the results for a sequence of samples show a regular patternor trend, then the underlying population may not be normally distributed,or there may be a time-dependent determinate error. For example, if werandomly select 20 pennies and find that the mass of each penny is largerthan that for the preceding penny, we might suspect that our balance isdrifting out of calibration.4EStatistical Analysis of DataA confidence interval is a useful way to report the result of an analysisbecause it sets limits on the expected result. In the absence of determinateerror, a confidence interval indicates the range of values in which we expectto find the population’s expected mean. When we report a 95% confidenceinterval for the mass of a penny as 3.117 g ± 0.047 g, for example, we areclaiming that there is only a 5% probability that the expected mass of pennyis less than 3.070 g or more than 3.164 g.

Chapter 4 Evaluating Analytical Data99Because a confidence interval is a statement of probability, it allows usto answer questions such as “Are the results for a newly developed methodfor determining cholesterol in blood significantly different from those obtainedusing a standard method?” or “Is there a significant variation inthe composition of rainwater collected at different sites downwind from acoal-burning utility plant?”. In this section we introduce a general approachto the statistical analysis of data. Specific statistical tests are presented inSection 4F.4E.1 Significance TestingLet’s consider the following problem. To determine if a medication is effectivein lowering blood glucose concentrations, we collect two sets of bloodsamples from a patient. We collect one set of samples immediately beforeadministering the medication, and collect the second set of samples severalhours later. After analyzing the samples, we report their respective meansand variances. How do we decide if the medication was successful in loweringthe patient’s concentration of blood glucose?One way to answer this question is to construct normal distributioncurves for each sample, and to compare them to each other. Three possibleoutcomes are shown in Figure 4.12. In Figure 4.12a, there is a completeseparation of the normal distribution curves, strongly suggesting that thesamples are significantly different. In Figure 4.12b, the normal distributionsfor the two samples almost completely overlap each other, suggestingthat any difference between the samples is insignificant. Figure 4.12c, however,presents a dilemma. Although the means for the two samples appearto be different, there is sufficient overlap of the normal distributions thata significant number of possible outcomes could belong to either distribution.In this case the best we can do is to make a statement about the probabilitythat the samples are significantly different.The process by which we determine the probability that there is a significantdifference between two samples is called significance testing orhypothesis testing. Before discussing specific examples we will first establisha general approach to conducting and interpreting significance tests.4E.2 Constructing a Significance TestThe purpose of a significance test is to determine whether the differencebetween two or more values is too large to be explained by indeterminateerror. The first step in constructing a significance test is to state the problemas a yes or no question, such as “Is this medication effective at loweringa patient’s blood glucose levels?”. A null hypothesis and an alternativehypothesis provide answers to the question. The null hypothesis, H 0 , isthat indeterminate error is sufficient to explain any differences in our data.The alternative hypothesis, H A , is that the differences are too great tobe explained by random error and, therefore, must be determinate. We test(a)(b)(c)ValuesValuesValuesFigure 4.12 Three examples showingpossible relationships betweenthe normal distribution curves fortwo samples. In (a) the curves arecompletely separate, suggestingthat the samples are significantlydifferent from each other. In (b)the two curves are almost identical,suggesting that the samplesare indistinguishable. The partialoverlap of the curves in (c) meansthat the best we can do is to indicatethe probability that there is adifference between the samples.

100 Analytical Chemistry 2.0The four steps for a statistical analysis ofdata:1. Pose a question, and state the null hypothesisand the alternative hypothesis.3. Choose a confidence level for the statisticalanalysis.3. Calculate an appropriate test statisticand compare it to a critical value.4. Either retain the null hypothesis, orreject it and accept the alternative hypothesis.In this textbook we use a to represent theprobability of incorrectly rejecting thenull hypothesis. In other textbooks thisprobability is given as p (often read as “pvalue”).Although the symbols differ, themeaning is the same.the null hypothesis, which we either retain or reject. If we reject the nullhypothesis, then we must accept the alternative hypothesis, concludingthat the difference is significant and that it cannot be explained by randomerror.Failing to reject a null hypothesis is not the same as accepting it. Weretain a null hypothesis because there is insufficient evidence to prove itincorrect. It is impossible to prove that a null hypothesis is true. This is animportant point that is easy to forget. To appreciate this point let’s returnto our sample of 100 pennies in Table 4.13. After looking at the data youmight propose the following null and alternative hypotheses.H 0 : The mass of any U.S. penny in circulation is in the range of 2.900g–3.200 g.H A : A U.S. penny in circulation may have a mass less than 2.900 g or amass of more than 3.200 g.To test the null hypothesis you reach into your pocket, retrieve a penny, anddetermine its mass. If the penny’s mass is 2.512 g then you reject the nullhypothesis, and accept the alternative hypothesis. Suppose that the penny’smass is 3.162 g. Although this result increases your confidence in the nullhypothesis, it does not prove it is correct because the next penny you pullfrom your pocket might weigh less than 2.900 g or more than 3.200 g.After stating the null and alternative hypotheses, the second step is tochoose a confidence level for the analysis. The confidence level defines theprobability that we will reject the null hypothesis when it is, in fact, true.We can express this as our confidence in correctly rejecting the null hypothesis(e.g. 95%), or as the probability that we are incorrectly rejecting thenull hypothesis. For the latter, the confidence level is given as a, whereα= 1−confidence level(%)100For a 95% confidence level, a is 0.05.The third step is to calculate an appropriate test statistic and to compareit to a critical value. The test statistic’s critical value defines a breakpointbetween values that lead us to reject or to retain the null hypothesis. Howwe calculate the test statistic depends on what we are comparing, a topicwe cover in section 4F. The last step is to either retain the null hypothesis,or to reject it and accept the alternative hypothesis.4E.3 One-Tailed and Two-Tailed Significance TestsSuppose we want to evaluate the accuracy of a new analytical method. Wemight use the method to analyze a Standard Reference Material containinga known concentration of analyte, m. We analyze the standard severaltimes, obtaining an mean value, X , for the analyte’s concentration. Ournull hypothesis is that there is no difference between X and m

Chapter 4 Evaluating Analytical Data101H : X =µ0If we conduct the significance test at a = 0.05, then we retain the null hypothesisif a 95% confidence interval around X contains m. If the alternativehypothesis isH : X ≠µAthen we reject the null hypothesis and accept the alternative hypothesis ifm lies in the shaded areas at either end of the sample’s probability distributioncurve (Figure 4.13a). Each of the shaded areas accounts for 2.5% ofthe area under the probability distribution curve. This is a two-tailedsignificance test because we reject the null hypothesis for values of m ateither extreme of the sample’s probability distribution curve.We also can write the alternative hypothesis in two additional waysH : X >µAH : X

102 Analytical Chemistry 2.0We reserve a one-tailed significance test for a situation where we arespecifically interested in whether one parameter is larger (or smaller) thanthe other parameter. For example, a one-tailed significance test is appropriateif we are evaluating a medication’s ability to lower blood glucose levels.In this case we are interested only in whether the result after administeringthe medication is less than the result before beginning treatment. If thepatient’s blood glucose level is greater after administering the medication,then we know the answer—the medication did not work—without conductinga statistical analysis.4E.4 Errors in Significance TestingBecause a significance test relies on probability, its interpretation is naturallysubject to error. In a significance test, a defines the probability ofrejecting a null hypothesis that is true. When we conduct a significancetest at a = 0.05, there is a 5% probability that we will incorrectly rejectthe null hypothesis. This is known as a type 1 error, and its risk is alwaysequivalent to a. Type 1 errors in two-tailed and one-tailed significance testscorrespond to the shaded areas under the probability distribution curvesin Figure 4.13.A second type of error occurs when we retain a null hypothesis eventhough it is false. This is as a type 2 error, and its probability of occurrenceis b. Unfortunately, in most cases we cannot calculate or estimate the valuefor b. The probability of a type 2 error, however, is inversely proportionalto the probability of a type 1 error.Minimizing a type 1 error by decreasing a increases the likelihood of atype 2 error. When we choose a value for a we are making a compromise betweenthese two types of error. Most of the examples in this text use a 95%confidence level (a = 0.05) because this provides a reasonable compromisebetween type 1 and type 2 errors. It is not unusual, however, to use morestringent (e.g. a = 0.01) or more lenient (e.g. a = 0.10) confidence levels.4FStatistical Methods for Normal DistributionsThe most common distribution for our results is a normal distribution.Because the area between any two limits of a normal distribution is welldefined, constructing and evaluating significance tests is straightforward.4F.1 Comparing X to μOne approach for validating a new analytical method is to analyze a samplecontaining a known amount of analyte, m. To judge the method’s accuracywe analyze several portions of the sample, determine the average amountof analyte in the sample, X , and use a significance test to compare it to m.Our null hypothesis, H : X =µ, is that any difference between X and m0is the result of indeterminate errors affecting the determination of X . The

Chapter 4 Evaluating Analytical Data103alternative hypothesis, H : X ≠µ A, is that the difference between X andm is too large to be explained by indeterminate error.The test statistic is t exp , which we substitute into the confidence intervalfor m (equation 4.12).µ= X ±Rearranging this equation and solving for t exptexpsn4.14texp =µ − X n4.15sgives the value of t exp when m is at either the right edge or the left edge ofthe sample’s confidence interval (Figure 4.14a).To determine if we should retain or reject the null hypothesis, we comparethe value of t exp to a critical value, t(a,n), where a is the confidencelevel and n is the degrees of freedom for the sample. The critical valuet(a,n) defines the largest confidence interval resulting from indeterminateerrors. If t exp > t(a,n), then our sample’s confidence interval is too large tobe explained by indeterminate errors (Figure 4.14b). In this case, we rejectthe null hypothesis and accept the alternative hypothesis. If t exp ≤ t(a,n),then the confidence interval for our sample can be explained indeterminateerror, and we retain the null hypothesis (Figure 4.14c).Example 4.16 provides a typical application of this significance test,which is known as a t-test of X to m.Values for t(a,n) are in Appendix 4.Another name for the t-test is Student’st-test. Student was the pen name for WilliamGossett (1876-1927) who developedthe t-test while working as a statisticianfor the Guiness Brewery in Dublin, Ireland.He published under the name Studentbecause the brewery did not wantits competitors to know they were usingstatistics to help improve the quality oftheir products.(a) (b) (c)Xt− expsnXt+ expsnXt− expsnt s t sX + exp X − expXnnt+ expsnt sX − ( αν , )nt sX + ( αν , )nt sX − ( αν , )nt sX + ( αν , )nFigure 4.14 Relationship between confidence intervals and the result of a significance test. (a) The shaded areaunder the normal distribution curve shows the confidence interval for the sample based on t exp . Based on thesample, we expect m to fall within the shaded area. The solid bars in (b) and (c) show the confidence intervals form that can be explained by indeterminate error given the choice of a and the available degrees of freedom, n. For(b) we must reject the null hypothesis because there are portions of the sample’s confidence interval that lie outsidethe confidence interval due to indeterminate error. In the case of (c) we retain the null hypothesis because theconfidence interval due to indeterminate error completely encompasses the sample’s confidence interval.

104 Analytical Chemistry 2.0Example 4.16Before determining the amount of Na 2 CO 3 in a sample, you decide tocheck your procedure by analyzing a sample known to be 98.76% w/wNa 2 CO 3 . Five replicate determinations of the %w/w Na 2 CO 3 in the standardgave the following results.98.71% 98.59% 98.62% 98.44% 98.58%Using a = 0.05, is there any evidence that the analysis is giving inaccurateresults?So l u t i o nThe mean and standard deviation for the five trials areX = 98.59 s = 0.0973Because there is no reason to believe that the results for the standard mustbe larger (or smaller) than m, a two-tailed t-test is appropriate. The nullhypothesis and alternative hypothesis areThe test statistic, t exp , isH : X =µ H : X ≠µ0 AThere is another way to interpret the resultof this t-test. Knowing that t exp is 3.91and that there are 4 degrees of freedom,we use Appendix 4 to estimate the a valuecorresponding to a t(a,4) of 3.91. FromAppendix 4, t(0.02,4) is 3.75 and t(0.01,4) is 4.60. Although we can reject the nullhypothesis at the 98% confidence level,we cannot reject it at the 99% confidencelevel.For a discussion of the advantages of thisapproach, see J. A. C. Sterne and G. D.Smith “Sifting the evidence—what’swrong with significance tests?” BMJ 2001,322, 226–231.texpu X n= − =s98. 76−98.59 5= 391 .0.0973The critical value for t(0.05,4) from Appendix 4 is 2.78. Since t exp is greaterthan t(0.05,4) we reject the null hypothesis and accept the alternative hypothesis.At the 95% confidence level the difference between X and m istoo large to be explained by indeterminate sources of error, suggesting thata determinate source of error is affecting the analysis.Practice Exercise 4.8To evaluate the accuracy of a new analytical method, an analyst determinesthe purity of a standard for which m is 100.0%, obtaining thefollowing results.99.28% 103.93% 99.43% 99.84% 97.60% 96.70% 98.02%Is there any evidence at a = 0.05 that there is a determinate error affectingthe results?Click here to review your answer to this exercise.Earlier we made the point that you need to exercise caution when interpretingthe results of a statistical analysis. We will keep returning to thispoint because it is an important one. Having determined that a result is

Chapter 4 Evaluating Analytical Data105inaccurate, as we did in Example 4.16, the next step is to identify and tocorrect the error. Before expending time and money on this, however, youshould first critically examine your data. For example, the smaller the valueof s, the larger the value of t exp . If the standard deviation for your analysis isunrealistically small, the probability of a type 2 error increases. Including afew additional replicate analyses of the standard and reevaluating the t-testmay strengthen your evidence for a determinate error, or it may show thatthere is no evidence for a determinate error.4F.2 Comparing s 2 to σ 2If we regularly analyze a particular sample, we may be able to establish theexpected variance, s 2 , for the analysis. This often is the case, for example,in clinical labs that routinely analyze hundreds of blood samples each day.A few replicate analyses of a single sample gives a sample variance, s 2 , whosevalue may or may not differ significantly from s 2 .We can use an F-test to evaluate whether a difference between s 2 ands 2 2 2is significant. The null hypothesis is H : s =σ and the alternative hypothesisis H : 2 20s ≠σ . The test statistic for evaluating the null hypothesisAis F exp , which is given as either2sF =exp 2σ2 2( s > σ )or2σF =exp 2s2 2( s < σ )4.16depending on whether s 2 is larger or smaller than s 2 . This way of definingF exp ensures that its value is always greater than or equal to one.If the null hypothesis is true, then F exp should equal one. Because ofindeterminate errors, however, F exp usually is greater than one. A criticalvalue, F(a, n num , n den ), gives the largest value of F exp that we can attributeto indeterminate error. It is chosen for a specified significance level, a, andfor the degrees of freedom for the variance in the numerator, n num , and thevariance in the denominator, n den . The degrees of freedom for s 2 is n – 1,where n is the number of replicates used to determine the sample’s variance,and the degrees of freedom for s 2 is ∞. Critical values of F for a = 0.05 arelisted in Appendix 5 for both one-tailed and two-tailed F-tests.Example 4.17A manufacturer’s process for analyzing aspirin tablets has a known varianceof 25. A sample of 10 aspirin tablets is selected and analyzed for theamount of aspirin, yielding the following results in mg aspirin/tablet.254 249 252 252 249 249 250 247 251 252Determine whether there is any evidence of a significant difference betweenthat the sample’s variance the expected variance at a=0.05.

106 Analytical Chemistry 2.0So l u t i o nThe variance for the sample of 10 tablets is 4.3. The null hypothesis andalternative hypotheses areH02 2: s =σ H s≠σA : 2 2The value for F exp isFexpσ 2 25= = = 58 .2s 43 .The critical value for F(0.05,∞,9) from Appendix 5 is 3.333. Since F expis greater than F(0.05,∞,9), we reject the null hypothesis and accept thealternative hypothesis that there is a significant difference between thesample’s variance and the expected variance. One explanation for the differencemight be that the aspirin tablets were not selected randomly.4F.3 Comparing Two Sample VariancesWe can extend the F-test to compare the variances for two samples, A andB, by rewriting equation 4.16 asFss2= Aexp 2Bdefining A and B so that the value of F exp is greater than or equal to 1.Example 4.18Table 4.11 shows results for two experiments to determine the mass ofa circulating U.S. penny. Determine whether there is a difference in theprecisions of these analyses at a = 0.05.So l u t i o nThe variances for the two experiments are 0.00259 for the first experiment(A) and 0.00138 for the second experiment (B). The null and alternativehypotheses areand the value of F exp is2 2H s = s H : s ≠ s:0 A BFexp2 2A A B2sA0.00259= = = 188 .2s 0.00138BFrom Appendix 5, the critical value for F(0.05,6,4) is 9.197. BecauseF exp < F(0.05,6,4), we retain the null hypothesis. There is no evidence ata = 0.05 to suggest that the difference in precisions is significant.

Chapter 4 Evaluating Analytical Data107Practice Exercise 4.9To compare two production lots of aspirin tablets, you collect samplesfrom each and analyze them, obtaining the following results (in mg aspirin/tablet).Lot 1: 256 248 245 245 244 248 261Lot 2: 241 258 241 244 256 254Is there any evidence at a = 0.05 that there is a significant difference inthe variance between the results for these two samples?Click here to review your answer to this exercise.4F.4 Comparing Two Sample MeansThree factors influence the result of an analysis: the method, the sample,and the analyst. We can study the influence of these factors by conductingexperiments in which we change one of the factors while holding the othersconstant. For example, to compare two analytical methods we can havethe same analyst apply each method to the same sample, and then examinethe resulting means. In a similar fashion, we can design experiments tocompare analysts or to compare samples.Before we consider the significance tests for comparing the means oftwo samples, we need to make a distinction between unpaired data andpaired data. This is a critical distinction and learning to distinguish betweenthe two types of data is important. Here are two simple examples thathighlight the difference between unpaired data and paired data. In eachexample the goal is to compare two balances by weighing pennies.• Example 1: Collect 10 pennies and weigh each penny on each balance.This is an example of paired data because we use the same 10 penniesto evaluate each balance.• Example 2: Collect 10 pennies and divide them into two groups offive pennies each. Weigh the pennies in the first group on one balanceand weigh the second group of pennies on the other balance.Note that no penny is weighed on both balances. This is an exampleof unpaired data because we evaluate each balance using a differentsample of pennies.In both examples the samples of pennies are from the same population. Thedifference is how we sample the population. We will learn why this distinctionis important when we review the significance test for paired data; first,however, we present the significance test for unpaired data.It also is possible to design experimentsin which we vary more than one of thesefactors. We will return to this point inChapter 14.One simple test for determining whetherdata are paired or unpaired is to look atthe size of each sample. If the samplesare of different size, then the data mustbe unpaired. The converse is not true. Iftwo samples are of equal size, they may bepaired or unpaired.Un p a i r e d Da t aConsider two analyses, A and B with means of X Aand X B, and standarddeviations of s A and s B . The confidence intervals for m A and for m B are

108 Analytical Chemistry 2.0tsAµ A= X A±nA4.17tsBµ B= X B±nB4.18where n A and n B are the sample sizes for A and B. Our null hypothesis,H : µ = µ0 A B, is that and any difference between m A and m B is the resultof indeterminate errors affecting the analyses. The alternative hypothesis,H : µ ≠ µA A B, is that the difference between m A and m B means is too largeto be explained by indeterminate error.To derive an equation for t exp , we assume that m A equals m B , and combineequations 4.17 and 4.18.XAt s t sexp Aexp± = X ±BnnABBProblem 9 asks you to use a propagationof uncertainty to show that equation 4.19is correct.Solving for XA− X and using a propagation of uncertainty, givesBs sAX − X = t × +A B expn n2 2BAB4.19Finally, we solve for t expSo how do you determine if it is okay topool the variances? Use an F-test.texp =XA− XB2 2s s4.20A B+n nAand compare it to a critical value, t(a,n), where a is the probability of atype 1 error, and n is the degrees of freedom.Thus far our development of this t-test is similar to that for comparingX to m, and yet we do not have enough information to evaluate the t-test.Do you see the problem? With two independent sets of data it is unclearhow many degrees of freedom we have.Suppose that the variances s A2 and s B2 provide estimates of the same s 2 .In this case we can replace s A2 and s B2 with a pooled variance, (s pool ) 2 , thatprovides a better estimate for the variance. Thus, equation 4.20 becomestexp =spoolwhere s pool , the pooled standard deviation, isXAB− X X − XBA B nnA B= ×1 1 s n + npoolA B4.21+n nAB

Chapter 4 Evaluating Analytical Data109spool=( n − 1) s + ( n −1)s2 2A A B BnA+ n −2B4.22The denominator of equation 4.22 shows us that the degrees of freedomfor a pooled standard deviation is n A + n B – 2, which also is the degrees offreedom for the t-test.If s A 2 and s B 2 are significantly different we must calculate t exp usingequation 4.20. In this case, we find the degrees of freedom using the followingimposing equation.⎡s s ⎤A B+ν= ⎣⎢n nA B ⎦⎥− 222⎛ 2 ⎞ 2s s ⎞AB⎝⎜n ⎠⎟nABn + + ⎝⎜⎠⎟1 n + 1A⎛B2 2 24.23Since the degrees of freedom must be an integer, we round to the nearestinteger the value of n obtained using equation 4.23.Regardless of whether we calculate t exp using equation 4.20 or equation4.21, we reject the null hypothesis if t exp is greater than t(a,n), and retainedthe null hypothesis if t exp is less than or equal to t(a,n). 4.Example 4.19Tables 4.11 provides results for two experiments to determine the mass ofa circulating U.S. penny. Determine whether there is a difference in themeans of these analyses at a=0.05.So l u t i o nFirst we must determine whether we can pool the variances. This is doneusing an F-test. We did this analysis in Example 4.18, finding no evidenceof a significant difference. The pooled standard deviation iss pool=( 7− 1)( 0. 00259) + ( 5−1)( 0. 00138)7+ 5−22 2=0.0459with 10 degrees of freedom. To compare the means the null hypothesis andalternative hypotheses areH : µ = µ H : µ ≠ µ0 A BA A BBecause we are using the pooled standard deviation, we calculate t exp usingequation 4.21.

110 Analytical Chemistry 2.03. 117 −3.081 7t exp=× × 5.0.0459 7+ 5= 134The critical value for t(0.05,10), from Appendix 4, is 2.23. Because t exp isless than t(0.05,10) we retain the null hypothesis. For a = 0.05 there is noevidence that the two sets of pennies are significantly different.Example 4.20One method for determining the %w/w Na 2 CO 3 in soda ash is an acid–base titration. When two analysts analyze the same sample of soda ash theyobtain the results shown here.Determine whether the difference in the mean values is significant ata=0.05.So l u t i o nAnalyst AAnalyst B86.82 81.0187.04 86.1586.93 81.7387.01 83.1986.20 80.2787.00 83.94We begin by summarizing the mean and standard deviation for each analyst.X = 86. 83% X = 82. 71%ABs = 032 . s = 2.16ATo determine whether we can use a pooled standard deviation, we firstcomplete an F-test of the following null and alternative hypotheses.2 2H s = s H : s ≠ s:0 A BCalculating F exp , we obtain a value ofB2 2A A B2( 216 . )F exp= = 45.62( 032 . )Because F exp is larger than the critical value of 7.15 for F(0.05,5,5) fromAppendix 5, we reject the null hypothesis and accept the alternative hypothesisthat there is a significant difference between the variances. As aresult, we cannot calculate a pooled standard deviation.

Chapter 4 Evaluating Analytical Data111To compare the means for the two analysts we use the following null andalternative hypotheses.H : µ = µ H : µ ≠ µ0 A B A A BBecause we cannot pool the standard deviations, we calculate t exp usingequation 4.20 instead of equation 4.2186. 83−82.71t exp=( 032 . ) ( 216 . )+6 62 2= 462 .and calculate the degrees of freedom using equation 4.23.⎡( 032 . ) ( 216 . ) ⎤+⎢⎥ν= ⎣ 6 6 ⎦2⎛ 2 ⎞( 032 . ) ⎛ 2 ( 216 . ) ⎞2⎝⎜6 ⎠⎟⎝⎜6 ⎠⎟+6+1 6+12 2 2− 2= 5.3≈5From Appendix 4, the critical value for t(0.05,5) is 2.57. Because t exp isgreater than t(0.05,5) we reject the null hypothesis and accept the alternativehypothesis that the means for the two analysts are significantly differentat a = 0.05.Practice Exercise 4.10To compare two production lots of aspirin tablets, you collect samplesfrom each and analyze them, obtaining the following results (in mg aspirin/tablet).Lot 1: 256 248 245 245 244 248 261Lot 2: 241 258 241 244 256 254Is there any evidence at a = 0.05 that there is a significant difference inthe variance between the results for these two samples? This is the samedata from Practice Exercise 4.9.Click here to review your answer to this exercise.Pa i r e d Da t aSuppose we are evaluating a new method for monitoring blood glucoseconcentrations in patients. An important part of evaluating a new methodis to compare it to an established method. What is the best way to gatherdata for this study? Because the variation in the blood glucose levelsamongst patients is large we may be unable to detect a small, but significantTypical blood glucose levels for mostnon-diabetic individuals ranges between80–120 mg/dL (4.4–6.7 mM), rising to ashigh as 140 mg/dL (7.8 mM) shortly aftereating. Higher levels are common for individualswho are pre-diabetic or diabetic.

112 Analytical Chemistry 2.0difference between the methods if we use different patients to gather datafor each method. Using paired data, in which the we analyze each patient’sblood using both methods, prevents a large variance within a populationfrom adversely affecting a t-test of means.When using paired data we first calculate the difference, d i , betweenthe paired values for each sample. Using these difference values, we thencalculate the average difference, d , and the standard deviation of the differences,s d . The null hypothesis, H : d = 0 , is that there is no difference0between the two samples, and the alternative hypothesis, H : d ≠ 0 , isAthat the difference between the two samples is significant.The test statistic, t exp , is derived from a confidence interval around dtexp =ds dnwhere n is the number of paired samples. As is true for other forms of thet-test, we compare t exp to t(a,n), where the degrees of freedom, n, is n – 1.If t exp is greater than t(a,n), then we reject the null hypothesis and acceptthe alternative hypothesis. We retain the null hypothesis if t exp is less thanor equal to t(a,n). This is known as a paired t-test.Example 4.21Marecek et. al. developed a new electrochemical method for rapidly determiningthe concentration of the antibiotic monensin in fermentation vats. 7The standard method for the analysis, a test for microbiological activity,is both difficult and time consuming. Samples were collected from thefermentation vats at various times during production and analyzed for theconcentration of monensin using both methods. The results, in parts perthousand (ppt), are reported in the following table.Sample Microbiological Electrochemical1 129.5 132.32 89.6 91.03 76.6 73.64 52.2 58.25 110.8 104.26 50.4 49.97 72.4 82.18 141.4 154.19 75.0 73.410 34.1 38.111 60.3 60.17 Marecek, V.; Janchenova, H.; Brezina, M.; Betti, M. Anal. Chim. Acta 1991, 244, 15–19.

Chapter 4 Evaluating Analytical Data113Is there a significant difference between the methods at a = 0.05?So l u t i o nAcquiring samples over an extended period of time introduces a substantialtime-dependent change in the concentration of monensin. Because thevariation in concentration between samples is so large, we use paired t-testwith the following null and alternative hypotheses.H : d = 0 H : d ≠ 00 ADefining the difference between the methods asd X X= ( ) −( )i elect i micro iwe can calculate the following differences for the samples.Sample 1 2 3 4 5 6 7 8 9 10 11d i 2.8 1.4 -3.0 6.0 -6.6 -0.5 9.7 12.7 -1.6 4.0 -0.2The mean and standard deviation for the differences are, respectively, 2.25and 5.63. The value of t exp is225 . 11t exp= = 133 .563 .which is smaller than the critical value of 2.23 for t(0.05,10) from Appendix4. We retain the null hypothesis and find no evidence for a significantdifference in the methods at a = 0.05.Practice Exercise 4.11Suppose you are studying the distribution of zinc in alake and want to know if there is a significant differencebetween the concentration of Zn 2+ at the sedimentwaterinterface and its concentration at the air-waterinterface. You collect samples from six locations—nearthe lake’s center, near its drainage outlet, etc.—obtainingthe results (in mg/L) shown in the table. Using thedata in the table shown to the right, determine if thereis a significant difference between the concentration ofZn 2+ at the two interfaces at a = 0.05.Complete this analysis treating the data as (a) unpaired, and (b) paired. Briefly comment on your results.Click here to review your answers to this exercise.One important requirement for a paired t-test is that determinate andindeterminate errors affecting the analysis must be independent of the ana-LocationAir-Water Sediment-WaterInterface Interface1 0.430 0.4152 0.266 0.2383 0.457 0.3904 0.531 0.4105 0.707 0.6056 0.716 0.609

114 Analytical Chemistry 2.0lyte’s concentration. If this is not the case, then a sample with an unusuallyhigh concentration of analyte will have an unusually large d i . Includingthis sample in the calculation of d and s d leads to a biased estimate ofthe expected mean and standard deviation. This is rarely a problem forsamples spanning a limited range of analyte concentrations, such as thosein Example 4.21 or Practice Exercise 4.11. When paired data span a widerange of concentrations, however, the magnitude of the determinate andindeterminate sources of error may not be independent of the analyte’s concentration.In such cases a paired t‐test may give misleading results since thepaired data with the largest absolute determinate and indeterminate errorswill dominate d . In this situation a regression analysis, which is the subjectof the next chapter, is more appropriate method for comparing the data.4F.5 OutliersEarlier in the chapter we examined several data sets consisting of the massof a circulating United States penny. Table 4.16 provides one more data set.Do you notice anything unusual in this data? Of the 112 pennies includedin Table 4.11 and Table 4.13, no penny weighed less than 3 g. In Table 4.16,however, the mass of one penny is less than 3 g. We might ask whether thispenny’s mass is so different from the other pennies that it is in error.Data that are not consistent with the remaining data are called outliers.An outlier might exist for many reasons: the outlier might be froma different population (Is this a Canadian penny?); the outlier might be acontaminated or otherwise altered sample (Is the penny damaged?); or theoutlier may result from an error in the analysis (Did we forget to tare thebalance?). Regardless of its source, the presence of an outlier compromisesany meaningful analysis of our data. There are many significance tests foridentifying potential outliers, three of which we present here.Di x o n’s Q-Te s tOne of the most common significance tests for outliers is Dixon’s Q‐test.The null hypothesis is that there are no outliers, and the alternative hypothesisis that there is an outlier. The Q-test compares the gap betweenthe suspected outlier and its nearest numerical neighbor to the range of theentire data set (Figure 4.15). The test statistic, Q exp , isgapQ exp= =rangeoutliersvalue ' − nearest valuelargest value−smallest valueTable 4.16 Mass (g) for Additional Sample of Circulating U. S. Pennies3.067 2.514 3.0943.049 3.048 3.1093.039 3.079 3.102

Chapter 4 Evaluating Analytical Data115d(a)Gap(b)RangeGapRangeFigure 4.15 Dotplots showing thedistribution of two data sets containinga possible outlier. In (a)the possible outlier’s value is largerthan the remaining data, and in(b) the possible outlier’s value issmaller than the remaining data.This equation is appropriate for evaluating a single outlier. Other forms ofDixon’s Q-test allow its extension to detecting multiple outliers. 8The value of Q exp is compared to a critical value, Q(a, n), where a isthe probability of rejecting a valid data point (a type 1 error) and n is thetotal number of data points. To protect against rejecting a valid data point,we usually apply the more conservative two-tailed Q-test, even though thepossible outlier is the smallest or the largest value in the data set. If Q exp isgreater than Q(a, n), then we reject the null hypothesis and may excludethe outlier. We retain the possible outlier when Q exp is less than or equalto Q(a, n). Table 4.17 provides values for Q(0.05, n) for a sample containing3–10 values. A more extensive table is in Appendix 6. Values for Q(a,n) assume an underlying normal distribution.Table 4.17 Dixon’s Q-Testn Q(0.05, n)3 0.9704 0.8295 0.7106 0.6257 0.5688 0.5269 0.49310 0.466Gr u b b’s Te s tAlthough Dixon’s Q-test is a common method for evaluating outliers, it is nolonger favored by the International Standards Organization (ISO), whichnow recommends Grubb’s test. 9 There are several versions of Grubb’s testdepending on the number of potential outliers. Here we will consider thecase where there is a single suspected outlier.The test statistic for Grubb’s test, G exp , is the distance between thesample’s mean, X , and the potential outlier, X out , in terms of the sample’sstandard deviation, s.Gexp =Xout− XsWe compare the value of G exp to a critical value G(a,n), where a is the probabilityof rejecting a valid data point and n is the number of data points inthe sample. If G exp is greater than G(a,n), we may reject the data point as an8 Rorabacher, D. B. Anal. Chem. 1991, 63, 139–146.9 International Standards ISO Guide 5752-2 “Accuracy (trueness and precision) of measurementmethods and results–Part 2: basic methods for the determination of repeatability and reproducibilityof a standard measurement method,” 1994.

116 Analytical Chemistry 2.0Table 4.18 Grubb’s Testn G(0.05, n)3 1.1154 1.4815 1.7156 1.8877 2.0208 2.1269 2.21510 2.290outlier, otherwise we retain the data point as part of the sample. Table 4.18provides values for G(0.05, n) for a sample containing 3–10 values. A moreextensive table is in Appendix 7. Values for G(a, n) assume an underlyingnormal distribution.Ch a u ve n e t’s Cr i t e r i o nOur final method for identifying outliers is Chauvenet’s criterion. UnlikeDixon’s Q-Test and Grubb’s test, you can apply this method to anydistribution as long as you know how to calculate the probability for aparticular outcome. Chauvenet’s criterion states that we can reject a datapoint if the probability of obtaining the data point’s value is less than (2n) –1 ,where n is the size of the sample. For example, if n = 10, a result with aprobability of less than (2×10) –1 , or 0.05, is considered an outlier.To calculate a potential outlier’s probability we first calculate its standardizeddeviation, zz =Xout− Xswhere X out is the potential outlier, X is the sample’s mean and s is thesample’s standard deviation. Note that this equation is identical to the equationfor G exp in the Grubb’s test. For a normal distribution, you can findthe probability of obtaining a value of z using the probability table in Appendix3.Example 4.22Table 4.16 contains the masses for nine circulating United States pennies.One of the values, 2.514 g, appears to be an outlier. Determine if thispenny is an outlier using the Q-test, Grubb’s test, and Chauvenet’s criterion.For the Q-test and Grubb’s test, let a = 0.05.So l u t i o nFor the Q-test the value for Q exp is2. 514−3.039Q exp=3. 109−2.514= 0.882From Table 4.17, the critical value for Q(0.05,9) is 0.493. Because Q expis greater than Q(0.05,9), we can assume that penny weighing 2.514 g isan outlier.For Grubb’s test we first need the mean and the standard deviation, whichare 3.011 g and 0.188 g, respectively. The value for G exp is2. 514−3.011G exp=0.188= 264 .

Chapter 4 Evaluating Analytical Data117Using Table 4.18, we find that the critical value for G(0.05,9) is 2.215. BecauseG exp is greater than G(0.05,9), we can assume that the penny weighing2.514 g is an outlier.For Chauvenet’s criterion, the critical probability is (2×9) –1 , or 0.0556.The value of z is the same as G exp , or 2.64. Using Appendix 3, the probabilityfor z = 2.64 is 0.00415. Because the probability of obtaining a mass of0.2514 g is less than the critical probability, we can assume that the pennyweighing 2.514 g is an outlier.You should exercise caution when using a significance test for outliersbecause there is a chance you will reject a valid result. In addition, youshould avoid rejecting an outlier if it leads to a precision that is unreasonablybetter than that expected based on a propagation of uncertainty. Giventhese two concerns it is not surprising that some statisticians caution againstthe removal of outliers. 10On the other hand, testing for outliers can provide useful information ifyou try to understand the source of the suspected outlier. For example, theoutlier in Table 4.16 represents a significant change in the mass of a penny(an approximately 17% decrease in mass), which is the result of a changein the composition of the U.S. penny. In 1982 the composition of a U.S.penny was changed from a brass alloy consisting of 95% w/w Cu and 5%w/w Zn, to a zinc core covered with copper. 11 The pennies in Table 4.16,therefore, were drawn from different populations.4GDetection LimitsThe International Union of Pure and Applied Chemistry (IUPAC) definesa method’s detection limit as the smallest concentration or absoluteamount of analyte that has a signal significantly larger than the signal froma suitable blank. 12 Although our interest is in the amount of analyte, in thissection we will define the detection limit in terms of the analyte’s signal.Knowing the signal you can calculate the analyte’s concentration, C A , orthe moles of analyte, n A , using the equationsS A = k A C A or S A = k A n Awhere k is the method’s sensitivity.Let’s translate the IUPAC definition of the detection limit into a mathematicalform by letting S mb represent the average signal for a method blank,and letting s mb represent the method blank’s standard deviation. The nullhypothesis is that the analyte is not present in the sample, and the alterna-You also can adopt a more stringent requirementfor rejecting data. When usingthe Grubb’s test, for example, the ISO5752 guidelines suggest retaining a valueif the probability for rejecting it is greaterthan a = 0.05, and flagging a value as a“straggler” if the probability for rejectingit is between a = 0.05 and 0.01. A “straggler”is retained unless there is compellingreason for its rejection. The guidelines recommendusing a = 0.01 as the minimumcriterion for rejecting a data point.See Chapter 3 for a review of these equations.10 Deming, W. E. Statistical Analysis of Data; Wiley: New York, 1943 (republished by Dover: NewYork, 1961); p. 171.11 Richardson, T. H. J. Chem. Educ. 1991, 68, 310–311.12 IUPAC Compendium of Chemical Technology, Electronic Version, http://goldbook.iupac.org/D01629.html

118 Analytical Chemistry 2.0dIf s mb is not known, we can replace itwith s mb ; equation 4.24 then becomes( S )tsmb= S +nA DL mbYou can make similar adjustments to otherequations in this section.See, for example, Kirchner, C. J. “Estimationof Detection Limits for EnvironmentalAnalytical Procedures,” in Currie, L.A. (ed) Detection in Analytical Chemistry:Importance, Theory, and Practice; AmericanChemical Society: Washington, D.C., 1988.tive hypothesis is that the analyte is present in the sample. To detect theanalyte, its signal must exceed S mb by a suitable amount; thus,( S ) = S + zσ 4.24A DL mb mbwhere (S A ) DL is the analyte’s detection limit .The value we choose for z depends on our tolerance for reporting theanalyte’s concentration even though it is absent from the sample (a type1 error). Typically, z is set to three, which, from Appendix 3, correspondsto a probability, a, of 0.00135. As shown in Figure 4.16a, there is only a0.135% probability of detecting the analyte in a sample that actually isanalyte-free.A detection limit also is subject to a type 2 error in which we fail to findevidence for the analyte even though it is present in the sample. Consider,for example, the situation shown in Figure 4.16b where the signal for asample containing the analyte is exactly equal to (S A ) DL . In this case theprobability of a type 2 error is 50% because half of the signals arising fromsuch samples are below the detection limit. We will correctly detect the analyteat the IUPAC detection limit only half the time. The IUPAC definitionfor the detection limit indicates the smallest signal for which we can say, ata significance level of a, that an analyte is present in the sample. Failing todetect the analyte does not imply that it is not present in the sample.The detection limit is often represented, particularly when discussingpublic policy issues, as a distinct line separating detectable concentrationsType 2 Error = 50.0%Type 1 Error = 0.135%Type 1 Error = 0.135%(a)S mb(S A) DL= S mb+ 3σ mb(b)S mb(S A) DL= S mb+ 3σ mbFigure 4.16 Normal distribution curves showing the probability of type 1 and type 2 errors for the IUPACdetection limit. (a) The normal distribution curve for the method blank, with S mb = 0 and s mb = 1. Theminimum detectable signal for the analyte, (S A ) DL , has a type 1 error of 0.135%. (b) The normal distributioncurve for the analyte at its detection limit, (S A ) DL = 3, is superimposed on the normal distribution curvefor the method blank. The standard deviation for the analyte’s signal, s A , is 0.8, The area in green representsthe probability of a type 2 error, which is 50%. The inset shows, in blue, the probability of a type 1 error,which is 0.135%.

Chapter 4 Evaluating Analytical Data119S mb(SA ) DLof analytes that concentrations that cannot be detected. This use of a detectionlimit is incorrect. 13 As suggested by Figure 4.16, for concentrationsof analyte near the detection limit there is a high probability of failing todetect the analyte.An alternative expression for the detection limit, the limit of identification,minimizes both type 1 and type 2 errors. 14 The analyte’s signalat the limit of identification, (S A ) LOI , includes an additional term, zs A , toaccount for the distribution of the analyte’s signal.( S ) = ( S ) + zσ = S + zσ + zσA LOI A DL A mb mb AAs shown in Figure 4.17, the limit of identification provides an equal probabilityfor type 1 and type 2 errors at the detection limit. When the analyte’sconcentration is at its limit of identification, there is only a 0.135%probability that its signal will be indistinguishable from that of the methodblank.The ability to detect the analyte with confidence is not the same asthe ability to report with confidence its concentration, or to distinguishbetween its concentration in two samples. For this reason the AmericanChemical Society’s Committee on Environmental Analytical Chemistryrecommends the limit of quantitation, (S A ) LOQ . 15( S )(S A) LOI= S +10σA LOQ mb mbFigure 4.17 Normal distribution curves for a method blank and for asample at the limit of identification: S mb = 0; s mb = 1; s A = 0.8; and(S A ) LOI = 0 + 3 × 1 + 3 × 0.8 = 5.4. The inset shows that the probabilityof a type 1 error (0.135%) is the same as the probability of atype 2 error (0.135%).13 Rogers, L. B. J. Chem. Educ. 1986, 63, 3–6.14 Long, G. L.; Winefordner, J. D. Anal. Chem. 1983, 55, 712A–724A.15 “Guidelines for Data Acquisition and Data Quality Evaluation in Environmental Chemistry,”Anal. Chem. 1980, 52, 2242–2249.

120 Analytical Chemistry 2.0Once you install the Analysis ToolPak, itwill continue to load each time you launchExcel.4HUsing Excel and R to Analyze DataAlthough the calculations in this chapter are relatively straightforward, itcan be tedious to work problems using nothing more than a calculator.Both Excel and R include functions for descriptive statistics, for findingprobabilities for different distributions, and for carrying out significancetests. In addition, R provides useful functions for visualizing your data.4H.1 ExcelExcel provides two methods for working with data: built-in functions forindividual statistical parameters and a package of data analysis tools in theAnalysis ToolPak. The ToolPak is not a standard part of Excel’s instillation.To see if you have access to the Analysis ToolPak on your computer, selectTools from the menu bar and look for the Data Analysis... option. If youdo not see Data Analysis..., select Add-ins... from the Tools menu. Checkthe box for the Analysis ToolPak and click on OK to install them.Descriptive St a t i s t i c sA B1 mass (g)2 3.0803 3.0944 3.1075 3.0566 3.1127 3.1748 3.198Figure 4.18 Portion of a spreadsheetcontaining data from Table4.1.Let’s use Excel to provide a statistical summary of the data in Table 4.1.Enter the data into a spreadsheet, as shown in Figure 4.18. Select DataAnalysis... from the Tools menu, which opens a window entitled “DataAnalysis.” Scroll through the window, select Descriptive Statistics from theavailable options, and click OK. Place the cursor in the box for the “InputRange” and then click and drag over the cells B1:B8. Check the box for“Labels in the first row.” Select the radio button for “Output range,” place thecursor in the box and click on an empty cell; this is where Excel will placethe results. Check the boxes for “Summary statistics” and for the “Confidencelevel for the mean.” Accept the default value of 95% for the confidence level.Clicking OK generates the information shown in Figure 4.19.mass (g)Figure 4.19 Output from Excel’s Descriptive Statistics command inthe Analysis TookPak. Note that Excel does not adjust for significantfigures. The mode is the most common result, which is not relevanthere. Kurtosis is a measure of the “peakedness” of the data’s distribution,and is zero for a normal distribution. Skewness is a measureof the symmetry of the data’s distribution and is zero for a normaldistribution. For a small sample size—such as the seven samples inthis data set—skewness and kurtosis are not particularly useful. Youmay consult the textbooks listed in the Additional Resources formore information about kurtosis and skewness.Mean 3.11728571Standard Error 0.01924369Median 3.107Mode#N/AStandard Deviation 0.05091403Sample Variance 0.00259224Kurtosis -0.59879248Skewness 0.72905145Range 0.142Minimum 3.056Maximum 3.198Sum 21.821Count 7Confidence Level(95.0%) 0.04708762

Chapter 4 Evaluating Analytical Data121Table 4.19 Excel Functions for Descriptive StatisticsParameterExcel Functionmean=average(number1,number2,...)median=median(number1,number2,...)sample standard deviation =stdev(number1,number2,...)population standard deviation =stdevp(number1,number2,...)sample variance=var(number1,number2,...)population variance=varp(number1,number2,...)range=max((number1,number2,...) – min(number1,number2,...)The Descriptive Statistics command provides a table of values for asample. If your interest is in a statistical summary for a population, or ifyou do not want a summary of all descriptive statistics, you will need to useExcel’s built-in functions (Table 4.19). To use a function, click on an emptycell, enter the formula, and press Return or Enter. To find, for example,the population variance for the data in Figure 4.18, place the cursor in anempty cell and enter the formula=varp(b2:b8)The contents of the cell are replaced with Excel’s exact calculation of thepopulation’s variance (0.002 221 918).Pr o b a b i l i t y DistributionsIn Example 4.11 we showed that 91.10% of a manufacturer’s analgesictablets contain between 243 and 262 mg of aspirin. We obtained this resultby calculating the deviation, z, of each limit from the population’s expectedmean, m, of 250 mg in terms of the population’s expected standarddeviation, s, of 5 mg. After calculating values for z, we used the table inAppendix 3 to find the area under the normal distribution curve betweenthe two limits.We can complete this calculation in Excel using the built-in normdistfunction. The function’s general format is=normdist(x, m, s, TRUE)where x is the limit of interest. The function returns the probability of obtaininga result of less than x from a normal distribution with a mean of mand a standard deviation of s (Figure 4.20). To solve Example 4.11 usingExcel enter the following formulas into separate cells=normdist(243, 250, 5, TRUE)Figure 4.20 Shown in blue is thearea returned by the function=normdist(x, m, s,TRUE)x=normdist(262, 250, 5, TRUE)

122 Analytical Chemistry 2.0obtaining results of 0.080 756 659 and 0.991 802 464. Subtract the smallervalue from the larger value and adjust to the correct number of significantfigures to arrive at a probability of 0.9910, or 99.10%.Excel also includes a function for binomial distributions. The function’sformat is=binomdist(X, N, p, T/F)where X is the number of times a particular outcome occurs in N trials, andp is the probability of X occurring in one trial. Setting the function’s lastterm to TRUE gives the total probability for any result up to X and settingit to FALSE to give the probability for X. Using Example 4.10 to test thisfunction, we find that the probability of finding no atoms of 13 C atoms ina molecule of cholesterol, C 27 H 44 O) using the formula=binomdist(0, 27, 0.0111, FALSE)which returns a value of 0.740 as an answer, after adjusting the significantfigures. Using the formula=binomdist(2, 27, 0.0111, TRUE)we find that 99.7% of cholesterol molecules contain two or fewer atomsof 13 C.A B C1 Set 1 Set 22 3.080 3.0523 3.094 3.1414 3.107 3.0835 3.056 3.0836 3.112 3.0487 3.1748 3.198Figure 4.21 Portion of a spreadsheet containingthe data in Table 4.11.Significance Te s t sExcel’s Analysis ToolPak includes tools to help you complete the followingsignificance tests covered in this chapter:• an F-test of variances• an unpaired t-test of sample means assuming equal variances• an unpaired t-test of sample means assuming unequal variances• a paired t-test for of sample meansLet’s use the ToolPak to complete a t-test on the data in Table 4.11, whichcontains results for two experiments to determine the mass of a circulatingU. S. penny. Enter the data from Table 4.11 into a spreadsheet as shownin Figure 4.21. Because the data in this case are unpaired, we will useExcel to complete an unpaired t-test. Before we can complete a t-test wemust use an F-test to determine whether the variances for the two data setsare equal or unequal. Our null hypothesis is that the variances are equal,2 2s = s , and our alternative hypothesis is that the variances are notSet1 Set22 2equal, s ≠ s .Set1 Set2To complete the F-test select Data Analysis... from the Tools menu,which opens a window entitled “Data Analysis.” Scroll through the window,select F-Test Two Sample Variance from the available options, and clickOK. Place the cursor in the box for the “Variable 1 range” and then clickand drag over the cells B1:B8. Next, place the cursor in the box for “Variable2 range” and then click and drag over the cells B1:B6. Check the box

Chapter 4 Evaluating Analytical Data123F-Test Two-Sample for VariancesData Set 1 Data Set 2Mean 3.11728571 3.0814Variance 0.00259224 0.0013843Observations 7 5df 6 4F 1.87259849P(F

124 Analytical Chemistry 2.0dt-Test: Two-Sample Assuming Equal VariancesData Set 1 Data Set 2Mean 3.11728571 3.0814Variance 0.00259224 0.0013843Observations 7 5Pooled Variance 0.00210906Hypothesized Mean Difference 0df 10t Stat 1.33450508P(T

Chapter 4 Evaluating Analytical Data125Table 4.20 R Functions for Descriptive StatisticsParameterR Functionmeanmean(object)medianmedian(object)sample standard deviation sd(object)population standard deviation sd(object)*((length(object)-1)/length(object))^0.5sample variancevar(object)population variancevar(object)*((length(object)-1)/length(object))rangemax(object)-min(object)Pr o b a b i l i t y DistributionsIn Example 4.11 we showed that 91.10% of a manufacturer’s analgesictables contain between 243 and 262 mg of aspirin. We obtained this resultby calculating the deviation, z, of each limit from the population’s expectedmean, m, of 250 mg in terms of the population’s expected standard deviation, s, of 5 mg. After calculating values for z, we used the table in Appendix3 to find the area under the normal distribution curve between thetwo limits.We can complete this calculation in R using the function pnorm. Thefunction’s general format ispnorm(x, m, s)where x is the limit of interest, m is the distribution’s expected mean and sis the distribution’s expected standard deviation. The function returns theprobability of obtaining a result of less than x (Figure 4.24). Here is theoutput of an R session for solving Example 4.11.> pnorm(243,250,5)[1] 0.08075666> pnorm(262,250,5)[1] 0.9918025Subtracting the smaller value from the larger value and adjusting to thecorrect number of significant figures gives the probability as 0.9910, or99.10%.R also includes functions for binomial distributions. To find the probabilityof obtaining a particular outcome, X, in N trials we use the dbinomfunction.dbinom(X, N, p)where p is the probability of X occurring in one trial. Using Example 4.10to test this function, we find that the probability of finding no atoms of13 C atoms in a molecule of cholesterol, C 27 H 44 O) isFigure 4.24 Shown in blue is thearea returned by the functionpnorm(x, m, s)x

126 Analytical Chemistry 2.0> dbinom(0,27,0.0111)[1] 0.73979970.740 after adjusting the significant figures. To find the probability of obtainingany outcome up to a maximum value of X, we use the pbinomfunction.pbinom(X, N, p)To find the percentage of cholesterol molecules containing 0, 1, or 2 atomsof 13 C, we enter> pbinom(2,27,0.0111)[1] 0.9967226and find that the answer is 99.7% of cholesterol molecules.Significance Te s t sFor a one-tailed F-test the command isone of the followingvar.test(X, Y, alternative = “greater”)var.test(X, Y, alternative = “less”)where “greater” is used when the alterna-2 2tive hypothesis is s > s , and “less” isX Yused when the alternative hypothesis iss< s .2 2XYR includes commands to help you complete the following significance testscovered in this chapter:• an F-test of variances• an unpaired t-test of sample means assuming equal variances• an unpaired t-test of sample means assuming unequal variances• a paired t-test for of sample means• Dixon’s Q-test for outliers• Grubb’s test for outliersLet’s use R to complete a t-test on the data in Table 4.11, which containsresults for two experiments to determine the mass of a circulating U. S.penny. To do this, enter the data from Table 4.11 into two objects.> penny1=c(3.080, 3.094, 3.107, 3.056, 3.112, 3.174, 3.198)> penny2=c(3.052, 3.141, 3.083, 3.083, 3.048)Because the data in this case are unpaired, we will use R to complete anunpaired t-test. Before we can complete a t-test we must use an F-test todetermine whether the variances for the two data sets are equal or unequal.2 2Our null hypothesis is that the variances are equal, s = s , and ourSet1 Set22 2alternative hypothesis is that the variances are not equal, s ≠ s .Set1 Set2The command for a two-tailed F-test in R, which is our choice for thisproblem, isvar.test(X, Y)where X and Y are the objects containing the data sets. Figure 4.25 showsthe output from an R session to solve this problem.R does not provide the critical value for F(0.05, 6, 4). Instead it reportsthe 95% confidence interval for F exp . Because this confidence interval of0.204 to 11.661 includes the expected value for F of 1.00, we retain thenull hypothesis and have no evidence for a difference between the variances.

Chapter 4 Evaluating Analytical Data127> var.test(penny1, penny2)F test to compare two variancesdata: penny1 and penny2F = 1.8726, num df = 6, denom df = 4, p-value = 0.5661alternative hypothesis: true ratio of variances is not equal to 195 percent confidence interval:0.2036028 11.6609726sample estimates:ratio of variances1.872598R calculates F exp as (s X ) 2 /(s Y ) 2 . If we usethe commandvar.test(penny2, penny1)the output will give R as 0.534 and the95% confidence interval as 0.0858 to4.912. Because the expected value forF exp of 1.00 falls within the confidenceinterval, we retain the null hypothesis ofequal variances.Figure 4.25 Output of an R session for an F-test of variances. The p-value of 0.5661 is the probability of incorrectlyrejecting the null hypothesis that the variances are equal (note: R identifies the value a as a p-value). The95% confidence interval is the range of values for F exp that can be explained by random error. If this range includesthe expected value for F, in this case 1.00, then there is insufficient evidence to reject the null hypothesis.Note that R does not adjust for significant figures.R also provides the probability of incorrectly rejecting the null hypothesis,which in this case is 0.5561.Having found no evidence suggesting unequal variances, we now completean unpaired t-test assuming equal variances. Our null hypothesis is thatthere is no difference between the means, X = X , and our alternativeSet1 Set2hypothesis is that there is a difference between the means, X ≠ X . InSet1 Set2R there is a single command for all two-sample t-tests. The basic syntax fora two-tailed unpaired t-test with unequal variances ist.test(X, Y, mu = 0, paired = FALSE, var.equal = FALSE)where X and Y are the objects containing the data sets. You can changethe underlined terms to alter the nature of the t-test. Replacing “var.equal= FALSE” to “var.equal = TRUE” makes this a two-tailed t-test with equalvariances, and replacing “paired = FALSE” with “paired = TRUE” makesthis a paired t-test. The term “mu = 0” is the expected difference betweenthe means, which for a null hypothesis of X = XSet1 Set2is 0. You can, ofcourse, change this to suit your needs. The underlined terms are defaultvalues; if you omit them, then R assumes that you intend an unpaired twotailedt-test of the null hypothesis that X = Y with unequal variances. Figure4.26 shows the output of an R session for this problem.The p-value of 0.2116 means that there is a 21.16% probability ofincorrectly rejecting the null hypothesis. The 95% confidence interval of-0.024 to 0.0958, which is for the difference between the sample means,includes the expected value of zero. Both ways of looking at the results ofthe t-test provide no evidence for rejecting the null hypothesis; thus, weretain the null hypothesis and find no evidence for a difference betweenthe two samples.To complete a one-sided t-test, includethe commandoralternative = “greater”alternative = “less”A one-sided paired t-test that the differencebetween two samples is greater than0 becomest.test(X, Y, paired = TRUE, alternative =“greater”)

128 Analytical Chemistry 2.0> t.test(penny1, penny2, var.equal=TRUE)Two Sample t-testdata: penny1 and penny2t = 1.3345, df = 10, p-value = 0.2116alternative hypothesis: true difference in means is not equal to 095 percent confidence interval:-0.02403040 0.09580182sample estimates:mean of x mean of y3.117286 3.081400Figure 4.26 Output of an R session for an unpaired t-test with equal variances. The p-value of 0.2116 is theprobability of incorrectly rejecting the null hypothesis that the means are equal (note: R identifies the value a asa p-value). The 95% confidence interval is the range of values for the difference between the means that can beexplained by random error. If this range includes the expected value for the difference, in this case zero, thenthere is insufficient evidence to reject the null hypothesis. Note that R does not adjust for significant figures.Practice Exercise 4.13Rework Example 4.20 and Example 4.21 using R.Click here to review your answers to this exercise.You need to install a package once, butyou need to load the package each timeyou plan to use it. There are ways to configureR so that it automatically loadscertain packages; see An Introduction to Rfor more information (click here to view aPDF version of this document).Unlike Excel, R also includes functions for evaluating outliers. Thesefunctions are not part of R’s standard installation. To install them enter thefollowing command within R (note: you will need an internet connection todownload the package of functions).> install.packages(“outliers”)After installing the package, you will need to load the functions into R usingthe following command (note: you will need to do this step each time you begina new R session as the package does not automatically load when you start R).> library(“outliers”)Let’s use this package to find the outlier in Table 4.16 using bothDixon’s Q-test and Grubb’s test. The commands for these tests aredixon.test(X, type = 10, two.sided = TRUE)grubbs.test(X, type = 10, two.sided = TRUE)where X is the object containing the data, “type = 10” specifies that weare looking for one outlier, and “two.sided=TRUE” indicates that we areusing the more conservative two-tailed test. Both tests have other variantsthat allow the testing of outliers on both ends of the data set (“type = 11”)or for more than one outlier (“type = 20”), but we will not consider these.Figure 4.27 shows the output of a session for this problem. For both teststhe very small p-value indicates that we can treat as an outlier the pennywith a mass of 2.514 g.

Chapter 4 Evaluating Analytical Data129> penny3=c(3.067,3.049, 3.039, 2.514, 3.048, 3.079, 3.094, 3.109, 3.102)> dixon.test(penny3, type=10, two.sided=TRUE)Dixon test for outliersdata: penny3Q = 0.8824, p-value < 2.2e-16alternative hypothesis: lowest value 2.514 is an outlier> grubbs.test(penny3, type=10, two.sided=TRUE)Grubbs test for one outlierdata: penny3G = 2.6430, U = 0.0177, p-value = 1.938e-06alternative hypothesis: lowest value 2.514 is an outlierFigure 4.27 Output of an R session for Dixon’s Q-test and Grubb’s test for outliers. The p-values for both testsshow that we can treat as an outlier the penny with a mass of 2.514 g.Visualizing Da t aOne of the more useful features of R is the ability to visualize your data.Visualizing your data is important because it provides you with an intuitivefeel for your data that can help you in applying and evaluating statisticaltests. It is tempting to believe that a statistical analysis is foolproof, particularlyif the probability for incorrectly rejecting the null hypothesis is small.Looking at a visual display of your data, however, can help you determinewhether your data is normally distributed—a requirement for most of thesignificance tests in this chapter—and can help you identify potential outliers.There are many useful ways to look at your data, four of which weconsider here.To plot data in R will use the package “lattice,” which you will need toload using the following command.> library(“lattice”)To demonstrate the types of plots we can generate, we will use the object“penny,” which contains the masses of the 100 pennies in Table 4.13.Our first display is a histogram. To construct the histogram we usemass to divide the pennies into bins and plot the number of pennies or thepercent of pennies in each bin on the y-axis as a function of mass on thex-axis. Figure 4.28a shows the result of entering> histogram(penny, type = “percent”, xlab = “Mass (g)”,ylab = “Percent of Pennies”, main = “Histogram of Data in Table4.13”)Visualizing data is important, a point wewill return to in Chapter 5 when we considerthe mathematical modeling of data.You do not need to use the command install.packagethis time because lattice wasautomatically installed on your computerwhen you downloaded R.You can download the file “Penny.Rdata”from the textbook’s web site.To create a histogram showing the numberof pennies in each bin, change “percent”to “count.”

130 Analytical Chemistry 2.0(a)Histogram of Data in Table 4.13(b)Kernel Density Plot of Data in Table 4.13123010Percent of Pennies2010Density8642003.00 3.05 3.10 3.15 3.20Mass of Pennies (g)2.95 3.00 3.05 3.10 3.15 3.20Mass of Pennies (g)(c)Dotchart of Data in Table 4.13(d)Boxplot of Data in Table 4.13Penny Number3.00 3.05 3.10 3.15 3.20Mass of Pennies (g)3.00 3.05 3.10 3.15 3.20Mass of Pennies (g)Figure 4.28 Four different ways to plot the data in Table 4.13: (a) histogram; (b) kernel density plotshowing smoothed distribution and individual data points; (c) dot chart; and (d) box plot.A histogram allows us to visualize the data’s distribution. In this examplethe data appear to follow a normal distribution, although the largestbin does not include the mean of 3.095 g and the distribution is notperfectly symmetric. One limitation of a histogram is that its appearancedepends on how we choose to bin the data. Increasing the number of binsand centering the bins around the data’s mean gives a histogram that moreclosely approximates a normal distribution (Figure 4.10).An alternative to the histogram is a kernel density plot, which isbasically a smoothed histogram. In this plot each value in the data set isreplaced with a normal distribution curve whose width is a function of thedata set’s standard deviation and size. The resulting curve is a summation

Chapter 4 Evaluating Analytical Data131of the individual distributions. Figure 4.28b shows the result of enteringthe command> densityplot(penny, xlab = “Mass of Pennies (g)”, main = “KernelDensity Plot of Data in Table 4.13”)The circles at the bottom of the plot show the mass of each penny in thedata set. This display provides a more convincing picture that the data inTable 4.13 are normally distributed, although we can see evidence of a smallclustering of pennies with a mass of approximately 3.06 g.We analyze samples to characterize the parent population. To reach ameaningful conclusion about a population, the samples must be representativeof the population. One important requirement is that the samplesmust be random. A dot chart provides a simple visual display that allowsus look for non-random trends. Figure 4.28c shows the result of entering> dotchart(penny, xlab = “Mass of Pennies (g)”, ylab = “PennyNumber”, main = “Dotchart of Data in Table 4.13”)In this plot the masses of the 100 pennies are arranged along the y-axis inthe order of sampling. If we see a pattern in the data along the y-axis, suchas a trend toward smaller masses as we move from the first penny to thelast penny, then we have clear evidence of non-random sampling. Becauseour data do not show a pattern, we have more confidence in the quality ofour data.The last plot we will consider is a box plot, which is a useful way toidentify potential outliers without making any assumptions about the data’sdistribution. A box plot contains four pieces of information about a dataset: the median, the middle 50% of the data, the smallest value and thelargest value within a set distance of the middle 50% of the data, and possibleoutliers. Figure 4.28d shows the result of entering> bwplot(penny, xlab = “Mass of Pennies (g)”, main = “Boxplot ofData in Table 4.13)”The black dot (•) is the data set’s median. The rectangular box shows therange of masses for the middle 50% of the pennies. This also is known as theinterquartile range, or IQR. The dashed lines, which are called “whiskers,”extend to the smallest value and the largest value that are within ±1.5×IQRof the rectangular box. Potential outliers are shown as open circles (º). Fornormally distributed data the median will be near the center of the box andthe whiskers will be equidistant from the box. As is often true in statistics,the converse is not true—finding that a boxplot is perfectly symmetric doesnot prove that the data are normally distributed.The box plot in Figure 4.28d is consistent with the histogram (Figure4.28a) and the kernel density plot (Figure 4.28b). Together, the three plotsprovide evidence that the data in Table 4.13 are normally distributed. Thepotential outlier, whose mass of 3.198 g, is not sufficiently far away fromthe upper whisker to be of concern, particularly as the size of the data setNote that the dispersion of points alongthe x-axis is not uniform, with morepoints occurring near the center of the x-axis than at either end. This pattern is asexpected for a normal distribution.To find the interquartile range you firstfind the median, which divides the datain half. The median of each half providesthe limits for the box. The IQR is the medianof the upper half of the data minusthe median for the lower half of the data.For the data in Table 4.13 the median is3.098. The median for the lower half ofthe data is 3.068 and the median for theupper half of the data is 3.115. The IQRis 3.115 – 3.068 = 0.047. You can use thecommand “summary(penny)” in R to obtainthese values.The lower “whisker” extend to the firstdata point with a mass larger than3.068 – 1.5 × IQR = 3.068 – 1.5 × 0.047= 2.9975which for this data is 2.998 g. The upper“whisker” extends to the last data pointwith a mass smaller than3.115+1.5×IQR = 3.115 + 1.5×0.047 =3.1855which for this data is 3.181 g.

Chapter 4 Evaluating Analytical Data133precision. A propagation of uncertainty allows us to estimate how thesedeterminate and indeterminate errors will affect our results.When we analyze a sample several times the distribution of the resultsis described by a probability distribution, two examples of which are thebinomial distribution and the normal distribution. Knowing the type ofdistribution allows us to determine the probability of obtaining a particularrange of results. For a normal distribution we express this range as a confidenceinterval.A statistical analysis allows us to determine whether our results are significantlydifferent from known values, or from values obtained by otheranalysts, by other methods of analysis, or for other samples. We can use at-test to compare mean values and an F-test to compare precisions. To comparetwo sets of data you must first determine whether the data is paired orunpaired. For unpaired data you must also decide if the standard deviationscan be pooled. A decision about whether to retain an outlying value canbe made using Dixon’s Q-test, Grubb’s test, or Chauvenet’s criterion. Youshould be sure to exercise caution when deciding to reject an outlier.Finally, the detection limit is a statistical statement about the smallestamount of analyte that we can detect with confidence. A detection limit isnot exact since its value depends on how willing we are to falsely report theanalyte’s presence or absence in a sample. When reporting a detection limityou should clearly indicate how you arrived at its value.4KProblems1. The following masses were recorded for 12 different U.S. quarters (allgiven in grams):5.683 5.549 5.548 5.5525.620 5.536 5.539 5.6845.551 5.552 5.554 5.632Report the mean, median, range, standard deviation and variance forthis data.2. A determination of acetaminophen in 10 separate tablets of ExcedrinExtra Strength Pain Reliever gives the following results (in mg). 16224.3 240.4 246.3 239.4 253.1261.7 229.4 255.5 235.5 249.7Answers, but not worked solutions, tomost end-of-chapter problems are availablehere.Many of the problems that follow require accessto statistical tables. For your convenience,here are hyperlinks to the appendices containingthese tables.Appendix 3: Single-Sided Normal DistributionAppendix 4: Critical Values for the t-TestAppendix 5: Critical Values for the F-TestAppendix 6: Critical Values for Dixon’s Q-TestAppendix 7: Critical Values for Grubb’s Test(a) Report the mean, median, range, standard deviation and variancefor this data. (b) Assuming that X and s 2 are good approximations form and s 2 , and that the population is normally distributed, what per-16 Simonian, M. H.; Dinh, S.; Fray, L. A. Spectroscopy 1993, 8(6), 37–47.

134 Analytical Chemistry 2.0centage of tablets contain more than the standard amount of 250 mgacetaminophen per tablet?3. Salem and Galan developed a new method for determining the amountof morphine hydrochloride in tablets. 17 An analysis of tablets with differentnominal dosages gave the following results (in mg/tablet).Many of the problems that follow require accessto statistical tables. For your convenience,here are hyperlinks to the appendices containingthese tables.Appendix 3: Single-Sided Normal DistributionAppendix 4: Critical Values for the t-TestAppendix 5: Critical Values for the F-TestAppendix 6: Critical Values for Dixon’s Q-TestAppendix 7: Critical Values for Grubb’s TestSee Chapter 15 to learn more about usinga spike recovery to evaluate an analyticalmethod.100-mg tablets 60-mg tablets 30-mg tablets 10-mg tablets99.17 54.21 28.51 9.0694.31 55.62 26.25 8.8395.92 57.40 25.92 9.0894.55 57.51 28.6293.83 52.59 24.93(a) For each dosage, calculate the mean and standard deviation for themg of morphine hydrochloride per tablet. (b) For each dosage level,assuming that X and s 2 are good approximations for m and s 2 , andthat the population is normally distributed, what percentage of tabletscontain more than the nominal amount of morphine hydrochloride pertablet?4. Daskalakis and co-workers evaluated several procedures for digestingoyster and mussel tissue prior to analyzing them for silver. 18 To evaluatethe procedures they spiked samples with known amounts of silverand analyzed the samples to determine the amount of silver, reportingresults as the percentage of added silver found in the analysis. A procedurewas judged acceptable is the spike recoveries fell within the range100±15%. The spike recoveries for one method are shown here.106% 108% 92% 99%101% 93% 93% 104%Assuming a normal distribution for the spike recoveries, what is theprobability that any single spike recovery will be within the acceptedrange?5. The formula weight (FW) of a gas can be determined using the followingform of the ideal gas lawFW= gRTPVwhere g is the mass in grams, R is the gas constant, T is the temperaturein Kelvin, P is the pressure in atmospheres, and V is the volume in liters.17 Salem, I. I.; Galan, A. C. Anal. Chim. Acta 1993, 283, 334–337.18 Daskalakis, K. D.; O’Connor, T. P.; Crecelius, E. A. Environ. Sci. Technol. 1997, 31, 2303–2306.

Chapter 4 Evaluating Analytical Data135In a typical analysis the following data are obtained (with estimateduncertainties in parentheses)g = 0.118 g (± 0.002 g)R = 0.082056 L atm mol –1 K –1 (± 0.000001 L atm mol –1 K –1 )T = 298.2 K (± 0.1 K)P = 0.724 atm (± 0.005 atm)V = 0.250 L (± 0.005 L)(a) What is the compound’s formula weight and its estimated uncertainty?(b) To which variable(s) should you direct your attention if youwish to improve the uncertainty in the compound’s molecular weight?6. To prepare a standard solution of Mn 2+ a 0.250 g sample of Mn is dissolvedin 10 mL of concentrated HNO 3 (measured with a graduatedcylinder). The resulting solution is quantitatively transferred to a 100-mL volumetric flask and diluted to volume with distilled water. A 10mL aliquot of the solution is pipeted into a 500-mL volumetric flaskand diluted to volume. (a) Express the concentration of Mn in mg/L,and estimate its uncertainty using a propagation of uncertainty. (b)Can you improve the concentration’s uncertainty by using a pipet tomeasure the HNO 3 , instead of a graduated cylinder?7. The mass of a hygroscopic compound is measured using the techniqueof weighing by difference. In this technique the compound is placed ina sealed container and weighed. A portion of the compound is removed,and the container and the remaining material are reweighed. The differencebetween the two masses gives the sample’s mass. A solution ofa hygroscopic compound with a gram formula weight of 121.34 g/mol(±0.01 g/mol) was prepared in the following manner. A sample of thecompound and its container has a mass of 23.5811 grams. A portionof the compound was transferred to a 100-mL volumetric flask anddiluted to volume. The mass of the compound and container after thetransfer is 22.1559 grams. Calculate the compound’s molarity and estimateits uncertainty by a propagation of uncertainty.Many of the problems that follow require accessto statistical tables. For your convenience,here are hyperlinks to the appendices containingthese tables.Appendix 3: Single-Sided Normal DistributionAppendix 4: Critical Values for the t-TestAppendix 5: Critical Values for the F-TestAppendix 6: Critical Values for Dixon’s Q-TestAppendix 7: Critical Values for Grubb’s Test8. Show using a propagation of uncertainty that the standard error of themean for n determinations is s n .9. Beginning with equation 4.17 and equation 4.18, use a propagation ofuncertainty to derive equation 4.19.10. What is the smallest mass that we can measure on an analytical balancethat has a tolerance of ±0.1 mg, if the relative error must be less than0.1%?

136 Analytical Chemistry 2.011. Which of the following is the best way to dispense 100.0 mL of a reagent:(a) use a 50‐mL pipet twice; (b) use a 25-mL pipet four times;or (c) use a 10-mL pipet ten times?12. You can dilute a solution by a factor of 200 using readily available pipets(1-mL to 100-mL) and volumetric flasks (10-mL to 1000-mL) in eitherone step, two steps, or three steps. Limiting yourself to the glasswarein Table 4.2, determine the proper combination of glassware to accomplisheach dilution, and rank them in order of their most probableuncertainties.13. Explain why changing all values in a data set by a constant amount willchange X but will have no effect on s.Many of the problems that follow require accessto statistical tables. For your convenience,here are hyperlinks to the appendices containingthese tables.Appendix 3: Single-Sided Normal DistributionAppendix 4: Critical Values for the t-TestAppendix 5: Critical Values for the F-TestAppendix 6: Critical Values for Dixon’s Q-TestAppendix 7: Critical Values for Grubb’s Test14. Obtain a sample of a metal from your instructor and determine itsdensity by one or both of the following methods:Method A: Determine the sample’s mass with a balance. Calculate thesample’s volume using appropriate linear dimensions.Method B: Determine the sample’s mass with a balance. Calculate thesample’s volume by measuring the amount of water that it displaces.This can be done by adding water to a graduated cylinder, reading thevolume, adding the sample, and reading the new volume. The differencein volumes is equal to the sample’s volume.Determine the density at least 5 times. (a) Report the mean, the standarddeviation, and the 95% confidence interval for your results. (b)Find the accepted value for the metal’s density and determine the absoluteand relative error for your determination of the metal’s density.(c) Use a propagation of uncertainty to determine the uncertainty foryour method of analysis. Is the result of this calculation consistent withyour experimental results? If not, suggest some possible reasons for thisdisagreement.15. How many carbon atoms must a molecule have if the mean numberof 13 C atoms per molecule is 1.00? What percentage of such moleculeswill have no atoms of 13 C?16. In Example 4.10 we determined the probability that a molecule ofcholesterol, C 27 H 44 O, had no atoms of 13 C. (a) Calculate the probabilitythat a molecule of cholesterol, has 1 atom of 13 C. (b) What is theprobability that a molecule of cholesterol will have two or more atomsof 13 C?17. Berglund and Wichardt investigated the quantitative determinationof Cr in high-alloy steels using a potentiometric titration of Cr(VI) 19 .19 Berglund, B.; Wichardt, C. Anal. Chim. Acta 1990, 236, 399–410.

Chapter 4 Evaluating Analytical Data137Before the titration, samples of the steel were dissolved in acid and thechromium oxidized to Cr(VI) using peroxydisulfate. Shown here arethe results ( as %w/w Cr) for the analysis of a reference steel.16.968 16.922 16.840 16.88316.887 16.977 16.857 16.728Calculate the mean, the standard deviation, and the 95% confidenceinterval about the mean. What does this confidence interval mean?18. Ketkar and co-workers developed an analytical method for determiningtrace levels of atmospheric gases. 20 An analysis of a sample containing40.0 parts per thousand (ppt) 2‐chloroethylsulfide yielded the followingresults43.3 34.8 31.937.8 34.4 31.942.1 33.6 35.3(a) Determine whether there is a significant difference between theexperimental mean and the expected value at a = 0.05. (b) As part ofthis study a reagent blank was analyzed 12 times, giving a mean of 0.16ppt and a standard deviation of 1.20 ppt. What are the IUPAC detectionlimit, the limit of identification, and limit of quantitation for thismethod assuming a = 0.05?19. To test a spectrophotometer’s accuracy a solution of 60.06 ppm K 2 Cr 2 O 7in 5.0 mM H 2 SO 4 is prepared and analyzed. This solution has an expectedabsorbance of 0.640 at 350.0 nm in a 1.0-cm cell when using5.0 mM H 2 SO 4 as a reagent blank. Several aliquots of the solutionproduce the following absorbance values.0.639 0.638 0.640 0.639 0.640 0.639 0.638Determine whether there is a significant difference between the experimentalmean and the expected value at a = 0.01.Many of the problems that follow require accessto statistical tables. For your convenience,here are hyperlinks to the appendices containingthese tables.Appendix 3: Single-Sided Normal DistributionAppendix 4: Critical Values for the t-TestAppendix 5: Critical Values for the F-TestAppendix 6: Critical Values for Dixon’s Q-TestAppendix 7: Critical Values for Grubb’s Test20. Monna and co-workers used radioactive isotopes to date sedimentsfrom lakes and estuaries. 21 To verify this method they analyzed a 20 8Postandard known to have an activity of 77.5 decays/min, obtaining thefollowing results.77.09 75.37 72.42 76.84 77.84 76.6978.03 74.96 77.54 76.09 81.12 75.75Determine whether there is a significant difference between the meanand the expected value at a = 0.05.20 Ketkar, S. N.; Dulak, J. G.; Dheandhanou, S.; Fite, W. L. Anal. Chim. Acta 1996, 330, 267–270.21 Monna, F.; Mathieu, D.; Marques, A. N.; Lancelot, J.; Bernat, M. Anal. Chim. Acta 1996, 330,107–116.

138 Analytical Chemistry 2.021. A 2.6540-g sample of an iron ore, known to be 53.51% w/w Fe, is dissolvedin a small portion of concentrated HCl and diluted to volume ina 250-mL volumetric flask. A spectrophotometric determination of theconcentration of Fe in this solution yields results of 5840, 5770, 5650,and 5660 ppm. Determine whether there is a significant differencebetween the experimental mean and the expected value at a = 0.05.22. Horvat and co-workers used atomic absorption spectroscopy to determinethe concentration of Hg in coal fly ash. 22 Of particular interestto the authors was developing an appropriate procedure for digestingsamples and releasing the Hg for analysis. As part of their study theytested several reagents for digesting samples. Results obtained usingHNO 3 and using a 1 + 3 mixture of HNO 3 and HCl are shown here.All concentrations are given as ng Hg/g sample.HNO 3 161 165 160 167 1661+3 HNO 3 –HCl 159 145 140 147 143 156Many of the problems that follow require accessto statistical tables. For your convenience,here are hyperlinks to the appendices containingthese tables.Appendix 3: Single-Sided Normal DistributionAppendix 4: Critical Values for the t-TestAppendix 5: Critical Values for the F-TestAppendix 6: Critical Values for Dixon’s Q-TestAppendix 7: Critical Values for Grubb’s TestDetermine whether there is a significant difference between these methodsat a = 0.05.23. Lord Rayleigh, John William Strutt (1842-1919), was one of the mostwell known scientists of the late nineteenth and early twentieth centuries,publishing over 440 papers and receiving the Nobel Prize in 1904for the discovery of argon. An important turning point in Rayleigh’sdiscovery of Ar was his experimental measurements of the density of N 2 .Rayleigh approached this experiment in two ways: first by taking atmosphericair and removing all O 2 and H 2 ; and second, by chemicallyproducing N 2 by decomposing nitrogen containing compounds (NO,N 2 O, and NH 4 NO 3 ) and again removing all O 2 and H 2 . Following arehis results for the density of N 2 , published in Proc. Roy. Soc. 1894, LV,340 (publication 210) (all values are for grams of gas at an equivalentvolume, pressure, and temperature). 23AtmosphericOrigin:ChemicalOrigin:2.31017 2.30986 2.31010 2.310012.31024 2.31010 2.310282.30143 2.29890 2.29816 2.301822.29869 2.29940 2.29849 2.29889Explain why this data led Rayleigh to look for, and discover Ar.24. Gács and Ferraroli reported a method for monitoring the concentrationof SO 2 in air. 24 They compared their method to the standard method byanalyzing urban air samples collected from a single location. Sampleswere collected by drawing air through a collection solution for 6 min.22 Horvat, M.; Lupsina, V.; Pihlar, B. Anal. Chim. Acta 1991, 243, 71–79.23 Larsen, R. D. J. Chem. Educ. 1990, 67, 925–928.24 Gács, I.; Ferraroli, R. Anal. Chim. Acta 1992, 269, 177 –185.

Chapter 4 Evaluating Analytical Data139Shown here is a summary of their results with SO 2 concentrationsreported in mL/m 3 .standardmethod:newmethod:21.62 22.20 24.27 23.5424.25 23.09 21.0221.54 20.51 22.31 21.3024.62 25.72 21.54Using an appropriate statistical test determine whether there is any significantdifference between the standard method and the new methodat a = 0.05.25. One way to check the accuracy of a spectrophotometer is to measureabsorbencies for a series of standard dichromate solutions obtainedfrom the National Institute of Standards and Technology. Absorbenciesare measured at 257 nm and compared to the accepted values. Theresults obtained when testing a newly purchased spectrophotometer areshown here. Determine if the tested spectrophotometer is accurate ata = 0.05.Standard Measured Absorbance Expected Absorbance1 0.2872 0.28712 0.5773 0.57603 0.8674 0.86774 1.1623 1.16085 1.4559 1.456526. Maskarinec and co-workers investigated the stability of volatile organicsin environmental water samples. 25 Of particular interest was establishingproper conditions for maintaining the sample’s integrity between itscollection and analysis. Two preservatives were investigated—ascorbicacid and sodium bisulfate—and maximum holding times were determinedfor a number of volatile organics and water matrices. The followingtable shows results (in days) for the holding time of nine organiccompounds in surface water.Ascorbic Acid Sodium Bisulfatemethylene chloride 77 62carbon disulfide 23 54trichloroethane 52 51benzene 62 421,1,2-trichloroethane 57 531,1,2,2-tetrachlorethane 33 85tetrachloroethene 41 63Many of the problems that follow require accessto statistical tables. For your convenience,here are hyperlinks to the appendices containingthese tables.Appendix 3: Single-Sided Normal DistributionAppendix 4: Critical Values for the t-TestAppendix 5: Critical Values for the F-TestAppendix 6: Critical Values for Dixon’s Q-TestAppendix 7: Critical Values for Grubb’s Test25 Maxkarinec, M. P.; Johnson, L. H.; Holladay, S. K.; Moody, R. L.; Bayne, C. K.; Jenkins, R. A.Environ. Sci. Technol. 1990, 24, 1665–1670.

140 Analytical Chemistry 2.0toluene 32 94chlorobenzene 36 86Determine whether there is a significant difference in the effectivenessof the two preservatives at a = 0.10.27. Using X-ray diffraction, Karstang and Kvalhein reported a new methodfor determining the weight percent of kalonite in complex clay mineralsusing X-ray diffraction. 26 To test the method, nine samples containingknown amounts of kalonite were prepared and analyzed. The results(as % w/w kalonite) are shown here.Actual: 5.0 10.0 20.0 40.0 50.0 60.0 80.0 90.0 95.0Found: 6.8 11.7 19.8 40.5 53.6 61.7 78.9 91.7 94.7Evaluate the accuracy of the method at a = 0.05.Many of the problems that follow require accessto statistical tables. For your convenience,here are hyperlinks to the appendices containingthese tables.Appendix 3: Single-Sided Normal DistributionAppendix 4: Critical Values for the t-TestAppendix 5: Critical Values for the F-TestAppendix 6: Critical Values for Dixon’s Q-TestAppendix 7: Critical Values for Grubb’s Test28. Mizutani, Yabuki and Asai developed an electrochemical method foranalyzing l-malate. 27 As part of their study they analyzed a series ofbeverages using both their method and a standard spectrophotometricprocedure based on a clinical kit purchased from Boerhinger Scientific.The following table summarizes their results. All values are in ppm.Sample Electrode SpectrophotometricApple juice 1 34.0 33.4Apple juice 2 22.6 28.4Apple juice 3 29.7 29.5Apple juice 4 24.9 24.8Grape juice 1 17.8 18.3Grape juice 2 14.8 15.4Mixed fruit juice 1 8.6 8.5Mixed fruit juice 2 31.4 31.9White wine 1 10.8 11.5White wine 2 17.3 17.6White wine 3 15.7 15.4White wine 4 18.4 18.3Determine whether there is a significant difference between the methodsat a = 0.05.29. Alexiev and colleagues describe an improved photometric method fordetermining Fe 3+ based on its ability to catalyze the oxidation of sulphanilicacid by KIO 4 . 28 As part of their study the concentration of Fe 3+26 Karstang, T. V.; Kvalhein, O. M. Anal. Chem. 1991, 63, 767–772.27 Mizutani, F.; Yabuki, S.; Asai, M. Anal. Chim. Acta 1991, 245,145–150.28 Alexiev, A.; Rubino, S.; Deyanova, M.; Stoyanova, A.; Sicilia, D.; Perez Bendito, D. Anal. Chim.Acta, 1994, 295, 211–219.

Chapter 4 Evaluating Analytical Data141in human serum samples was determined by the improved method andthe standard method. The results, with concentrations in mmol/L, areshown in the following table.Sample Improved Method Standard Method1 8.25 8.062 9.75 8.843 9.75 8.364 9.75 8.735 10.75 13.136 11.25 13.657 13.88 13.858 14.25 13.53Determine whether there is a significant difference between the twomethods at a = 0.05.30. Ten laboratories were asked to determine an analyte’s concentration ofin three standard test samples. Following are the results, in mg/mL. 29Laboratory Sample 1 Sample 2 Sample 31 22.6 13.6 16.02 23.0 14.2 15.93 21.5 13.9 16.94 21.9 13.9 16.95 21.3 13.5 16.76 22.1 13.5 17.47 23.1 13.9 17.58 21.7 13.5 16.89 22.2 12.9 17.210 21.7 13.8 16.7Determine if there are any potential outliers in Sample 1, Sample 2 orSample 3 at a significance level of a = 0.05. Use all three methods—Dixon’s Q-test, Grubb’s test, and Chauvenet’s criterion—and comparethe results to each other.Many of the problems that follow require accessto statistical tables. For your convenience,here are hyperlinks to the appendices containingthese tables.Appendix 3: Single-Sided Normal DistributionAppendix 4: Critical Values for the t-TestAppendix 5: Critical Values for the F-TestAppendix 6: Critical Values for Dixon’s Q-TestAppendix 7: Critical Values for Grubb’s Test31. When copper metal and powdered sulfur are placed in a crucible andignited, the product is a sulfide with an empirical formula of Cu x S. Thevalue of x can be determined by weighing the Cu and S before ignition,and finding the mass of Cu x S when the reaction is complete (any excesssulfur leaves as SO 2 ). The following table shows the Cu/S ratios from62 such experiments.29 Data adapted from Steiner, E. H. “Planning and Analysis of Results of Collaborative Tests,” inStatistical Manual of the Association of Official Analytical Chemists, Association of Official AnalyticalChemists: Washington, D. C., 1975.

142 Analytical Chemistry 2.0Many of the problems that follow require accessto statistical tables. For your convenience,here are hyperlinks to the appendices containingthese tables.Appendix 3: Single-Sided Normal DistributionAppendix 4: Critical Values for the t-TestAppendix 5: Critical Values for the F-TestAppendix 6: Critical Values for Dixon’s Q-TestAppendix 7: Critical Values for Grubb’s Test1.764 1.838 1.865 1.866 1.872 1.8771.890 1.891 1.891 1.897 1.899 1.9001.906 1.908 1.910 1.911 1.916 1.9191.920 1.922 1.927 1.931 1.935 1.9361.936 1.937 1.939 1.939 1.940 1.9411.941 1.942 1.943 1.948 1.953 1.9551.957 1.957 1.957 1.959 1.962 1.9631.963 1.963 1.966 1.968 1.969 1.9731.975 1.976 1.977 1.981 1.981 1.9881.993 1.993 1.995 1.995 1.995 2.0172.029 2.042(a) Calculate the mean and standard deviation for this data. (b)Construct a histogram for this data. From a visual inspection ofyour histogram, does the data appear to be normally distributed?(c) In a normally distributed population 68.26% of all members liewithin the range m ± 1s. What percentage of the data lies withinthe range ± 1s? Does this support your answer to the previous question?(d) Assuming that X and s 2 are good approximations for mand s 2 , what percentage of all experimentally determined Cu/Sratios will be greater than 2? How does this compare with the experimentaldata? Does this support your conclusion about whetherthe data is normally distributed? (e) It has been reported that thismethod of preparing copper sulfide results in a non-stoichiometriccompound with a Cu/S ratio of less than 2. Determine if the meanvalue for this data is significantly less than 2 at a significance levelof a = 0.01.32. Real-time quantitative PCR is an analytical method for determiningtrace amounts of DNA. During the analysis, each cycle doubles theamount of DNA. A probe species that fluoresces in the presence ofDNA is added to the reaction mixture and the increase in fluorescenceis monitored during the cycling. The cycle threshold, C t , is the cyclewhen the fluorescence exceeds a threshold value. The data in the followingtable shows C t values for three samples using real-time quantitativePCR. 30 Each sample was analyzed 18 times.Sample X Sample Y Sample Z24.24 25.14 24.41 28.06 22.97 23.4323.97 24.57 27.21 27.77 22.93 23.6624.44 24.49 27.02 28.74 22.95 28.7924.79 24.68 26.81 28.35 23.12 23.7730 Burns, M. J.; Nixon, G. J.; Foy, C. A.; Harris, N. BMC Biotechnol. 2005, 5:31 (open accesspublication).

Chapter 4 Evaluating Analytical Data14323.92 24.45 26.64 28.80 23.59 23.9824.53 24.48 27.63 27.99 23.37 23.5624.95 24.30 28.42 28.21 24.17 22.8024.76 24.60 25.16 28.00 23.48 23.2925.18 24.57 28.53 28.21 23.80 23.86Examine this data statistically and write a brief report on your conclusions.Issues you may wish to address include the presence of outliersin the samples, a summary of the descriptive statistics for each sample,and any evidence for a difference between the samples.Many of the problems that follow require accessto statistical tables. For your convenience,here are hyperlinks to the appendices containingthese tables.Appendix 3: Single-Sided Normal DistributionAppendix 4: Critical Values for the t-TestAppendix 5: Critical Values for the F-TestAppendix 6: Critical Values for Dixon’s Q-TestAppendix 7: Critical Values for Grubb’s Test4LSolutions to Practice ExercisesPractice Exercise 4.1Mean: To find the mean we sum up the individual measurements and divideby the number of measurements. The sum of the 10 concentrationsis 1405. Dividing the sum by 10, gives the mean as 140.5, or 1.40×10 2mmol/L.Median: To find the mean we arrange the 10 measurements from thesmallest concentration to the largest concentration; thus118 132 137 140 141 143 143 145 149 157The median for a data set with 10 members is the average of the fifth andsixth values; thus, the median is (141 + 143)/2, or 141 mmol/L.Range: The range is the difference between the largest value and the smallestvalue; thus, the range is 157 – 118 = 39 mmol/L.Standard Deviation: To calculate the standard deviation we first calculatethe difference between each measurement and the mean value (140.5),square the resulting differences, and add them together. The differencesare-0.5 2.5 0.5 -3.5 -8.5 16.5 2.5 8.5 -22.5 4.5and the squared differences are0.25 6.25 0.25 12.25 72.25 272.25 6.25 72.25 506.25 20.25The total sum of squares, which is the numerator of equation 4.1, is968.50. The standard deviation iss =968.5010 − 1= 10. 37 ≈ 10.4Variance: The variance is the square of the standard deviation, or 108.

144 Analytical Chemistry 2.0Click here to return to the chapter.Practice Exercise 4.2The first step is to determine the concentration of Cu 2+ in the final solution.The mass of copper is74.2991 g – 73.3216 g = 0.9775 g CuThe 10 mL of HNO 3 used to dissolve the copper does not factor into ourcalculation. The concentration of Cu 2+ is0.9775 gCu 1.000 10 30.5000 L× mL mg250.0 mL× g= 7.8202+mg Cu /LHaving found the concentration of Cu 2+ we continue on to complete thepropagation of uncertainty. The absolute uncertainty in the mass of Cuwire isu gCu2 2= ( 0. 0001) + ( 0. 0001) = 0.00014 gThe relative uncertainty in the concentration of Cu 2+ isu mg /L27.820 mg/L = ⎛ 0.00014⎞+ ⎛ 020 . ⎞⎝⎜0.9775 ⎠⎟⎝⎜⎠⎟ +⎛ 0.006⎞500.0 ⎝⎜1.000⎟ ⎠+⎛ 012 . ⎞⎝⎜250 0⎠⎟ = 0.00603.22 2Solving for u mg/L gives the uncertainty as 0.0472. The concentration anduncertainty for Cu 2+ is 7.820 mg/L ± 0.047 mg/L.Click here to return to the chapter.Practice Exercise 4.3The first step is to calculate the absorbance, which isPA =− ⎛ ⎞⎝P ⎜ ⎠⎟ =− ⎛2150 . × 10 ⎞log log⎝⎜380 . × 10 ⎠⎟ =− log( 0 . 3947 ) = 0 . 4037 ≈ 0 . 404oHaving found the absorbance we continue on to complete the propagationof uncertainty. First, we find the uncertainty for the ratio P/P o .uP/PoP Po=2⎛15 ⎞⎝⎜. × ⎠⎟ + ⎛ 15 ⎞22380 10 ⎝⎜150 . × 10 ⎠⎟ = 1.075 × 10−22Finally, from Table 4.10 the uncertainty in the absorbance isuAuP P−2= 0. 4343× = ( 0. 4343) × (. 1 075× 10 ) = 4.669× 10 −3P P/ ooThe absorbance and uncertainty is 0.404 ± 0.005 absorbance units.

Chapter 4 Evaluating Analytical Data145Click here to return to the chapter.Practice Exercise 4.4An uncertainty of 0.8% is a relative uncertainty in the concentration of0.008; thus20.0280.008 = ⎛ ⎞ ⎛⎝⎜23.41⎠⎟ + u ⎞kA⎜⎜0.186 ⎝ ⎠⎟Squaring both sides of the equation gives25 0.02864 . × 10 = ⎛ ⎞ ⎛⎝⎜23.41⎠⎟ + u ⎞−kA⎜⎝0.186⎠⎟2⎛6. 257× 10 5 u ⎞−=kA⎝⎜0.186⎠⎟Sovling for u k A gives its value as 1.47×10–3 , or ±0.0015 ppm –1 .Click here to return to the chapter.Practice Exercise 4.5To find the percentage of tablets containing less than 245 mg of aspirinwe calculate the deviation, z,245−250z = =−100.5and look up the corresponding probability in Appendix 3A, obtaining avalue of 15.87%. To find the percentage of tablets containing less than240 mg of aspirin we find that240−250z = =−200.5which corresponds to 2.28%. The percentage of tablets containing between240 and 245 mg of aspiring is 15.87% – 2.28% = 13.59%.Click here to return to the chapter.Practice Exercise 4.6The mean is 249.9 mg aspirin/tablet for this sample of seven tablets. Fora 95% confidence interval the value of z is 1.96. The confidence intervalis22

146 Analytical Chemistry 2.01.96× 5249. 9 ± = 249. 9 ± 3.7 ≈ 250 mg ± 4mg7Click here to return to the chapter.Practice Exercise 4.7With 100 pennies, we have 99 degrees of freedom for the mean. AlthoughTable 4.15 does not include a value for t(0.05, 99), we can approximateits value using the values for t(0.05, 60) and t(0.05, 100) and assuming alinear change in its value.39t( 005 . , 99) = t( 0. 05, 60) − t( 005 . , 60) − t(0. 05,100)4039t( 005 . , 99) = 2. 000 − { . .40 2 000 − 19 84}= 1.9844The 95% confidence interval for the pennies is{ }1. 9844×0.03463.095±= 3.095 g ± 0.007 g100From Example 4.15, the 95% confidence intervals for the two samples inTable 4.11 are 3.117 g ± 0.047 g and 3.081 g ± 0.046 g. As expected, theconfidence interval for the sample of 100 pennies is much smaller thanthat for the two smaller samples of pennies. Note, as well, that the confidenceinterval for the larger sample fits within the confidence intervalsfor the two smaller samples.Click here to return to the chapter.Practice Exercise 4.8The null hypothesis is H : X =µand the alternative hypothesis is0H : X ≠µ.The mean and standard deviation for the data are 99.26%Aand 2.35%, respectively. The value for t exp is100. 0−9926 . 7t exp=235 .= 0.833and the critical value for t(0.05, 6) is 0.836. Because t exp is less thant(0.05, 6) we retain the null hypothesis and have no evidence for a significantdifference between X and m.Click here to return to the chapter.

Chapter 4 Evaluating Analytical Data147Practice Exercise 4.9The standard deviations for Lot 1 is 6.451 mg, and 7.849 mg for Lot 2.The null and alternative hypotheses are2 22H s = s H : s ≠ s0 : Lot1and the value of F exp isLot2A Lot 12( 7. 849)F exp= = 1.4802( 6. 451)2Lot2The critical value for F(0.05, 5, 6) is 5.988. Because F exp < F(0.05, 5, 6),we retain the null hypothesis. There is no evidence at a = 0.05 to suggestthat the difference in the variances is significant.Click here to return to the chapter.Practice Exercise 4.10To compare the means for the two lots, we will use an unpaired t-testof the null hypothesis H : X = X and the alternative hypothesis0 Lot1 Lot2H : X ≠ X . Because there is no evidence suggesting a differenceA Lot 1 Lot 2in the variances (see Practice Exercise 4.9) we pool the standard deviations,obtaining an s pool ofs pool=( 7− 1)( 6. 451) + ( 6−1)( 7. 849)7+ 6−22 2= 712 . 1The means for the two samples are 249.57 mg for Lot 1 and 249.00 mgfor Lot 2. The value for t exp is249. 57 − 249.00 7t exp=× × 6.7.121 7+ 6= 0 1439The critical value for t(0.05, 11) is 2.204. Because t exp is less than t(0.05,11), we retain the null hypothesis and find no evidence at a = 0.05 fora significant difference between the means for the two lots of aspirintablets.Click here to return to the chapter.Practice Exercise 4.11Treating as Unpaired Data: The mean and standard deviation for the concentrationof Zn 2+ at the air-water interface are 0.5178 mg/L and 0.1732mg/L respectively, and the values for the sediment-water interface are0.4445 mg/L and 0.1418 mg/L. An F-test of the variances gives an F exp of1.493 and an F(0.05, 5, 5) of 7.146. Because F exp is smaller than F(0.05,5,5) we have no evidence at a = 0.05 to suggest that the difference in variancesis significant. Pooling the standard deviations gives an s pool of 0.1582

148 Analytical Chemistry 2.0mg/L. An unpaired t-test gives t exp as 0.8025. Because t exp is smaller thant(0.05, 11), which is 2.204, we have no evidence that there is a differencein the concentration of Zn 2+ between the two interfaces.Treating as Paired Data: To treat as paired data we need to calculate thedifference, d i , between the concentration of Zn 2+ at the air-water interfaceand at the sediment-water interface for each location.d i[Zn 2+2+] [Zn ]air-waterised-wateri= ( ) −( )Location 1 2 3 4 5 6d i (mg/L) 0.015 0.028 0.067 0.121 0.102 0.107The mean difference is 0.07333 mg/L with a standard deviation of 0.0441mg/L. The null hypothesis and alternative hypothesis areand the value of t exp isH : d = 0 H : d ≠ 00 A0.07333 6t exp= = 4.0730.04410Because t exp is greater than t(0.05, 5), which is 2.571, we reject the nullhypothesis and accept the alternative hypothesis that there is a significantdifference in the concentration of Zn 2+ between the air-water interfaceand the sediment-water interface.The difference in the concentration of Zn 2+ between locations is muchlarger than the difference in the concentration of Zn 2+ between the interfaces.Because out interest is in studying differences between the interfaces,the larger standard deviation when treating the data as unpaired increasesthe probability of incorrectly retaining the null hypothesis, a type 2 error.Click here to return to the chapter.Practice Exercise 4.12You will find small differences between the values given here for t exp andF exp , and for those values shown with the worked solutions in the chapter.These differences arise because Excel does not round off the results ofintermediate calculations.The two snapshots of Excel spreadsheets shown in Figure 4.29 providesolutions to these two examples.Click here to return to the chapter.

Chapter 4 Evaluating Analytical Data149Example 4.20F-Test Two-Sample for VariancesPractice Exercise 4.13Shown here are copies of R sessions for each problem. You will find smalldifferences between the values given here for t exp and F exp , and for thosevalues shown with the worked solutions in the chapter. These differencesarise because R does not round off the results of intermediate calculations.Example 4.20Analyst A Analyst B86.82 81.0187.04 86.1586.93 81.7387.01 83.1986.20 80.2787.00 83.94t-Test: Two-Sample Assuming Unequal VariancesAnalyst A Analyst BMean 86.8333333 82.715Variance 0.10246667 4.67615Observations 6 6Hypothesized Mean Difference 0df 5t Stat 4.6147271P(T AnalystB=c(81.01, 86.15, 81.73, 83.19, 80.27, 83.94)> var.test(AnalystB, AnalystA)F test to compare two variancesdata: AnalystB and AnalystAF = 45.6358, num df = 5, denom df = 5, p-value = 0.0007148alternative hypothesis: true ratio of variances is not equal to 195 percent confidence interval:6.385863 326.130970sample estimates:ratio of variances45.63582> t.test(AnalystA, AnalystB, var.equal=FALSE)Example 4.21Microbiological Electrochemical129.5 132.389.6 91.076.6 73.652.2 58.2110.8 104.250.4 49.972.4 82.1141.4 154.175.0 73.434.1 38.160.3 60.1Analyst B Analyst AMean 82.715 86.8333333Variancedf4.6761550.102466675ObservationsF645.63581656t-Test: Paired Two Sample for MeansMicrobiologicalP(F

150 Analytical Chemistry 2.0Welch Two Sample t-testdata: AnalystA and AnalystBt = 4.6147, df = 5.219, p-value = 0.005177alternative hypothesis: true difference in means is not equal to 095 percent confidence interval:1.852919 6.383748sample estimates:mean of x mean of y86.83333 82.71500Example 4.21> micro=c(129.5, 89.6, 76.6, 52.2, 110.8, 50.4, 72.4, 141.4, 75.0, 34.1,60.3)> elect=c(132.3, 91.0, 73.6, 58.2, 104.2, 49.9, 82.1, 154.1, 73.4, 38.1,60.1)> t.test(micro,elect,paired=TRUE)Paired t-testdata: micro and electt = -1.3225, df = 10, p-value = 0.2155alternative hypothesis: true difference in means is not equal to 095 percent confidence interval:-6.028684 1.537775sample estimates:mean of the differences-2.245455Click here to return to the chapter.Practice Exercise 4.14Because we are selecting a random sample of 100 members from a uniformdistribution, you will see subtle differences between your plots andthe plots shown as part of this answer. Here is a record of my R sessionand the resulting plots.> data=runif(100, min=0, max=0)> data[1] 18.928795 80.423589 39.399693 23.757624 30.088554[6] 76.622174 36.487084 62.186771 81.115515 15.726404[11] 85.765317 53.994179 7.919424 10.125832 93.153308[16] 38.079322 70.268597 49.879331 73.115203 99.329723[21] 48.203305 33.093579 73.410984 75.128703 98.682127[26] 11.433861 53.337359 81.705906 95.444703 96.843476

Chapter 4 Evaluating Analytical Data151[31] 68.251721 40.567993 32.761695 74.635385 70.914957[36] 96.054750 28.448719 88.580214 95.059215 20.316015[41] 9.828515 44.172774 99.648405 85.593858 82.745774[46] 54.963426 65.563743 87.820985 17.791443 26.417481[51] 72.832037 5.518637 58.231329 10.213343 40.581266[56] 6.584000 81.261052 48.534478 51.830513 17.214508[61] 31.232099 60.545307 19.197450 60.485374 50.414960[66] 88.908862 68.939084 92.515781 72.414388 83.195206[71] 74.783176 10.643619 41.775788 20.464247 14.547841[76] 89.887518 56.217573 77.606742 26.956787 29.641171[81] 97.624246 46.406271 15.906540 23.007485 17.715668[86] 84.652814 29.379712 4.093279 46.213753 57.963604[91] 91.160366 34.278918 88.352789 93.004412 31.055807[96] 47.822329 24.052306 95.498610 21.089686 2.629948> histogram(data, type=”percent”)> densityplot(data)> dotchart(data)> bwplot(data)Figure 4.30 shows the four plots. The histogram divides the data into eightbins, each containing between 10 and 15 members. As we expect for auniform distribution, the histogram’s overall pattern suggests that eachoutcome is equally probable. In interpreting the kernel density plot it isimportant to remember that it treats each data point as if it is from a normallydistributed population (even though, in this case, the underlyingpopulation is uniform). Although the plot appears to suggest that thereare two normally distributed populations, the individual results shown atthe bottom of the plot provide further evidence for a uniform distribution.The dot chart shows no trend along the y-axis, indicating that the individualmembers of this sample were drawn randomly from the population.The distribution along the x-axis also shows no pattern, as expected for auniform distribution, Finally, the box plot shows no evidence of outliers.DensityPercent of Total1510500 20 40 60 80 100data0.0100.0080.0060.0040.0020.0000 50 100data0 20 40 60 80 100Click here to return to the chapter.0 20 40 60 80 100dataFigure 4.30 Plots generated usingR to solve Practice Exercise 4.13.


DRAFTChapter 5Standardizing AnalyticalChapter OverviewMethods5A Analytical Standards5B Calibrating the Signal (S total )5C Determining the Sensitivity (k A )5D Linear Regression and Calibration Curves5E Compensating for the Reagent Blank (S reag )5F Using Excel and R for a Regression Analysis5G Key Terms5H Chapter Summary5I Problems5J Solutions to Practice ExercisesThe American Chemical Society’s Committee on Environmental Improvement definesstandardization as the process of determining the relationship between the signal and theamount of analyte in a sample. 1 In Chapter 3 we defined this relationship asS = k n + S or S = k C + Stotal A A reag total A A r eagwhere S total is the signal, n A is the moles of analyte, C A is the analyte’s concentration, k A is themethod’s sensitivity for the analyte, and S reag is the contribution to S total from sources otherthan the sample. To standardize a method we must determine values for k A and S reag . Strategiesfor accomplishing this are the subject of this chapter.1 ACS Committee on Environmental Improvement “Guidelines for Data Acquisition and Data Quality Evaluation inEnvironmental Chemistry,” Anal. Chem. 1980, 52, 2242–2249.Copyright: David Harvey, 2009153

154 Analytical Chemistry 2.0See Chapter 9 for a thorough discussion oftitrimetric methods of analysis.The base NaOH is an example of a secondarystandard. Commercially availableNaOH contains impurities of NaCl,Na 2 CO 3 , and Na 2 SO 4 , and readilyabsorbs H 2 O from the atmosphere. Todetermine the concentration of NaOH ina solution, it is titrated against a primarystandard weak acid, such as potassium hydrogenphthalate, KHC 8 H 4 O 4 .5AAnalytical StandardsTo standardize an analytical method we use standards containing knownamounts of analyte. The accuracy of a standardization, therefore, dependson the quality of the reagents and glassware used to prepare these standards.For example, in an acid–base titration the stoichiometry of the acid–base reactiondefines the relationship between the moles of analyte and the molesof titrant. In turn, the moles of titrant is the product of the titrant’s concentrationand the volume of titrant needed to reach the equivalence point.The accuracy of a titrimetric analysis, therefore, can be no better than theaccuracy to which we know the titrant’s concentration.5A.1 Primary and Secondary StandardsWe divide analytical standards into two categories: primary standards andsecondary standards. A primary standard is a reagent for which we candispense an accurately known amount of analyte. For example, a 0.1250-gsample of K 2 Cr 2 O 7 contains 4.249 × 10 –4 moles of K 2 Cr 2 O 7 . If we placethis sample in a 250-mL volumetric flask and dilute to volume, the concentrationof the resulting solution is 1.700 × 10 –3 M. A primary standardmust have a known stoichiometry, a known purity (or assay), and it mustbe stable during long-term storage. Because of the difficulty in establishingthe degree of hydration, even after drying, a hydrated reagent usually is nota primary standard.Reagents that do not meet these criteria are secondary standards.The concentration of a secondary standard must be determined relativeto a primary standard. Lists of acceptable primary standards are available. 2Appendix 8 provides examples of some common primary standards.5A.2 Other ReagentsPreparing a standard often requires additional reagents that are not primarystandards or secondary standards. Preparing a standard solution, forexample, requires a suitable solvent, and additional reagents may be needto adjust the standard’s matrix. These solvents and reagents are potentialsources of additional analyte, which, if not accounted for, produce a determinateerror in the standardization. If available, reagent grade chemicalsconforming to standards set by the American Chemical Society should beused. 3 The label on the bottle of a reagent grade chemical (Figure 5.1) listseither the limits for specific impurities, or provides an assay for the impurities.We can improve the quality of a reagent grade chemical by purifying it,or by conducting a more accurate assay. As discussed later in the chapter, we2 (a) Smith, B. W.; Parsons, M. L. J. Chem. Educ. 1973, 50, 679–681; (b) Moody, J. R.; Greenburg,P. R.; Pratt, K. W.; Rains, T. C. Anal. Chem. 1988, 60, 1203A–1218A.3 Committee on Analytical Reagents, Reagent Chemicals, 8th ed., American Chemical Society:Washington, D. C., 1993.

Chapter 5 Standardizing Analytical Methods155can correct for contributions to S total from reagents used in an analysis byincluding an appropriate blank determination in the analytical procedure.5A.3 Preparing Standard SolutionsIt is often necessary to prepare a series of standards, each with a differentconcentration of analyte. We can prepare these standards in two ways. If therange of concentrations is limited to one or two orders of magnitude, theneach solution is best prepared by transferring a known mass or volume ofthe pure standard to a volumetric flask and diluting to volume.When working with larger ranges of concentration, particularly thoseextending over more than three orders of magnitude, standards are best preparedby a serial dilution from a single stock solution. In a serial dilutionwe prepare the most concentrated standard and then dilute a portion of itto prepare the next most concentrated standard. Next, we dilute a portionof the second standard to prepare a third standard, continuing this processuntil all we have prepared all of our standards. Serial dilutions must be preparedwith extra care because an error in preparing one standard is passedon to all succeeding standards.(a)(b)Figure 5.1 Examples of typical packaging labels for reagent grade chemicals. Label(a) provides the manufacturer’s assay for the reagent, NaBr. Note that potassiumis flagged with an asterisk (*) because its assay exceeds the limits established bythe American Chemical Society (ACS). Label (b) does not provide an assay forimpurities, but indicates that the reagent meets ACS specifications. An assay forthe reagent, NaHCO 3 is provided.

156 Analytical Chemistry 2.0See Section 2D.1 to review how an electronicbalance works. Calibrating a balanceis important, but it does not eliminate allsources of determinate error in measuringmass. See Appendix 9 for a discussion ofcorrecting for the buoyancy of air.5B Calibrating the Signal (S total )The accuracy of our determination of k A and S reag depends on how accuratelywe can measure the signal, S total . We measure signals using equipment, suchas glassware and balances, and instrumentation, such as spectrophotometersand pH meters. To minimize determinate errors affecting the signal,we first calibrate our equipment and instrumentation. We accomplish thecalibration by measuring S total for a standard with a known response of S std ,adjusting S total untilStotal= SHere are two examples of how we calibrate signals. Other examples areprovided in later chapters focusing on specific analytical methods.When the signal is a measurement of mass, we determine S total usingan analytical balance. To calibrate the balance’s signal we use a referenceweight that meets standards established by a governing agency, such as theNational Institute for Standards and Technology or the American Societyfor Testing and Materials. An electronic balance often includes an internalcalibration weight for routine calibrations, as well as programs for calibratingwith external weights. In either case, the balance automatically adjustsS total to match S std .We also must calibrate our instruments. For example, we can evaluatea spectrophotometer’s accuracy by measuring the absorbance of a carefullyprepared solution of 60.06 mg/L K 2 Cr 2 O 7 in 0.0050 M H 2 SO 4 , using0.0050 M H 2 SO 4 as a reagent blank. 4 An absorbance of 0.640 ± 0.010absorbance units at a wavelength of 350.0 nm indicates that the spectrometer’ssignal is properly calibrated. Be sure to read and carefully follow thecalibration instructions provided with any instrument you use.std5C Determining the Sensitivity (k A )To standardize an analytical method we also must determine the value ofk A in equation 5.1 or equation 5.2.S = k n + Stotal A A reag5.1S = k C + Stotal A A reag5.2In principle, it should be possible to derive the value of k A for any analyticalmethod by considering the chemical and physical processes generatingthe signal. Unfortunately, such calculations are not feasible when we lacka sufficiently developed theoretical model of the physical processes, or arenot useful because of nonideal chemical behavior. In such situations wemust determine the value of k A by analyzing one or more standard solutions,each containing a known amount of analyte. In this section we consider4 Ebel, S. Fresenius J. Anal. Chem. 1992, 342, 769.

Chapter 5 Standardizing Analytical Methods157several approaches for determining the value of k A . For simplicity we willassume that S reag has been accounted for by a proper reagent blank, allowingus to replace S total in equation 5.1 and equation 5.2 with the analyte’ssignal, S A .SS= k n5.3A A A= k C5.4A A A5C.1 Single-Point versus Multiple-Point StandardizationsThe simplest way to determine the value of k A in equation 5.4 is by a single-pointstandardization in which we measure the signal for a standard,S std , containing a known concentration of analyte, C std . Substituting thesevalues into equation 5.4kASstd= 5.5Cgives the value for k A . Having determined the value for k A , we can calculatethe concentration of analyte in any sample by measuring its signal, S samp ,and calculating C A using equation 5.6.CAstdSsamp= 5.6kA single-point standardization is the least desirable method for standardizinga method. There are at least two reasons for this. First, any errorin our determination of k A carries over into our calculation of C A . Second,our experimental value for k A is for a single concentration of analyte. Extendingthis value of k A to other concentrations of analyte requires us toassume a linear relationship between the signal and the analyte’s concentration,an assumption that often is not true. 5 Figure 5.2 shows how assuminga constant value of k A may lead to a determinate error in the analyte’sconcentration. Despite these limitations, single-point standardizations findroutine use when the expected range for the analyte’s concentrations issmall. Under these conditions it is often safe to assume that k A is constant(although you should verify this assumption experimentally). This is thecase, for example, in clinical labs where many automated analyzers use onlya single standard.The preferred approach to standardizing a method is to prepare a seriesof standards, each containing the analyte at a different concentration.Standards are chosen such that they bracket the expected range for the analyte’sconcentration. A multiple-point standardization should includeat least three standards, although more are preferable. A plot of S std versusAEquation 5.3 and equation 5.4 are essentiallyidentical, differing only in whetherwe choose to express the amount of analytein moles or as a concentration. For theremainder of this chapter we will limit ourtreatment to equation 5.4. You can extendthis treatment to equation 5.3 by replacingC A with n A .Linear regression, which also is known asthe method of least squares, is one such algorithm.Its use is covered in Section 5D.5 Cardone, M. J.; Palmero, P. J.; Sybrandt, L. B. Anal. Chem. 1980, 52, 1187–1191.

158 Analytical Chemistry 2.0assumed relationshipactual relationshipS sampFigure 5.2 Example showing how a single-pointstandardization leads to a determinate error in ananalyte’s reported concentration if we incorrectlyassume that the value of k A is constant.S stdC std(C A ) reported(C A ) actualC std is known as a calibration curve. The exact standardization, or calibrationrelationship is determined by an appropriate curve-fitting algorithm.There are at least two advantages to a multiple-point standardization.First, although a determinate error in one standard introduces a determinateerror into the analysis, its effect is minimized by the remaining standards.Second, by measuring the signal for several concentrations of analyte weno longer must assume that the value of k A is independent of the analyte’sconcentration. Constructing a calibration curve similar to the “actual relationship”in Figure 5.2, is possible.Appending the adjective “external” to thenoun “standard” might strike you as oddat this point, as it seems reasonable to assumethat standards and samples mustbe analyzed separately. As you will soonlearn, however, we can add standards toour samples and analyze them simultaneously.5C.2 External StandardsThe most common method of standardization uses one or more externalstandards, each containing a known concentration of analyte. We callthem “external” because we prepare and analyze the standards separate fromthe samples.Si n g l e Ex t e r n a l St a n d a r dA quantitative determination using a single external standard was describedat the beginning of this section, with k A given by equation 5.5. After determiningthe value of k A , the concentration of analyte, C A , is calculatedusing equation 5.6.Example 5.1A spectrophotometric method for the quantitative analysis of Pb 2+ inblood yields an S std of 0.474 for a single standard whose concentration oflead is 1.75 ppb What is the concentration of Pb 2+ in a sample of bloodfor which S samp is 0.361?

Chapter 5 Standardizing Analytical Methods159So l u t i o nEquation 5.5 allows us to calculate the value of k A for this method usingthe data for the standard.kASstd0.474= = = 0.2709 ppbC 175 . ppbstdHaving determined the value of k A , the concentration of Pb 2+ in the sampleof blood is calculated using equation 5.6.CASsamp0.361= = = 133 . ppb-1k 0.2709 ppbA-1Mu l t i p l e Ex t e r n a l St a n d a r d sFigure 5.3 shows a typical multiple-point external standardization. Thevolumetric flask on the left is a reagent blank and the remaining volumetricflasks contain increasing concentrations of Cu 2+ . Shown below thevolumetric flasks is the resulting calibration curve. Because this is the mostcommon method of standardization the resulting relationship is called anormal calibration curve.When a calibration curve is a straight-line, as it is in Figure 5.3, theslope of the line gives the value of k A . This is the most desirable situationsince the method’s sensitivity remains constant throughout the analyte’sconcentration range. When the calibration curve is not a straight-line, the0.250.20S std0.150.100.0500 0.0020 0.0040 0.0060 0.0080C std(M)Figure 5.3 Shown at the top is a reagentblank (far left) and a set of five externalstandards for Cu 2+ with concentrationsincreasing from left to right.Shown below the external standards isthe resulting normal calibration curve.The absorbance of each standard, S std ,is shown by the filled circles.

160 Analytical Chemistry 2.0method’s sensitivity is a function of the analyte’s concentration. In Figure5.2, for example, the value of k A is greatest when the analyte’s concentrationis small and decreases continuously for higher concentrations of analyte.The value of k A at any point along the calibration curve in Figure 5.2 is givenby the slope at that point. In either case, the calibration curve provides ameans for relating S samp to the analyte’s concentration.Example 5.2A second spectrophotometric method for the quantitative analysis of Pb 2+in blood has a normal calibration curve for whichSstd-1= ( 0. 296 ppb )× C + 0.003What is the concentration of Pb 2+ in a sample of blood if S samp is 0.397?So l u t i o nTo determine the concentration of Pb 2+ in the sample of blood we replaceS std in the calibration equation with S samp and solve for C A .CAS −0.003samp 0. 397 −0.003=== 133 . ppb-1-10.296 ppb 0. 296 ppbIt is worth noting that the calibration equation in this problem includesan extra term that does not appear in equation 5.6. Ideally we expectthe calibration curve to have a signal of zero when C A is zero. This is thepurpose of using a reagent blank to correct the measured signal. The extraterm of +0.003 in our calibration equation results from the uncertainty inmeasuring the signal for the reagent blank and the standards.stdThe one-point standardization in this exerciseuses data from the third volumetricflask in Figure 5.3.Practice Exercise 5.1Figure 5.3 shows a normal calibration curve for the quantitative analysisof Cu 2+ . The equation for the calibration curve isS std = 29.59 M –1 × C std + 0.0015What is the concentration of Cu 2+ in a sample whose absorbance, S samp ,is 0.114? Compare your answer to a one-point standardization where astandard of 3.16 × 10 –3 M Cu 2+ gives a signal of 0.0931.Click here to review your answer to this exercise.An external standardization allows us to analyze a series of samplesusing a single calibration curve. This is an important advantage when wehave many samples to analyze. Not surprisingly, many of the most commonquantitative analytical methods use an external standardization.There is a serious limitation, however, to an external standardization.When we determine the value of k A using equation 5.5, the analyte is pres-

Chapter 5 Standardizing Analytical Methods161S sampstandard’smatrixsample’smatrix(C A ) reported(C A ) actualFigure 5.4 Calibration curves for an analyte in thestandard’s matrix and in the sample’s matrix. If thematrix affects the value of k A , as is the case here, thenwe introduce a determinate error into our analysis ifwe use a normal calibration curve.ent in the external standard’s matrix, which usually is a much simpler matrixthan that of our samples. When using an external standardization weassume that the matrix does not affect the value of k A . If this is not true,then we introduce a proportional determinate error into our analysis. Thisis not the case in Figure 5.4, for instance, where we show calibration curvesfor the analyte in the sample’s matrix and in the standard’s matrix. In thisexample, a calibration curve using external standards results in a negativedeterminate error. If we expect that matrix effects are important, then wetry to match the standard’s matrix to that of the sample. This is known asmatrix matching. If we are unsure of the sample’s matrix, then we mustshow that matrix effects are negligible, or use an alternative method of standardization.Both approaches are discussed in the following section.The matrix for the external standards inFigure 5.3, for example, is dilute ammonia,which is added because the Cu(NH 3 ) 42+complex absorbs more strongly thanCu 2+ . If we fail to add the same amountof ammonia to our samples, then we willintroduce a proportional determinate errorinto our analysis.5C.3 Standard AdditionsWe can avoid the complication of matching the matrix of the standards tothe matrix of the sample by conducting the standardization in the sample.This is known as the method of standard additions.Si n g l e St a n d a r d Ad d i t i o nThe simplest version of a standard addition is shown in Figure 5.5. First weadd a portion of the sample, V o , to a volumetric flask, dilute it to volume,V f , and measure its signal, S samp . Next, we add a second identical portionof sample to an equivalent volumetric flask along with a spike, V std , of anexternal standard whose concentration is C std . After diluting the spikedsample to the same final volume, we measure its signal, S spike . The followingtwo equations relate S samp and S spike to the concentration of analyte, C A , inthe original sample.

162 Analytical Chemistry 2.0add V o of C Aadd V std of C stdFigure 5.5 Illustration showing the method of standardadditions. The volumetric flask on the left containsa portion of the sample, V o , and the volumetricflask on the right contains an identical portion of thesample and a spike, V std , of a standard solution of theanalyte. Both flasks are diluted to the same final volume,V f . The concentration of analyte in each flask isshown at the bottom of the figure where C A is the analyte’sconcentration in the original sample and C std isthe concentration of analyte in the external standard.Concentrationof AnalyteCAdilute to V fVo× CVfAV Vo× + C ×stdV VfstdfThe ratios V o /V f and V std /V f account forthe dilution of the sample and the standard,respectively.Sk C V o= 5.7Vsamp A A⎛S k C V o= + Cspike A A⎝⎜VAs long as V std is small relative to V o , the effect of the standard’s matrix onthe sample’s matrix is insignificant. Under these conditions the value of k Ais the same in equation 5.7 and equation 5.8. Solving both equations fork A and equating givesSsampC V AVof=C V AVfofSspikef+ Cwhich we can solve for the concentration of analyte, C A , in the originalsample.Example 5.3A third spectrophotometric method for the quantitative analysis of Pb 2+ inblood yields an S samp of 0.193 when a 1.00 mL sample of blood is dilutedto 5.00 mL. A second 1.00 mL sample of blood is spiked with 1.00 mL ofa 1560-ppb Pb 2+ external standard and diluted to 5.00 mL, yielding anstdstdV VV Vstdfstdf⎞⎠⎟5.85.9

Chapter 5 Standardizing Analytical Methods163S spike of 0.419. What is the concentration of Pb 2+ in the original sampleof blood?So l u t i o nWe begin by making appropriate substitutions into equation 5.9 and solvingfor C A . Note that all volumes must be in the same units; thus, we firstcovert V std from 1.00 mL to 1.00 × 10 –3 mL.CA0.193=⎛100. mL ⎞C⎝⎜500 . mL⎠⎟A0.419⎛10. 0 mL ⎞100 . 10 3 m1560 ppb⎝⎜500 . mL⎠⎟ + ⎛ × − L⎞⎝⎜500 . mL ⎠⎟0.1930.419=0.200C0. 200C+ 0.3120 ppbA A0.0386C A + 0.0602 ppb = 0.0838C A0.0452C A = 0.0602 ppbC A = 1.33 ppbThe concentration of Pb 2+ in the original sample of blood is 1.33 ppb.It also is possible to make a standard addition directly to the sample,measuring the signal both before and after the spike (Figure 5.6). In thiscase the final volume after the standard addition is V o + V std and equation5.7, equation 5.8, and equation 5.9 becomeadd V std of C stdV oV oConcentrationof AnalyteC ACAVoVo+ Vstd+ CstdVoVstd+ VstdFigure 5.6 Illustration showing an alternative form of the method of standard additions.In this case we add a spike of the external standard directly to the samplewithout any further adjust in the volume.

164 Analytical Chemistry 2.0S= k Csamp A A⎛S = k C⎝⎜spike A AVoVo+ Vstd+ CstdVoVstd+ Vstd⎞⎠⎟5.10SsampCA=CAVoVo+ VstdSspike+ CstdVoVstd+ Vstd5.11Example 5.4A fourth spectrophotometric method for the quantitative analysis of Pb 2+in blood yields an S samp of 0.712 for a 5.00 mL sample of blood. After spikingthe blood sample with 5.00 mL of a 1560-ppb Pb 2+ external standard,an S spike of 1.546 is measured. What is the concentration of Pb 2+ in theoriginal sample of blood.V o + V std = 5.00 mL + 5.00×10 –3 mL= 5.005 mLSo l u t i o nTo determine the concentration of Pb 2+ in the original sample of blood, wemake appropriate substitutions into equation 5.11 and solve for C A .0. 712 1.546=CA ⎛ 500 . mL ⎞C1A⎝⎜5.005 mL⎠⎟ + 560⎛−3500 . × 10 mL⎞ppb⎝⎜5.005 mL ⎠⎟0. 712 1.546=C 0. 9990C+ 1.558 ppbA A0.7113C A + 1.109 ppb = 1.546C AC A = 1.33 ppbThe concentration of Pb 2+ in the original sample of blood is 1.33 ppb.Mu l t i p l e St a n d a r d Ad d i t i o n sWe can adapt the single-point standard addition into a multiple-pointstandard addition by preparing a series of samples containing increasingamounts of the external standard. Figure 5.7 shows two ways to plot astandard addition calibration curve based on equation 5.8. In Figure 5.7awe plot S spike against the volume of the spikes, V std . If k A is constant, thenthe calibration curve is a straight-line. It is easy to show that the x-interceptis equivalent to –C A V o /C std .

Chapter 5 Standardizing Analytical Methods165(a)0.60S spike6.00C std0.50y-intercept = k AC A V oV f0.400.30slope = k AC stdV f0.200.100-2.00 0 2.00 4.00V stdx-intercept = -C (mL)AV oC std(b)S spike0.600.500.400.30y-intercept = k AC A V oV fslope = k A0.200.100-4.00 -2.00 0 2.00 4.00 6.00 8.00 10.00 12.00V stdx-intercept = -C AV oV f×Vf(mg/L)Figure 5.7 Shown at the top is a set ofsix standard additions for the determinationof Mn 2+ . The flask on the leftis a 25.00 mL sample diluted to 50.00mL. The remaining flasks contain 25.00mL of sample and, from left to right,1.00, 2.00, 3.00, 4.00, and 5.00 mLof an external standard of 100.6 mg/LMn 2+ . Shown below are two ways toplot the standard additions calibrationcurve. The absorbance for each standardaddition, S spike , is shown by thefilled circles.Example 5.5Beginning with equation 5.8 show that the equations in Figure 5.7a forthe slope, the y-intercept, and the x-intercept are correct.So l u t i o nWe begin by rewriting equation 5.8 asSspikekCV kCA A o A std= + × VV Vwhich is in the form of the equation for a straight-linefY = y-intercept + slope × Xfstd

166 Analytical Chemistry 2.0where Y is S spike and X is V std . The slope of the line, therefore, is k A C std /V fand the y-intercept is k A C A V o /V f . The x-intercept is the value of X whenY is zero, orkCV kCA A o A std0 = + × x-interceptV VffkCVA A oV CVfAx-intercept=− =−kCA stdCstdVfoPractice Exercise 5.2Beginning with equation 5.8 show that the equations in Figure 5.7b forthe slope, the y-intercept, and the x-intercept are correct.Click here to review your answer to this exercise.Because we know the volume of the original sample, V o , and the concentrationof the external standard, C std , we can calculate the analyte’s concentrationsfrom the x-intercept of a multiple-point standard additions.Example 5.6A fifth spectrophotometric method for the quantitative analysis of Pb 2+in blood uses a multiple-point standard addition based on equation 5.8.The original blood sample has a volume of 1.00 mL and the standard usedfor spiking the sample has a concentration of 1560 ppb Pb 2+ . All sampleswere diluted to 5.00 mL before measuring the signal. A calibration curveof S spike versus V std has the following equationS spike = 0.266 + 312 mL –1 × V stdWhat is the concentration of Pb 2+ in the original sample of blood.So l u t i o nTo find the x-intercept we set S spike equal to zero.0 = 0.266 + 312 mL –1 × V stdSolving for V std , we obtain a value of –8.526 × 10 –4 mL for the x-intercept.Substituting the x-interecpt’s value into the equation from Figure 5.7a− × =− =− ×−4CV C 100 . mLA oA8.526 10 mLC 1560 ppband solving for C A gives the concentration of Pb 2+ in the blood sampleas 1.33 ppb.std

Chapter 5 Standardizing Analytical Methods167Practice Exercise 5.3Figure 5.7 shows a standard additions calibration curve for the quantitativeanalysis of Mn 2+ . Each solution contains 25.00 mL of the originalsample and either 0, 1.00, 2.00, 3.00, 4.00, or 5.00 mL of a 100.6 mg/Lexternal standard of Mn 2+ . All standard addition samples were diluted to50.00 mL before reading the absorbance. The equation for the calibrationcurve in Figure 5.7a isSince we construct a standard additions calibration curve in the sample,we can not use the calibration equation for other samples. Each sample,therefore, requires its own standard additions calibration curve. This is aserious drawback if you have many samples. For example, suppose you needto analyze 10 samples using a three-point calibration curve. For a normalcalibration curve you need to analyze only 13 solutions (three standardsand ten samples). If you use the method of standard additions, however,you must analyze 30 solutions (each of the ten samples must be analyzedthree times, once before spiking and after each of two spikes).Us i n g a St a n d a r d Ad d i t i o n t o Id e n t i f y Ma t r i x Ef f e c t sWe can use the method of standard additions to validate an external standardizationwhen matrix matching is not feasible. First, we prepare a normalcalibration curve of S std versus C std and determine the value of k A fromits slope. Next, we prepare a standard additions calibration curve usingequation 5.8, plotting the data as shown in Figure 5.7b. The slope of thisstandard additions calibration curve provides an independent determinationof k A . If there is no significant difference between the two values ofk A , then we can ignore the difference between the sample’s matrix and thatof the external standards. When the values of k A are significantly different,then using a normal calibration curve introduces a proportional determinateerror.5C.4 Internal StandardsS std = 0.0854 × V std + 0.1478What is the concentration of Mn 2+ in this sample? Compare your answerto the data in Figure 5.7b, for which the calibration curve isS std = 0.0425 × C std (V std /V f ) + 0.1478Click here to review your answer to this exercise.To successfully use an external standardization or the method of standardadditions, we must be able to treat identically all samples and standards.When this is not possible, the accuracy and precision of our standardizationmay suffer. For example, if our analyte is in a volatile solvent, then itsconcentration increases when we lose solvent to evaporation. Suppose we

168 Analytical Chemistry 2.0have a sample and a standard with identical concentrations of analyte andidentical signals. If both experience the same proportional loss of solventthen their respective concentrations of analyte and signals continue to beidentical. In effect, we can ignore evaporation if the samples and standardsexperience an equivalent loss of solvent. If an identical standard and samplelose different amounts of solvent, however, then their respective concentrationsand signals will no longer be equal. In this case a simple externalstandardization or standard addition is not possible.We can still complete a standardization if we reference the analyte’ssignal to a signal from another species that we add to all samples and standards.The species, which we call an internal standard, must be differentthan the analyte.Because the analyte and the internal standard in any sample or standardreceive the same treatment, the ratio of their signals is unaffected by anylack of reproducibility in the procedure. If a solution contains an analyte ofconcentration C A , and an internal standard of concentration, C IS , then thesignals due to the analyte, S A , and the internal standard, S IS , areS= k CA A AS= k CIS IS ISwhere k A and k IS are the sensitivities for the analyte and internal standard.Taking the ratio of the two signals gives the fundamental equation for aninternal standardization.SSAISkCKC A AA= = × 5.12kC CISBecause K is a ratio of the analyte’s sensitivity and the internal standard’ssensitivity, it is not necessary to independently determine values for eitherk A or k IS .Si n g l e In t e r n a l St a n d a r dIn a single-point internal standardization, we prepare a single standard containingthe analyte and the internal standard, and use it to determine thevalue of K in equation 5.12.C SK = ⎛ ⎞× ⎛ ⎞ISA5.13⎝⎜C ⎠⎟⎝⎜S ⎠⎟A std IS stdHaving standardized the method, the analyte’s concentration is given byC SIS AC = × ⎛ ⎞AK ⎝⎜S ⎠⎟ISISsampIS

Chapter 5 Standardizing Analytical Methods169Example 5.7A sixth spectrophotometric method for the quantitative analysis of Pb 2+ inblood uses Cu 2+ as an internal standard. A standard containing 1.75 ppbPb 2+ and 2.25 ppb Cu 2+ yields a ratio of (S A /S IS ) std of 2.37. A sample ofblood is spiked with the same concentration of Cu 2+ , giving a signal ratio,(S A /S IS ) samp , of 1.80. Determine the concentration of Pb 2+ in the sampleof blood.So l u t i o nEquation 5.13 allows us to calculate the value of K using the data for thestandardKC= ⎛ ⎝⎜CISA⎞⎠⎟stdS× ⎛ ⎝⎜SThe concentration of Pb 2+ , therefore, isC SIS Appb CuC = × ⎛ ⎞= 225AK ⎝⎜SIS⎠⎟ppb Cusamp 305 .ppb PbAIS⎞⎠⎟std225 . ppb Cuppb Cu= × 237 . = 3.052+175 . ppb Pbppb Pb. 2+2+2+× 180 . = 1.33 ppb Cu2+2+2+Mu l t i p l e In t e r n a l St a n d a r d sA single-point internal standardization has the same limitations as a singlepointnormal calibration. To construct an internal standard calibrationcurve we prepare a series of standards, each containing the same concentrationof internal standard and a different concentrations of analyte. Underthese conditions a calibration curve of (S A /S IS ) std versus C A is linear witha slope of K/C IS .Example 5.8A seventh spectrophotometric method for the quantitative analysis of Pb 2+in blood gives a linear internal standards calibration curve for which⎛S⎝⎜SAIS⎞⎠⎟std-1= ( 211 . ppb ) × C −0.006AAlthough the usual practice is to preparethe standards so that each contains anidentical amount of the internal standard,this is not a requirement.What is the ppb Pb 2+ in a sample of blood if (S A /S IS ) samp is 2.80?So l u t i o nTo determine the concentration of Pb 2+ in the sample of blood we replace(S A /S IS ) std in the calibration equation with (S A /S IS ) samp and solve for C A .

170 Analytical Chemistry 2.0CA⎛S⎞A+ 0.006⎝⎜SIS⎠⎟samp=-12.11 ppb2. 80 + 0.006== 133 . ppb-1211 . ppbThe concentration of Pb 2+ in the sample of blood is 1.33 ppb.In some circumstances it is not possible to prepare the standards so thateach contains the same concentration of internal standard. This is the case,for example, when preparing samples by mass instead of volume. We canstill prepare a calibration curve, however, by plotting (S A /S IS ) std versus C A /C IS , giving a linear calibration curve with a slope of K.5DLinear Regression and Calibration CurvesIn a single-point external standardization we determine the value of k A bymeasuring the signal for a single standard containing a known concentrationof analyte. Using this value of k A and the signal for our sample, wethen calculate the concentration of analyte in our sample (see Example5.1). With only a single determination of k A , a quantitative analysis usinga single-point external standardization is straightforward.A multiple-point standardization presents a more difficult problem.Consider the data in Table 5.1 for a multiple-point external standardization.What is our best estimate of the relationship between S std and C std ?It is tempting to treat this data as five separate single-point standardizations,determining k A for each standard, and reporting the mean value.Despite it simplicity, this is not an appropriate way to treat a multiple-pointstandardization.So why is it inappropriate to calculate an average value for k A as donein Table 5.1? In a single-point standardization we assume that our reagentblank (the first row in Table 5.1) corrects for all constant sources of determinateerror. If this is not the case, then the value of k A from a single-pointstandardization has a determinate error. Table 5.2 demonstrates how anTable 5.1 Data for a Hypothetical Multiple-Point ExternalStandardizationC std (arbitrary units) S std (arbitrary units) k A = S std / C std0.000 0.00 —0.100 12.36 123.60.200 24.83 124.20.300 35.91 119.70.400 48.79 122.00.500 60.42 122.8mean value for k A = 122.5

172 Analytical Chemistry 2.05D.1 Linear Regression of Straight Line Calibration CurvesWhen a calibration curve is a straight-line, we represent it using the followingmathematical equationy = β + β x0 15.14where y is the signal, S std , and x is the analyte’s concentration, C std . Theconstants b 0 and b 1 are, respectively, the calibration curve’s expected y-interceptand its expected slope. Because of uncertainty in our measurements,the best we can do is to estimate values for b 0 and b 1 , which we representas b 0 and b 1 . The goal of a linear regression analysis is to determine thebest estimates for b 0 and b 1 . How we do this depends on the uncertaintyin our measurements.5D.2 Unweighted Linear Regression with Errors in yThe most common approach to completing a linear regression for equation5.14 makes three assumptions:(1) that any difference between our experimental data and the calculatedregression line is the result of indeterminate errors affecting y,(2) that indeterminate errors affecting y are normally distributed, and(3) that the indeterminate errors in y are independent of the value of x.Because we assume that the indeterminate errors are the same for all standards,each standard contributes equally in estimating the slope and they-intercept. For this reason the result is considered an unweighted linearregression.The second assumption is generally true because of the central limit theorem,which we considered in Chapter 4. The validity of the two remainingassumptions is less obvious and you should evaluate them before acceptingthe results of a linear regression. In particular the first assumption is alwayssuspect since there will certainly be some indeterminate errors affecting thevalues of x. When preparing a calibration curve, however, it is not unusualfor the uncertainty in the signal, S std , to be significantly larger than that forthe concentration of analyte in the standards C std . In such circumstancesthe first assumption is usually reasonable.Ho w a Li n e a r Re g r e s s i o n Wo r k sTo understand the logic of an linear regression consider the example shownin Figure 5.9, which shows three data points and two possible straight-linesthat might reasonably explain the data. How do we decide how well thesestraight-lines fits the data, and how do we determine the best straightline?Let’s focus on the solid line in Figure 5.9. The equation for this line isŷ b bx = + 0 15.15

Chapter 5 Standardizing Analytical Methods173Figure 5.9 Illustration showing three data points and twopossible straight-lines that might explain the data. The goalof a linear regression is to find the mathematical model, inthis case a straight-line, that best explains the data.where b 0 and b 1 are our estimates for the y-intercept and the slope, and ŷ isour prediction for the experimental value of y for any value of x. Because weassume that all uncertainty is the result of indeterminate errors affecting y,the difference between y and ŷ for each data point is the residual error,r, in the our mathematical model for a particular value of x.r = ( y − yˆ )i i iFigure 5.10 shows the residual errors for the three data points. The smallerthe total residual error, R, which we define asR = ∑( y − yˆ ) 2 i i5.16ithe better the fit between the straight-line and the data. In a linear regressionanalysis, we seek values of b 0 and b 1 that give the smallest total residualerror.ŷ 3r = ( y − yˆ )2 2 2ŷ ŷ 2 1y 2y 3r = ( y − yˆ )1 1 1y 1ŷ = b 0+ bx 1r = ( y − yˆ )3 3 3If you are reading this aloud, you pronounceŷ as y-hat.The reason for squaring the individual residualerrors is to prevent positive residualerror from canceling out negative residualerrors. You have seen this before in theequations for the sample and populationstandard deviations. You also can seefrom this equation why a linear regressionis sometimes called the method of leastsquares.Figure 5.10 Illustration showing the evaluation of a linear regression in which we assume that all uncertaintyis the result of indeterminate errors affecting y. The points in blue, y i , are the original data and thepoints in red, ŷ i, are the predicted values from the regression equation, ŷ = b 0+ bx 1.The smaller thetotal residual error (equation 5.16), the better the fit of the straight-line to the data.

174 Analytical Chemistry 2.0Fi n d i n g t h e Sl o p e a n d y-In t e r ce p tAlthough we will not formally develop the mathematical equations for alinear regression analysis, you can find the derivations in many standardstatistical texts. 6 The resulting equation for the slope, b 1 , isb1∑n x y − x yi iiii i=22n x − ⎛ ⎞∑xii⎝⎜∑⎠⎟and the equation for the y-intercept, b 0 , isi∑∑ ∑ii5.17y −b xi 1 iiib =5.180nAlthough equation 5.17 and equation 5.18 appear formidable, it is onlynecessary to evaluate the following four summations∑ix i∑iy i∑i∑xy i i∑i2x iSee Section 5F in this chapter for detailson completing a linear regression analysisusing Excel and R.Equations 5.17 and 5.18 are written interms of the general variables x and y. Asyou work through this example, rememberthat x corresponds to C std , and that ycorresponds to S std .Many calculators, spreadsheets, and other statistical software packages arecapable of performing a linear regression analysis based on this model. Tosave time and to avoid tedious calculations, learn how to use one of thesetools. For illustrative purposes the necessary calculations are shown in detailin the following example.Example 5.9Using the data from Table 5.1, determine the relationship between S stdand C std using an unweighted linear regression.So l u t i o nWe begin by setting up a table to help us organize the calculation.x i y i x i y i2x i0.000 0.00 0.000 0.0000.100 12.36 1.236 0.0100.200 24.83 4.966 0.0400.300 35.91 10.773 0.0900.400 48.79 19.516 0.1600.500 60.42 30.210 0.250Adding the values in each column gives2∑ x i= 1.500 ∑ y i= 182.31 ∑ xy i i= 66.701 ∑ x i= 0.550iiii6 See, for example, Draper, N. R.; Smith, H. Applied Regression Analysis, 3rd ed.; Wiley: NewYork, 1998.

Chapter 5 Standardizing Analytical Methods175Substituting these values into equation 5.17 and equation 5.18, we findthat the slope and the y-intercept are( 6 66 701 1 500 182 31b 1= × . )−( . × . )= 120. 706 ≈120.712( 6×0. 550)−( 1.500)( ) = ≈182. 31− 120. 706×1.500b 0=60. 209 021 .The relationship between the signal and the analyte, therefore, isS std = 120.71 × C std + 0.21For now we keep two decimal places to match the number of decimal placesin the signal. The resulting calibration curve is shown in Figure 5.11.Un c e r t a i n t y in t h e Re g r e s s i o n An a l y s i sAs shown in Figure 5.11, because of indeterminate error affecting our signal,the regression line may not pass through the exact center of each data point.The cumulative deviation of our data from the regression line—that is, thetotal residual error—is proportional to the uncertainty in the regression.We call this uncertainty the standard deviation about the regression,s r , which is equal tosr =∑( y − y ˆ i i)2in − 25.19Did you notice the similarity between thestandard deviation about the regression(equation 5.19) and the standard deviationfor a sample (equation 4.1)?605040S std 30201000.0 0.1 0.2 0.3 0.4 0.5C stdFigure 5.11 Calibration curve for the data in Table 5.1 and Example 5.9.

176 Analytical Chemistry 2.0where y i is the i th experimental value, and ŷ iis the corresponding value predictedby the regression line in equation 5.15. Note that the denominatorof equation 5.19 indicates that our regression analysis has n–2 degrees offreedom—we lose two degree of freedom because we use two parameters,the slope and the y-intercept, to calculate ŷ i.A more useful representation of the uncertainty in our regression isto consider the effect of indeterminate errors on the slope, b 1 , and the y-intercept, b 0 , which we express as standard deviations.sb1=ns2n x − ⎛ ⎞∑xii⎝⎜∑⎠⎟i2ri2=2sr2∑( x − xi ) 5.20isb0=sx2 2r ii2n x − ⎛ ⎞∑xii⎝⎜∑⎠⎟i∑i2=sr∑x2 2ii∑( −i )n x xi2 5.21We use these standard deviations to establish confidence intervals for theexpected slope, b 1 , and the expected y-intercept, b 0β 1= b 1± ts b 5.221β 0= b 0± ts b 5.230You might contrast this with equation4.12 for the confidence interval around asample’s mean value.As you work through this example, rememberthat x corresponds to C std , andthat y corresponds to S std .where we select t for a significance level of a and for n–2 degrees of freedom.Note that equation 5.22 and equation 5.23 do not contain a factor of( n)−1 because the confidence interval is based on a single regression line.Again, many calculators, spreadsheets, and computer software packagesprovide the standard deviations and confidence intervals for the slope andy-intercept. Example 5.10 illustrates the calculations.Example 5.10Calculate the 95% confidence intervals for the slope and y-intercept fromExample 5.9.So l u t i o nWe begin by calculating the standard deviation about the regression. To dothis we must calculate the predicted signals, ŷ i, using the slope and y‐interceptfrom Example 5.9, and the squares of the residual error, ( y − yˆ )2 .i iUsing the last standard as an example, we find that the predicted signal isyˆ = b + bx = 0. 209 + 120. 706×0. 500 60.5626 0 1 6 ( )=and that the square of the residual error is

Chapter 5 Standardizing Analytical Methods1772 2( y − yˆ ) = ( 60. 42 − 60. 562) = 0. 2016 ≈ 0.202iiThe following table displays the results for all six solutions.x i y iŷ i( y yˆ )− 2i i0.000 0.00 0.209 0.04370.100 12.36 12.280 0.00640.200 24.83 24.350 0.23040.300 35.91 36.421 0.26110.400 48.79 48.491 0.08940.500 60.42 60.562 0.0202Adding together the data in the last column gives the numerator of equation5.19 as 0.6512. The standard deviation about the regression, therefore,iss r=0.65126− 2= 0.4035Next we calculate the standard deviations for the slope and the y-interceptusing equation 5.20 and equation 5.21. The values for the summationterms are from in Example 5.9.sb1=ns2n x − ⎛ ⎞∑xii⎝⎜∑⎠⎟i2ri2=26×( 04035 . )2( 6×0 550)−( 1 550) =. .0.965sb0=2 2s x2r i( 0. 4035) × 0.550i22n x − ⎛ =⎞∑6×0. 550 1.550xii⎝⎜∑⎠⎟i∑i( )−( )2Finally, the 95% confidence intervals (a = 0.05, 4 degrees of freedom) forthe slope and y-intercept areYou can find values for t in Appendix 4.= b ± ts b= 120. 706 ± ( 278 . × 0. 965)= 120. 7±2.7β 1 1 1= b ± ts b= 0. 209 ± ( 278 . × 0. 292)= 02 . ± 08 .β 0 0 0The standard deviation about the regression, s r , suggests that the signal, S std ,is precise to one decimal place. For this reason we report the slope and they-intercept to a single decimal place.

178 Analytical Chemistry 2.0Minimizing Un c e r t a i n t y in Calibration Cu r ve sTo minimize the uncertainty in a calibration curve’s slope and y-intercept,you should evenly space your standards over a wide range of analyte concentrations.A close examination of equation 5.20 and equation 5.21 willhelp you appreciate why this is true. The denominators of both equationsinclude the term ∑( x − x)2 i. The larger the value of this term—whichyou accomplish by increasing the range of x around its mean value—thesmaller the standard deviations in the slope and the y-intercept. Furthermore,to minimize the uncertainty in the y‐intercept, it also helps to decreasethe value of the term ∑ x iin equation 5.21, which you accomplishby including standards for lower concentrations of the analyte.Ob t a i n i n g t h e An a l y t e’s Co n c e n t r a t i o n Fr o m a Re g r e s s i o n Eq u a t i o nEquation 5.25 is written in terms of a calibrationexperiment. A more general formof the equation, written in terms of x andy, is given here.sxs 1 1r= + +b m n12( Y − y)∑2 2( b) ( x − x )A close examination of equation 5.25should convince you that the uncertaintyin C A is smallest when the sample’s averagesignal, Ssamp, is equal to the averagesignal for the standards, Sstd. When practical,you should plan your calibrationcurve so that S samp falls in the middle ofthe calibration curve.1iiOnce we have our regression equation, it is easy to determine the concentrationof analyte in a sample. When using a normal calibration curve, forexample, we measure the signal for our sample, S samp , and calculate theanalyte’s concentration, C A , using the regression equation.CAS=sampb−b015.24What is less obvious is how to report a confidence interval for C A thatexpresses the uncertainty in our analysis. To calculate a confidence intervalwe need to know the standard deviation in the analyte’s concentration, s CA,which is given by the following equationsCAsr1 1= + +b m n1( S −Ssamp std )(2( ) −1 ∑b C Cistdi2std)25.25where m is the number of replicate used to establish the sample’s averagesignal ( S samp), n is the number of calibration standards, S stdis the averagesignal for the calibration standards, and C stdiand C stdare the individual andmean concentrations for the calibration standards. 7 Knowing the value ofs CA, the confidence interval for the analyte’s concentration isµ C= C ± tsA A C Awhere m C A is the expected value of C A in the absence of determinate errors,and with the value of t based on the desired level of confidence and n–2degrees of freedom.7 (a) Miller, J. N. Analyst 1991, 116, 3–14; (b) Sharaf, M. A.; Illman, D. L.; Kowalski, B. R. Chemometrics,Wiley-Interscience: New York, 1986, pp. 126-127; (c) Analytical Methods Committee“Uncertainties in concentrations estimated from calibration experiments,” AMC TechnicalBrief, March 2006 (http://www.rsc.org/images/Brief22_tcm18-51117.pdf)

Chapter 5 Standardizing Analytical Methods179Example 5.11Three replicate analyses for a sample containing an unknown concentrationof analyte, yield values for S samp of 29.32, 29.16 and 29.51. Usingthe results from Example 5.9 and Example 5.10, determine the analyte’sconcentration, C A , and its 95% confidence interval.So l u t i o nThe average signal, S samp, is 29.33, which, using equation 5.24 and theslope and the y-intercept from Example 5.9, gives the analyte’s concentrationasCAS=samp− b029. 33−0.209= = 0.241b 120.7061To calculate the standard deviation for the analyte’s concentration we must∑( )determine the values for S stdand C −Cstdistd. The former is just theaverage signal for the calibration standards, which, using the data in Table∑( )5.1, is 30.385. Calculating C −Cstdistdlooks formidable, but we can simplifyits calculation by recognizing that this sum of squares term is thenumerator in a standard deviation equation; thus,∑22( C −C s nstd std ) = ( C ) × −1i2std2( )where s Cstdis the standard deviation for the concentration of analyte inthe calibration standards. Using the data in Table 5.1 we find that s Cstdis0.1871 and22∑( C −Cstd std ) = ( 0. 1871) × ( 6− 1) = 0.175iSubstituting known values into equation 5.25 gives20.4035 1 1 ( 29. 33−30.385)s CA= + += 0.00242120.706 3 6 120. 706 0.175( ) ×Finally, the 95% confidence interval for 4 degrees of freedom is= ± = 0. 241± ( 278 . × 0. 0024)= 0. 241±0.007µ CC tsA A C AFigure 5.12 shows the calibration curve with curves showing the 95%confidence interval for C A .You can find values for t in Appendix 4.

180 Analytical Chemistry 2.0605040S std 300.0 0.1 0.2 0.3 0.4 0.5Figure 5.12 Example of a normal calibration curve witha superimposed confidence interval for the analyte’s concentration.The points in blue are the original data fromTable 5.1. The black line is the normal calibration curveas determined in Example 5.9. The red lines show the95% confidence interval for C A assuming a single determinationof S samp .20100C stdPractice Exercise 5.4Figure 5.3 shows a normal calibration curve for the quantitative analysisof Cu 2+ . The data for the calibration curve are shown here.[Cu 2+ ] (M)Absorbance0 01.55×10 –3 0.0503.16×10 –3 0.0934.74×10 –3 0.1436.34×10 –3 0.1887.92×10 –3 0.236Complete a linear regression analysis for this calibration data, reportingthe calibration equation and the 95% confidence interval for the slopeand the y-intercept. If three replicate samples give an S samp of 0.114,what is the concentration of analyte in the sample and its 95% confidenceinterval?Click here to review your answer to this exercise.In a standard addition we determine the analyte’s concentration byextrapolating the calibration curve to the x-intercept. In this case the valueof C A isand the standard deviation in C A isC = bx =− 0-interceptAb1

Chapter 5 Standardizing Analytical Methods181sCAsr1= +b n1( S )std2 21 ∑( stdistd )i( ) −b C C2where n is the number of standard additions (including the sample with noadded standard), and S stdis the average signal for the n standards. Becausewe determine the analyte’s concentration by extrapolation, rather than byinterpolation, s CAfor the method of standard additions generally is largerthan for a normal calibration curve.Ev a l u a t i n g a Li n e a r Re g r e s s i o n Mod e lYou should never accept the result of a linear regression analysis withoutevaluating the validity of the your model. Perhaps the simplest way to evaluatea regression analysis is to examine the residual errors. As we saw earlier,the residual error for a single calibration standard, r i , isr = ( y − yˆ )i i iIf your regression model is valid, then the residual errors should be randomlydistributed about an average residual error of zero, with no apparenttrend toward either smaller or larger residual errors (Figure 5.13a). Trendssuch as those shown in Figure 5.13b and Figure 5.13c provide evidence thatat least one of the model’s assumptions is incorrect. For example, a trendtoward larger residual errors at higher concentrations, as shown in Figure5.13b, suggests that the indeterminate errors affecting the signal are notindependent of the analyte’s concentration. In Figure 5.13c, the residual(a) (b) (c)residual errorresidual errorresidual error0.0 0.1 0.2 0.3 0.4 0.5C std0.0 0.1 0.2 0.3 0.4 0.5C std0.0 0.1 0.2 0.3 0.4 0.5C stdFigure 5.13 Plot of the residual error in the signal, S std , as a function of the concentration of analyte, C std for anunweighted straight-line regression model. The red line shows a residual error of zero. The distribution of the residualerror in (a) indicates that the unweighted linear regression model is appropriate. The increase in the residual errors in(b) for higher concentrations of analyte, suggest that a weighted straight-line regression is more appropriate. For (c), thecurved pattern to the residuals suggests that a straight-line model is inappropriate; linear regression using a quadraticmodel might produce a better fit.

182 Analytical Chemistry 2.0Practice Exercise 5.5Using your results from Practice Exercise 5.4, construct a residual plotand explain its significance.Click here to review your answer to this exercise.errors are not random, suggesting that the data can not be modeled with astraight-line relationship. Regression methods for these two cases are discussedin the following sections.5D.3 Weighted Linear Regression with Errors in yOur treatment of linear regression to this point assumes that indeterminateerrors affecting y are independent of the value of x. If this assumption isfalse, as is the case for the data in Figure 5.13b, then we must include thevariance for each value of y into our determination of the y-intercept, b o ,and the slope, b 1 ; thusb1b0=∑iwy−b w xiin1∑∑ ∑ ∑iii5.26n w xy−wx wyi i ii i i iii i=2n w x − ⎛ 2⎞5.27∑w xi i∑ i i⎝⎜⎠⎟iiThis is the same data used in Example 5.9with additional information about thestandard deviations in the signal.where w i is a weighting factor that accounts for the variance in y ins ( )−2yiw =i−2 5.28s∑i( )and s yiis the standard deviation for y i . In a weighted linear regression,each xy-pair’s contribution to the regression line is inversely proportionalto the precision of y i —that is, the more precise the value of y, the greaterits contribution to the regression.Example 5.12Shown here are data for an external standardization in which s std is thestandard deviation for three replicate determination of the signal.C std (arbitrary units) S std (arbitrary units) s std0.000 0.00 0.020.100 12.36 0.020.200 24.83 0.07yi

Chapter 5 Standardizing Analytical Methods1830.300 35.91 0.130.400 48.79 0.220.500 60.42 0.33Determine the calibration curve’s equation using a weighted linear regression.As you work through this example, rememberthat x corresponds to C std , andthat y corresponds to S std .So l u t i o nWe begin by setting up a table to aid in calculating the weighting factors.x i y is yi ( s yi ) −2 w i0.000 0.00 0.02 2500.00 2.83390.100 12.36 0.02 2500.00 2.83390.200 24.83 0.07 204.08 0.23130.300 35.91 0.13 59.17 0.06710.400 48.79 0.22 20.66 0.02340.500 60.42 0.33 9.18 0.0104Adding together the values in the forth column givesAs a check on your calculations, the sumof the individual weights must equal thenumber of calibration standards, n. Thesum of the entries in the last column is6.0000, so all is well.i( ) =∑ 2 5293.09s y iwhich we use to calculate the individual weights in the last column. Aftercalculating the individual weights, we use a second table to aid in calculatingthe four summation terms in equation 5.26 and equation 5.27.x i y i w i w i x i w i y i w i x 2 i w i x i y i0.000 0.00 2.8339 0.0000 0.0000 0.0000 0.00000.100 12.36 2.8339 0.2834 35.0270 0.0283 3.50270.200 24.83 0.2313 0.0463 5.7432 0.0093 1.14860.300 35.91 0.0671 0.0201 2.4096 0.0060 0.72290.400 48.79 0.0234 0.0094 1.1417 0.0037 0.45670.500 60.42 0.0104 0.0052 0.6284 0.0026 0.3142Adding the values in the last four columns gives∑i∑iwxiwxii2i∑= 0. 3644 wy=44.9499i= 0.0499 ∑ wxy = 6.1451i i iSubstituting these values into the equation 5.26 and equation 5.27 givesthe estimated slope and estimated y-intercept asiii

184 Analytical Chemistry 2.0( 6 6 1451 0 3644 44 9499b 1= × . ) − ( . × . )= 122.9852( 6× 0. 0499) −( 0. 3644)44. 9499− ( 122. 985×0. 3644)b 0== 0.02246The calibration equation isS std = 122.98 × C std + 0.02Figure 5.14 shows the calibration curve for the weighted regression and thecalibration curve for the unweighted regression in Example 5.9. Althoughthe two calibration curves are very similar, there are slight differences in theslope and in the y-intercept. Most notably, the y-intercept for the weightedlinear regression is closer to the expected value of zero. Because the standarddeviation for the signal, S std , is smaller for smaller concentrations ofanalyte, C std , a weighted linear regression gives more emphasis to thesestandards, allowing for a better estimate of the y-intercept.Equations for calculating confidence intervals for the slope, the y-intercept,and the concentration of analyte when using a weighted linearregression are not as easy to define as for an unweighted linear regression. 8The confidence interval for the analyte’s concentration, however, is at its8 Bonate, P. J. Anal. Chem. 1993, 65, 1367–1372.6050weighted linear regressionunweighted linear regression40S std 30201000.0 0.1 0.2 0.3 0.4 0.5C stdFigure 5.14 A comparison of unweighted and weighted normal calibration curves.See Example 5.9 for details of the unweighted linear regression and Example 5.12for details of the weighted linear regression.

Chapter 5 Standardizing Analytical Methods185optimum value when the analyte’s signal is near the weighted centroid, y c, of the calibration curve.y= 1∑wxnc i ii5D.4 Weighted Linear Regression with Errors in Both x and yIf we remove our assumption that the indeterminate errors affecting a calibrationcurve exist only in the signal (y), then we also must factor intothe regression model the indeterminate errors affecting the analyte’s concentrationin the calibration standards (x). The solution for the resultingregression line is computationally more involved than that for either theunweighted or weighted regression lines. 9 Although we will not considerthe details in this textbook, you should be aware that neglecting the presenceof indeterminate errors in x can bias the results of a linear regression.5D.5 Curvilinear and Multivariate RegressionA straight-line regression model, despite its apparent complexity, is thesimplest functional relationship between two variables. What do we do ifour calibration curve is curvilinear—that is, if it is a curved-line instead ofa straight-line? One approach is to try transforming the data into a straightline.Logarithms, exponentials, reciprocals, square roots, and trigonometricfunctions have been used in this way. A plot of log(y) versus x is a typicalexample. Such transformations are not without complications. Perhaps themost obvious complication is that data with a uniform variance in y will notmaintain that uniform variance after the transformation.Another approach to developing a linear regression model is to fit apolynomial equation to the data, such as y = a + bx + cx 2 . You can uselinear regression to calculate the parameters a, b, and c, although the equationsare different than those for the linear regression of a straight line. 10If you cannot fit your data using a single polynomial equation, it may bepossible to fit separate polynomial equations to short segments of the calibrationcurve. The result is a single continuous calibration curve known asa spline function.The regression models in this chapter apply only to functions containinga single independent variable, such as a signal that depends upon the analyte’sconcentration. In the presence of an interferent, however, the signalmay depend on the concentrations of both the analyte and the interferentSee Figure 5.2 for an example of a calibrationcurve that deviates from a straightlinefor higher concentrations of analyte.It is worth noting that in mathematics, theterm “linear” does not mean a straightline.A linear function may contain manyadditive terms, but each term can haveone and only one adjustable parameter.The functiony = ax + bx 2is linear, but the functiony = ax bis nonlinear. This is why you can use linearregression to fit a polynomial equation toyour data.Sometimes it is possible to transform anonlinear function. For example, takingthe log of both sides of the nonlinear functionshown above gives a linear function.log(y) = log(a) + blog(x)9 See, for example, Analytical Methods Committee, “Fitting a linear functional relationship todata with error on both variable,” AMC Technical Brief, March, 2002 (http://www.rsc.org/images/brief10_tcm18-25920.pdf).10 For details about curvilinear regression, see (a) Sharaf, M. A.; Illman, D. L.; Kowalski, B. R.Chemometrics, Wiley-Interscience: New York, 1986; (b) Deming, S. N.; Morgan, S. L. ExperimentalDesign: A Chemometric Approach, Elsevier: Amsterdam, 1987.

186 Analytical Chemistry 2.0Check out the Additional Resources at theend of the textbook for more informationabout linear regression with errors in bothvariables, curvilinear regression, and multivariateregression.S = k C + kC + SA A I I reagwhere k I is the interferent’s sensitivity and C I is the interferent’s concentration.Multivariate calibration curves can be prepared using standards thatcontain known amounts of both the analyte and the interferent, and modeledusing multivariate regression. 115EBlank CorrectionsThus far in our discussion of strategies for standardizing analytical methods,we have assumed the use of a suitable reagent blank to correct for signalsarising from sources other than the analyte. We did not, however ask animportant question—“What constitutes an appropriate reagent blank?”Surprisingly, the answer is not immediately obvious.In one study, approximately 200 analytical chemists were asked to evaluatea data set consisting of a normal calibration curve, a separate analyte-freeblank, and three samples of different size but drawn from the same source. 12The first two columns in Table 5.3 shows a series of external standards andtheir corresponding signals. The normal calibration curve for the data isS std = 0.0750 × W std + 0.1250where the y-intercept of 0.1250 is the calibration blank. A separate reagentblank gives the signal for an analyte-free sample.In working up this data, the analytical chemists used at least four differentapproaches for correcting signals: (a) ignoring both the calibrationblank, CB, and the reagent blank, RB, which clearly is incorrect; (b) usingthe calibration blank only; (c) using the reagent blank only; and (d) usingboth the calibration blank and the reagent blank. Table 5.4 shows the equa-11 Beebe, K. R.; Kowalski, B. R. Anal. Chem. 1987, 59, 1007A–1017A.12 Cardone, M. J. Anal. Chem. 1986, 58, 433–438.Table 5.3Data Used to Study the Blank in an Analytical MethodW std S std Sample Number W samp S samp1.6667 0.2500 1 62.4746 0.80005.0000 0.5000 2 82.7915 1.00008.3333 0.7500 3 103.1085 1.200011.6667 0.841318.1600 1.4870 reagent blank 0.100019.9333 1.6200Calibration equation: S std = 0.0750 × W std + 0.1250W std: weight of analyte used to prepare the external standard; diluted to volume, V.W samp: weight of sample used to prepare sample; diluted to volume, V.

Chapter 5 Standardizing Analytical Methods187Table 5.4 Equations and Resulting Concentrations of Analyte for DifferentApproaches to Correcting for the BlankConcentration of Analyte in...Approach for Correcting Signal Equation Sample 1 Sample 2 Sample 3ignore calibration and reagent blankCAW SAsamp= = 0.1707 0.1610 0.1552W kWsampAsampuse calibration blank onlyuse reagent blank onlyCCAAW S −CBA samp= =W kWsampAsampW S −RBA samp= =W kWsampAsamp0.1441 0.1409 0.13900.1494 0.1449 0.1422use both calibration and reagent blank CAW SA= =Wsampsamp−CB−RBkWAsamp0.1227 0.1248 0.1261use total Youden blankCAW S − TYBA samp= =W kWsampAsamp0.1313 0.1313 0.1313C A = concentration of analyte; W A = weight of analyte; W samp = weight of sample; k A = slope of calibration curve (0.075–seeTable 5.3); CB = calibration blank (0.125–see Table 5.3); RB = reagent blank (0.100–see Table 5.3); TYB = total Youden blank(0.185–see text)tions for calculating the analyte’s concentration using each approach, alongwith the resulting concentration for the analyte in each sample.That all four methods give a different result for the analyte’s concentrationunderscores the importance of choosing a proper blank, but does nottell us which blank is correct. Because all four methods fail to predict thesame concentration of analyte for each sample, none of these blank correctionsproperly accounts for an underlying constant source of determinateerror.To correct for a constant method error, a blank must account for signalsfrom any reagents and solvents used in the analysis, as well as any bias resultingfrom interactions between the analyte and the sample’s matrix. Boththe calibration blank and the reagent blank compensate for signals fromreagents and solvents. Any difference in their values is due to indeterminateerrors in preparing and analyzing the standards.Unfortunately, neither a calibration blank nor a reagent blank can correctfor a bias resulting from an interaction between the analyte and thesample’s matrix. To be effective, the blank must include both the sample’smatrix and the analyte and, consequently, must be determined using thesample itself. One approach is to measure the signal for samples of differentsize, and to determine the regression line for a plot of S samp versus theBecause we are considering a matrix effectof sorts, you might think that the methodof standard additions is one way to overcomethis problem. Although the methodof standard additions can compensate forproportional determinate errors, it cannotcorrect for a constant determinate error;see Ellison, S. L. R.; Thompson, M. T.“Standard additions: myth and reality,”Analyst, 2008, 133, 992–997.

188 Analytical Chemistry 2.0amount of sample. The resulting y-intercept gives the signal in the absenceof sample, and is known as the total Youden blank. 13 This is the trueblank correction. The regression line for the three samples in Table 5.3 isS samp = 0.009844 × W samp + 0.185giving a true blank correction of 0.185. As shown by the last row of Table5.4, using this value to correct S samp gives identical values for the concentrationof analyte in all three samples.The use of the total Youden blank is not common in analytical work,with most chemists relying on a calibration blank when using a calibrationcurve, and a reagent blank when using a single-point standardization.As long we can ignore any constant bias due to interactions between theanalyte and the sample’s matrix, which is often the case, the accuracy of ananalytical method will not suffer. It is a good idea, however, to check forconstant sources of error before relying on either a calibration blank or areagent blank.5FUsing Excel and R for a Regression AnalysisAlthough the calculations in this chapter are relatively straightforward—consisting, as they do, mostly of summations—it can be quite tedious towork through problems using nothing more than a calculator. Both Exceland R include functions for completing a linear regression analysis and forvisually evaluating the resulting model.A B1 Cstd Sstd2 0.000 0.003 0.100 12.364 0.200 24.835 0.300 35.916 0.400 48.797 0.500 60.42Figure 5.15 Portion of a spreadsheetcontaining data from Example5.9 (Cstd = C std ; Sstd =S std ).5F.1 ExcelLet’s use Excel to fit the following straight-line model to the data in Example5.9.y = β + β x0 1Enter the data into a spreadsheet, as shown in Figure 5.15. Dependingupon your needs, there are many ways that you can use Excel to completea linear regression analysis. We will consider three approaches here.Us e Ex c e l ’s Bu i l t -In Fu n c t i o n sIf all you need are values for the slope, b 1 , and the y-intercept, b 0 , you canuse the following functions:=intercept(known_y’s, known_x’s)=slope(known_y’s, known_x’s)where known_y’s is the range of cells containing the signals (y), and known_x’sis the range of cells containing the concentrations (x). For example, clickingon an empty cell and entering13 Cardone, M. J. Anal. Chem. 1986, 58, 438–445.

Chapter 5 Standardizing Analytical Methods189=slope(B2:B7, A2:A7)returns Excel’s exact calculation for the slope (120.705 714 3).Us e Ex c e l ’s Da t a An a l y s i s To o l sTo obtain the slope and the y-intercept, along with additional statisticaldetails, you can use the data analysis tools in the Analysis ToolPak. TheToolPak is not a standard part of Excel’s instillation. To see if you haveaccess to the Analysis ToolPak on your computer, select Tools from themenu bar and look for the Data Analysis... option. If you do not see DataAnalysis..., select Add-ins... from the Tools menu. Check the box for theAnalysis ToolPak and click on OK to install them.Select Data Analysis... from the Tools menu, which opens the DataAnalysis window. Scroll through the window, select Regression from theavailable options, and press OK. Place the cursor in the box for Input Yrange and then click and drag over cells B1:B7. Place the cursor in the boxfor Input X range and click and drag over cells A1:A7. Because cells A1 andB1 contain labels, check the box for Labels. Select the radio button forOutput range and click on any empty cell; this is where Excel will place theresults. Clicking OK generates the information shown in Figure 5.16.There are three parts to Excel’s summary of a regression analysis. At thetop of Figure 5.16 is a table of Regression Statistics. The standard error is thestandard deviation about the regression, s r . Also of interest is the value forMultiple R, which is the model’s correlation coefficient, r, a term with whichyou may already by familiar. The correlation coefficient is a measure of theextent to which the regression model explains the variation in y. Values of rrange from –1 to +1. The closer the correlation coefficient is to ±1, the betterthe model is at explaining the data. A correlation coefficient of 0 meansthat there is no relationship between x and y. In developing the calculationsfor linear regression, we did not consider the correlation coefficient. ThereOnce you install the Analysis ToolPak, itwill continue to load each time you launchExcel.Including labels is a good idea. Excel’ssummary output uses the x-axis label toidentify the slope.SUMMARY OUTPUTRegression StatisticsMultiple R 0.99987244R Square 0.9997449Adjusted R Square 0.99968113Standard Error 0.40329713Observations 6ANOVAdf SS MS F Significance FRegression 1 2549.727156 2549.72716 15676.296 2.4405E-08Residual 4 0.650594286 0.16264857Total 5 2550.37775Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0%Intercept 0.20857143 0.29188503 0.71456706 0.51436267 -0.60183133 1.01897419 -0.60183133 1.01897419Cstd 120.705714 0.964064525 125.205016 2.4405E-08 118.029042 123.382387 118.029042 123.382387Figure 5.16 Output from Excel’s Regression command in the Analysis ToolPak. See the text for a discussion of how tointerpret the information in these tables.

190 Analytical Chemistry 2.0y1086420r = 0.9930 2 4 6 8 10xFigure 5.17 Example of fitting astraight-line to curvilinear data.See Section 4F.2 and Section 4F.3 for areview of the F-test.See Section 4F.1 for a review of the t-test.is a reason for this. For most straight-line calibration curves the correlationcoefficient will be very close to +1, typically 0.99 or better. There isa tendency, however, to put too much faith in the correlation coefficient’ssignificance, and to assume that an r greater than 0.99 means the linearregression model is appropriate. Figure 5.17 provides a counterexample.Although the regression line has a correlation coefficient of 0.993, the dataclearly shows evidence of being curvilinear. The take-home lesson here is:don’t fall in love with the correlation coefficient!The second table in Figure 5.16 is entitled ANOVA, which stands foranalysis of variance. We will take a closer look at ANOVA in Chapter 14.For now, it is sufficient to understand that this part of Excel’s summaryprovides information on whether the linear regression model explains asignificant portion of the variation in the values of y. The value for F is theresult of an F-test of the following null and alternative hypotheses.H 0 : regression model does not explain the variation in yH A : regression model does explain the variation in yThe value in the column for Significance F is the probability for retainingthe null hypothesis. In this example, the probability is 2.5×10 –6 %, suggestingthat there is strong evidence for accepting the regression model. Asis the case with the correlation coefficient, a small value for the probabilityis a likely outcome for any calibration curve, even when the model is inappropriate.The probability for retaining the null hypothesis for the data inFigure 5.17, for example, is 9.0×10 –7 %.The third table in Figure 5.16 provides a summary of the model itself.The values for the model’s coefficients—the slope, b 1 , and the y-intercept,b 0 —are identified as intercept and with your label for the x-axis data, whichin this example is Cstd. The standard deviations for the coefficients, s b 0 ands b , are in the column labeled Standard error. The column t Stat and the1column P-value are for the following t-tests.slope H 0 : b 1 = 0, H A : b 1 ≠ 0y-intercept H 0 : b 0 = 0, H A : b 0 ≠ 0The results of these t-tests provide convincing evidence that the slope isnot zero, but no evidence that the y-intercept significantly differs fromzero. Also shown are the 95% confidence intervals for the slope and they-intercept (lower 95% and upper 95%).Pr o g r a m t h e Fo r m u l a s Yo u r s e l fA third approach to completing a regression analysis is to program a spreadsheetusing Excel’s built-in formula for a summation=sum(first cell:last cell)and its ability to parse mathematical equations. The resulting spreadsheetis shown in Figure 5.18.

Chapter 5 Standardizing Analytical Methods191A B C D E F1 x y xy x^2 n = 62 0.000 0.00 =A2*B2 =A2^2 slope = =(F1*C8 - A8*B8)/(F1*D8-A8^2)3 0.100 12.36 =A3*B3 =A3^2 y-int = =(B8-F2*A8)/F14 0.200 24.83 =A4*B4 =A4^25 0.300 35.91 =A5*B5 =A5^26 0.400 48.79 =A6*B6 =A6^27 0.500 60.42 =A7*B7 =A7^289 =sum(A2:A7) =sum(B2:B7) =sum(C2:C7) =sum(D2:D7)

192 Analytical Chemistry 2.00.60.4Residuals0.200 0.1 0.2 0.3 0.4 0.5 0.6Residuals-0.2-0.4Figure 5.20 Example of Excel’s plot of a regressionmodel’s residual errors.-0.6CstdExcel also will create a plot of the regression model’s residual errors. Tocreate the plot, build the regression model using the Analysis ToolPak, asdescribed earlier. Clicking on the option for Residual plots creates the plotshown in Figure 5.20.Practice Exercise 5.6Use Excel to complete theregression analysis in PracticeExercise 5.4.Click here to review your answerto this exercise.Limitations t o Us i n g Ex c e l f o r a Re g r e s s i o n An a l y s i sExcel’s biggest limitation for a regression analysis is that it does not providea function for calculating the uncertainty when predicting values ofx. In terms of this chapter, Excel can not calculate the uncertainty for theanalyte’s concentration, C A , given the signal for a sample, S samp . Anotherlimitation is that Excel does not include a built-in function for a weightedlinear regression. You can, however, program a spreadsheet to handle thesecalculations.5F.2 RLet’s use Excel to fit the following straight-line model to the data in Example5.9.y = β + β x0 1En t e r i n g Da t a a n d Cr e a t i n g t h e Re g r e s s i o n Mod e lTo begin, create objects containing the concentration of the standards andtheir corresponding signals.> conc = c(0, 0.1, 0.2, 0.3, 0.4, 0.5)> signal = c(0, 12.36, 24.83, 35.91, 48.79, 60.42)The command for creating a straight-line linear regression model isAs you might guess, lm is short for linearmodel.lm(y ~ x)

Chapter 5 Standardizing Analytical Methods193where y and x are the objects containing our data. To access the results ofthe regression analysis, we assign them to an object using the followingcommand> model = lm(signal ~ conc)where model is the name we assign to the object.You can choose any name for the objectcontaining the results of the regressionanalysis.Ev a l u a t i n g t h e Li n e a r Re g r e s s i o n Mod e lTo evaluate the results of a linear regression we need to examine the dataand the regression line, and to review a statistical summary of the model. Toexamine our data and the regression line, we use the plot command, whichtakes the following general formplot(x, y, optional arguments to control style)where x and y are objects containing our data, and the abline commandabline(object, optional arguments to control style)where object is the object containing the results of the linear regression.Entering the commands> plot(conc, signal, pch = 19, col = “blue”, cex = 2)> abline(model, col = “red”)creates the plot shown in Figure 5.21.To review a statistical summary of the regression model, we use thesummary command.> summary(model)The resulting output, shown in Figure 5.22, contains three sections.The first section of R’s summary of the regression model lists the residualerrors. To examine a plot of the residual errors, use the commandsignal0 10 20 30 40 50 60> plot(model, which=1)0.0 0.1 0.2 0.3 0.4 0.5concFigure 5.21 Example of a regression plot in R showing the data andthe regression line. You can customize your plot by adjusting theplot command’s optional arguments. The argument pch controlsthe symbol used for plotting points, the argument col allows you toselect a color for the points or the line, and the argument cex setsthe size for the points. You can use the commandhelp(plot)The name abline comes from the followingcommon form for writing the equationof a straight-line.y = a + bxThe reason for including the argumentwhich=1 is not immediately obvious.When you use R’s plot command on anobject created by the lm command, the defaultis to create four charts summarizingthe model’s suitability. The first of thesecharts is the residual plot; thus, which=1limits the output to this plot.to learn more about the options for plotting data in R.

194 Analytical Chemistry 2.0> model=lm(signal~conc)> summary(model)Call:lm(formula = signal ~ conc)Residuals:1 2 3 4 5 6-0.20857 0.08086 0.48029 -0.51029 0.29914 -0.14143Figure 5.22 The summary of R’s regression analysis. See thetext for a discussion of how to interpret the information in theoutput’s three sections.Coefficients:Estimate Std. Error t value Pr(>|t|)(Intercept) 0.2086 0.2919 0.715 0.514conc 120.7057 0.9641 125.205 2.44e-08 ***---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1Residual standard error: 0.4033 on 4 degrees of freedomMultiple R-Squared: 0.9997, Adjusted R-squared: 0.9997F-statistic: 1.568e+04 on 1 and 4 DF, p-value: 2.441e-08See Section 4F.1 for a review of the t-test.which produces the result shown in Figure 5.23. Note that R plots the residualsagainst the predicted (fitted) values of y instead of against the knownvalues of x. The choice of how to plot the residuals is not critical, as you cansee by comparing Figure 5.23 to Figure 5.20. The line in Figure 5.23 is asmoothed fit of the residuals.The second section of Figure 5.22 provides the model’s coefficients—the slope, b 1 , and the y-intercept, b 0 —along with their respective standarddeviations (Std. Error). The column t value and the column Pr(>|t|) are forthe following t-tests.slope H 0 : b 1 = 0, H A : b 1 ≠ 0y-intercept H 0 : b 0 = 0, H A : b 0 ≠ 0The results of these t-tests provide convincing evidence that the slope is notzero, but no evidence that the y-intercept significantly differs from zero.Residuals-0.6 -0.4 -0.2 0.0 0.2 0.4 0.6Residuals vs Fitted345Figure 5.23 Example showing R’s plot of a regression model’sresidual error.0 10 20 30 40 50 60Fitted valueslm(signal ~ conc)

Chapter 5 Standardizing Analytical Methods195The last section of the regression summary provides the standard deviationabout the regression (residual standard error), the square of the correlationcoefficient (multiple R-squared), and the result of an F-test on themodel’s ability to explain the variation in the y values. For a discussion ofthe correlation coefficient and the F-test of a regression model, as well astheir limitations, refer to the section on using Excel’s data analysis tools.See Section 4F.2 and Section 4F.3 for areview of the F-test.Predicting t h e Un c e r t a i n t y in C A Gi v e n S s a m pUnlike Excel, R includes a command for predicting the uncertainty in ananalyte’s concentration, C A , given the signal for a sample, S samp . This commandis not part of R’s standard installation. To use the command you needto install the “chemCal” package by entering the following command (note:you will need an internet connection to download the package).> install.packages(“chemCal”)After installing the package, you will need to load the functions into Rusing the following command. (note: you will need to do this step each timeyou begin a new R session as the package does not automatically load when youstart R).> library(“chemCal”)The command for predicting the uncertainty in C A is inverse.predict,which takes the following form for an unweighted linear regressioninverse.predict(object, newdata, alpha = value)where object is the object containing the regression model’s results, newdatais an object containing values for S samp , and value is the numerical value forthe significance level. Let’s use this command to complete Example 5.11.First, we create an object containing the values of S samp> sample = c(29.32, 29.16, 29.51)and then we complete the computation using the following command> inverse.predict(model, sample, alpha = 0.05)producing the result shown in Figure 5.24. The analyte’s concentration, C A ,is given by the value $Prediction, and its standard deviation, s C , is shownAas $`Standard Error`. The value for $Confidence is the confidence interval,±ts C , for the analyte’s concentration, and $`Confidence Limits` providesAthe lower limit and upper limit for the confidence interval for C A .You need to install a package once, butyou need to load the package each timeyou plan to use it. There are ways to configureR so that it automatically loadscertain packages; see An Introduction to Rfor more information (click here to view aPDF version of this document).Us i n g R f o r a We i g h t e d Li n e a r Re g r e s s i o nR’s command for an unweighted linear regression also allows for a weightedlinear regression by including an additional argument, weights, whose valueis an object containing the weights.lm(y ~ x, weights = object)

196 Analytical Chemistry 2.0> inverse.predict(model, sample, alpha = 0.05)$Prediction[1] 0.2412597$`Standard Error`[1] 0.002363588$Confidence[1] 0.006562373Figure 5.24 Output from R’s command for predicting the analyte’sconcentration, C A , from the sample’s signal, S samp .$`Confidence Limits`[1] 0.2346974 0.2478221You may have noticed that this way ofdefining weights is different than thatshown in equation 5.28. In deriving equationsfor a weighted linear regression, youcan choose to normalize the sum of theweights to equal the number of points, oryou can choose not to—the algorithm inR does not normalize the weights.Practice Exercise 5.7Use Excel to complete theregression analysis in PracticeExercise 5.4.Click here to review your answerto this exercise.Let’s use this command to complete Example 5.12. First, we need to createan object containing the weights, which in R are the reciprocals of the standarddeviations in y, (s y i )–2 . Using the data from Example 5.12, we enter> syi=c(0.02, 0.02, 0.07, 0.13, 0.22, 0.33)> w=1/syi^2to create the object containing the weights. The commands> modelw = lm(signal ~ conc, weights = w)> summary(modelw)generate the output shown in Figure 5.25. Any difference between theresults shown here and the results shown in Example 5.12 are the result ofround-off errors in our earlier calculations.> modelw=lm(signal~conc, weights = w)> summary(modelw)Call:lm(formula = signal ~ conc, weights = w)Residuals:1 2 3 4 5 6-2.223 2.571 3.676 -7.129 -1.413 -2.864Figure 5.25 The summary of R’s regression analysis fora weighted linear regression. The types of informationshown here is identical to that for the unweighted linearregression in Figure 5.22.Coefficients:Estimate Std. Error t value Pr(>|t|)(Intercept) 0.04446 0.08542 0.52 0.63conc 122.64111 0.93590 131.04 2.03e-08 ***---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1Residual standard error: 4.639 on 4 degrees of freedomMultiple R-Squared: 0.9998, Adjusted R-squared: 0.9997F-statistic: 1.717e+04 on 1 and 4 DF, p-value: 2.034e-08

Chapter 5 Standardizing Analytical Methods1975GKey Termsexternal standard internal standard linear regressionmatrix matchingmethod of standardadditionsmultiple-pointstandardizationnormal calibration curve primary standard reagent graderesidual error secondary standard serial dilutionsingle-pointstandardizationunweighted linearregressionstandard deviation aboutthe regressionweighted linear regressiontotal Youden blankAs you review this chapter, try to define akey term in your own words. Check youranswer by clicking on the key term, whichwill take you to the page where it was firstintroduced. Clicking on the key termthere, will bring you back to this page sothat you can continue with another keyterm.5HChapter SummaryIn a quantitative analysis we measure a signal, S total , and calculate theamount of analyte, n A or C A , using one of the following equations.S = k n + Stotal A A reagS = k C + Stotal A A reagTo obtain an accurate result we must eliminate determinate errors affectingthe signal, S total , the method’s sensitivity, k A , and the signal due to thereagents, S reag .To ensure that we accurately measure S total , we calibrate our equipmentand instruments. To calibrate a balance, for example, we a standard weightof known mass. The manufacturer of an instrument usually suggests appropriatecalibration standards and calibration methods.To standardize an analytical method we determine its sensitivity. Thereare several standardization strategies, including external standards, themethod of standard addition and internal standards. The most commonstrategy is a multiple-point external standardization, resulting in a normalcalibration curve. We use the method of standard additions, in whichknown amounts of analyte are added to the sample, when the sample’smatrix complicates the analysis. When it is difficult to reproducibly handlesamples and standards, we may choose to add an internal standard.Single-point standardizations are common, but are subject to greateruncertainty. Whenever possible, a multiple-point standardization is preferred,with results displayed as a calibration curve. A linear regressionanalysis can provide an equation for the standardization.A reagent blank corrects for any contribution to the signal from thereagents used in the analysis. The most common reagent blank is one inwhich an analyte-free sample is taken through the analysis. When a simplereagent blank does not compensate for all constant sources of determinateerror, other types of blanks, such as the total Youden blank, can be used.

198 Analytical Chemistry 2.05IProblemsAnswers, but not worked solutions, tomost end-of-chapter problems are availablehere.1. Describe how you would use a serial dilution to prepare 100 mL eachof a series of standards with concentrations of 1.00×10 –5 , 1.00×10 –4 ,1.00×10 –3 , and 1.00×10 –2 M from a 0.100 M stock solution. Calculatethe uncertainty for each solution using a propagation of uncertainty,and compare to the uncertainty if you were to prepare each solution bya single dilution of the stock solution. You will find tolerances for differenttypes of volumetric glassware and digital pipets in Table 4.2 andTable 4.3. Assume that the uncertainty in the stock solution’s molarityis ±0.002.2. Three replicate determinations of S total for a standard solution that is10.0 ppm in analyte give values of 0.163, 0.157, and 0.161 (arbitraryunits). The signal for the reagent blank is 0.002. Calculate the concentrationof analyte in a sample with a signal of 0.118.3. A 10.00-g sample containing an analyte is transferred to a 250-mLvolumetric flask and diluted to volume. When a 10.00 mL aliquot ofthe resulting solution is diluted to 25.00 mL it gives signal of 0.235(arbitrary units). A second 10.00-mL portion of the solution is spikedwith 10.00 mL of a 1.00-ppm standard solution of the analyte and dilutedto 25.00 mL. The signal for the spiked sample is 0.502. Calculatethe weight percent of analyte in the original sample.4. A 50.00 mL sample containing an analyte gives a signal of 11.5 (arbitraryunits). A second 50 mL aliquot of the sample, which is spiked with1.00 mL of a 10.0-ppm standard solution of the analyte, gives a signalof 23.1. What is the analyte’s concentration in the original sample?5. An appropriate standard additions calibration curve based on equation5.10 places S spike ×(V o + V std ) on the y-axis and C std ×V std on the x-axis.Clearly explain why you can not plot S spike on the y-axis and C std ×[V std /(V o + V std )] on the x-axis. In addition, derive equations for the slopeand y-intercept, and explain how you can determine the amount ofanalyte in a sample from the calibration curve.6. A standard sample contains 10.0 mg/L of analyte and 15.0 mg/L ofinternal standard. Analysis of the sample gives signals for the analyteand internal standard of 0.155 and 0.233 (arbitrary units), respectively.Sufficient internal standard is added to a sample to make its concentration15.0 mg/L Analysis of the sample yields signals for the analyte andinternal standard of 0.274 and 0.198, respectively. Report the analyte’sconcentration in the sample.

Chapter 5 Standardizing Analytical Methods199(a)SignalSignalC AC A(b)SignalSignal(c)C AC ASignalSignalC A7. For each of the pair of calibration curves shown in Figure 5.26, selectthe calibration curve using the more appropriate set of standards.Briefly explain the reasons for your selections. The scales for the x-axisand y-axis are the same for each pair.8. The following data are for a series of external standards of Cd 2+ bufferedto a pH of 4.6. 14C A[Cd 2+ ] (nM) 15.4 30.4 44.9 59.0 72.7 86.0S total (nA) 4.8 11.4 18.2 25.6 32.3 37.7Figure 5.26 Calibration curves to accompanyProblem 7.14 Wojciechowski, M.; Balcerzak, J. Anal. Chim. Acta 1991, 249, 433–445.

200 Analytical Chemistry 2.0(a) Use a linear regression to determine the standardization relationshipand report confidence intervals for the slope and the y-intercept.(b) Construct a plot of the residuals and comment on their significance.At a pH of 3.7 the following data were recorded for the same set ofexternal standards.[Cd 2+ ] (nM) 15.4 30.4 44.9 59.0 72.7 86.0S total (nA) 15.0 42.7 58.5 77.0 101 118(c) How much more or less sensitive is this method at the lower pH?(d) A single sample is buffered to a pH of 3.7 and analyzed for cadmium,yielding a signal of 66.3. Report the concentration of Cd 2+ in thesample and its 95% confidence interval.9. To determine the concentration of analyte in a sample, a standard additionswas performed. A 5.00-mL portion of sample was analyzed andthen successive 0.10-mL spikes of a 600.0-mg/L standard of the analytewere added, analyzing after each spike. The following table shows theresults of this analysis.V spike (mL) 0.00 0.10 0.20 0.30S total (arbitrary units) 0.119 0.231 0.339 0.442Construct an appropriate standard additions calibration curve and usea linear regression analysis to determine the concentration of analyte inthe original sample and its 95% confidence interval.10. Troost and Olavsesn investigated the application of an internal standardizationto the quantitative analysis of polynuclear aromatic hydrocarbons.15 The following results were obtained for the analysis ofphenanthrene using isotopically labeled phenanthrene as an internalstandard. Each solution was analyzed twice.C A /C IS0.50 1.25 2.00 3.00 4.00S A /S IS0.5140.5220.9931.0241.4861.4712.04420.802.3422.550(a) Determine the standardization relationship using a linear regression,and report confidence intervals for the slope and the y-intercept.Average the replicate signals for each standard before completingthe linear regression analysis.(b) Based on your results explain why the authors concluded that theinternal standardization was inappropriate.15 Troost, J. R.; Olavesen, E. Y. Anal. Chem. 1996, 68, 708–711.

Chapter 5 Standardizing Analytical Methods20111. In Chapter 4 we used a paired t-test to compare two analytical methodsused to independently analyze a series of samples of variable composition.An alternative approach is to plot the results for one method versusthe results for the other method. If the two methods yield identicalresults, then the plot should have an expected slope, b 1 , of 1.00 andan expected y-intercept, b 0 , of 0.0. We can use a t-test to compare theslope and the y-intercept from a linear regression to the expected values.The appropriate test statistic for the y-intercept is found by rearrangingequation 5.23.bt = − b=exps sβ 0 0 0b0 b0Rearranging equation 5.22 gives the test statistic for the slope.texpβ b . b= − 1 1100−1=s sb1 b1Reevaluate the data in problem 25 from Chapter 4 using the samesignificance level as in the original problem.Although this is a common approach forcomparing two analytical methods, itdoes violate one of the requirements foran unweighted linear regression—that indeterminateerrors affect y only. Becauseindeterminate errors affect both analyticalmethods, the result of unweighted linearregression is biased. More specifically, theregression underestimates the slope, b 1 ,and overestimates the y-intercept, b 0 . Wecan minimize the effect of this bias byplacing the more precise analytical methodon the x-axis, by using more samplesto increase the degrees of freedom, andby using samples that uniformly cover therange of concentrations.For more information, see Miller, J. C.;Miller, J. N. Statistics for Analytical Chemistry,3rd ed. Ellis Horwood PTR Prentice-Hall:New York, 1993. Alternativeapproaches are found in Hartman, C.;Smeyers-Verbeke, J.; Penninckx, W.; Massart,D. L. Anal. Chim. Acta 1997, 338,19–40, and Zwanziger, H. W.; Sârbu, C.Anal. Chem. 1998, 70, 1277–1280.12. Consider the following three data sets, each containing value of y forthe same values of x.Data Set 1 Data Set 2 Data Set 3x y 1 y 2 y 310.00 8.04 9.14 7.468.00 6.95 8.14 6.7713.00 7.58 8.74 12.749.00 8.81 8.77 7.1111.00 8.33 9.26 7.8114.00 9.96 8.10 8.846.00 7.24 6.13 6.084.00 4.26 3.10 5.3912.00 10.84 9.13 8.157.00 4.82 7.26 6.425.00 5.68 4.74 5.73(a) An unweighted linear regression analysis for the three data sets givesnearly identical results. To three significant figures, each data sethas a slope of 0.500 and a y-intercept of 3.00. The standard deviationsin the slope and the y-intercept are 0.118 and 1.125 for each

202 Analytical Chemistry 2.0data set. All three standard deviations about the regression are 1.24,and all three data regression lines have a correlation coefficients of0.816. Based on these results for a linear regression analysis, commenton the similarity of the data sets.(b) Complete a linear regression analysis for each data set and verifythat the results from part (a) are correct. Construct a residual plotfor each data set. Do these plots change your conclusion from part(a)? Explain.(c) Plot each data set along with the regression line and comment onyour results.(d) Data set 3 appears to contain an outlier. Remove this apparentoutlier and reanalyze the data using a linear regression. Commenton your result.(e) Briefly comment on the importance of visually examining yourdata.13. Fanke and co-workers evaluated a standard additions method for a voltammetricdetermination of Tl. 16 A summary of their results is tabulatedin the following table.ppm TladdedInstrument Response (mA)0.000 2.53 2.50 2.70 2.63 2.70 2.80 2.520.387 8.42 7.96 8.54 8.18 7.70 8.34 7.981.851 29.65 28.70 29.05 28.30 29.20 29.95 28.955.734 84.8 85.6 86.0 85.2 84.2 86.4 87.8Use a weighted linear regression to determine the standardization relationshipfor this data.5JSolutions to Practice ExercisesPractice Exercise 5.1Substituting the sample’s absorbance into the calibration equation andsolving for C A giveS samp = 0.114 = 29.59 M –1 × C A + 0.015C A = 3.35 × 10 -3 MFor the one-point standardization, we first solve for k A16 Franke, J. P.; de Zeeuw, R. A.; Hakkert, R. Anal. Chem. 1978, 50, 1374–1380.

Chapter 5 Standardizing Analytical Methods203kASstd0.0931= =−3C 316 . × 10std= 29.46 MM−1and then use this value of k A to solve for C A .CASsamp0.114= = = 387 . × 10−1k 29.46 MA−3MWhen using multiple standards, the indeterminate errors affecting thesignal for one standard are partially compensated for by the indeterminateerrors affecting the other standards. The standard selected for the onepointstandardization has a signal that is smaller than that predicted by theregression equation, which underestimates k A and overestimates C A .Click here to return to the chapter.Practice Exercise 5.2We begin with equation 5.8rewriting it as⎛S k C V o= + Cspike A A⎝⎜VfstdV VkCV0 = + k × ⎧ ⎨ ⎪ C V A A oA stdV⎩⎪Vfwhich is in the form of the linear equationstdfstdf⎞⎠⎟⎫⎬⎪⎭⎪Y = y-intercept + slope × Xwhere Y is S spike and X is C std ×V std /V f . The slope of the line, therefore,is k A , and the y-intercept is k A C A V o /V f . The x-intercept is the value of Xwhen Y is zero, orkCVClick here to return to the chapter.Practice Exercise 5.3A A o0 = + k ×{ x-intercept}VfkCVA A oV CVfAx-intercept=− =−k VAUsing the calibration equation from Figure 5.7a, we find that the x-interceptisAfo

204 Analytical Chemistry 2.00.1478x-intercept =− =−1.731 mL-10.0854 mLPlugging this into the equation for the x-intercept and solving for C A givesthe concentration of Mn 2+ asC mLx-intercept=− 3 478 mL =− × 25 00.A.100.6 mg/ LFor Figure 7b, the x-intercept is= 696 .0.1478x-intercept =− =−3.478 mL-10.0425 mLand the concentration of Mn 2+ ismg/LC mLx-intercept=− mL =− × 25 003.478A.= 6 .96 mg/L50.00 LClick here to return to the chapter.Practice Exercise 5.4We begin by setting up a table to help us organize the calculation.x i y i x i y i2x i0.000 0.00 0.000 0.0001.55×10 –3 0.050 7.750×10 –5 2.403×10 –63.16×10 –3 0.093 2.939×10 –4 9.986×10 –64.74×10 –3 0.143 6.778×10 –4 2.247×10 –56.34×10 –3 0.188 1.192×10 –3 4.020×10 –57.92×10 –3 0.236 1.869×10 –3 6.273×10 –5Adding the values in each column gives∑ x i= 2.371×10 –2i∑ y i= 0.710i∑ xy i i= 4.110×10 –3 2∑ x i= 1.278×10 –4iSubstituting these values into equation 5.17 and equation 5.18, we findthat the slope and the y-intercept are6 4 110 10 2 371 10 0 710= × ( . × ) − ( . × ) × ( . )= 29.57−4 −2 2( 6×1.378× 10 ) − ( 2. 371×10 )−3 −2b 1i

Chapter 5 Standardizing Analytical Methods2050. 710− 29. 57× ( 2. 371×10 )= 0.00156−2b 0=The regression equation isS std = 29.57 × C std + 0.0015To calculate the 95% confidence intervals, we first need to determinethe standard deviation about the regression. The following table willhelp us organize the calculation.x i y iŷ i( y yˆ )− 2i i0.000 0.00 0.0015 2.250×10 –61.55×10 –3 0.050 0.0473 7.110×10 –63.16×10 –3 0.093 0.0949 3.768×10 –64.74×10 –3 0.143 0.1417 1.791×10 –66.34×10 –3 0.188 0.1890 9.483×10 –77.92×10 –3 0.236 0.2357 9.339×10 –8Adding together the data in the last column gives the numerator ofequation 5.19 as 1.596×10 –5 . The standard deviation about the regression,therefore, iss r=1.596×106−2−6= 1.997×10−3Next, we need to calculate the standard deviations for the slope and they-intercept using equation 5.20 and equation 5.21.−3 26× (. 1997×10 )s b1== 0.3007−4− 2 26× (. 1 378× 10 ) − ( 2.371×10 )−3 2 −4(. 1 997× 10 ) × ( 1. 378×10 )s b0=−4−6× (. 1 378×10 ) − ( 2. 371×10 )2 2= 1.441×10−3The 95% confidence intervals areβ 1 1 1= b ± ts b= 29. 57 ± ( 278 . × 0. 3007) = 29. 57 M -1 ± 0.85 M -1−3β 0 0 0= b ± ts b= 0. 0015 ± { 278 . × ( 1. 441× 10 } = 0. 0015± 0.0040With an average S samp of 0.114, the concentration of analyte, C A , isCAS=samp− b00. 114−0.0015= = 380 . × 10-1b 29.57 M1−3 M

206 Analytical Chemistry 2.0The standard deviation in C A is1.997×10s CA=29.571 1 ( 0. 114 −0. 1183)+ += 4.778× 10−23 6 ( 29. 57) × ( 4.408×10 -5 )−3 2and the 95% confidence interval isµ CC tsA A C A−3 −5= ± = 380 . × 10 ± { 2. 78× ( 4. 778×10 )}−= 3. 80× 10 M± 013 . × 10Click here to return to the chapter.Practice Exercise 5.53 −3To create a residual plot, we need to calculate the residual error for eachstandard. The following table contains the relevant information.M5residual error0.0100.000-0.0100.000 0.002 0.004 0.006 0.008C AFigure 5.27 Plot of the residual errors forthe data in Practice Exercise 5.5.x i y iŷ iy − yˆ0.000 0.00 0.0015 -0.00151.55×10 –3 0.050 0.0473 0.00273.16×10 –3 0.093 0.0949 -0.00194.74×10 –3 0.143 0.1417 0.00136.34×10 –3 0.188 0.1890 -0.00107.92×10 –3 0.236 0.2357 0.0003Figure 5.27 shows a plot of the resulting residual errors is shown here. Theresidual errors appear random and do not show any significant dependenceon the analyte’s concentration. Taken together, these observationssuggest that our regression model is appropriate.iiClick here to return to the chapterPractice Exercise 5.6Begin by entering the data into an Excel spreadsheet, following the formatshown in Figure 5.15. Because Excel’s Data Analysis tools provide most ofthe information we need, we will use it here. The resulting output, whichis shown in Figure 5.28, contains the slope and the y-intercept, alongwith their respective 95% confidence intervals. Excel does not provide afunction for calculating the uncertainty in the analyte’s concentration, C A ,given the signal for a sample, S samp . You must complete these calculationsby hand. With an S samp . of 0.114, C ACAS=samp− b00. 114−0.0014= = 380 . × 10-1b 29.59 M1−3 M

Chapter 5 Standardizing Analytical Methods207SUMMARY OUTPUTRegression StatisticsMultiple R 0.99979366R Square 0.99958737Adjusted R Square 0.99948421Standard Error 0.00199602Observations 6ANOVAdf SS MS F Significance FRegression 1 0.0386054 0.0386054 9689.9103 6.3858E-08Residual 4 1.5936E-05 3.9841E-06Total 5 0.03862133Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0%Intercept 0.00139272 0.00144059 0.96677158 0.38840479 -0.00260699 0.00539242 -0.00260699 0.00539242Cstd 29.5927329 0.30062507 98.437342 6.3858E-08 28.7580639 30.4274019 28.7580639 30.4274019Figure 5.28 Excel’s summary of the regression results for Practice Exercise 5.6.The standard deviation in C A is1.996×10s CA=29.591 1 ( 0. 114−0. 1183)+ += 4.772× 10−23 6 ( 29. 59) × ( 4.408×10 -5 )−3 2and the 95% confidence interval isµ CC tsA A C A−3 −5= ± = 380 . × 10 ± { 278 . × ( 4. 772×10 )}−= 3. 80× 10 M± 013 . × 10Click here to return to the chapterPractice Exercise 5.73 −3Figure 5.29 shows an R session for this problem, including loading thechemCal package, creating objects to hold the values for C std , S std , andSsamp. Note that for Ssamp, we do not have the actual values for thethree replicate measurements. In place of the actual measurements, wejust enter the average signal three times. This is okay because the calculationdepends on the average signal and the number of replicates, and noton the individual measurements.Click here to return to the chapterM5

208 Analytical Chemistry 2.0> library("chemCal")> conc=c(0, 1.55e-3, 3.16e-3, 4.74e-3, 6.34e-3, 7.92e-3)> signal=c(0, 0.050, 0.093, 0.143, 0.188, 0.236)> model=lm(signal~conc)> summary(model)Call:lm(formula = signal ~ conc)Residuals:1 2 3 4 5 6-0.0013927 0.0027385 -0.0019058 0.0013377 -0.0010106 0.0002328Coefficients:Estimate Std. Error t value Pr(>|t|)(Intercept) 0.001393 0.001441 0.967 0.388conc 29.592733 0.300625 98.437 6.39e-08 ***---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1Residual standard error: 0.001996 on 4 degrees of freedomMultiple R-Squared: 0.9996, Adjusted R-squared: 0.9995F-statistic: 9690 on 1 and 4 DF, p-value: 6.386e-08> samp=c(0.114, 0.114, 0.114)> inverse.predict(model,samp,alpha=0.05)$Prediction[1] 0.003805234$`Standard Error`[1] 4.771723e-05$Confidence[1] 0.0001324843$`Confidence Limits`[1] 0.003672750 0.003937719Figure 5.29 R session for completing Practice Exercise 5.7.

DRAFTChapter 6Equilibrium ChemistryChapter Overview6A Reversible Reactions and Chemical Equilibria6B Thermodynamics and Equilibrium Chemistry6C Manipulating Equilibrium Constants6D Equilibrium Constants for Chemical Reactions6E Le Châtelier’s Principle6F Ladder Diagrams6G Solving Equilibrium Problems6H Buffer Solutions6I Activity Effects6J Using Excel and R to Solve Equilibrium Problems6K Three Final Thoughts About Equilibrium Chemistry6L Key Terms6M Chapter Summary6N Problems6O Solutions to Practice ExercisesRegardless of the problem on which an analytical chemist is working, its solution requiresa knowledge of chemistry and the ability to apply that knowledge. For example, an analyticalchemist studying the effect of pollution on spruce trees needs to know, or know where tofind, the chemical differences between p‐hydroxybenzoic acid and p‐hydroxyacetophenone,two common phenols found in the needles of spruce trees.Your ability to “think as a chemist” is a product of your experience in the classroom andin the laboratory. The material in this text assumes your familiarity with topics from earliercourses. Because of its importance to analytical chemistry, this chapter provides a review ofequilibrium chemistry. Much of the material in this chapter should be familiar to you, althoughsome topics—ladder diagrams and activity, for example—afford you with new ways to look atequilibrium chemistry.Copyright: David Harvey, 2009209

210 Analytical Chemistry 2.0Napoleon’s expedition to Egypt was thefirst to include a significant scientific presence.The Commission of Sciences andArts, which included Claude Berthollet,began with 151 members, and operatedin Egypt for three years. In addition toBerthollet’s work, other results includeda publication on mirages, and detailedcatalogs of plant and animal life, mineralogy,and archeology. For a review of theCommission’s contributions, see Gillispie,C. G. “Scientific Aspects of the FrenchEgyptian Expedition, 1798-1801,” Proc.Am. Phil. Soc. 1989, 133, 447–474.6AReversible Reactions and Chemical EquilibriaIn 1798, the chemist Claude Berthollet accompanied Napoleon’s militaryexpedition to Egypt. While visiting the Natron Lakes, a series of salt waterlakes carved from limestone, Berthollet made an observation that led himto an important discovery. When exploring the lake’s shore Berthollet founddeposits of Na 2 CO 3 , a result he found surprising. Why did Berthollet findthis result surprising and how did it contribute to an important discovery?Answering these questions provides an example of chemical reasoning andintroduces us to the topic of this chapter.At the end of the 18th century, chemical reactivity was explained interms of elective affinities. 1 If, for example, substance A reacts with substanceBC to form ABA+ BC → AB+Cthen A and B were said to have an elective affinity for each other. With electiveaffinity as the driving force for chemical reactivity, reactions were understoodto proceed to completion and to proceed in one direction. Onceformed, the compound AB could not revert to A and BC.AB + C → A+BCFrom his experience in the laboratory, Berthollet knew that addingsolid Na 2 CO 3 to a solution of CaCl 2 produces a precipitate of CaCO 3 .Na CO () s + CaCl ( aq) → 2NaCl( aq) + CaCO () s2 3 2 3Natron is another name for the mineralsodium carbonate, Na 2 CO 3 •10H 2 O. Innature, it usually contains impurities ofNaHCO 3 , and NaCl. In ancient Egypt,natron was mined and used for a varietyof purposes, including as a cleaning agentand in mummification.Understanding this, Berthollet was surprised to find solid Na 2 CO 3 formingon the edges of the lake, particularly since the deposits formed only whenthe lake’s salt water was in contact with limestone, CaCO 3 . Where the lakewas in contact with clay soils, there was little or no Na 2 CO 3 .Berthollet’s important insight was recognizing that the chemistry leadingto the formation of Na 2 CO 3 is the reverse of that seen in the laboratory.2NaCl( aq) + CaCO () s → NaCO () s + CaCl ( aq)3 2 3 2Using this insight Berthollet reasoned that the reaction is reversible, andthat the relative amounts of NaCl, CaCO 3 , Na 2 CO 3 , and CaCl 2 determinethe direction in which the reaction occurs and the final composition of thereaction mixture. We recognize a reaction’s ability to move in both directionsby using a double arrow when writing the reaction.Na CO () s + CaCl ( aq) 2NaCl( aq) + CaCO () s2 3 2 3Berthollet’s reasoning that reactions are reversible was an importantstep in understanding chemical reactivity. When we mix together solutionsof Na 2 CO 3 and CaCl 2 they react to produce NaCl and CaCO 3 . If during1 Quilez, J. Chem. Educ. Res. Pract. 2004, 5, 69–87 (http://www.uoi.gr/cerp).

Chapter 6 Equilibrium Chemistry211the reaction we monitor the mass of Ca 2+ remaining in solution and themass of CaCO 3 that precipitates, the result looks something like Figure6.1. At the start of the reaction the mass of Ca 2+ decreases and the mass ofCaCO 3 increases. Eventually the reaction reaches a point after which thereis no further change in the amounts of these species. Such a condition iscalled a state of equilibrium.Although a system at equilibrium appears static on a macroscopic level,it is important to remember that the forward and reverse reactions continueto occur. A reaction at equilibrium exists in a steady-state, in which therate at which a species forms equals the rate at which it is consumed.6BThermodynamics and Equilibrium ChemistryThermodynamics is the study of thermal, electrical, chemical, and mechanicalforms of energy. The study of thermodynamics crosses many disciplines,including physics, engineering, and chemistry. Of the various branchesof thermodynamics, the most important to chemistry is the study of thechange in energy during a chemical reaction.Consider, for example, the general equilibrium reaction shown in equation6.1, involving the species A, B, C, and D, with stoichiometric coefficientsa, b, c, and d.aA+ bB cC+dD6.1By convention, we identify species on the left side of the equilibrium arrowas reactants, and those on the right side of the equilibrium arrow asproducts. As Berthollet discovered, writing a reaction in this fashion doesnot guarantee that the reaction of A and B to produce C and D is favorable.Depending on initial conditions, the reaction may move to the left, moveto the right, or be in a state of equilibrium. Understanding the factors thatdetermine the reaction’s final, equilibrium position is one of the goals ofchemical thermodynamics.The direction of a reaction is that which lowers the overall free energy.At a constant temperature and pressure, typical of many bench-top chemicalreactions, a reaction’s free energy is given by the Gibb’s free energyfunctionMassCaCO 3Ca 2+Timeequilibrium reachedFigure 6.1 Graph showing howthe masses of Ca 2+ and CaCO 3change as a function of time duringthe precipitation of CaCO 3 .The dashed line indicates whenthe reaction reaches equilibrium.Prior to equilibrium the masses ofCa 2+ and CaCO 3 are changing;after reaching equilibrium, theirmasses remain constant.For obvious reasons, we call the double arrow, , an equilibrium arrow.∆G = ∆H −T∆S6.2where T is the temperature in kelvin, and ∆G, ∆H, and ∆S are the differencesin the Gibb's free energy, the enthalpy, and the entropy between theproducts and the reactants.Enthalpy is a measure of the flow of energy, as heat, during a chemicalreaction. Reactions releasing heat have a negative ∆H and are called exothermic.Endothermic reactions absorb heat from their surroundings andhave a positive ∆H. Entropy is a measure of energy that is unavailable foruseful, chemical work. The entropy of an individual species is always posi-For many students, entropy is the mostdifficult topic in thermodynamics to understand.For a rich resource on entropy,visit the following web site: http://www.entropysite.com/.

212 Analytical Chemistry 2.0Equation 6.2 shows that the sign of DGdepends on the signs of DH and DS, andthe temperature, T. The following tablesummarizes the possibilities.DH DS DG- + DG < 0 at all temperatures- - DG < 0 at low temperatures+ + DG < 0 at high temperatures+ - DG > 0 at all temperaturestive and tends to be larger for gases than for solids, and for more complexmolecules than for simpler molecules. Reactions producing a large numberof simple, gaseous products usually have a positive ∆S.The sign of ∆G indicates the direction in which a reaction moves toreach its equilibrium position. A reaction is thermodynamically favorablewhen its enthalpy, ∆H, decreases and its entropy, ∆S, increases. Substitutingthe inequalities ∆H < 0 and ∆S > 0 into equation 6.2 shows that areaction is thermodynamically favorable when ∆G is negative. When ∆G ispositive the reaction is unfavorable as written (although the reverse reactionis favorable). A reaction at equilibrium has a ∆G of zero.As a reaction moves from its initial, non-equilibrium condition to itsequilibrium position, the value of ∆G approaches zero. At the same time,the chemical species in the reaction experience a change in their concentrations.The Gibb's free energy, therefore, must be a function of the concentrationsof reactants and products.As shown in equation 6.3, we can split the Gibb’s free energy into twoterms.∆G = ∆G o + RT lnQ6.3Although not shown here, each concentrationterm in equation 6.4 is divided bythe corresponding standard state concentration;thus, the term [C] c really means⎧⎨⎪⎩⎪[ C][ ]C o⎫⎬⎪⎭⎪where [C] o is the standard state concentrationfor C. There are two importantconsequences of this: (1) the value of Q isunitless; and (2) the ratio has a value of 1for a pure solid or a pure liquid. This is thereason that pure solids and pure liquids donot appear in the reaction quotient.cThe first term, ∆G o , is the change in Gibb’s free energy when each speciesin the reaction is in its standard state, which we define as follows: gaseswith partial pressures of 1 atm, solutes with concentrations of 1 mol/L, andpure solids and pure liquids. The second term, which includes the reactionquotient, Q, accounts for non-standard state pressures or concentrations.For reaction 6.1 the reaction quotient iscQ = [ C ][ D ]a[ A] [ B]where the terms in brackets are the concentrations of the reactants andproducts. Note that we define the reaction quotient with the products arein the numerator and the reactants are in the denominator. In addition, weraise the concentration of each species to a power equivalent to its stoichiometryin the balanced chemical reaction. For a gas, we use partial pressurein place of concentration. Pure solids and pure liquids do not appear in thereaction quotient.At equilibrium the Gibb’s free energy is zero, and equation 6.3 simplifiesto∆G o =−RT ln Kdb6.4where K is an equilibrium constant that defines the reaction’s equilibriumposition. The equilibrium constant is just the numerical value of thereaction quotient, Q, when substituting equilibrium concentrations intoequation 6.4.

Chapter 6 Equilibrium Chemistry213Kc d= [ C ] [ D ] eq eqa b[ A] [ B]eq eqHere we include the subscript “eq” to indicate a concentration at equilibrium.Although we usually will omit the “eq” when writing equilibriumconstant expressions, it is important to remember that the value of K isdetermined by equilibrium concentrations.6.5As written, equation 6.5 is a limiting lawthat applies only to infinitely dilute solutionswhere the chemical behavior of onespecies is unaffected by the presence ofother species. Strictly speaking, equation6.5 should be written in terms of activitiesinstead of concentrations. We will returnto this point in Section 6I. For now, wewill stick with concentrations as this conventionis already familiar to you.6CManipulating Equilibrium ConstantsWe will take advantage of two useful relationships when working with equilibriumconstants. First, if we reverse a reaction’s direction, the equilibriumconstant for the new reaction is simply the inverse of that for the originalreaction. For example, the equilibrium constant for the reactionis the inverse of that for the reaction[ AB ]2A+ B ABK =[ A][ B]22 1 2AB A 2BK K + =( ) =2 2 1− 1 [ A][B][ AB ]Second, if we add together two reactions to obtain a new reaction, theequilibrium constant for the new reaction is the product of the equilibriumconstants for the original reactions.[AC]A+ C AC K =3[A][C][AC ]2AC + C AC K =2 4[AC][C][AC] [AC ] [AC]22A+ 2C AC K = K × K = × =2 5 3 42[A][C] [AC][C] [A][C]Example 6.1Calculate the equilibrium constant for the reactiongiven the following information2A+ BC+3D22

214 Analytical Chemistry 2.0Rxn1: A+ B DK = 040 .Rxn2: A+ EC+ D+ F K = 010 .Rxn 3: C+ E BK = 20 .Rxn4: F+ C D+ B K = 50 .1234So l u t i o nThe overall reaction is equivalent toRxn1+ Rxn2− Rxn3+Rxn4Subtracting a reaction is equivalent to adding the reverse reaction; thus,the overall equilibrium constant isKK K K= × × 1 2 4040 . × 0. 10×50 .== 010 .K20 .3Practice Exercise 6.1Calculate the equilibrium constant for the reactionC+ D+ F 2A+3Busing the equilibrium constants from Example 6.1.Click here to review your answer to this exercise.Another common name for an oxidation–reduction reaction is a redox reaction,where “red” is short for reduction and “ox”is short for oxidation.6DEquilibrium Constants for Chemical ReactionsSeveral types of chemical reactions are important in analytical chemistry,either in preparing a sample for analysis or during the analysis. The mostsignificant of these are: precipitation reactions, acid–base reactions, complexationreactions, and oxidation–reduction reactions. In this section wereview these reactions and their equilibrium constant expressions.6D.1 Precipitation ReactionsIn a precipitation reaction, two or more soluble species combine to forman insoluble precipitate. The most common precipitation reaction is ametathesis reaction, in which two soluble ionic compounds exchange parts.For example, if we add a solution of lead nitrate, Pb(NO 3 ) 2 , to a solution ofpotassium chloride, KCl, the result is a precipitate of lead chloride, PbCl 2 .We usually write a precipitation reaction as a net ionic equation, showingonly the precipitate and those ions forming the precipitate. Thus, the precipitationreaction for PbCl 2 is2+ −Pb ( aq) + Cl ( aq) PbCl () s22

Chapter 6 Equilibrium Chemistry215When writing an equilibrium constant for a precipitation reaction, we focuson the precipitate’s solubility. Thus, for PbCl 2 , the solubility reaction is2PbCl () s Pb + ( aq) + Cl− ( aq )22and its equilibrium constant, which we call the solubility product, K sp ,is2+ − 2 −5K sp= [ Pb ][ Cl ] = 17 . × 10 6.6Even though it does not appear in the K sp expression, it is important toremember that equation 6.6 is valid only if PbCl 2 (s) is present and in equilibriumwith Pb 2+ and Cl – . You will find values for selected solubility productsin Appendix 10.6D.2 Acid–Base ReactionsA useful definition of acids and bases is that independently introduced in1923 by Johannes Brønsted and Thomas Lowry. In the Brønsted-Lowrydefinition, an acid is a proton donor and a base is a proton acceptor. Notethe connection in these definitions—defining a base as a proton acceptorimplies that there is an acid available to donate the proton. For example, inreaction 6.7 acetic acid, CH 3 COOH, donates a proton to ammonia, NH 3 ,which serves as the base.+ −CH COOH( aq) + NH ( aq) NH 4( aq) + CH COO ( aq)6.73 3 3When an acid and a base react, the products are a new acid and a newbase. For example, the acetate ion, CH 3 COO – , in reaction 6.7 is a base thatcan accept a proton from the acidic ammonium ion, NH 4 + , forming aceticacid and ammonia. We call the acetate ion the conjugate base of acetic acid,and the ammonium ion is the conjugate acid of ammonia.St r o n g a n d We a k Ac i d sThe reaction of an acid with its solvent (typically water) is an acid dissociationreaction. We divide acids into two categories—strong and weak—based on their ability to donate a proton to the solvent. A strong acid, suchas HCl, almost completely transfers its proton to the solvent, which actsas the base.+ −HCl( aq) + H O() l → H O ( aq) + Cl ( aq )2 3We use a single arrow (→ ) in place of the equilibrium arrow ( ) becausewe treat HCl as if it completely dissociates in aqueous solutions. Inwater, the common strong acids are hydrochloric acid (HCl), hydroiodicacid (HI), hydrobromic acid (HBr), nitric acid (HNO 3 ), perchloric acid(HClO 4 ), and the first proton of sulfuric acid (H 2 SO 4 ).In a different solvent, HCl may not be astrong acid. For example, HCl does notact as a strong acid in methanol. In thiscase we use the equilibrium arrow whenwriting the acid–base reaction.+ −3 3 2HCl( aq ) + CHOH( l) CHOH ( aq) + Cl ( aq )

216 Analytical Chemistry 2.0A weak acid, of which aqueous acetic acid is one example, does notcompletely donate its acidic proton to the solvent. Instead, most of the acidremains undissociated, with only a small fraction present as the conjugatebase.+ −CH COOH( aq) + HO() l HO ( aq) + CH COO ( aq )3 2 3 3Earlier we noted that we omit pure solidsand pure liquids from equilibriumconstant expressions. Because the solvent,H 2 O, is not pure, you might wonder whywe have not included it in acetic acid’sK a expression. Recall that we divide eachterm in the equilibrium constant expressionby its standard state value. Becausethe concentration of H 2 O is so large—itis approximately 55.5 mol/L—its concentrationas a pure liquid and as a solvent arevirtually identical. The ratiois essentially 1.00.[H O][H O]22oThe equilibrium constant for this reaction is an acid dissociation constant,K a , which we write as[ CH COO ][ H O ]−= = 175 . × 10 5[ CH COOH]− +3 3K a3The magnitude of K a provides information about a weak acid’s relativestrength, with a smaller K a corresponding to a weaker acid. The ammoniumion, NH 4 + , for example, with a K a of 5.702 × 10 –10 , is a weaker acidthan acetic acid.Monoprotic weak acids, such as acetic acid, have only a single acidicproton and a single acid dissociation constant. Other acids, such as phosphoricacid, have more than one acidic proton, each characterized by anacid dissociation constant. We call such acids polyprotic weak acids. Phosphoricacid, for example, has three acid dissociation reactions and three aciddissociation constants.+ −HPO ( aq) + H O() l H O ( aq) + H PO ( aq )3 4 2 3 2 4[ HPO ][ H O ]− +2 4 3−3K a1= = 711 . × 10[H PO3 4]− + 2−HPO ( aq) + H O() l H O ( aq) + HPO ( aq )2 4 2 34= [ HPO ][ H O ]= 632 . × 102− +4 3K a2 −[H PO ]2 4−82− + 3−HPO ( aq) + H O() l H O ( aq) + PO ( aq )42 3= [ PO ][ HO ]= 45 . × 103− +4 3K a3 2−[HPO ]4The decrease in the acid dissociation constants from K a1 to K a3 tells usthat each successive proton is harder to remove. Consequently, H 3 PO 4 is astronger acid than H 2 PO 4 – , and H 2 PO 4 – is a stronger acid than HPO 4 2– .−134

Chapter 6 Equilibrium Chemistry217St r o n g a n d We a k Ba s e sThe most common example of a strong base is an alkali metal hydroxide,such as sodium hydroxide, NaOH, which completely dissociates to producehydroxide ion.+ −NaOH() s → Na ( aq) + OH ( aq )A weak base, such as the acetate ion, CH 3 COO – , only partially acceptsa proton from the solvent, and is characterized by a base dissociationconstant, K b . For example, the base dissociation reaction and the basedissociation constant for the acetate ion are−−CH COO ( aq) + H O() l OH ( aq) + CHCOOH( aq )3 2 3−= [ CH COOH][ OH ]CH COO= × −571 . 10 10−[ ]K b33A polyprotic weak base, like a polyprotic acid, has more than one base dissociationreaction and more than one base dissociation constant.Am p h i p r o t i c Sp e c i e sSome species can behave as either a weak acid or as a weak base. For example,the following two reactions show the chemical reactivity of the bicarbonateion, HCO 3 – , in water.− + 2−HCO ( aq) + H O() l H O ( aq) + CO ( aq ) 6.83 2 33−−HCO ( aq) + H O() l OH ( aq) + H CO ( aq ) 6.93 2 2 3A species that is both a proton donor and a proton acceptor is called amphiprotic.Whether an amphiprotic species behaves as an acid or as a basedepends on the equilibrium constants for the competing reactions. Forbicarbonate, the acid dissociation constant for reaction 6.8= [ CO ][ HO ]= 469 . × 102− +3 3K a2 −[ HCO ]3−11is smaller than the base dissociation constant for reaction 6.9.−= [ HCO ][ OH ]HCO= 225 . × 10−[ ]K b22 33Because bicarbonate is a stronger base than it is an acid, we expect an aqueoussolution of HCO 3 – to be basic.−8

218 Analytical Chemistry 2.0Dissociation o f Wa t e rWater is an amphiprotic solvent because it can serve as an acid or as a base.An interesting feature of an amphiprotic solvent is that it is capable of reactingwith itself in an acid–base reaction.2H O() l H O + ( aq) + OH− ( aq )6.102 3We identify the equilibrium constant for this reaction as water’s dissociationconstant, K w ,K w 3= [ HO+ ][ OH− ] = 100 . × 10 −146.11which has a value of 1.0000 × 10 –14 at a temperature of 24 o C. The valueof K w varies substantially with temperature. For example, at 20 o CK w is 6.809 × 10 –15 , while at 30 o C K w is 1.469 × 10 –14 . At 25 o C, K w is1.008 × 10 –14 , which is sufficiently close to 1.00 × 10 –14 that we can usethe latter value with negligible error.An important consequence of equation 6.11 is that the concentrationof H 3 O + and the concentration of OH – are related. If we know [H 3 O + ] fora solution, then we can calculate [OH – ] using equation 6.11.Example 6.2What is the [OH – ] if the [H 3 O + ] is 6.12 × 10 -5 M?So l u t i o n−Kw100 . × 10[ OH ] = =+[ HO ] 612 . × 103−14−5= 163 . × 10−10Th e pH Sc a l epH = –log[H 3 O + ]Equation 6.11 allows us to develop a pH scale that indicates a solution’sacidity. When the concentrations of H 3 O + and OH – are equal a solution isneither acidic nor basic; that is, the solution is neutral. Lettingsubstituting into equation 6.11+ −[ HO ] = [ OH ]3K w 3and solving for [H 3 O + ] gives= [ HO ] = 100 . × 10+ 2 − 14[ HO ] = 100 . × 10 = 1.00×10+ −14 −73

Chapter 6 Equilibrium Chemistry219A neutral solution has a hydronium ion concentration of 1.00 × 10 -7 Mand a pH of 7.00. For a solution to be acidic the concentration of H 3 O +must be greater than that for OH – , which means that+ −[ HO ] > 100 . × 10 7 M3The pH of an acidic solution, therefore, must be less than 7.00. A basicsolution, on the other hand, has a pH greater than 7.00. Figure 6.2 showsthe pH scale and pH values for some representative solutions.Ta b u l a t i n g Va l u e s f o r K aa n d K bA useful observation about acids and bases is that the strength of a base isinversely proportional to the strength of its conjugate acid. Consider, forexample, the dissociation reactions of acetic acid and acetate.+ −CH COOH( aq) + HO() l HO ( aq) + CH COO ( aq ) 6.123 2 3 3−−CH COO ( aq) + H O() l OH ( aq) + CHCOOH( aq ) 6.133 2 3Adding together these two reactions gives the reaction2H O() l H O + ( aq) + OH− ( aq )2 3for which the equilibrium constant is K w . Because adding together tworeactions is equivalent to multiplying their respective equilibrium constants,we may express K w as the product of K a for CH 3 COOH and K b forCH 3 COO – .K = K × K−w a,CH3COOH b,CH3COOFor any weak acid, HA, and its conjugate weak base, A – , we can generalizethis to the following equation.K = K × Kw a,HA − 6.14b,AThe relationship between K a and K b for a conjugate acid–base pair simplifiesour tabulation of acid and base dissociation constants. Appendix 11includes acid dissociation constants for a variety of weak acids. To find thevalue of K b for a weak base, use equation 6.14 and the K a value for its correspondingweak acid.Example 6.3Using Appendix 11, calculate values for the following equilibrium constants.(a) K b for pyridine, C 5 H 5 N(b) K b for dihydrogen phosphate, H 2 PO 4–pH1234567891011121314Gastric JuiceVinegar“Pure” RainNeutralSeawaterMilkBloodMilk of MagnesiaHousehold BleachFigure 6.2 Scale showing the pHvalue for representative solutions.Milk of Magnesia is a saturatedsolution of Mg(OH) 2 .A common mistake when using equation6.14 is to forget that it applies only to aconjugate acid–base pair.

220 Analytical Chemistry 2.0When finding the K b value for polyproticweak base, you must be careful to choosethe correct K a value. Remember thatequation 6.14 applies only to a conjugateacid–base pair. The conjugate acid ofSo l u t i o n(a) K(b) Kb, CHN 5 5b, HPO 2Kw100 . × 10= =K 590 . × 10+a, CHNH 5 5Kw100 . × 10−= =4 K 711 . × 10a, HPO 3 4−14−3−14−6== 169 . × 10 −9141 . × 10 −12H 2 PO 4– is H 3 PO 4 , not HPO 42– .Practice Exercise 6.2Using Appendix 11, calculate the K b values for hydrogen oxalate, HC 2 O 4 – ,and oxalate, C 2 O 4 2– .Click here to review your answer to this exercise.6D.3 Complexation ReactionsA more general definition of acids and bases was proposed in1923 by G.N. Lewis. The Brønsted-Lowry definition of acids and bases focuses on anacid’s proton-donating ability and a base’s proton-accepting ability. Lewistheory, on the other hand, uses the breaking and forming of covalent bondsto describe acid–base characteristics. In this treatment, an acid is an electronpair acceptor and a base in an electron pair donor. Although we canapply Lewis theory to the treatment of acid–base reactions, it is more usefulfor treating complexation reactions between metal ions and ligands.The following reaction between the metal ion Cd 2+ and the ligandNH 3 is typical of a complexation reaction.2+ 2+Cd ( aq) + 4:NH ( aq) Cd:NH ( ) ( aq)6.153 3 4The product of this reaction is a metal–ligand complex. In writing thisreaction we show ammonia as :NH 3 , using a pair of dots to emphasize thepair of electrons it donates to Cd 2+ . In subsequent reactions we will omitthis notation.Me t a l-Li g a n d Fo r m a t i o n Co n s t a n t sWe characterize the formation of a metal–ligand complex by a formationconstant, K f . The complexation reaction between Cd 2+ and NH 3 , for example,has the following equilibrium constant.K f2+= [ Cd( NH ) ]3 4Cd NH= × 755 . 102+46.16[ ][ ]The reverse of reaction 6.15 is a dissociation reaction, which we characterizeby a dissociation constant, K d , that is the reciprocal of K f .Many complexation reactions occur in a stepwise fashion. For example,the reaction between Cd 2+ and NH 3 involves four successive reactions.3

Chapter 6 Equilibrium Chemistry2212+ 2+Cd ( aq) + NH ( aq) Cd( NH ) ( aq)6.173 32+ 2+Cd( NH ) ( aq) + NH ( aq) Cd( NH ) ( aq)6.1833 3 22+ 2+Cd( NH ) ( aq) + NH ( aq) Cd( NH ) ( aq)6.193 23 3 32+ 2+Cd( NH ) ( aq) + NH ( aq) Cd( NH ) ( aq)6.203 33 3 4To avoid ambiguity, we divide formation constants into two categories.Stepwise formation constants, which we designate as K i for the ith step,describe the successive addition of one ligand to the metal–ligand complexfrom the previous step. Thus, the equilibrium constants for reactions6.17–6.20 are, respectively, K 1 , K 2 , K 3 , and K 4 . Overall, or cumulativeformation constants, which we designate as b i , describe the addition ofi ligands to the free metal ion. The equilibrium constant in equation 6.16is correctly identified as b 4 , whereβ 4= K 1× K 2× K 3× K 4In generalβ i= K × K × ×K1 2iStepwise and overall formation constants for selected metal–ligand complexesare in Appendix 12.Me t a l-Li g a n d Co m p l e x a t i o n a n d So l u b i l i t yA formation constant characterizes the addition of one or more ligands toa free metal ion. To find the equilibrium constant for a complexation reactioninvolving a solid, we combine appropriate K sp and K f expressions. Forexample, the solubility of AgCl increases in the presence of excess chlorideas the result of the following complexation reaction.AgCl() s + Cl − ( aq) AgCl−2( aq )6.21We can write this reaction as the sum of three other reactions with knownequilibrium constants—the solubility of AgCl, described by its K spAgCl() s Ag + ( aq) + Cl− ( aq )and the stepwise formation of AgCl 2 – , described by K 1 and K 2 .+ −Ag ( aq) + Cl ( aq) AgCl( aq)AgCl( aq) + Cl − ( aq) AgCl− ( aq) 2The equilibrium constant for reaction 6.21, therefore, is K sp × K 1 × K 2 .

222 Analytical Chemistry 2.0Example 6.4Determine the value of the equilibrium constant for the reactionPbCl() s PbCl ( aq )2 2So l u t i o nWe can write this reaction as the sum of three other reactions. The first ofthese reactions is the solubility of PbCl 2 (s), described by its K sp reaction.2PbCl () s Pb + ( aq) + 2Cl− ( aq )2The remaining two reactions are the stepwise formation of PbCl 2 (aq), describedby K 1 and K 2 .2+ − +Pb ( aq) + Cl ( aq) PbCl ( aq)+ −PbCl ( aq) + Cl ( aq) PbCl 2( aq)Using values for K sp , K 1 , and K 2 from Appendix 10 and Appendix 12, wefind that the equilibrium constant is−K = K × K × K = (. 17× 10 ) × 38. 9× 1. 62 = 11 . × 10sp 1 25 −3Practice Exercise 6.3What is the equilibrium constant for the following reaction? You will findappropriate equilibrium constants in Appendix 10 and Appendix 11.2− 3− −AgBr() s + SO ( aq) Ag(S O ) ( aq) + Br ( aq )22 32Click here to review your answer to this exercise.6D.4 Oxidation–Reduction (Redox) ReactionsAn oxidation–reduction reaction occurs when electrons move from onereactant to another reactant. As a result of this electron transfer, these reactantsundergo a change in oxidation state. Those reactants that experiencean increase in oxidation state undergo oxidation, and those experiencing adecrease in oxidation state undergo reduction. For example, in the followingredox reaction between Fe 3+ and oxalic acid, H 2 C 2 O 4 , iron is reducedbecause its oxidation state changes from +3 to +2.3+2Fe ( aq) + HCO ( aq) + 2H O()l 2 2 4 22++2Fe( aq)+ 2CO( g) + 2H O6.223( aq )Oxalic acid, on the other hand, undergoes oxidation because the oxidationstate for carbon increases from +3 in H 2 C 2 O 4 to +4 in CO 2 .32

Chapter 6 Equilibrium Chemistry223We can divide a redox reaction, such as reaction 6.22, into separatehalf-reactions that show the oxidation and the reduction processes.HCO ( aq) + 2H O() l 2CO ( g) + 2H O ( aq)+ 2e2 2 4 2 2 3Fe( aq) + e Fe ( aq )3 + − 2 ++ −It is important to remember, however, that an oxidation reaction and areduction reaction occur as a pair. We formalize this relationship by identifyingas a reducing agent the reactant undergoing oxidation, becauseit provides the electrons for the reduction half-reaction. Conversely, thereactant undergoing reduction is an oxidizing agent. In reaction 6.22,Fe 3+ is the oxidizing agent and H 2 C 2 O 4 is the reducing agent.The products of a redox reaction also have redox properties. For example,the Fe 2+ in reaction 6.22 can be oxidized to Fe 3+ , while CO 2 can be reducedto H 2 C 2 O 4 . Borrowing some terminology from acid–base chemistry, Fe 2+is the conjugate reducing agent of the oxidizing agent Fe 3+ , and CO 2 is theconjugate oxidizing agent of the reducing agent H 2 C 2 O 4 .Th e r m o d y n a m ic s o f Re d o x Re a c t i o n sUnlike precipitation reactions, acid–base reactions, and complexation reactions,we rarely express the equilibrium position of a redox reaction usingan equilibrium constant. Because a redox reaction involves a transfer ofelectrons from a reducing agent to an oxidizing agent, it is convenient toconsider the reaction’s thermodynamics in terms of the electron.For a reaction in which one mole of a reactant undergoes oxidation orreduction, the net transfer of charge, Q, in coulombs isQ= nFwhere n is the moles of electrons per mole of reactant, and F is Faraday’sconstant (96,485 C/mol). The free energy, ∆G, to move this charge, Q, overa change in potential, E, is∆G = EQThe change in free energy (in kJ/mole) for a redox reaction, therefore, is∆G =−nFE6.23where ∆G has units of kJ/mol. The minus sign in equation 6.23 is the resultof a difference in the conventions for assigning a reaction’s favorable direction.In thermodynamics, a reaction is favored when ∆G is negative, buta redox reaction is favored when E is positive. Substituting equation 6.23into equation 6.3o− nFE =− nFE + RT lnQand dividing by -nF, leads to the well-known Nernst equation

224 Analytical Chemistry 2.0ln(x) = 2.303log(x)RTE = Eo −nFlnQwhere E o is the potential under standard-state conditions. Substituting appropriatevalues for R and F, assuming a temperature of 25 o C (298 K), andswitching from ln to log gives the potential in volts as0.05916E = Eo − logQ6.24nSt a n d a r d Po t e n t i a l sA standard potential is the potential whenall species are in their standard states. Youmay recall that we define standard stateconditions as: all gases have partial pressuresof 1 atm, all solutes have concentrationsof 1 mol/L, and all solids and liquidsare pure.A redox reaction’s standard potential, E o , provides an alternative way ofexpressing its equilibrium constant and, therefore, its equilibrium position.Because a reaction at equilibrium has a ∆G of zero, the potential, E, alsomust be zero at equilibrium. Substituting these values into equation 6.24and rearranging provides a relationship between E o and K.Eo = 0. 05916 log K6.25nWe generally do not tabulate standard potentials for redox reactions.Instead, we calculate E o using the standard potentials for the correspondingoxidation half-reaction and reduction half-reaction. By convention,standard potentials are provided for reduction half-reactions. The standardpotential for a redox reaction, E o , iso oE = E −Eredooxwhere E o red and E o ox are the standard reduction potentials for the reductionhalf-reaction and the oxidation half-reaction.Because we cannot measure the potential for a single half-reaction, wearbitrarily assign a standard reduction potential of zero to a reference halfreactionand report all other reduction potentials relative to this reference.The reference half-reaction is+ −2HO ( aq) + 2e 2HO() l + H ( g)3 2 2Appendix 13 contains a list of selected standard reduction potentials. Themore positive the standard reduction potential, the more favorable the reductionreaction under standard state conditions. Thus, under standardstate conditions the reduction of Cu 2+ to Cu (E o = +0.3419 V) is morefavorable than the reduction of Zn 2+ to Zn (E o = –0.7618 V).Example 6.5Calculate (a) the standard potential, (b) the equilibrium constant, and (c)the potential when [Ag + ] = 0.020 M and [Cd 2+ ] = 0.050 M, for the followingreaction at 25 o C.

Chapter 6 Equilibrium Chemistry225+ 2+Cd() s + 2Ag ( aq) 2Ag() s + Cd ( aq)So l u t i o n(a) In this reaction Cd is undergoing oxidation and Ag + is undergoingreduction. The standard cell potential, therefore, isoE = ooE − E = − − =+ 2+0. 7996 ( 0. 4030) 1.2026 VAg/ Ag Cd / Cd(b) To calculate the equilibrium constant we substitute appropriate valuesinto equation 6.25.Eo= 1.2026 V =0.059162Vlog KSolving for K gives the equilibrium constant aslog K = 40.6558K = 4.527×10 40(c) To calculate the potential when [Ag + ] is 0.020 M and [Cd 2+ ] is0.050 M, we use the appropriate relationship for the reaction quotient,Q, in equation 6.24.2+o 0.05916 V CdE = E − log [ ]+ 2n [ Ag ]0.05916 V 0 050E = 1.2606 − log ( . )V2 ( 0. 020)2E = 114 . VPractice Exercise 6.4For the following reaction at 25 o C2+ − +5Fe ( aq) + MnO ( aq) + 8H( aq)45Fe( aq) + Mn ( aq) + 4H O( l )3+ 2+calculate (a) the standard potential, (b) the equilibrium constant, and (c)the potential under these conditions: [Fe 2+ ] = 0.50 M, [Fe 3+ ] = 0.10 M,[MnO 4 – ] = 0.025 M, [Mn 2+ ] = 0.015 M, and a pH of 7.00. See Appendix13 for standard state reduction potentials.Click here to review your answer to this exercise.2When writing precipitation, acid–base,and metal–ligand complexation reaction,we represent acidity as H 3 O + . Redox reactionsare more commonly written usingH + instead of H 3 O + . For the reaction inPractice Exercise 6.4, we could replace H +with H 3 O + and increase the stoichiometriccoefficient for H 2 O from 4 to 12.

226 Analytical Chemistry 2.0So what is the effect on the solubility ofAgCl of adding AgNO 3 ? Adding AgNO 3increases the concentration of Ag + in solution.To reestablish equilibrium, some ofthe Ag + and Cl – react to form additionalAgCl; thus, the solubility of AgCl decreases.The solubility product, K sp , of course,remains unchanged.6ELe Châtelier’s PrincipleAt a temperature of 25 o C, acetic acid’s dissociation reaction+ −CH COOH( aq) + HO() l HO ( aq) + CH COO ( aq )3 2 3 3has an equilibrium constant of[ CH COO ][ H O ]−= = 175 . × 10 5 6.26[ CH COOH]− +3 3K a3Because equation 6.26 has three variables—[CH 3 COOH], [CH 3 COO – ],and [H 3 O + ]—it does not have a unique mathematical solution. Nevertheless,although two solutions of acetic acid may have different values for[CH 3 COOH], [CH 3 COO – ], and [H 3 O + ], each solution must have thesame value of K a .If we add sodium acetate to a solution of acetic acid, the concentrationof CH 3 COO – increases, suggesting an apparent increase in the value of K a .Because K a must remain constant, the concentration of all three species inequation 6.26 must change to restore K a to its original value. In this case,a partial reaction of CH 3 COO – and H 3 O + decreases their concentrations,producing additional CH 3 COOH and reestablishing the equilibrium.The observation that a system at equilibrium responds to an externalstress by reequilibrating in a manner that diminishes the stress, is formalizedas Le Châtelier’s principle. One of the most common stresses to a systemat equilibrium is to change the concentration of a reactant or product. Wealready have seen, in the case of adding sodium acetate to acetic acid, thatif we add a product to a reaction at equilibrium the system responds byconverting some of the products into reactants. Adding a reactant has theopposite effect, resulting in the conversion of reactants to products.When we add sodium acetate to a solution of acetic acid, we are directlyapplying the stress to the system. It is also possible to indirectly apply aconcentration stress. Consider, for example, the solubility of AgCl.AgCl() s Ag + ( aq) + Cl− ( aq )6.27The effect on the solubility of AgCl of adding AgNO 3 is obvious, but whatis the effect of adding a ligand that forms a stable, soluble complex withAg + ? Ammonia, for example, reacts with Ag + as shown here+ +Ag ( aq) + NH ( aq) Ag(NH) ( aq)6.2823 3 2Adding ammonia decreases the concentration of Ag + as the Ag(NH 3 ) 2+complex forms. In turn, decreasing the concentration of Ag + increases thesolubility of AgCl as reaction 6.27 reestablishes its equilibrium position.Adding together reaction 6.27 and reaction 6.28 clarifies the effect of ammoniaon the solubility of AgCl, by showing ammonia as a reactant.+ −AgCl() s + NH ( aq) Ag(NH ) ( aq) + Cl ( aq ) 6.2923 3 2

Chapter 6 Equilibrium Chemistry227Example 6.6What happens to the solubility of AgCl if we add HNO 3 to the equilibriumsolution defined by reaction 6.29?So l u t i o nNitric acid is a strong acid, which reacts with ammonia as shown here+ −HNO ( aq) + NH ( aq) NH ( aq) + NO ( aq)3 3 4 3Adding nitric acid lowers the concentration of ammonia. Decreasing ammonia’sconcentration causes reaction 6.29 to move from products to reactants,decreasing the solubility of AgCl.Increasing or decreasing the partial pressure of a gas is the same as increasingor decreasing its concentration. Because the concentration of a gasdepends on its partial pressure, and not on the total pressure of the system,adding or removing an inert gas has no effect on a reaction’s equilibriumposition.Most reactions involve reactants and products dispersed in a solvent.If we change the amount of solvent by diluting or concentrating the solution,then the concentrations of all reactants and products either decreaseor increase. The effect of simultaneously changing the concentrations of allreactants and products is not as intuitively obvious as when changing theconcentration of a single reactant or product. As an example, let’s considerhow diluting a solution affects the equilibrium position for the formationof the aqueous silver-amine complex (reaction 6.28). The equilibrium constantfor this reaction isThe relationship between pressure andconcentration can be deduced using theideal gas law. Starting with PV = nRT, wesolve for the molar concentrationnmolarconcentration = =VPRTOf course, this assumes that the gas is behavingideally, which usually is a reasonableassumption under normal laboratoryconditions.[ Ag(NH )]+3 2 eqβ 2=+2[ Ag ] [ NH ]eq 3 eq6.30where we include the subscript “eq” for clarification. If we dilute a portionof this solution with an equal volume of water, each of the concentrationterms in equation 6.30 is cut in half. The reaction quotient, Q, becomes+05 .[ Ag(NH )]3 2 eq 05 .Q = =+ 2 205 .[ Ag ] ( 05 .)[ NH ] ( 05 . )eq3eq3+× [ Ag(NH )]3 2 eq4[ ] [ ]= β+Ag NHeq32 2eqBecause Q is greater than β 2 , equilibrium is reestablished by shifting thereaction to the left, decreasing the concentration of Ag(NH 3 ) 2 + . Note thatthe new equilibrium position lies toward the side of the equilibrium reactionhaving the greatest number of solute particles (one Ag + ion and twomolecules of NH 3 versus a single metal-ligand complex). If we concentratethe solution of Ag(NH 3 ) 2 + by evaporating some of the solvent, equilibriumis reestablished in the opposite direction. This is a general conclusion thatwe can apply to any reaction. Increasing volume always favors the direc-

228 Analytical Chemistry 2.0One of the primary sources of determinateerrors in many analytical methods isfailing to account for potential chemicalinterferences.Ladder diagrams are a great tool for helpingyou to think intuitively about analyticalchemistry. We will make frequent useof them in the chapters to follow.tion producing the greatest number of particles, and decreasing volumealways favors the direction producing the fewest particles. If the numberof particles is the same on both sides of the reaction, then the equilibriumposition is unaffected by a change in volume.6FLadder DiagramsWhen developing or evaluating an analytical method, we often need tounderstand how the chemistry taking place affects our results. Suppose wewish to isolate Ag + by precipitating it as AgCl. If we also a need to controlpH, then we must use a reagent that will not adversely affects the solubilityof AgCl. It is a mistake to add NH 3 to the reaction mixture, for example,because it increases the solubility of AgCl (reaction 6.29).In this section we introduce the ladder diagram as a simple graphicaltool for evaluating the equilibrium chemistry. 2 Using ladder diagramswe will be able to determine what reactions occur when combining severalreagents, estimate the approximate composition of a system at equilibrium,and evaluate how a change to solution conditions might affect an analyticalmethod.6F.1 Ladder Diagrams for Acid–Base EquilibriaLet’s use acetic acid, CH 3 COOH, to illustrate the process of drawing andinterpreting an acid–base ladder diagram. Before drawing the diagram,however, let’s consider the equilibrium reaction in more detail. The equilibriumconstant expression for acetic acid’s dissociation reactionis+ −CH COOH( aq) + HO() l HO ( aq) + CH COO ( aq )3 2 3 3[ CH COO ][ H O ]−= = 175 . × 10 5[ CH COOH]− +3 3K a3Taking the logarithm of each term in this equation, and multiplying throughby –1 gives−+− log =−log[ ] − log [ CH COO ]3K aHO 3= .[ CH COOH]476Replacing the negative log terms with p-functions and rearranging theequation, leaves us with the result shown here.32 Although not specifically on the topic of ladder diagrams as developed in this section, the followingsources provide appropriate background information: (a) Runo, J. R.; Peters, D. G. J. Chem.Educ. 1993, 70, 708–713; (b) Vale, J.; Fernández-Pereira, C.; Alcalde, M. J. Chem. Educ. 1993,70, 790–795; (c) Fernández-Pereira, C.; Vale, J. Chem. Educator 1996, 6, 1–18; (d) Fernández-Pereira, C.; Vale, J.; Alcalde, M. Chem. Educator 2003, 8, 15–21; (e) Fernández-Pereira, C.;Alcalde, M.; Villegas, R.; Vale, J. J. Chem. Educ. 2007, 84, 520–525.

Chapter 6 Equilibrium Chemistry229−CH COO3pH = pK + log [ ] = 476 .a6.31[ CH COOH]Equation 6.31 tells us a great deal about the relationship between pHand the relative amounts of acetic acid and acetate at equilibrium. If theconcentrations of CH 3 COOH and CH 3 COO – are equal, then equation6.31 reduces topH = pK+ log( 1) = pK= 4.76aIf the concentration of CH 3 COO – is greater than that of CH 3 COOH,then the log term in equation 6.31 is positive andpH > pK or pH > 476 .aThis is a reasonable result because we expect the concentration of the conjugatebase, CH 3 COO – , to increase as the pH increases. Similar reasoningshows that the concentration of CH 3 COOH exceeds that of CH 3 COO –whenpH < pK or pH < 476 .aNow we are ready to construct acetic acid’s ladder diagram (Figure 6.3).First, we draw a vertical arrow representing the solution’s pH, with smaller(more acidic) pH levels at the bottom and larger (more basic) pH levels atthe top. Second, we draw a horizontal line at a pH equal to acetic acid’spK a value. This line, or step on the ladder, divides the pH axis into regionswhere either CH 3 COOH or CH 3 COO – is the predominate species. Thiscompletes the ladder diagram.Using the ladder diagram, it is easy to identify the predominate form ofacetic acid at any pH. At a pH of 3.5, for example, acetic acid exists primarilyas CH 3 COOH. If we add sufficient base to the solution such that thepH increases to 6.5, the predominate form of acetic acid is CH 3 COO – .more basic3a[CH 3COO – ] > [CH 3COOH]pH pH = pK a= 4.76[CH 3COO – ] = [CH 3COOH][CH 3COOH] > [CH 3COO – ]more acidicFigure 6.3 Acid–base ladder diagram for acetic acid showing the relative concentrations of CH 3 COOHand CH 3 COO – . A simpler version of this ladder diagram dispenses with the equalities and shows onlythe predominate species in each region.

230 Analytical Chemistry 2.0more basicO 2NO –Example 6.7Draw a ladder diagram for the weak base p-nitrophenolate and identify itspredominate form at a pH of 6.00.pH pK a= 7.15O 2Nmore acidicOHSo l u t i o nFigure 6.4 Acid–base ladder diagramfor p-nitrophenolate.Practice Exercise 6.5To draw a ladder diagram for a weak base, we simply draw the ladder diagramfor its conjugate weak acid. From Appendix 12, the pK a for p-nitrophenolis 7.15. The resulting ladder diagram is shown in Figure 6.4. At apH of 6.00, p-nitrophenolate is present primarily in its weak acid form.Draw a ladder diagram for carbonic acid, H 2 CO 3 . Because H 2 CO 3 isa diprotic weak acid, your ladder diagram will have two steps. What isthe predominate form of carbonic acid when the pH is 7.00? Relevantequilibrium constants are in Appendix 11.Click here to review your answer to this exercise.Ladder diagrams are particularly useful for evaluating the reactivity betweena weak acid and a weak base. Figure 6.5 shows a single ladder diagramfor acetic acid/acetate and p-nitrophenol/p-nitrophenolate. An acid and abase can not co-exist if their respective areas of predominance do not overlap.If we mix together solutions of acetic acid and sodium p-nitrophenolate,the reaction6.32occurs because the areas of predominance for acetic acid and p-nitrophenolatedo not overlap. The solution’s final composition depends on which spe-O 2NO –pHpK a= 7.15CH 3COO −O 2NOHpK a= 4.74CH 3COOHFigure 6.5 Acid–base ladder diagram showing the areas of predominance for acetic acid/acetate and for p-nitrophenol/p-nitrophenolate. The areas in blue shading show the pH range where the weak bases are the predominate species;the weak acid forms are the predominate species in the areas shown in pink shading.

Chapter 6 Equilibrium Chemistry231cies is the limiting reagent. The following example shows how we can usethe ladder diagram in Figure 6.5 to evaluate the result of mixing togethersolutions of acetic acid and p-nitrophenolate.Example 6.8Predict the approximate pH and the final composition of mixing together0.090 moles of acetic acid and 0.040 moles of p-nitrophenolate.So l u t i o nThe ladder diagram in Figure 6.5 indicates that the reaction between aceticacid and p-nitrophenolate is favorable. Because acetic acid is in excess, weassume that the reaction of p-nitrophenolate to p-nitrophenol is complete.At equilibrium essentially no p-nitrophenolate remains and there are 0.040mol of p-nitrophenol. Converting p-nitrophenolate to p-nitrophenol consumes0.040 moles of acetic acid; thusmoles CH 3 COOH = 0.090 – 0.040 = 0.050 molmoles CH 3 COO – = 0.040 molAccording to the ladder diagram, the pH is 4.76 when there are equalamounts of CH 3 COOH and CH 3 COO – . Because we have slightly moreCH 3 COOH than CH 3 COO – , the pH is slightly less than 4.76.Practice Exercise 6.6Using Figure 6.5, predict the approximate pH and the composition of asolution formed by mixing together 0.090 moles of p-nitrophenolate and0.040 moles of acetic acid.Click here to review your answer to this exercise.If the areas of predominance for an acid and a base overlap, then practicallyno reaction occurs. For example, if we mix together solutions ofCH 3 COO – and p-nitrophenol, there is no significant change in the molesof either reagent. Furthermore, the pH of the mixture must be between4.76 and 7.15, with the exact pH depending upon the relative amounts ofCH 3 COO – and p-nitrophenol.We also can use an acid–base ladder diagram to evaluate the effect ofpH on other equilibria. For example, the solubility of CaF 2CaF Ca F2 () s 2 + ( aq ) −+ 2 ( aq )is affected by pH because F – is a weak base. Using Le Châtelier’s principle,converting F – to HF increases the solubility of CaF 2 . To minimize thesolubility of CaF 2 we need to maintain the solution’s pH so that F – is thepredominate species. The ladder diagram for HF (Figure 6.6) shows us thatmaintaining a pH of more than 3.17 minimizes solubility losses.more basicF –pH pK a= 3.17HFmore acidicFigure 6.6 Acid–base ladder diagramfor HF. To minimize the solubilityof CaF 2 , we need to keepthe pH above 3.17, with morebasic pH levels leading to smallersolubility losses. See Chapter 8 fora more detailed discussion.

232 Analytical Chemistry 2.0less ligandCd 2+logK 1 = 2.55Cd(NH 3 ) 2+logK 2 = 2.01pNH 2+3 Cd(NH 3 ) 2logK 3 = 1.34Cd(NH 3 ) 32+logK 4 = 0.84Cd(NH 3 ) 42+more ligandFigure 6.7 Metal–ligand ladderdiagram for Cd 2+ –NH 3 complexationreactions. Note that higherordercomplexes form when pNH 3is smaller (which corresponds tolarger concentrations of NH 3 ).6F.2 Ladder Diagrams for Complexation EquilibriaWe can apply the same principles for constructing and interpreting acid–base ladder diagrams to equilibria involving metal–ligand complexes. Fora complexation reaction we define the ladder diagram’s scale using theconcentration of uncomplexed, or free ligand, pL. Using the formation ofCd(NH 3 ) 2+ as an example2+ 2+Cd ( aq) + NH ( aq) Cd(NH ) ( aq)3 3we can easily show that log K 1 is the dividing line between the areas of predominancefor Cd 2+ and Cd(NH 3 ) 2+ .[ Cd(NH ) ]= = 355 . × 10[ ][ ]2+3K 1 2+Cd NH3log log [ ) 2+K = Cd(NH ] 31log[ ] 255 .2 3[ ]− NH =+Cdlog log [ ) 2+K = Cd(NH ] 31255 .2 3[ ]+ pNH =+CdpNH2+= + [ Cd ]logKlog2552Cd(NH= .+[ ) ]3 1Thus, Cd 2+ is the predominate species when pNH 3 is greater than 2.55 (aconcentration of NH 3 smaller than 2.82 × 10 –3 M) and for pNH 3 valuesless than 2.55, Cd(NH 3 ) 2+ is the predominate species. Figure 6.7 shows acomplete metal–ligand ladder diagram for Cd 2+ and NH 3 .Example 6.9Draw a single ladder diagram for the Ca(EDTA) 2– and Mg(EDTA) 2–metal–ligand complexes. Using your ladder diagram, predict the result ofadding 0.080 moles of Ca 2+ to 0.060 moles of Mg(EDTA) 2– . EDTA is anabbreviation for the ligand ethylenediaminetetraacetic acid.32So l u t i o nFigure 6.8 shows the ladder diagram for this system of metal–ligand complexes.Because the predominance regions for Ca 2+ and Mg(EDTA) 2- donot overlap, the reactionCa 2 +Mg(EDTA) 2 −Ca(EDTA) 2 −Mg2 +( aq) + ( aq) ( aq)+ ( aq)takes place. Because Ca 2+ is the excess reagent, the composition of the finalsolution is approximatelymoles Ca 2+ = 0.080 – 0.060 = 0.020 mol

Chapter 6 Equilibrium Chemistry233pEDTACa 2+logK Ca(EDTA) 2- = 10.69Ca(EDTA) 2– Mg 2+logK Mg(EDTA) 2- = 8.79moles Ca(EDTA) 2– = 0.060 molmoles Mg 2+ = 0.060 molmoles Mg(EDTA) 2– = 0 molThe metal–ligand ladder diagram in Figure 6.7 uses stepwise formationconstants. We can also construct ladder diagrams using cumulative formationconstants. The first three stepwise formation constants for the reactionof Zn 2+ with NH 32+ 2+Zn ( aq) + NH ( aq) Zn(NH ) ( aq) K = 16 . × 103Mg(EDTA) 2–Figure 6.8 Metal–ligand ladder diagram for Ca(EDTA) 2– and for Mg(EDTA) 2– . The areas with blueshading shows the pEDTA range where the free metal ions are the predominate species; the metal–ligand complexes are the predominate species in the areas shown with pink shading.2+2+Zn(NH ) ( aq) + NH ( aq) Zn(NH ) ( aq) K = 195 . ×10 23 3 3 223122+2+Zn(NH ) ( aq) + NH ( aq) Zn(NH ) ( aq) K = . ×10 23 2233 3 33show that the formation of Zn(NH 3 ) 3 2+ is more favorable than the formationof Zn(NH 3 ) 2+ or Zn(NH 3 ) 2 2+ . For this reason, the equilibrium is bestrepresented by the cumulative formation reaction shown here.2+2+Zn ( aq) + 3NH ( aq) Zn(NH ) ( aq) β = 7.2×103 3 33To see how we incorporate this cumulative formation constant into a ladderdiagram, we begin with the reaction’s equilibrium constant expression.6Because K 3 is greater than K 2 , which isgreater than K 1 , the formation of themetal-ligand complex Zn(NH 3 ) 32+ ismore favorable than the formation of theother metal ligand complexes. For thisreason, at lower values of pNH 3 the concentrationof Zn(NH 3 ) 32+ is larger thanthat for Zn(NH 3 ) 22+ and Zn(NH 3 ) 2+ .The value of b 3 isb 3 = K 1 × K 2 × K 3

234 Analytical Chemistry 2.0[ Zn(NH ) ]2+3 3β 3= 2 +3[ Zn ][ NH ]3Taking the log of each side givesless ligandZn 2+logβ 3 = 2.29orlog log [ 2+Zn(NH ) ] 3 3β 3= log[ ][ 2−3 NH+3Zn ]pNH1 1 [ Zn]= logβ+ log2+3 3 [ Zn(NH ) ]3 33 313EpNH 2+3 Zn(NH 3 ) 3logK 4 = 2.03Zn(NH 3 ) 42+more ligandFigure 6.9 Ladder diagram forZn 2+ –NH 3 metal–ligand complexationreactions showing both astep based on a cumulative formationconstant, and a step based ona stepwise formation constant.When the concentrations of Zn 2+ and Zn(NH 3 ) 32+are equal, then1pNH 3= log β33=229 .In general, for the metal–ligand complex ML n , the step for a cumulativeformation constant ispL = 1 logβnnFigure 6.9 shows the complete ladder diagram for the Zn 2+ –NH 3 system.6F.3 Ladder Diagram for Oxidation/Reduction EquilibriaWe also can construct ladder diagrams to help evaluate redox equilibria.Figure 6.10 shows a typical ladder diagram for two half-reactions in whichthe scale is the potential, E. The Nernst equation defines the areas of predominance.Using the Fe 3+ /Fe 2+ half-reaction as an example, we writemore positiveFe 3+Figure 6.10 Redox ladder diagram for Fe 3+ /Fe 2+ and for Sn 4+ /Sn 2+ . The areas with blue shading show the potential rangewhere the oxidized forms are the predominate species; the reducedforms are the predominate species in the areas shownwith pink shading. Note that a more positive potential favorsthe oxidized form.E o Fe3+/Fe2+ = +0.771VSn 4+Fe 2+ E o Sn4+/Sn2+ = +0.154 VSn 2+more negative

Chapter 6 Equilibrium Chemistry235+RTE = oE − FenF=+ −Fln [ 2] . . log [ 2+e ]0 771 0 059163+3+[ Fe ][ Fe ]At potentials more positive than the standard state potential, the predominatespecies is Fe 3+ , whereas Fe 2+ predominates at potentials more negativethan E o . When coupled with the step for the Sn 4+ /Sn 2+ half-reaction wesee that Sn 2+ is a useful reducing agent for Fe 3+ . If Sn 2+ is in excess, thepotential of the resulting solution is near +0.151 V.Because the steps on a redox ladder diagram are standard state potentials,complications arise if solutes other than the oxidizing agent and reducingagent are present at non-standard state concentrations. For example, thepotential for the half-reaction2+ + − 4+UO ( aq) + 4HO ( aq) + 2e U ( aq) + 6H O()l23 2depends on the solution’s pH. To define areas of predominance in this casewe begin with the Nernst equation4+0.05916 [ U ]E =+ 0.327 − log+ +2 [ UO ][ HO ]and factor out the concentration of H 3 O + .0.05916+ 4E =+ 0.327 + log[ HO ] −322 42 30.0591624+log [ U ]2+[ UO 2]From this equation we see that the area of predominance for UO 2 2+ andU 4+ is defined by a step whose potential is0.05916+ 4E =+ 0.327 + log[ HO ] =+ 0. 327 −0.1183pH32Figure 6.11 shows how pH affects the step for the UO 2 2+ /U 4+ half-reaction.more positiveEUO 22+E o = +0.327 V (pH = 0)E o = +0.209 V (pH = 1)E o = +0.090 V (pH = 2)U 4+more negativeFigure 6.11 Redox ladder diagram2+for the UO 2 /U4+ half-reactionshowing the effect of pH on thestep.6GSolving Equilibrium ProblemsLadder diagrams are a useful tool for evaluating chemical reactivity, usuallyproviding a reasonable approximation of a chemical system’s compositionat equilibrium. If we need a more exact quantitative description of theequilibrium condition, then a ladder diagram is insufficient. In this casewe need to find an algebraic solution. In this section we will learn how toset-up and solve equilibrium problems. We will start with a simple problemand work toward more complex problems.

236 Analytical Chemistry 2.0When we first add solid Pb(IO 3 ) 2 to water,the concentrations of Pb 2+ and IO 3–are zero and the reaction quotient, Q, isQ = [Pb 2+ ][IO 3 – ] 2 = 0As the solid dissolves, the concentrationsof these ions increase, but Q remainssmaller than K sp . We reach equilibriumand “satisfy the solubility product” whenQ = K spBecause a solid, such as Pb(IO 3 ) 2 , doesnot appear in the solubility product expression,we do not need to keep track ofits concentration. Remember, however,that the K sp value applies only if there issome Pb(IO 3 ) 2 present at equilibrium.6G.1 A Simple Problem—Solubility of Pb(IO 3 ) 2If we place an insoluble compound such as Pb(IO 3 ) 2 in deionized water, thesolid dissolves until the concentrations of Pb 2+ and IO 3 – satisfy the solubilityproduct for Pb(IO 3 ) 2 . At equilibrium the solution is saturated withPb(IO 3 ) 2 , which simply means that no more solid can dissolve. How dowe determine the equilibrium concentrations of Pb 2+ and IO 3 – , and whatis the molar solubility of Pb(IO 3 ) 2 in this saturated solution?We begin by writing the equilibrium reaction and the solubility productexpression for Pb(IO 3 ) 2 .Pb(IO ) () s Pb 2 + ( ) −aq + 2IO( aq )3 2 32+ − 2 −13K sp= [ Pb ][ IO ] = 25 . × 10 6.333As Pb(IO 3 ) 2 dissolves, two IO 3 – ions are produced for each ion of Pb 2+ . Ifwe assume that the change in the molar concentration of Pb 2+ at equilibriumis x, then the change in the molar concentration of IO 3 – is 2x. Thefollowing table helps us keep track of the initial concentrations, the changein concentrations, and the equilibrium concentrations of Pb 2+ and IO 3 – .Concentrations Pb(IO 3 ) 2 (s) Pb 2+ (aq) + 2IO 3– (aq)Initial solid 0 0Change solid +x +2xEquilibrium solid x 2xSubstituting the equilibrium concentrations into equation 6.33 and solvinggives( x)( 2x) = 4x= 25 . × 102 3 −13x = 397 . × 10Substituting this value of x back into the equilibrium concentration expressionsfor Pb 2+ and IO 3 – gives their concentrations as−5We can express a compound’s solubilityin two ways: molar solubility (mol/L) ormass solubility (g/L). Be sure to expressyour answer clearly.[ Pb ] = x = 40 . × 102+ −5−[IO ] = 2x= 7.9×103Because one mole of Pb(IO 3 ) 2 contains one mole of Pb 2+ , the molar solubilityof Pb(IO 3 ) 2 is equal to the concentration of Pb 2+ , or 4.0 × 10 –5 M.Practice Exercise 6.7Calculate the molar solubility and the mass solubility for Hg 2 Cl 2 , giventhe following solubility reaction and K sp value.−5Hg Cl () s Hg ( aq) + 2Cl( aq ) K = 1.2×102 18 + − −2 2 2spClick here to review your answer to this exercise.MM

Chapter 6 Equilibrium Chemistry2376G.2 A More Complex Problem—The Common Ion EffectCalculating the solubility of Pb(IO 3 ) 2 in deionized water is a straightforwardproblem since the solid’s dissolution is the only source of Pb 2+ andIO 3 – .But what if we add Pb(IO 3 ) 2 to a solution of 0.10 M Pb(NO 3 ) 2 , whichprovides a second source of Pb 2+ ? Before we set-up and solve this problemalgebraically, think about the system’s chemistry and decide whether thesolubility of Pb(IO 3 ) 2 will increase, decrease or remain the same.We begin by setting up a table to help us keep track of the concentrationsof Pb 2+ and IO 3 – as this system moves toward and reaches equilibrium.Concentrations Pb(IO 3 ) 2 (s) Pb 2+ (aq) + 2IO 3– (aq)Initial solid 0.10 0Change solid +x +2xEquilibrium solid 0.10 + x 2xBeginning a problem by thinking aboutthe likely answer is a good habit to develop.Knowing what answers are reasonablewill help you spot errors in your calculationsand give you more confidence thatyour solution to a problem is correct.Because the solution already contains asource of Pb 2+ , we can use Le Châtelier’sprinciple to predict that the solubility ofPb(IO 3 ) 2 is smaller than that in our previousproblem.Substituting the equilibrium concentrations into equation 6.33( 010 . + x)( 2x) = 25 . × 10−2 13and multiplying out the terms on the equation’s left side leaves us with4x3 + 0. 40x2 = 25 . × 10−136.34This is a more difficult equation to solve than that for the solubility ofPb(IO 3 ) 2 in deionized water, and its solution is not immediately obvious.We can find a rigorous solution to equation 6.34 using available computersoftware packages and spreadsheets, some of which are described in Section6.J.How might we solve equation 6.34 if we do not have access to a computer?One approach is to use our understanding of chemistry to simplifythe problem. From Le Châtelier’s principle we know that a large initialconcentration of Pb 2+ significantly decreases the solubility of Pb(IO 3 ) 2 .One reasonable assumption is that the equilibrium concentration of Pb 2+is very close to its initial concentration. If this assumption is correct, thenthe following approximation is reasonable2[ Pb+ ] = 010 . + x ≈ 0.10 MThere are several approaches to solvingcubic equations, but none are computationallyeasy.Substituting our approximation into equation 6.33 and solving for x gives( 01 . )( 2x) = 2.5×102 −1304 . x = 25 . × 102 −13x = 791 . × 10 −7Before accepting this answer, we must verify that our approximation is reasonable.The difference between the calculated concentration of Pb 2+ , 0.10 + xM, and our assumption that it is 0.10 M is 7.9 × 10 –7 , or 7.9 × 10 –4 % of( 010 . + x) −0.10% error =× 100010 .791 . × 10=010 .−7791 10 4= × − . %× 100

238 Analytical Chemistry 2.0the assumed concentration. This is a negligible error. Accepting the resultof our calculation, we find that the equilibrium concentrations of Pb 2+ andIO 3 – are2+[ Pb ] = 010 . + x ≈ 0.10 M−[IO ] = 2x= 16 . × 103The molar solubility of Pb(IO 3 ) 2 is equal to the additional concentration ofPb 2+ in solution, or 7.9 × 10 –4 mol/L. As expected, Pb(IO 3 ) 2 is less solublein the presence of a solution that already contains one of its ions. This isknown as the common ion effect.As outlined in the following example, if an approximation leads to anunacceptably large error we can extend the process of making and evaluatingapproximations.Example 6.10Calculate the solubility of Pb(IO 3 ) 2 in 1.0 × 10 –4 M Pb(NO 3 ) 2 .So l u t i o nLetting x equal the change in the concentration of Pb 2+ , the equilibriumconcentrations of Pb 2+ and IO 3 – are−6M2[ Pb+ 4] = 10 . × 10 − + x [ IO− ] = 2xSubstituting these concentrations into equation 6.33 leaves us with−(. 10× 10 + x)( 2x) = 2.5×104 2 −13To solve this equation for x, we make the following assumption3[ Pb ] = 10 . × 10 + x ≈ 10 . × 102 + − 4 − 4Mobtaining a value for x of 2.50× 10 –4 . Substituting back, gives the calculatedconcentration of Pb 2+ at equilibrium as[ Pb ] = 10 . × 10 + 250 . × 10 = 1.25×102+ −4 −5 −4Ma value that differs by 25% from our assumption that the equilibriumconcentration is 1.0× 10 –4 M. This error seems unreasonably large. Ratherthan shouting in frustration, we make a new assumption. Our first assumption—thatthe concentration of Pb 2+ is 1.0× 10 –4 M—was too small.The calculated concentration of 1.25× 10 –4 M, therefore, is probably a bittoo large. For our second approximation, let’s assume that[ ] = 10 . × 10 + x ≈ 125 . × 10Pb 2+ −4 −4Substituting into equation 6.33 and solving for x gives its value as2.24× 10 –5 . The resulting concentration of Pb 2+ is

Chapter 6 Equilibrium Chemistry239[ Pb ] = 10 . × 10 + 224 . × 10 = 1.22×102+ −4 −5 −4Mwhich differs from our assumption of 1.25× 10 –4 M by 2.4%. Because theoriginal concentration of Pb 2+ is given to two significant figure, this is amore reasonable error. Our final solution, to two significant figures, is2+ −4− −5[ Pb ] = 12 . × 10 M [ IO ] = 45 . × 10 Mand the molar solubility of Pb(IO 3 ) 2 is 2.2× 10 –5 mol/L. This iterativeapproach to solving an equation is known as the method of successiveapproximations.Practice Exercise 6.8Calculate the molar solubility for Hg 2 Cl 2 in 0.10 M NaCl and compareyour answer to its molar solubility in deionized water (see Practice Exercise6.7).Click here to review your answer to this exercise.6G.3 A Systematic Approach to Solving Equilibrium ProblemsCalculating the solubility of Pb(IO 3 ) 2 in a solution of Pb(NO 3 ) 2 is morecomplicated than calculating its solubility in deionized water. The calculation,however, is still relatively easy to organize, and the simplifying assumptionfairly obvious. This problem is reasonably straightforward because itinvolves only one equilibrium reaction and one equilibrium constant.Determining the equilibrium composition of a system with multipleequilibrium reactions is more complicated. In this section we introduce asystematic approach to setting-up and solving equilibrium problems. Asshown in Table 6.1, this approach involves four steps.3Table 6.1 Systematic Approach to Solving Equilibrium ProblemsStep 1: Write all relevant equilibrium reactions and equilibrium constant expressions.Step 2:Step 3:Step 4:Count the unique species appearing in the equilibrium constant expressions;these are your unknowns. You have enough information to solve the problemif the number of unknowns equals the number of equilibrium constant expressions.If not, add a mass balance equation and/or a charge balance equation.Continue adding equations until the number of equations equals the numberof unknowns.Combine your equations and solve for one unknown. Whenever possible, simplifythe algebra by making appropriate assumptions. If you make an assumption,set a limit for its error. This decision influences your evaluation of theassumption.Check your assumptions. If any assumption proves invalid, return to the previousstep and continue solving. The problem is complete when you have ananswer that does not violate any of your assumptions.

240 Analytical Chemistry 2.0You may recall from Chapter 2 that this isthe difference between a formal concentrationand a molar concentration. Thevariable C represents a formal concentration.We use absolute values because we arebalancing the concentration of charge andconcentrations cannot be negative.There are situations where it is impossibleto write a charge balance equation becausewe do not have enough information aboutthe solution’s composition. For example,suppose we fix a solution’s pH using a buffer.If the buffer’s composition is not specified,then a charge balance equation cannot be written.In addition to equilibrium constant expressions, two other equationsare important to the systematic approach for solving equilibrium problems.The first of these is a mass balance equation, which is simply a statementthat matter is conserved during a chemical reaction. In a solution of a aceticacid, for example, the combined concentrations of the conjugate weak acid,CH 3 COOH, and the conjugate weak base, CH 3 COO – , must equal aceticacid’s initial concentration, C CH 3COOH .C CH COOH= CH33COOH + 3−[ ] [ CH COO ]The second equation is a charge balance equation, which requiresthat total charge from the cations equal the total charge from the anions.Mathematically, the charge balance equation is∑i+ z + − z−( z ) [ C ] = ( z ) [ A ]iiwhere [C z+ ] i and [A z– ] j are, respectively, the concentrations of the ith cationand the jth anion, and |(z + ) i | and |(z – ) j | are the absolute values of theith cation’s charge and the jth anion’s charge. Every ion in solution, evenif it does not appear in an equilibrium reaction, must appear in the chargebalance equation. For example, the charge balance equation for an aqueoussolution of Ca(NO 3 ) 2 is∑22× [ Ca + ] + [ HO + ] = [ OH − ] + [ NO− ]3Note that we multiply the concentration of Ca 2+ by two, and that we includethe concentrations of H 3 O + and OH – .Example 6.11Write mass balance equations and a charge balance equation for a 0.10 Msolution of NaHCO 3 .jjj3So l u t i o nIt is easier to keep track of the species in solution if we write down thereactions controlling the solution’s composition. These reactions are thedissolution of a soluble saltNaHCO Na HCO3 () s → + ( aq ) + −( aq3)and the acid–base dissociation reactions of HCO 3 – and H 2 O− + 2−HCO ( aq) + H O() l H O ( aq) + CO ( aq )3 2 33−−HCO ( aq) + H O() l OH ( aq) + H CO ( aq )3 2 2 32H O() l H O + ( aq) + OH− ( aq )2 3

Chapter 6 Equilibrium Chemistry241The mass balance equations are−2−010 M= H CO + HCO + CO2 3 3 3. [ ] [ ] [ ]+010 . M=[ Na ]and the charge balance equation is+ + − − 2−[ Na ] + [ HO ] = [ OH ] + [ HCO ] + 2×[ CO ]33 3Practice Exercise 6.9Write appropriate mass balance and charge balance equations for a solutioncontaining 0.10 M KH 2 PO 4 and 0.050 M Na 2 HPO 4 .Click here to review your answer to this exercise.6G.4 pH of a Monoprotic Weak AcidTo illustrate the systematic approach to solving equilibrium problems, let’scalculate the pH of 1.0 M HF. Two equilibrium reactions affect the pH.The first, and most obvious, is the acid dissociation reaction for HF+ −HF( aq) + HO() l HO ( aq) + ( aq )2 3FStep 1: Write all relevant equilibrium reactionsand equilibrium constant expressions.for which the equilibrium constant expression is+ −[ HO ][ F ]−= = 68 . × 10 4 6.35[ HF]K a3The second equilibrium reaction is the dissociation of water, which is anobvious yet easily neglected reaction2H O() l H O + ( aq) + OH− ( aq )2 3K w 3= [ HO+ ][ OH− ] = 100 . × 10 −146.36Counting unknowns, we find four: [HF], [F – ], [H 3 O + ], and [OH – ]. Tosolve this problem we need two additional equations. These equations area mass balance equation on hydrofluoric acidand a charge balance equationC HF= [ HF] + [ F− ] 6.37+ − −[ HO ] = [ OH ] + [ F ]6.383With four equations and four unknowns, we are ready to solve theproblem. Before doing so, let’s simplify the algebra by making twoassumptions.Step 2: Count the unique species appearingin the equilibrium constant expressions;these are your unknowns. You haveenough information to solve the problemif the number of unknowns equals thenumber of equilibrium constant expressions.If not, add a mass balance equationand/or a charge balance equation. Continueadding equations until the numberof equations equals the number of unknowns.

242 Analytical Chemistry 2.0Step 3: Combine your equations andsolve for one unknown. Whenever possible,simplify the algebra by making appropriateassumptions. If you make anassumption, set a limit for its error. Thisdecision influences your evaluation theassumption.Assumption One. Because HF is a weak acid, the solution must be acidic.For an acidic solution it is reasonable to assume that[H 3 O + ] >> [OH – ]which simplifies the charge balance equation to+ −[ HO ] = [ F ]6.393Assumption Two. Because HF is a weak acid, very little dissociation occurs.Most of the HF remains in its conjugate weak acid form and it isreasonable to assume that[HF] >> [F – ]which simplifies the mass balance equation toC HF= [ HF] 6.40For this exercise let’s accept an assumption if it introduces an error of lessthan ±5%.Substituting equation 6.39 and equation 6.40 into equation 6.35, andsolving for the concentration of H 3 O + gives usKa+ + + 2[ HO ][ HO ] [ HO ]3 33= = = 68 . × 10CCHFHF−4Step 4: Check your assumptions. If anyassumption proves invalid, return to theprevious step and continue solving. Theproblem is complete when you have ananswer that does not violate any of yourassumptions.[ HO ] = KC = ( 68 . × 10 )( 10 . ) = 26 . × 10+ −4 −23 a HFBefore accepting this answer, we must verify our assumptions. The first assumptionis that [OH – ] is significantly smaller than [H 3 O + ]. Using equation6.36, we find that−Kw100 . × 10[ OH ] = =+[ HO ] 26 . × 103−14−2= 38 . × 10−13Clearly this assumption is acceptable. The second assumption is that [F – ] issignificantly smaller than [HF]. From equation 6.39 we have[F – ] = 2.6 × 10 –2 MBecause [F – ] is 2.60% of C HF , this assumption is also acceptable. Given that[H 3 O + ] is 2.6 × 10 –2 M, the pH of 1.0 M HF is 1.59.How does the calculation change if we limit an assumption’s error toless than ±1%? In this case we can no longer assume that [HF] >> [F – ] andwe cannot simplify the mass balance equation. Solving the mass balanceequation for [HF][ HF] = C − [ F − ] = C −[ H O+ ]HF HF 3and substituting into the K a expression along with equation 6.39 gives

Chapter 6 Equilibrium Chemistry243Ka=C+[ HO ]3+−[ HO ]HF 3Rearranging this equation leaves us with a quadratic equation+ 2+[ HO ] + K [ HO ] − K C = 03 a 3 a HFwhich we solve using the quadratic formulab b acx = − ± 2 − 42awhere a, b, and c are the coefficients in the quadratic equationax 2 + bx + c = 0Solving a quadratic equation gives two roots, only one of which has chemicalsignificance. For our problem, the equation’s roots arex = − × −4 ± × −4 2 − × −468 . 10 ( 68 . 10 ) ( 4)()( 1 68 . 10 )( 1)( 2)()14 268 . 10 522 . 10x = − × −± × −2−2x = 257 . × 10 or − 263 . × 10 −22Only the positive root is chemically significant because the negative rootgives a negative concentration for H 3 O + . Thus, [H 3 O + ] is 2.6 × 10 –2 Mand the pH is 1.59.You can extend this approach to calculating the pH of a monoproticweak base by replacing K a with K b , replacing C HF with the weak base’sconcentration, and solving for [OH – ] in place of [H 3 O + ].Practice Exercise 6.10Calculate the pH of 0.050 M NH 3 . State any assumptions you make insolving the problem, limiting the error for any assumption to ±5%. TheK b value for NH 3 is 1.75 × 10 –5 .Click here to review your answer to this exercise.6G.5 pH of a Polyprotic Acid or BaseA more challenging problem is to find the pH of a solution containing apolyprotic weak acid or one of its conjugate species. As an example, considerthe amino acid alanine, whose structure is shown in Figure 6.12. Theladder diagram in Figure 6.13 shows alanine’s three acid–base forms andtheir respective areas of predominance. For simplicity, we identify thesespecies as H 2 L + , HL, and L – .OH 2 N CH COHCH 3Figure 6.12 Structure of the aminoacid alanine, which has pK avalues of 2.348 and 9.867.

244 Analytical Chemistry 2.0pHOH 2 N CH C O–CH 3pK a1= 9.867O+ H3 N CH C O–CH 3pK a2= 2.348O+H 3 N CH C OHCH 3L –HLH 2 L +Figure 6.13 Ladder diagram for alanine.pH o f 0.10 M Al a n i n e Hy d r o c h l o r i d e (H 2 L + )Alanine hydrochloride is a salt of the diprotic weak acid H 2 L + and Cl – .Because H 2 L + has two acid dissociation reactions, a complete systematicsolution to this problem is more complicated than that for a monoproticweak acid. The ladder diagram in Figure 6.13 helps us simplify the problem.Because the areas of predominance for H 2 L + and L – are so far apart, we canassume that a solution of H 2 L + is not likely to contain significant amountsof L – . As a result, we can treat H 2 L + as though it is a monoprotic weak acid.Calculating the pH of 0.10 M alanine hydrochloride, which is 1.72, is leftto the reader as an exercise.pH o f 0.10 M So d i u m Al a n i n a t e (L – )The alaninate ion is a diprotic weak base. Because L – has two base dissociationreactions, a complete systematic solution to this problem is morecomplicated than that for a monoprotic weak base. Once again, the ladderdiagram in Figure 6.13 helps us simplify the problem. Because the areasof predominance for H 2 L + and L – are so far apart, we can assume that asolution of L – is not likely to contain significant amounts of H 2 L + . As aresult, we can treat L – as though it is a monoprotic weak base. Calculatingthe pH of 0.10 M sodium alaninate, which is 11.42, is left to the reader asan exercise.pH o f 0.1 M Al a n i n e (HL)Finding the pH of a solution of alanine is more complicated than our previoustwo examples because we cannot ignore the presence of both H 2 L +and L – . To calculate the solution’s pH we must consider alanine’s acid dissociationreaction+ −HL( aq) + HO() l HO ( aq) + ( aq )2 3Land its base dissociation reaction− +HL( aq) + HO() l OH ( aq) + HL( aq )2 2As always, we must also consider the dissociation of water2H O() l H O + ( aq) + OH− ( aq )2 3This leaves us with five unknowns—[H 2 L + ], [HL], [L – ], [H 3 O + ], and[OH – ]—for which we need five equations. These equations are K a2 andK b2 for alanine+ −[ HO ][ L ]=[ HL]K a23

Chapter 6 Equilibrium Chemistry245the K w equationKb2− +K [ OH ][ HL ]w2= =K [ HL]a1K w= 3+ −[ HO ][ OH ]a mass balance equation for alanineand a charge balance equation+ −C HL= [ HL ] + [ HL] + [ L ]2+ + − −[ HL ] + [ HO ] = [ OH ] + [ L ]2 3Because HL is a weak acid and a weak base, it seems reasonable to assumethat[HL] >> [H 2 L + ] + [L – ]which allows us to simplify the mass balance equation toNext we solve K b2 for [H 2 L + ]C HL= [ HL]++K [ HL][ HO ][ HL] C [ HOw3HL 3[ HL ] = = =2−K [ OH ] Ka1a1K a1+ ]and K a2 for [L – ]−K [ HL]K Ca2a2 HL[ L ] = =+ +[ HO ] [ HO ]33Substituting these equations for [H 2 L + ] and [L – ], along with the equationfor K w , into the charge balance equation give usC+[ HO ]+K K Cwa2 HL+ [ HO ] = +3+ +K[H O ] [ H O ]HL 3which we simplify toa1⎛+[ HO ]C3⎝⎜K+[ HO ] =HL⎞++⎠⎟ = 11K[ HO ]+ K Ca1 3( K C + K )=⎛C ⎞HL+ 1⎝⎜K ⎠⎟2 a2 HL w3a13( )3w a2 HL( K C + K )C + KKa 1 a2 HL w( )HLa1

246 Analytical Chemistry 2.0[ HO ]K K C+ a1 a2 HL a1 w3=CHL+ K K+ Ka1We can further simplify this equation if K a1 K w

Chapter 6 Equilibrium Chemistry247+ −[ NH ][ OH ]4−5K b= = 175 . × 10 6.42[ NH ]3K w 3= [ HO+ ][ OH− ] = 100 . × 10 −146.43[ Ag(NH ) ]= = 17 . × 10+3 2β 2 +2[ Ag ][ NH ]376.44We still need three additional equations. The first of these equation is a massbalance for NH 3 .+ +C = [ NH ] + [ NH ] + 2×[ Ag(NH) ]NH 3 3 4 3 2 6.45In writing this mass balance equation we multiply the concentration ofAg(NH 3 ) 2 + by two since there are two moles of NH 3 per mole of Ag(NH 3 ) 2 + .The second additional equation is a mass balance between iodide and silver.Because AgI is the only source of I - and Ag + , each iodide in solution musthave an associated silver ion, which may be Ag + or Ag(NH 3 ) 2 + ; thusFinally, we include a charge balance equation.[ I − ] = [ Ag + ] + [ Ag(NH ) + ]3 26.46+ + + + − −[ Ag ] + [ Ag(NH ) ] + [ NH ] + [ H O ] = [ OH ] + [ I ]3 2 4 36.47Although the problem looks challenging, three assumptions greatlysimplify the algebra.Assumption One. Because the formation of the Ag(NH 3 ) 2 + complex is sofavorable (b 2 is 1.7 × 10 7 ), there is very little free Ag + and it is reasonableto assume that[Ag + ]

248 Analytical Chemistry 2.0C NH3= [ NH ]3 6.48Finally, using assumption one, which suggests that the concentration ofAg(NH 3 ) 2 + is much larger than the concentration of Ag + , we simplify themass balance equation for I – to− +[ I ] = [ Ag(NH ) ]3 26.49Now we are ready to combine equations and solve the problem. Webegin by solving equation 6.41 for [Ag + ] and substitute it into b 2 (equation6.44), leaving us with[ Ag(NH ) ][ I ]+ −3 2β 2=2K sp[ NH ]36.50Next we substitute equation 6.48 and equation 6.49 into equation 6.50,obtaining[ I ]− 2β 2=Solving equation 6.51 for [I – ] givesK( C )sp NH 326.51−7 −17[ ] = C β 2K = ( 010 . ) ( 1. 7× 10 )(. 8 3× 10 ) = 3.76× 10 −6MINH3spBecause one mole of AgI produces one mole of I – , the molar solubility ofAgI is the same as the [I – ], or 3.8 × 10 –6 mol/L.Before accepting this answer we need to check our assumptions. Substituting[I – ] into equation 6.41, we find that the concentration of Ag + isK−+ sp 83 . × 10[ Ag ] = =−[ I ] 376 . × 1017−6= 22 . × 10− 11 MDid you notice that our solution to thisproblem did not make use of equation6.47, the charge balance equation? Thereason for this is that we did not try tosolve for the concentration of all sevenspecies. If we need to know the completeequilibrium composition of the reactionmixture, then we would need to incorporatethe charge balance equation into oursolution.Substituting the concentrations of I – and Ag + into the mass balance equationfor iodide (equation 6.46), gives the concentration of Ag(NH 3 ) 2 + as[Ag(NH 3 ) 2 + ] = [I – ] – [Ag + ] = 3.76 × 10 –6 – 2.2 × 10 –11 =3 .8 × 10 –6 MOur first assumption that [Ag + ] is significantly smaller than the [Ag(NH 3 ) 2 + ]is reasonable.Substituting the concentrations of Ag + and Ag(NH 3 ) 2 + into equation6.44 and solving for [NH 3 ], gives+−6[ Ag(NH ) ]3 238 . × 10[ NH ] = == 010 . M3+−117[ Ag ] β ( 22 . × 10 )( 17 . × 10 )2From the mass balance equation for NH 3 (equation 6.44) we see that[NH 4 + ] is negligible, verifying our second assumption that [NH 4 + ] is sig-

Chapter 6 Equilibrium Chemistry249nificantly smaller than [NH 3 ]. Our third assumption that [Ag(NH 3 ) 2 + ] issignificantly smaller than [NH 3 ] also is reasonable.6HBuffer SolutionsAdding as little as 0.1 mL of concentrated HCl to a liter of H 2 O shifts thepH from 7.0 to 3.0. Adding the same amount of HCl to a liter of a solutionthat is 0.1 M in acetic acid and 0.1 M in sodium acetate, however, resultsin a negligible change in pH. Why do these two solutions respond so differentlyto the addition of HCl?A mixture of acetic acid and sodium acetate is one example of an acid–base buffer. To understand how this buffer works to limit the change inpH, we need to consider its acid dissociation reaction+ −CH COOH( aq) + HO() l HO ( aq) + CH COO ( aq )3 2 3 3and its corresponding acid dissociation constant[ CH COO ][ H O ]−= = 175 . × 10 5 6.52[ CH COOH]− +3 3K a3Taking the negative log of the terms in equation 6.52 and solving for pHleaves us with the result shown here.−CH COOCH CO33pH = pK + log [ ] = . + log [ O− ]476a[ CH COOH][ CH COOH]336.53Buffering occurs because of the logarithmic relationship between pH andthe ratio of the concentrations of acetate and acetic acid. Here is an exampleto illustrate this point. If the concentrations of acetic acid and acetate areequal, the buffer’s pH is 4.76. If we convert 10% of the acetate to acetic acid,by adding a strong acid, the ratio [CH 3 COO – ]/[CH 3 COOH] changesfrom 1.00 to 0.818, and the pH decreases from 4.76 to 4.67—a decreaseof only 0.09 pH units.6H.1 Systematic Solution to Buffer ProblemsEquation 6.53 is written in terms of the equilibrium concentrations ofCH 3 COOH and CH 3 COO – . A more useful relationship relates a buffer’spH to the initial concentrations of the weak acid and the weak base. Wecan derive a general buffer equation by considering the following reactionsfor a weak acid, HA, and the salt of its conjugate weak base, NaA.+ −NaA( aq) → Na ( aq) + A ( aq)+−HA( aq) + HO() l HO ( aq ) + A ( aq)2 3+ −2HO() l HO ( aq) + OH ( aq )2 3

250 Analytical Chemistry 2.0Because the concentrations of Na + , A – , HA, H 3 O + , and OH – are unknown,we need five equations to uniquely define the solution’s composition. Twoof these equations are the equilibrium constant expressions for HA andH 2 O.+ −[ HO ][ A ]=[ HA]K a36.54K w= 3+ −[ HO ][ OH ]The remaining three equations are mass balance equations for HA and forNa + C + C = [ HA] + [ A− ] 6.55HA NaALawrence Henderson (1878-1942)first developed a relationship between[H 3 O + ], [HA], and [A – ] while studyingthe buffering of blood. Kurt Hasselbalch(1874-1962) modified Henderson’s equationby transforming it to the logarithmicform shown in equation 6.60.The assumptions leading to equation 6.60produce a minimal error in pH (

Chapter 6 Equilibrium Chemistry251Example 6.12Calculate the pH of a buffer that is 0.020 M in NH 3 and 0.030 M inNH 4 Cl. What is the pH after adding 1.0 mL of 0.10 M NaOH to 0.10L of this buffer?So l u t i o nThe acid dissociation constant for NH 4 + is 5.70 × 10 –10 , or a pK a of 9.24.Substituting the initial concentrations of NH 3 and NH 4 Cl into equation6.60 and solving, we find that the buffer’s pH is0 020pH = 924 . + log . = 906 .0.030Adding NaOH converts a portion of the NH 4 + to NH 3 as a result of thefollowing reaction+ −NH + OH HO+NH4 23With a pH of 9.06, the concentration ofH 3 O + is 8.71×10 –10 and the concentrationof OH – is 1.15×10 –5 . Because bothof these concentrations are much smallerthan either C NH3 or C NH4 Cl , the approximationsleading to equation 6.60 arereasonable.Because this reaction’s equilibrium constant is so large (it is 5.7 × 10 4 ), wemay treat the reaction as if it goes to completion. The new concentrationsof NH 4 + and NH 3 areC+=NH4molNH+ −4V− mol OHtotal−3( 0. 030 M)(0.10 L) − ( 010 . M)(1.0×10 L)== 0. 029 M−3 0.10 L + 1.0×10 LCNH3molNH + mol OH3=Vtotal−−3( 0. 020 M)(0.10 L) + ( 010 . M)(1.0×10 L)== 0.021 M−3 0.10 L + 1.0×10 LSubstituting these concentrations into the equation 6.60 gives a pH ofPractice Exercise 6.120 021pH = 924 . + log . = 910 .0.029Calculate the pH of a buffer that is 0.10 M in KH 2 PO 4 and 0.050 M inNa 2 HPO 4 . What is the pH after adding 5.0 mL of 0.20 M HCl to 0.10L of this buffer. Use Appendix 11 to find the appropriate K a value.Click here to review your answer to this exerciseNote that adding NaOH increases the pHfrom 9.06 to 9.10. As we expect, adding abase makes the pH more basic. Checkingto see that the pH changes in the rightdirection is one way to catch a calculationerror.

252 Analytical Chemistry 2.0We can use a multiprotic weak acid to prepare buffers at as many differentpH’s as there are acidic protons, with the Henderson–Hasselbalch equationapplying in each case. For example, using malonic acid (pK a1 = 2.85and pK a1 = 5.70) we can prepare buffers with pH values ofCpH = 285 . + logCCpH = 570 . + logC−HMHM 22−M−HMwhere H 2 M, HM – and M 2– are malonic acid’s different acid–base forms.Although our treatment of buffers relies on acid–base chemistry, wecan extend the use of buffers to equilibria involving complexation or redoxreactions. For example, the Nernst equation for a solution containing Fe 2+and Fe 3+ is similar in form to the Henderson-Hasselbalch equation.+oFeE = E3+ +− . log [ 2]20 05916Fe / Fe3+[ Fe ]A solution containing similar concentrations of Fe 2+ and Fe 3+ is bufferedto a potential near the standard state reduction potential for Fe 3+ . We callsuch solutions redox buffers. Adding a strong oxidizing agent or a strongreducing agent to a redox buffer results in a small change in potential.6H.2 Representing Buffer Solutions with Ladder DiagramsA ladder diagram provides a simple graphical description of a solution’spredominate species as a function of solution conditions. It also providesa convenient way to show the range of solution conditions over which abuffer is effective. For example, an acid–base buffer exists when the concentrationsof the weak acid and its conjugate weak base are similar. Forconvenience, let’s assume that an acid–base buffer exists when110−[ CH COO ]3≤≤[ CH COOH]Substituting these ratios into the Henderson–Hasselbalch equation1pH = pK+ log = pK−1aa1010pH = pK+ log = pK+ 1aa13shows that an acid–base buffer works over a pH range of pK a ± 1.Using the same approach, it is easy to show that a metal-ligand complexationbuffer for ML n exists when101

Chapter 6 Equilibrium Chemistry253pHpLE4.17F –11.69Ca 2+0.184Sn 4+pK a= 3.17logK1 = 10.69E o = 0.1542.17HF9.69Ca(EDTA) 2-0.124Sn 2+(a) (b) (c)Figure 6.14 Ladder diagrams showing buffer regions for (a) an acid–base buffer forHF and F – ; (b) a metal–ligand complexation buffer for Ca 2+ and Ca(EDTA) 2– ;and (c) an oxidation–reduction (redox) buffer for Sn 4+ and Sn 2+ .1 1pL = logK± 1 or pL = log β ±nnn npL = –log[L]where K n or b n is the relevant stepwise or overall formation constant. Foran oxidizing agent and its conjugate reducing agent, a redox buffer existswhen1 RT 0.05916E = Eo ± × = Eo ± (at25C)on FnLadder diagrams showing buffer regions for several equilibria are shown inFigure 6.14.6H.3 Preparing BuffersBuffer capacity is the ability of a buffer to resist a change in pH whenadding a strong acid or a strong base. A buffer’s capacity to resist a change inpH is a function of the concentrations of the weak acid and the weak base,as well as their relative proportions. The importance of the weak acid’s concentrationand the weak base’s concentration is obvious. The more moles ofweak acid and weak base a buffer has, the more strong base or strong acidit can neutralize without significantly changing its pH.The relative proportions of a weak acid and a weak base also affects howmuch the pH changes when adding a strong acid or a strong base. Buffersthat are equimolar in weak acid and weak base require a greater amount ofstrong acid or strong base to bring about a one unit change in pH. Consequently,a buffer is most effective against the addition of strong acids orstrong bases when its pH is near the weak acid’s pK a value.Although higher concentrations of bufferingagents provide greater buffer capacity,there are reasons for using smaller concentrations,including the formation of unwantedprecipitates and the tolerance ofcells for high concentrations of dissolvedsalts.A good “rule of thumb” when choosing abuffer is to select one whose reagents havea pK a value close to your desired pH.

254 Analytical Chemistry 2.0The 1mM FeCl 3 also contains a few dropsof concentrated HNO 3 to prevent theprecipitation of Fe(OH) 3 .Buffer solutions are often prepared using standard “recipes” found inthe chemical literature. 3 In addition, there are computer programs and onlinecalculators to aid in preparing buffers. 4 Perhaps the simplest way tomake a buffer, however, is to prepare a solution containing an appropriateconjugate weak acid and weak base and measure its pH. You can then adjustthe pH to the desired value by adding small portions of either a strong acidor a strong base.6IActivity EffectsCareful measurements on the metal–ligand complex Fe(SCN) 2+ suggestthat its stability decreases in the presence of inert ions. 5 We can demonstratethis by adding an inert salt to an equilibrium mixture of Fe 3+ and SCN – .Figure 6.15a shows the result of mixing together equal volumes of 1.0mM FeCl 3 and 1.5 mM KSCN, both of which are colorless. The solution’sreddish–orange color is due to the formation of Fe(SCN) 2+ .Fe 3 + −SCN Fe(SCN)2 +( aq) + ( aq) ( aq)6.61Adding 10 g of KNO 3 to the solution and stirring to dissolve the solid, producesthe result shown in Figure 6.15b. The solution’s lighter color suggeststhat adding KNO 3 shifts reaction 6.61 to the left, decreasing the concentrationof Fe(SCN) 2+ and increasing the concentrations of Fe 3+ and SCN – .The result is a decrease in the complex’s formation constant, K 1 .[ Fe(SCN) ]2+K 1=3+ −[ Fe ][ SCN ]6.623 See, for example, (a) Bower, V. E.; Bates, R. G. J. Res. Natl. Bur. Stand. (U. S.) 1955, 55, 197–200; (b) Bates, R. G. Ann. N. Y. Acad. Sci. 1961, 92, 341–356; (c) Bates, R. G. Determinationof pH, 2nd ed.; Wiley-Interscience: New York, 1973.4 (a) Lambert, W. J. J. Chem. Educ. 1990, 67, 150–153; (b) http://www.bioinformatics.org/JaMBW/5/4/index.html.5 Lister, M. W.; Rivington, D. E. Can. J. Chem. 1995, 33, 1572–1590.Figure 6.15 The effect of a inert salt on a reaction’sequilibrium position is shown by the solutionsin these two beakers. The beaker on theleft contains equal volumes of 1.0 mM FeCl 3and 1.5 mM KSCN. The solution’s color is dueto the formation of the metal–ligand complexFe(SCN) 2+ . Adding 10 g of KNO 3 to the beakeron the left produces the result shown onthe right. The lighter color suggests that thereis less Fe(SCN) 2+ as a result of the equilibriumin reaction 6.61 shifting to the left. (a) (b)

Chapter 6 Equilibrium Chemistry255ionic atmosphereδ – δ +chargeFe 3+ SCN –charge+30distancedistance–10Figure 6.16 Ions of Fe 3+ and SCN – aresurrounded by ionic atmospheres withnet charges of d – and d + . Because of theseionic atmospheres, each ion’s apparentcharge at the edge of its ionic atmosphereis less than the ion’s actual charge.Why should adding an inert electrolyte affect a reaction’s equilibriumposition? We can explain the effect of KNO 3 on the formation ofFe(SCN) 2+ by considering the reaction on a microscopic scale. The solutionin Figure 6.15b contains a variety of cations and anions—Fe 3+ , SCN – , K + ,NO 3 – , H 3 O + , and OH – . Although the solution is homogeneous, on average,there are slightly more anions in regions near Fe 3+ ions, and slightlymore cations in regions near SCN – ions. As shown in Figure 6.16, eachFe 3+ ion and SCN – ion is surrounded by an ionic atmosphere of oppositecharge (d – and d + ) that partially screen the ions from each other. Becauseeach ion’s apparent charge at the edge of its ionic atmosphere is less thanits actual charge, the force of attraction between the two ions is smaller. Asa result, the formation of the Fe(SCN) 2+ is slightly less favorable and theformation constant in equation 6.62 is slightly smaller. Higher concentrationsof KNO 3 increase d – and d + , resulting in even smaller values for theformation constant.Io n i c St r e n g t hTo factor the concentration of ions into the formation constant forFe(SCN) 2+ , we need a way to express that concentration in a meaningfulway. Because both an ion’s concentration and its charge are important, wedefine the solution’s ionic strength, µ asµ= 1 ∑2where c i and z i are the concentration and charge of the ith ion.icz i2i

256 Analytical Chemistry 2.0In calculating the ionic strengths of thesesolutions we are ignoring the presenceof H 3 O + and OH – , and, in the case ofNa 2 SO 4 , the presence of HSO 4– fromthe base dissociation reaction of SO 42– .In the case of 0.10 M NaCl, the concentrationsof H 3 O + and OH – are 1.0 × 10 –7 ,which is significantly smaller than theconcentrations of Na + and Cl – .Because SO2– 4 is a very weak base(K b = 1.0 × 10 –12 ), the solution is onlyslightly basic (pH = 7.5), and the concentrationsof H 3 O + , OH – , and HSO– 4 arenegligible.Although we can ignore the presence ofH 3 O + , OH – , and HSO 4– when calculatingthe ionic strength of these two solutions,be aware that an equilibrium reactionmay well generate ions that affect thesolution’s ionic strength.Example 6.13Calculate the ionic strength for a solution of 0.10 M NaCl. Repeat thecalculation for a solution of 0.10 M Na 2 SO 4 .So l u t i o nThe ionic strength for 0.10 M NaCl is1 + 2 −2µ= [ Na ] ×+ ( 1) + [ Cl ] × ( −1)21= ( × +2 010 . ) ( 1 ) + ( 010 . ) × ( −1)= 010 . M{ }2 2{ }For 0.10 M Na 2 SO 4 the ionic strength is1 + 2−µ= [ Na ] ×+ ( 1) + [ SO ] × ( −2)421= ( × +2 010 2 2. ) ( 1) + ( 0. 20) × ( −2)= 030 . M2 2{ }{ }Note that the unit for ionic strength is molarity, but that a salt’s ionicstrength need not match its molar concentration. For a 1:1 salt, such asNaCl, ionic strength and molar concentration are identical. The ionicstrength of a 2:1 electrolyte, such as Na 2 SO 4 , is three times larger than theelectrolyte’s molar concentration.Activity a n d Activity Co e f f i c i e n t sFigure 6.15 shows that adding KNO 3 to a mixture of Fe 3+ and SCN – decreasesthe formation constant for Fe(SCN) 2+ . This creates a contradiction.Earlier in this chapter we showed that there is a relationship between a reaction’sstandard-state free energy, DG o , and its equilibrium constant, K.∆G o =−RT ln KBecause a reaction has only one standard-state, its equilibrium constantmust be independent of solution conditions. Although ionic strength affectsthe apparent formation constant for Fe(SCN) 2+ , reaction 6.61 musthave an underlying thermodynamic formation constant that is independentof ionic strength.The apparent formation constant for Fe(SCN) 2+ , as shown in equation6.62, is a function of concentrations. In place of concentrations, we definethe true thermodynamic equilibrium constant using activities. The activityof species A, a A , is the product of its concentration, [A], and a solutiondependentactivity coefficient, γ A .

Chapter 6 Equilibrium Chemistry257a A= [ A]γThe true thermodynamic formation constant for Fe(SCN) 2+ , therefore, isAK12+a2+[ Fe(SCN) ]γ2Fe(SCN)Fe(SCN)= =3+ −a a [ Fe ] γ [ SCN ] γ3+ −Fe SCN3+ −FeSCNThe activity coefficient for a species corrects for any deviation betweenits physical concentration and its ideal value. For a gas, a pure solid,a pure liquid, or a non-ionic solute, the activity coefficient is approximatelyone under most reasonable experimental conditions. For reactions involvingonly these species, the difference between activity and concentration isnegligible. The activity coefficient for an ion, however, depends on the solution’sionic strength, the ion’s charge, and the ion’s size. It is possible to calculateactivity coefficients using the extended Debye-Hückel equation+Unless otherwise specified, the equilibriumconstants in the appendices are thermodynamicequilibrium constants.For a gas the proper terms are fugacity andfugacity coefficient, instead of activity andactivity coefficient.log γA. µA= − 051 × z 2×1+ 3.3× α × µA6.63where z A is the ion’s charge, α A is the effective diameter of the hydrated ionin nanometers (Table 6.2), m is the solution’s ionic strength, and 0.51 and3.3 are constants appropriate for an aqueous solution at 25 o C. An ion’s effectivehydrated radius is the radius of the ion plus those water moleculesclosely bound to the ion. The effective radius is greater for smaller, morehighly charged ions than it is for larger, less highly charged ions.Table 6.2 Effective Diameters (a) for Selected IonsIonEffective Diameter (nm)H 3 O + 0.9Li + 0.6Na + , IO – 3 , HSO – 3 , HCO – –3 , H 2 PO 4 0.45OH – , F – , SCN – , HS – , ClO – 3 , ClO – –4 , MnO 4 0.35K + , Cl – , Br – , I – , CN – , NO – –2 , NO 3 0.3Cs + , Tl + , Ag + +, NH 4 0.25Mg 2+ , Be 2+ 0.8Ca 2+ , Cu 2+ , Zn 2+ , Sn 2+ , Mn 2+ , Fe 2+ , Ni 2+ , Co 2+ 0.6Sr 2+ , Ba 2+ , Cd 2+ , Hg 2+ , S 2– 0.5Pb 2+ , CO 2– 2–3 , SO 3 0.45Hg 2+ 2 , SO 2– 4 , S 2 O 2– 3 , CrO 2– 2–4 , HPO 4 0.40Al 3+ , Fe 3+ , Cr 3+ 0.9PO 3– 3–4 , Fe(CN) 6 0.4Zr 4+ , Ce 4+ , Sn 4+ 1.14–Fe(CN) 6 0.5Source: Kielland, J. J. Am. Chem. Soc. 1937, 59, 1675–1678.

258 Analytical Chemistry 2.0Several features of equation 6.63 deserve mention. First, as the ionicstrength approaches zero an ion’s activity coefficient approaches a value ofone. In a solution where m= 0, an ion’s activity and its concentration areidentical. We can take advantage of this fact to determine a reaction’s thermodynamicequilibrium constant by measuring the apparent equilibriumconstant for several increasingly smaller ionic strengths and extrapolatingback to an ionic strength of zero. Second, an activity coefficient is smaller,and the effect of activity is more important, for an ion with a higher chargeand a smaller effective radius. Finally, the extended Debye-Hückel equationprovides a reasonable value for an ion’s activity coefficient when the ionicstrength is less than 0.1. Modifications to equation 6.63 extend the calculationof activity coefficients to higher ionic strengths. 6In c l u d i n g Activity Co e f f i c i e n t s Wh e n So l v i n g Equilibrium Pr o b l e m sEarlier in this chapter we calculated the solubility of Pb(IO 3 ) 2 in deionizedwater, obtaining a result of 4.0 × 10 -5 mol/L. Because the only significantsource of ions is from the solubility reaction, the ionic strength is very lowand we can assume that g ≈ 1 for both Pb 2+ and IO 3 – . In calculating thesolubility of Pb(IO 3 ) 2 in deionized water, we do not need to account forionic strength.But what if the we need to know the solubility of Pb(IO 3 ) 2 in a solutioncontaining other, inert ions? In this case we need to include activitycoefficients in our calculation.Example 6.14Calculate the solubility of Pb(IO 3 ) 2 in a matrix of 0.020 M Mg(NO 3 ) 2 .As is true for any assumption, we will needto verify that it does not introduce toomuch error into our calculation.So l u t i o nWe begin by calculating the solution’s ionic strength. Since Pb(IO 3 ) 2 isonly sparingly soluble, we will assume that we can ignore its contributionto the ionic strength; thus12 2µ= {( 0. 020 M)( + 2) + ( 0. 040 M) ( −1) }= 0.060 M2Next, we use equation 6.63 to calculate the activity coefficients for Pb 2+and IO 3 – .logγ Pb22051 . ( 2) 0.060+= − × + ×1+ 3. 3× 0. 45×0.060=−0.366γ Pb2+ = 0.4316 Davies, C. W. Ion Association, Butterworth: London, 1962.

Chapter 6 Equilibrium Chemistry2592051 . ( 1) 0.060log γ− IO3= − × − ×1+ 3. 3× 0. 45×0.060=−0.0916γ− IO3= 0.810Defining the equilibrium concentrations of Pb 2+ and IO 3 – in terms of thevariable xConcentrations Pb(IO 3 ) 2 (s) Pb 2+ (aq) + 2IO 3– (aq)Initial solid 0 0Change solid +x +2xEquilibrium solid x 2xand substituting into the thermodynamic solubility product for Pb(IO 3 ) 2leaves us with2 2+ 2 − 2K = a a = γ [ Pb ] γ [ IO ] = 2.5× 10 −13sp 2+ − 2+ −Pb IO3Pb IO3K = ( 0. 431)( x)( 0. 810) ( 2x) = 2.5 × 10−sp32 2 13K = 1. 131x= 25 . × 10−sp3 13Solving for x gives 6.0 × 10 –5 , or a molar solubility of 6.0 × 10 –5 mol/L.Ignoring activity, as we did in our earlier calculation, gives the molar solubilityas 4.0 × 10 -5 mol/L. Failing to account for activity in this case underestimatesthe molar solubility of Pb(IO 3 ) 2 by 33%.As this example shows, failing to correct for the effect of ionic strengthcan lead to a significant error in an equilibrium calculation. Nevertheless,it is not unusual to ignore activities and to assume that the equilibriumconstant is expressed in terms of concentrations. There is a practical reasonfor this—in an analysis we rarely know the exact composition, much lessthe ionic strength of aqueous samples or of solid samples brought intosolution. Equilibrium calculations are a useful guide when developing ananalytical method; however, only by completing an analysis and evaluatingthe results can we judge whether our theory matches reality. In the end,our work in the laboratory is the most critical step in developing a reliableanalytical method.Practice Exercise 6.13Calculate the molar solubility of Hg 2 Cl 2 in 0.10 M NaCl, taking intoaccount the effect of ionic strength. Compare your answer to that fromPractice Exercise 6.8 in which you ignored the effect of ionic strength.Click here to review your answer to this exercise.The solution’s equilibrium composition is[Pb 2+ ] = 6.0×10 –5 M[IO 3– ] = 1.2×10–4 M[Mg 2+ ] = 0.020 M[NO 3– ] = 0.040 MBecause the concentrations of Pb 2+ andIO 3– are much smaller than the concentrationsof Mg 2+ and NO 3– , our decisionto ignore the contribution of Pb 2+ andIO 3– to the ionic strength is reasonable.How do we handle the calculation if wecan not ignore the concentrations of Pb 2+and IO 3– ? One approach is to use themethod of successive approximations.First, we recalculate the ionic strengthusing the concentrations of all ions, includingPb 2+ and IO 3– . Next, we recalculatethe activity coefficients for Pb 2+and IO 3– , and then recalculate the molarsolubility. We continue this cycle untiltwo successive calculations yield the samemolar solubility within an acceptable marginof error.This is a good place to revisit the meaningof pH. In Chapter 2 we defined pH aspH =−log[ HO ]Now we see that the correct definition is=pH =−loga+ −log γ + [ HO ]3+HO HO 33 3Failing to account for the effect of ionicstrength can lead to a significant error inthe reported concentration of H 3 O + . Forexample, if the pH of a solution is 7.00and the activity coefficient for H 3 O + is0.90, then the concentration of H 3 O + is1.11 × 10 –7 M, not 1.00 × 10 –7 M, anerror of +11%. Fortunately, in developingand carrying out analytical methods, weare more interested in controlling pH thanin calculating [H 3 O + ]. As a result, the differencebetween the two definitions of pHis rarely a significant concern.+

260 Analytical Chemistry 2.06JUsing Excel and R to Solve Equilibrium ProblemsIn solving equilibrium problems we typically make one or more assumptionsto simplify the algebra. These assumptions are important because theyallow us to reduce the problem to an equation in x that we can solve bysimply taking a square-root, a cube-root, or by using the quadratic equation.Without these assumptions, most equilibrium problems result in acubic equation (or a higher-order equation) that is harder to solve. BothExcel and R are useful tools for solving such equations.6J.1 ExcelExcel offers a useful tool—the Solver function—for finding a polynomialequation’s chemically significant roo. In addition, it is easy to solve a systemof simultaneous equations by constructing a spreadsheet that allows you totest and evaluate multiple solutions. Let’s work through two examples.Ex a m p l e 1: So l u b i l i t y o f Pb(IO 3 ) 2 in 0.10 M Pb(NO 3 ) 2In our earlier treatment of this problem we arrived at the following cubicequation4x+ 0. 40x= 25 . × 10−3 2 13where x is the equilibrium concentration of Pb 2+ . Although there are severalapproaches for solving cubic equations, none are computationally easy.One approach is to iterate in on the answer by finding two values of x, oneof which leads to a result larger than 2.5×10 –13 and one of which gives aresult smaller than 2.5×10 –13 . Having established boundaries for the valueof x, we then shift the upper limit and the lower limit until the precisionof our answer is satisfactory. Without going into details, this is how Excel’sSolver function works.To solve this problem, we first rewrite the cubic equation so that itsright-side equals zero.4 3 0 40 2 −x + . x − 25 . × 10 13 = 0Next, we set up the spreadsheet shown in Figure 6.17a, placing the formulafor the cubic equation in cell B2, and entering our initial guess for x inFigure 6.17 Spreadsheet demonstrating the use ofExcel’s Solver function to find the root of a cubicequation. The spreadsheet in (a) shows the cubicequation in cell B2 and the initial guess for the valueof x in cell B1; Excel replaces the formula with itsequivalent value. The spreadsheet in (b) shows theresults of running Excel’s Solver function.(a)(b)AB1 x = 02 function = 4*b1^3 + 0.4*b1^2 – 2.5e–13AB1 x = 7.90565E–072 function –5.71156E–19

Chapter 6 Equilibrium Chemistry261cell B1. We expect x to be small—because Pb(IO 3 ) 2 is not very soluble—so setting our initial guess to 0 seems reasonable. Finally, we access theSolver function by selecting Solver... from the Tools menu, which opensthe Solver Parameters window.To define the problem, place the cursor in the box for Set Target Celland then click on cell B2. Select the Value of: radio button and enter 0 inthe box. Place the cursor in the box for By Changing Cells: and click on cellB1. Together, these actions instruct the Solver function to change the valueof x, which is in cell B1, until the cubic equation in cell B2 equals zero.Before we actually solve the function, we need to consider whetherthere are any limitations for an acceptable result. For example, we knowthat x cannot be smaller than 0 because a negative concentration is notpossible. We also want to ensure that the solution’s precision is acceptable.Click on the button labeled Options... to open the Solver Options window.Checking the option for Assume Non-Negative forces the Solver to maintaina positive value for the contents of cell B1, meeting one of our criteria. Settingthe precision takes a bit more thought. The Solver function uses theprecision to decide when to stop its search, doing so whenexpected value − calculated value × 100 ≤ precision (%)where expected value is the target cell’s desired value (0 in this case), calculatedvalue is the function’s current value (cell B1 in this case), and precisionis the value we enter in the box for Precision. Because our initial guess ofx = 0 gives a calculated result of 2.5×10 –13 , accepting the Solver’s defaultprecision of 1×10 –6 stops the search after one cycle. To be safe, set the precisionto 1×10 –18 . Click OK and then Solve. When the Solver functionfinds a solution, the results appear in your spreadsheet (see Figure 6.17b).Click OK to keep the result, or Cancel to return to the original values.Note that the answer here agrees with our earlier result of 7.91×10 –7 Mfor the solubility of Pb(IO 3 ) 2 .Be sure to evaluate the reasonableness ofSolver’s answer. If necessary, repeat theprocess using a smaller value for the precision.Ex a m p l e 2: pH o f 1.0 M HFIn developing our solution to this problem we began by identifying fourunknowns and writing out the following four equations.+ −[ HO ][ F ]−= = 68 . × 10 4[ HF]K a3K w 3= [ HO+ ][ OH− ] = 100 . × 10 −14C HF= [ HF] + [ F− ]+ − −[ HO ] = [ OH ] + [ F ]3

262 Analytical Chemistry 2.0From this point, we made two assumptions, simplifying the problem toone that was easy to solve.[ HO ] = KC = ( 68 . × 10 )( 10 . ) = 26 . × 10+ −4 −23 a HFYou also can solve this set of simultaneousequations using Excel’s Solver function.To do so, create the spreadsheet in Figure6.18a, but omit all columns other thanA and B. Select Solver... from the Toolsmenu and define the problem by using B6for Set Target Cell and setting its desiredvalue to 0, and selecting B1 for By ChangingCells:. You may need to play with theSolver’s options to find a suitable solutionto the problem, and it is wise to try severaldifferent initial guesses.The Solver function works well for relativelysimple problems, such as findingthe pH of 1.0 M HF. As problems becomemore complex—such as solving an equilibriumproblem with lots of unknowns—the Solver function becomes less reliablein finding a solution.Although we did not note this at the time, without making assumptionsthe solution to our problem is a cubic equation+ +[ HO ] + K [ HO ]3 23 a 3+− ( KC + K )[ HO ] − K K = 0a HA w 3 a w6.64that we can solve using Excel’s Solver function. Of course, this assumes thatwe successfully complete the derivation!Another option is to use Excel to solve the equations simultaneouslyby iterating in on values for [HF], [F – ], [H 3 O + ], and [OH – ]. Figure 6.18ashows a spreadsheet for this purpose. The cells in the first row contain ourinitial guesses for the equilibrium pH. Using the ladder diagram in Figure6.14, choosing pH values between 1 and 3 seems reasonable. You can addadditional columns if you wish to include more pH values. The formulasin rows 2–5 use the definition of pH to calculate [H 3 O + ], K w to calculate[OH – ], the charge balance equation to calculate [F – ], and K a to calculate[HF]. To evaluate the initial guesses, we use the mass balance expressionfor HF, rewriting it as−−[ HF] + [ F ] − C = [ HF] + [ F ] − 10 . = 0HFand entering it in the last row. This cell gives the calculation’s errorFigure 6.18b shows the actual values for the spreadsheet in Figure 6.18a.The negative value in cells B6 and C6 means that the combined concentrationsof HF and F – are too small, and the positive value in cell D6 meansthat the combined concentrations are too large. The actual pH, therefore,must lie between 2.00 and 1.00. Using these pH values as new limits forthe spreadsheet’s first row, we continue to narrow the range for the actualpH. Figure 6.18c shows a final set of guesses, with the actual pH fallingbetween 1.59 and 1.58. Because the error for 1.59 is smaller than that for1.58, we will accept a pH of 1.59 as the answer. Note that this is an agreementwith our earlier result.Practice Exercise 6.14Using Excel, calculate the solubility of AgI in 0.10 M NH 3 without makingany assumptions. See our earlier treatment of this problem for therelevant equilibrium reactions and constants.Click here to review your answer to this exercise.

Chapter 6 Equilibrium Chemistry263A B C D1 pH = 3.00 2.00 1.00(a)2 [H3O+] = = 10^–b1 = 10^–c1 = 10^–d13 [OH-] = = 1e–14/b2 = 1e–14/c2 = 1e–14/d24 [F-] = = b2 – b3 = c2 – c3 = d2 – d35 [HF] = = (b2 * b4)/6.8e–4 = (c2 * c4)/6.8e–4 = (d2 * d4)/6.8e–46 error = b5 + b4 – 1 = c5 + c4 – 1 = d5 + d4 – 1A B C D1 pH = 3.00 2.00 1.002 [H3O+] = 1.00E-03 1.00E-02 1.00E-13 [OH-] = 1.00E–11 1.00E–12 1.00E–134 [F-] = 1.00E–03 1.00E–02 1.00E–015 [HF] = 0.001470588 0.147058824 14.705882356 error -9.98E-01 -8.43E-01 1.38E+01A B C D1 pH = 1.59 1.58 1.572 [H3O+] = 2.57E-02 2.63E-02 2.69E-023 [OH-] = 3.89E-13 3.80E-13 3.72E-134 [F-] = 2.57E-02 2.63E-02 2.69E-025 [HF] = 0.971608012 1.017398487 1.0653476 error -2.69E-03 4.37E-02 9.23E-02(b)(c)Figure 6.18 Spreadsheet demonstrating the use of Excel to solve a set of simultaneous equations. The spreadsheetin (a) shows the initial guess for [H 3 O + ] in the first row, and the formulas that you must enter in rows 2–6. Enterthe formulas in cells B2–B6 and then copy and paste them into the appropriate cells in the remaining columns.As shown in (b), Excel replaces the formulas with their equivalent values. The spreadsheet in (c) shows the resultsafter our final iteration. See the text for further details.6J.2 RR has a simple command—uniroot—for finding the chemically significantroot of a polynomial equation. In addition, it is easy to write a functionto solve a set of simultaneous equations by iterating in on a solution. Let’swork through two examples.Ex a m p l e 1: So l u b i l i t y o f Pb(IO 3 ) 2 in 0.10 M Pb(NO 3 ) 2In our earlier treatment of this problem we arrived at the following cubicequation4x+ 0. 40x= 25 . × 10−3 2 13

264 Analytical Chemistry 2.0For example, entering> eqn(2)passes the value x = 2 to the function andreturns an answer of 33.6.$root[1] 7.905663e-07$f.root[1] 0$iter[1] 46$estim.prec[1] 1.827271e-12Figure 6.19 The summary of R’soutput from the uniroot command.See the text for a discussionof how to interpret the results.where x is the equilibrium concentration of Pb 2+ . Although there are severalapproaches for solving cubic equations, none are computationally easy.One approach to solving the problem is to iterate in on the answer by findingtwo values of x, one of which leads to a result larger than 2.5×10 –13and one of which gives a result smaller than 2.5×10 –13 . Having establishedboundaries for the value of x, we then shift the upper limit and the lowerlimit until the precision of our answer is satisfactory. Without going intodetails, this is how the uniroot command works.The general form of the uniroot command isuniroot(function, lower, upper, tol)where function is an object containing the equation whose root we areseeking, lower and upper and boundaries for the root, and tol is the desiredaccuracy for the root. To create an object containing the equation, we mustrewrite it so that its right-side equals zero.4 3 0 40 2 −x + . x − 25 . × 10 13 = 0Next, we enter the following code, which defines our cubic equation as afunction with the name eqn.> eqn = function(x) 4*x^3 + 0.4*x^2 – 2.5e–13Because our equation is a function, the uniroot command can send a valueof x to eqn and receive back the equation’s corresponding value. Finally, weuse the uniroot command to find the root.> uniroot(eqn, lower = 0, upper = 0.1, tol = 1e–18)We expect x to be small—because Pb(IO 3 ) 2 is not very soluble—so settingthe lower limit to 0 is a reasonable choice. The choice for the upperlimit is less critical. To ensure that the solution has sufficient precision, thetolerance should be smaller than the expected root. Figure 6.19 shows theresulting output. The value $root is the equation’s root, which is in goodagreement with our earlier result of 7.91×10 –7 for the molar solubility ofPb(IO 3 ) 2 . The other results are the equation’s value for the root, the numberof iterations needed to find the root, and the root’s estimated precision.Ex a m p l e 2: pH o f 1.0 M HFIn developing our solution to this problem we began by identifying fourunknowns and writing out the following four equations.+ −[ HO ][ F ]−= = 68 . × 10 4[ HF]K a3K w 3= [ HO+ ][ OH− ] = 100 . × 10 −14C HF= [ HF] + [ F− ]

Chapter 6 Equilibrium Chemistry265+ − −[ HO ] = [ OH ] + [ F ]3From this point, we made two assumptions, simplifying the problem toone that was easy to solve.[ HO ] = KC = ( 68 . × 10 )( 10 . ) = 26 . × 10+ −4 −23 a HFAlthough we did not note this at the time, without making assumptionsthe solution to our problem is a cubic equation+ +[ HO ] + K [ HO ]3 23 a 3+− ( KC + K )[ HO ] − K K = 0a HA w 3 a wthat we can solve using the uniroot command. Of course, this assumes thatwe successfully complete the derivation!Another option is to use write a function to solve simultaneously thefour equations for the variables [HF], [F – ], [H 3 O + ], and [OH – ]. Here isthe code for this function, which we will call eval.> eval = function(pH){+ h3o =10^–pH+ oh = 1e–14/h3o+ hf = (h3o*f)/6.8e–4+ error = hf + f – 1+ output = data.frame(pH, error)+ print(output)+ }Let’s examine more closely how this function works. The function acceptsa guess for the pH and uses the definition of pH to calculate [H 3 O + ], K wto calculate [OH – ], the charge balance equation to calculate [F – ], and K ato calculate [HF]. The function then evaluates the solution using the massbalance expression for HF, rewriting it as−−[ HF] + [ F ] − C = [ HF] + [ F ] − 10 . = 0HFThe open { tells R that we intend to enterour function over several lines. When wepress enter at the end of a line, R changesits prompt from > to +, indicating thatwe are continuing to enter the same command.The close } on the last line indicatesthat we are done entering the function.The command data.frame combines twoor more objects into a table.You can adapt this function to other problemsby changing the variable you pass tothe function and the equations you includewithin the function.The function then gathers together the initial guess for the pH and the errorand prints them as a table.The beauty of this function is that the object we pass to it, pH, cancontain many values, allowing us to easily search for a solution. BecauseHF is an acid, we know that the solution must be acidic. This sets a lowerlimit of 7 for the pH. We also know that the pH of 1.0 M HF can be nolarger than 1.0 M as this would be the upper limit if HF was a strong acid.For our first pass, let’s enter the following code> pH = c(7, 6, 5, 4, 3, 2, 1)> func(pH)

266 Analytical Chemistry 2.0(a) pH error(b) pH error (c)1 7 -1.00000002 6 -0.99999903 5 -0.99998994 4 -0.99988535 3 -0.99752946 2 -0.84294127 1 13.80588248 0 1470.5882353Figure 6.20 The output of three iterations to find the pH for a solution of 1.0 M HF. The results are for pHvalues between (a) 7 and 0, (b) 2.0 and 1.0, and (c) 1.60 M and 1.50. The columns labeled “error” show anevaluation of the mass balance equation for HF, with positive values indicating that the pH is too low andnegative values indicating that the pH is too high.which varies the pH within these limits. The result, which is shown inFigure 6.20a, indicates that the pH must be less than 2 and greater than 1because it is in this interval that the error changes sign.For our second pass, we explore pH values between 2.0 and 1.0 tofurther narrow down the problem’s solution.> pH = c(2.0, 1.9, 1.8, 1.7, 1.6, 1.5, 1.4, 1.3, 1.2, 1.1, 1.0)> func(pH)The result in Figure 6.20b show that the pH must be less than 1.6 andgreater than 1.5. A third pass between these limits gives the result shown inFigure 6.20c, which is consistent with our earlier result of a pH 1.59.Practice Exercise 6.15Using R, calculate the solubility of AgI in 0.10 M NH 3 without makingany assumptions. See our earlier treatment of this problem for therelevant equilibrium reactions and constantsClick here to review your answer to this exercise.6K1 2.0 -0.842941182 1.9 -0.754338223 1.8 -0.614756004 1.7 -0.394595665 1.6 -0.047002696 1.5 0.502211017 1.4 1.370536008 1.3 2.744069369 1.2 4.9176129510 1.1 8.3582173011 1.0 13.80588235pH error1 1.60 -0.0470026882 1.59 -0.0026880303 1.58 0.0437011674 1.57 0.0922623485 1.56 0.1430975446 1.55 0.1963135867 1.54 0.2520223318 1.53 0.3103409019 1.52 0.37139192810 1.51 0.43530381611 1.50 0.502211012Some Final Thoughts on Equilibrium CalculationsIn this chapter we have developed several tools for evaluating the compositionof a system at equilibrium. These tools differ in how accurately theyallow us to answer questions involving equilibrium chemistry. They alsodiffer in their ease of use. An important part of having several tools availableto you is knowing when to use them. If you need to know whethera reaction if favorable, or to estimate the pH of a solution, then a ladderdiagram will meet your needs. On the other hand, if you require a more

Chapter 6 Equilibrium Chemistry267accurate estimate of a compound’s solubility, then a rigorous calculationthat includes activity coefficients is necessary.A critical part of solving an equilibrium problem is knowing what equilibriumreactions to include. The importance of including all relevant reactionsis obvious, and at first glance this does not appear to be a significantproblem—it is, however, a potential source of significant errors. The tablesof equilibrium constants in this textbook, although extensive, are a smallsubset of all known equilibrium constants, making it easy to overlook animportant equilibrium reaction. Commercial and freeware computationalprograms with extensive databases are available for equilibrium modeling.Two excellent freeware programs are Visual Minteq (Windows only) andChemEQL (Windows, Mac, Linux, and Solaris). These programs also includethe ability to account for ionic strength.Finally, a consideration of equilibrium chemistry can only help us decideif a reaction is favorable. It does not, however, guarantee that the reactionoccurs. How fast a reaction approaches its equilibrium position doesnot depend on the equilibrium constant. The rate of a chemical reactionis a kinetic, not a thermodynamic, phenomenon. We will consider kineticeffects and their application in analytical chemistry in Chapter 13.6LKey Termsacid acid dissociation constant activityactivity coefficient amphiprotic basebase dissociation constant buffer buffer capacitycharge balance equation common ion effect cumulative formationconstantdissociation constant enthalpy entropyequilibrium equilibrium constant extended Debye-Hückelequationformation constant Gibb’s free energy half-reactionHenderson–Hasselbalchequationionic strengthladder diagramLe Châtelier’s principle ligand mass balance equationmetal–ligand complexmethod of successiveapproximationsmonoproticNernst equation oxidation oxidizing agentpH scale polyprotic potentialprecipitate redox reaction reducing agentreduction standard-state standard potentialsteady statestepwise formationconstantsolubility productAs you review this chapter, try to define akey term in your own words. Check youranswer by clicking on the key term, whichwill take you to the page where it was firstintroduced. Clicking on the key termthere, will bring you back to this page sothat you can continue with another keyterm.

268 Analytical Chemistry 2.06MChapter SummaryAnalytical chemistry is more than a collection of techniques; it is the applicationof chemistry to the analysis of samples. As we will see in laterchapters, almost all analytical methods use chemical reactivity to accomplishone or more of the following: dissolve the sample, separate analytesfrom interferents, transform the analyte to a more useful form, or providea signal. Equilibrium chemistry and thermodynamics provide us with ameans for predicting which reactions are likely to be favorable.The most important types of reactions are precipitation reactions, acid–base reactions, metal-ligand complexation reactions, and redox reactions.In a precipitation reaction two or more soluble species combine to producean insoluble product called a precipitate, which we characterize using asolubility product.An acid–base reaction occurs when an acid donates a proton to a base.The reaction’s equilibrium position is described using either an acid dissociationconstant, K a , or a base dissociation constant, K b . The product ofK a and K b for an acid and its conjugate base is the dissociation constant forwater, K w .When a ligand donates one or more pairs of electron to a metal ion,the result is a metal–ligand complex. Two types of equilibrium constantsare used to describe metal–ligand complexation—stepwise formation constantsand overall formation constants. There are two stepwise formationconstants for the metal–ligand complex ML 2 , each describing the additionof one ligand; thus, K 1 represents the addition of the first ligand to M, andK 2 represents the addition of the second ligand to ML. Alternatively, we canuse a cumulative, or overall formation constant, b 2 , for the metal–ligandcomplex ML 2 , in which both ligands are added to M.In a redox reaction, one of the reactants undergoes oxidation and anotherreactant undergoes reduction. Instead of using an equilibrium constantsto characterize a redox reactions, we use the potential, positive valuesof which indicate a favorable reaction. The Nernst equation relates thispotential to the concentrations of reactants and products.Le Châtelier’s principle provides a means for predicting how a systemat equilibrium responds to a change in conditions. If we apply a stress toa system at equilibrium—by adding a reactant or product, by adding areagent that reacts with one of the reactants or products, or by changingthe volume—the system responds by moving in the direction that relievesthe stress.You should be able to describe a system at equilibrium both qualitativelyand quantitatively. You can develop a rigorous solution to an equilibriumproblem by combining equilibrium constant expressions with appropriatemass balance and charge balance equations. Using this systematic approach,you can solve some quite complicated equilibrium problems. If a less rigor-

Chapter 6 Equilibrium Chemistry269ous answer is acceptable, then a ladder diagram may help you estimate theequilibrium system’s composition.Solutions containing relatively similar amounts of a weak acid and itsconjugate base experience only a small change in pH upon adding a smallamount of a strong acid or a strong base. We call these solutions buffers. Abuffer can also be formed using a metal and its metal–ligand complex, oran oxidizing agent and its conjugate reducing agent. Both the systematicapproach to solving equilibrium problems and ladder diagrams are usefultools for characterizing buffers.A quantitative solution to an equilibrium problem may give an answerthat does not agree with experimental results if we do not consider the effectof ionic strength. The true, thermodynamic equilibrium constant is afunction of activities, a, not concentrations. A species’ activity is related toits molar concentration by an activity coefficient, γ. Activity coefficients canbe calculated using the extended Debye-Hückel equation, making possiblea more rigorous treatment of equilibria.6NProblems1. Write equilibrium constant expressions for the following reactions.What is the value for each reaction’s equilibrium constant?+ −a. NH ( aq) + HCl( aq) NH ( aq) + Cl ( aq)3 42−−b. PbI () s + S ( aq) PbS() s + 2I( aq)22− − −c Cd(EDTA) ( aq) + 4CN ( aq) Cd(CN) ( aq)+ EDTA− ( aq)+ −d. AgCl() s + NH ( aq) + Ag(NH ) ( aq) + Cl ( aq )23 3 22 44+ 2+e. BaCO () s + HO ( aq) Ba ( aq) + HCO ( aq) + H O()l33 2 32Answers, but not worked solutions, tomost end-of-chapter problems are availablehere.Most of the problems that follow require one ormore equilibrium constants or standard state potentials.For your convenience, here are hyperlinksto the appendices containing these constantsAppendix 10: Solubility ProductsAppendix 11: Acid Dissociation ConstantsAppendix 12: Metal-Ligand Formation ConstantsAppendix 13: Standard State Reduction Potentials2. Using a ladder diagram, explain why the first reaction is favorable andthe second reaction is unfavorable.−−HPO ( aq) + F ( aq) HF( aq) + HPO ( aq)3 4 2 4−2−HPO ( aq) + 2F ( aq) 2HF( aq) + HPO ( aq)3 4 4Determine the equilibrium constant for these reactions and verify thatthey are consistent with your ladder diagram.3. Calculate the potential for the following redox reaction for a solutionin which [Fe 3+ ] = 0.050 M, [Fe 2+ ] = 0.030 M, [Sn 2+ ] = 0.015 M and[Sn 4+ ] = 0.020 M.3+ 2+ 4+ 2+2Fe ( aq) + Sn ( aq) Sn ( aq) + 2Fe( aq)

270 Analytical Chemistry 2.0These redox reactions in this problem arenot balanced. You will need to balance thereactions before calculating their standardstate potentials. Although you may recallhow to do this from another course, thereis a much easier approach that you can usehere. Identify the oxidizing agent and thereducing agent and divide the reaction intotwo unbalanced half-reactions. Using Appendix13, find the appropriate balancedhalf-reactions. Add the two half-reactionstogether and simplify the stoichiometry toarrive at the balanced redox reaction.As an example, in (a) the oxidizing agent isMnO 4– and its unbalanced half-reactionisMnO− − 2+4( aq ) + 5e Mn ( aq )The corresponding balanced half-reactionfrom Appendix 13 isMnO− + −( aq ) + 8H( aq ) + 5e4Mn2+( aq ) + 4HO()lMost of the problems that follow require one ormore equilibrium constants or standard state potentials.For your convenience, here are hyperlinksto the appendices containing these constantsAppendix 10: Solubility ProductsAppendix 11: Acid Dissociation ConstantsAppendix 12: Metal-Ligand Formation ConstantsAppendix 13: Standard State Reduction Potentials24. Calculate the standard state potential and the equilibrium constant foreach of the following redox reactions. Assume that [H 3 O + ] is 1.0 M foracidic solutions, and that [OH – ] is 1.0 M for basic solutions.− 2+ 2−a. MnO ( aq) + H SO ( aq) Mn ( aq) + SO ( aq)acidicsolution4 2 34− −b. IO ( aq) + I ( aq) I () s acidic solution3 2c. ClO − ( aq) + I − ( aq) IO − ( ) + Cl −3aq ( aq)basicsolution5. One analytical method for determining the concentration of sulfur isto oxidize it to SO 4 2- and then precipitate it as BaSO 4 by adding BaCl 2 .The mass of the resulting precipitate is proportional to the amount ofsulfur in the original sample. The accuracy of this method depends onthe solubility of BaSO 4 , the reaction for which is shown here.22BaSO () s Ba + ( aq) + SO− ( aq )4How do the following affect the solubility of BaSO 4 and, therefore, theaccuracy of the analytical method?a. decreasing the solution’s pHb. adding more BaCl 2c. increasing the volume of the solution by adding H 2 O6. Write a charge balance equation and mass balance equations for the followingsolutions. Some solutions may have more than one mass balanceequation.a. 0.10 M NaClb. 0.10 M HClc. 0.10 M HFd. 0.10 M NaH 2 PO 4e. MgCO 3 (saturated solution)f. 0.10 M Ag(CN) 2–g. 0.10 M HCl and 0.050 M NaNO 247. Using the systematic approach to equilibrium problems, calculate thepH of the following solutions. Be sure to state and justify any assumptionsyou make in solving the problems.a. 0.050 M HClO 4b. 1.00 × 10 –7 M HClc. 0.025 M HClOd. 0.010 M HCOOHe. 0.050 M Ba(OH) 2f. 0.010 M C 5 H 5 N

Chapter 6 Equilibrium Chemistry2718. Construct ladder diagrams for the following diprotic weak acids (H 2 L)and estimate the pH of 0.10 M solutions of H 2 L, HL – and L 2– .a. maleic acidb. malonic acidc. succinic acid9. Using the systematic approach to solving equilibrium problems, calculatethe pH of the diprotic weak acid in problem 8. Be sure to state andjustify any assumptions you make in solving the problems.10. Ignoring activity effects, calculate the concentration of Hg 2 2+ in thefollowing solutions. Be sure to state and justify any assumption youmake in solving the problems.a. a saturated solution of Hg 2 Br 2b. 0.025 M Hg 2 (NO 3 ) 2 saturated with Hg 2 Br 2c. 0.050 M NaBr saturated with Hg 2 Br 211. The solubility of CaF 2 is controlled by the following two reactions2CaF () s Ca + ( aq) + 2F− ( aq )2+ −HF( aq) + HO() l HO ( aq) + F ( aq )2 3Calculate the solubility of CaF 2 in a solution buffered to a pH of 7.00.Use a ladder diagram to help simplify the calculations. How would yourapproach to this problem change if the pH is buffered to 2.00? Whatis the solubility of CaF 2 at this pH? Be sure to state and justify any assumptionsyou make in solving the problems.Most of the problems that follow require one ormore equilibrium constants or standard state potentials.For your convenience, here are hyperlinksto the appendices containing these constantsAppendix 10: Solubility ProductsAppendix 11: Acid Dissociation ConstantsAppendix 12: Metal-Ligand Formation ConstantsAppendix 13: Standard State Reduction Potentials12. Calculate the solubility of Mg(OH) 2 in a solution buffered to a pH of7.00. How does this compare to its solubility in unbuffered deionizedwater? Be sure to state and justify any assumptions you make in solvingthe problem.13. Calculate the solubility of Ag 3 PO 4 in a solution buffered to a pH of7.00. Be sure to state and justify any assumptions you make in solvingthe problem.14. Determine the equilibrium composition of saturated solution of AgCl.Assume that the solubility of AgCl is influenced by the following reactions.AgCl() s Ag + ( aq) + Cl− ( aq )+ −Ag ( aq) + Cl ( aq) AgCl( aq)AgCl( aq) + Cl − ( aq) AgCl− ( aq) 2Be sure to state and justify any assumptions you make in solving theproblem.

272 Analytical Chemistry 2.015. Calculate the ionic strength of the following solutionsa. 0.050 M NaClb. 0.025 M CuCl 2c. 0.10 M Na 2 SO 416. Repeat the calculations in problem 10, this time correcting for the effectof ionic strength. Be sure to state and justify any assumptions you makein solving the problems.17. Over what pH range do you expect Ca 3 (PO4) 2 to have its minimumsolubility?18. Construct ladder diagrams for the following systems, each consistingof two or three equilibria. Using your ladder diagrams, what reactionsare likely to occur in each system?a. HF and H 3 PO 4b. Ag(CN) 2 – , Ni(CN) 4 2– and Fe(CN) 63–c. Cr 2 O 7 2– /Cr 3+ and Fe 3+ /Fe 2+Most of the problems that follow require one ormore equilibrium constants or standard state potentials.For your convenience, here are hyperlinksto the appendices containing these constantsAppendix 10: Solubility ProductsAppendix 11: Acid Dissociation ConstantsAppendix 12: Metal-Ligand Formation ConstantsAppendix 13: Standard State Reduction Potentials19. Calculate the pH of the following acid–base buffers. Be sure to stateand justify any assumptions you make in solving the problems.a. 100 mL of 0.025 M formic acid and 0.015 M sodium formateb. 50.00 mL of 0.12 M NH 3 and 5.30 mL of 1.0 M HClc. 5.00 g of Na 2 CO 3 and 5.00 g of NaHCO 3 diluted to 100 mL20. Calculate the pH of the buffers in problem 19 after adding 5.0 mL of0.10 M HCl. Be sure to state and justify any assumptions you make insolving the problems.21. Calculate the pH of the buffers in problem 19 after adding 5.0 mL of0.10 M NaOH. Be sure to state and justify any assumptions you makein solving the problems.22. Consider the following hypothetical complexation reaction between ametal, M, and a ligand, LM( aq) + L( aq) ML( aq)with a formation constant of 1.5 × 10 8 . Derive an equation, similar tothe Henderson–Hasselbalch equation, relating pM to the concentrationsof L and ML. What is the pM for a solution containing 0.010mol of M and 0.020 mol of L? What is pM be if you add 0.002 molof M to this solution? Be sure to state and justify any assumptions youmake in solving the problem.23. A redox buffer contains an oxidizing agent and its conjugate reducingagent. Calculate the potential of a solution containing 0.010 mol ofFe 3+ and 0.015 mol of Fe 2+ . What is the potential if you add sufficient

Chapter 6 Equilibrium Chemistry273oxidizing agent to convert 0.002 mol of Fe 2+ to Fe 3+ ? Be sure to stateand justify any assumptions you make in solving the problem.24. Use either Excel or R to solve the following problems. For these problems,make no simplifying assumptions.a. the solubility of CaF 2 in deionized waterb. the solubility of AgCl in deionized water (see Problem 14 for therelevant equilibria)Most of the problems that follow require one ormore equilibrium constants or standard state potentials.For your convenience, here are hyperlinksto the appendices containing these constantsAppendix 10: Solubility ProductsAppendix 11: Acid Dissociation ConstantsAppendix 12: Metal-Ligand Formation ConstantsAppendix 13: Standard State Reduction Potentialsc. the pH of 0.10 M fumaric acid25. Beginning with the relevant equilibrium reactions, derive equation6.64 for the rigorous solution to the pH of 0.1 M HF.6OSolutions to Practice ExercisesPractice Exercise 6.1The overall reaction is equivalent toRxn4− 2×Rxn 1Subtracting a reaction is equivalent to adding the reverse reaction; thus,the overall equilibrium constant isKK4= = = ≈( )K150 .( 040 . )2 2Click here to return to the chapter.Practice Exercise 6.2The K b for hydrogen oxalate isKb,HC2Oand the K b for oxalate isK2b,C2O4Kw100 . × 10−= =4 K 560 . × 10a,H2CO2 4Kw100 . × 10−= =K 542 . × 10-a,HC2O431.25 31−14−2−14−5= 1.79× 10 −13= 1.85× 10 −10As we expect, the K b value for C 2 O 42–is larger than that for HC 2 O 4 – .Click here to return to the chapter.Practice Exercise 6.3We can write the reaction as the sum of three other reactions. The first ofthese reactions is the solubility of AgCl(s), which we characterize by itsK sp .

274 Analytical Chemistry 2.0AgBr() s Ag + ( aq) + Br− ( aq )The remaining two reactions are the stepwise formation of Ag(S 2 O 3 ) 2 3– ,which we characterize by K 1 and K 2 .+ 2− −Ag ( aq) + SO ( aq) Ag( SO)( aq)2 32 3Ag( SO) − 2( ) SO − 3aq + ( aq) Ag( SO)− ( aq)2 3 2 32 3 2Using values for K sp , K 1 , and K 2 from Appendix 10 and Appendix 11, wefind that the equilibrium constant for our reaction is−13 8 4K = K × K × K = ( 50 . × 10 )( 66 . × 10 )( 71 . × 10 ) = 23sp 1 2Click here to return to the chapter.Practice Exercise 6.4The two half-reactions are the oxidation of Fe 2+Feand the reduction of MnO 4 – .( aq) Fe ( aq)+e2+ 3+ −− + − +MnO ( aq) + 8H ( aq) + 5e Mn ( aq) + 4HO()l42From Appendix 13, the standard state reduction potentials for these halfreactionsareEo= 0. 771 V E = 151 . Vo3+ 2+ − 2+Fe / FeMnO4/ Mn(a) The standard state potential for the reaction iso ooE = E− 2+ − E3+ 2+= 151 . V− 0. 771 V = 074 . VMnO4/ MnFe / Fe(b) To calculate the equilibrium constant we substitute appropriate valuesinto equation 6.25.Eo= 074 . V =Solving for K gives its value as0.059165log K = 62.5K = 32 . × 10 62log K(c) To calculate the potential under these non-standard state conditions,we make appropriate substitutions into the Nernst equation.2

Chapter 6 Equilibrium Chemistry2752+ 3+5o RT [ Mn ][ Fe ]E = E − ln− + +nF [ MnO ][ Fe ] [ H ]42 5 850.05916 ( 0. 015)( 010 . )E = 074 . − log5 ( 0. 025)( 05 . 0)( 1× 10− )5 7 850.05916 ( 0. 015)( 010 . )E = 074 . − log−5 ( 0. 025)( 05 . 0)( 1×10 )Click here to return to the chapter.5 7 8= 012 . VpH2–Practice Exercise 6.5From Appendix 11, the pK a values for H 2 CO 3 are 6.352 and 10.329. Theladder diagram for H 2 CO 3 is shown to the side. The predominate format a pH of 7.00 is HCO 3 – .Click here to return to the chapter.Practice Exercise 6.6The ladder diagram in Figure 6.5 indicates that the reaction between aceticacid and p-nitrophenolate is favorable. Because p-nitrophenolate is inexcess, we assume that the reaction of acetic acid to acetate is complete. Atequilibrium essentially no acetic acid remains and there are 0.040 molesof acetate. Converting acetic acid to acetate consumes 0.040 moles ofp-nitrophenolate; thusmolesp -nitrophenolate = 0. 090− 0. 040 = 0. 050 molCO 3--pK a2= 10.329HCO 3pK a1= 6.352H 2CO 3molesp -nitrophenol= 0.040 molAccording to the ladder diagram for this system, the pH is 7.15 whenthere are equal concentrations of p-nitrophenol and p-nitrophenolate. Becausewe have slightly more p-nitrophenolate than we have p-nitrophenol,the pH is slightly greater than 7.15.Click here to return to the chapter.Practice Exercise 6.7As Hg 2 Cl 2 dissolves, two Cl – are produced for each ion of Hg 2 2+ . If weassume that x is the change in the molar concentration of Hg 2 2+ , then thechange in the molar concentration of Cl – is 2x. The following table helpsus keep track of our solution to this problem.Concentrations Hg 2 Cl 2 (s) Hg 22+ (aq) + 2Cl – (aq)Initial solid 0 0Change solid +x +2x

276 Analytical Chemistry 2.0Equilibrium solid x 2xSubstituting the equilibrium concentrations into the K sp expression forHg 2 Cl 2 givesK = [ Hg ][ Cl ] = ( x)( 2x) = 4x= 12 . × 10sp 22+ − 2 2 3 −18x = 669 . × 10 −7Substituting x back into the equilibrium expressions for Hg 2+ 2 and Cl –gives their concentrations as2[ Hg + 7 6] = x = 67 . × 10 − M [ Cl − ] = 2x= 13 . × 10− M2The molar solubility is equal to [Hg 2 2+ ], or 6.7 × 10 –7 mol/L.Click here to return to the chapter.Practice Exercise 6.8We begin by setting up a table to help us keep track of the concentrationsof Hg 2 2+ and Cl – as this system moves toward and reaches equilibrium.Concentrations Hg 2 Cl 2 (s) Hg 22+ (aq) + 2Cl – (aq)Initial solid 0 0.10Change solid +x +2xEquilibrium solid x 0.10 + 2xSubstituting the equilibrium concentrations into the K sp expression forHg 2 Cl 2 leaves us with a difficult to solve cubic equation.2+ − 2 2 3 2K = [ Hg ][ Cl ] = ( x)( 010 . ) = 4x + 040 . x + 0.10xsp 2Let’s make an assumption to simplify this problem. Because we expect thevalue of x to be small, let’s assume thatThis simplifies our problems to[ Cl− ] = 010 . + x ≈ 0.10 M2+ − 2 2−K = [ Hg ][ Cl ] = ( x)( 010 . ) = 0. 010x= 12 . × 10 18sp 2which gives the value of x asx = 12 . × 10 −16MThe difference between the actual concentration of Cl – , which is(0.10 + x) M, and our assumption that it is 0.10 M introduces an error of1.2 × 10 –14 %. This is a negligible error. The molar solubility of Hg 2 Cl 2is the same as the concentration of Hg 2 2+ , or 1.2 × 10 –16 M. As expected,the molar solubility in 0.10 M NaCl is less than 6.7 × 10 –7 mol/L, whichis its solubility in water (see solution to Practice Exercise 6.7).Click here to return to the chapter.

Chapter 6 Equilibrium Chemistry277Practice Exercise 6.9To help us in determining what ions are in solution, let’s write down allthe reaction leading to the preparation of the solutions and the equilibriawithin the solutions. These reactions are the dissolution of two solublesalts+ −KH PO () s → K ( aq) + H PO ( aq )2 4 2 4+ 2−Na HPO () s → 2Na ( aq) + HPO ( aq )2 4 4and the acid–base dissociation reactions for H 2 PO 4 – , HPO 4 2– , andH 2 O.− + 2−HPO ( aq) + H O() l H O ( aq) + HPO ( aq )2 4 2 34−−HPO ( aq) + H O() l OH ( aq) + H PO ( aq )2 42 3 42− + 3−HPO ( aq) + H O() l H O ( aq) + PO ( aq )42 32H O() l H O + ( aq) + OH− ( aq )2 3Note that we did not include the base dissociation reaction for HPO 42–because we already have accounted for its product, H 2 PO 4 – , in another reaction.The mass balance equations for K + and Na + are straightforward[ K + ] = 010 . M [ Na + ] = 0.10 Mbut the mass balance equation for the phosphate takes a little bit ofthought. Both H 2 PO 4 – andHPO 4 2– produce the same ions in solution.We can, therefore, imagine that the solution initially contains 0.15 MKH 2 PO 4 , which gives the following mass balance equation.− 2− 3−0.15 M= [ H PO ] + [ HPO ] + [ HPO ] + [ PO ]3 4 2The charge balance equation is+ + +[ HO ] + [ K ] + [ Na ] =34 4−2−[ HPO ] + 2× [ HPO ] + 3×[ PO2 4 44Click here to return to the chapter.Practice Exercise 6.10443−−] + [ OH ]In determining the pH of 0.050 M NH 3 , we need to consider two equilibriumreactions—the base dissociation reaction for NH 3NH ( ) HO () −OH ( ) +aq + l aq + NH ( aq )and water’s dissociation reaction.3 242H O() l H O + ( aq) + OH− ( aq )2 3

278 Analytical Chemistry 2.0These two reactions contain four species whose concentrations we need toconsider: NH 3 , NH 4 + , H 3 O + , and OH – . We need four equations to solvethe problem—these equations are the K b equation for NH 3the K w equation for H 2 O− +[ OH ][ NH ]4K b= = 175 . × 10[ NH ]K w 3a mass balance on ammoniaand a charge balance equation3−5= [ HO+ ][ OH− ] = 100 . × 10 −14+C NH= 0. 050 M= [ NH ] + [ NH ]3 43+ + −[ HO ] + [ NH ] = [ OH ]3To solve this problem, we will make two assumptions. Because NH 3 is abase, our first assumption is− +[ OH ] >> [ HO ]which simplifies the charge balance equation to+ −[ NH ] = [ OH ]Because NH 3 is a weak base, our second assumption is4+[ NH ] >> [ NH ]433 4which simplifies the mass balance equation toC NH= 0. 050 M=[ NH ]33Substituting the simplified charge balance equation and mass balanceequation into the K b equation leave us withKb− − − 2[ OH ][ OH ] [ OH ]= = = 175 . × 10C CNH3 NH3[ ] = KC = (. 175× 10 )( 0. 050) = 935 . × 10OH − − 5 − 4b NH3Before accepting this answer, we must verify our two assumptions. Thefirst assumption is that the concentration of OH – is significantly greaterthan the concentration of H 3 O + . Using K w , we find that+Kw100 . × 10[ HO ] = =3−[ OH ] 935 . × 10−14−4−5= 107 . × 10−11

Chapter 6 Equilibrium Chemistry279Clearly this assumption is acceptable. Our second assumption is that theconcentration of NH 3 is scientifically greater than the concentration ofNH 4 + . Using our simplified charge balance equation, we find that[ NH ] = [ OH ] = 935 . × 10+ − −44Because the concentration of NH + 4 is 1.9% of C NH , our second assumptionalso is reasonable. Given that [H 3 O + ] is 1.07 × 10 –11 , the pH3is 10.97.Click here to return to the chapter.Practice Exercise 6.11In solving for the pH of 0.10 M alanine, we made the following threeassumptions.Assumption One: [ HL] >> [ HL + ] + [ L− ]Assumption Two: K K

280 Analytical Chemistry 2.0Adding HCl converts a portion of HPO 4 2– to H 2 PO 4 – as a result of thefollowing reaction2− + −HPO ( aq ) + H O ( aq ) H O() l + H PO ( aq )43 2 2Because this reaction’s equilibrium constant is so large (it is 1.59 × 10 7 ),we may treat the reaction as if it goes to completion. The new concentrationsof H 2 PO 4 – and HPO 4 2– areCC−=HPO 2 42− =HPO4molH PO2−4V+ molHCltotal−3( 010 . M)( 0. 10 L) + ( 020 . M)( 5. 0×10 L)=3010 . L + 5.0×10− LmolHPOV2−4total-mol HCl−3( 005 . M)( 0. 10 L) − ( 020 . M)( 5. 0×10 L)=3010 . L + 5.0×10− L4= 0.105M= 0.0381 MSubstituting these concentrations into equation 6.60 gives a pH of2−HPO40 038pH = 7. 199 + log [ ] = 7. 199 + log . 1 = 6. 759 ≈676.−[ HPO ]0.10524As we expect, adding HCl decreases the buffer’s pH by a small amount,dropping from 6.90 to 6.76.Click here to return to the chapter.Practice Exercise 6.13We begin by calculating the solution’s ionic strength. Because NaCl is a 1:1ionic salt, the ionic strength is the same as the concentration of NaCl; thusm = 0.10 M. This assumes, of course, that we can ignore the contributionsof Hg 2 2+ and Cl – from the solubility of Hg 2 Cl 2 .Next we use equation 6.63 to calculate the activity coefficients for Hg 22+and Cl – .2051 . ( 2) 010 .log γ2+ Hg2= − × + ×1+ 3. 3× 0. 40×010 .γ Hg+= 0.35122=−0.455. ( ) .log γ Cl−= − × − 2051 1 × 010=−01. 21+ 3. 3× 0. 3×0.10γ Cl−= 075 .Defining the equilibrium concentrations of Hg 2 2+ and Cl – in terms ofthe variable x

Chapter 6 Equilibrium Chemistry281Concentrations Hg 2 Cl 2 (s) Hg 22+ (aq) + 2Cl – (aq)Initial solid 0 0.10Change solid +x +2xEquilibrium solid x 0.10 + 2xand substituting into the thermodynamic solubility product for Hg 2 Cl 2 ,leave us with22+ 2 − 2K = a a = γ [ Hg ] γ [ Cl ] = 1.2× 10 −18sp 2+ − 2+ −Hg2Cl Hg 22Cl12 . × 10 − = ( 0. 351)( x)( 075 . )( 0. 10 + x)18 2 2Because the value of x is likely to be small, let’s simplify this equation to12 . × 10 − = ( 0. 351)( x)( 075 . )( 0. 10)18 2 2Solving for x gives its value as 6.1 × 10 –16 . Because x is the concentrationof Hg 2 2+ and 2x is the concentration of Cl – , our decision to ignore theircontributions to the ionic strength is reasonable. The molar solubility ofHg 2 Cl 2 in 0.10 M NaCl is 6.1 × 10 –16 mol/L. In Practice Exercise 6.8,where we ignored ionic strength, we determined that the molar solubilityof Hg 2 Cl 2 is 1.2 × 10 –16 mol/L, a result that is 5× smaller than the itsactual value.Click here to return to the chapter.Practice Exercise 6.14To solve this problem, let’s set up the following spreadsheetA1 pI = 32 [I-] = = 10^–b13 [Ag+] = = 8.3e–17/b24 [Ag(NH3)2+] = = b2 – b35 [NH3] = = (b4/(b3*1.7e7))^0.56 [NH4+] = = 0.10 – b5-2*b47 [OH-] = = 1.75e–5*b5/b68 [H3O+] = = 1.00e–14/b79 error = b3 + b4 + b6 + b8 – b2 – b7copying the contents of cells B1-B9 into several additional columns. Seeour earlier treatment of this problem for the relevant equilibrium reactionsand equilibrium constants. The initial guess for pI in cell B1 givesthe concentration of I – in cell B2. Cells B3–B8 calculate the remainingconcentrations, using the K sp to obtain [Ag + ], using the mass balance oniodide and silver to obtain [Ag(NH 3 ) 2 + ], using b 2 to calculate [NH 3 ], usingthe mass balance on ammonia to find [NH 4 + ], using K b to calculateB

282 Analytical Chemistry 2.0[OH – ], and using K w to calculate [H 3 O + ]. The system’s charge balanceequation provides a means for determining the calculation’s error.+ + + + − −[ Ag ] + [ Ag(NH ) ] + [ NH ] + [ H O ] −[ I ] − [ OH ] =03 2 4 3The largest possible value for pI—corresponding to the smallest concentrationof I – and the lowest possible solubility—occurs for a simple, saturatedsolution of AgI. When [Ag + ] = [I – ], the concentration of iodide is[ I ] = K = 83 . × 10 = 91 . × 10− −17 −9spcorresponding to a pI of 8.04. Entering initial guesses for pI of 4, 5, 6, 7,and 8 shows that the error changes sign between a pI of 5 and 6. Continuingin this way to narrow down the range for pI, we find that theerror function is closest to zero at a pI of 5.42. The concentration of I – atequilibrium, and the molar solubility of AgI, is 3.8 × 10 –6 mol/L, whichagrees with our earlier solution to this problem.Click here to return to the chapterPractice Exercise 6.15To solve this problem, let’s use the following function> eval = function(pI){+ I =10^–pI+ Ag = 8.3e–17/I+ AgNH3 = Ag – I+ NH3 =(AgNH3/(1.7e7*Ag))^0.5+ NH4 =0.10-NH3 – 2*AgNH3+ OH =1.75e–5*NH3/NH4+ H3O =1e–14/OH+ error = Ag + AgNH3 + NH4 + H3O – OH – I+ output = data.frame(pI, error)+ print(output)+ }The function accepts an initial guess for pI and calculates the concentrationsof species in solution using the definition of pI to calculate [I – ],using the K sp to obtain [Ag + ], using the mass balance on iodide and silverto obtain [Ag(NH 3 ) 2 + ], using b 2 to calculate [NH 3 ], using the mass balanceon ammonia to find [NH 4 + ], using K b to calculate [OH – ], and usingK w to calculate [H 3 O + ]. The system’s charge balance equation provides ameans for determining the calculation’s error.+ + + + − −[ Ag ] + [ Ag(NH ) ] + [ NH ] + [ H O ] −[ I ] − [ OH ] =M03 2 4 3The largest possible value for pI—corresponding to the smallest concentrationof I – and the lowest possible solubility—occurs for a simple, saturatedsolution of AgI. When [Ag + ] = [I – ], the concentration of iodide is

Chapter 6 Equilibrium Chemistry283[ I ] = K = 83 . × 10 = 91 . × 10− −17 −9spcorresponding to a pI of 8.04. The following session shows the functionin action.> pI =c(4, 5, 6, 7, 8)> eval(pI)pI error1 4 -2.562356152 5 -0.166209303 6 0.073371014 7 0.097348245 8 0.09989073> pI =c(5.1, 5.2, 5.3, 5.4, 5.5, 5.6, 5.7, 5.8, 5.9, 6.0)> eval(pI)pI error1 5.1 -0.111446582 5.2 -0.067941053 5.3 -0.033364754 5.4 -0.005681165 5.5 0.015715496 5.6 0.033089297 5.7 0.046859378 5.8 0.057792149 5.9 0.0664747510 6.0 0.07337101> pI =c(5.40, 5.41, 5.42, 5.43, 5.44, 5.45, 5.46, 5.47, 5.48, 5.49, 5.50)> eval(pI)pI error1 5.40 -0.00568116052 5.41 -0.00307154843 5.42 0.00023103694 5.43 -0.00051348985 5.44 0.00282818786 5.45 0.00523709807 5.46 0.00747581818 5.47 0.00962603709 5.48 0.011710549810 5.49 0.013738729111 5.50 0.0157154889The error function is closest to zero at a pI of 5.42. The concentration ofI – at equilibrium, and the molar solubility of AgI, is 3.8 × 10 –6 mol/L,which agrees with our earlier solution to this problem.Click here to return to the chapterM


DRAFTChapter 7Chapter Overview7A7B7C7D7E7F7G7H7I7J7K7LCollecting andPreparing SamplesThe Importance of SamplingDesigning a Sampling PlanImplementing the Sampling PlanSeparating The Analyte From InterferentsGeneral Theory of Separation EfficiencyClassifying Separation TechniquesLiquid –Liquid ExtractionsSeparation Versus PreconcentrationKey TermsChapter SummaryProblemsSolutions to Practice ExercisesWhen we use an analytical method to solve a problem, there is no guarantee that our resultswill be accurate or precise. In designing an analytical method we consider potential sources ofdeterminate error and indeterminate error, and take appropriate steps to minimize their effect,such as including reagent blanks and calibrating instruments. Why might a carefully designedanalytical method give poor results? One possibility is that we may have failed to account forerrors associated with the sample. If we collect the wrong sample, or if we lose analyte whilepreparing the sample for analysis, then we introduce a determinate source of error. If we fail tocollect enough samples, or if we collect samples of the wrong size, then our precision may suffer.In this chapter we consider how collecting samples and preparing them for analysis affects theaccuracy and precision of our results.Copyright: David Harvey, 2009285

286 Analytical Chemistry 2.0Figure 7.1 Certificate of analysisfor a production lot of NaBr.The result for iron meets the ACSspecifications, but the result forpotassium does not.Equation 7.1 should be familiar to you.See Chapter 4 to review confidence intervalsand see Appendix 4 for values of t.For a review of the propagation of uncertainty,see Chapter 4C and Appendix 2.Although equation 7.1 is written in termsof a standard deviation, s, a propagationof uncertainty is written in terms of variances,s 2 . In this section, and those thatfollow, we will use both standard deviationsand variances to discuss samplinguncertainty.7AThe Importance of SamplingWhen a manufacturer lists a chemical as ACS Reagent Grade, they mustdemonstrate that it conforms to specifications set by the American ChemicalSociety (ACS). For example, the ACS specifications for NaBr requirethat the concentration of iron be ≤5 ppm. To verify that a production lotmeets this standard, the manufacturer collects and analyzes several samples,reporting the average result on the product’s label (Figure 7.1).If the individual samples do not accurately represent the populationfrom which they are drawn—what we call the target population—theneven a careful analysis must yield an inaccurate result. Extrapolating thisresult from a sample to its target population introduces a determinate samplingerror. To minimize this determinate sampling error, we must collectthe right sample.Even if we collect the right sample, indeterminate sampling errors maylimit the usefulness of our analysis. Equation 7.1 shows that a confidenceinterval about the mean, X , is proportional to the standard deviation, s, ofthe analysisµ= X ±where n is the number of samples and t is a statistical factor that accountsfor the probability that the confidence interval contains the true value, m.Each step of an analysis contributes random error that affects the overallstandard deviation. For convenience, let’s divide an analysis into twosteps—collecting the samples and analyzing the samples—each characterizedby a standard deviation. Using a propagation of uncertainty, the relationshipbetween the overall variance, s 2 , and the variances due to sampling,2s samp2, and the analytical method, s meth, istsns = s + s7.12 2 2samp meth 7.2Equation 7.2 shows that the overall variance for an analysis may belimited by either the analytical method or the collecting of samples. Unfortunately,analysts often try to minimize the overall variance by improvingonly the method’s precision. This is a futile effort, however, if the standarddeviation for sampling is more than three times greater than that for themethod. 1 Figure 7.2 shows how the ratio s samp /s meth affects the method’scontribution to the overall variance. As shown by the dashed line, if thesample’s standard deviation is 3× the method’s standard deviation, thenindeterminate method errors explain only 10% of the overall variance. If indeterminatesampling errors are significant, decreasing s meth provides onlya nominal change in the overall precision.1 Youden, Y. J. J. Assoc. Off. Anal. Chem. 1981, 50, 1007–1013.

Chapter 7 Collecting and Preparing Samples287Percent of Overall Variance Due to the Method1008060402000 1 2 3 4 5s samp /s methFigure 7.2 The blue curve shows the method’s contributionto the overall variance, s 2 , as a function of the relativemagnitude of the standard deviation in sampling, s samp ,and the method’s standard deviation, s meth . The dashedred line shows that the method accounts for only 10% ofthe overall variance when s samp = 3 × s meth . Understandingthe relative importance of potential sources of indeterminateerror is important when considering how toimprove the overall precision of the analysis.Example 7.1A quantitative analysis gives a mean concentration of 12.6 ppm for ananalyte. The method’s standard deviation is 1.1 ppm and the standarddeviation for sampling is 2.1 ppm. (a) What is the overall variance for theanalysis? (b) By how much does the overall variance change if we improves meth by 10% to 0.99 ppm? (c) By how much does the overall variancechange if we improve s samp by 10% to 1.89 ppm?So l u t i o n(a) The overall variance is2 2 2 2 2s = s + s = ( 21 . ppm) + ( 11 . ppm) = 56 . ppm 2samp meth(b) Improving the method’s standard deviation changes the overall variancetos 2 = ( 21 . ppm) 2 + ( 099 . ppm) 2 = 5.4 ppm 2Improving the method’s standard deviation by 10% improves theoverall variance by approximately 4%.(c) Changing the standard deviation for samplings 2 2 2= (. 19ppm) + ( 11 . ppm) = 48 . ppm 2improves the overall variance by almost 15%. As expected, becauses samp is larger than s meth , we obtain a bigger improvement in the overallvariance when we focus our attention on sampling problems.Practice Exercise 7.1Suppose you wish to reduce theoverall variance in Example 7.1to 5.0 ppm 2 . If you focus on themethod, by what percentage doyou need to reduce s meth ? If youfocus on the sampling, by whatpercentage do you need to reduces samp ?Click here to review your answerto this exerciseTo determine which step has the greatest effect on the overall variance,we need to measure both s samp and s meth . The analysis of replicate samples

288 Analytical Chemistry 2.0There are several ways to minimize thestandard deviation for sampling. Here aretwo examples. One approach is to use astandard reference material (SRM) thathas been carefully prepared to minimizeindeterminate sampling errors. When thesample is homogeneous—as is the case,for example, with aqueous samples—auseful approach is to conduct replicateanalyses on a single sample.provides an estimate of the overall variance. To determine the method’svariance we analyze samples under conditions where we may assume thatthe sampling variance is negligible. The sampling variance is determinedby difference.Example 7.2The following data were collected as part of a study to determine the effectof sampling variance on the analysis of drug-animal feed formulations. 2% Drug (w/w) % Drug (w/w)0.0114 0.0099 0.0105 0.0105 0.0109 0.01070.0102 0.0106 0.0087 0.0103 0.0103 0.01040.0100 0.0095 0.0098 0.0101 0.0101 0.01030.0105 0.0095 0.0097The data on the left were obtained under conditions where both s samp ands meth contribute to the overall variance. The data on the right were obtainedunder conditions where s samp is known to be insignificant. Determine theoverall variance, and the standard deviations due to sampling and the analyticalmethod. To which factor—sampling or the method—should youturn your attention if you want to improve the precision of the analysis?See Chapter 4 for a review of how to calculatethe variance.So l u t i o nUsing the data on the left, the overall variance, s 2 , is 4.71 × 10 –7 . To find2the method’s contribution to the overall variance, s meth, we use the data onthe right, obtaining a value of 7.00 × 10 –8 . The variance due to sampling,, is2s samp2 2 2 −7 −8s = s − s = 471 . × 10 − 7. 00× 10 = 401 . × 10 −7sampmethConverting variances to standard deviations gives s samp as 6.32 × 10 –4 ands meth as 2.65 × 10 –4 . Because s samp is more than twice as large as s meth , improvingthe precision of the sampling process has the greatest impact onthe overall precision.2 Fricke, G. H.; Mischler, P. G.; Staffieri, F. P.; Houmyer, C. L. Anal. Chem. 1987, 59, 1213–1217.Practice Exercise 7.2A polymer’s density provides a measure of its crystallinity. The standarddeviation for the determination of density using a single sample of a polymeris 1.96 × 10 –3 g/cm 3 . The standard deviation when using differentsamples of the polymer is 3.65 × 10 –2 g/cm 3 . Determine the standarddeviations due to sampling and the analytical method.Click here to review your answer to this exercise.

Chapter 7 Collecting and Preparing Samples2897BDesigning A Sampling PlanA sampling plan must support the goals of an analysis. For example, amaterial scientist interested in characterizing a metal’s surface chemistry ismore likely to choose a freshly exposed surface, created by cleaving the sampleunder vacuum, than a surface previously exposed to the atmosphere. Ina qualitative analysis, a sample does not need to be identical to the originalsubstance, provided that there is sufficient analyte to ensure its detection.In fact, if the goal of an analysis is to identify a trace-level component, itmay be desirable to discriminate against major components when collectingsamples.For a quantitative analysis, the sample’s composition must accuratelyrepresent the target population, a requirement that necessitates a carefulsampling plan. Among the issues to consider are these five questions.1. From where within the target population should we collect samples?2. What type of samples should we collect?3. What is the minimum amount of sample for each analysis?4. How many samples should we analyze?5. How can we minimize the overall variance for the analysis?7B.1 Where to Sample the Target PopulationA sampling error occurs whenever a sample’s composition is not identicalto its target population. If the target population is homogeneous, thenwe can collect individual samples without giving consideration to where tosample. Unfortunately, in most situations the target population is heterogeneous.Due to settling, a medication available as an oral suspension mayhave a higher concentration of its active ingredients at the bottom of thecontainer. The composition of a clinical sample, such as blood or urine, maydepend on when it is collected. A patient’s blood glucose level, for instance,changes in response to eating and exercise. Other target populations showboth a spatial and a temporal heterogeneity. The concentration of dissolvedO 2 in a lake is heterogeneous due both to the changing seasons and to pointsources of pollution.If the analyte’s distribution within the target population is a concern,then our sampling plan must take this into account. When feasible, homogenizingthe target population is a simple solution—in most cases, however,this is impracticable. Additionally, homogenization destroys informationabout the analyte’s spatial or temporal distribution within the target population,information that may be of importance.For an interesting discussion of the importanceof a sampling plan, see Buger, J. etal. “Do Scientists and Fishermen Collectthe Same Size Fish? Possible Implicationsfor Exposure Assessment,” Environ. Res.2006, 101, 34–41.The composition of a homogeneous targetpopulation is the same regardless ofwhere we sample, when we sample, or thesize of our sample. For a heterogeneoustarget population, the composition is notthe same at different locations, at differenttimes, or for different sample sizes.Ra n d o m Sa m p l i n gThe ideal sampling plan provides an unbiased estimate of the target population’sproperties. A random sampling is the easiest way to satisfy this

290 Analytical Chemistry 2.0Appendix 14 provides a random numbertable that you can use for designing samplingplans.requirement. 3 Despite its apparent simplicity, a truly random sample isdifficult to collect. Haphazard sampling, in which samples are collectedwithout a sampling plan, is not random and may reflect an analyst’s unintentionalbiases.Here is a simple method for ensuring that we collect random samples.First, we divide the target population into equal units and assign a uniquenumber to each unit. Then, we use a random number table to select theunits to sample. Example 7.3 provides an illustrative example.Example 7.3To analyze a polymer’s tensile strength, individual samples of the polymerare held between two clamps and stretched. In evaluating a productionlot, the manufacturer’s sampling plan calls for collecting ten 1 cm × 1 cmsamples from a 100 cm × 100 cm polymer sheet. Explain how we can usea random number table to ensure that our samples are random.So l u t i o nAs shown by the grid, we divide the polymer sheet into 10 000 1 cm × 1cm squares, each identified by its row number and its column number,with numbers running from 0 to 99.For example, the blue square is in row98 and column 1. To select ten squaresat random, we enter the random numbertable in Appendix 14 at an arbitrarypoint, and let the entry’s last four digitsrepresent the row and the column for thefirst sample. We then move through thetable in a predetermined fashion, selectingrandom numbers until we have 10samples. For our first sample, let’s use thesecond entry in the third column, which is 76831. The first sample, therefore,is row 68 and column 31. If we proceed by moving down the thirdcolumn, then the 10 samples are as follows:Sample Number Row Column Sample Number Row Column1 76831 68 31 6 41701 17 012 66558 65 58 7 38605 86 053 33266 32 66 8 64516 45 164 12032 20 32 9 13015 30 155 14063 40 63 10 12138 21 3801298990 1 2 98 99In collecting a random sample we make no assumptions about the targetpopulation, making this the least biased approach to sampling. On the3 Cohen, R. D. J. Chem. Educ. 1991, 68, 902–903.

Chapter 7 Collecting and Preparing Samples291other hand, a random sample often requires more time and expense thanother sampling strategies because we need a greater number of samples toensure that we adequately sample the target population. 4Ju d g m e n t a l Sa m p l i n gThe opposite of random sampling is selective, or judgmental sampling inwhich we use prior information about the target population to help guideour selection of samples. Judgmental sampling is more biased than randomsampling, bur requires fewer samples. Judgmental sampling is useful if wewish to limit the number of independent variables influencing our results.For example, if we are studying the bioaccumulation of PCB’s in fish, wemay choose to exclude fish that are too small or that appear diseased.Sy s t e m a t ic Sa m p l i n gRandom sampling and judgmental sampling represent extremes in bias andin the number of samples needed to characterize the target population. Systematicsampling falls in between these extremes. In systematic samplingwe sample the target population at regular intervals in space or time. Figure7.3 shows an aerial photo of the Great Salt Lake in Utah. A railroad linedivides the lake into two sections with different chemical compositions. Tocompare the lake’s two sections—and to evaluation spatial variations withineach section—we use a two-dimensional grid to define sampling locations.4 Borgman, L. E.; Quimby, W. F. in Keith, L. H., ed. Principles of Environmental Sampling, AmericanChemical Society: Washington, D. C., 1988, 25–43.railroad lineFigure 7.3 Aerial photo of the Great Salt Lake in Utah,taken from the International Space Station at a distanceof approximately 380 km. The railroad line divides thelake into two sections that differ in chemical composition.Superimposing a two-dimensional grid divides eachsection of the lake into sampling units. The red dots atthe center of each unit represent sampling sites. Photocourtesy of the Image Science and Analysis Laboratory,NASA Johnson Space Center, Photo Number ISS007-E-13002 (eol.jsc.nasa.gov).

292 Analytical Chemistry 2.0-1.0 -0.5 0.0 0.5 1.0(a)(b)0 2 4 6 8 10Figure 7.4 Effect of sampling frequencywhen monitoring a periodic signal. Individualsamples are shown by the reddots (•). In (a) the sampling frequencyis approximately 1.5 samples per period.The dashed red line shows the apparentsignal based on five samples and the solidblue line shows the true signal. In (b) asampling frequency of approximately 5samples per period accurately reproducesthe true signal.When a population’s heterogeneity is time-dependent, as is common inclinical studies, samples are drawn at regular intervals in time.If a target population’s properties have a periodic trend, a systematicsampling will lead to a significant bias if our sampling frequency is toosmall. This is a common problem when sampling electronic signals wherethe problem is known as aliasing. Consider, for example, a signal consistingof a simple sign wave. Figure 7.4a shows how an insufficient samplingfrequency underestimates the signal’s true frequency. The apparent signal,shown by the dashed red line passing through the five data points, is significantlydifferent from the true signal shown by the solid blue line.According to the Nyquist theorem, to accurately determine a periodicsignal’s true frequency, we must sample the signal at least twice during eachcycle or period. If we collect samples at an interval of Dt, the highest frequencywe can accurately monitor is (2Dt) –1 . For example, if our samplingrate is 1 sample/hr, the highest frequency we can monitor is (2×1 hr) –1 or0.5 hr –1 , corresponding to a period of less than 2 hr. If our signal’s periodis less than 2 hours (a frequency of more than 0.5 hr –1 ), then we must usea faster sampling rate. Ideally, the sampling rate should be at least 3-4 timesgreater than the highest frequency signal of interest. If our signal has a periodof one hour, we should collect a new sample every 15-20 minutes.Sy s t e m a t ic –Ju d g m e n t a l Sa m p l i n gCombinations of the three primary approaches to sampling are also possible.5 One such combination is systematic–judgmental sampling, inwhich we use prior knowledge about a system to guide a systematic samplingplan. For example, when monitoring waste leaching from a landfill,we expect the plume to move in the same direction as the flow of groundwater—thishelps focus our sampling, saving money and time. The systematic–judgmentalsampling plan in Figure 7.5 includes a rectangular grid formost of the samples and linear transects to explore the plume’s limits. 65 Keith, L. H. Environ. Sci. Technol. 1990, 24, 610–617.6 Flatman, G. T.; Englund, E. J.; Yfantis, A. A. in Keith, L. H., ed. Principles of EnvironmentalSampling, American Chemical Society: Washington, D. C., 1988, 73–84.Direction ofgroundwater flowFigure 7.5 Systematic–judgmental sampling plan for monitoring the leachingof pollutants from a landfill. The sampling sites, shown as red dots (•), are ona systematic grid straddling the direction of the groundwater’s flow. Samplingalong linear transects helps establish the plume’s limits.Landfill

Chapter 7 Collecting and Preparing Samples293St r a t i f i e d Sa m p l i n gAnother combination of the three primary approaches to sampling is judgmental–random,or stratified sampling. Many target populations consistof distinct units, or strata. For example, suppose we are studying particulatePb in urban air. Because particulates come in a range of sizes—some visibleand some microscopic—and from many sources—road dust, diesel soot,and fly ash to name a few—we can subdivide the target population by sizeor source. If we choose a random sampling plan, then we collect sampleswithout considering the different strata. For a stratified sampling, we dividethe target population into strata and collect random samples from withineach stratum. After analyzing the samples from each stratum, we pool theirrespective means to give an overall mean for the target population. Theadvantage of stratified sampling is that individual strata usually are morehomogeneous than the target population. The overall sampling variancefor stratified sampling is always at least as good, and often better than thatobtained by simple random sampling.Co n v e n i e n c e Sa m p l i n gOne additional method of sampling deserves brief mention. In conveniencesampling we select sample sites using criteria other than minimizingsampling error and sampling variance. In a survey of rural groundwaterquality, for example, we can choose to drill wells at randomly selected sitesor we can make use of existing wells, which is usually the preferred choice.In this case cost, expedience, and accessibility are more important thanensuring a random sample.7B.2 What Type of Sample to CollectAfter determining where to collect samples, the next step in designing asampling plan is to decide what type of sample to collect. There are threecommon methods for obtaining samples: grab sampling, composite sampling,and in situ sampling.The most common type of sample is a grab sample, in which we collecta portion of the target population at a specific time and/or location, providinga “snapshot” of the target population. If our target population is homogeneous,a series of random grab samples allows us to establish its properties.For a heterogeneous target population, systematic grab sampling allows usto characterize how its properties change over time and/or space.A composite sample is a set of grab samples that we combine into asingle sample before analysis. Because information is lost when we combineindividual samples, we normally analyze grab sample separately. Insome situations, however, there are advantages to working with a compositesample.One situation where composite sampling is appropriate is when ourinterest is in the target population’s average composition over time or space.

294 Analytical Chemistry 2.0For example, wastewater treatment plants must monitor and report theaverage daily composition of the treated water they release to the environment.The analyst can collect and analyze individual grab samples using asystematic sampling plan, reporting the average result, or she can combinethe grab samples into a single composite sample. Analyzing a single compositesample instead of many individual grab samples, saves time andmoney.Composite sampling is also useful when a single sample can not supplysufficient material for the analysis. For example, analytical methodsfor determining PCB’s in fish often require as much as 50 g of tissue, anamount that may be difficult to obtain from a single fish. By combiningand homogenizing tissue samples from several fish, it is easy to obtain thenecessary 50-g sample.A significant disadvantage of grab samples and composite samples is thatwe cannot use them to continuously monitor a time-dependent change inthe target population. In situ sampling, in which we insert an analyticalsensor into the target population, allows us to continuously monitor thetarget population without removing individual grab samples. For example,we can monitor the pH of a solution moving through an industrial productionline by immersing a pH electrode in the solution’s flow.Example 7.4A study of the possible relationship between traffic density and the concentrationsof lead, cadmium, and zinc in roadside soils, made use of thefollowing sampling plan. 7 Samples of surface soil (0–10 cm) were collectedat perpendicular distances of 1, 5, 10, 20, and 30 m from the roadway. Ateach distance, 10 samples were taken from different locations and mixedto form a single sample. What type of sampling plan is this? Explain whythis is an appropriate sampling plan.So l u t i o nThis is an example of a systematic–judgemental sampling plan using compositesamples. These are good choices given the goals of the study. Automobileemissions release particulates containing elevated concentrationsof lead, cadmium, and zinc—this study was conducted in Uganda whereleaded gasoline is still in use—which settle out on the surrounding roadsidesoils as “dry rain.” Sampling in areas near roadways, and samplingat fixed distances from the roadway provides sufficient data for the study,while limiting the total number of samples. Combining samples from thesame distance into a single, composite sample has the advantage of decreasingsampling uncertainty. Because variations in metal concentrationsperpendicular to the roadway is not of interest, the composite samples donot result in a loss of information.7 Nabulo, G.; Oryem-Origa, H.; Diamond, M. Environ. Res. 2006, 101, 42–52.

Chapter 7 Collecting and Preparing Samples2957B.3 How Much Sample to CollectTo minimize sampling errors, samples must be of an appropriate size. If asample is too small, its composition may differ substantially from that ofthe target population, introducing a sampling error. Samples that are toolarge, however, require more time and money to collect and analyze, withoutproviding a significant improvement in the sampling error.Let’s assume that our target population is a homogeneous mixture oftwo types of particles. Particles of type A contain a fixed concentration ofanalyte, and particles without analyte are of type B. Samples from this targetpopulation follow a binomial distribution. If we collect a sample containingn particles, the expected number of particles containing analyte, n A , isnA =npFor a review of the binomial distribution,see Chapter 4.where p is the probability of selecting a particle of type A. The standarddeviation for sampling iss np ( 1−p )7.3samp =relTo calculate the relative standard deviation for sampling, s samp, we divideequation 7.3 by n A , obtainingsrelsamp=np( 1−p)npSolving for n allows us to calculate the number of particles providing thedesired relative sampling variance.Example 7.51 pn = − 1×p ( s )rel 27.4sampSuppose we are analyzing a soil where the particles containing analyterepresent only 1 × 10 –7 % of the population. How many particles must wecollect to give a percent relative standard deviation for sampling of 1%?So l u t i o nSince the particles of interest account for 1 × 10 –7 % of all particles, theprobability, p, of selecting one of these particles is only 1 × 10 –9 . Substitutinginto equation 7.4 givesn = − × −91 ( 1 10 ) 1× = 1×10−9 21×10 ( 001 . )To obtain a relative standard deviation for sampling of 1%, we need tocollect 1 × 10 13 particles.13

296 Analytical Chemistry 2.0A sample containing 10 13 particles may be fairly large. Suppose this isequivalent to a mass of 80 g. Working with a sample this large clearly is notpractical. Does this mean we must work with a smaller sample and accepta larger relative standard deviation for sampling? Fortunately the answeris no. An important feature of equation 7.4 is that the relative standarddeviation for sampling is a function of the number of particles, not thecombined mass of the particles. If we crush and grind the particles to makethem smaller, then a sample containing 10 13 particles will have a smallermass. If we assume that a particle is spherical, then its mass is proportionalto the cube of its radius.mass ∝ r 3Decreasing a particle’s radius by a factor of 2, for example, decreases its massby a factor of 2 3 , or 8.Example 7.6Assume that a sample of 10 13 particles from Example 7.5 weighs 80 g andthat the particles are spherical. By how much must we reduce a particle’sradius if we wish to work with 0.6-g samples?So l u t i o nTo reduce the sample’s mass from 80 g to 0.6 g, we must change its massby a factor of80= 133×06 .To accomplish this we must decrease a particle’s radius by a factor ofr3= 133×r = 51 . ×Decreasing the radius by a factor of approximately 5 allows us to decreasethe sample’s mass from 80 g to 0.6 g.Treating a population as though it contains only two types of particlesis a useful exercise because it shows us that we can improve the relativestandard deviation for sampling by collecting more particles. Of course, areal population contains more than two types of particles, with the analytepresent at several levels of concentration. Nevertheless, many well-mixedpopulations, in which the population’s composition is homogeneous on thescale at which we sample, approximate binomial sampling statistics. Underthese conditions the following relationship between the mass of a randomgrab sample, m, and the percent relative standard deviation for sampling,R, is often valid

Chapter 7 Collecting and Preparing Samples297mR2 = K s7.5where K s is a sampling constant equal to the mass of a sample producing apercent relative standard deviation for sampling of ±1%. 8Example 7.7The following data were obtained in a preliminary determination of theamount of inorganic ash in a breakfast cereal.Mass of Cereal (g) 0.9956 0.9981 1.0036 0.9994 1.0067% w/w Ash 1.34 1.29 1.32 1.26 1.28What is the value of K s and what size samples are needed to give a percentrelative standard deviation for sampling of ±2.0%. Predict the percentrelative standard deviation and the absolute standard deviation if we collect5.00-g samples.Problem 8 in the end of chapter problemsasks you to derive equation 7.5.So l u t i o nTo determine the sampling constant, K s , we need to know the average massof the cereal samples and the relative standard deviation for the amount ofash. The average mass of the cereal samples is 1.0007 g. The average % w/wash and its absolute standard deviation are, respectively, 1.298% w/w and0.03194% w/w. The percent relative standard deviation, R, therefore, isssamp0. 03194%w/wR = × 100 = × 100 = 24 .6%X 1. 298%w/wSolving for K s gives its value as2 2K = mR = (. 1 0007 g)( 246 . ) = 6.06 gsTo obtain a percent relative standard deviation of ±2%, samples need tohave a mass of at leastm = K sR= 606 . g= 15 . g22(2.0)If we use 5.00-g samples, then the expected percent relative standard deviationisKs606 . gR = = = 110 . %m 5.00 gand the expected absolute standard deviation isssampRX (. 110)( 1. 298% w/w)= = = 0. 0143% w/w100 1008 Ingamells, C. O.; Switzer, P. Talanta 1973, 20, 547–568.

298 Analytical Chemistry 2.0Practice Exercise 7.3Olaquindox is a synthetic growth promoter in medicated feeds for pigs.In an analysis of a production lot of feed, five samples with nominalmasses of 0.95 g were collected and analyzed, with the results shown inthe following table.mass (g) 0.9530 0.9728 0.9660 0.9402 0.9576mg olaquindox/kg feed 23.0 23.8 21.0 26.5 21.4What is the value of K s and what size samples are needed to obtain apercent relative deviation for sampling of 5.0%? By how much do youneed to reduce the average particle size if samples must weigh no morethan 1 g?Click here to review your answer to this exercise.When we use equation 7.7, the standarddeviation for sampling, s samp , and the error,e, must be expressed in the same way.Because s samp is given as a percent relativestandard deviation, the error, e, is givenas a percent relative error. When you useequation 7.7, be sure to check that you areexpressing s samp and e in the same way.7B.4 How Many Samples to CollectIn the previous section we considered how much sample we need to minimizethe standard deviation due to sampling. Another important considerationis the number of samples to collect. If our samples are normallydistributed, then the confidence interval for the sampling error isµ= X ±where n samp is the number of samples and s samp is the standard deviationfor sampling. Rearranging equation 7.6 and substituting e for the quantityX −µ, gives the number of samples asnsamptsts=esampnsamp2 2sampBecause the value of t depends on n samp , the solution to equation 7.7 isfound iteratively.Example 7.8In Example 7.7 we determined that we need 1.5-g samples to establish ans samp of ±2.0% for the amount of inorganic ash in cereal. How many 1.5-gsamples do we need to obtain a percent relative sampling error of ±0.80%at the 95% confidence level?27.67.7So l u t i o nBecause the value of t depends on the number of samples—a result we haveyet to calculate—we begin by letting n samp = ∞ and using t(0.05, ∞) for t.From Appendix 4, the value for t(0.05, ∞) is 1.960. Substituting knownvalues into equation 7.7 gives the number of samples as

Chapter 7 Collecting and Preparing Samples2992 2(. 1 960)( 20 . )n samp= = 24.0≈242( 080 . )Letting n samp = 24, the value of t(0.05, 23) from Appendix 4 is 2.073.Recalculating n samp givesWith 24 samples, the degrees of freedomfor t is 23.2 2( 2. 073)( 20 . )n samp= = 26.9≈272( 080 . )When n samp = 27, the value of t(0.05, 26) from Appendix 4 is 2.060. Recalculatingn samp gives2 2( 2. 060)( 20 . )n samp= = 26.52 ≈ 272( 080 . )Because two successive calculations give the same value for n samp , we havean iterative solution to the problem. We need 27 samples to achieve a percentrelative sampling error of ±0.80% at the 95% confidence level.Practice Exercise 7.4Assuming that the percent relative standard deviation for sampling in thedetermination of olaquindox in medicated feed is 5.0% (see Practice Exercise7.3), how many samples do we need to analyze to obtain a percentrelative sampling error of ±2.5% at a = 0.05?Click here to review your answer to this exercise.Equation 7.7 provides an estimate for the smallest number of samplesthat will produce the desired sampling error. The actual sampling errormay be substantially larger if s samp for the samples we collect during thesubsequent analysis is greater than s samp used to calculate n samp . This is notan uncommon problem. For a target population with a relative samplingvariance of 50 and a desired relative sampling error of ±5%, equation 7.7predicts that 10 samples are sufficient. In a simulation using 1000 samplesof size 10, however, only 57% of the trials resulted in a sampling error ofless than ±5%. 9 Increasing the number of samples to 17 was sufficient toensure that the desired sampling error was achieved 95% of the time.7B.5 Minimizing the Overall VarianceA final consideration when developing a sampling plan is to minimize theoverall variance for the analysis. Equation 7.2 shows that the overall varianceis a function of the variance due to the method, s meth2, and the variance2due to sampling, s samp. As we have seen, we can improve the sampling varianceby collecting more samples of the proper size. Increasing the numberFor an interesting discussion of why thenumber of samples is important, see Kaplan,D.; Lacetera, N.; Kaplan, C. “SampleSize and Precision in NIH Peer Review,”Plos One, 2008, 3(7), 1–3. When reviewinggrants, individual reviewers report ascore between 1.0 and 5.0 (two significantfigure). NIH reports the average score tothree significant figures, implying that differencesof 0.01 are significant. If the individualscores have a standard deviation of0.1, then a difference of 0.01 is significantat a = 0.05 only if there are 384 reviews.The authors conclude that NIH reviewpanels are too small to provide a statisticallymeaningful separation between proposalsreceiving similar scores.9 Blackwood, L. G. Environ. Sci. Technol. 1991, 25, 1366–1367.

300 Analytical Chemistry 2.02of times we analyze each sample improves the method’s variance. If s sampis2significantly greater than s meth, we can ignore the method’s contribution tothe overall variance and use equation 7.7 to estimate the number of samplesto analyze. Analyzing any sample more than once will not improve theoverall variance, because the method’s variance is insignificant.22If s methis significantly greater than s samp, then we need to collect andanalyze only one sample. The number of replicate analyses, n rep , needed tominimize the error due to the method is given by an equation similar toequation 7.7.nrepts=e2 2meth2Unfortunately, the simple situations described above are often the exception.For many analyses, both the sampling variance and the methodvariance are significant, and both multiple samples and replicate analysesof each sample are necessary. The overall error in this case isssampse = t +n n2 2methsampEquation 7.8 does not have a unique solution as different combinations ofn samp and n rep give the same overall error. How many samples we collect andhow many times we analyze each sample is determined by other concerns,such as the cost of collecting and analyzing samples, and the amount ofavailable sample.Example 7.9An analytical method has a percent relative sampling variance of 0.40%and a percent relative method variance of 0.070%. Evaluate the percentrelative error (a = 0.05) if you collect 5 samples, analyzing each twice, andif you collect 2 samples, analyzing each 5 times.So l u t i o nBoth sampling strategies require a total of 10 analyses. From Appendix 4we find that the value of t(0.05, 9) is 2.262. Using equation 7.8, the relativeerror for the first sampling strategy issampe = 2 262 040 . 0.070. + = 067 . %5 5 × 2and that for the second sampling strategy ise = 2 262 040 . 0.070. + = 10 .%2 2 × 5nrep7.8

Chapter 7 Collecting and Preparing Samples301Because the method variance is smaller than the sampling variance, weobtain a smaller relative error if we collect more samples, analyzing eachfewer times.Practice Exercise 7.5An analytical method has a percent relative sampling variance of 0.10%and a percent relative method variance of 0.20%. The cost of collecting asample is $20 and the cost of analyzing a sample is $50. Propose a samplingstrategy that provides a maximum percent relative error of ±0.50%(a = 0.05) and a maximum cost of $700.Click here to review your answer to this exercise.7CImplementing the Sampling PlanImplementing a sampling plan normally involves three steps: physicallyremoving the sample from its target population, preserving the sample, andpreparing the sample for analysis. Except for in situ sampling, we analyzea sample after removing it from its target population. Because samplingexposes the target population to potential contamination, the samplingdevice must be inert and clean.After removing a sample from its target population, there is a dangerthat it will undergo a chemical or physical change before we can completeits analysis. This is a serious problem because the sample’s properties nolonger are representative of the target population. To prevent this problem,we often preserve samples before transporting them to the laboratory foranalysis. Even when analyzing samples in the field, preservation may stillbe necessary.The initial sample is called the primary or gross sample, and may be asingle increment drawn from the target population, or a composite of severalincrements. In many cases we cannot analyze the gross sample withoutfirst reducing the sample’s particle size, converting the sample into a morereadily analyzable form, or improving its homogeneity.7C.1 SolutionsTypical examples of solution samples include those drawn from containersof commercial solvents; beverages, such as milk or fruit juice; natural waters,including lakes, streams, seawater and rain; bodily fluids, such as bloodand urine; and, suspensions; such as those found in many oral medications.Let’s use the sampling of natural waters and wastewaters as a case study inhow to sample solutions.Although you may never work with thespecific samples highlighted in this section,the case studies presented here mayhelp you in envisioning potential problemsassociated with your samples.Sa m p l e Co l l e c t i o nThe chemical composition of a surface water—such as a stream, river, lake,estuary, or ocean—is influenced by flow rate and depth. Rapidly flow-

302 Analytical Chemistry 2.0capcapwinch linespringFigure 7.6 A Niskin samplingbottle for collecting water samplesfrom lakes and oceans. Afterlowering the bottle to the desireddepth, a weight is sent down thewinch line, tripping a spring thatcloses the bottle. Source: NOAA(photolib.noaa.gov).ing shallow streams and rivers, and shallow (

Chapter 7 Collecting and Preparing Samples303Table 7.1 Preservation Methods and Maximum Holding Times for SelectedAnalytes in Natural Waters and WastewatersAnalyte Preservation Method Maximum Holding Timeammonia cool to 4 o C; add H 2 SO 4 to pH < 2 28 dayschloride none required 28 daysmetals—Cr(VI) cool to 4 o C 24 hoursmetals—Hg HNO 3 to pH < 2 28 daysmetals—all others HNO 3 to pH < 2 6 monthsnitrate none required 48 hoursorganochlorine pesticides 1 mL of 10 mg/mL HgCl 2 orimmediate extraction with a7 days without extraction40 days with extractionsuitable non-aqueous solventpH none required analyze immediatelyIn most cases the sample bottle has a wide mouth, making it easy to filland remove the sample. A narrow-mouth sample bottle is used if exposingthe sample to the container’s cap or to the outside environment is a problem.Unless exposure to plastic is a problem, caps for sample bottles aremanufactured from polyethylene. When polyethylene must be avoided, thecontainer cap includes an inert interior liner of neoprene or Teflon.Sa m p l e Pr e s e r v a t i o n a n d Pr e p a r a t i o nAfter removing a sample from its target population, its chemical compositionmay change as a result of chemical, biological, or physical processes.To prevent a change in composition, samples are preserved by controllingthe solution’s pH and temperature, by limiting its exposure to light or tothe atmosphere, or by adding a chemical preservative. After preserving asample, it may be safely stored for later analysis. The maximum holdingtime between preservation and analysis depends on the analyte’s stabilityand the effectiveness of sample preservation. Table 7.1 provides a listof representative methods for preserving samples and maximum holdingtimes for several analytes of importance in the analysis of natural watersand wastewaters.Most solution samples do not need additional preparation before analysis.This is the case for samples of natural waters and wastewaters. Solutionsamples with particularly complex matricies—blood and milk are twoexamples—may need additional processing to separate the analytes frominterferents, a topic covered later in this chapter.7C.2 GasesTypical examples of gaseous samples include automobile exhaust, emissionsfrom industrial smokestacks, atmospheric gases, and compressed gases. AlsoHere our concern is only with the need toprepare the gross sample by converting itinto a form suitable for analysis. Analyticalmethods may include additional samplepreparation steps, such as concentratingor diluting the analyte, or adjusting theanalyte’s chemical form. We will considerthese forms of sample preparation in laterchapters focusing on specific analyticalmethods.

304 Analytical Chemistry 2.0included in this category are aerosol particulates—the fine solid particlesand liquid droplets that form smoke and smog. Let’s use the sampling ofurban air as a case study in how to sample gases.Sa m p l e Co l l e c t i o n1 m 3 is equivalent to 10 3 L.One approach for collecting a sample of urban air is to fill a stainless steelcanister or a Tedlar/Teflon bag. A pump pulls the air into the containerand, after purging, the container is sealed. This method has the advantageof simplicity and of collecting a representative sample. Disadvantages includethe tendency for some analytes to adsorb to the container’s walls, thepresence of analytes at concentrations too low to detect with accuracy andprecision, and the presence of reactive analytes, such as ozone and nitrogenoxides, that may react with the container or that may otherwise alter thesample’s chemical composition during storage. When using a stainless steelcanister, cryogenic cooling, which changes the sample from a gaseous stateto a liquid state, may limit some of these disadvantages.Most urban air samples are collected using a trap containing a solidsorbent or by filtering. Solid sorbents are used for volatile gases (vapor pressuresmore than 10 –6 atm) and semi-volatile gases (vapor pressures between10 –6 atm and 10 –12 atm). Filtration is used to collect aerosol particulates.Trapping and filtering allows for sampling larger volumes of gas—an importantconcern for an analyte with a small concentration—and stabilizesthe sample between its collection and its analysis.In solid sorbent sampling, a pump pulls the urban air through a canisterpacked with sorbent particles. Typically 2–100 L of air are sampledwhen collecting volatile compounds, and 2–500 m 3 when collecting semivolatilegases. A variety of inorganic, organic polymer, and carbon sorbentshave been used. Inorganic sorbents, such as silica gel, alumina, magnesiumaluminum silicate, and molecular sieves, are efficient collectors for polarcompounds. Their efficiency for collecting water, however, limits their sorptioncapacity for many organic compounds.Organic polymeric sorbents include polymeric resins of 2,4-diphenylp-phenyleneoxide or styrene-divinylbenzene for volatile compounds, andpolyurethane foam for semi-volatile compounds. These materials have alow affinity for water, and are efficient collectors for all but the most highlyvolatile organic compounds, and some lower molecular weight alcoholsand ketones. The adsorbing ability of carbon sorbents is superior to that oforganic polymer resins, which makes them useful for highly volatile organiccompounds that can not be collected by polymeric resins. The adsorbingability of carbon sorbents may be a disadvantage, however, since the sorbedcompounds may be difficult to desorb.Non-volatile compounds are normally present either as solid particulates,or are bound to solid particulates. Samples are collected by pullinglarge volumes of urban air through a filtering unit, collecting the particulateson glass fiber filters.

Chapter 7 Collecting and Preparing Samples305The short term exposure of humans, animals, and plants to atmosphericpollutants is more severe than that for pollutants in other matrices. Becausethe composition of atmospheric gases can vary significantly over a time, thecontinuous monitoring of atmospheric gases such as O 3 , CO, SO 2 , NH 3 ,H 2 O 2 , and NO 2 by in situ sampling is important. 11Sa m p l e Pr e s e r v a t i o n a n d Pr e p a r a t i o nAfter collecting a gross sample of urban air, there is generally little needfor sample preservation or preparation. The chemical composition of a gassample is usually stable when it is collected using a solid sorbent, a filter, orby cryogenic cooling. When using a solid sorbent, gaseous compounds arereleased for analysis by thermal desorption or by extracting with a suitablesolvent. If the sorbent is selective for a single analyte, the increase in the sorbent’smass can be used to determine the amount of analyte in the sample.7C.3 SolidsTypical examples of solid samples include large particulates, such as thosefound in ores; smaller particulates, such as soils and sediments; tablets, pellets,and capsules used for dispensing pharmaceutical products and animalfeeds; sheet materials, such as polymers and rolled metals; and tissuesamples from biological specimens. Solids are usually heterogeneous andsamples must be collected carefully if they are to be representative of thetarget population. Let’s use the sampling of sediments, soils, and ores as acase study in how to sample solids.Sa m p l e Co l l e c t i o nSediments from the bottom of streams, rivers, lakes, estuaries, and oceansare collected with a bottom grab sampler, or with a corer. A bottom grabsampler (Figure 7.7) is equipped with a pair of jaws that close when theycontact the sediment, scooping up sediment in the process. Its principaladvantages are ease of use and the ability to collect a large sample. Disadvantagesinclude the tendency to lose finer grain sediment particles as waterflows out of the sampler, and the loss of spatial information—both laterallyand with depth—due to mixing of the sample.An alternative method for collecting sediments is a cylindrical coringdevice (Figure 7.8). The corer is dropped into the sediment, collecting acolumn of sediment and the water in contact with the sediment. With thepossible exception of sediment at the surface, which may experience mixing,samples collected with a corer maintain their vertical profile, preservinginformation about how the sediment’s composition changes with depth.Collecting soil samples at depths of up to 30 cm is easily accomplishedwith a scoop or shovel, although the sampling variance is generally high. AFigure 7.7 Bottom grab samplerbeing prepared for deployment.Source: NOAA (photolib.noaa.gov).11 Tanner, R. L. in Keith, L. H., ed. Principles of Environmental Sampling, American ChemicalSociety: Washington, D. C., 1988, 275–286.

306 Analytical Chemistry 2.0cable to shipweightcore linerFigure 7.8 Schematic diagram of a gravity corer in operation. The corer’sweight is sufficient to penetrate into the sediment to a depth of approximately2 m. Flexible metal leaves on the bottom of the corer are pushedaside by the sediment, allowing it to enter the corer. The leaves bend backand hold the core sample in place as it is hauled back to the surface.better tool for collecting soil samples near the surface is a soil punch, whichis a thin-walled steel tube that retains a core sample after it is pushed intothe soil and removed. Soil samples from depths greater than 30 cm are collectedby digging a trench and collecting lateral samples with a soil punch.Alternatively, an auger may be used to drill a hole to the desired depth andthe sample collected with a soil punch.For particulate materials, particle size often determines the samplingmethod. Larger particulate solids, such as ores, are sampled using a riffle(Figure 7.9), which is a trough containing an even number of compartments.Because adjoining compartments empty onto opposite sides of thegross sampleFigure 7.9 Example of a four-unit riffle. Passing thegross sample, shown within the circle, through the riffledivides it into four piles, two on each side. Combiningthe piles from one side of the riffle provides a newsample, which may be passed through the riffle again orkept as the final sample. The piles from the other side ofthe riffle are discarded.separated portions of sample

Chapter 7 Collecting and Preparing Samples307riffle, dumping a gross sample into the riffle divides it in half. By repeatedlypassing half of the separated material back through the riffle, a sample ofany desired size may be collected.A sample thief (Figure 7.10) is used for sampling smaller particulatematerials, such as powders. A typical sample thief consists of two tubesthat are nestled together. Each tube has one or more slots aligned down thesample thief’s length. Before inserting the sample thief into the materialbeing sampled, the slots are closed by rotating the inner tube. When thesample thief is in place, rotating the inner tube opens the slots, which fillwith individual samples. The inner tube is then rotated to the closed positionand the sample thief withdrawn.slotsSa m p l e Pr e s e r v a t i o nWithout preservation, a solid sample may undergo a change in compositiondue to the loss of volatile material, biodegradation, and chemical reactivity(particularly redox reactions). Storing samples at lower temperatures makesthem less prone to biodegradation and to the loss of volatile material, butfracturing of solids and phase separations may present problems. To minimizethe loss of volatiles, the sample container is filled completely, eliminatinga headspace where gases collect. Samples that have not been exposedto O 2 are particularly susceptible to oxidation reactions. For example, thecontact of air with anaerobic sediments must be prevented.openclosedFigure 7.10 Sample thief showingits open and closed positions.Sa m p l e Pr e p a r a t i o nUnlike gases and liquids, which generally require little sample preparation,a solid sample usually needs some processing before analysis. There are tworeasons for this. First, as discussed in section 7B.3, the standard deviationfor sampling, s samp , is a function of the number of particles in the sample,not the combined mass of the particles. For a heterogeneous material consistingof large particulates, the gross sample may be too large to analyze.For example, a Ni-bearing ore with an average particle size of 5 mm mayrequire a sample weighing one ton to obtain a reasonable s samp . Reducingthe sample’s average particle size allows us to collect the same number ofparticles with a smaller, more manageable mass. Second, many analyticaltechniques require that the analyte be in solution.Re d u c i n g Pa r t i c l e Si z eA reduction in particle size is accomplished by a combination of crushingand grinding the gross sample. The resulting particulates are then thoroughlymixed and divided into subsamples of smaller mass. This processseldom occurs in a single step. Instead, subsamples are cycled through theprocess several times until a final laboratory sample is obtained.Crushing and grinding uses mechanical force to break larger particlesinto smaller particles. A variety of tools are used depending on the particle’s

308 Analytical Chemistry 2.0size and hardness. Large particles are crushed using jaw crushers capable ofreducing particles to diameters of a few millimeters. Ball mills, disk mills,and mortars and pestles are used to further reduce particle size.A significant change in the gross sample’s composition may occur duringcrushing and grinding. Decreasing particle size increases the availablesurface area, which increases the risk of losing volatile components. Thisproblem is made worse by the frictional heat that accompanies crushingand grinding. Increasing the surface area also exposes interior portions ofthe sample to the atmosphere where oxidation may alter the gross sample’scomposition. Other problems include contamination from the materialsused to crush and grind the sample, and differences in the ease with whichparticles are reduced in size. For example, softer particles are easier to reducein size and may be lost as dust before the remaining sample is processed.This is a particular problem if the analyte’s distribution between differenttypes of particles is not uniform.The gross sample is reduced to a uniform particle size by intermittentlypassing it through a sieve. Those particles not passing through the sievereceive additional processing until the entire sample is of uniform size. Theresulting material is mixed thoroughly to ensure homogeneity and a subsampleobtained with a riffle, or by coning and quartering. As shownin Figure 7.11, the gross sample is piled into a cone, flattened, and dividedinto four quarters. After discarding two diagonally opposed quarters, theremaining material is cycled through the process of coning and quarteringuntil a suitable laboratory sample remains.gather material into a coneflatten the conelaboratory samplenewcyclediscarddivide into quartersdivide in halfFigure 7.11 Illustration showing the method of coning and quartering for reducing sample size. After gatheringthe gross sample into a cone, the cone is flattened, divided in half, and then divided into quarters. Two opposingquarters are combined to form the laboratory sample, or the subsample is sent back through another cycle. Thetwo remaining quarters are discarded.

Chapter 7 Collecting and Preparing Samples309Br i n g i n g So l i d Sa m p l e s In t o So l u t i o nIf you are fortunate, your sample will easily dissolve in a suitable solvent,requiring no more effort than gently swirling and heating. Distilled water isusually the solvent of choice for inorganic salts, but organic solvents, suchas methanol, chloroform, and toluene are useful for organic materials.When a sample is difficult to dissolve, the next step is to try digestingit with an acid or a base. Table 7.2 lists several common acids and bases,and summarizes their use. Digestions are carried out in an open container,usually a beaker, using a hot-plate as a source of heat. The main advantageof an open-vessel digestion is cost because it requires no special equipment.Volatile reaction products, however, are lost, resulting in a determinate errorif they include the analyte.Many digestions are now carried out in a closed container using microwaveradiation as the source of energy. Vessels for microwave digestionare manufactured using Teflon (or some other fluoropolymer) or fusedsilica. Both materials are thermally stable, chemically resistant, transparentto microwave radiation, and capable of withstanding elevated pressures. Atypical microwave digestion vessel, as shown in Figure 7.12, consists of aninsulated vessel body and a cap with a pressure relief valve. The vessels areplaced in a microwave oven (typically 6–14 vessels can be accommodated)and microwave energy is controlled by monitoring the temperature or pressurewithin one of the vessels.Table 7.2 Acids and Bases Used for Digesting SamplesSolutionUses and PropertiesHCl (37% w/w) • dissolves metals more easily reduced than H2 (E o < 0)• dissolves insoluble carbonate, sulfides, phosphates, fluorides,sulfates, and many oxidesHNO 3 (70% w/w) • strong oxidizing agent• dissolves most common metals except Al, Au, Pt, and Cr• decomposes organics and biological samples (wet ashing)H 2 SO 4 (98% w/w) • dissolves many metals and alloys• decomposes organics by oxidation and dehydrationHF (50% w/w) • dissolves silicates by forming volatile SiF4HClO 4 (70% w/w) • hot, concentrated solutions are strong oxidizing agents• dissolves many metals and alloys• decomposes organics ( Caution: reactions with organics often areexplosive; use only in a specially equipped hood with a blast shieldand after prior decomposition with HNO 3 )HCl:HNO 3 (3:1 v/v) • also known as aqua regia• dissolves Au and PtNaOH• dissolves Al and amphoteric oxides of Sn, Pb, Zn, and Cr

310 Analytical Chemistry 2.0(a)(b)pressure relief valvedigestion vesselFigure 7.12 Microwave digestion unit. (a) View of the unit’s interior showing the carousel holding the digestionvessels. (b) Close-up of a Teflon digestion vessel, which is encased in a thermal sleeve. The pressure relief value,which is part of the vessel’s cap, contains a membrane that ruptures if the internal pressure becomes too high.A microwave digestion has several important advantages over an openvesseldigestion, including higher temperatures (200–300 o C) and pressures(40–100 bar). As a result, digestions requiring several hours in anopen-vessel may need no more than 30 minutes when using a microwavedigestion. In addition, a closed container prevents the loss of volatile gases.Disadvantages include the inability to add reagents during the digestion,limitations on the sample’s size (typically < 1 g), and safety concerns dueto the high pressures and corrosive reagents.Inorganic samples that resist decomposition by digesting with acidsor bases often can be brought into solution by fusing with a large excessof an alkali metal salt, called a flux. After mixing the sample and the fluxin a crucible, they are heated to a molten state and allowed to cool slowlyto room temperature. The melt usually dissolves readily in distilled wateror dilute acid. Table 7.3 summarizes several common fluxes and their uses.Fusion works when other methods of decomposition do not because of thehigh temperature and the flux’s high concentration in the molten liquid.Table 7.3 Common Fluxes for Decomposing Inorganic SamplesFluxMeltingTemperature ( o C) Crucible Typical SamplesNa 2 CO 3 851 Pt silicates, oxides, phosphates, sulfidesLi 2 B 4 O 7 930 Pt, graphite aluminosilicates, carbonatesLiBO 2 845 Pt, graphite aluminosilicates, carbonatesNaOH 318 Au, Ag silicates, silicon carbideKOH 380 Au, Ag silicates, silicon carbideNa 2 O 2 — Ni silicates, chromium steels, Pt alloysK 2 S 2 O 7 300 Ni, porcelain oxidesB 2 O 3 577 Pt silicates, oxides

Chapter 7 Collecting and Preparing Samples311Disadvantages include contamination from the flux and the crucible, andthe loss of volatile materials.Finally, we can decompose organic materials by dry ashing. In thismethod the sample is placed in a suitable crucible and heated over a flameor in a furnace. The carbon present in the sample oxidizes to CO 2 , andhydrogen, sulfur, and nitrogen leave as H 2 O, SO 2 , and N 2 . These gases canbe trapped and weighed to determine their concentration in the organicmaterial. Often the goal of dry ashing is to remove the organic material,leaving behind an inorganic residue, or ash, that can be further analyzed.7DSeparating the Analyte from InterferentsWhen an analytical method is selective for the analyte, analyzing samples isa relatively simple task. For example, a quantitative analysis for glucose inhoney is relatively easy to accomplish if the method is selective for glucose,even in the presence of other reducing sugars, such as fructose. Unfortunately,few analytical methods are selective toward a single species.In the absence of interferents, the relationship between the sample’ssignal, S samp , and the concentration of analyte, C A , isS= k C7.9samp A Awhere k A is the analyte’s sensitivity. If an interferent, is present, then equation7.9 becomesS = k C + kCsamp A A I I7.10where k I and C I are, respectively, the interferent’s sensitivity and concentration.A method’s selectivity for the analyte is determined by the relativedifference in its sensitivity toward the analyte and the interferent. If k Ais greater than k I , then the method is more selective for the analyte. Themethod is more selective for the interferent if k I is greater than k A .Even if a method is more selective for an interferent, we can use it todetermine C A if the interferent’s contribution to S samp is insignificant. Theselectivity coefficient, K A,I , which we introduced in Chapter 3, providesa way to characterize a method’s selectivity.KA,IkI= 7.11kSolving equation 7.11 for k I , substituting into equation 7.10, and simplifying,givesAS = k ( C + K × C ) 7.12samp A A A,I IAn interferent, therefore, does not pose a problem as long as the product ofits concentration and its selectivity coefficient is significantly smaller thanthe analyte’s concentration.In equation 7.9, and the equations thatfollow, you can replace the analyte’s concentration,C A , with the moles of analyte,n A when working with methods, such asgravimetry, that respond to the absoluteamount of analyte in a sample. In this casethe interferent also is expressed in termsof moles.

312 Analytical Chemistry 2.0K × C

Chapter 7 Collecting and Preparing Samples313tions of Cu or Zn are prepared and analyzed. When a sample containing128.6 ppm Cu is taken through the separation, the concentration of Curemaining is 127.2 ppm. Taking a 134.9 ppm solution of Zn through theseparation leaves behind a concentration of 4.3 ppm Zn. Calculate therecoveries for Cu and Zn, and the separation factor.So l u t i o nUsing equation 7.13 and equation 7.14, the recoveries for the analyte andinterferent areR CuR Znand the separation factor is127.2 ppm= = 0. 9891, or 98. 91%128.6 ppm4.3 ppm= = 0. 032, or 3.2%134.9 ppmSZn,CuRZn0.032= = = 0.032R 0.9891CuRecoveries and separation factors are useful for evaluating a separation’spotential effectiveness. They do not, however, provide a direct indicationof the error that results from failing to remove all the interferent or fromfailing to completely recover the analyte. The relative error due to the separation,E, isSE =SS− ∗samp samp∗samp7.16∗where S sampis the sample’s signal for an ideal separation in which we completelyrecover the analyte.∗S = k ( C )7.17samp A A oSubstituting equation 7.12 and equation 7.17 into equation 7.16, and rearrangingk ( C + K × C ) −k ( C )E =k ( C )A A A,I I A A oA A oC + K × C −( C )E =( C )A A,I I A oC ( C ) K × CA A o A,IE = − +( C ) ( C ) ( C )AoAAooAoI

314 Analytical Chemistry 2.0leaves us withK × CA,I IE = ( R − 1)+A7.18( C )A more useful equation is obtained by solving equation 7.14 for C I andsubstituting into equation 7.18.⎡ K × ( C ) ⎤E = R − +A,I I o( 1)× RAI⎣⎢C7.19( )A o ⎦⎥The first term of equation 7.19 accounts for the analyte’s incomplete recovery,and the second term accounts for failing to remove all the interferent.Example 7.11Following the separation outlined in Example 7.10, an analysis is carriedout to determine the concentration of Cu in an industrial plating bath.Analysis of standard solutions containing either Cu or Zn give the followinglinear calibrations.SCu−1= 1250 ppm × CAoCuSZn−1= 2310 ppm × C(a) What is the relative error if we analyze samples without removing theZn? Assume that the initial concentration ratio, Cu:Zn, is 7:1. (b) What isthe relative error if we first complete the separation, obtaining the recoveriesfrom Example 7.10? (c) What is the maximum acceptable recovery forZn if the recovery for Cu is 1.00 and the error due to the separation mustbe no greater than 0.10%?So l u t i o n(a) If we complete the analysis without separating Cu and Zn, then R Cuand R Zn are exactly 1 and equation 7.19 simplifies toKE =Cu,Zn× ( C )( C )CuoZn oUsing equation 7.11, we find that the selectivity coefficient isKCu,ZnZn−1= kZnppmk= 23101250 ppm= 185 .−1 CuGiven the initial concentration ratio, Cu:Zn, of 7:1, the relative errorif we do not attempt a separation of Cu and Zn is185 . × 1E = = 0. 264, or 26.4%7

Chapter 7 Collecting and Preparing Samples315(b) To calculate the relative error we substitute the recoveries from Example7.10 into equation 7.19, obtaining⎡185 . × 1 ⎤E = ( 0. 9891− 1)+× 0.032⎢⎥⎣ 7 ⎦E =−0.0109 + 0. 0085 =−0.0024or –0.24%. Note that the negative determinate error due to the analyte’sincomplete recovery is partially offset by a positive determinateerror from failing to remove all the interferent.(c) To determine the maximum recovery for Zn, we make appropriatesubstitutions into equation 7.19⎡185 . × 1E = 0. 0010 = ( 1− 1)+× R⎢⎣ 7and solve for R Zn , obtaining a recovery of 0.0038, or 0.38%. Thus,we must remove at least100.00% − 0.38% = 99.62%of the Zn to obtain an error of 0.10% when R Cu is exactly 1.Zn⎤⎥⎦7FClassifying Separation TechniquesWe can separate an analyte and an interferent if there is a significant differencein at least one of their chemical or physical properties. Table 7.4provides a partial list of separation techniques, classified by the chemical orphysical property being exploited.Table 7.4 Classification of Separation TechniquesBasis of SeparationSeparation Techniquesizefiltrationdialysissize-exclusion chromatographymass or densitycentrifugationcomplex formationmaskingchange in physical statedistillationsublimationrecrystallizationchange in chemical state precipitationelectrodepositionvolatilizationpartitioning between phases extractionchromatography

316 Analytical Chemistry 2.0(a) (b) (c)membranemembranemembraneFigure 7.13 Three types of membrane filters for separating analytes from interferents. (a) A centrifugal filter for concentratingand desalting macromolecular solutions. The membrane has a nominal molecular weight cut-off of 1 × 10 6g/mol. The sample is placed in the upper reservoir and the unit is placed in a centrifuge. Upon spinning the unit at2000×g–5000×g, the filtrate collects in the bottom reservoir and the retentate remains in the upper reservoir. (b) A 0.45mm membrane syringe filter. The photo on the right shows the membrane filter in its casing. In the photo on the left,the filter is attached to a syringe. Samples are placed in the syringe and pushed through the filter. The filtrate is collectedin a test tube or other suitable container. (c) A disposable filter system with a 0.22 mm cellulose acetate membrane filter.The sample is added to the upper unit and vacuum suction is used to draw the filtrate through the membrane and intothe lower unit. To store the filtrate, the top half of the unit is removed and a cap placed on the lower unit. The filter unitshown here has a capacity of 150 mL.For applications of gravity filtration andsuction filtration in gravimetric methodsof analysis, see Chapter 8.7F.1 Separations Based on SizeSize is the simplest physical property we can exploit in a separation. Toaccomplish the separation we use a porous medium through which onlythe analyte or the interferent can pass. Examples of size-based separationsinclude filtration, dialysis, and size-exclusion.In a filtration we separate a particulate interferent from dissolvedanalytes using a filter whose pore size retains the interferent. The solutionpassing through the filter is called the filtrate, and the material retainedby the filter is the retentate. Gravity filtration and suction filtration usingfilter paper are techniques with which you should already be familiar. Membranefilters, available in a variety of micrometer pores sizes, are the methodof choice for particulates that are too small to be retained by filter paper.Figure 7.13 provides information about three types of membrane filters.Dialysis is another example of a separation technique that uses sizeto separate the analyte and the interferent. A dialysis membrane is usuallyconstructed from cellulose and fashioned into tubing, bags, or cassettes.Figure 7.14 shows an example of a commercially available dialysis cassette.The sample is injected into the dialysis membrane, which is sealed tightly

Chapter 7 Collecting and Preparing Samples317injection portdialysismembranefoam buoyFigure 7.14 Example of a dialysis cassette. The dialysis membrane in this unit hasa molecular weight cut-off of 10 000 g/mol. Two sheets of the membrane are separatedby a gasket and held in place by the plastic frame. Four ports, one of which islabeled, provide a means for injecting the sample between the dialysis membranes.The cassette is inverted and submerged in a beaker containing the external solution,which is stirred using a stir bar. A foam buoy, used as a stand in the photo, servesas a float so that the unit remains suspended above the stir bar. The external solutionis usually replaced several time during dialysis. When dialysis is complete, thesolution remaining in the cassette is removed through an injection port.by a gasket, and the unit is placed in a container filled with a solutionwhose composition is different from the sample. If the concentration ofa particular species is different on the membrane’s two sides, the resultingconcentration gradient provides a driving force for its diffusion across themembrane. While small species freely pass through the membrane, largerspecies are unable to pass. Dialysis is frequently used to purify proteins,hormones, and enzymes. During kidney dialysis, metabolic waste products,such as urea, uric acid, and creatinine, are removed from blood by passingit over a dialysis membrane.Size-exclusion chromatography is a third example of a separationtechnique that uses size as a means for effecting a separation. In this techniquea column is packed with small, approximately 10-mm, porous polymerbeads of cross-linked dextrin or polyacrylamide. The pore size of theparticles is controlled by the degree of cross-linking, with more cross-linkingproducing smaller pore sizes. The sample is placed into a stream of solventthat is pumped through the column at a fixed flow rate. Those species toolarge to enter the pores pass through the column at the same rate as thesolvent. Species entering into the pores take longer to pass through the column,with smaller species needing more time to pass through the column.Size-exclusion chromatography is widely used in the analysis of polymers,and in biochemistry, where it is used for the separation of proteins.A more detailed treatment of size-exclusionchromatography, which also is calledgel permeation chromatography, is inChapter 12.7F.2 Separations Based on Mass or DensityIf the analyte and the interferent differ in mass or density, then a separationusing centrifugation may be possible. The sample is placed in a centrifugetube and spun at a high angular velocity, measured in revolutions perminute (rpm). The sample’s constituents experience a centrifugal force thatpulls them toward the bottom of the centrifuge tube. The species experiencingthe greatest centrifugal force has the fastest sedimentation rate, and

318 Analytical Chemistry 2.0Table 7.5 Conditions for Separating Selected CellularComponents by CentrifugationComponents Centrifugal Force (× g) Time (min)eukaryotic cells 1000 5cell membranes, nuclei 4000 10mitochondria, bacterial cells 15000 20lysosomes, bacterial membranes 30000 30ribosomes 100000 180Source: Adapted from Zubay, G. Biochemistry, 2nd ed. Macmillan: New York, 1988, p.120.is the first to reach the bottom of the centrifuge tube. If two species haveequal density, their separation is based on mass, with the heavier specieshaving the greater sedimentation rate. If the species are of equal mass, thenthe species with the largest density has the greatest sedimentation rate.Centrifugation is an important separation technique in biochemistry.Table 7.5, for example, lists conditions for separating selected cellular components.We can separate lysosomes from other cellular components byseveral differential centrifugations, in which we divide the sample into asolid residue and a supernatant solution. After destroying the cells, the solutionis centrifuged at 15 000 × g (a centrifugal force that is 15 000 timesthat of the earth’s gravitational force) for 20 minutes, leaving a solid residueof cell membranes and mitochondria. The supernatant, which contains thelysosomes, is isolated by decanting it from the residue and centrifuged at30 000 × g for 30 minutes, leaving a solid residue of lysosomes. Figure 7.15shows a typical centrifuge capable of produce the centrifugal forces neededfor biochemical separations.An alternative approach to differential centrifugation is density gradientcentrifugation. To prepare a sucrose density gradient, for example,a solution with a smaller concentration of sucrose—and, thus, of lowerdensity—is gently layered upon a solution with a higher concentration ofFigure 7.15 Bench-top centrifuge capable of reaching speeds upto 14 000 rpm and centrifugal forces of 20 800 × g. This particularcentrifuge is refrigerated, allowing samples to be cooled totemperatures as low as –4 o C.

Chapter 7 Collecting and Preparing Samples319Figure 7.16 Example of a sucrose density gradient centrifugation of thylakoid membranes fromwild type (WT) and lut2 plants. The thylakoid membranes were extracted from the plant’sleaves and separated by centrifuging in a 0.1–1 M sucrose gradient for 22 h at 280 000 × g at4 o C. Six bands and their chlorophyll contents are shown. Adapted from Dall’Osto, L.; Lico,C.; Alric, J.; Giuliano, G.; Havaux, M.; Bassi, R. BMC Plant Biology 2006, 6:32.sucrose. Repeating this process several times, fills the centrifuge tube witha multi-layer density gradient. The sample is placed on top of the densitygradient and centrifuged using a force greater than 150 000 × g. Duringcentrifugation, each of the sample’s components moves through the gradientuntil it reaches a position where its density matches the surroundingsucrose solution. Each component is isolated as a separate band positionedwhere its density is equal to that of the local density within the gradient.Figure 7.16 provides an example of a typical sucrose density centrifugationfor separating plant thylakoid membranes.7F.3 Separations Based on Complexation Reactions (Masking)One widely used technique for preventing an interference is to bind theinterferent in a strong, soluble complex that prevents it from interfering inthe analyte’s determination. This process is known as masking. As shown inTable 7.6, a wide variety of ions and molecules are useful masking agents,and, as a result, selectivity is usually not a problem.Example 7.12Using Table 7.6, suggest a masking agent for an analysis of aluminum inthe presence of iron.So l u t i o nA suitable masking agent must form a complex with the interferent, butnot with the analyte. Oxalate, for example, is not a suitable masking agentTechnically, masking is not a separationtechnique because we do not physicallyseparate the analyte and the interferent.We do, however, chemically isolate theinterferent from the analyte, resulting ina pseudo-separation.

320 Analytical Chemistry 2.0Table 7.6 Selected Inorganic and Organic Masking Agents for Metal IonsMasking Agent Elements Whose Ions Can Be MaskedCN –Ag, Au, Cd, Co, Cu, Fe, Hg, Mn, Ni, Pd, Pt, ZnSCN –Ag, Cd, Co, Cu, Fe, Ni, Pd, Pt, ZnNH 3Ag, Co, Ni, Cu, ZnF –Al, Co, Cr, Mg, Mn, Sn, Zn2–S 2 O 3 Au, Ce, Co, Cu, Fe, Hg, Mn, Pb, Pd, Pt, Sb, Sn, ZntartrateAl, Ba, Bi, Ca, Ce, Co, Cr, Cu, Fe, Hg, Mn, Pb, Pd, Pt, Sb, Sn, ZnoxalateAl, Fe, Mg, Mnthioglycolic acid Cu, Fe, SnSource: Meites, L. Handbook of Analytical Chemistry, McGraw-Hill: New York, 1963.because it binds both Al and Fe. Thioglycolic acid, on the other hand, isa selective masking agent for Fe in the presence of Al. Other acceptablemasking agents are cyanide (CN – ) thiocyanate (SCN – ), and thiosulfate(S 2 O 3 2– ).Practice Exercise 7.6Using Table 7.6, suggest a masking agent for the analysis of Fe in thepresence of Al.Click here to review your answer to this exercise.As shown in Example 7.13, we can judge a masking agent’s effectivenessby considering the relevant equilibrium constants.Example 7.13Show that CN – is an appropriate masking agent for Ni 2+ in a methodwhere nickel’s complexation with EDTA is an interference.You will find the formation constants forthese reactions in Appendix 12.So l u t i o nThe relevant reactions and formation constants are+ − −Ni ( aq) + Y ( aq) NiY ( aq) K = 42 . × 1012 4 2 182Ni + ( aq) + 4CN − ( aq) Ni(CN) − ( aq) β = 1.7×10442 30where Y 4– is an abbreviation for EDTA. Cyanide is an appropriate maskingagent because the formation constant for the Ni(CN) 4 2– is greater thanthat for the Ni–EDTA complex. In fact, the equilibrium constant for thereaction in which EDTA displaces the masking agent2− 4− 2+−Ni(CN) ( aq) + Y ( aq) NiY ( aq) + 4CN( aq)4

Chapter 7 Collecting and Preparing Samples321KK 42 . × 10= =β 17 . × 101 41830= 25 . × 10−12is very small, indicating that Ni(CN) 4 2– is relatively inert in the presenceof EDTA.Practice Exercise 7.7Use the formation constants in Appendix 12 to show that 1,10-phenanthroline is a suitable maskingagent for Fe 2+ in the presence of Fe 3+ . Use a ladder diagram to define any limitations on using1,10-phenanthroline as a masking agent. See Chapter 6 for a review of ladder diagrams.Click here to review your answer to this exercise.7F.4 Separations Based on a Change of StateBecause an analyte and its interferent are usually in the same phase, we canachieve a separation if one of them undergoes a change in its physical stateor its chemical state.Ch a n g e s in Ph y s i c a l St a t eWhen the analyte and the interferent are miscible liquids, a distillationmay be possible if their boiling points are significantly different. Figure7.17 shows the progress of a distillation as a plot of temperature versus themixture’s vapor-phase and liquid-phase composition. The initial mixture(point A), contains more interferent than analyte. When this solution isbrought to its boiling point, the vapor phase in equilibrium with the liquidphase is enriched in analyte (point B). The horizontal line connectingpoints A and B represents this vaporization equilibrium. Condensing thevapor phase at point B, by lowering the temperature, creates a new liquidphase whose composition is identical to that of the vapor phase (point C).The vertical line connecting points B and C represents this condensationTemperatureVapor PhaseDLiquid Phase100 Mole % Analyte0BCAFigure 7.17 Boiling point versus compositiondiagram for a near-ideal solution consistingof a low-boiling analyte and a highboilinginterferent. The horizontal linesrepresent vaporization equilibria and thevertical lines represent condensation equilibria.See the text for additional details.

322 Analytical Chemistry 2.0(a)(b)thermometerthermometerdistillationadaptordistillationadaptorcondensercondenserdistillation flaskfractionatingcolumnreceiving flaskreceiving flaskdistillation flaskFigure 7.18 Typical experimental set-up for (a) a simple distillation, and (b) a fractional distillation.sublimationchambercrude sampleheatcooling watervacuum inletsublimed analyteFigure 7.19 Typical experimental setupfor a sublimation. The sample isplaced in the sublimation chamber,which may be evacuated. Heating thesample causes the analyte to vaporizeand sublime onto a cooled probe.Modified from Slashme (commons.wikipedia.org).equilibrium. The liquid phase at point C has a lower boiling point thanthe original mixture, and is in equilibrium with the vapor phase at point D.This process of repeated vaporization and condensation gradually separatesthe analyte and interferent.Two experimental set-ups for distillations are shown in Figure 7.18. Thesimple distillation apparatus shown in Figure 7.18a does not produce a veryefficient separation, and is useful only for separating a volatile analyte (orinterferent) from a non-volatile interferent (or analyte), or for an analyteand interferent whose boiling points differ by more than 150 o C. A moreefficient separation is achieved by a fractional distillation (Figure 7.18b).Packing the distillation column with a high surface area material, such asa steel sponge or glass beads, provides more opportunity for the repeatedprocess of vaporization and condensation necessary to effect a completeseparation.When the sample is a solid, sublimation may provide a useful separationof the analyte and the interferent. The sample is heated at a temperatureand pressure below the analyte’s triple point, allowing it to vaporizewithout passing through the liquid state. Condensing the vapor recoversthe purified analyte (Figure 7.19). A useful analytical example of sublimationis the isolation of amino acids from fossil mollusk shells and deep-seasediments. 1313 Glavin, D. P.; Bada, J. L. Anal. Chem. 1998, 70, 3119–3122.

Chapter 7 Collecting and Preparing Samples323Recrystallization is another method for purifying a solid. A solventis chosen in which the analyte’s solubility is significant when the solvent ishot, and minimal when the solvent is cold. The interferents must be lesssoluble in the hot solvent than the analyte, or present in much smalleramounts. After heating a portion of the solvent in an Erlenmeyer flask,small amounts of the sample are added until undissolved sample is visible.Additional hot solvent is added until the sample redissolves, or until onlyinsoluble impurities remain. This process of adding sample and solvent isrepeated until the entire sample has been added to the Erlenmeyer flask.Any insoluble impurities are removed by filtering the hot solution. Thesolution is allowed to cool slowly, promoting the growth of large, purecrystals, and then cooled in an ice bath to minimize solubility losses. Thepurified sample is isolated by filtration and rinsed to remove any solubleimpurities. Finally, the sample is dried to remove any remaining traces ofthe solvent. Further purification, if necessary, can be accomplished by additionalrecrystallizations.Ch a n g e s in Ch e m ic a l St a t eDistillation, sublimation, and recrystallization use a change in physicalstate as a means of separation. Chemical reactivity also can be a useful toolfor separating analytes and interferents. For example, we can separate SiO 2from a sample by reacting it with HF to form SiF 4 . Because SiF 4 is volatile,it is easy to remove by evaporation. If we wish to collect the reaction’s volatileproduct, then a distillation is possible. For example, we can isolate theNH 4 + in a sample by making the solution basic and converting it to NH 3 .The ammonia is then removed by distillation. Table 7.7 provides additionalexamples of this approach for isolating inorganic ions.Another reaction for separating analytes and interferents is precipitation.Two important examples of using a precipitation reaction in a separationare the pH-dependent solubility of metal oxides and hydroxides, andthe solubility of metal sulfides.Table 7.7 Examples of Using a Chemical Reaction and a Distillation toSeparate an Inorganic Analyte From InterferentsAnalyte Treatment Isolated Species2– 2− +CO 3CO ( aq) + 2HO ( aq) → CO ( g) + 3HO()l33 22CO 2+ + −NH 4NH ( aq) + OH ( aq) → NH ( g) + HO()l4 3 2NH 32– 2− +SO 3SO ( aq) + 2HO ( aq) → SO ( g) + 3HO()l33 22SO 2S 2– 2-+S ( aq) + 2H O ( aq) → H S( g) + 2H O()l3 2 2H 2 S

324 Analytical Chemistry 2.0Separations based on the pH-dependent solubility of oxides and hydroxidesusually use strong acids, strong bases, or an NH 3 /NH 4 Cl buffer.Most metal oxides and hydroxides are soluble in hot concentrated HNO 3 ,although a few oxides, such as WO 3 , SiO 2 , and SnO 2 remain insoluble evenunder these harsh conditions. In determining the amount of Cu in brass,for example, we can avoid an interference from Sn by dissolving the samplewith a strong acid and filtering to remove the solid residue of SnO 2 .Most metals form a hydroxide precipitate in the presence of concentratedNaOH. Those metals forming amphoteric hydroxides, however, donot precipitate because they react to form higher-order hydroxo-complexes.For example, Zn 2+ and Al 3+ do not precipitate in concentrated NaOHbecause the form the soluble complexes Zn(OH) 3 – and Al(OH) 4 – . Thesolubility of Al 3+ in concentrated NaOH allows us to isolate aluminumfrom impure samples of bauxite, an ore of Al 2 O 3 . After crushing the ore,we place it in a solution of concentrated NaOH, dissolving the Al 2 O 3 andforming Al(OH) 4 – . Other oxides in the ore, such as Fe 2 O 3 and SiO 2 , remaininsoluble. After filtering, we recover the aluminum as a precipitate ofAl(OH) 3 by neutralizing some of the OH – with acid.The pH of an NH 3 /NH 4 Cl buffer (pK a = 9.26) is sufficient to precipitatemost metals as the hydroxide. The alkaline earths and alkaline metals,however, do not precipitate at this pH. In addition, metal ions that formsoluble complexes with NH 3 , such as Cu 2+ , Zn 2+ , Ni 2+ , and Co 2+ also donot precipitate under these conditions.The use of S 2– as a precipitating reagent is one of the earliest examplesof a separation technique. In Fresenius’s 1881 text A System of Instruction inQuantitative Chemical Analysis, sulfide is frequently used to separate metalions from the remainder of the sample’s matrix. 14 Sulfide is a useful reagentfor separating metal ions for two reasons: (1) most metal ions, except for thealkaline earths and alkaline metals, form insoluble sulfides; and (2) thesemetal sulfides show a substantial variation in solubility. Because the concentrationof S 2– is pH-dependent, we can control which metal precipitatesby adjusting the pH. For example, in Fresenius’s gravimetric procedure fordetermining Ni in ore samples (see Figure 1.1 in Chapter 1 for a schematicdiagram of this procedure), sulfide is used three times as a means of separatingCo 2+ and Ni 2+ from Cu 2+ and, to a lesser extent, from Pb 2+ .7F.5 Separations Based on a Partitioning Between PhasesThe most important group of separation techniques uses a selective partitioningof the analyte or interferent between two immiscible phases. If webring a phase containing a solute, S, into contact with a second phase, thesolute partitions itself between the two phases, as shown by the followingequilibrium reaction.14 Fresenius. C. R. A System of Instruction in Quantitative Chemical Analysis, John Wiley and Sons:New York, 1881.

Chapter 7 Collecting and Preparing Samples325Sphase1The equilibrium constant for reaction 7.20 S7.20phase2Sphase2K D= [ ][ S ]phase1is called the distribution constant or partition coefficient. If K D is sufficientlylarge, then the solute moves from phase 1 to phase 2. The soluteremains in phase 1 if the partition coefficient is sufficiently small. Whenwe bring a phase containing two solutes into contact with a second phase,if K D is favorable for only one of the solutes a separation of the solutes ispossible. The physical states of the phases are identified when describingthe separation process, with the phase containing the sample listed first. Forexample, if the sample is in a liquid phase and the second phase is a solid,then the separation involves liquid–solid partitioning.Ex t r a c t i o n Be t we e n Tw o Ph a s e sWe call the process of moving a species from one phase to another phasean extraction. Simple extractions are particularly useful for separationswhere only one component has a favorable partition coefficient. Several importantseparation techniques are based on a simple extraction, includingliquid–liquid, liquid–solid, solid–liquid, and gas–solid extractions.Liquid–Liquid Ex t r a c t i o n sA liquid–liquid extraction is usually accomplished with a separatory funnel(Figure 7.20). After placing the two liquids in the separatory funnel, weshake the funnel to increase the surface area between the phases. When the(a)(b)Phase 2Phase 1Figure 7.20 Example of a liquid–liquid extraction using a separatory funnel.(a) Before the extraction, 100% of the analyte is in phase 1. (b) After theextraction, most of the analyte is in phase 2, although some analyte remainsin phase 1. Although one liquid–liquid extraction can result in the completetransfer of analyte, a single extraction usually is not sufficient. See Section7G for a discussion of extraction efficiency and multiple extractions.

326 Analytical Chemistry 2.0μL drop ofextracting solventSyringe needleFigure 7.21 Schematic of a liquid–liquid microextraction showing asyringe needle with a mL drop ofthe extracting solvent.extraction is complete, we allow the liquids to separate. The stopcock at thebottom of the separatory funnel allows us to remove the two phases.We also can carry out a liquid–liquid extraction without a separatoryfunnel by adding the extracting solvent to the sample container. Pesticidesin water, for example, are preserved in the field by extracting them into asmall volume of hexane. Liquid–liquid microextractions, in which the extractingphase is a 1-mL drop suspended from a microsyringe (Figure 7.21),also have been described. 15 Because of its importance, a more thoroughdiscussion of liquid–liquid extraction is in section 7G.So l i d Ph a s e Ex t r a c t i o n sIn a solid phase extraction of a liquid sample, we pass the sample through acartridge containing a solid adsorbent, several examples of which are shownin Figure 7.22. The choice of adsorbent is determined by the species wewish to separate. Table 7.8 provides several representative examples of solidadsorbents and their applications.As an example, let’s examine a procedure for isolating the sedativessecobarbital and phenobarbital from serum samples using a C-18 solidadsorbent. 16 Before adding the sample, the solid phase cartridge is rinsedwith 6 mL each of methanol and water. Next, a 500-mL sample of serumis pulled through the cartridge, with the sedatives and matrix interferentsretained by a liquid–solid extraction (Figure 7.23a). Washing the cartridgewith distilled water removes any interferents (Figure 7.23b). Finally, weelute the sedatives using 500 mL of acetone (Figure 7.23c). In comparisonto a liquid–liquid extraction, a solid phase extraction has the advantage ofbeing easier, faster, and requiring less solvent.15 Jeannot, M. A.; Cantwell, F. F. Anal. Chem. 1997, 69, 235–239.16 Alltech Associates Extract-Clean SPE Sample Preparation Guide, Bulletin 83.Figure 7.22 Selection of solid phase extraction cartridges forliquid samples. The solid adsorbent is the white material ineach cartridge. Source: Jeff Dahl (commons.wikipedia.org).

Chapter 7 Collecting and Preparing Samples327Table 7.8 Representative Adsorbents for the Solid Phase Extraction of Liquid SamplesAdsorbent Structure Properties and Uses• retains low to moderate polarity species from organicsilica Si OHmatricies• fat soluble vitamins, steroidsaminopropylSiNH 2• retains polar compounds• carbohydrates, organic acids• retains wide variety of species from aqueous and organicCNcyanopropylSimatricies• pesticides, hydrophobic peptidesOH • retains wide variety of species from aqueous and organicdiolSiOH matricies• proteins, peptides, fungicidesoctadecyl (C-18) —C 18 H 37• retains hydrophobic species from aqueous matricies• caffeine, sedatives, polyaromatic hydrocarbons, carbohydrates,pesticidesoctyl (C-8) —C 8 H 17 • similar to C-18Co n t i n u o u s Ex t r a c t i o n sAn extraction is possible even if the analyte has an unfavorable partitioncoefficient, provided that the sample’s other components have significantlysmaller partition coefficients. Because the analyte’s partition coefficient isunfavorable, a single extraction will not recover all the analyte. Instead wecontinuously pass the extracting phase through the sample until a quantitativeextraction is achieved.Add the SampleinterferentanalyteWash the Sampleto Remove InterferentsElute the Analytes(a) (b) (c)Figure 7.23 Steps in a typical solid phase extraction. After preconditioning thesolid phase cartridge with solvent, (a) the sample is added to the cartridge, (b) thesample is washed to remove interferents, and (c) the analytes eluted.

328 Analytical Chemistry 2.0samplethimbleupperreservoirsolventwater-cooledcondenserreturntubelowerreservoirFigure 7.24 Soxhlet extractor. Seetext for details.A continuous extraction of a solid sample is carried out using a Soxhletextractor (Figure 7.24). The extracting solvent is placed in the lowerreservoir and heated to its boiling point. Solvent in the vapor phase movesupwards through the tube on the far right side of the apparatus, reachingthe condenser where it condenses back to the liquid state. The solvent thenpasses through the sample, which is held in a porous cellulose filter thimble,collecting in the upper reservoir. When the solvent in the upper reservoirreaches the return tube’s upper bend, the solvent and extracted analyte aresiphoned back to the lower reservoir. Over time the analyte’s concentrationin the lower reservoir increases.Microwave-assisted extractions have replaced Soxhlet extractions insome applications. 17 The process is the same as that described earlier for amicrowave digestion. After placing the sample and the solvent in a sealeddigestion vessel, a microwave oven is used to heat the mixture. Using asealed digestion vessel allows the extraction to take place at a higher temperatureand pressure, reducing the amount of time needed for a quantitativeextraction. In a Soxhlet extraction the temperature is limited by thesolvent’s boiling point at atmospheric pressure. When acetone is the solvent,for example, a Soxhlet extraction is limited to 56 o C, but a microwave extractioncan reach 150 o C.Two other continuous extractions deserve mention. Volatile organiccompounds (VOCs) can be quantitatively removed from liquid samplesby a liquid–gas extraction. As shown in Figure 7.25, an inert purging gas,such as He, is passed through the sample. The purge gas removes the VOCs,which are swept to a primary trap where they collect on a solid absorbent.A second trap provides a means for checking to see if the primary trap’scapacity is exceeded. When the extraction is complete, the VOCs are re-17 Renoe, B. W. Am. Lab August 1994, 34–40.primaryadsorbenttrapsamplesecondaryadsorbenttrapFigure 7.25 Schematic diagram of a purge-and-trap system for extractingvolatile analytes. The purge gas releases the analytes, which subsequentlycollect in the primary adsorbent trap. The secondary adsorptiontrap is monitored for evidence of the analyte’s breakthrough.purge gas

Chapter 7 Collecting and Preparing Samples329moved from the primary trap by rapidly heating the tube while flushingwith He. This technique is known as a purge-and-trap. Because the analyte’srecovery may not be reproducible, an internal standard is necessaryfor quantitative work.Continuous extractions also can be accomplished using supercriticalfluids. 18 If we heat a substance above its critical temperature and pressure,it forms a supercritical fluid whose properties are between those of agas and a liquid. A supercritical fluid is a better solvent than a gas, whichmakes it a better reagent for extractions. In addition, a supercritical fluid’sviscosity is significantly less than that of a liquid, making it easier to push itthrough a particulate sample. One example of a supercritical fluid extractionis the determination of total petroleum hydrocarbons (TPHs) in soils,sediments, and sludges using supercritical CO 2 . 19 An approximately 3-gsample is placed in a 10-mL stainless steel cartridge and supercritical CO 2at a pressure of 340 atm and a temperature of 80 o C is passed through thecartridge for 30 minutes at flow rate of 1–2 mL/min. To collect the TPHs,the effluent from the cartridge is passed through 3 mL of tetrachloroethyleneat room temperature. At this temperature the CO 2 reverts to the gasphase and is released to the atmosphere.Ch r o m a t o g r a p h i c Se p a r a t i o n sIn an extraction, the sample is one phase and we extract the analyte or theinterferent into a second phase. We also can separate the analyte and interferentsby continuously passing one sample-free phase, called the mobilephase, over a second sample-free phase that remains fixed or stationary. Thesample is injected into the mobile phase and the sample’s components partitionthemselves between the mobile phase and the stationary phase. Thosecomponents with larger partition coefficients are more likely to move intothe stationary phase, taking a longer time to pass through the system. Thisis the basis of all chromatographic separations. Chromatography providesboth a separation of analytes and interferents, and a means for performinga qualitative or quantitative analysis for the analyte. For this reason a morethorough treatment of chromatography is found in Chapter 12.7GLiquid–Liquid ExtractionsA liquid–liquid extraction is an important separation technique for environmental,clinical, and industrial laboratories. A standard environmentalanalytical method illustrates the importance of liquid–liquid extractions.Municipal water departments routinely monitor public water supplies fortrihalomethanes (CHCl 3 , CHBrCl 2 , CHBr 2 Cl, and CHBr 3 ) because theyare known or suspected carcinogens. Before their analysis by gas chroma-18 McNally, M. E. Anal. Chem. 1995, 67, 308A–315A.19 “TPH Extraction by SFE,” ISCO, Inc. Lincoln, NE, Revised Nov. 1992.The Environmental Protection Agency(EPA) also publishes two additional methodsfor trihalomethanes. Method 501.1and Method 501.3 use a purge-and-trapto collect the trihalomethanes prior toa gas chromatographic analysis with ahalide-specific detector (Method 501.1)or a mass spectrometer as the detector(Method 501.3). You will find more detailsabout gas chromatography, includingdetectors, in Chapter 12.

330 Analytical Chemistry 2.0tography, trihalomethanes are separated from their aqueous matrix by extractingwith pentane. 20In a simple liquid–liquid extraction the solute partitions between twoimmiscible phases. One phase usually is aqueous and the other phase is anorganic solvent, such as the pentane used in extracting trihalomethanesfrom water. Because the phases are immiscible they form two layers, withthe denser phase on the bottom. The solute is initially present in one of thetwo phases; after the extraction it is present in both phases. Extractionefficiency—that is, the percentage of solute moving from one phase tothe other—is determined by the equilibrium constant for the solute’s partitioningbetween the phases and any other reactions involving the solute.Examples of other reactions affecting extraction efficiency include acid–base reactions and complexation reactions.7G.1 Partition Coefficients and Distribution RatiosAs we learned earlier in this chapter, a solute’s partitioning between twophases is described by a partition coefficient, K D . If we extract a solute froman aqueous phase into an organic phasethen the partition coefficient isSaq SorgSorgK D= [ ][ S ]aqA large value for K D . indicates that the solute’s extraction into the organicphase is favorable.To evaluate an extraction’s efficiency we must consider the solute’s totalconcentration in each phase. We define the distribution ratio, D, as theratio of the solute’s total concentration in each phase.D = [ S ][ S ]org totalaq totalThe partition coefficient and the distribution ratio are identical if the solutehas only one chemical form in each phase. If the solute exists in more thanone chemical form in either phase, then K D and D usually have differentvalues. For example, if the solute exists in two forms in the aqueous phase,A and B, only one of which, A, partitions between the two phases, then[ S ]org AD=[ S ] + [ S ]aq A aq B[ S ]≤ K =D[ S ]org AaqA20 “The Analysis of Trihalomethanes in Drinking Water by Liquid Extraction,” EPA Method 501.2(EPA 500-Series, November 1979).

Chapter 7 Collecting and Preparing Samples331This distinction between K D and D is important. The partition coefficientis an equilibrium constant and has a fixed value for the solute’s partitioningbetween the two phases. The distribution ratio’s value, however,changes with solution conditions if the relative amounts of A and B change.If we know the equilibrium reactions taking place within each phase andbetween the phases, we can derive an algebraic relationship between K Dand D.7G.2 Liquid–Liquid Extraction With No Secondary ReactionsIn a simple liquid–liquid extraction, the only reaction affecting extractionefficiency is the solute’s partitioning between the two phases (Figure 7.26).In this case the distribution ratio and the partition coefficient are equal.[ S ] [ S ]org totalorgD= = K =D[ S ] [ S ]aq totalaq7.21Let’s assume that the solute is initially present in the aqueous phaseand that we are extracting it into the organic phase. A conservation of massrequires that the moles of solute initially present in the aqueous phase equalthe combined moles of solute in the aqueous phase and the organic phaseafter the extraction.( molesS ) = ( molesS ) + ( molesS ) 7.22aq 0 aq 1 org 1where the subscripts indicate the extraction number. After the extraction,the solute’s concentration in the aqueous phase is[ S ]aq 1and its concentration in the organic phase is[ S ]org 1( molesS )aq 1= 7.23Vaq( molesS )org 1= 7.24Vwhere V aq and V org are the volumes of the aqueous phase and the organicphase. Solving equation 7.22 for (moles S org ) 1 and substituting into equation7.24 leave us with[ S ]org 1org( molesS ) −( molesS )aqaq=V0 1Substituting equation 7.23 and equation 7.25 into equation 7.21 givesorg7.25K DS orgS aqorganic phaseaqueous phaseFigure 7.26 Scheme for a simpleliquid–liquid extraction in whichthe solute’s partitioning dependsonly on the K D equilibrium.The subscript 0 represents the systembefore the extraction and the subscript 1represents the system after the first extraction.

332 Analytical Chemistry 2.0D =( molesS ) −( molesS )aq0 aq 1Vorg( molesS )Vaqaq1( molesS ) × V − ( molesS ) × V0 1=( mole ) 1×Vaq aq aq aqsS aq orgRearranging, and solving for the fraction of solute remaining in the aqueousphase after one extraction, (q aq ) 1 , gives( q )aq1( molesS ) Vaq 1aq= =( molesS ) DV + Vaq 0 org a qThe fraction present in the organic phase after one extraction, (q org ) 1 , is( q )org1( molesS )DVorg 1aq= = 1− ( q ) =aq 1( molesS )DV + Vorg0orgaq7.26Example 7.14 shows how we can use equation 7.26 to calculate the efficiencyof a simple liquid-liquid extraction.Example 7.14A solute has a K D between water and chloroform of 5.00. Suppose weextract a 50.00-mL sample of a 0.050 M aqueous solution of the solutewith 15.00 mL of chloroform. (a) What is the separation’s extraction efficiency?(b) What volume of chloroform must we use to extract 99.9%of the solute?This is large volume of chloroform. Clearly,a single extraction is not reasonableunder these conditions.So l u t i o nFor a simple liquid–liquid extraction the distribution ratio, D, and thepartition coefficient, K D , are identical.(a) The fraction of solute remaining in the aqueous phase after the extractionis given by equation 7.26.( q )aq1Vaq=DV + Vorgaq50.00 mL== 0.400( 500 . )( 1500 . mL) + 50.00 mLThe fraction of solute in the organic phase is 1– 0.400, or 0.600.Extraction efficiency is the percentage of solute moving into the extractingphase. The extraction efficiency, therefore, is 60.0%.(b) To extract 99.9% of the solute (q aq ) 1 must be 0.001. Solving equation7.26 for V org , and making appropriate substitutions for (q aq ) 1 and V aqgivesVorgV −( q ) V1=( q ) Daq aq aqaq150. 00 mL −( 0. 001)(50. 00 mL)== 9990 mL( 0. 001)( 500 . )

Chapter 7 Collecting and Preparing Samples333In Example 7.14, a single extraction provides an extraction efficiencyof only 60%. If we carry out a second extraction, the fraction of soluteremaining in the aqueous phase, (q aq ) 2 , is( q )aq2( molesS ) Vaq 2aq= =( molesS ) DV + Vaq 1 org a qIf V aq and V org are the same for both extractions, then the cumulative fractionof solute remaining in the aqueous layer after two extractions, (Q aq ) 2 ,is2( molesS )⎛aq 2V ⎞aq( Q ) = = ( q ) × ( q )aq 2aq 1 aq( molesS )2=+aq 0⎝⎜DV Vorg aq ⎠⎟In general, for a series of n identical extractions, the fraction of analyteremaining in the aqueous phase after the last extraction is( Q )aqn⎛Vaq=⎝⎜DV + Vorgaq⎞n⎠⎟7.27Example 7.15For the extraction described in Example 7.14, determine (a) the extractionefficiency for two extractions and for three extractions; and (b) the numberof extractions required to ensure that we extract 99.9% of the solute.So l u t i o n(a) The fraction of solute remaining in the aqueous phase after two extractionsand three extractions isQ aq( )Q aq( )32⎛ 50.00 mL ⎞=⎝⎜( 500 . )( 1500 . mL) + 5000 . mL ⎠⎟ =⎛ 50.00 mL ⎞=⎝⎜( 500 . )( 1500 . mL) + 5000 . mL ⎠⎟ =320.1600.0640The extraction efficiencies are 84.0% with two extractions and 93.6%with three extractions.(b) To determine the minimum number of extractions for an efficiencyof 99.9%, we set (Q aq ) n to 0.001 and solve for n in equation 7.27.⎛ 50.000.001=mL ⎞⎜⎝( 500 . )( 1500 . mL) + 5000 . mL ⎠⎟ = ( 0. 400)Taking the log of both sides and solving for nnn

334 Analytical Chemistry 2.0Extraction Efficiency1008060402000 2 4 6 8 10Number of ExtractionsFigure 7.27 Plot of extractionefficiency versus the number ofextractions for the liquid–liquidextraction in Example 7.15.log( 0. 001) = nlog( 0. 400)n = 754 .we find that a minimum of 8 extractions is necessary.The last two examples provide us with an important observation—forany extraction efficiency, we need less solvent if we complete several extractionsusing smaller portions of solvent instead of one extraction using alarger volume of solvent. For the conditions in Example 7.14 and Example7.15, an extraction efficiency of 99.9% requires one extraction with 9990mL of chloroform, or 120 mL when using eight 15-mL portions of chloroform.Although extraction efficiency increases dramatically with the firstfew multiple extractions, the effect quickly diminishes as we increase thenumber of extractions (Figure 7.27). In most cases there is little improvementin extraction efficiency after five or six extractions. For the conditionsin Example 7.15, we reach an extraction efficiency of 99% after fiveextractions and need three additional extractions to obtain the extra 0.9%increase in extraction efficiency.Practice Exercise 7.8To plan a liquid–liquid extraction we need to know the solute’s distributionratio between the two phases. One approach is to carry out theextraction on a solution containing a known amount of solute. Afterextracting the solution, we isolate the organic phase and allow it to evaporate,leaving behind the solute. In one such experiment, 1.235 g of asolute with a molar mass of 117.3 g/mol is dissolved in 10.00 mL ofwater. After extracting with 5.00 mL of toluene, 0.889 g of the solute isrecovered in the organic phase. (a) What is the solute’s distribution ratiobetween water and toluene? (b) If we extract 20.00 mL of an aqueoussolution containing the solute with 10.00 mL of toluene, what is theextraction efficiency? (c) How many extractions will we need to recover99.9% of the solute?Click here to review your answer to this exercise.7G.3 Liquid–Liquid Extractions Involving Acid–Base EquilibriaAs shown by equation 7.21, in a simple liquid–liquid extraction the distributionratio and the partition coefficient are identical. As a result, the distributionratio does not depend on the composition of the aqueous phaseor the organic phase. Changing the pH of the aqueous phase, for example,will not affect the solute’s extraction efficiency.If the solute participates in an additional equilibrium reaction withina phase, then the distribution ratio and the partition coefficient may notbe the same. For example, Figure 7.28 shows the equilibrium reactions affectingthe extraction of the weak acid, HA, by an organic phase in which

Chapter 7 Collecting and Preparing Samples335organic phaseHA orgK D+ H 2 O H 3 O + + A –HA aqK aFigure 7.28 Scheme for a liquid–liquid extraction of the weak acid, HA. Although the weak acid issoluble in both phases, its conjugate weak base, A – , is soluble only in the aqueous phase. The K a reaction,which is called a secondary equilibrium reaction, affects the extraction efficiency because itcontrols the relative abundance of HA in solution.ionic species are not soluble. In this case the partition coefficient and thedistribution ratio areHAorgK D= [ ][ HA ]aq[ HA ] [ HA ]org totalorgD = =[ HA ] [ HA ] + [ Aaq totalaq− ]aq7.287.29Because the position of an acid–base equilibrium depends on the pH, thedistribution ratio is pH-dependent. To derive an equation for D that showsthis dependency, we begin with the acid dissociation constant for HA.[ HO ][ A ]+ −3 aq aqK a=[ HA ]Solving equation 7.30 for the concentration of A –aq[ HA ]− a aq[ A ] = K ×aq+[ HO ]and substituting into equation 7.29 givesD =3 aq[ HA ]orgK+ × [ HA ]a aq[ HA ]aq+[ HO ]aqueous phase3 aq7.30Factoring [HA aq ] from the denominator, replacing [HA org ]/[HA aq ] withK D (equation 7.28), and simplifying leaves us with a relationship betweenthe distribution ratio, D, and the pH of the aqueous solution.

336 Analytical Chemistry 2.0+K [ HO ]D 3 aqD =+[ HO ] + K3 aq a7.31Example 7.16An acidic solute, HA, has a K a of 1.00 × 10 –5 and a K D between water andhexane of 3.00. Calculate the extraction efficiency if we extract a 50.00 mLsample of a 0.025 M aqueous solution of HA, buffered to a pH of 3.00,with 50.00 mL of hexane. Repeat for pH levels of 5.00 and 7.00.Extraction Efficiency80604020HA A –02 4 6 8 10 12pHFigure 7.29 Plot of extractionefficiency versus pH of the aqueousphase for the extraction of theweak acid HA in Example 7.16. Aladder diagram for HA is superimposedalong the x-axis, dividingthe pH scale into regions whereHA and A – are the predominateaqueous phase species. The greatestchange in extraction efficiencyoccurs as the pH moves throughHA’s buffer region.So l u t i o n+When the pH is 3.00, [ HO ] is 1.0 × 10 –3 and the distribution ratio is3 aq−3( 300 . )( 1. 0×10 )D =−10 . × 10 + 10 . × 103 −5= 297 .The fraction of solute remaining in the aqueous phase isQ aq( )150.00 mL== 0.252( 297 . )( 5000 . mL) + 50.00 mLThe extraction efficiency, therefore, is almost 75%. The same calculation ata pH of 5.00 gives the extraction efficiency as 60%. At a pH of 7.00 theextraction efficiency is only 3% .The extraction efficiency in Example 7.16 is greater at more acidic pHlevels because HA is the solute’s predominate form in the aqueous phase. Ata more basic pH, where A – is the solute’s predominate form, the extractionefficiency is smaller. A graph of extraction efficiency versus pH is shownin Figure 7.29. Note that extraction efficiency is essentially independent ofpH for pHs more acidic than the weak acid’s pK a , and that it is essentiallyzero for pHs more basic than the pK a . The greatest change in extraction efficiencyoccurs at pHs where both HA and A – are predominate species. Theladder diagram for HA along the graph’s x-axis helps illustrate this effect.Practice Exercise 7.9The liquid–liquid extraction of the weak base B is governed by the followingequilibrium reactions:B( aq) B( org )−B( aq) + H O() l OH ( aq ) + HB2K D= 500 .( aq) K b= 10 . × 10 4+ −Derive an equation for the distribution ratio, D, and calculate the extractionefficiency if 25.0 mL of a 0.025 M solution of B, buffered to a pHof 9.00, is extracted with 50.0 mL of the organic solvent.Click here to review your answer to this exercise.

Chapter 7 Collecting and Preparing Samples337organic phaseHL orgK D,HLHL aq+H 2 OK an L –+H 3 O +(ML n ) orgK D,cβ n+ M n+(ML n ) aqaqueous phaseFigure 7.30 Scheme for the liquid–liquid extractionof a metal ion, M n+ , by the ligand L – . Theligand initially is present in the organic phase asHL. Four equilibrium reactions are needed to explainthe extraction efficiency.7G.4 Liquid–Liquid Extraction of a Metal–Ligand ComplexOne important application of liquid–liquid extractions is the selective extractionof metal ions using a ligand. Unfortunately, many ligands are notvery soluble in water or undergo hydrolysis or oxidation in aqueous solutions.For these reasons the ligand is added to the organic solvent insteadof the aqueous phase. Figure 7.30 shows the relevant equilibria (and equilibriumconstants) for the extraction of M n+ by the ligand HL, includingthe ligand’s extraction into the aqueous phase (K D,HL ), the ligand’s aciddissociation reaction (K a ), the formation of the metal–ligand complex (b n ),and the complex’s extraction into the organic phase (K D,c ).If the ligand’s concentration is much greater than the metal ion’s concentration,then the distribution ratio isn nβ K ( K ) ( C )n D,c a HLD =n + nn( K ) [ HO ] + β ( K ) (C ) 7.32nD,HL 3 n a HLwhere C HL is the ligand’s initial concentration in the organic phase. Asshown in Example 7.17, the extraction efficiency for metal ions shows amarked pH dependency.Example 7.17A liquid–liquid extraction of the divalent metal ion, M 2+ , uses the schemeoutlined in Figure 7.30. The partition coefficients for the ligand, K D,HL ,and for the metal–ligand complex, K D,c , are 1.0 × 10 4 and 7.0 × 10 4 , respectively.The ligand’s acid dissociation constant, K a , is 5.0 × 10 –5 , andthe formation constant for the metal–ligand complex, b 2 , is 2.5 × 10 16 .Problem 32 in the end-of-chapter problemsasks you to derive equation 7.32.

338 Analytical Chemistry 2.0Extraction Efficiency1008060402001 2 3 4 5 6 7pHFigure 7.31 Plot of extraction efficiencyversus pH for the extractionof the metal ion, M 2+ , in Example7.17.SHNNNHNdithizoneCalculate the extraction efficiency if we extract 100.0 mL of a 1.0 × 10 –6M aqueous solution of M 2+ , buffered to a pH of 1.00, with 10.00 mLof an organic solvent that is 0.1 mM in the chelating agent. Repeat thecalculation at a pH of 3.00.So l u t i o nWhen the pH is 1.00 the distribution ratio is6 4)−5 2 −4 2( 25 . × 10 )( 70 . × 10 )( 50 . × 10 )(. 10×10 )D =4 2 2 6 5 2(. 10× 10 )( 010 . ) + ( 2. 5× 10 )( 50 . × 10 − ) (. 1 0×10D = 0.0438and the fraction of metal ion remaining in the aqueous phase isQ aq( )1100.0 mL=( 0. 0438)( 10. 00 mL) + 100.0 mL= 0.996)−4 2At a pH of 1.00, we extract only 0.40% of the metal into the organic phase.Changing the pH to 3.00, however, increases the extraction efficiency to97.8%. Figure 7.31 shows how the pH of the aqueous phase affects theextraction efficiency for M 2+ .One advantage of using a ligand to extract a metal ion is the high degreeof selectivity that it brings to a liquid–liquid extraction. As seen in Figure7.31, a divalent metal ion’s extraction efficiency increases from approximately0% to 100% over a range of 2 pH units. Because a ligand’s ability toform a metal–ligand complex varies substantially from metal ion to metalion, significant selectivity is possible by carefully controlling pH. Table 7.9shows the minimum pH for extracting 99% of a metal ion from an aqueoussolution using an equal volume of 4 mM dithizone in CCl 4 .Example 7.18Using Table 7.9, explain how you can separate the metal ions in an aqueousmixture of Cu 2+ , Cd 2+ and Ni 2+ by extracting with an equal volumeof dithizone in CCl 4 .So l u t i o nFrom Table 7.9, a quantitative separation of Cu 2+ from Cd 2+ and Ni 2+ ispossible if we acidify the aqueous phase to a pH of less than 1. This pHis greater than the minimum pH for extracting Cu 2+ and significantlysmaller than the minimum pH for extracting either Cd 2+ or Ni 2+ Afterthe extraction is complete, buffering the aqueous phase to 4.0 allows us toextract the Cd 2+ leaving the Ni 2+ in the aqueous phase.

Chapter 7 Collecting and Preparing Samples339Table 7.9 Minimum pH for Extracting 99%of an Aqueous Metal Ion Using4.0 mM Dithizone in CCl 4 (V aq = V org )Metal IonMinimum pHHg 2+ -8.7Ag + -1.7Cu 2+ -0.8Bi 3+ 0.9Zn 2+ 2.3Cd 2+ 3.6Co 2+ 3.6Pb 2+ 4.1Ni 2+ 6.0Tl + 8.7Source: Kolthoff, I. M.; Sandell, E. B.; Meehan, E. J.; Bruckenstein, S.Quantitative Chemical Analysis, Macmillan: New York, 1969, p. 353.7HSeparation Versus PreconcentrationTwo common analytical problems are: (1) matrix components that interferewith an analyte’s analysis; and (2) an analyte with a concentration that is toosmall to analyze accurately. We have seen that we can use a separation tosolve the first problem. Interestingly, we often can use a separation to solvethe second problem as well. For a separation in which we recover the analytein a new phase, it may be possible to increase the analyte’s concentration.This step in an analytical procedure is known as a preconcentration.An example from the analysis of water samples illustrate how we cansimultaneously accomplish a separation and a preconcentration. In the gaschromatographic analysis for organophosphorous pesticides in environmentalwaters, the analytes in a 1000-mL sample may be separated fromtheir aqueous matrix by a solid-phase extraction using 15 mL of ethyl acetate.21 After the extraction, the analytes in the ethyl acetate have a concentrationthat is 67 times greater than that in the original sample (assumingthe extraction is 100% efficient).7IKey Termscentrifugation composite sample coning and quarteringconvenience samplingdensity gradientcentrifugationdialysisdistillation distribution ratio extractionextraction efficiency filtrate filtration1000 mL≈ 67 ×15 mLAs you review this chapter, try to define akey term in your own words. Check youranswer by clicking on the key term, whichwill take you to the page where it was firstintroduced. Clicking on the key termthere, will bring you back to this page sothat you can continue with another keyterm.21 Aguilar, C.; Borrul, F.; Marcé, R. M. LC•GC 1996, 14, 1048 –1054.

340 Analytical Chemistry 2.0grab sample gross sample heterogeneoushomogeneous in situ sampling judgmental samplinglaboratory sample masking masking agentsNyquist theorem partition coefficient preconcentrationpurge-and-trap random sampling recoveryrecrystallization retentate sampling plansecondary equilibriumreactionsize exclusionchromatographyselectivity coefficientSoxhlet extractorseparation factorstratified samplingsublimation subsamples supercritical fluidsystematic–judgmentalsamplingsystematic samplingtarget population7J Chapter SummaryAn analysis requires a sample. How we acquire that sample is critical. Thesamples we collect must accurately represent their target population, andour sampling plan must provide a sufficient number of samples of appropriatesize so that the uncertainty in sampling does not limit the precisionof our analysis.A complete sampling plan requires several considerations, including:the type of sample (random, judgmental, systematic, systematic–judgmental,stratified, or convenience); whether to collect grab samples, compositesamples, or in situ samples; whether the population is homogeneous orheterogeneous; the appropriate size for each sample; and, the number ofsamples to collect.Removing a sample from its population may induce a change in itscomposition due to a chemical or physical process. For this reason, we collectsamples in inert containers and we often preserve them at the time ofcollection.When an analytical method’s selectivity is insufficient, we may needto separate the analyte from potential interferents. Such separations cantake advantage of physical properties—such as size, mass or density—orchemical properties. Important examples of chemical separations includemasking, distillation, and extractions.7KProblemsAnswers, but not worked solutions, tomost end-of-chapter problems are availablehere.1. Because of the risk of lead poisoning, the exposure of children to leadbasedpaint is a significant public health concern. The first step in thequantitative analysis of lead in dried paint chips is to dissolve the sample.Corl has evaluated several dissolution techniques. 22 In this study,samples of paint were collected and pulverized with a Pyrex mortar andpestle. Replicate portions of the powdered paint were then taken for22 Corl, W. E. Spectroscopy 1991, 6(8), 40–43.

Chapter 7 Collecting and Preparing Samples341analysis. The following table shows results for a paint sample and for astandard reference material. Both samples and standards were digestedwith HNO 3 on a hot plate.Replicate% w/w Pbin Sample% w/w Pbin Standard1 5.09 11.482 6.29 11.623 6.64 11.474 4.63 11.86(a) Determine the overall variance, the variance due to the method andthe variance due to sampling. (b) What percentage of the overall varianceis due to sampling? (c) How might you decrease the variance dueto sampling?2. To analyze a shipment of 100 barrels of an organic solvent, you plan tocollect and analyze single samples from ten barrels using a random sampling.From which barrels should you collect samples if the first barrelis given by the 12th entry in the random number table in Appendix 14,with subsequent barrels given by every third entry? Assume that entriesin the random number table are arranged by rows.3. The concentration of dissolved O 2 in a lake shows a daily cycle fromthe effect of photosynthesis, and a yearly cycle due to seasonal changesin temperature. Suggest an appropriate systematic sampling plan formonitoring the daily change in dissolved O 2 . Suggest an appropriatesystematic sampling plan for monitoring the yearly change in dissolvedO 2 .4. The data in the following table were collected during a preliminarystudy of the pH of an industrial wastewater stream.Time (h) pH Time (h) pH0.5 4.4 9.0 5.71.0 4.8 9.5 5.51.5 5.2 10.0 6.52.0 5.2 10.5 6.02.5 5.6 11.0 5.83.0 5.4 11.5 6.03.5 5.4 12.0 5.64.0 4.4 12.5 5.64.5 4.8 13.0 5.45.0 4.8 13.5 4.9

342 Analytical Chemistry 2.0Time (h) pH Time (h) pH5.5 4.2 14.0 5.26.0 4.2 14.5 4.46.5 3.8 15.0 4.07.0 4.0 15.5 4.57.5 4.0 16.0 4.08.0 3.9 16.5 5.08.5 4.7 17.0 5.0Construct a graph of pH as a function of time and suggest an appropriatesampling frequency for a long-term monitoring program.5. You have been asked to monitor the daily fluctuations in atmosphericozone in the downtown area of a city to determine if there is relationshipbetween daily traffic patterns and ozone levels. (a) Which of thefollowing sampling plan will you use: random, systematic, judgmental,systematic–judgmental, or stratified? (b) Do you plan to collect andanalyze a series of grab samples, or will you form a single compositesample? (c) Will your answers to these questions change if your goal isto determine if the average daily ozone level exceeds a threshold value?If your answer is yes, then what is your new sampling strategy?6. The distinction between a homogeneous population and a heterogeneouspopulation is important when you are developing a samplingplan. (a) Define homogeneous and heterogeneous. (b) If you collectand analyze a single sample, can you determine if the population ishomogeneous or heterogeneous?7. The sampling of a heterogeneous population is particularly challenging.Explain why you can minimize the sampling variance for a heterogeneouspopulation by increasing the number of samples, but not byincreasing the mass of individual samples.8. Beginning with equation 7.4, derive equation 7.5. Assume that theparticles are spherical with a radius of r and a density of d.9. The sampling constant for the radioisotope 24 Na in homogenized humanliver has been reported as approximately 35 g. 23 (a) What is theexpected relative standard deviation for sampling if you analyze 1.0-gsamples? (b) How many 1.0-g samples must you analyze to obtain amaximum sampling error of ±5% at the 95% confidence level?23 Kratochvil, B.; Taylor, J. K. Anal. Chem. 1981, 53, 924A–938A.

Chapter 7 Collecting and Preparing Samples34310. Engels and Ingamells reported the following results for the % w/w K 2 Oin a mixture of amphibolite and orthoclase. 240.247 0.300 0.236 0.258 0.304 0.3300.247 0.275 0.212 0.311 0.258 0.187Each of the 12 samples had a nominal weight of 0.1 g. Determine theapproximate value for K s and the mass of sample needed to achieve apercent relative standard deviation of 2%.11. The following data has been reported for the determination of KH 2 PO 4in a mixture of KH 2 PO 4 and NaCl. 25Nominal Mass (g) Actual Mass (g) % w/w KH 2 PO 40.10 0.1039 0.0850.1015 1.0780.1012 0.4130.1010 1.2480.1060 0.6540.0997 0.5070.25 0.2515 0.8470.2465 0.5980.2770 0.4310.2460 0.8420.2485 0.9640.2590 1.1780.50 0.5084 1.0090.4954 0.9470.5286 0.6180.5232 0.7440.4965 0.5720.4995 0.7091.00 1.027 0.6960.987 0.8430.991 0.5350.998 0.7500.997 0.7111.001 0.63924 Engels, J. C.; Ingamells, C. O. Geochim. Cosmochim. Acta 1970, 34, 1007–1017.25 Guy, R. D.; Ramaley, L.; Wentzell, P. D. J. Chem. Educ. 1998, 75, 1028–1033.

344 Analytical Chemistry 2.0Nominal Mass (g) Actual Mass (g) % w/w KH 2 PO 42.50 2.496 0.7662.504 0.7692.496 0.6822.496 0.6092.557 0.5892.509 0.617(a) Prepare a graph of % w/w KH 2 PO 4 versus the actual sample massand discuss how this graph is consistent with your understanding of thefactors affecting sampling variance. (b) For each nominal mass, calculatethe percent relative standard deviation for the analysis. The value ofK s for this analysis has been estimated as 350. For each nominal mass,use this K s to determine the percent relative standard deviation dueto sampling. Considering these calculations, what is your conclusionabout the importance of indeterminate sampling errors for this analysis?(c) For each nominal mass, convert the percent relative standarddeviation to an absolute standard deviation. Plot points on your graphcorresponding to ±1 absolute standard deviations about the overallaverage % w/w KH 2 PO 4 . Draw smooth curves through these two setsof points. Considering these curves, does the sample appear to be homogeneouson the scale at which it is sampled?12. In this problem you will collect and analyze data in a simulation of thesampling process. Obtain a pack of M&M’s (or other similar candy).Collect a sample of five candies and count the number that are red (orany other color of your choice). Report the result of your analysis as% red. Return the candies to the bag, mix thoroughly, and repeat theanalysis for a total of 20 determinations. Calculate the mean and thestandard deviation for your data. Remove all candies from the bag anddetermine the true % red for the population. Sampling in this exerciseshould follow binomial statistics. Calculate the expected mean valueand the expected standard deviation, and compare to your experimentalresults.13. Determine the error (a = 0.05) for the following situations. In each caseassume that the variance for a single determination is 0.0025 and thatthe variance for collecting a single sample is 0.050. (a) Nine samples arecollected, each analyzed once. (b) One sample is collected and analyzednine times. (c) Three samples are collected, each analyzed three times.14. Which of the sampling schemes in problem 13 is best if you wish tolimit the overall error to less than ±0.25, and the cost of collecting asingle sample is 1 (arbitrary units), and the cost of analyzing a single

Chapter 7 Collecting and Preparing Samples345sample is 10? Which is the best sampling scheme if the cost of collectinga single sample is 7, and the cost of analyzing a single sample is 3?15. Maw, Witry, and Emond evaluated a microwave digestion method forHg against the standard open-vessel digestion method. 26 The standardmethod requires a 2-h digestion and is operator-intensive. The microwavedigestion is complete in approximately 0.5 h and requires littlemonitoring by the operator. Samples of baghouse dust from air-pollution-controlequipment were collected from a hazardous waste incineratorand digested in triplicate before determining the concentrationof Hg in ppm. Results are summarized in the following tables.ppm Hg Following Microwave DigestionSample Replicate 1 Replicate 2 Replicate 31 7.12 7.66 7.172 16.1 15.7 15.63 4.89 4.62 4.284 9.64 9.03 8.445 6.76 7.22 7.506 6.19 6.61 7.617 9.44 9.56 10.78 30.8 29.0 26.2ppm Hg Following Standard DigestionSample Replicate 1 Replicate 2 Replicate 31 5.50 5.54 5.402 13.1 12.8 13.03 5.39 5.12 5.364 6.59 6.52 7.205 6.20 6.03 5.776 6.25 5.65 5.617 15.0 13.9 14.08 20.4 16.1 20.0Does the microwave digestion method yields acceptable results in comparisonto the standard digestion method?16. Simpson, Apte, and Batley investigated methods for preserving watersamples collected from anoxic (O 2 -poor) environments that containhigh concentrations of dissolved sulfide. 27 They found that preservingwater samples with HNO 3 (a common method for preserving aerobicsamples) gave significant negative determinate errors when analyzing26 Maw, R.; Witry, L.; Emond, T. Spectroscopy 1994, 9, 39–41.27 Simpson, S. L.: Apte, S. C.; Batley, G. E. Anal. Chem. 1998, 70, 4202–4205.

346 Analytical Chemistry 2.0for Cu 2+ . Preserving samples by first adding H 2 O 2 and then addingHNO 3 , eliminated the determinate error. Explain their observations.17. In a particular analysis the selectivity coefficient, K A,I , is 0.816. Whena standard sample with an analyte-to-interferent ratio of 5:1 is carriedthrough the analysis, the error in determining the analyte is +6.3%.(a) Determine the apparent recovery for the analyte if R I = 0. (b) Determinethe apparent recovery for the interferent if R A = 0.18. The amount of Co in an ore is to be determined using a procedure forwhich Fe in an interferent. To evaluate the procedure’s accuracy, a standardsample of ore known to have a Co/Fe ratio of 10.2:1 is analyzed.When pure samples of Co and Fe are taken through the procedure thefollowing calibration relationships are obtainedSSCoFe= 0.786×m= 0.669×mwhere S is the signal and m is the mass of Co or Fe. When 278.3 mg ofCo are taken through the separation step, 275.9 mg are recovered. Only3.6 mg of Fe are recovered when a 184.9 mg sample of Fe is carriedthrough the separation step. Calculate (a) the recoveries for Co and Fe;(b) the separation factor; (c) the error if no attempt is made to separatethe Co and Fe before the analysis; (d) the error if the separation stepis carried out; and (e) the maximum recovery for Fe if the recovery forCo is 1.00, and the maximum allowed error is 0.05%.19. The amount of calcium in a sample of urine was determined by a methodfor which magnesium is an interferent. The selectivity coefficient,K Ca,Mg , for the method is 0.843. When a sample with a Mg/Ca ratioof 0.50 was carried through the procedure, an error of –3.7% was obtained.The error was +5.5% when using a sample with a Mg/Ca ratioof 2.0. (a) Determine the recoveries for Ca and Mg. (b) What is theexpected error for a urine sample in which the Mg/Ca ratio is 10.0?20. Using the formation constants in Appendix 12, show that F – is aneffective masking agent for preventing a reaction between Al 3+ andEDTA. Assume that the only significant forms of fluoride and EDTAare F – and Y 4- .21. Cyanide is frequently used as a masking agent for metal ions. Its effectivenessas a masking agent is better in more basic solutions. Explainthe reason for this pH dependency.CoFe

Chapter 7 Collecting and Preparing Samples34722. Explain how an aqueous sample consisting of Cu 2+ , Sn 4+ , Pb 2+ , andZn 2+ can be separated into its component parts by adjusting the pH ofthe solution.23. A solute, S, has a distribution ratio between water and ether of 7.5.Calculate the extraction efficiency if you extract a 50.0-mL aqueoussample of S using 50.0 mL of ether as: (a) a single portion of 50.0 mL;(b) two portions, each of 25.0 mL; (c) four portions, each of 12.5 mL;and (d) five portions, each of 10.0 mL. Assume that the solute is notinvolved in any secondary equilibria.24. What volume of ether is needed to extract 99.9% of the solute in problem23 when using: (a) 1 extraction; (b) 2 extractions; (c) four extractions;and (d) five extractions.25. What is the minimum distribution ratio if 99% of the solute in a 50.0-mL sample is to be extracted with a single 50.0-mL portion of an organicsolvent? Repeat for the case where two 25.0-mL portions of theorganic solvent are used.26. A weak acid, HA, with a K a of 1.0 × 10 –5 has a partition coefficient, K D ,of 1.2 × 10 3 between water and an organic solvent. What restrictionon the sample’s pH is necessary to ensure that 99.9% of the weak acidin a 50.0-mL sample is extracted with a single 50.0-mL portion of theorganic solvent?27. For problem 26, how many extractions are needed if the sample’s pHcannot be decreased below 7.0?28. A weak base, B, with a K b of 1.0 × 10 –3 has a partition coefficient, K D ,of 5.0 × 10 2 between water and an organic solvent. What restrictionon the sample’s pH is necessary to ensure that 99.9% of the weak baseis in a 50.0-mL sample is extracted with two 25.0-mL portions of theorganic solvent?29. A sample contains a weak acid analyte, HA, and a weak acid interferent,HB. The acid dissociation constants and the partition coefficients forthe weak acids are as follows: K a,HA = 1.0 × 10 –3 , K a,HB = 1.0 × 10 –7 ,K D,HA = K D,HB = 5.0 × 10 2 . (a) Calculate the extraction efficiency forHA and HB when a 50.0-mL sample, buffered to a pH of 7.0, is extractedwith 50.0 mL of the organic solvent. (b) Which phase is enrichedin the analyte? (c) What are the recoveries for the analyte andthe interferent in this phase? (d) What is the separation factor? (e) Aquantitative analysis is conducted on the phase enriched in analyte.What is the expected relative error if the selectivity coefficient, K HA,HB ,is 0.500 and the initial ratio of HB/HA is 10.0?

348 Analytical Chemistry 2.0organic phase(I 2 ) orgK D(I 2 ) aq + I – K fI – 3Figure 7.32 Extraction scheme forProblem 7.30.(ML 2 ) orgaqueous phaseorganic phaseK Dβ 2(ML 2 ) aq M 2+ + 2L –aqueous phaseFigure 7.33 Extraction scheme forProblem 7.31.30. The relevant equilibria for the extraction of I 2 from an aqueous solutionof KI into an organic phase are shown in Figure 7.32. (a) Is the extractionefficiency for I 2 better at higher or at lower concentrations of I – ?(b) Derive an expression for the distribution ratio for this extraction.31. The relevant equilibria for extracting the metal-ligand complex ML 2from an aqueous solution into an organic phase are shown in Figure7.31. (a) Derive an expression for the distribution ratio for this extraction.(b) Calculate the extraction efficiency when a 50.0-mL aqueoussample that is 0.15 mM in M 2+ and 0.12 M in L – is extracted with 25.0mL of the organic phase. Assume that K D is 10.3 and that b 2 is 560.32. Derive equation 7.32.33. The following information is available for the extraction of Cu 2+ byCCl 4 and dithizone: K D,c = 7× 10 4 ; b 2 = 5 × 10 22 ; K a,HL = 2 × 10 –5 ;K D,HL = 1.01 × 10 4 ; and n = 2. What is the extraction efficiency if a100-mL sample of an aqueous solution that is 1.0 × 10 –7 M Cu 2+ and1 M in HCl is extracted with 10 mL of CCl 4 containing 4.0 × 10 –4 Mdithizone (HL)?34. Cupferron is a ligand whose strong affinity for metal ions makes it usefulas a chelating agent in liquid–liquid extractions. The following tableprovides pH-dependent distribution ratios for the extraction of Hg 2+ ,Pb 2+ , and Zn 2+ from an aqueous solution to an organic solvent.Distribution Ratio forpH Hg 2+ Pb 2+ Zn 2+1 3.3 0.0 0.02 10.0 0.43 0.03 32.3 999 0.04 32.3 9999 0.05 19.0 9999 0.186 4.0 9999 0.337 1.0 9999 0.828 0.54 9999 1.509 0.15 9999 2.5710 0.05 9999 2.57(a) Suppose you have a 50.0-mL sample of an aqueous solution containingHg 2+ , Pb 2+ , and Zn 2+ . Describe how you can effect a separationof these metal ions. (b) Under the conditions of your extractionfor Hg 2+ , what percent of the Hg 2+ remains in the aqueous phase afterthree 50.0-mL extractions with the organic solvent? (c) Under the conditionsof your extraction for Pb 2+ , what is the minimum volume of

Chapter 7 Collecting and Preparing Samples3497Lorganic solvent needed to extract 99.5% of the Pb 2+ in a single extraction?(d) Under the conditions of your extraction for Zn 2+ , how manyextractions are needed to remove 99.5% of the Zn 2+ if each extractionuses 25.0 mL of organic solvent?Solutions to Practice ExercisesPractice Exercise 7.1To reduce the overall variance by improving the method’s standard deviationrequires thats 2 2500 s 2 s 22= . ppm = + = ( 21 . ppm) + ssampmeth2methSolving for s meth gives its value as 0.768 ppm. Relative to its original valueof 1.1 ppm, this is reduction of 3.0 × 10 1 %. To reduce the overall varianceby improving the standard deviation for sampling requires that2 2 2 2 2s = 500 . ppm = s + s = s + ( 1.1 ppm)samp meth sampSolving for s samp gives its value as 1.95 ppm. Relative to its original valueof 2.1 ppm, this is reduction of 7.1%.Click here to return to the chapter.Practice Exercise 7.2The analytical method’s standard deviation is 1.96 × 10 –3 g/cm 3 as this isthe standard deviation for the analysis of a single sample of the polymer.The sampling variance is2 2 2 2 2 3 2s = s − s = 365× 10 −−( . ) − ( 1. 96× 10 ) = 133 . × 10 −3sampmethConverting the variance to a standard deviation gives s meth as 3.64 × 10 –2g/cm 3 .Click here to return to the chapter.Practice Exercise 7.3To determine the sampling constant, K s , we need to know the averagemass of the samples and the percent relative standard deviation for theconcentration of olaquindox in the feed. The average mass for the fivesamples is 0.95792 g. The average concentration of olaquindox in thesamples is 23.14 mg/kg with a standard deviation of 2.200 mg/kg. Thepercent relative standard deviation, R, isssamp2.200 mg/kgR = × 100 = × 100 = 95 .07 ≈ 951 .X 23.14 mg/kgSolving for K s gives its value asKs22= mR = ( 0. 95792 g)(9.507) = 86.58 g ≈86.6 g2

350 Analytical Chemistry 2.0To obtain a percent relative standard deviation of 5.0%, individual samplesneed to have a mass of at leastm = K sR= 86.58 g= 35 . g22(5.0)To reduce the sample’s mass from 3.5 g to 1 g, we must change the massby a factor of35 . g35 .1 g = ×If we assume that the sample’s particles are spherical, then we must reducea particle’s radius by a factor ofClick here to return to the chapter.r3= 35 . ×r = 15 . ×Practice Exercise 7.4Because the value of t depends on the number of samples—a result wehave yet to calculate—we begin by letting n samp = ∞ and using t(0.05,∞) for the value of t. From Appendix 4, the value for t(0.05, ∞) is 1.960.Our first estimate for n samp isnsamp2 2ts2 2samp (. 1 960)( 50 . )= = = 15. 4 ≈1522e ( 25 . )Letting n samp = 15, the value of t(0.05, 14) from Appendix 4 is 2.145.Recalculating n samp givesnsamp2 2ts2 2samp ( 2. 145)( 50 . )= = = 18. 4 ≈1822e ( 25 . )Letting n samp = 18, the value of t(0.05, 17) from Appendix 4 is 2.103.Recalculating n samp givesnsamp2 2ts2 2samp ( 2. 103)( 50 . )= = = 17. 7 ≈1822e ( 25 . )Because two successive calculations give the same value for n samp , we need18 samples to achieve a sampling error of ±2.5% at the 95% confidenceinterval.Click here to return to the chapter.

Chapter 7 Collecting and Preparing Samples351Practice Exercise 7.5If we collect a single sample (cost $20), then we can analyze that sample13 times (cost $650) and stay within our budget. For this scenario, thepercent relative error is2 2s se = samptn+ methn n= 2 179 010 .1+ 0.20..1× 13= 074sampsamprepwhere t(0.05, 12) is 2.179. Because this percent relative error is larger than±0.50%, this is not a suitable sampling strategy.Next, we try two samples (cost $40), analyzing each six times (cost $600).For this scenario, the percent relative error is2 2s se = samptn+ methn n= 2 2035 010 ..2+ 020 .057 .2× 6=sampsamprepwhere t(0.05, 11) is 2.2035. Because this percent relative error is largerthan ±0.50%, this also is not a suitable sampling strategy.Next we try three samples (cost $60), analyzing each four times (cost$600). For this scenario, the percent relative error is2 2s se = samptn+ methn n= 2 2035 010 ..3+ 020 .049 .3× 4=sampsamprepwhere t(0.05, 11) is 2.2035. Because both the total cost ($660) and thepercent relative error meet our requirements, this is a suitable samplingstrategy.There are other suitable sampling strategies that meet both goals. Thestrategy requiring the least expense is to collect eight samples, analyzingeach once for a total cost of $560 and a percent relative error of ±0.46%.Collecting 10 samples and analyzing each one time, gives a percent relativeerror of ±0.39% at a cost of $700.Click here to return to the chapter.Practice Exercise 7.6The fluoride ion, F – , is a suitable masking agent as it binds with Al 3+ toform the stable AlF 6 3– complex, leaving iron in solution.Click here to return to the chapter.Practice Exercise 7.7The relevant reactions and equilibrium constants are2Fe + 2( aq) + 3phen( aq) Fe( phen)+ ( aq)β = 5×103320

352 Analytical Chemistry 2.0Fe 2+p(phen)13logβ 3 = 6.9Fe 3+Fe(phen)2+ 313logβ 3 = 4.6Fe(phen) 33+Figure 7.34 Ladder diagram forPractice Exercise 7.7.3Fe + 3( aq) + 3phen( aq) Fe( phen)+ ( aq)β = 6×10where phen is an abbreviation for 1,10-phenanthroline. Because b 3 islarger for the complex with Fe 2+ than it is for the complex with Fe 3+ ,1,10-phenanthroline will bind Fe 2+ before it binds Fe 3+ . A ladder diagramfor this system (Figure 6.34) suggests that an equilibrium p(phen) between5.6 and 5.9 will fully complex Fe 2+ without any significant formationof the Fe(phen) 3 3+ complex. Adding a stoichiometrically equivalentamount of 1,10-phenanthroline to a solution of Fe 2+ will be sufficient tomask Fe 2+ in the presence of Fe 3+ . Adding a large excess of 1,10-phenanthroline,however, will decrease p(phen) and allow the formation of bothmetal–ligand complexes.Click here to return to the chapter.Practice Exercise 7.8(a) The solute’s distribution ratio between water and toluene is1mol10.889 g× ×[ S ]org117.3 g 0.00500 LD = =[ S ]1mol1aq(. 1 235 g− 0.889 g) × ×117.3 g 0.01000 L3313= 514 .(b) The fraction of solute remaining in the aqueous phase after one extractionis( q )aq1Vaq=DV + VorgThe extraction efficiency, therefore, is 72.0%.(c) To extract 99.9% of the solute requiresaq20.00 mL== 0.280( 514 . )( 1000 . mL) + 20.00 mL⎛ 20.00 mL ⎞( Q aq) n= 0.001=⎝⎜( 514 . )( 1000 . mL) + 2000 . mL⎠⎟ = 0.280nna minimum of six extractions.Click here to return to the chapter.Practice Exercise 7.9log( 0. 001) = nlog( 0. 280)n = 54 .Because the weak base exists in two forms, only one of which extracts intothe organic phase, the partition coefficient, K D , and the distribution ratio,D, are not identical.

Chapter 7 Collecting and Preparing Samples353BorgK D= [ ][ B ]aq[ B ] [ B ]org totalorgD = =+[ B ] [ B ] + [ HB ]aq totalUsing the K b expression for the weak baseaq[ OH ][ HB ]− +aq aqK b=[ B ]we solve for the concentration of HB + and substitute back into the equationfor D, obtaining−[ B ][ B ] KorgorgD[ OH ]aqD =K+ × ==[ B ] ⎧b aqK⎫[ OH[ B ]baqaq− [ B ] 1+aq ⎨⎪−⎬⎪− ] + K b[ OH ]aq[ OH ]⎩⎪aq ⎭⎪At a pH of 9.0, the [OH – ] is 1 × 10 –5 M and the distribution ratio hasa value of−K [ OH ]D aqD =−[ OH ] + Kaqbaqaq−5( 500 . )( 1. 0×10 )=−10 . × 10 + 10 . × 105 −4= 0.455After one extraction, the fraction of B remaining in the aqueous phase isq aq( )125.00 mL== 0.524( 0. 455)( 50. 00 mL) + 25.00 mLThe extraction efficiency, therefore, is 47.6%. At a pH of 9, most of theweak base is present as HB + , which explains why the overall extractionefficiency is so poor.Click here to return to the chapter.


DRAFTChapter 8Gravimetric MethodsChapter Overview8A Overview of Gravimetric Methods8B Precipitation Gravimetry8C Volatilization Gravimetry8D Particulate Gravimetry8E Key Terms8F Chapter Summary8G Problems8H Solutions to Practice ExercisesGravimetry includes all analytical methods in which the analytical signal is a measurementof mass or a change in mass. When you step on a scale after exercising you are making, in asense, a gravimetric determination of your mass. Mass is the most fundamental of all analyticalmeasurements, and gravimetry is unquestionably our oldest quantitative analytical technique.The publication in 1540 of Vannoccio Biringuccio’s Pirotechnia is an early example of applyinggravimetry—although not yet known by this name—to the analysis of metals and ores. 1Although gravimetry no longer is the most important analytical method, it continues to finduse in specialized applications.1 Smith, C. S.; Gnodi, M. T. translation of Biringuccio, V. Pirotechnia, MIT Press: Cambridge, MA, 1959.Copyright: David Harvey, 2009355

356 Analytical Chemistry 2.0Method 2540D in Standard Methods forthe Examination of Waters and Wastewaters,20th Edition (American Public HealthAssociation, 1998) provides an approvedmethod for determining total suspendedsolids. The method uses a glass-fiber filterto retain the suspended solids. After filteringthe sample, the filter is dried to aconstant weight at 103–105 o C.Method 925.10 in Official Methods ofAnalysis, 18th Edition (AOAC International,2007) provides an approvedmethod for determining the moisturecontent of flour. A preweighed sample isheated for one hour in a 130 o C oven andtransferred to a desiccator while it cools toroom temperature. The loss in mass givesthe amount of water in the sample.8AOverview of Gravimetric MethodsBefore we consider specific gravimetric methods, let’s take a moment todevelop a broad survey of gravimetry. Later, as you read through the descriptionsof specific gravimetric methods, this survey will help you focuson their similarities instead of their differences. You will find that it is easierto understand a new analytical method when you can see its relationship toother similar methods.8A.1 Using Mass as an Analytical SignalSuppose you are to determine the total suspended solids in the water releasedby a sewage-treatment facility. Suspended solids are just that—solidmatter that has yet to settle out of its solution matrix. The analysis is easy.After collecting a sample, you pass it through a preweighed filter that retainsthe suspended solids, and dry the filter and solids to remove any residualmoisture. The mass of suspended solids is the difference between the filter’sfinal mass and its original mass. We call this a direct analysis becausethe analyte—the suspended solids in this example—is the species that isweighed.What if our analyte is an aqueous ion, such as Pb 2+ ? Because the analyteis not a solid, we cannot isolate it by filtration. We can still measure theanalyte’s mass directly if we first convert it into a solid form. If we suspend apair of Pt electrodes in the sample and apply a sufficiently positive potentialbetween them for a long enough time, we can force the following reactionto completion.2+ +Pb ( aq) + 4HO() l PbO () s + H ( g) + 2HO( aq )2 2 23Oxidizing Pb 2+ deposits PbO 2 on the Pt anode. If we weigh the anode beforeand after applying the potential, the change in its mass gives the massof PbO 2 and, from the reaction’s stoichiometry, the amount of Pb 2+ in thesample. This is a direct analysis because PbO 2 contains the analyte.Sometimes it is easier to remove the analyte and let a change in massserve as the analytical signal. Suppose you need to determine a food’s moisturecontent. One approach is to heat a sample of the food to a temperaturethat vaporizes the water, capturing it in a preweighed absorbent trap.The change in the absorbent’s mass provides a direct determination of theamount of water in the sample. An easier approach is to weigh the sample offood before and after heating, using the change in its mass as an indicationof the amount of water originally present. We call this an indirect analysisbecause we determine the analyte using a signal that is proportional itsdisappearance.The indirect determination of a sample’s moisture content is done bydifference. The sample’s initial mass includes the water, but its final massdoes not. We can also determine an analyte indirectly without its ever being

Chapter 8 Gravimetric Methods357weighed. For example, phosphite, PO 3 3– , reduces Hg 2+ to Hg 2 2+ , which inthe presence of Cl – precipitates as Hg 2 Cl 2 .3−2HgCl ( aq) + PO ( aq) + 3HO()l 2 3+ − 3−Hg Cl () s + 2HO( aq) + 2Cl( aq) + 2PO( aq)342 22If we add HgCl 2 in excess, each mole of PO 3 3– produces one mole ofHg 2 Cl 2 . The precipitate’s mass, therefore, provides an indirect measurementof the amount of PO 3 3– in the original sample.8A.2 Types of Gravimetric MethodsThe four examples in the previous section illustrate different ways in whichthe measurement of mass may serve as an analytical signal. When the signalis the mass of a precipitate, we call the method precipitation gravimetry.The indirect determination of PO 3 3– by precipitating Hg 2 Cl 2 is an example,as is the direct determination of Cl – by precipitating AgCl.In electrogravimetry, we deposit the analyte as a solid film an electrodein an electrochemical cell. The deposition as PbO 2 at a Pt anode isone example of electrogravimetry. The reduction of Cu 2+ to Cu at a Ptcathode is another example of electrogravimetry.When we use thermal or chemical energy to remove a volatile species,we call the method volatilization gravimetry. In determining the moisturecontent of bread, for example, we use thermal energy to vaporize thewater in the sample. To determine the amount of carbon in an organic compound,we use the chemical energy of combustion to convert it to CO 2 .Finally, in particulate gravimetry we determine the analyte byseparating it from the sample’s matrix using a filtration or an extraction.The determination of total suspended solids is one example of particulategravimetry.8A.3 Conservation of MassAn accurate gravimetric analysis requires that the analytical signal—whetherit is a mass or a change in mass—be proportional to the amount of analytein our sample. For all gravimetric methods this proportionality involves aconservation of mass. If the method relies on one or more chemical reactions,then the stoichiometry of the reactions must be known. Thus, forthe analysis of PO 3 3– described earlier, we know that each mole of Hg 2 Cl 2corresponds to a mole of PO 33–in our sample. If we remove the analytefrom its matrix, then the separation must be selective for the analyte. Whendetermining the moisture content in bread, for example, we know that themass of H 2 O in the bread is the difference between the sample’s final massand its initial mass.We will not consider electrogravimetry inthis chapter. See Chapter 11 on electrochemicalmethods of analysis for a furtherdiscussion of electrogravimetry.We will return to this concept of applyinga conservation of mass later in the chapterwhen we consider specific examples ofgravimetric methods.

358 Analytical Chemistry 2.0Other examples of definitive techniquesare coulometry and isotope-dilution massspectrometry. Coulometry is discussed inChapter 11. Isotope-dilution mass spectrometryis beyond the scope of an introductorytextbook; however, you will findsome suggested readings in this chapter’sAdditional Resources.Most precipitation gravimetric methodswere developed in the nineteenth century,or earlier, often for the analysis of ores.Figure 1.1 in Chapter 1, for example, illustratesa precipitation gravimetric methodfor the analysis of nickel in ores.8A.4 Why Gravimetry is ImportantExcept for particulate gravimetry, which is the most trivial form of gravimetry,you probably will not use gravimetry after you complete this course.Why, then, is familiarity with gravimetry still important? The answer is thatgravimetry is one of only a small number of definitive techniques whosemeasurements require only base SI units, such as mass or the mole, and definedconstants, such as Avogadro’s number and the mass of 12 C. Ultimately,we must be able to trace the result of an analysis to a definitive technique,such as gravimetry, that we can relate to fundamental physical properties. 2Although most analysts never use gravimetry to validate their results, theyoften verifying an analytical method by analyzing a standard reference materialwhose composition is traceable to a definitive technique. 38BPrecipitation GravimetryIn precipitation gravimetry an insoluble compound forms when we add aprecipitating reagent, or precipitant, to a solution containing our analyte.In most methods the precipitate is the product of a simple metathesis reactionbetween the analyte and the precipitant; however, any reaction generatinga precipitate can potentially serve as a gravimetric method.8B.1 Theory and PracticeAll precipitation gravimetric analysis share two important attributes. First,the precipitate must be of low solubility, of high purity, and of known compositionif its mass is to accurately reflect the analyte’s mass. Second, theprecipitate must be easy to separate from the reaction mixture.So l u b i l i t y Co n s i d e r at i o n sA total analysis technique is one in whichthe analytical signal—mass in this case—is proportional to the absolute amount ofanalyte in the sample. See Chapter 3 fora discussion of the difference between totalanalysis techniques and concentrationtechniques.To provide accurate results, a precipitate’s solubility must be minimal. Theaccuracy of a total analysis technique typically is better than ±0.1%, whichmeans that the precipitate must account for at least 99.9% of the analyte.Extending this requirement to 99.99% ensures that the precipitate’s solubilitydoes not limit the accuracy of a gravimetric analysis.We can minimize solubility losses by carefully controlling the conditionsunder which the precipitate forms. This, in turn, requires that weaccount for every equilibrium reaction affecting the precipitate’s solubility.For example, we can determine Ag + gravimetrically by adding NaCl as aprecipitant, forming a precipitate of AgCl.+ −Ag ( aq) + Cl ( aq) AgCl()s8.1If this is the only reaction we consider, then we predict that the precipitate’ssolubility, S AgCl , is given by the following equation.2 Valacárcel, M.; Ríos, A. Analyst 1995, 120, 2291–2297.3 (a) Moody, J. R.; Epstein, M. S. Spectrochim. Acta 1991, 46B, 1571–1575; (b) Epstein, M. S.Spectrochim. Acta 1991, 46B, 1583–1591.

Chapter 8 Gravimetric Methods359-2-3log(SAgCl)-4-5-6-7Ag + – 2–AgCl(aq) AgCl 2 AgCl 37 6 5 4 3 2 1 0pClFigure 8.1 Solubility of AgCl as a function of pCl. The dashed red line shows our predictionfor S AgCl if we incorrectly assume that only reaction 8.1 and equation 8.2 affect silver chloride’ssolubility. The solid blue curve is calculated using equation 8.7, which accounts for reaction 8.1and reactions 8.3–8.5. Because the solubility of AgCl spans several orders of magnitude, SAgClis displayed on the y-axis in logarithmic form.SAgClKsp= [ Ag+ ] =−[ Cl ]Equation 8.2 suggests that we can minimize solubility losses by adding alarge excess of Cl – . In fact, as shown in Figure 8.1, adding a large excess ofCl – increases the precipitate’s solubility.To understand why the solubility of AgCl is more complicated thanthe relationship suggested by equation 8.2, we must recognize that Ag + alsoforms a series of soluble silver-chloro metal–ligand complexes.8.2+ −Ag ( aq) + Cl ( aq) AgCl( aq) log K = 370 . 8.31−−AgCl( aq) + Cl ( aq) AgCl 2( aq) log K 2= 192 . 8.4− − 2−AgCl ( aq) + Cl ( aq) AgCl ( aq) log K = 078 .2 338.5The actual solubility of AgCl is the sum of the equilibrium concentrationsfor all soluble forms of Ag + , as shown by the following equation.+ − 2−S = [ Ag ] + [ AgCl( aq) ] + [ AgCl ] + [ AgCl ]AgCl 2 3By substituting into equation 8.6 the equilibrium constant expressions forreaction 8.1 and reactions 8.3–8.5, we can define the solubility of AgCl asSAgCl8.6Ksp−= + KK + KKK [ Cl ] + K KKK [ Cl − ]2 8.7− 1 sp 1 2 sp 1 2 3 sp[ Cl ]Equation 8.7 explains the solubility curve for AgCl shown in Figure 8.1.As we add NaCl to a solution of Ag + , the solubility of AgCl initially decreas-Problem 1 in the end-of-chapter problemsasks you to show that equation 8.7 is correctby completing the derivation.

360 Analytical Chemistry 2.0The predominate silver-chloro complexesfor different values of pCl are shown bythe ladder diagram along the x-axis in Figure8.1 Note that the increase in solubilitybegins when the higher-order solublecomplexes, AgCl 2– and AgCl32– , becomethe predominate species.Be sure that equation 8.10 makes sense toyou. Reaction 8.8 tells us that the dissolutionof CaF 2 produces one mole of Ca 2+for every two moles of F – , which explainsthe term of 1/2 in equation 8.10. BecauseF – is a weak base, we need to account forboth of its chemical forms in solution,which explains why we include HF.Problem 4 in the end-of-chapter problemsasks you to show that equation 8.11 is correctby completing the derivation.es because of reaction 8.1. Under these conditions, the final three termsin equation 8.7 are small and equation 8.1 is sufficient to describe AgCl’ssolubility. At higher concentrations of Cl – , reaction 8.4 and reaction 8.5increase the solubility of AgCl. Clearly the equilibrium concentration ofchloride is important if we want to determine the concentration of silver byprecipitating AgCl. In particular, we must avoid a large excess of chloride.Another important parameter that may affect a precipitate’s solubilityis pH. For example, a hydroxide precipitates such as Fe(OH) 3 is moresoluble at lower pH levels where the concentration of OH – is small. Becausefluoride is a weak base, the solubility of calcium fluoride, S CaF 2 , alsois pH-dependent. We can derive an equation for S CaF by considering the2following equilibrium reactions2 11CaF () s Ca + ( aq) + 2F − ( aq ) K = 3.9× 10−2sp8.8+ − −HF( aq) + HO() l HO ( aq) + F ( aq ) K = 68 . × 10 4 8.92 3 aand the following equation for the solubility of CaF 2 .{ }+ −S CaF2= [ Ca ] = [ F ] + [ HF]8.102Substituting the equilibrium constant expressions for reaction 8.8 and reaction8.9 into equation 8.10 defines the solubility of CaF 2 in terms of theequilibrium concentration of H 3 O + .2 1⎧2K ⎛+sp HO ⎞ ⎫+3S = Ca = [ ]2[ ]+CaF⎨⎪24 ⎝⎜1 ⎬⎪8.11Ka ⎠⎟⎩⎪⎭⎪Figure 8.2 shows how pH affects the solubility of CaF 2 . Depending on thesolution’s pH, the predominate form of fluoride is either HF or F – . Whenthe pH is greater than 4.17, the predominate species is F – and the solubilityof CaF 2 is independent of pH because only reaction 8.8 occurs to an ap-13 /-1.0-1.5Figure 8.2 Solubility of CaF 2 as a function of pH. The solid bluecurve is a plot of equation 8.11. The predominate form of fluoridein solution is shown by the ladder diagram along the x-axis, with theblack rectangle showing the region where both HF and F – are importantspecies. Note that the solubility of CaF 2 is independent ofpH for pH levels greater than 4.17, and that its solubility increasesdramatically at lower pH levels where HF is the predominate species.Because the solubility of CaF 2 spans several orders of magnitude,its solubility is shown in logarithmic form.logS(CaF 2)-2.0-2.5-3.0-3.5-4.0HF F –0 2 4 6 8 10 12 14pH

Chapter 8 Gravimetric Methods361preciable extent. At more acidic pH levels, the solubility of CaF 2 increasesbecause of the contribution of reaction 8.9.When solubility is a concern, it may be possible to decrease solubilityby using a non-aqueous solvent. A precipitate’s solubility is generally greaterin an aqueous solution because of water’s ability to stabilize ions throughsolvation. The poorer solvating ability of non-aqueous solvents, even thosewhich are polar, leads to a smaller solubility product. For example, the K spof PbSO 4 is 2 × 10 –8 in H 2 O and 2.6 × 10 –12 in a 50:50 mixture of H 2 Oand ethanol.Avoiding ImpuritiesIn addition to having a low solubility, the precipitate must be free fromimpurities. Because precipitation usually occurs in a solution that is richin dissolved solids, the initial precipitate is often impure. We must removethese impurities before determining the precipitate’s mass.The greatest source of impurities is the result of chemical and physicalinteractions occurring at the precipitate’s surface. A precipitate is generallycrystalline—even if only on a microscopic scale—with a well-defined latticeof cations and anions. Those cations and anions at the precipitate’s surfacecarry, respectively, a positive or a negative charge because they have incompletecoordination spheres. In a precipitate of AgCl, for example, each silverion in the precipitate’s interior is bound to six chloride ions. A silver ion atthe surface, however, is bound to no more than five chloride ions and carriesa partial positive charge (Figure 8.3). The presence of these partial chargesmakes the precipitate’s surface an active site for the chemical and physicalinteractions that produce impurities.One common impurity is an inclusion. A potential interfering ionwhose size and charge is similar to a lattice ion, may substitute into theinterior Cl –surrounded by six Ag +interior Ag +surrounded by six Cl –Practice Exercise 8.1You can use a ladder diagram topredict the conditions for minimizinga precipitate’s solubility.Draw a ladder diagram for oxalicacid, H 2 C 2 O 4 , and use it to establisha suitable range of pH valuesfor minimizing the solubility ofCaC 2 O 4 . Relevant equilibriumconstants may be found in the appendices.Click here to review your answer tothis exercise.xyzCl – on faceAg + on edgesurrounded by five Ag +surrounded by four Cl – Cl –Ag +Figure 8.3 Ball-and-stick diagram showing the lattice structureof AgCl. Each silver ion in the lattice’s interior binds with sixchloride ions, and each chloride ion in the interior binds withsix silver ions. Those ions on the lattice’s surface or edges bindto fewer than six ions and carry a partial charge. A silver ionon the surface, for example, carries a partial positive charge.These charges make the surface of a precipitate an active site forchemical and physical interactions.

362 Analytical Chemistry 2.0(a) (b) (c)zzzyxyFigure 8.4 Three examples of impurities that may form during precipitation. The cubic frame represents the precipitateand the blue marks are impurities: (a) inclusions, (b) occlusions, and (c) surface adsorbates. Inclusions are randomly distributedthroughout the precipitate. Occlusions are localized within the interior of the precipitate and surface adsorbatesare localized on the precipitate’s exterior. For ease of viewing, in (c) adsorption is shown on only one surface.xyxSuppose that 10% of an interferent formsan inclusion during each precipitation.When we initially form the precipitate,10% of the original interferent is presentas an inclusion. After the first reprecipitation,10% of the included interferent remains,which is 1% of the original interferent.A second reprecipitation decreasesthe interferent to 0.1% of the originalamount.lattice structure, provided that the interferent precipitates with the samecrystal structure (Figure 8.4a). The probability of forming an inclusionis greatest when the concentration of the interfering ion is substantiallygreater than the lattice ion’s concentration. An inclusion does not decreasethe amount of analyte that precipitates, provided that the precipitant ispresent in sufficient excess. Thus, the precipitate’s mass is always larger thanexpected.An inclusion is difficult to remove since it is chemically part of theprecipitate’s lattice. The only way to remove an inclusion is through reprecipitation.After isolating the precipitate from its supernatant solution,we dissolve it by heating in a small portion of a suitable solvent. We thenallow the solution to cool, reforming the precipitate. Because the interferent’sconcentration is less than that in the original solution, the amount ofincluded material is smaller. We can repeat the process of reprecipitationuntil the inclusion’s mass is insignificant. The loss of analyte during reprecipitation,however, can be a significant source of error.Occlusions form when interfering ions become trapped within thegrowing precipitate. Unlike inclusions, which are randomly dispersed withinthe precipitate, an occlusion is localized, either along flaws within theprecipitate’s lattice structure or within aggregates of individual precipitateparticles (Figure 8.4b). An occlusion usually increases a precipitate’s mass;however, the mass is smaller if the occlusion includes the analyte in a lowermolecular weight form than that of the precipitate.We can minimize occlusions by maintaining the precipitate in equilibriumwith its supernatant solution for an extended time. This process iscalled a digestion. During digestion, the dynamic nature of the solubility–precipitation equilibrium, in which the precipitate dissolves and reforms,ensures that the occlusion is reexposed to the supernatant solution. Because

Chapter 8 Gravimetric Methods363the rates of dissolution and reprecipitation are slow, there is less opportunityfor forming new occlusions.After precipitation is complete the surface continues to attract ionsfrom solution (Figure 8.4c). These surface adsorbates comprise a thirdtype of impurity. We can minimize surface adsorption by decreasing theprecipitate’s available surface area. One benefit of digesting a precipitateis that it increases the average particle size. Because the probability of aparticle completely dissolving is inversely proportional to its size, duringdigestion larger particles increase in size at the expense of smaller particles.One consequence of forming a smaller number of larger particles is anoverall decrease in the precipitate’s surface area. We also can remove surfaceadsorbates by washing the precipitate, although the potential loss of analytecan not be ignored.Inclusions, occlusions, and surface adsorbates are examples of coprecipitates—otherwisesoluble species that form within the precipitate containingthe analyte. Another type of impurity is an interferent that forms anindependent precipitate under the conditions of the analysis. For example,the precipitation of nickel dimethylglyoxime requires a slightly basic pH.Under these conditions, any Fe 3+ in the sample precipitates as Fe(OH) 3 .In addition, because most precipitants are rarely selective toward a singleanalyte, there is always a risk that the precipitant will react with both theanalyte and an interferent.We can minimize the formation of additional precipitates by carefullycontrolling solution conditions. If an interferent forms a precipitate thatis less soluble than the analyte’s precipitate, we can precipitate the interferentand remove it by filtration, leaving the analyte behind in solution.Alternatively, we can mask the analyte or the interferent to prevent itsprecipitation.Both of the above-mentioned approaches are illustrated in Fresenius’analytical method for determining Ni in ores containing Pb 2+ , Cu 2+ , andFe 3+ (see Figure 1.1 in Chapter 1). Dissolving the ore in the presence ofH 2 SO 4 selectively precipitates Pb 2+ as PbSO 4 . Treating the supernatantwith H 2 S precipitates the Cu 2+ as CuS. After removing the CuS by filtration,adding ammonia precipitates Fe 3+ as Fe(OH) 3 . Nickel, which formsa soluble amine complex, remains in solution.In addition to forming a precipitate withNi 2+ , dimethylglyoxime also forms precipitateswith Pd 2+ and Pt 2+ . These cationsare potential interferents in an analysisfor nickel.Co n t r o l l i n g Pa r t i c l e Si z eSize matters when it comes to forming a precipitate. Larger particles areeasier to filter, and, as noted earlier, a smaller surface area means there isless opportunity for surface adsorbates to form. By carefully controllingthe reaction conditions we can significantly increase a precipitate’s averageparticle size.Precipitation consists of two distinct events: nucleation, the initial formationof smaller stable particles of precipitate, and particle growth. Largerparticles form when the rate of particle growth exceeds the rate of nucle-

364 Analytical Chemistry 2.0A supersaturated solution is one thatcontains more dissolved solute than thatpredicted by equilibrium chemistry. Thesolution is inherently unstable and precipitatessolute to reach its equilibriumposition. How quickly this process occursdepends, in part, on the value of RSS.ation. Understanding the conditions favoring particle growth is importantwhen designing a gravimetric method of analysis.We define a solute’s relative supersaturation, RSS, asQ SRSS = − 8.12Swhere Q is the solute’s actual concentration and S is the solute’s concentrationat equilibrium. 4 The numerator of equation 8.12, Q – S, is a measureof the solute’s supersaturation. A solution with a large, positive value ofRSS has a high rate of nucleation, producing a precipitate with many smallparticles. When the RSS is small, precipitation is more likely to occur byparticle growth than by nucleation.Examining equation 8.12 shows that we can minimize RSS by decreasingthe solute’s concentration, Q, or by increasing the precipitate’s solubility,S. A precipitate’s solubility usually increases at higher temperatures, andadjusting pH may affect a precipitate’s solubility if it contains an acidic ora basic ion. Temperature and pH, therefore, are useful ways to increase thevalue of S. Conducting the precipitation in a dilute solution of analyte, oradding the precipitant slowly and with vigorous stirring are ways to decreasethe value of Q.There are practical limitations to minimizing RSS. Some precipitates,such as Fe(OH) 3 and PbS, are so insoluble that S is very small and a largeRSS is unavoidable. Such solutes inevitably form small particles. In addition,conditions favoring a small RSS may lead to a relatively stable supersaturatedsolution that requires a long time to fully precipitate. For example,almost a month is required to form a visible precipitate of BaSO 4under conditions in which the initial RSS is 5. 5A visible precipitate takes longer to form when RSS is small both becausethere is a slow rate of nucleation and because there is a steady decreasein RSS as the precipitate forms. One solution to the latter problem is togenerate the precipitant in situ as the product of a slow chemical reaction.This maintains the RSS at an effectively constant level. Because the precipitateforms under conditions of low RSS, initial nucleation produces asmall number of particles. As additional precipitant forms, particle growthsupersedes nucleation, resulting in larger precipitate particles. This processis called homogeneous precipitation. 6Two general methods are used for homogeneous precipitation. If theprecipitate’s solubility is pH-dependent, then we can mix the analyte andthe precipitant under conditions where precipitation does not occur, andthen increase or decrease the pH by chemically generating OH – or H 3 O + .For example, the hydrolysis of urea is a source of OH – .4 Von Weimarn, P. P. Chem. Revs. 1925, 2, 217–242.5 Bassett, J.; Denney, R. C.; Jeffery, G. H. Mendham. J. Vogel’s Textbook of Quantitative InorganicAnalysis, Longman: London, 4th Ed., 1981, p. 408.6 Gordon, L.; Salutsky, M. L.; Willard, H. H. Precipitation from Homogeneous Solution, Wiley:NY, 1959.

Chapter 8 Gravimetric Methods365CO(NH ) ( aq) + H O() l 2NH ( aq) + CO ( g)2 2 23 2NH ( ) HO () −OH ( ) +aq + l aq + NH ( aq )3 24Because the hydrolysis of urea is temperature-dependent—it is negligibleat room temperature—we can use temperature to control the rate of hydrolysisand the rate of precipitate formation. Precipitates of CaC 2 O 4 , forexample, have been produced by this method. After dissolving the samplecontaining Ca 2+ , the solution is made acidic with HCl before adding a solutionof 5% w/v (NH 4 ) 2 C 2 O 4 . Because the solution is acidic, a precipitateof CaC 2 O 4 does not form. The solution is heated to approximately 50 o Cand urea is added. After several minutes, a precipitate of CaC 2 O 4 begins toform, with precipitation reaching completion in about 30 min.In the second method of homogeneous precipitation, the precipitant isgenerated by a chemical reaction. For example, Pb 2+ is precipitated homogeneouslyas PbCrO 4 by using bromate, BrO 3 – , to oxidize Cr 3+ to CrO 4 2– .− 3+6BrO ( aq) + 10Cr ( aq) + 22H O()l 33Br( aq)+ 10CrO22( aq) + 44H( aq)2− +4Figure 8.5 shows the result of preparing PbCrO 4 by the direct addition ofKCrO 4 (Beaker A) and by homogenous precipitation (Beaker B). Both beakerscontain the same amount of PbCrO 4 . Because the direct addition ofKCrO 4 leads to rapid precipitation and the formation of smaller particles,the precipitate remains less settled than the precipitate prepared homogeneously.Note, as well, the difference in the color of the two precipitates.A homogeneous precipitation produces large particles of precipitatethat are relatively free from impurities. These advantages, however, are offsetby requiring more time to produce the precipitate and a tendency for theprecipitate to deposit as a thin film on the container’s walls. The latter problemis particularly severe for hydroxide precipitates generated using urea.An additional method for increasing particle size deserves mention.When a precipitate’s particles are electrically neutral they tend to coagulateThe effect of particle size on color is wellknownto geologists, who use a streaktest to help identify minerals. The colorof a bulk mineral and its color whenpowdered are often different. Rubbing amineral across an unglazed porcelain plateleaves behind a small streak of the powderedmineral. Bulk samples of hematite,Fe 2 O 3 , are black in color, but its streakis a familiar rust-red. Crocite, the mineralPbCrO 4 , is red-orange in color; its streakis orange-yellow.Beaker ABeaker BFigure 8.5 Two precipitates of PbCrO 4 . In Beaker A,combining 0.1 M Pb(NO 3 ) 2 and 0.1 M K 2 CrO 4 formsthe precipitate under conditions of high RSS. The precipitateforms rapidly and consists of very small particles.In Beaker B, heating a solution of 0.1 M Pb(NO 3 ) 2 , 0.1M Cr(NO 3 ) 3 , and 0.1 M KBrO 3 slowly oxidizes Cr 3+ toCrO 4 2– , precipitating PbCrO 4 under conditions of lowRSS. The precipitate forms slowly and consists of muchlarger particles.

366 Analytical Chemistry 2.0primaryadsorption layerinert cation+ –– –+ –++–+ –+ + AgCl–+ –+ + +––+ – –inert anionchemicallyadsorbed Ag +secondaryadsorption layerprecipitateparticle(a)(b)(c)+ –– –+ –++–+ –+ + AgCl–+ –+ + +––+ – –+ – +– –+ + ––+–+ AgCl +––++ +––+ – + –+ ––– +++ ––++ –+ AgCl–+–+ + +– –+ – –+– +– +–++––++AgCl+–––+ ++––+ +––– +– ++++ +– + AgCl–– + ++– –––+ –+ ––+ +––– –+ –+–+ –– + +–AgCl +– ++++ + ––+––++––+ –+–+– ++–AgCl–++––– ++– +–– – + –+–+ –– + +AgCl + –– + +– + + +–+ – –precipitatecoagulatedprecipitateparticleparticlesparticleFigure 8.6 Two methods for coagulating a precipitate of AgCl. (a) Coagulation does not happen due to the electrostaticrepulsion between the two positively charged particles. (b) Decreasing the charge within the primary adsorption layer, byadding additional NaCl, decreases the electrostatic repulsion, allowing the particles to coagulate. (c) Adding additionalinert ions decreases the thickness of the secondary adsorption layer. Because the particles can approach each other moreclosely, they are able to coagulate.into larger particles that are easier to filter. Surface adsorption of excess latticeions, however, provides the precipitate’s particles with a net positive ora net negative surface charge. Electrostatic repulsion between the particlesprevents them from coagulating into larger particles.Let’s use the precipitation of AgCl from a solution of AgNO 3 usingNaCl as a precipitant to illustrate this effect. Early in the precipitation,when NaCl is the limiting reagent, excess Ag + ions chemically adsorb tothe AgCl particles, forming a positively charged primary adsorption layer(Figure 8.6a). The solution in contact with this layer contains more inertanions, NO 3 – in this case, than inert cations, Na + , giving a secondary adsorptionlayer with a negative charge that balances the primary adsorptionlayer’s positive charge. The solution outside the secondary adsorption layerremains electrically neutral. Coagulation cannot occur if the secondaryadsorption layer is too thick because the individual particles of AgCl areunable to approach each other closely enough.We can induce coagulation in three ways: by decreasing the number ofchemically adsorbed Ag + ions, by increasing the concentration of inert ions,or by heating the solution. As we add additional NaCl, precipitating moreof the excess Ag + , the number of chemically adsorbed silver ions decreases

Chapter 8 Gravimetric Methods367and coagulation occurs (Figure 8.6b). Adding too much NaCl, however,creates a primary adsorption layer of excess Cl – with a loss of coagulation.A second way to induce coagulation is to add an inert electrolyte,which increases the concentration of ions in the secondary adsorptionlayer. With more ions available, the thickness of the secondary absorptionlayer decreases. Particles of precipitate may now approach each other moreclosely, allowing the precipitate to coagulate. The amount of electrolyteneeded to cause spontaneous coagulation is called the critical coagulationconcentration.Heating the solution and precipitate provides a third way to inducecoagulation. As the temperature increases, the number of ions in the primaryadsorption layer decreases, lowering the precipitate’s surface charge.In addition, heating increases the particles’ kinetic energy, allowing themto overcome the electrostatic repulsion that prevents coagulation at lowertemperatures.The coagulation and decoagulation ofAgCl as we add NaCl to a solution ofAgNO 3 can serve as an endpoint for atitration. See Chapter 9 for additionaldetails.Filtering t h e PrecipitateAfter precipitating and digesting the precipitate, we separate it from solutionby filtering. The most common filtration method uses filter paper,which is classified according to its speed, its size, and its ash content onignition. Speed, or how quickly the supernatant passes through the filterpaper, is a function of the paper’s pore size. A larger pore allows the supernatantto pass more quickly through the filter paper, but does not retainsmall particles of precipitate. Filter paper is rated as fast (retains particleslarger than 20–25 mm), medium–fast (retains particles larger than 16 mm),medium (retains particles larger than 8 mm), and slow (retains particleslarger than 2–3 mm). The proper choice of filtering speed is important. Ifthe filtering speed is too fast, we may fail to retain some of the precipitate,causing a negative determinate error. On the other hand, the precipitatemay clog the pores if we use a filter paper that is too slow.Because filter paper is hygroscopic, it is not easy to dry it to a constantweight. When accuracy is important, the filter paper is removed beforedetermining the precipitate’s mass. After transferring the precipitate andfilter paper to a covered crucible, we heat the crucible to a temperature thatcoverts the paper to CO 2 (g) and H 2 O(g), a process called ignition.Gravity filtering is accomplished by folding the filter paper into a coneand placing it in a long-stem funnel (Figure 8.7). A seal between the filtercone and the funnel is formed by dampening the paper with water or supernatant,and pressing the paper to the wall of the funnel. When properlyprepared, the funnel’s stem fills with the supernatant, increasing the rateof filtration.The precipitate is transferred to the filter in several steps. The first step isto decant the majority of the supernatant through the filter paper withouttransferring the precipitate (Figure 8.8). This prevents the filter paper fromclogging at the beginning of the filtration process. The precipitate is rinsedA filter paper’s size is just its diameter. Filterpaper comes in many sizes, including4.25 cm, 7.0 cm, 11.0 cm, 12.5 cm, 15.0cm, and 27.0 cm. Choose a size that fitscomfortably into your funnel. For a typical65-mm long-stem funnel, 11.0 cm and12.5 cm filter paper are good choices.Igniting a poor quality filter paper leavesbehind a residue of inorganic ash. Forquantitative work, use a low-ash filter paper.This grade of filter paper is pretreatedwith a mixture of HCl and HF to removeinorganic materials. Quantitative filterpaper typically has an ash content of lessthan 0.010% w/w.

368 Analytical Chemistry 2.0(a)(b)(c)(f)Figure 8.7 Preparing a filter paper cone. The filterpaper circle in (a) is folded in half (b), andfolded in half again (c). The folded filter paper isparted (d) and a small corner is torn off (e). Thefilter paper is opened up into a cone and placedin the funnel (f).(d)(e)glass stirring rodsupernatantprecipitateFigure 8.8 Proper procedure fortransferring the supernatant to thefilter paper cone.while it remains in its beaker, with the rinsings decanted through the filterpaper. Finally, the precipitate is transferred onto the filter paper using astream of rinse solution. Any precipitate clinging to the walls of the beakeris transferred using a rubber policeman (a flexible rubber spatula attachedto the end of a glass stirring rod).An alternative method for filtering a precipitate is a filtering crucible.The most common is a fritted-glass crucible containing a porous glass diskfilter. Fritted-glass crucibles are classified by their porosity: coarse (retainingparticles larger than 40–60 mm), medium (retaining particles greater than10–15 mm), and fine (retaining particles greater than 4–5.5 mm). Anothertype of filtering crucible is the Gooch crucible, which is a porcelain cruciblewith a perforated bottom. A glass fiber mat is placed in the crucible to retainthe precipitate. For both types of crucibles, the precipitate is transferred inthe same manner described earlier for filter paper. Instead of using gravity,the supernatant is drawn through the crucible with the assistance of suctionfrom a vacuum aspirator or pump (Figure 8.9).Ri n s i n g t h e PrecipitateBecause the supernatant is rich with dissolved inert ions, we must removeany residual traces of supernatant to avoid a positive determinate errorwithout incurring solubility losses. In many cases this simply involves theuse of cold solvents or rinse solutions containing organic solvents such asethanol. The pH of the rinse solution is critical if the precipitate containsan acidic or basic ion. When coagulation plays an important role in determiningparticle size, adding a volatile inert electrolyte to the rinse solutionprevents the precipitate from reverting into smaller particles that might passthrough the filter. This process of reverting to smaller particles is called peptization.The volatile electrolyte is removed when drying the precipitate.

Chapter 8 Gravimetric Methods369crucibleto vacuumsuctionflaskvacuumtrapFigure 8.9 Procedure for filtering a precipitate through a filtering crucible. Thetrap prevents water from an aspirator from back-washing into the suction flask.In general, we can minimize the loss of analyte by using several smallportions of rinse solution instead of a single large volume. Testing the usedrinse solution for the presence of impurities is another way to guard againstover rinsing the precipitate. For example, if Cl – is a residual ion in the supernatant,we can test for its presence using AgNO 3 . After collecting a smallportion of the rinse solution, we add a few drops of AgNO 3 and look for thepresence or absence of a precipitate of AgCl. If a precipitate forms, then weknow that Cl – is present and continue to rinse the precipitate. Additionalrinsing is not needed if the AgNO 3 does not produce a precipitate.Dr y i n g t h e PrecipitateAfter separating the precipitate from its supernatant solution, the precipitateis dried to remove residual traces of rinse solution and any volatile impurities.The temperature and method of drying depend on the method offiltration and the precipitate’s desired chemical form. Placing the precipitatein a laboratory oven and heating to a temperature of 110 o C is sufficientwhen removing water and other easily volatilized impurities. Higher temperaturesrequire a muffle furnace, a Bunsen burner, or a Meker burner,and are necessary if we need to thermally decompose the precipitate beforeweighing.Because filter paper absorbs moisture, we must remove it before weighingthe precipitate. This is accomplished by folding the filter paper overthe precipitate and transferring both the filter paper and the precipitateto a porcelain or platinum crucible. Gentle heating first dries and thenchars the filter paper. Once the paper begins to char, we slowly increase thetemperature until all traces of the filter paper are gone and any remainingcarbon is oxidized to CO 2 .

370 Analytical Chemistry 2.0Fritted-glass crucibles can not withstand high temperatures and mustbe dried in an oven at temperatures below 200 o C. The glass fiber matsused in Gooch crucibles can be heated to a maximum temperature of approximately500 o C.Co m p o s i t i o n o f t h e Fi n a l PrecipitateFor a quantitative application, the final precipitate must have a welldefinedcomposition. Precipitates containing volatile ions or substantialamounts of hydrated water, are usually dried at a temperature that completelyremoves these volatile species. For example, one standard gravimetricmethod for the determination of magnesium involves its precipitationas MgNH 4 PO 4 •6H 2 O. Unfortunately, this precipitate is difficult to dryat lower temperatures without losing an inconsistent amount of hydratedwater and ammonia. Instead, the precipitate is dried at temperatures above1000 o C where it decomposes to magnesium pyrophosphate, Mg 2 P 2 O 7 .An additional problem is encountered if the isolated solid is nonstoichiometric.For example, precipitating Mn 2+ as Mn(OH) 2 and heating frequentlyproduces a nonstoichiometric manganese oxide, MnO x , where xvaries between one and two. In this case the nonstoichiometric product isthe result of forming of a mixture of oxides with different oxidation state ofmanganese. Other nonstoichiometric compounds form as a result of latticedefects in the crystal structure. 7The best way to appreciate the theoreticaland practical details discussed in this sectionis to carefully examine a typical precipitationgravimetric method. Althougheach method is unique, the determinationof Mg 2+ in water and wastewaterby precipitating MgNH 4 PO 4 • 6H 2 Oand isolating Mg 2 P 2 O 7 provides an instructiveexample of a typical procedure.The description here is based on Method3500-Mg D in Standard Methods for theExamination of Water and Wastewater,19th Ed., American Public Health Association:Washington, D. C., 1995. Withthe publication of the 20th Edition in1998, this method is no longer listed asan approved method.7 Ward, R., ed., Non-Stoichiometric Compounds (Ad. Chem. Ser. 39), American Chemical Society:Washington, D. C., 1963.Representative Method 8.1Determination of Mg 2+ in Water and WastewaterDescription o f Me t h o dMagnesium is precipitated as MgNH 4 PO 4 • 6H 2 O using (NH 4 ) 2 HPO 4as the precipitant. The precipitate’s solubility in neutral solutions is relativelyhigh (0.0065 g/100 mL in pure water at 10 o C), but it is much lesssoluble in the presence of dilute ammonia (0.0003 g/100 mL in 0.6 MNH 3 ). Because the precipitant is not selective, a preliminary separation ofMg 2+ from potential interferents is necessary. Calcium, which is the mostsignificant interferent, is removed by precipitating it as CaC 2 O 4 . Thepresence of excess ammonium salts from the precipitant, or the additionof too much ammonia leads to the formation of Mg(NH 4 ) 4 (PO 4 ) 2 , whichforms Mg(PO 3 ) 2 after drying. The precipitate is isolated by filtering, usinga rinse solution of dilute ammonia. After filtering, the precipitate isconverted to Mg 2 P 2 O 7 and weighed.Pr o c e d u r eTransfer a sample containing no more than 60 mg of Mg 2+ into a 600-mLbeaker. Add 2–3 drops of methyl red indicator, and, if necessary, adjust

Chapter 8 Gravimetric Methods371the volume to 150 mL. Acidify the solution with 6 M HCl and add 10mL of 30% w/v (NH 4 ) 2 HPO 4 . After cooling and with constant stirring,add concentrated NH 3 dropwise until the methyl red indicator turns yellow(pH > 6.3). After stirring for 5 min, add 5 mL of concentrated NH 3and continue stirring for an additional 10 min. Allow the resulting solutionand precipitate to stand overnight. Isolate the precipitate by filteringthrough filter paper, rinsing with 5% v/v NH 3 . Dissolve the precipitate in50 mL of 10% v/v HCl, and precipitate a second time following the sameprocedure. After filtering, carefully remove the filter paper by charring.Heat the precipitate at 500 o C until the residue is white, and then bringthe precipitate to constant weight at 1100 o C.Qu e s t i o n s1. Why does the procedure call for a sample containing no more than60 mg of Mg 2+ ?A 60-mg portion of Mg 2+ generates approximately 600 mg ofMgNH 4 PO 4 • 6H 2 O. This is a substantial amount of precipitate. Alarger quantity of precipitate may be difficult to filter and difficult toadequately rinse free of impurities.2. Why is the solution acidified with HCl before adding the precipitant?The HCl ensures that MgNH 4 PO 4 • 6H 2 O does not immediatelyprecipitate when adding the precipitant. Because PO 4 3– is a weakbase, the precipitate is soluble in a strongly acidic solution. If theprecipitant is added under neutral or basic conditions (high RSS) theresulting precipitate consists of smaller, less pure particles. Increasingthe pH by adding base allows the precipitate to form under morefavorable (low RSS) conditions.3. Why is the acid–base indicator methyl red added to the solution?The indicator’s color change, which occurs at a pH of approximately6.3, indicates when there is sufficient NH 3 to neutralize the HCladded at the beginning of the procedure. The amount of NH 3 is crucialto this procedure. If we add insufficient NH 3 , then the solutionis too acidic, which increases the precipitate’s solubility and leads to anegative determinate error. If we add too much NH 3 , the precipitatemay contain traces of Mg(NH 4 ) 4 (PO 4 ) 2 , which, on drying, formsMg(PO 3 ) 2 instead of Mg 2 P 2 O 7 . This increases the mass of the ignitedprecipitate, giving a positive determinate error. After adding enoughNH 3 to neutralize the HCl, we add the additional 5 mL of NH 3 toquantitatively precipitate MgNH 4 PO 4 • 6H 2 O.4. Explain why forming Mg(PO 3 ) 2 instead of Mg 2 P 2 O 7 increases theprecipitate’s mass.

372 Analytical Chemistry 2.0Each mole of Mg 2 P 2 O 7 contains two moles of phosphorous and eachmole of Mg(PO 3 ) 2 contains only one mole of phosphorous A conservationof mass, therefore, requires that two moles of Mg(PO 3 ) 2 formin place of each mole of Mg 2 P 2 O 7 . One mole of Mg 2 P 2 O 7 weighs222.6 g. Two moles of Mg(PO 3 ) 2 weigh 364.5 g. Any replacementof Mg 2 P 2 O 7 with Mg(PO 3 ) 2 must increase the precipitate’s mass.5. What additional steps, beyond those discussed in questions 2 and 3,help improve the precipitate’s purity?Two additional steps in the procedure help in forming a precipitatethat is free of impurities: digestion and reprecipitation.6. Why is the precipitate rinsed with a solution of 5% v/v NH 3 ?This is done for the same reason that precipitation is carried out inan ammonical solution; using dilute ammonia minimizes solubilitylosses when rinsing the precipitate.8B.2 Quantitative ApplicationsAlthough no longer a commonly used technique, precipitation gravimetrystill provides a reliable means for assessing the accuracy of other methodsof analysis, or for verifying the composition of standard reference materials.In this section we review the general application of precipitation gravimetryto the analysis of inorganic and organic compounds.In o r g a n i c An a l y s i sTable 8.1 provides a summary of some precipitation gravimetric methodsfor inorganic cations and anions. Several methods for the homogeneousgeneration of precipitants are shown in Table 8.2. The majority of inorganicprecipitants show poor selectivity for the analyte. Many organic precipitants,however, are selective for one or two inorganic ions. Table 8.3 listsseveral common organic precipitants.Precipitation gravimetry continues to be listed as a standard method forthe determination of SO 4 2– in water and wastewater analysis. 8 Precipitationis carried out using BaCl 2 in an acidic solution (adjusted with HCl to a pHof 4.5–5.0) to prevent the possible precipitation of BaCO 3 or Ba 3 (PO 4 ) 2 ,and near the solution’s boiling point. The precipitate is digested at 80–90 o Cfor at least two hours. Ashless filter paper pulp is added to the precipitate toaid in filtration. After filtering, the precipitate is ignited to constant weightat 800 o C. Alternatively, the precipitate is filtered through a fine porosityfritted glass crucible (without adding filter paper pulp), and dried toconstant weight at 105 o C. This procedure is subject to a variety of errors,including occlusions of Ba(NO 3 ) 2 , BaCl 2 , and alkali sulfates.8 Method 4500-SO 42– C and Method 4500-SO42– D as published in Standard Methods for theExamination of Waters and Wastewaters, 20th Ed., American Public Health Association: Washington,D. C., 1998.

Chapter 8 Gravimetric Methods373Table 8.1 Selected Precipitation Gravimetric Methods for InorganicCations and Anions (Arranged by Precipitant)Analyte Precipitant Precipitate Formed Precipitate WeighedBa 2+ (NH 4 ) 2 CrO 4 BaCrO 4 BaCrO 4Pb 2+ K 2 CrO 4 PbCrO 4 PbCrO 4Ag + HCl AgCl AgCl2+Hg 2 HCl Hg 2 Cl 2 Hg 2 Cl 2Al 3+ NH 3 Al(OH) 3 Al 2 O 3Be 2+ NH 3 Be(OH) 2 BeOFe 3+ NH 3 Fe(OH) 3 Fe 2 O 3Ca 2+ (NH 4 ) 2 C 2 O 4 CaC 2 O 4 CaCO 3 or CaOSb 3+ H 2 S Sb 2 S 3 Sb 2 S 3As 3+ H 2 S As 2 S 3 As 2 S 3Hg 2+ H 2 S HgS HgSBa 2+ H 2 SO 4 BaSO 4 BaSO 4Pb 2+ H 2 SO 4 PbSO 4 PbSO 4Sr 2+ H 2 SO 4 SrSO 4 SrSO 4Be 3+ (NH 4 ) 2 HPO 4 NH 4 BePO 4 Be 2 P 2 O 7Mg 2+ (NH 4 ) 2 HPO 4 NH 4 MgPO 4 Mg 2 P 2 O 7Zn 2+ (NH 4 ) 2 HPO 4 NH 4 ZnPO 4 Zn 2 P 2 O 7Sr 2+ KH 2 PO 4 SrHPO 4 Sr 2 P 2 O 7CN – AgNO 3 AgCN AgCNI – AgNO 3 AgI AgIBr – AgNO 3 AgBr AgBrCl – AgNO 3 AgCl AgCl–ClO 3 FeSO 4 /AgNO 3 AgCl AgClSCN – SO 2 /CuSO 4 CuSCN CuSCN2–SO 4 BaCl 2 BaSO 4 BaSO 4Or g a n i c An a l y s i sSeveral organic functional groups or heteroatoms can be determined usingprecipitation gravimetric methods. Table 8.4 provides a summary of severalrepresentative examples. Note that the procedure for the determination ofalkoxy functional groups is an indirect analysis.Qu a n t i t a t i ve Ca l c u l at i o n sThe stoichiometry of a precipitation reaction provides a mathematical relationshipbetween the analyte and the precipitate. Because a precipitationgravimetric method may involve several chemical reactions before the precipitationreaction, knowing the stoichiometry of the precipitation reaction

374 Analytical Chemistry 2.0Table 8.2 Reactions for the Homogeneous Preparation of SelectedInorganic PrecipitantsPrecipitantReactionOH – + −( NH )CO( aq) + 3H O() l 2NH ( aq) + CO ( g) + 2OH( aq)2 2 24 22– + + 2−SO 4NH HSO ( aq) + 2H O() l NH ( aq) + H O ( aq ) + SO ( aq)2 3 2 4 34S 2– CH CSNH ( aq) + HO() l CH CONH ( aq) + HS( aq )3 2 2 3 22– −IO 3HOCH CH OH( aq) + IO ( aq) HCHO( aq) + HO()l + IO− ( aq) 22 2 4 233–PO 4(CH O)PO( aq) + 3H O() l 3CHOH( aq) + H PO ( aq )3 3 2 3 3 42–C 2 O 4(C H)CO( aq) + 2HO() l 2CHOH( aq ) + HCO ( aq)2 5 2 2 4 2 2 5 2 2 42– −2−CO 3Cl CCOOH( aq) + 2OH ( aq) CHCl ( aq) + CO ( aq)+ H3 3 32 O() lTable 8.3 Selected Precipitation Gravimetric Methods for Inorganic Ions Using anOrganic PrecipitantAnalyte Precipitant StructurePrecipitateFormedPrecipitateWeighedNi 2+dimethylglyoximeHONNOHNi(C 4 H 7 O 2 N 2 ) 2 Ni(C 4 H 7 O 2 N 2 ) 2NOFe 3+cupferronNONH 4+Fe(C 6 H 5 N 2 O 2 ) 3 Fe 2 O 3Cu 2+Co 2+cupron1-nitrso-2-naptholC 6 H 5 C 6 H 5CuC 14 H 11 O 2 N CuC 14 H 11 O 2 NHON OHNOOHCo(C 10 H 6 O 2 N) 3 Co or CoSO 4K + sodium tetraphenylborate Na[B(C 6 H 5 ) 4 ] K[B(C 6 H 5 ) 4 K[B(C 6 H 5 ) 4NO 3–nitronNC 6 H 5NCC 20 H 16 N 4 HNO 20 H 16 N 4 H-3NNO 3C 6 H 5 C 6 H 5

Chapter 8 Gravimetric Methods375Table 8.4 Selected Precipitation Gravimetric Methods for the Analysis ofOrganic Functional Groups and HeteroatomsAnalyte Treatment Precipitant PrecipitateOrganic halidesR–XX = Cl, Br, IOxidation with HNO 3 in the presence of Ag + AgNO 3 AgXOrganic halidesR–XX = Cl, Br, Imay not be sufficient. Even if you do not have a complete set of balancedchemical reactions, you can deduce the mathematical relationship betweenthe analyte and the precipitate using a conservation of mass. The followingexample demonstrates the application of this approach to the direct analysisof a single analyte.Example 8.1To determine the amount of magnetite, Fe 3 O 4 , in an impure ore, a 1.5419-g sample is dissolved in concentrated HCl, giving a mixture of Fe 2+ andFe 3+ . After adding HNO 3 to oxidize Fe 2+ to Fe 3+ and diluting with water,Fe 3+ is precipitated as Fe(OH) 3 by adding NH 3 . Filtering, rinsing, andigniting the precipitate provides 0.8525 g of pure Fe 2 O 3 . Calculate the%w/w Fe 3 O 4 in the sample.So l u t i o nA conservation of mass requires that all the iron from the ore is found in theFe 2 O 3 . We know there are 2 moles of Fe per mole of Fe 2 O 3 (FW = 159.69g/mol) and 3 moles of Fe per mole of Fe 3 O 4 (FW = 231.54 g/mol); thus0.8525 gFeO232mol Fe 231.54gFeO3× ×159.69 gFeO 3 molFeThe % w/w s Fe 3 O 4 in the sample, therefore, is0.82405 gFeO1.5419 gsampleCombustion in O 2 (with a Pt catalyst) in thepresence of Ag + AgNO 3 AgXOrganic sulfurOxidation with HNO 3 in the presence ofBa 2+ BaCl 2 BaSO 4Organic sulfurCombustion in O 2 (with Pt catalyst) withSO 2 and SO 3 collected in dilute H 2 O 2BaCl 2 BaSO 4Alkoxy groups–O–R or –COO–RR= CH 3 or C 2 H 5Reaction with HI to produce RI AgNO 3 AgI3243× 100 = 53. 44%w/w Fe O4= 0.82405 gFeO3 434

376 Analytical Chemistry 2.0Practice Exercise 8.2A 0.7336-g sample of an alloy containing copper and zinc is dissolved in 8M HCl and diluted to 100 mL in a volumetric flask. In one analysis, thezinc in a 25.00-mL portion of the solution is precipitated as ZnNH 4 PO 4 ,and subsequently isolated as Zn 2 P 2 O 7 , yielding 0.1163 g. The copperin a separate 25.00-mL portion of the solution is treated to precipitateCuSCN, yielding 0.2931 g. Calculate the %w/w Zn and the %w/w Cuin the sample.Click here to review your answer to this exercise.In Practice Exercise 8.2 the sample contains two analytes. Because wecan selectively precipitate each analyte, finding their respective concentrationsis a straightforward stoichiometric calculation. But what if we cannotprecipitate the two analytes separately? To find the concentrations of bothanalytes, we still need to generate two precipitates, at least one of whichmust contain both analytes. Although this complicates the calculations, wecan still use a conservation of mass to solve the problem.Example 8.2A 0.611-g sample of an alloy containing Al and Mg is dissolved and treatedto prevent interferences by the alloy’s other constituents. Aluminum andmagnesium are precipitated using 8-hydroxyquinoline, providing a mixedprecipitate of Al(C 9 H 6 NO) 3 and Mg(C 9 H 6 NO) 2 that weighs 7.815 g.Igniting the precipitate converts it to a mixture of Al 2 O 3 and MgO thatweighs 1.002 g. Calculate the %w/w Al and %w/w Mg in the alloy.So l u t i o nThe masses of the solids provide us with the following two equations.gAl(C HNO) + gMg(C HNO) = 7.815 g9 6 3 9 6 2gAlO + gMgO =1 . 002 g2 3With two equations and four unknowns, we need two additional equationsto solve the problem. A conservation of mass requires that all thealuminum in Al(C 9 H 6 NO) 3 is found in Al 2 O 3 ; thusgAlO1 molAl= gAl(C HNO) 3×459.45 gAl(C HNO)2 3 9 6gAlO9 6= 0.11096×gAl(C HNO)32 3 9 63101.96 gAlO×2 molAlO2 32 3Using the same approach, a conservation of mass for magnesium gives

Chapter 8 Gravimetric Methods3771 molMggMgO = gMg(C HNO) ×9 6 2312 61 gMg(C HNO)gMgO= 0.12893 ×.9 6 2gMg(C HNO)9 6 240 304× .gMgOmolMgSubstituting the equations for g MgO and g Al 2 O 3 into the equation forthe combined weights of MgO and Al 2 O 3 leaves us with two equationsand two unknowns.gAl(C HNO) + gMg(C HNO) = 7.815 g9 6 3 9 6 20. 11096× gAl(C HNO) + 0.12893 × gMg(C HNO)= 1.002 g9 6 39 6 2Multiplying the first equation by 0.11096 and subtracting the secondequation gives− 0. 01797× gMg(C HNO) =−0.1348 g9 6 2gMg(C HNO) = 7.501 g9 6 2gAl(C HNO) = 7. 815 g− 7. 501 gMg(C HNO) = 0.314 g9 6 3 9 6 2Now we can finish the problem using the approach from Example 8.1. Aconservation of mass requires that all the aluminum and magnesium in thesample of Dow metal is found in the precipitates of Al(C 9 H 6 NO) 3 and theMg(C 9 H 6 NO) 2 . For aluminum, we find that1 molAl0. 314 gAl(C HNO) 3×9 6459.45 gAl(C H NO)9 6 326.982 gAl× =0.01844 gAlmolAland for magnesium we have0.01844 gAl× 100 = 302 . % w/wAl0.611 gsample1 molMg7. 501 gMg(C HNO) 2×9 6312.61 gMg(C H NO)9 6 224.305 gMg× =0.5832 gMgmolMg0.5832 gMg× 100 = 95.%50.611 gsamplew/wMg

378 Analytical Chemistry 2.0Although you can write the balanced reactionsfor any analysis, applying conservationprinciples can save you a significantamount of time!Practice Exercise 8.3A sample of a silicate rock weighing 0.8143 g is brought into solutionand treated to yield a 0.2692-g mixture of NaCl and KCl. The mixture ofchloride salts is subsequently dissolved in a mixture of ethanol and water,and treated with HClO 4 , precipitating 0.5713 g of KClO 4 . What is the%w/w Na 2 O in the silicate rock?Click here to review your answer to this exercise.The previous problems are examples of direct methods of analysis becausethe precipitate contains the analyte. In an indirect analysis the precipitateforms as a result of a reaction with the analyte, but the analyte isnot part of the precipitate. As shown by the following example, despite theadditional complexity, we can use conservation principles to organize ourcalculations.Example 8.3An impure sample of Na 3 PO 3 weighing 0.1392 g is dissolved in 25 mLof water. A solution containing 50 mL of 3% w/v HgCl 2 , 20 mL of 10%w/v sodium acetate, and 5 mL of glacial acetic acid is then prepared. Addingthe solution of Na 3 PO 3 to the solution containing HgCl 2 , oxidizesPO 3 3– to PO 4 3– , precipitating Hg 2 Cl 2 . After digesting, filtering, and rinsingthe precipitate, 0.4320 g of Hg 2 Cl 2 is obtained. Report the purity ofthe original sample as % w/w Na 3 PO 3 .So l u t i o nThis is an example of an indirect analysis because the precipitate, Hg 2 Cl 2 ,does not contain the analyte, Na 3 PO 3 . Although the stoichiometry of thereaction between Na 3 PO 3 and HgCl 2 is given earlier in the chapter, let’ssee how we can solve the problem using conservation principles.The reaction between Na 3 PO 3 and HgCl 2 is a redox reaction in whichphosphorous increases its oxidation state from +3 in Na 3 PO 3 to +5 inNa 3 PO 4 , and in which mercury decreases its oxidation state from +2 inHgCl 2 to +1 in Hg 2 Cl 2 . A redox reaction must obey a conservation ofelectrons—all the electrons released by the reducing agent, Na 3 PO 3 , mustbe accepted by the oxidizing agent, HgCl 2 . Knowing this, we write thefollowing stoichiometric conversion factors:−2mol emolNaPO−1mol eandmolHgCl3 42Now we are ready to solve the problem. First, we use a conservation of massfor mercury to convert the precipitate’s mass to the moles of HgCl 2 .

Chapter 8 Gravimetric Methods3790.4320 gHgCl222 molHg× ×472.09 gHgCl1 mol HgClmolHg222= 18302 . × 10 −3molHgCl2Next, we use the conservation of electrons to find the mass of Na 3 PO 3 .−−31 mol e 1mol Na PO3 318302 . × 10 molHgCl × ××2molHgCl 2mol e −147.94 gNaPO23molNaPOFinally, we calculate the %w/w Na 3 PO 3 in the sample.0.13538 gNaPO0.1392 gsample33333= 0.13538× 100 = 97. 26%w/wNaPO3 3gNaPO33As you become comfortable with usingconservation principles, you will see opportunitiesfor further simplifying problems.For example, a conservation of electronsrequires that the electrons released byNa 3 PO 3 end up in the product, Hg 2 Cl 2 ,yielding the following stoichiometric conversionfactor:2mole −molHgCl2 2This conversion factor provides a directlink between the mass of Hg 2 Cl 2 and themass of Na 3 PO 3 .Practice Exercise 8.4One approach for determining phosphate, PO 4 3– , is to precipitate it asammonium phosphomolybdate, (NH 4 ) 3 PO 4 •12MoO 3 . After isolatingthe precipitate by filtration, it is dissolved in acid and the molybdate precipitatedand weighed as PbMoO 3 . Suppose we know that our samplescontain at least 12.5% Na 3 PO 4 and we need to recover a minimum of0.600 g of PbMoO 3 ? What is the minimum amount of sample neededfor each analysis?Click here to review your answer to this exercise.8B.2 Qualitative ApplicationsA precipitation reaction is a useful method for identifying inorganic andorganic analytes. Because a qualitative analysis does not require quantitativemeasurements, the analytical signal is simply the observation that aprecipitate has formed. Although qualitative applications of precipitationgravimetry have been largely replaced by spectroscopic methods of analysis,they continue to find application in spot testing for the presence of specificanalytes. 9Any of the precipitants listed in Table 8.1,Table 8.3, and Table 8.4 can be used for aqualitative analysis.8B.3 Evaluating Precipitation GravimetrySc a l e o f Op e r a t i o nThe scale of operation for precipitation gravimetry is limited by the sensitivityof the balance and the availability of sample. To achieve an accuracy9 Jungreis, E. Spot Test Analysis; 2nd Ed., Wiley: New York, 1997.

380 Analytical Chemistry 2.0of ±0.1% using an analytical balance with a sensitivity of ±0.1 mg, wemust isolate at least 100 mg of precipitate. As a consequence, precipitationgravimetry is usually limited to major or minor analytes, in macro or mesosamples (see Figure 3.5 in Chapter 3). The analysis of trace level analytes ormicro samples usually requires a microanalytical balance.Ac c u r a c yProblem 8.27 provides an example of howto determine an analyte’s concentration byestablishing an empirical relationship betweenthe analyte and the precipitate.For a macro sample containing a major analyte, a relative error of 0.1–0.2%is routinely achieved. The principle limitations are solubility losses, impuritiesin the precipitate, and the loss of precipitate during handling. Whenit is difficult to obtain a precipitate that is free from impurities, it may bepossible to determine an empirical relationship between the precipitate’smass and the mass of the analyte by an appropriate calibration.PrecisionThe relative precision of precipitation gravimetry depends on the sample’ssize and the precipitate’s mass. For a smaller amount of sample or precipitate,a relative precision of 1–2 ppt is routinely obtained. When workingwith larger amounts of sample or precipitate, the relative precision can beextended to several ppm. Few quantitative techniques can achieve this levelof precision.SensitivityEquation 8.13 assumes that we have useda suitable blank to correct the signal forany contributions of the reagent to theprecipitate’s mass.For any precipitation gravimetric method we can write the following generalequation relating the signal (grams of precipitate) to the absolute amountof analyte in the samplegramsprecipitate= k × grams analyte8.13where k, the method’s sensitivity, is determined by the stoichiometry betweenthe precipitate and the analyte. Consider, for example, the determinationof Fe as Fe 2 O 3 . Using a conservation of mass for iron, the precipitate’smass isgFeO231 molFe FW Fe O2= gFe× ×AW Fe 2 molFe3and the value of k isk = 1 ×2FW Fe O2 38.14AW FeAs we can see from equation 8.14, there are two ways to improve a method’ssensitivity. The most obvious way to improve sensitivity is to increase theratio of the precipitate’s molar mass to that of the analyte. In other words,it helps to form a precipitate with the largest possible formula weight. Aless obvious way to improve a method’s sensitivity is indicated by the term

Chapter 8 Gravimetric Methods381of 1/2 in equation 8.14, which accounts for the stoichiometry betweenthe analyte and precipitate. We can also improve sensitivity by forming aprecipitate that contains fewer units of the analyte.Practice Exercise 8.5Suppose you wish to determine the amount of iron in a sample. Which ofthe following compounds—FeO, Fe 2 O 3 , or Fe 3 O 4 —provides the greatestsensitivity?Click here to review your answer to this exercise.Se l e c t i v i t yDue to the chemical nature of the precipitation process, precipitants areusually not selective for a single analyte. For example, silver is not a selectiveprecipitant for chloride because it also forms precipitates with bromide andiodide. Interferents are often a serious problem and must be considered ifaccurate results are to be obtained.Tim e , Co s t, a n d Eq u i pm e n tPrecipitation gravimetry is time intensive and rarely practical if you havea large number of samples to analyze. However, because much of the timeinvested in precipitation gravimetry does not require an analyst’s immediatesupervision, it may be a practical alternative when working with onlya few samples. Equipment needs are few—beakers, filtering devices, ovensor burners, and balances—inexpensive, routinely available in most laboratories,and easy to maintain.8CVolatilization GravimetryA second approach to gravimetry is to thermally or chemically decomposethe sample and measure the resulting change in its mass. Alternatively, wecan trap and weigh a volatile decomposition product. Because the releaseof a volatile species is an essential part of these methods, we classify themcollectively as volatilization gravimetric methods of analysis.8C.1 Theory and PracticeWhether an analysis is direct or indirect, volatilization gravimetry usuallyrequires that we know the products of the decomposition reaction. Thisis rarely a problem for organic compounds, which typically decompose toform simple gases such as CO 2 , H 2 O, and N 2 . For an inorganic compound,however, the products often depend on the decomposition temperature.ThermogravimetryOne method for determining the products of a thermal decomposition isto monitor the sample’s mass as a function of temperature, a process called

382 Analytical Chemistry 2.0thermogravimetry. Figure 8.10 shows a typical thermogram in whicheach change in mass—each “step” in the thermogram—represents the lossof a volatile product. As shown in Example 8.4, we can use a thermogramto identify a compound’s decomposition reactions.Example 8.4The thermogram in Figure 8.10 shows the mass of a sample of calciumoxalate monohydrate, CaC 2 O 4 •H 2 O, as a function of temperature. Theoriginal sample weighing 17.61 mg was heated from room temperature to1000 o C at a rate of 20 o C per minute. For each step in the thermogram,identify the volatilization product and the solid residue that remains.So l u t i o nFrom 100–250 o C the sample loses 17.61 mg – 15.44 mg, or 2.17 mg,which is217 . mg100 12.%317.61mg × =of the sample’s original mass. In terms of CaC 2 O 4 •H 2 O, this correspondsto a loss of0. 123× 146. 11 g/mol=18.0 g/mol1817.61 mgCaC 2 O 4 •H 2 O1615.44 mgWeight (mg)141212.06 mg1086.76 mg60 200 400 600 800 1000Temperature (°C)Figure 8.10 Thermogram for CaC 2 O 4 •H 2 O obtained by heating a sample from roomtemperature to 1000 o C at a rate of 20 o C/min. Each change in mass results fromthe loss of a volatile product. The sample’s initial mass and its mass after each lossare shown by the dotted lines. See Example 8.4 for information on interpreting thisthermogram.

Chapter 8 Gravimetric Methods383The product’s molar mass and the temperature range for the decomposition,suggest that this is a loss of H 2 O(g), leaving a residue of CaC 2 O 4 .The loss of 3.38 mg from 350–550 o C is a 19.2% decrease in the sample’soriginal mass, or a loss of0. 192× 146. 11 g/mol=28.1 g/molwhich is consistent with the loss of CO(g) and a residue of CaCO 3 .Finally, the loss of 5.30 mg from 600-800 o C is a 30.1% decrease in thesample’s original mass, or a loss of0. 301× 146. 11 g/mol=44.0 g/molThis loss in molar mass is consistent with the release of CO 2 (g), leaving afinal residue of CaO.Identifying the products of a thermal decomposition provides informationthat we can use to develop an analytical procedure. For example,the thermogram in Figure 8.10 shows that we must heat a precipitate ofCaC 2 O 4 •H 2 O to a temperature between 250 and 400 o C if we wish toisolate and weigh CaC 2 O 4 . Alternatively, heating the sample to 1000 o Callows us to isolate and weigh CaO.Practice Exercise 8.6Under the same conditions as Figure 8.10, the thermogram for a 22.16mg sample of MgC 2 O 4 •H 2 O shows two steps: a loss of 3.06 mg from100–250 o C and a loss of 12.24 mg from 350–550 o C. For each step,identify the volatilization product and the solid residue that remains.Using your results from this exercise and the results from Example 8.4,explain how you can use thermogravimetry to analyze a mixture containingCaC 2 O 4 •H 2 O and MgC 2 O 4 •H 2 O. You may assume that other componentsin the sample are inert and thermally stable below 1000 o C.Click here to review your answer to this exercise.Eq u i pm e n tDepending on the method of analysis, the equipment for volatilizationgravimetry may be simple or complex. In the simplest experimental design,we place the sample in a crucible and decompose it at a fixed temperatureusing a Bunsen burner, a Meker burner, a laboratory oven, or a muffle furnace.The sample’s mass and the mass of the residue are measured using ananalytical balance.Trapping and weighing the volatile products of a thermal decompositionrequires specialized equipment. The sample is placed in a closed containerand heated. As decomposition occurs, a stream of an inert purge-gas sweepsthe volatile products through one or more selective absorbent traps.

384 Analytical Chemistry 2.0(a)balance(b)heatexchangerhookbalance panfurnacegas linethermocouplefurnaceplatformFigure 8.11 (a) Instrumentation for conducting a thermogravimetric analysis. The balance sits on the top of theinstrument with the sample suspended below. A gas line supplies an inert gas that sweeps the volatile decompositionproducts out of the furnace. The heat exchanger dissipates the heat from the furnace to a reservoir of water. (b) Closeupshowing the balance pan, which sits on a moving platform, the thermocouple for monitoring temperature, ahook for lowering the sample pan into the furnace, and the opening to the furnace. After placing a small portion ofthe sample on the balance pan, the platform rotates over the furnace and transfers the balance pan to a hook that issuspended from the balance. Once the balance pan is in place, the platform rotates back to its initial position. Thebalance pan and the thermocouple are then lowered into the furnace.The best way to appreciate the theoreticaland practical details discussed in this sectionis to carefully examine a typical volatilizationgravimetric method. Althougheach method is unique, the determinationof Si in ores and alloys by forming volatileSiF 4 provides an instructive example of atypical procedure. The description here isbased on a procedure from Young, R. S.Chemical Analysis in Extractive Metallurgy,Griffen: London, 1971, pp. 302–304.In a thermogravimetric analysis, the sample is placed on a small balancepan attached to one arm of an electromagnetic balance (Figure 8.11). Thesample is lowered into an electric furnace and the furnace’s temperatureis increased at a fixed rate of few degrees per minute while continuouslymonitoring the sample’s weight. The instrument usually includes a gas linefor purging the volatile decomposition products out of the furnace, and aheat exchanger to dissipate the heat emitted by the furnace.Representative Method 8.2Description o f Me t h o dDetermination of Si in Ores and AlloysSilicon is determined by dissolving the sample in acid and dehydratingto precipitate SiO 2 . Because a variety of other insoluble oxides also form,the precipitate’s mass is not a direct measure of the amount of silicon inthe sample. Treating the solid residue with HF results in the formation ofvolatile SiF 4 . The decrease in mass following the loss of SiF 4 provides anindirect measure of the amount of silicon in the original sample.Pr o c e d u r eTransfer a sample of between 0.5 and 5.0 g to a platinum crucible alongwith an excess of Na 2 CO 3 , and heat until a melt forms. After cooling,

Chapter 8 Gravimetric Methods385dissolve the residue in dilute HCl. Evaporate the solution to dryness ona steam bath and heat the residue, which contains SiO 2 and other solids,for one hour at 110 o C. Moisten the residue with HCl and repeat the dehydration.Remove any acid soluble materials from the residue by adding50 mL of water and 5 mL of concentrated HCl. Bring the solution to aboil and filter through #40 filter paper. Wash the residue with hot 2% v/vHCl followed by hot water. Evaporate the filtrate to dryness twice and,following the same procedure, treat to remove any acid-soluble materials.Combine the two precipitates, and dry and ignite to a constant weight at1200 o C. After cooling, add 2 drops of 50% v/v H 2 SO 4 and 10 mL of HF.Remove the volatile SiF 4 by evaporating to dryness on a hot plate. Finally,bring the residue to constant weight by igniting at 1200 o C.#40 filter paper is a medium speed, ashlessfilter paper for filtering crystalline solids.Qu e s t i o n s1. According to the procedure the sample should weigh between 0.5 and5.0 g. How should you decide upon the amount of sample to use?In this procedure the critical measurement is the decrease in massfollowing the volatilization of SiF 4 . The reaction responsible for theloss of mass isSiO () s + 4HF( aq) → SiF ( g) + 2HO()l2 42Water and any excess HF are removed during the final ignition, anddo not contribute to the change in mass. The loss in mass, therefore,is equivalent to the mass of SiO 2 present after the dehydration step.Every 0.1 g of Si in the original sample results in the loss of 0.21 g ofSiO 2 .How much sample we use depends on what is an acceptable uncertaintywhen measuring mass. A 0.5-g sample that is 50% w/w in Si,for example, will lose 0.53 g. If we are using a balance that measuresmass to the nearest ±0.1 mg, then the relative uncertainty in mass isapproximately ±0.02%; this is a reasonable level of uncertainty fora gravimetric analysis. A 0.5 g sample that is only 5% w/w Si experiencesa weight loss of only 0.053 g and has a relative uncertainty of±0.2%. In this case a larger sample is needed.2. Why are acid-soluble materials removed before treating the dehydratedresidue with HF?Any acid-soluble materials in the sample will react with HF or H 2 SO 4 .If the products of these reactions are volatile, or if they decomposeat 1200 o C, then the change in mass is not due solely to the volatilizationof SiF 4 . As a result, we overestimate the amount of Si in oursample.Problem 8.31 asks you to verify that thisloss of mass is correct.

386 Analytical Chemistry 2.03. Why is H 2 SO 4 added with the HF?Many samples containing silicon also contain aluminum and iron,which, when dehydrating the sample, form Al 2 O 3 and Fe 2 O 3 . Theseoxides are potential interferents because they also form volatile fluorides.In the presence of H 2 SO 4 , however, aluminum and iron preferentiallyform non-volatile sulfates. These sulfates eventually decomposeback to their respective oxides when we heat the residue to1200 o C. As a result, the change in weight after treating with HF andH 2 SO 4 is due only to the loss of SiF 4 .8C.2 Quantitative ApplicationsUnlike precipitation gravimetry, which is rarely used as a standard methodof analysis, volatilization gravimetric methods continue to play an importantrole in chemical analysis. Several important examples are discussedbelow.In o r g a n i c An a l y s i sDetermining the inorganic ash content of an organic material, such as apolymer, is an example of a direct volatilization gravimetric analysis. Afterweighing the sample, it is placed in an appropriate crucible and the organicmaterial carefully removed by combustion, leaving behind the inorganicash. The crucible containing the residue is heated to a constant weight usingeither a burner or an oven before determining the mass of the inorganicash.Another example of volatilization gravimetry is the determination ofdissolved solids in natural waters and wastewaters. In this method, a sampleof water is transferred to a weighing dish and dried to a constant weight ateither 103–105 o C or at 180 o C. Samples dried at the lower temperatureretain some occluded water and lose some carbonate as CO 2 . The loss of organicmaterial, however, is minimal. At the higher temperature, the residueis free from occluded water, but the loss of carbonate is greater. In addition,some chloride, nitrate, and organic material is lost through thermal decomposition.In either case, the residue that remains after drying to a constantweight at 500 o C is the amount of fixed solids in the sample, and the loss inmass provides an indirect measure of the sample’s volatile solids.Indirect analyses based on the weight of residue remaining after volatilizationare commonly used in determining moisture in a variety of products,and in determining silica in waters, wastewaters, and rocks. Moistureis determined by drying a preweighed sample with an infrared lamp or alow temperature oven. The difference between the original weight and theweight after drying equals the mass of water lost.

Chapter 8 Gravimetric Methods387Or g a n i c An a l y s i sThe most important application of volatilization gravimetry is for the elementalanalysis of organic materials. During combustion with pure O 2 ,many elements, such as carbon and hydrogen, are released as gaseous combustionproducts, such as CO 2 (g) and H 2 O(g). Passing the combustionproducts through preweighed tubes containing selective absorbents andmeasuring the increases in mass provides a direct analysis for the mass ofcarbon and hydrogen in the organic material.Alkaline metals and earths in organic materials can be determined byadding H 2 SO 4 to the sample before combustion. After combustion is complete,the metal remains behind as a solid residue of metal sulfate. Silver,gold, and platinum can be determined by burning the organic sample,leaving a metallic residue of Ag, Au, or Pt. Other metals are determined byadding HNO 3 before combustion, leaving a residue of the metal oxide.Volatilization gravimetry is also used to determine biomass in watersand wastewaters. Biomass is a water quality index that provides an indicationof the total mass of organisms contained within a sample of water.A known volume of the sample is passed through a preweighed 0.45-mmmembrane filter or a glass-fiber filter, and dried at 105 o C for 24 h. Theresidue’s mass provides a direct measure of biomass. If samples are knownto contain a substantial amount of dissolved inorganic solids, the residuecan be ignited at 500 o C for one hour, volatilizing the biomass. The resultinginorganic residue is wetted with distilled water to rehydrate any clayminerals and dried to a constant weight at 105 o C. The difference in massbefore and after ignition provides an indirect measure of biomass.Instead of measuring mass, modern instrumentsfor completing an elementalanalysis use gas chromatography (Chapter12) or infrared spectroscopy (Chapter 10)to monitor the gaseous decompositionproducts.Qu a n t i t a t i ve Ca l c u l at i o n sFor some applications, such as determining the amount of inorganic ashin a polymer, a quantitative calculation is straightforward and does notrequire a balanced chemical reaction. For other applications, however, therelationship between the analyte and the analytical signal depends upon thestoichiometry of any relevant reactions. Once again, a conservation of massis useful when solving problems.Example 8.5A 101.3-mg sample of an organic compound containing chlorine is combustedin pure O 2 and the volatile gases collected in absorbent traps. Thetrap for CO 2 increases in mass by 167.6 mg and the trap for H 2 O showsa 13.7-mg increase. A second sample of 121.8 mg is treated with concentratedHNO 3 producing Cl 2 , which subsequently reacts with Ag + , forming262.7 mg of AgCl. Determine the compound’s composition, as well as itsempirical formula.

388 Analytical Chemistry 2.0So l u t i o nA conservation of mass requires that all the carbon in the organic compoundmust be in the CO 2 produced during combustion; thus167. 6 mgCO21 gCO21 molC× ×1000 mg CO 44.011 gCO212.011 gC 1000 mg C× =molC gC2×45.4mgC745.74 mg C× 100 = 45. 15%w/wC101.3 mg sampleUsing the same approach for hydrogen and chlorine, we find that1 gHO22mol H13. 7 mgHO× ××21000 mg HO 18.015gHO221.008 gH 1000 mg H× = 1. 533 mg HmolH gH1.533 mg H× 100 = 151 . % w/wH101.3 mg sample1262. 7 mgAgCl × gAgCl 1mol Cl1000 mg AgCl× ×143.32 gAgCl35.453 gCl 1000 mg Cl×molCl gCl64.98 mg Cl× 100 = 53. 35%w/wCl121.8 mg sample= 64.98mg ClAdding together the weight percents for C, H, and Cl gives a total of100.01%; thus, the compound contains only these three elements. To determinethe compound’s empirical formula we note that a gram of samplecontains 0.4515 g of C, 0.0151 g of H and 0.5335 g of Cl. Expressing eachelement in moles gives 0.0376 moles C, 0.0150 moles H and 0.0150 molesCl. Hydrogen and chlorine are present in a 1:1 molar ratio. The molar ratioof C to moles of H or Cl ismolesCmolesHmolesC= = 0 . 0376molesCl 0.0150= 251 . ≈2.5Thus, the simplest, or empirical formula for the compound is C 5 H 2 Cl 2 .

Chapter 8 Gravimetric Methods389In an indirect volatilization gravimetric analysis, the change in thesample’s weight is proportional to the amount of analyte. Note that in thefollowing example it is not necessary to apply a conservation of mass torelate the analytical signal to the analyte.Example 8.6A sample of slag from a blast furnace is analyzed for SiO 2 by decomposinga 0.5003-g sample with HCl, leaving a residue with a mass of 0.1414 g.After treating with HF and H 2 SO 4 , and evaporating the volatile SiF 4 , aresidue with a mass of 0.0183 g remains. Determine the %w/w SiO 2 inthe sample.So l u t i o nIn this procedure the difference in the residue’s mass before and after volatilizingSiF 4 gives the mass of SiO 2 in the sample; thus the sample containsand the %w/w SiO 2 is0. 1414 g− 0. 0183 g=0.1231 gSiO 20.1231 gSiO0.5003 g2× 100 = 24. 61%w/wSiO2Practice Exercise 8.7Heating a 0.3317-g mixture of CaC 2 O 4 and MgC 2 O 4 yields a residue of0.1794 g at 600 o C and a residue of 0.1294 g at 1000 o C. Calculate the%w/w CaC 2 O 4 in the sample. You may wish to review your answer toPractice Exercise 8.6 as you consider this problem.Click here to review your answer to this exercise.Finally, in some quantitative applications we can compare the result fora sample to a similar result obtained using a standard.Example 8.7A 26.23-mg sample of MgC 2 O 4 •H 2 O and inert materials is heated to constantweight at 1200 o C, leaving a residue weighing 20.98 mg. A sampleof pure MgC 2 O 4 •H 2 O, when treated in the same fashion, undergoes a69.08% change in its mass. Determine the %w/w MgC 2 O 4 •H 2 O in thesample.So l u t i o nThe change in the sample’s mass is 5.25 mg, which corresponds to

390 Analytical Chemistry 2.0Alternatively, you can determine that thefinal product of the decomposition isMgO (see Practice Exercise 8.6) and usea conservation of mass for Mg to arrive atthe same answer.100.0 mg MgC OiHO2 4 2525 . mg lost×= 760 . mg MgC OiHO2 4 269.08 mg lostThe %w/w MgC 2 O 4 •H 2 O in the sample is760 . mg MgC OiHO2 4 2× 100 = 29.%0 w/wMgC O iH O2 4 226.23 mg sample8C.3 Evaluating Volatilization GravimetryThe scale of operation, accuracy, and precision of a gravimetric volatilizationmethod is similar to that described in Section 8B.4 for precipitationgravimetry. The sensitivity of a direct analysis is fixed by the analyte’schemical form following combustion or volatilization. We can improve thesensitivity of an indirect analysis by choosing conditions that give the largestpossible change in mass. For example, the thermogram in Figure 8.10shows us that an indirect analysis for CaC 2 O 4 •H 2 O is more sensitive if wemeasure the change in mass following ignition at 1000 o C than if we ignitethe sample at 300 o C.Selectivity is not a problem for a direct analysis if we trap the analyteusing a selective absorbent trap. A direct analysis based on the residue’sweight following combustion or volatilization is possible if the residue onlycontains the analyte of interest. As noted earlier, an indirect analysis is onlyfeasible when the change in mass results from the loss of a single volatileproduct containing the analyte.Volatilization gravimetric methods are time and labor intensive. Equipmentneeds are few, except when combustion gases must be trapped, or fora thermogravimetric analysis, when specialized instrumentation is needed.A particulate is any tiny portion of matter,whether it is a speck of dust, a globuleof fat, or a molecule of ammonia. Forparticulate gravimetry we simply need amethod for collecting the particles and abalance for measuring their mass.8DParticulate GravimetryPrecipitation and volatilization gravimetric methods require that the analyte,or some other species in the sample, participate in a chemical reaction.In a direct precipitation gravimetric analysis, for example, we convert asoluble analyte into an insoluble form that precipitates from solution. Insome situations, however, the analyte is already present as in a particulateform that is easy to separate from its liquid, gas, or solid matrix. When sucha separation is possible, we can determine the analyte’s mass without relyingon a chemical reaction.8D.1 Theory and PracticeThere are two methods for separating a particulate analyte from its matrix.The most common method is filtration, in which we separate solid par-

Chapter 8 Gravimetric Methods391ticulates from their gas, liquid, or solid matrix. A second method, which isuseful for gas particles, solutes, and solids, is an extraction.Fi l t r a t i o nTo separate solid particulates from their matrix we use gravity or applysuction from a vacuum pump or aspirator to pull the sample through afilter. The type of filter we use depends upon the size of the solid particlesand the sample’s matrix. Filters for liquid samples are constructed from avariety of materials, including cellulose fibers, glass fibers, cellulose nitrate,and polytetrafluoroethylene (PTFE). Particle retention depends on the sizeof the filter’s pores. Cellulose fiber filter papers range in pore size from 30mm to 2–3 mm. Glass fiber filters, manufactured using chemically inertborosilicate glass, are available with pore sizes between 2.5 mm and 0.3 mm.Membrane filters, which are made from a variety of materials, includingcellulose nitrate and PTFE, are available with pore sizes from 5.0 mm to0.1 mm.Solid aerosol particulates are collected using either a single-stage or amultiple-stage filter. In a single-stage system, we pull the gas through a singlefilter, retaining particles larger than the filter’s pore size. When collectingsamples from a gas line, we place the filter directly in the line. Atmosphericgases are sampled with a high volume sampler that uses a vacuum pumpto pull air through the filter at a rate of approximately 75 m 3 /h. In eithercase, the filtering media for liquid samples also can be used to collect aerosolparticulates. In a multiple-stage system, a series of filtering units separatesand the particles in two or more size ranges.The particulates in a solid matrix are separated by size using one or moresieves (Figure 8.12). Sieves are available in a variety of mesh sizes rangingfrom approximately 25 mm to 40 mm. By stacking sieves of different meshsize, we can isolate particulates into several narrow size ranges. Using thesieves in Figure 8.12, for example, we can separate a solid into particles withdiameters >1700 mm, with diameters between 1700 mm and 500 mm, withdiameters between 500 mm and 250mm, and those with a diameter

392 Analytical Chemistry 2.0(a)(b)(c)(d)Figure 8.13 Four possible mechanisms for the solid-state extraction of an analyte:(a) adsorption onto a solid substrate; (b) absorption into a thin polymer film orchemical film coated on a solid substrate; (c) metal–ligand complexation in whichthe ligand is covalently bound to the solid substrate using an organic tether; and(d) antibody–antigen binding in which the receptor is covalently bound to thesolid substrate using an organic tether.If you own a wristwatch, there is a goodchance that its operation relies on a quartzcrystal. The piezoelectric properties ofquartz were discovered in 1880 by Paul-Jacques Currie and Pierre Currie. Becausethe oscillation frequency of a quartz crystalis so precise, it quickly found use inthe keeping of time. The first quartz clockwas built in 1927 at the Bell Telephonelabs, and Seiko introduced the first quartzwristwatches in 1969.Another method for extracting an analyte from its matrix is by adsorptiononto a solid substrate, by absorption into a thin polymer or chemicalfilm coated on a solid substrate, or by chemically binding to a suitablereceptor that is covalently bound to a solid substrate (Figure 8.13). Adsorption,absorption, and binding occur at the interface between the solutioncontaining the analyte and the substrate’s surface, the thin film, or thereceptor. Although the amount of extracted analyte is too small to measureusing a conventional balance, it can be measured using a quartz crystalmicrobalance.The measurement of mass using a quartz crystal microbalancetakes advantage of the piezoelectric effect. 10 The application of an alternatingelectrical field across a quartz crystal induces an oscillatory vibrationalmotion in the crystal. Every quartz crystal vibrates at a characteristic resonantfrequency that depends on the crystal’s properties, including the massper unit area of any material coated on the crystal’s surface. The change inmass following adsorption, absorption, or binding of the analyte, therefore,can be determined by monitoring the change in the quartz crystal’s characteristicresonant frequency. The exact relationship between the change infrequency and mass is determined by a calibration curve.8D.2 Quantitative ApplicationsParticulate gravimetry is important in the environmental analysis of water,air, and soil samples. The analysis for suspended solids in water samples,for example, is accomplished by filtering an appropriate volume of a well-10 (a) Ward, M. D.; Buttry, D. A. Science 1990, 249, 1000–1007; (b) Grate, J. W.; Martin, S. J. ;White, R. M. Anal. Chem. 1993, 65, 940A–948A; (c) Grate, J. W.; Martin, S. J. ; White, R. M.Anal. Chem. 1993, 65, 987A–996A.

Chapter 8 Gravimetric Methods393mixed sample through a glass fiber filter and drying the filter to constantweight at 103–105 o C.The microbiological testing of water also uses particulate gravimetry.One example is the analysis for coliform bacteria in which an appropriatevolume of sample is passed through a sterilized 0.45-mm membrane filter.The filter is placed on a sterilized absorbent pad saturated with a culturingmedium and incubated for 22–24 hours at 35 ± 0.5 o C. Coliform bacteriaare identified by the presence of individual bacterial colonies that form duringthe incubation period (Figure 8.14). As with qualitative applications ofprecipitation gravimetry, the signal in this case is a visual observation ratherthan a measurement of mass.Total airborne particulates are determined using a high-volume airsampler equipped with either cellulose fiber or glass fiber filters. Samplesfrom urban environments require approximately 1 h of sampling time, butsamples from rural environments require substantially longer times.Grain size distributions for sediments and soils are used to determinethe amount of sand, silt, and clay in a sample. For example, a grain size of2 mm serves as the boundary between gravel and sand. The grain size forthe sand–silt and the silt–clay boundaries are 1/16 mm and 1/256 mm,respectively.Several standard quantitative analytical methods for agricultural productsare based on measuring the sample’s mass following a selective solventextraction. For example, the crude fat content in chocolate can be determinedby extracting with ether for 16 hours in a Soxhlet extractor. Afterthe extraction is complete, the ether is allowed to evaporate and the residueis weighed after drying at 100 o C. This analysis has also been accomplishedindirectly by weighing a sample before and after extracting with supercriticalCO 2 .Quartz crystal microbalances equipped with thin film polymer filmsor chemical coatings have found numerous quantitative applications inenvironmental analysis. Methods have been reported for the analysis of avariety of gaseous pollutants, including ammonia, hydrogen sulfide, ozone,sulfur dioxide, and mercury. Biochemical particulate gravimetric sensorsalso have been developed. For example, a piezoelectric immunosensor hasbeen developed that shows a high selectivity for human serum albumin, andis capable of detecting microgram quantities. 11Figure 8.14 Colonies of fecal coliformbacteria from a water supply.Source: Susan Boyer (www.ars.usda.gov).Qu a n t i t a t i ve Ca l c u l at i o n sThe result of a quantitative analysis by particulate gravimetry is just the ratio,using appropriate units, of the amount of analyte relative to the amountof sample.11 Muratsugu, M.; Ohta, F.; Miya, Y.; Hosokawa, T.; Kurosawa, S.; Kamo, N.; Ikeda, H. Anal.Chem. 1993, 65, 2933–2937.

394 Analytical Chemistry 2.0Example 8.8A 200.0-mL sample of water was filtered through a pre-weighed glass fiberfilter. After drying to constant weight at 105 o C, the filter was foundto have increased in mass by 48.2 mg. Determine the sample’s total suspendedsolids.So l u t i o nA ppm is equivalent to a mg of analyte per liter of solution; thus, the totalsuspended solids for the sample is48.2 mg solids= 241 ppm solids0.2000 Lsample8D.3 Evaluating Particulate GravimetryThe scale of operation and detection limit for particulate gravimetrycan be extended beyond that of other gravimetric methods by increasingthe size of the sample taken for analysis. This is usually impracticable forother gravimetric methods because of the difficulty of manipulating a largersample through the individual steps of the analysis. With particulate gravimetry,however, the part of the sample that is not analyte is removedwhen filtering or extracting. Consequently, particulate gravimetry is easilyextended to the analysis of trace-level analytes.Except for methods relying on a quartz crystal microbalance, particulategravimetry uses the same balances as other gravimetric methods, andis capable of achieving similar levels of accuracy and precision. Since particulategravimetry is defined in terms of the mass of the particle itself, thesensitivity of the analysis is given by the balance’s sensitivity. Selectivity, onthe other hand, is determined either by the filter’s pore size, or by the propertiesof the extracting phase. Because it requires a single step, particulategravimetric methods based on filtration generally require less time, laborand capital than other gravimetric methods.As you review this chapter, try to define akey term in your own words. Check youranswer by clicking on the key term, whichwill take you to the page where it was firstintroduced. Clicking on the key termthere, will bring you back to this page sothat you can continue with another keyterm.8E Key Termscoagulation conservation of mass coprecipitatedefinitive technique digestion direct analysiselectrogravimetry gravimetry homogeneous precipitationignition inclusion indirect analysisocclusion particulate gravimetry peptizationprecipitant precipitation gravimetry quartz crystal microbalancerelative supersaturation reprecipitation supernatantsurface adsorbate thermogram thermogravimetryvolatilization gravimetry

Chapter 8 Gravimetric Methods3958FChapter SummaryIn a gravimetric analysis, a measurement of mass or a change in mass providesquantitative information about the analyte. The most common formof gravimetry uses a precipitation reaction to generate a product whosemass is proportional to the amount of analyte. In many cases the precipitateincludes the analyte; however, an indirect analysis in which the analytecauses the precipitation of another compound also is possible. Precipitationgravimetric procedures must be carefully controlled to produce precipitatesthat are easy to filter, free from impurities, and of known stoichiometry.In volatilization gravimetry, thermal or chemical energy decomposesthe sample containing the analyte. The mass of residue remaining afterdecomposition, the mass of volatile products collected with a suitable trap,or a change in mass due to the loss of volatile material are all gravimetricmeasurements.When the analyte is already present in a particulate form that is easyto separate from its matrix, then a particulate gravimetric analysis may befeasible. Examples include the determination of dissolved solids and thedetermination of fat in foods.8GProblems1. Starting with the equilibrium constant expressions for reaction 8.1, andreactions 8.3–8.5, verify that equation 8.7 is correct.Answers, but not worked solutions, tomost end-of-chapter problems are availablehere.2. Equation 8.7 explains how the solubility of AgCl varies as a functionof the equilibrium concentration of Cl – . Derive a similar equation thatdescribes the solubility of AgCl as a function of the equilibrium concentrationof Ag + . Graph the resulting solubility function and compareit to that shown in Figure 8.1.3. Construct a solubility diagram for Zn(OH) 2 that takes into accountthe following soluble zinc-hydroxide complexes: Zn(OH) + , Zn(OH) 3 – ,and Zn(OH) 4 2– . What is the optimum pH for quantitatively precipitatingZn(OH) 2 ? For your solubility diagram, plot log(S) on the y-axisand pH on the x-axis. See the appendices for relevant equilibrium constants.4. Starting with equation 8.10, verify that equation 8.11 is correct.5. For each of the following precipitates, use a ladder diagram to identifythe pH range where the precipitates has its lowest solubility? See theappendices for relevant equilibrium constants.a. CaC 2 O 4 b. PbCrO 4 c. BaSO 4d. SrCO 3 e. ZnS

396 Analytical Chemistry 2.06. Mixing solutions of 1.5 M KNO 3 and 1.5 M HClO 4 produces a whiteprecipitate of KClO 4 . If permanganate ions are present, an inclusionof KMnO 4 is possible. Impure precipitates of KClO 4 are purple if aninclusion of KMnO 4 is present. Shown below are descriptions of twoexperiments in which KClO 4 is precipitated in the presence of MnO 4 – .Explain why the experiments lead to the different results shown in Figure8.15.(a)(b)Figure 8.15 Results for the experimentsin Problem 8.6. (a) Experiment1; (b) Experiment 2.Experiment 1. Place 1 mL of 1.5 M KNO 3 in a test tube, add 3drops of 0.1 M KMnO 4 , and swirl to mix. Add 1 mL of 1.5 MHClO 4 dropwise, agitating the solution between drops. Destroythe excess KMnO 4 by adding 0.1 M NaHSO 3 dropwise. The resultingprecipitate of KClO 4 has an intense purple color.Experiment 2. Place 1 mL of 1.5 M HClO 4 in a test tube, add3 drops of 0.1 M KMnO 4 , and swirl to mix. Add 1 mL of 1.5 MKNO 3 dropwise, agitating the solution between drops. Destroy theexcess KMnO 4 by adding 0.1 M NaHSO 3 dropwise. The resultingprecipitate of KClO 4 has a pale purple in color.7. Mixing solutions of Ba(SCN) 2 and MgSO 4 produces a precipitate ofBaSO 4 . Shown below are the descriptions and results for three experimentsusing different concentrations of Ba(SCN) 2 and MgSO 4 . Explainwhy these experiments produce different results.Experiment 1. When equal volumes of 3.5 M Ba(SCN) 2 and 3.5M MgSO 4 are mixed, a gelatinous precipitate immediately forms.Experiment 2. When equal volumes of 1.5 M Ba(SCN) 2 and 1.5M MgSO 4 are mixed, a curdy precipitate immediately forms. Individualparticles of BaSO 4 can be seen as points under a magnificationof 1500(a particle size less than 0.2 mm).Experiment 3. When equal volumes of 0.5 mM Ba(SCN) 2 and0.5 mM MgSO 4 are mixed, the complete precipitation of BaSO 4requires 2–3 h. Individual crystals of BaSO 4 obtain lengths of approximately5 mm.8. Aluminum is determined gravimetrically by precipitating Al(OH) 3 andisolating Al 2 O 3 . A sample containing approximately 0.1 grams of Al isdissolved in 200 mL of H 2 O, and 5 grams of NH 4 Cl and a few dropsof methyl red indicator are added (methyl red is red at pH levels below4 and yellow at pH levels above 6). The solution is heated to boiling and1:1 NH 3 is added dropwise till the indicator turns yellow, precipitatingAl(OH) 3 . The precipitate is held at the solution’s boiling point for severalminutes before filtering and rinsing with a hot solution of 2% w/vNH 4 NO 3 . The precipitate is then ignited at 1000–1100 o C, formingAl 2 O 3 .

Chapter 8 Gravimetric Methods397(a) Cite two ways in which this procedure encourages the formation oflarger particles of precipitate.(b) The ignition step must be carried out carefully to ensure the quantitativeconversion of Al(OH) 3 to Al 2 O 3 . What is the effect of anincomplete conversion on the %w/w Al?(c) What is the purpose of adding NH 4 Cl and methyl red indicator?(d) An alternative procedure involves isolating and weighing the precipitateas the 8-hydroxyquinolate, Al(C 9 H 6 ON) 3 . Why might thisbe a more advantageous form of Al for a gravimetric analysis? Arethere any disadvantages?9. Calcium is determined gravimetrically by precipitating CaC 2 O 4 •H 2 Oand isolating CaCO 3 . After dissolving a sample in 10 mL of water and15 mL of 6 M HCl, the resulting solution is heated to boiling and awarm solution of excess ammonium oxalate is added. The solution ismaintained at 80 o C and 6 M NH 3 is added dropwise, with stirring,until the solution is faintly alkaline. The resulting precipitate and solutionare removed from the heat and allowed to stand for at least onehour. After testing the solution for completeness of precipitation, thesample is filtered, rinsed with 0.1% w/v ammonium oxalate, and driedat 100–120 o C for 1 hour. The precipitate is transferred to a mufflefurnace where it is converted to CaCO 3 by drying at 500 ± 25 o C untilconstant weight.(a) Why is the precipitate of CaC 2 O 4 •H 2 O converted to CaCO 3 ?(b) In the final step, if the sample is heated at too high of a temperaturesome CaCO 3 may be converted to CaO. What effect would thishave on the reported %w/w Ca?(c) Why is the precipitant, (NH 4 ) 2 C 2 O 4 , added to a hot, acidic solutioninstead of a cold, alkaline solution?10. Iron is determined gravimetrically by precipitating as Fe(OH) 3 andigniting to Fe 2 O 3 . After dissolving a sample in 50 mL of H 2 O and 10mL of 6 M HCl, any Fe 2+ is converted Fe 3+ by oxidizing with 1–2 mLof concentrated HNO 3 . The sample is heated to remove the oxides ofnitrogen and the solution is diluted to 200 mL. After bringing the solutionto a boil, Fe(OH) 3 is precipitated by slowly adding 1:1 NH 3 untilthe odor of NH 3 is detected. The solution is boiled for an additionalminute and the precipitate is allowed to settler. The precipitate is thenfiltered and rinsed with several portions of hot 1% w/v NH 4 NO 3 untilno Cl – is found in the wash water. Finally, the precipitate is ignited toconstant weight at 500–550 o C and weighed as Fe 2 O 3 .

398 Analytical Chemistry 2.0(a) If ignition is not carried out under oxidizing conditions (plentyof O 2 present), the final product may contain Fe 3 O 4 . What effectwould this have on the reported %w/w Fe?(b) The precipitate is washed with a dilute solution of NH 4 NO 3 . Whyis NH 4 NO 3 added to the wash water?(c) Why does the procedure call for adding NH 3 until the odor of ammoniais detected?(d) Describe how you might test the filtrate for Cl – .11. Sinha and Shome described a gravimetric method for molybdenum inwhich it is precipitated as MoO 2 (C 13 H 10 NO 2 ) 2 using n-benzoylphenylhydroxylamine,C 13 H 11 NO 2 , as a precipitant. 12 The precipitate isweighed after igniting to MoO 3 . As part of their study, the authors determinedthe optimum conditions for the analysis. Samples containing0.0770 g of Mo were taken through the procedure while varying thetemperature, the amount of precipitant added, and the pH of the solution.The solution volume was held constant at 300 mL for all experiments.A summary of their results are shown in the following table.Temperature( o C)Mass (g)of precipitantVolume (mL)of 10 M HClMass (g)of MoO 330 0.20 0.9 0.067530 0.30 0.9 0.101430 0.35 0.9 0.114030 0.42 0.9 0.115530 0.42 0.3 0.115030 0.42 18.0 0.115230 0.42 48.0 0.116030 0.42 75.0 0.115950 0.42 0.9 0.115675 0.42 0.9 0.115880 0.42 0.9 0.1129Considering these results, discuss the optimum conditions for determiningMo by this method. Express your results for the precipitant asthe minimum %w/v in excess, needed to ensure a quantitative precipitation.12. A sample of an impure iron ore is approximately 55% w/w Fe. Theamount of Fe in the sample is to be determined gravimetrically byisolating it as Fe 2 O 3 . What mass of sample do you need to ensure thatyou isolate at least 1 g of Fe 2 O 3 ?12 Sinha, S. K.; Shome, S. C. Anal. Chim. Acta 1960, 24, 33–36.

Chapter 8 Gravimetric Methods39913. The concentration of arsenic in an insecticide is determined gravimetricallyby precipitating MgNH 4 AsO 4 and isolating Mg 2 As 2 O 7 . Determinethe %w/w As 2 O 3 in a 1.627-g sample of insecticide if it yields106.5 mg of Mg 2 As 2 O 7 .14. After preparing a sample of alum, K 2 SO 4 •Al 2 (SO 4 ) 3 •24H 2 O, a studentdetermined its purity by dissolving a 1.2391-g sample and precipitatingthe aluminum as Al(OH) 3 . After filtering, rinsing, and igniting, 0.1357g of Al 2 O 3 is obtained. What is the purity of the alum preparation?15. To determine the amount of iron in a dietary supplement, a randomsample of 15 tablets weighing a total of 20.505 g was ground into a finepowder. A 3.116-g sample was dissolved and treated to precipitate theiron as Fe(OH) 3 . The precipitate was collected, rinsed, and ignited toa constant weight as Fe 2 O 3 , yielding 0.355 g. Report the iron contentof the dietary supplement as g FeSO 4 •7H 2 O per tablet.16. A 1.4639-g sample of limestone was analyzed for Fe, Ca, and Mg. Theiron was determined as Fe 2 O 3 yielding 0.0357 g. Calcium was isolatedas CaSO 4 , yielding a precipitate of 1.4058 g, and Mg was isolated as0.0672 g of Mg 2 As 2 O 7 . Report the amount of Fe, Ca, and Mg in thelimestone sample as %w/w Fe 2 O 3 , %w/w CaO, and %w/w MgO.17. The number of ethoxy groups (CH 3 CH 2 O–) in an organic compoundis determined by the following two reactions.R(CH CH ) + xHI→ R(OH) + xCH CH I2 3 xx3 2CH CH I+ Ag + H O→ AgI()s + CH CH OH3 2+ 2 3 2A 36.92-mg sample of an organic compound with an approximate molecularweight of 176 was treated in this fashion, yielding 0.1478 g ofAgI. How many ethoxy groups are there in each molecule of the compound?18. A 516.7-mg sample containing a mixture of K 2 SO 4 and (NH 4 ) 2 SO 4was dissolved in water and treated with BaCl 2 , precipitating the SO 42–as BaSO 4 . The resulting precipitate was isolated by filtration, rinsedfree of impurities, and dried to a constant weight, yielding 863.5 mg ofBaSO 4 . What is the %w/w K 2 SO 4 in the sample?19. The amount of iron and manganese in an alloy can be determinedby precipitating the metals with 8-hydroxyquinoline, C 9 H 7 NO. Afterweighing the mixed precipitate, the precipitate is dissolved and theamount of 8-hydroxyquinoline determined by another method. In atypical analysis a 127.3-mg sample of an alloy containing iron, manga-

400 Analytical Chemistry 2.0nese, and other metals was dissolved in acid and treated with appropriatemasking agents to prevent an interference from other metals. Theiron and manganese were precipitated and isolated as Fe(C 9 H 6 NO) 3and Mn(C 9 H 6 NO) 2 , yielding a total mass of 867.8 mg. The amountof 8-hydroxyquinolate in the mixed precipitate was determined to be5.276 mmol. Calculate the %w/w Fe and %w/w Mn in the alloy.20. A 0.8612-g sample of a mixture of NaBr, NaI, and NaNO 3 was analyzedby adding AgNO 3 and precipitating a 1.0186-g mixture of AgBrand AgI. The precipitate was then heated in a stream of Cl 2 , convertingit to 0.7125 g of AgCl. Calculate the %w/w NaNO 3 in the sample.A B C D EA NP Y NP WB Y W WC NP NPDWFigure 8.16 Results of the binary mixingof solutions A–E for Problem 8.22.20. The earliest determinations of elemental atomic weights were accomplishedgravimetrically. In determining the atomic weight of manganese,a carefully purified sample of MnBr 2 weighing 7.16539 g wasdissolved and the Br – precipitated as AgBr, yielding 12.53112 g. Whatis the atomic weight for Mn if the atomic weights for Ag and Br aretaken to be 107.868 and 79.904, respectively?22. While working as a laboratory assistant you prepared 0.4 M solutionsof AgNO 3 , Pb(NO 3 ) 2 , BaCl 2 , KI and Na 2 SO 4 . Unfortunately, youbecame distracted and forgot to label the solutions before leaving thelaboratory. Realizing your error, you label the solutions A–E and performall possible binary mixings of the five solutions, obtaining theresults shown in Figure 8.16 (key: NP means no precipitate formed,W means a white precipitate formed, and Y means a yellow precipitateformed). Identify solutions A–E.23. A solid sample has approximately equal amounts of two or more of thefollowing soluble salts: AgNO 3 , ZnCl 2 , K 2 CO 3 , MgSO 4 , Ba(C 2 H 3 O 2 ) 2 ,and NH 4 NO 3 . A sample of the solid, sufficient to give at least 0.04moles of any single salt, was added to 100 mL of water, yielding awhite precipitate and a clear solution. The precipitate was collectedand rinsed with water. When a portion of the precipitate was placedin dilute HNO 3 it completely dissolved, leaving a colorless solution.A second portion of the precipitate was placed in dilute HCl, yieldinga precipitate and a clear solution. Finally, the filtrate from the originalprecipitate was treated with excess NH 3 , yielding a white precipitate.Identify the salts that must be present in the sample, the salts that mustbe absent and the salts for which there is insufficient information tomake this determination. 1324. Two methods have been proposed for the analysis of sulfur in impuresamples of pyrite, FeS 2 . Sulfur can be determined in a direct analysis by13 Adapted from Sorum, C. H.; Lagowski, J. J. Introduction to Semimicro Qualitative Analysis,Prentice-Hall: Englewood Cliffs, N. J., 5th Ed., 1977, p. 285.

Chapter 8 Gravimetric Methods401oxidizing it to SO 4 2– and precipitating it as BaSO 4 . An indirect analysisis possible if the iron is precipitated as Fe(OH) 3 and isolated as Fe 2 O 3 .Which of these methods will provide a more sensitive determinationfor sulfur? What other factors should you consider in choosing betweenthese methods?25. A sample of impure pyrite known to be approximately 90–95% w/wFeS 2 is to be analyzed by oxidizing the sulfur to SO 4 2– and precipitatingit as BaSO 4 . How many grams of the sample must you take to form atleast 1 g of BaSO 4 ?26. A series of samples containing any possible combination of KCl, NaCl,and NH 4 Cl is to be analyzed by adding AgNO 3 and precipitatingAgCl. What is the minimum volume of 5% w/v AgNO 3 necessary tocompletely precipitate the chloride in any 0.5-g sample?27. If a precipitate of known stoichiometry does not form, a gravimetricanalysis is still feasible if we can establish experimentally the mole ratiobetween the analyte and the precipitate. Consider, for example, theprecipitation gravimetric analysis of Pb as PbCrO 4 . 14(a) For each gram of Pb, how many grams of PbCrO 4 should form?(b) In a study of this procedure, Grote found that 1.568 g of PbCrO 4formed for each gram of Pb. What is the apparent stoichiometrybetween Pb and PbCrO 4 ?(c) Does failing to account for the actual stoichiometry lead to a positivedeterminate error or a negative determinate error?28. Determine the uncertainty for the gravimetric analysis described inExample 8.1. The expected accuracy for a gravimetric method is 0.1–0.2%. What additional sources of error might account for the differencebetween your estimated uncertainty and the expected accuracy?29. A 38.63-mg sample of potassium ozonide, KO 3 , was heated to 70 o Cfor 1 h, undergoing a weight loss of 7.10 mg. A 29.6-mg sample ofimpure KO 3 experiences a 4.86-mg weight loss when treated undersimilar condition. What is the %w/w KO 3 in the sample?30. The water content of an 875.4-mg sample of cheese was determinedwith a moisture analyzer. What is the %w/w H 2 O in the cheese if thefinal mass was found to be 545.8 mg?31. Representative Method 8.2 describes a procedure for determining Si inores and alloys. In this analysis a weight loss of 0.21 g corresponds to0.1 g of Si. Show that this relationship is correct.14 Grote, F. Z. Anal. Chem. 1941, 122, 395–398.

402 Analytical Chemistry 2.032. The iron in an organometallic compound was determined by treatinga 0.4873-g sample with HNO 3 and heating to volatilize the organicmaterial. After ignition, the residue of Fe 2 O 3 weighed 0.2091 g.(a) What is the %w/w Fe in this compound?(b) The carbon and hydrogen in a second sample of the compoundwere determined by a combustion analysis. When a 0.5123-g samplewas carried through the analysis, 1.2119 g of CO 2 and 0.2482g of H 2 O were collected. What are the %w/w C and %w/w H inthis compound and what is the compound’s empirical formula?33. A polymer’s ash content is determined by placing a weighed sample in aPt crucible that has been previously brought to a constant weight. Thepolymer is melted under gentle heating from a Bunsen burner untilthe volatile vapor ignites. The polymer is allowed to burn until only anon-combustible residue remains. The residue is brought to constantweight at 800 o C in a muffle furnace. The following data were collectedduring the analysis of two samples of a polymer resin.Polymer A g crucible g crucible + polymer g crucible + ashreplicate 1 19.1458 21.2287 19.7717replicate 2 15.9193 17.9522 16.5310replicate 3 15.6992 17.6660 16.2909Polymer B g crucible g crucible + polymer g crucible + ashreplicate 1 19.1457 21.0693 19.7187replicate 2 15.6991 12.8273 16.3327replicate 3 15.9196 17.9037 16.5110(a) For each polymer, determine the mean and the standard deviationfor the %w/w ash.(b) Is there any evidence at a = 0.05 for a significant difference betweenthe two polymers? See the appendices for statistical tables.34. In the presence of water vapor the surface of zirconia, ZrO 2 , chemicallyadsorbs H 2 O, forming surface hydroxyls, ZrOH (additional water isphysically adsorbed as H 2 O). When heated above 200 o C, the surfacehydroxyls convert to H 2 O(g), releasing one molecule of water for everytwo surface hydroxyls. Below 200 o C only physically absorbed water islost. Nawrocki, et al. used thermogravimetry to determine the densityof surface hydroxyls on a sample of zirconia that was heated to 700 o Cand cooled in a desiccator containing humid N 2 . 15 Heating the samplefrom 200 o C to 900 o C released 0.006 g of H 2 O for every gram of dehydroxylatedZrO 2 . Given that the zirconia had a surface area of 33 m 2 /g15 Nawrocki, J.; Carr, P. W.; Annen, M. J.; Froelicher, S. Anal. Chim. Acta 1996, 327, 261–266.

Chapter 8 Gravimetric Methods403and that one molecule of H 2 O forms two surface hydroxyls, calculatethe density of surface hydroxyls in mmol/m 2 .35. The concentration of airborne particulates in an industrial workplacewas determined by pulling the air through a single-stage air samplerequipped with a glass-fiber filter. The air was sampled for 20 min ata rate of 75 m 3 /h. At the end of the sampling period, the filter’s masswas found to have increased by 345.2 mg. What is the concentrationof particulates in the air sample in mg/m 3 and mg/L?36. The fat content of potato chips is determined indirectly by weighing asample before and after extracting the fat with supercritical CO 2 . Thefollowing data were obtained for the analysis of potato chips. 16Sample Number Initial Mass (g) Final Mass (g)1 1.1661 0.92532 1.1723 0.92523 1.2525 0.98504 1.2280 0.95625 1.2837 1.0119(a) Determine the mean and standard deviation for the %w/w fat.(b) This sample of potato chips is known to have a fat content of 22.7%w/w. Is there any evidence for a determinate error at a = 0.05? Seethe appendices for statistical tables.37. Delumyea and McCleary reported results for the %w/w organic materialin sediment samples collected at different depths from a cove on theSt. Johns River in Jacksonville, FL. 17 After collecting a sediment core,they sectioned it into 2-cm increments. Each increment was treatedusing the following procedure:(a) The sediment was placed in 50 mL of deionized water and the resultingslurry filtered through preweighed filter paper;(b) The filter paper and the sediment were placed in a preweighed evaporatingdish and dried to a constant weight in an oven at 110 o C;(c) The evaporating dish containing the filter paper and the sedimentwere transferred to a muffle furnace where the filter paper and anyorganic material in the sample were removed by ashing;(d) Finally, the inorganic residue remaining after ashing was weighed.Using the following data, determine the %w/w organic material as afunction of the average depth for each increment.16 Fat Determination by SFE, ISCO, Inc. Lincoln, NE.17 Delumyea, R. D.; McCleary, D. L. J. Chem. Educ. 1993, 70, 172–173.

404 Analytical Chemistry 2.0Mass (g) of Filter Paper,Mass (g) of... Dish, and Sediment...Depth (cm) Filter Paper DishAfterDryingAfterAshing0–2 1.590 43.21 52.10 49.492–4 1.745 40.62 48.83 46.004–6 1.619 41.23 52.86 47.846–8 1.611 42.10 50.59 47.138–10 1.658 43.62 51.88 47.5310–12 1.628 43.24 49.45 45.3112–14 1.633 43.08 47.92 44.2014–16 1.630 43.96 58.31 55.5316–18 1.636 43.36 54.37 52.7538. Yao, et al. recently described a method for the quantitative analysis ofthiourea based on its reaction with I 2 . 18CS(NH ) + 4I + 6H O → (NH )SO + 8HI+CO2 2 2 2 4 2 4 2The procedure calls for placing a 100-mL aqueous sample containingthiourea in a 60-mL separatory funnel and adding 10 mL of a pH 7buffer and 10 mL of 12 mM I 2 in CCl 4 . The contents of the separatoryfunnel are shaken and the organic and aqueous layers allowed toseparate. The organic layer, which contains the excess I 2 , is transferredto the surface of a piezoelectric crystal on which a thin layer of Au hasbeen deposited. After allowing the I 2 to adsorb to the Au, the CCl 4 isremoved and the crystal’s frequency shift, Df, measured. The followingdata is reported for a series of thiourea standards.[thiourea] (M) Df (Hz) [thiourea] (M) Df (Hz)3.0010 –7 74.6 1.5010 –6 3275.0010 –7 120 2.5010 –6 5437.0010 –7 159 3.5010 –6 7899.0010 –7 205 5.0010 –6 1089(a) Characterize this method with respect to the scale of operationshown in Figure 3.5 of Chapter 3.(b) Using a regression analysis, determine the relationship between thecrystal’s frequency shift and the concentration of thiourea.(c) If a sample containing an unknown amount of thiourea gives aDf of 176 Hz, what is the molar concentration of thiourea in thesample?18 Yao, S. F.; He, F. J. Nie, L. H. Anal. Chim. Acta 1992, 268, 311–314.

Chapter 8 Gravimetric Methods4058H(d) What is the 95% confidence interval for the concentration of thioureain this sample assuming one replicate? See the appendices forstatistical tables.Solutions to Practice ExercisesPractice Exercise 8.1The solubility reaction for CaC 2 O 4 is+ 2−CaC O() s Ca ( aq) + CO ( aq ) 2 2 4 2 4pH2–C 2O 4--pK a2= 4.266HC 2O 4To minimize solubility, the pH needs to be basic enough that oxalate,C 2 O 4 2– , does not react to form HC 2 O 4 – or H 2 C 2 O 4 . The ladder diagramfor oxalic acid, including approximate buffer ranges, is shown in Figure8.17. Maintaining a pH greater than 5.3 ensures that C 2 O 4 2– is the onlyimportant form of oxalic acid in solution, minimizing the solubility ofCaC 2 O 4 .pK a1= 1.252H 2C 2O 4Figure 8.17 pH ladder diagramfor oxalic acid, H 2 C 2 O 4 .Click here to return to the chapter.Practice Exercise 8.2A conservation of mass requires that all the zinc in the alloy is found inthe final product, Zn 2 P 2 O 7 . We know that there are 2 moles of Zn permole of Zn 2 P 2 O 7 ; thus0.1163 gZnPO2 2 72 molZn× ×304.72 gZnPO2 2 76539. gZn= 4.991× 10 −2gZnmolZnThis is the mass of Zn in 25% of the sample (a 25.00 mL portion of the100.0 mL total volume). The %w/w Zn, therefore, isFor copper, we find that4.991× 10 −2gZn× 4 × 100 = 27 . 22 % w/wZn0.7336 gsample1mol Cu0.1163 gCuSCN× ×121.64 gCuSCN63.55 gCumolCu= 0.12490.1249 gCu × 4 × 100 = 67 . 88 % w/w Cu0.7336 gsampleClick here to return to the chapter.gCu

406 Analytical Chemistry 2.0Practice Exercise 8.3The masses of the solids provide us with the following equationsgNaCl+ gKCl=0.2692 g(g KClO ) + (gKClO ) =0.5713 g4 NaCl 4 KClwhere (g KClO 4 ) NaCl and (g KClO 4 ) KCl are the masses of KClO 4 fromthe reaction of HClO 4 with NaCl and KCl. With two equations andfour unknowns, we need two additional equations to solve the problem.A conservation of mass requires that all the chlorine in NaCl is found inthe (KClO 4 ) NaCl ; thusmolCl(g KClO ) = gNaCl× 1 138× . 55 gKClO4 NaCl58.44 gNaCl molCl(g KClO ) = 2.3708×gNaCl4 NaClUsing the same approach for KCl gives1 molCl 138. 55 gKClO(g KClO ) = gKCl× ×4 KCl74.55 gKCl molCl(g KClO ) = 1.8584×gKCl4 KClSubstituting the equations for (g KClO 4 ) NaCl and (g KClO 4 ) KCl into theequation for their combined masses leaves us with two equations and twounknowns.gNaCl+ gKCl=0.2692 g2. 3708× gNaCl+ 1. 8584× gKCl=0.5713 gMultiplying the first equation by 1.8584 and subtracting the second equationgives− 0. 5124× gNaCl=−0.07102 ggNaCl= 0.1386To report the %w/w Na 2 O in the sample, we use a conservation of masson sodium.1 molNa 61.98 g Na O20.1386 gNaCl× ×58.44 gNaCl 2mol Nag= 0.073500.07350 gNaO2× 100 = 9. 026%w/ wNaO0.8143 gsample2Click here to return to the chapter.44gNaO2

Chapter 8 Gravimetric Methods407Practice Exercise 8.4To find the mass of (NH 4 ) 3 PO 4 •12MoO 3 producing 0.600 g of Pb-MoO 3 , we first use a conservation of mass for molybdenum; thus0.600 gPbMoO31 molMo× ×351.14 gPbMoO31876. 38 g(NH ) PO i12MoO4 3 4 3=12 molMo0.2672 g( NH ) 3PO i12MoO4 4 3Next, to convert this mass of (NH 4 ) 3 PO 4 •12MoO 3 to a mass of Na 3 PO 4 ,we use a conservation of mass on PO 4 3– .0. 2672 g(NH ) 3PO i 12MoO×4 4 31 molPO3-41876.38 g(NH ) PO i12MoO4 3 4 3163.94 gNaPO×molPO= 0.02335 gNaPO3 43- 3 44Finally, we convert this mass of Na 3 PO 4 to the corresponding mass ofsample.0.02335 gNaPO3 4100 gsample× =0. 187 gsample12.5 gNaPO3 4A sample of 0.187 g is sufficient to guarantee 0.600 g of PbMoO 3 . If asample contains more than 12.5% Na 3 PO 4 , then a 0.187-g sample willproduce more than 0.600 g of PbMoO 3 .Click here to return to the chapter.Practice Exercise 8.5To determine which form has the greatest sensitivity, we use a conservationof mass for iron to find the relationship between the precipitate’s massand the mass of iron.1 molFe 71.85 gFeOgFeO = gFe× ×55.85 gFe mol Fe= 1.286×gFegFeO = molFegFe × 1 gFe2 355.85 gFe× 159 70 O2mol Fe.2 3gFeO = molFegFe × 231.55 gFe3 455.85 gFe×O3mol Fe13 4= 1.430×gFe= 1.382×gFe

408 Analytical Chemistry 2.0Of the three choices, the greatest sensitivity is obtained with Fe 2 O 3 becauseit provides the largest value for k.Click here to return to the chapter.Practice Exercise 8.6From 100–250 o C the sample loses 13.8% of its mass, or a loss of0. 138× 130. 35 g/mol=18.0 g/molconsistent with the loss of H 2 O(g), leaving a residue of MgC 2 O 4 . From350–550 o C the sample loses 55.23% of its original mass, or a loss of0. 5523× 130. 35 g/mol=71.99 g/molThis weight loss is consistent with the simultaneous loss of CO(g) andCO 2 (g), leaving a residue of MgO.We can analyze the mixture by heating a portion of the sample to 300 o C,600 o C, and 1000 o C, recording the mass at each temperature. The loss ofmass between 600 o C and 1000 o C, Dm 2 , is due to the loss of CO 2 (g) fromthe decomposition of CaCO 3 to CaO, and is proportional to the mass ofCaC 2 O 4 •H 2 O in the sample.1 molCO 146.11gCaC OiHO22 4 2gCaC O i H O = ∆m × ×2 4 2244.01 gCO molCOThe change in mass between 300 o C and 600 o C, Dm 1 , is due to the lossof CO(g) from CaC 2 O 4 •H 2 O and the loss of CO(g) and CO 2 (g) fromMgC 2 O 4 •H 2 O. Because we already know the amount of CaC 2 O 4 •H 2 Oin the sample, we can calculate its contribution to Dm 1 .1 molCO( ∆m 1) = gCaC O iH O××Ca 2 4 2146.11gCaC O iH O22 42228 . 01gCOmolCOThe change in mass due to the decomposition of MgC 2 O 4 •H 2 O( ∆m ) = ( ∆m ) −( ∆m)1 Mg1 1 Caprovides the mass of MgC 2 O 4 •H 2 O in the sample.gMgC O i H O = ( ∆m ) ×2 4 2 1 Mg1 mol(CO + CO ) 78. 02 g(CO + CO )22×130.35gMgC OiHO1mol (CO+CO )2 4 2Click here to return to the chapter.2

Chapter 8 Gravimetric Methods409Practice Exercise 8.7In Practice Exercise 8.6 we developed an equation for the mass ofCaC 2 O 4 •H 2 O in a mixture of CaC 2 O 4 •H 2 O, MgC 2 O 4 •H 2 O, and inertmaterials. Adapting this equation to a sample containing CaC 2 O 4 ,CaC 2 O 4 , and inert materials is easy; thusgCaC O = ( 0. 1794 g− 0.1294 g) ×2 41 molCO244.01 gCOThe %w/w CaC 2 O 4 in the sample is2128.10 gCaC O2 4× =0 1455 gCaC OmolCO2.2 40.1455 gCaC O2 4× 100 = 43. 86%w/w CaC O0.3317 gsampleClick here to return to the chapter.2 4


DRAFTChapter 9Titrimetric MethodsChapter OverviewSection 9A Overview of TitrimetrySection 9B Acid–Base TitrationsSection 9C Complexation TitrationsSection 9D Redox TitrationsSection 9E Precipitation TitrationsSection 9F Key TermsSection 9G Chapter SummarySection 9H ProblemsSection 9I Solutions to Practice ExercisesTitrimetry, in which volume serves as the analytical signal, made its first appearance as ananalytical method in the early eighteenth century. Titrimetric methods were not well received bythe analytical chemists of that era because they could not duplicate the accuracy and precisionof a gravimetric analysis. Not surprisingly, few standard texts from the 1700s and 1800s includetitrimetric methods of analysis.Precipitation gravimetry developed as an analytical method without a general theory ofprecipitation. An empirical relationship between a precipitate’s mass and the mass of analyte—what analytical chemists call a gravimetric factor—was determined experimentally by takinga known mass of analyte through the procedure. Today, we recognize this as an early exampleof an external standardization. Gravimetric factors were not calculated using the stoichiometryof a precipitation reaction because chemical formulas and atomic weights were not yetavailable! Unlike gravimetry, the development and acceptance of titrimetry required a deeperunderstanding of stoichiometry, of thermodynamics, and of chemical equilibria. By the 1900s,the accuracy and precision of titrimetric methods were comparable to that of gravimetricmethods, establishing titrimetry as an accepted analytical technique.Copyright: David Harvey, 2009411

412 Analytical Chemistry 2.0We are deliberately avoiding the termanalyte at this point in our introductionto titrimetry. Although in most titrationsthe analyte is the titrand, there are circumstanceswhere the analyte is the titrant.When discussing specific methods, we willuse the term analyte where appropriate.9AOverview of TitrimetryIn titrimetry we add a reagent, called the titrant, to a solution containinganother reagent, called the titrand, and allow them to react. The typeof reaction provides us with a simple way to divide titrimetry into thefollowing four categories: acid–base titrations, in which an acidic or basictitrant reacts with a titrand that is a base or an acid; complexometric titrationsbased on metal–ligand complexation; redox titrations, in which thetitrant is an oxidizing or reducing agent; and precipitation titrations, inwhich the titrand and titrant form a precipitate.Despite the difference in chemistry, all titrations share several commonfeatures. Before we consider individual titrimetric methods in greaterdetail, let’s take a moment to consider some of these similarities. As youwork through this chapter, this overview will help you focus on similaritiesbetween different titrimetric methods. You will find it easier to understanda new analytical method when you can see its relationship to other similarmethods.9A.1 Equivalence Points and End pointsIf a titration is to be accurate we must combine stoichiometrically equivalentamount of titrant and titrand. We call this stoichiometric mixture theequivalence point. Unlike precipitation gravimetry, where we add theprecipitant in excess, an accurate titration requires that we know the exactvolume of titrant at the equivalence point, V eq . The product of the titrant’sequivalence point volume and its molarity, M T , is equal to the moles oftitrant reacting with the titrand.molesoftitrant = M × VT eqIf we know the stoichiometry of the titration reaction, then we can calculatethe moles of titrand.Unfortunately, for most titrations there is no obvious sign when wereach the equivalence point. Instead, we stop adding titrant when at an endpoint of our choosing. Often this end point is a change in the color of asubstance, called an indicator, that we add to the titrand’s solution. Thedifference between the end point volume and the equivalence point volumeis a determinate titration error. If the end point and the equivalencepoint volumes coincide closely, then the titration error is insignificant andit is safely ignored. Clearly, selecting an appropriate end point is criticallyimportant.Instead of measuring the titrant’s volume,we may choose to measure its mass. Althoughwe generally can measure massmore precisely than we can measure volume,the simplicity of a volumetric titrationmakes it the more popular choice.9A.2 Volume as a SignalAlmost any chemical reaction can serve as a titrimetric method providedit meets the following four conditions. The first condition is that we mustknow the stoichiometry between the titrant and the titrand. If this is not

Chapter 9 Titrimetric Methods413the case, then we cannot convert the moles of titrant consumed in reachingthe end point to the moles of titrand in our sample. Second, the titrationreaction must effectively proceed to completion; that is, the stoichiometricmixing of the titrant and the titrand must result in their reaction. Third, thetitration reaction must occur rapidly. If we add the titrant faster than it canreact with the titrand, then the end point and the equivalence point willdiffer significantly. Finally, there must be a suitable method for accuratelydetermining the end point. These are significant limitations and, for thisreason, there are several common titration strategies.A simple example of a titration is an analysis for Ag + using thiocyanate,SCN – , as a titrant.+ −Ag ( aq) + SCN ( aq) Ag(SCN) () sThis reaction occurs quickly and with a known stoichiometry, satisfying twoof our requirements. To indicate the titration’s end point, we add a smallamount of Fe 3+ to the analyte’s solution before beginning the titration.When the reaction between Ag + and SCN – is complete, formation of thered-colored Fe(SCN) 2+ complex signals the end point. This is an exampleof a direct titration since the titrant reacts directly with the analyte.If the titration’s reaction is too slow, if a suitable indicator is not available,or if there is no useful direct titration reaction, then an indirect analysismay be possible. Suppose you wish to determine the concentration offormaldehyde, H 2 CO, in an aqueous solution. The oxidation of H 2 COby I 3–HCO( aq) + I − ( aq) + 3OH − ( aq) HCO − ( aq) + 3I− ( aq)+ 2H O2 3 22() lis a useful reaction, but it is too slow for a titration. If we add a known excessof I 3 – and allow its reaction with H 2 CO to go to completion, we can titratethe unreacted I 3 – with thiosulfate, S 2 O 3 2– .− 2− 2− −I ( aq) + 2S O ( aq) S O ( aq) + 3I( aq)3 2 34The difference between the initial amount of I 3 – and the amount in excessgives us the amount of I 3 – reacting with the formaldehyde. This is an exampleof a back titration.Calcium ion plays an important role in many environmental systems. Adirect analysis for Ca 2+ might take advantage of its reaction with the ligandethylenediaminetetraacetic acid (EDTA), which we represent here as Y 4– .Ca 2 +Y 4 −CaY2 −( aq) + ( aq) ( aq)Unfortunately, for most samples this titration does not have a useful indicator.Instead, we react the Ca 2+ with an excess of MgY 2–Ca 2 +MgY 2 −CaY 2 −Mg2 +( aq) + ( aq) ( aq) + ( aq)6Depending on how we are detecting theendpoint, we may stop the titration tooearly or too late. If the end point is a functionof the titrant’s concentration, thenadding the titrant too quickly leads to anearly end point. On the other hand, if theend point is a function of the titrant’s concentration,then the end point exceeds theequivalence point.This is an example of a precipitation titration.You will find more information aboutprecipitation titrations in Section 9E.This is an example of a redox titration. Youwill find more information about redoxtitrations in Section 9D.MgY 2– is the Mg 2+ –EDTA metal–ligandcomplex. You can prepare a solution ofMgY 2– by combining equimolar solutionsof Mg 2+ and EDTA.

414 Analytical Chemistry 2.0This is an example of a complexationtitration. You will find more informationabout complexation titrations in Section9C.releasing an amount of Mg 2+ equivalent to the amount of Ca 2+ in thesample. Because the titration of Mg 2+ with EDTAMg 2 +Y 4 −Y2 −( aq) + ( aq) Mg ( aq)has a suitable end point, we can complete the analysis. The amount ofEDTA used in the titration provides an indirect measure of the amount ofCa 2+ in the original sample. Because the species we are titrating was displacedby the analyte, we call this a displacement titration.If a suitable reaction involving the analyte does not exist it may be possibleto generate a species that we can titrate. For example, we can determinethe sulfur content of coal by using a combustion reaction to convertsulfur to sulfur dioxideS() s + O ( g) →SO( g )2 2and then convert the SO 2 to sulfuric acid, H 2 SO 4 , by bubbling it throughan aqueous solution of hydrogen peroxide, H 2 O 2 .SO HO HSO2 ( g) + 2 2 ( aq) → 2 4( aq )This is an example of an acid–base titration.You will find more informationabout acid–base titrations in Section 9B.Why a pH of 7.0 is the equivalence pointfor this titration is a topic we will cover inSection 9B.For the titration curve in Figure 9.1,the volume of titrant to reach a pH of6.8 is 24.99995 mL, a titration error of–2.0010 –4 %. Typically, we can onlyread the volume to the nearest ±0.01 mL,which means this uncertainty is too smallto affect our results.The volume of titrant to reach a pH of11.6 is 27.07 mL, or a titration error of+8.28%. This is a significant error.Titrating H 2 SO 4 with NaOHHSO ( aq) + 2NaOH( aq) 2H O() l + NaSO ( aq )2 4 2 2 4provides an indirect determination of sulfur.9A.3 Titration CurvesTo find a titration’s end point, we need to monitor some property of thereaction that has a well-defined value at the equivalence point. For example,the equivalence point for a titration of HCl with NaOH occurs at a pH of7.0. A simple method for finding the equivalence point is to continuouslymonitor the titration mixture’s pH using a pH electrode, stopping the titrationwhen we reach a pH of 7.0. Alternatively, we can add an indicator tothe titrand’s solution that changes color at a pH of 7.0.Suppose the only available indicator changes color at an end point pHof 6.8. Is the difference between the end point and the equivalence pointsmall enough that we can safely ignore the titration error? To answer thisquestion we need to know how the pH changes during the titration.A titration curve provides us with a visual picture of how a propertyof the titration reaction changes as we add the titrant to the titrand. Thetitration curve in Figure 9.1, for example, was obtained by suspending a pHelectrode in a solution of 0.100 M HCl (the titrand) and monitoring thepH while adding 0.100 M NaOH (the titrant). A close examination of thistitration curve should convince you that an end point pH of 6.8 produces anegligible titration error. Selecting a pH of 11.6 as the end point, however,produces an unacceptably large titration error.

Chapter 9 Titrimetric Methods4151412end pointpH of 11.610pH8642pH at V eq = 7.000 10 20 30 40 50V NaOH (mL)V eq = 25.0 mLend pointpH of 6.8Figure 9.1 Typical acid–base titration curve showing howthe titrand’s pH changes with the addition of titrant. Thetitrand is a 25.0 mL solution of 0.100 M HCl and the titrantis 0.100 M NaOH. The titration curve is the solid blue line,and the equivalence point volume (25.0 mL) and pH (7.00)are shown by the dashed red lines. The green dots show twoend points. The end point at a pH of 6.8 has a small titrationerror, and the end point at a pH of 11.6 has a largertitration error.The titration curve in Figure 9.1 is not unique to an acid–base titration.Any titration curve that follows the change in concentration of a species inthe titration reaction (plotted logarithmically) as a function of the titrant’svolume has the same general sigmoidal shape. Several additional examplesare shown in Figure 9.2.The titrand’s or the titrant’s concentration is not the only property wecan use when recording a titration curve. Other parameters, such as thetemperature or absorbance of the titrand’s solution, may provide a usefulend point signal. Many acid–base titration reactions, for example, areexothermic. As the titrant and titrand react the temperature of the titrand’ssolution steadily increases. Once we reach the equivalence point, furtheradditions of titrant do not produce as exothermic a response. Figure 9.3shows a typical thermometric titration curve with the intersection ofthe two linear segments indicating the equivalence point.(a)15(b)1.6(c) 101.48101.26pCd50E (V)1.040.80.620 10 20 30 40 50 0 10 20 30 40 50 0 10 20 30 40 50V EDTA (mL) V Ce 4+ (mL) V AgNO3 (mL)Figure 9.2 Additional examples of titration curves. (a) Complexation titration of 25.0 mL of 1.0 mM Cd 2+ with 1.0 mMEDTA at a pH of 10. The y-axis displays the titrand’s equilibrium concentration as pCd. (b) Redox titration of 25.0 mLof 0.050 M Fe 2+ with 0.050 M Ce 4+ in 1 M HClO 4 . The y-axis displays the titration mixture’s electrochemical potential,E, which, through the Nernst equation is a logarithmic function of concentrations. (c) Precipitation titration of 25.0 mLof 0.10 M NaCl with 0.10 M AgNO 3 . The y-axis displays the titrant’s equilibrium concentration as pAg.pAg

416 Analytical Chemistry 2.0Temperature ( o C)equivalencepointFigure 9.3 Example of a thermometric titration curveshowing the location of the equivalence point.9A.4 The BuretVolume of titrant (mL)The only essential equipment for an acid–base titration is a means for deliveringthe titrant to the titrand’s solution. The most common method fordelivering titrant is a buret (Figure 9.4). A buret is a long, narrow tube withgraduated markings, equipped with a stopcock for dispensing the titrant.The buret’s small internal diameter provides a better defined meniscus, makingit easier to read the titrant’s volume precisely. Burets are available in avariety of sizes and tolerances (Table 9.1), with the choice of buret determinedby the needs of the analysis. You can improve a buret’s accuracy bycalibrating it over several intermediate ranges of volumes using the methoddescribed in Chapter 5 for calibrating pipets. Calibrating a buret correctsfor variations in the buret’s internal diameter.A titration can be automated by using a pump to deliver the titrant ata constant flow rate (Figure 9.5). Automated titrations offer the additionaladvantage of using a microcomputer for data storage and analysis.stopcockFigure 9.4 Typical volumetric buret.The stopcock is in the openposition, allowing the titrant toflow into the titrand’s solution.Rotating the stopcock controls thetitrant’s flow rate.Table 9.1 Specifications for Volumetric BuretsVolume (mL) Class Subdivision (mL) Tolerance (mL)5 A 0.01 ±0.01B 0.01 ±0.0110 AB25 AB50 AB100 AB0.020.020.10.10.10.10.20.2±0.02±0.04±0.03±0.06±0.05±0.10±0.10±0.20

Chapter 9 Titrimetric Methods417pumptitranttitrandFigure 9.5 Typical instrumentation for an automated acid–base titration showing the titrant, the pump, andthe titrand. The pH electrode in the titrand’s solution is used to monitor the titration’s progress. You can seethe titration curve in the lower-left quadrant of the computer’s display. Modified from: Datamax (commons.wikipedia.org).9BAcid–Base TitrationsBefore 1800, most acid–base titrations used H 2 SO 4 , HCl, or HNO 3 asacidic titrants, and K 2 CO 3 or Na 2 CO 3 as basic titrants. A titration’s endpoint was determined using litmus as an indicator, which is red in acidicsolutions and blue in basic solutions, or by the cessation of CO 2 effervescencewhen neutralizing CO 3 2– . Early examples of acid–base titrimetryinclude determining the acidity or alkalinity of solutions, and determiningthe purity of carbonates and alkaline earth oxides.Three limitations slowed the development of acid–base titrimetry: thelack of a strong base titrant for the analysis of weak acids, the lack of suitableindicators, and the absence of a theory of acid–base reactivity. Theintroduction, in 1846, of NaOH as a strong base titrant extended acid–base titrimetry to the determination of weak acids. The synthesis of organicdyes provided many new indicators. Phenolphthalein, for example, wasfirst synthesized by Bayer in 1871 and used as an indicator for acid–basetitrations in 1877.Despite the increasing availability of indicators, the absence of a theoryof acid–base reactivity made it difficult to select an indicator. The developmentof equilibrium theory in the late 19th century led to significantimprovements in the theoretical understanding of acid–base chemistry, and,in turn, of acid–base titrimetry. Sørenson’s establishment of the pH scalein 1909 provided a rigorous means for comparing indicators. The determinationof acid–base dissociation constants made it possible to calculatea theoretical titration curve, as outlined by Bjerrum in 1914. For the firstThe determination of acidity and alkalinitycontinue to be important applicationsof acid–base titrimetry. We will take acloser look at these applications later inthis section.

418 Analytical Chemistry 2.0time analytical chemists had a rational method for selecting an indicator,establishing acid–base titrimetry as a useful alternative to gravimetry.9B.1 Acid–Base Titration CurvesIn the overview to this chapter we noted that a titration’s end point shouldcoincide with its equivalence point. To understand the relationship betweenan acid–base titration’s end point and its equivalence point we must knowhow the pH changes during a titration. In this section we will learn how tocalculate a titration curve using the equilibrium calculations from Chapter6. We also will learn how to quickly sketch a good approximation of anyacid–base titration curve using a limited number of simple calculations.Ti t r a t i n g St r o n g Ac i d s a n d St r o n g Ba s e sAlthough we have not written reaction 9.1as an equilibrium reaction, it is at equilibrium;however, because its equilibriumconstant is large—it is equal to (K w ) –1 or1.00 × 10 14 —we can treat reaction 9.1 asthough it goes to completion.Step 1: Calculate the volume of titrantneeded to reach the equivalence point.For our first titration curve, let’s consider the titration of 50.0 mL of 0.100M HCl using a titrant of 0.200 M NaOH. When a strong base and a strongacid react the only reaction of importance is+ −HO ( aq) + OH ( aq) →2HO()l9.13 2The first task in constructing the titration curve is to calculate the volumeof NaOH needed to reach the equivalence point, V eq . At the equivalencepoint we know from reaction 9.1 thatmoles HCl = moles NaOHM × V = M × Va a b bwhere the subscript ‘a’ indicates the acid, HCl, and the subscript ‘b’ indicatesthe base, NaOH. The volume of NaOH needed to reach the equivalencepoint isVeq= MVV = a aM)(50.0 mL)bM= ( 0.100. M= 25.0mL0 200bStep 2: Calculate pH values before theequivalence point by determining theconcentration of unreacted titrand.Before the equivalence point, HCl is present in excess and the pH isdetermined by the concentration of unreacted HCl. At the start of thetitration the solution is 0.100 M in HCl, which, because HCl is a strongacid, means that the pH is+pH =− log[ HO ] =− log[ HCl] =− log( 0. 100) = 100 .3After adding 10.0 mL of NaOH the concentration of excess HCl isinitialmoles HCl−moles NaOH added MV − MV[ HCl]==total volumeV + Va a b b( 0.100 M)(50.0 mL) −(0.200 M)(10.0 mL)== 0.0500 M50.0 mL + 10.0 mLab

Chapter 9 Titrimetric Methods419and the pH increases to 1.30.At the equivalence point the moles of HCl and the moles of NaOH areequal. Since neither the acid nor the base is in excess, the pH is determinedby the dissociation of water.Step 3: The pH at the equivalence pointfor the titration of a strong acid with astrong base is 7.00.− 14 + − + 2K w= 100 . × 10 = [ HO 3][ OH ] = [ HO 3]+[ HO ] = 10 .0× 10 −7M3Thus, the pH at the equivalence point is 7.00.For volumes of NaOH greater than the equivalence point, the pH isdetermined by the concentration of excess OH – . For example, after adding30.0 mL of titrant the concentration of OH – is[ OH ]− =molesNaOHadded−initial molesHCl MV − MVb b a a=total volumeV + V( 0.200 M)(30.0 mL) −(0.100 M)(50.0 mL)== 0.0125 M50.0 mL + 30.0 mLabStep 4: Calculate pH values after theequivalence point by determining theconcentration of excess titrant.To find the concentration of H 3 O + we use the K w expression−+Kw100 . × 10[ HO ] = =3−[ OH ] 0.0125 M14= 800 . × 10− 13 Mgiving a pH of 12.10. Table 9.2 and Figure 9.6 show additional results forthis titration curve. You can use this same approach to calculate the titrationcurve for the titration of a strong base with a strong acid, except thestrong base is in excess before the equivalence point and the strong acid isin excess after the equivalence point.Practice Exercise 9.1Construct a titration curve for the titration of 25.0 mL of 0.125 MNaOH with 0.0625 M HCl.Click here to review your answer to this exercise.Table 9.2 Titration of 50.0 mL of 0.100 M HCl with 0.200 M NaOHVolume of NaOH (mL) pH Volume of NaOH (mL) pH0.00 1.00 26.0 11.425.00 1.14 28.0 11.8910.0 1.30 30.0 12.1015.0 1.51 35.0 12.3720.0 1.85 40.0 12.5222.0 2.08 45.0 12.6224.0 2.57 50.0 12.7025.0 7.00

420 Analytical Chemistry 2.0141210pH86Figure 9.6 Titration curve for the titration of 50.0 mL of 0.100 MHCl with 0.200 M NaOH. The red points correspond to the datain Table 9.2. The blue line shows the complete titration curve.4200 10 20 30 40 50Volume of NaOH (mL)Ti t r a t i n g a We a k Ac i d w it h a St r o n g Ba s eStep 1: Calculate the volume of titrantneeded to reach the equivalence point.For this example, let’s consider the titration of 50.0 mL of 0.100 M aceticacid, CH 3 COOH, with 0.200 M NaOH. Again, we start by calculatingthe volume of NaOH needed to reach the equivalence point; thusmolesCHCOOH3=molesNaOHVeqM × V = M × Va a b b= MVV = a aM)(50.0 mL)bM= ( 0.100. M= 25.0mL0 200bStep 2: Before adding the titrant, the pHis determined by the titrand, which in thiscase is a weak acid.Before adding NaOH the pH is that for a solution of 0.100 M aceticacid. Because acetic acid is a weak acid, we calculate the pH using themethod outlined in Chapter 6.+ −CH COOH( aq) + HO() l HO ( aq) + CH COO ( aq )3 2 3 3Ka+ −[ HO ][ CH COO ]3 3( x)( x)= =[ CH COOH]0.100− x= 1.75× 10 −53Because the equilibrium constant for reaction9.2 is quite largeK = K a /K w = 1.75 10 9we can treat the reaction as if it goes tocompletion.x = [ HO+ ] = 132 . × 10 −3M3At the beginning of the titration the pH is 2.88.Adding NaOH converts a portion of the acetic acid to its conjugatebase, CH 3 COO – .−−CH COOH( aq) + OH ( aq) → HO() l + CH COO ( aq )3 2 3Any solution containing comparable amounts of a weak acid, HA, and itsconjugate weak base, A – , is a buffer. As we learned in Chapter 6, we cancalculate the pH of a buffer using the Henderson–Hasselbalch equation.9.2

Chapter 9 Titrimetric Methods421−ApH = pK + log [ ]a[ HA]Before the equivalence point the concentration of unreacted acetic acid isinitialmoles CH COOH − molesNaOH added3[ CH COOH]=3totalvolumeMV − MVa a b b=V + VabStep 3: Before the equivalence point, thepH is determined by a buffer containingthe titrand and its conjugate form.and the concentration of acetate ismolesNaOHadded[ CH COO− b b] = = MV3totalvolume V + VFor example, after adding 10.0 mL of NaOH the concentrations ofCH 3 COOH and CH 3 COO – are(0.100 M)(50.0 mL) −(0.200 M)(10.0 mL)[ CH COOH ] =350.0 mL + 10.0 mL= 0.0500 M[ CH COO ]which gives us a pH of3− =( 0.200 M)(10.0 mL)50.0 mL + 10.0 mLa= 0.03330 0333 MpH = 476 . + log . = 458 .0.0500 MbMAt the equivalence point the moles of acetic acid initially present andthe moles of NaOH added are identical. Because their reaction effectivelyproceeds to completion, the predominate ion in solution is CH 3 COO – ,which is a weak base. To calculate the pH we first determine the concentrationof CH 3 COO –molesNaOHadded[ CH COO− ] =3totalvolume( .= 0200 M)(25.0 mL)= 0.0667 M50.0 mL + 25.0 mLNext, we calculate the pH of the weak base as shown earlier in Chapter 6.−−CH COO ( aq) + H O() l OH ( aq) + CHCOOH( aq )3 2 3Step 4: The pH at the equivalence pointis determined by the titrand’s conjugateform, which in this case is a weak base.Alternatively, we can calculate acetate’s concentrationusing the initial moles of acetic acid;thus−initialmoles CH COOH3[ CH COO ]3 =totalvolume( 0.100 M)(50.0 mL)=50.0 mL + 25.0 mL= 0.0667 M

422 Analytical Chemistry 2.0Kb−[ OH ][ CH COOH]3( x)( x)= =CH COO 0.0667 − x=−[ ]35.71× 10 −10x = −−[ OH ] = 617 . × 10 6M+Kw100 . × 10[ HO ] = =3−[ OH ] 617 . × 10−14−6= 162 . × 10− 9 MStep 5: Calculate pH values after theequivalence point by determining theconcentration of excess titrant.The pH at the equivalence point is 8.79.After the equivalence point, the titrant is in excess and the titration mixtureis a dilute solution of NaOH. We can calculate the pH using the samestrategy as in the titration of a strong acid with a strong base. For example,after adding 30.0 mL of NaOH the concentration of OH – is[ OH ]− =( 0. 200 M)(30.0 mL) −(0.100 M)(50.0 mL)50.0 mL + 30.0 mL= 0.0125M−+Kw100 . × 10[ HO ] = =3−[ OH ] 0.0125 M14= 800 . × 10− 13 Mgiving a pH of 12.10. Table 9.3 and Figure 9.7 show additional resultsfor this titration. You can use this same approach to calculate the titrationcurve for the titration of a weak base with a strong acid, except the initialpH is determined by the weak base, the pH at the equivalence point byits conjugate weak acid, and the pH after the equivalence point by excessstrong acid.Practice Exercise 9.2Construct a titration curve for the titration of 25.0 mL of 0.125 M NH 3with 0.0625 M HCl.Click here to review your answer to this exercise.Table 9.3 Titration of 50.0 mL of 0.100 M Acetic Acid with 0.200 M NaOHVolume of NaOH (mL) pH Volume of NaOH (mL) pH0.00 2.88 26.0 11.425.00 4.16 28.0 11.8910.0 4.58 30.0 12.1015.0 4.94 35.0 12.3720.0 5.36 40.0 12.5222.0 5.63 45.0 12.6224.0 6.14 50.0 12.7025.0 8.79

Chapter 9 Titrimetric Methods423141210pH864200 10 20 30 40 50Volume of NaOH (mL)We can extend our approach for calculating a weak acid–strong basetitration curve to reactions involving multiprotic acids or bases, and mixturesof acids or bases. As the complexity of the titration increases, however,the necessary calculations become more time consuming. Not surprisingly,a variety of algebraic 1 and computer spreadsheet 2 approaches have beendescribed to aid in constructing titration curves.Sk e t c h i n g An Ac i d–Ba s e Ti t r a t i o n Cu r v eFigure 9.7 Titration curve for the titration of 50.0 mL of 0.100M CH 3 COOH with 0.200 M NaOH. The red points correspondto the data in Table 9.3. The blue line shows the completetitration curve.To evaluate the relationship between a titration’s equivalence point and itsend point, we need to construct only a reasonable approximation of theexact titration curve. In this section we demonstrate a simple method forsketching an acid–base titration curve. Our goal is to sketch the titrationcurve quickly, using as few calculations as possible. Let’s use the titrationof 50.0 mL of 0.100 M CH 3 COOH with 0.200 M NaOH to illustrateour approach.We begin by calculating the titration’s equivalence point volume, which,as we determined earlier, is 25.0 mL. Next we draw our axes, placing pH onthe y-axis and the titrant’s volume on the x-axis. To indicate the equivalencepoint volume, we draw a vertical line corresponding to 25.0 mL of NaOH.Figure 9.8a shows the result of the first step in our sketch.Before the equivalence point the titration mixture’s pH is determined bya buffer of acetic acid, CH 3 COOH, and acetate, CH 3 COO – . Although wecan easily calculate a buffer’s pH using the Henderson–Hasselbalch equation,we can avoid this calculation by making a simple assumption. Youmay recall from Chapter 6 that a buffer operates over a pH range extend-This is the same example that we used indeveloping the calculations for a weakacid–strong base titration curve. You canreview the results of that calculation inTable 9.3 and Figure 9.7.1 (a) Willis, C. J. J. Chem. Educ. 1981, 58, 659–663; (b) Nakagawa, K. J. Chem. Educ. 1990, 67,673–676; (c) Gordus, A. A. J. Chem. Educ. 1991, 68, 759–761; (d) de Levie, R. J. Chem. Educ.1993, 70, 209–217; (e) Chaston, S. J. Chem. Educ. 1993, 70, 878–880; (f) de Levie, R. Anal.Chem. 1996, 68, 585–590.2 (a) Currie, J. O.; Whiteley, R. V. J. Chem. Educ. 1991, 68, 923–926; (b) Breneman, G. L.; Parker,O. J. J. Chem. Educ. 1992, 69, 46–47; (c) Carter, D. R.; Frye, M. S.; Mattson, W. A. J. Chem.Educ. 1993, 70, 67–71; (d) Freiser, H. Concepts and Calculations in Analytical Chemistry, CRCPress: Boca Raton, 1992.

424 Analytical Chemistry 2.014 (a)1210141210(b)pH86pH864422000 10 20 30 40 50Volume of NaOH (mL)0 10 20 30 40 50Volume of NaOH (mL)14(c)14(d)12121010pH86pH864422000 10 20 30 40 50Volume of NaOH (mL)0 10 20 30 40 50Volume of NaOH (mL)14(e)14(f)12121010pH86pH864422000 10 20 30 40 50Volume of NaOH (mL)0 10 20 30 40 50Volume of NaOH (mL)Figure 9.8 Illustrations showing the steps in sketching an approximate titration curve for the titration of 50.0mL of 0.100 M CH 3 COOH with 0.200 M NaOH: (a) locating the equivalence point volume; (b) plottingtwo points before the equivalence point; (c) plotting two points after the equivalence point; (d) preliminaryapproximation of titration curve using straight-lines; (e) final approximation of titration curve using a smoothcurve; (f) comparison of approximate titration curve (solid black line) and exact titration curve (dashed red line).See the text for additional details.

Chapter 9 Titrimetric Methods425ing approximately ±1 pH unit on either side of the weak acid’s pK a value.The pH is at the lower end of this range, pH = pK a – 1, when the weakacid’s concentration is 10 greater than that of its conjugate weak base.The buffer reaches its upper pH limit, pH = pK a + 1, when the weak acid’sconcentration is 10 smaller than that of its conjugate weak base. Whentitrating a weak acid or a weak base, the buffer spans a range of volumesfrom approximately 10% of the equivalence point volume to approximately90% of the equivalence point volume.Figure 9.8b shows the second step in our sketch. First, we superimposeacetic acid’s ladder diagram on the y-axis, including its buffer range, usingits pK a value of 4.76. Next, we add points representing the pH at 10% ofthe equivalence point volume (a pH of 3.76 at 2.5 mL) and at 90% of theequivalence point volume (a pH of 5.76 at 22.5 mL).The third step in sketching our titration curve is to add two points afterthe equivalence point. The pH after the equivalence point is fixed by theconcentration of excess titrant, NaOH. Calculating the pH of a strong baseis straightforward, as we have seen earlier. Figure 9.8c shows the pH afteradding 30.0 mL and 40.0 mL of NaOH.Next, we draw a straight line through each pair of points, extending thelines through the vertical line representing the equivalence point’s volume(Figure 9.8d). Finally, we complete our sketch by drawing a smooth curvethat connects the three straight-line segments (Figure 9.8e). A comparisonof our sketch to the exact titration curve (Figure 9.8f) shows that they arein close agreement.Practice Exercise 9.3The actual values are 9.09% and 90.9%,but for our purpose, using 10% and 90%is more convenient; that is, after all, oneadvantage of an approximation! Problem9.4 in the end-of-chapter problems asksyou to verify these percentages.See Table 9.3 for the values.Sketch a titration curve for the titration of 25.0 mL of 0.125 M NH 3with 0.0625 M HCl and compare to the result from Practice Exercise9.2.Click here to review your answer to this exercise.As shown by the following example, we can adapt this approach toacid–base titrations, including those involving polyprotic weak acids andbases, or mixtures of weak acids and bases.Example 9.1Sketch titration curves for the following two systems: (a) the titration of50.0 mL of 0.050 M H 2 A, a diprotic weak acid with a pK a1 of 3 and a pK a2of 7; and (b) the titration of a 50.0 mL mixture containing 0.075 M HA,a weak acid with a pK a of 3, and 0.025 M HB, a weak acid with a pK a of7. For both titrations the titrant is 0.10 M NaOH.So l u t i o nFigure 9.9a shows the titration curve for H 2 A, including the ladder diagramon the y-axis, the equivalence points at 25.0 mL and 50.0 mL, twopoints before each equivalence point, two points after the last equivalence

426 Analytical Chemistry 2.014 (a)1210141210(b)pH86pH86442200 20 40 60 80 100 0 20 40 60 80 100Volume of NaOH (mL)Volume of NaOH (mL)Figure 9.9 Titration curves for Example 9.1. The red arrows show the locations of the equivalence points.point, and the straight-lines that help in sketching the final curve. Beforethe first equivalence point the pH is controlled by a buffer consistingof H 2 A and HA – . An HA – /A 2– buffer controls the pH between the twoequivalence points. After the second equivalence point the pH reflects theconcentration of excess NaOH.Figure 9.9b shows the titration curve for the mixture of HA and HB.Again, there are two equivalence points. In this case, however, the equivalencepoints are not equally spaced because the concentration of HA isgreater than that for HB. Since HA is the stronger of the two weak acidsit reacts first; thus, the pH before the first equivalence point is controlledby a buffer consisting of HA and A – . Between the two equivalence pointsthe pH reflects the titration of HB and is determined by a buffer consistingof HB and B – . After the second equivalence point excess NaOH isresponsible for the pH.Practice Exercise 9.4Sketch the titration curve for 50.0 mL of 0.050 M H 2 A, a diprotic weakacid with a pK a1 of 3 and a pK a2 of 4, using 0.100 M NaOH as thetitrant. The fact that pK a2 falls within the buffer range of pK a1 presents achallenge that you will need to consider.Click here to review your answer to this exercise.9B.2 Selecting and Evaluating the End point0Earlier we made an important distinction between a titration’s end pointand its equivalence point. The difference between these two terms is importantand deserves repeating. An equivalence point, which occurs whenwe react stoichiometrically equal amounts of the analyte and the titrant, isa theoretical not an experimental value. A titration’s end point is an experimentalresult, representing our best estimate of the equivalence point. Any

Chapter 9 Titrimetric Methods427difference between an equivalence point and its corresponding end point isa source of determinate error. It is even possible that an equivalence pointdoes not have a useful end point.Wh e r e is Th e Eq u i v a l e n c e Po i nt ?Earlier we learned how to calculate the pH at the equivalence point for thetitration of a strong acid with a strong base, and for the titration of a weakacid with a strong base. We also learned to quickly sketch a titration curvewith only a minimum of calculations. Can we also locate the equivalencepoint without performing any calculations. The answer, as you might guess,is often yes!For most acid–base titrations the inflection point, the point on a titrationcurve having the greatest slope, very nearly coincides with the equivalencepoint. 3 The red arrows in Figure 9.9, for example, indicate the equivalencepoints for the titration curves from Example 9.1. An inflection pointactually precedes its corresponding equivalence point by a small amount,with the error approaching 0.1% for weak acids or weak bases with dissociationconstants smaller than 10 –9 , or for very dilute solutions.The principal limitation to using an inflection point to locate the equivalencepoint is that the inflection point must be present. For some titrationsthe inflection point may be missing or difficult to find. Figure 9.10, forexample, demonstrates the affect of a weak acid’s dissociation constant, K a ,on the shape of titration curve. An inflection point is visible, even if barelyso, for acid dissociation constants larger than 10 –9 , but is missing when K ais 10 –11 .An inflection point also may be missing or difficult to detect if theanalyte is a multiprotic weak acid or weak base with successive dissociationconstants that are similar in magnitude. To appreciate why this is true let’sconsider the titration of a diprotic weak acid, H 2 A, with NaOH. Duringthe titration the following two reactions occur.3 Meites, L.; Goldman, J. A. Anal. Chim. Acta 1963, 29, 472–479.1412pH1086(f)(e)(d)420(c)(b)(a)0 10 20 30 40 50 60 70Volume of NaOH (mL)Figure 9.10 Weak acid–strong base titration curves for the titrationof 50.0 mL of 0.100 M HA with 0.100 M NaOH. The pK avalues for HA are (a) 1, (b) 3, (c) 5, (d) 7, (e) 9, and (f) 11.

428 Analytical Chemistry 2.0141210Maleic AcidpK a1 = 1.91pK a2 = 6.33141210Malonic AcidpK a1 = 2.85pK a2 = 5.70141210Succinic AcidpK a1 = 4.21pK a2 = 5.64pH86pH86pH8644422200 20 40 60 80Volume of NaOH (mL)00 20 40 60 80Volume of NaOH (mL)Figure 9.11 Titration curves for the diprotic weak acids maleic acid, malonic acid, and succinic acid. Each titrationcurve is for 50.0 mL of 0.0500 M weak acid using 0.100 M NaOH. Although each titration curve has equivalencepoints at 25.0 mL and 50.0 mL of NaOH, the titration curve for succinic acid shows only one inflection point.00 20 40 60 80Volume of NaOH (mL)−−HA( aq) + OH ( aq) → HA ( aq) + HO()l2 29.3− − 2−HA ( aq) + OH ( aq) → A ( aq) + H O()lTo see two distinct inflection points, reaction 9.3 must be essentially completebefore reaction 9.4 begins.Figure 9.11 shows titration curves for three diprotic weak acids. Thetitration curve for maleic acid, for which K a1 is approximately 20,000larger than K a2 , has two distinct inflection points. Malonic acid, on theother hand, has acid dissociation constants that differ by a factor of approximately690. Although malonic acid’s titration curve shows two inflectionpoints, the first is not as distinct as that for maleic acid. Finally, the titrationcurve for succinic acid, for which the two K a values differ by a factor of only27, has only a single inflection point corresponding to the neutralizationof HC 4 H 4 O 4 – to C 4 H 4 O 4 2– . In general, we can detect separate inflectionpoints when successive acid dissociation constants differ by a factor of atleast 500 (a DpK a of at least 2.7).Fi n d i n g t h e En d p o i n t w it h a n In d i c a t o rOne interesting group of weak acids and weak bases are organic dyes. Becausean organic dye has at least one highly colored conjugate acid–basespecies, its titration results in a change in both pH and color. We can usethis change in color to indicate the end point of a titration, provided thatit occurs at or near the titration’s equivalence point.Let’s use an indicator, HIn, to illustrate how an acid–base indicatorworks. Because the indicator’s acid and base forms have different colors—the weak acid, HIn, is yellow and the weak base, In – , is red—the color of asolution containing the indicator depends on their relative concentrations.The indicator’s acid dissociation reaction+ −HIn( aq) + H O() l H O ( aq) + In ( aq )2 329.4

Chapter 9 Titrimetric Methods429has an equilibrium constant of+ −[ HO ][ In ]=[ HIn]K a3Taking the negative log of each side of equation 9.5, and rearranging tosolve for pH leaves with a equation−InpH = pK + log [ ]a[ HIn]relating the solution’s pH to the relative concentrations of HIn and In – .If we can detect HIn and In – with equal ease, then the transition fromyellow to red (or from red to yellow) reaches its midpoint, which is orange,when their concentrations are equal, or when the pH is equal to the indicator’spK a . If the indicator’s pK a and the pH at the equivalence point areidentical, then titrating until the indicator turns orange is a suitable endpoint. Unfortunately, we rarely know the exact pH at the equivalence point.In addition, determining when the concentrations of HIn and In – are equalmay be difficult if the indicator’s change in color is subtle.We can establish the range of pHs over which the average analyst observesa change in the indicator’s color by making the following assumptions—theindicator’s color is yellow if the concentration of HIn is 10 greater thanthat of In – , and its color is red if the concentration of HIn is 10 smallerthan that of In – . Substituting these inequalities into equation 9.6pH = pK+ log 1 = pK−1aa109.59.6pH = pK+ log 10 = pK+ 1aa1shows that the indicator changes color over a pH range extending ±1 uniton either side of its pK a . As shown in Figure 9.12, the indicator is yellowwhen the pH is less than pK a – 1, and it is red for pHs greater than pK a + 1.pHIn –indicatoris color of In –pH = pK a,HInHInindicator’scolor transitionrangeindicatoris color of HInFigure 9.12 Diagram showing the relationship between pHand an indicator’s color. The ladder diagram defines pH valueswhere HIn and In – are the predominate species. Theindicator changes color when the pH is between pK a – 1 andpK a + 1.

430 Analytical Chemistry 2.0You may wonder why an indicator’s pHrange, such as that for phenolphthalein, isnot equally distributed around its pK a value.The explanation is simple. Figure 9.12presents an idealized view of an indicatorin which our sensitivity to the indicator’stwo colors is equal. For some indicatorsonly the weak acid or the weak base is colored.For other indicators both the weakacid and the weak base are colored, butone form is easier to see. In either case,the indicator’s pH range is skewed in thedirection of the indicator’s less coloredform. Thus, phenolphthalein’s pH rangeis skewed in the direction of its colorlessform, shifting the pH range to values lowerthan those suggested by Figure 9.12.Table 9.4 Properties of Selected Acid–Base IndicatorsIndicatorAcidColorBaseColor pH Range pK acresol red red yellow 0.2–1.8 –thymol blue red yellow 1.2–2.8 1.7bromophenol blue yellow blue 3.0–4.6 4.1methyl orange red yellow 3.1–4.4 3.7Congo red blue red 3.0–5.0 –bromocresol green yellow blue 3.8–5.4 4.7methyl red red yellow 4.2–6.3 5.0bromocresol purple yellow purple 5.2–6.8 6.1litmus red blue 5.0–8.0 –bromothymol blue yellow blue 6.0–7.6 7.1phenol red yellow blue 6.8–8.4 7.8cresol red yellow red 7.2–8.8 8.2thymol blue yellow red 8.0–9.6 8.9phenolphthalein colorless red 8.3–10.0 9.6alizarin yellow R yellow orange–red 10.1–12.0 –For pHs between pK a – 1 and pK a + 1 the indicator’s color passes throughvarious shades of orange. The properties of several common acid–base indicatorsare shown in Table 9.4.The relatively broad range of pHs over which an indicator changes colorplaces additional limitations on its feasibility for signaling a titration’s endpoint. To minimize a determinate titration error, an indicator’s entire pHrange must fall within the rapid change in pH at the equivalence point.For example, in Figure 9.13 we see that phenolphthalein is an appropriateindicator for the titration of 50.0 mL of 0.050 M acetic acid with 0.10 MNaOH. Bromothymol blue, on the other hand, is an inappropriate indicatorbecause its change in color begins before the initial sharp rise in pH,and, as a result, spans a relatively large range of volumes. The early changein color increases the probability of obtaining inaccurate results, while therange of possible end point volumes increases the probability of obtainingimprecise results.Practice Exercise 9.5Suggest a suitable indicator for the titration of 25.0 mL of 0.125 M NH 3with 0.0625 M NaOH. You constructed a titration curve for this titrationin Practice Exercise 9.2 and Practice Exercise 9.3.Click here to review your answer to this exercise.

Chapter 9 Titrimetric Methods4311211pH109phenolphthalein’spH range87623 24 25 26 27Volume of NaOH (mL)Fi n d i n g t h e En d p o i n t b y Mo n i t o r i n g pHbromothymol blue’spH rangeFigure 9.13 Portion of the titration curve for50.0 mL of 0.050 M CH 3 COOH with 0.10 MNaOH, highlighting the region containing theequivalence point. The end point transitions forthe indicators phenolphthalein and bromothymolblue are superimposed on the titration curve.An alternative approach for locating a titration’s end point is to continuouslymonitor the titration’s progress using a sensor whose signal is a functionof the analyte’s concentration. The result is a plot of the entire titrationcurve, which we can use to locate the end point with a minimal error.The obvious sensor for monitoring an acid–base titration is a pH electrodeand the result is a potentiometric titration curve. For example,Figure 9.14a shows a small portion of the potentiometric titration curve forthe titration of 50.0 mL of 0.050 M CH 3 COOH with 0.10 M NaOH, focusingon the region containing the equivalence point. The simplest methodfor finding the end point is to locate the titration curve’s inflection point,as shown by the arrow. This is also the least accurate method, particularly ifthe titration curve has a shallow slope at the equivalence point.Another method for locating the end point is to plot the titration curve’sfirst derivative, which gives the titration curve’s slope at each point alongthe x-axis. Examine Figure 9.14a and consider how the titration curve’sslope changes as we approach, reach, and pass the equivalence point. Becausethe slope reaches its maximum value at the inflection point, the firstderivative shows a spike at the equivalence point (Figure 9.14b).The second derivative of a titration curve may be more useful than thefirst derivative because the equivalence point intersects the volume axis.Figure 9.14c shows the resulting titration curve.Derivative methods are particularly useful when titrating a sample thatcontains more than one analyte. If we rely on indicators to locate the endpoints, then we usually must complete separate titrations for each analyte.If we record the titration curve, however, then a single titration is sufficient.See Chapter 11 for more details about pHelectrodes.

432 Analytical Chemistry 2.0Suppose we have the following threepoints on our titration curve:volume (mL)pH23.65 6.0023.91 6.1024.13 6.20Mathematically, we can approximate thefirst derivative as DpH/DV, where DpH isthe change in pH between successive additionsof titrant. Using the first two points,the first derivative is∆pH=610 . − 600 .= 0.385∆V23. 91−23.65which we assign to the average of the twovolumes, or 23.78 mL. For the second andthird points, the first derivative is 0.455and the average volume is 24.02 mL.volume (mL)DpH/DV23.78 0.38524.02 0.455We can approximate the second derivativeas D(DpH/DV)/DV, or D 2 pH/DV 2 . Usingthe two points from our calculation of thefirst derivative, the second derivative is2∆ pH 0. 455 − 0.385== 0.2922∆V23. 78 − 24.02Note that calculating the first derivativecomes at the expense of losing one pieceof information (three points become twopoints), and calculating the second derivativecomes at the expense of losing twopieces of information.The precision with which we can locate the end point also makes derivativemethods attractive for an analyte with a poorly defined normal titrationcurve.Derivative methods work well only if we record sufficient data duringthe rapid increase in pH near the equivalence point. This is usually not aproblem if we use an automatic titrator, such as that seen earlier in Figure9.5. Because the pH changes so rapidly near the equivalence point—achange of several pH units with the addition of several drops of titrant isnot unusual—a manual titration does not provide enough data for a usefulderivative titration curve. A manual titration does contain an abundanceof data during the more gently rising portions of the titration curve beforeand after the equivalence point. This data also contains information aboutthe titration curve’s equivalence point.Consider again the titration of acetic acid, CH 3 COOH, with NaOH.At any point during the titration acetic acid is in equilibrium with H 3 O +and CH 3 COO –+ −CH COOH( aq) + HO() l HO ( aq) + CH COO ( aq )3 2 3 3for which the equilibrium constant ispH1211109876(a)[ HO ][ CH COO ]+ −3 3K a=[ CH COOH]3ΔpH/ΔV605040302010(b)5023 24 25 26 27Volume of NaOH (mL)23 24 25 26 27Volume of NaOH (mL)4000(c)5e-05(d)4e-05Figure 9.14 Titration curves for the titrationof 50.0 mL of 0.050 M CH 3 COOHwith 0.10 M NaOH: (a) normal titrationcurve; (b) first derivative titration curve;(c) second derivative titration curve; (d)Gran plot. The red arrow shows the locationof the titration’s end point.Δ 2 pH/ΔV 220000-200023 24 25 26 27Volume of NaOH (mL)Vb×[H3O + ]3e-052e-051e-050e+0023 24 25 26 27Volume of NaOH (mL)

Chapter 9 Titrimetric Methods433Before the equivalence point the concentrations of CH 3 COOH andCH 3 COO – areinitialmoles CH COOH − molesNaOH added3[ CH COOH]=3totalvolumeMV − MVa a b b=V + VabmolesNaOHadded[ CH COO− b b] = = MV3totalvolume V + VSubstituting these equations into the K a expression and rearranging leavesus withKa+[ HO ]( MV )3 b b=MV − MVa a b b+KMV − K MV = [ HO ]( MV )a a a a b b 3 b babKMVa a aMb− KV = [ HO+ ] × Va b 3 bFinally, recognizing that the equivalence point volume isleaves us with the following equation.VeqMVa=M[ HO+ ] × V = K V − K V3 b a eq a bFor volumes of titrant before the equivalence point, a plot of V b [H 3 O + ]versus V b is a straight-line with an x-intercept of V eq and a slope of –K a .Figure 9.14d shows a typical result. This method of data analysis, whichconverts a portion of a titration curve into a straight-line, is a Gran plot.baFi n d i n g t h e En d p o i n t b y Mo n i t o r i n g Te m p e ra t u r eThe reaction between an acid and a base is exothermic. Heat generated bythe reaction is absorbed by the titrand, increasing its temperature. Monitoringthe titrand’s temperature as we add the titrant provides us with anothermethod for recording a titration curve and identifying the titration’s endpoint (Figure 9.15).Before adding titrant, any change in the titrand’s temperature is the resultof warming or cooling as it equilibrates with the surroundings. Addingtitrant initiates the exothermic acid–base reaction, increasing the titrand’stemperature. This part of a thermometric titration curve is called the titra-

434 Analytical Chemistry 2.0Excess Titrant BranchTemperatureTitration BranchFigure 9.15 Typical thermometric titration curve. The endpoint,shown by the red arrow, is found by extrapolating the titrationbranch and the excess titration branch.tion branch. The temperature continues to rise with each addition of titrantuntil we reach the equivalence point. After the equivalence point, anychange in temperature is due to the titrant’s enthalpy of dilution, and thedifference between the temperatures of the titrant and titrand. Ideally, theequivalence point is a distinct intersection of the titration branch and theexcess titrant branch. As shown in Figure 9.15, however, a thermometrictitration curve usually shows curvature near the equivalence point due toan incomplete neutralization reaction, or to the excessive dilution of thetitrand and the titrant during the titration. The latter problem is minimizedby using a titrant that is 10–100 times more concentrated than the analyte,although this results in a very small end point volume and a larger relativeerror. If necessary, the end point is found by extrapolation.Although not a particularly common method for monitoring acid–basetitrations, a thermometric titration has one distinct advantage over thedirect or indirect monitoring of pH. As discussed earlier, the use of anindicator or the monitoring of pH is limited by the magnitude of the relevantequilibrium constants. For example, titrating boric acid, H 3 BO 3 , withNaOH does not provide a sharp end point when monitoring pH because,boric acid’s K a of 5.8 10 –10 is too small (Figure 9.16a). Because boricacid’s enthalpy of neutralization is fairly large, –42.7 kJ/mole, however, itsthermometric titration curve provides a useful endpoint (Figure 9.16b).9B.3 Titrations in Nonaqueous SolventsVolume of TitrantThus far we have assumed that the titrant and the titrand are aqueous solutions.Although water is the most common solvent in acid–base titrimetry,switching to a nonaqueous solvent can improve a titration’s feasibility.For an amphoteric solvent, SH, the autoprotolysis constant, K s , relatesthe concentration of its protonated form, SH 2 + , to that of its deprotonatedform, S –0

Chapter 9 Titrimetric Methods43514(a)25.6(b)1225.5pH10864Temperature ( o C)25.425.325.2225.100 2 4 6 8 10Volume of NaOH (mL)25.00 2 4 6 8 10Volume of NaOH (mL)Figure 9.16 Titration curves for the titration of 50.0 mL of 0.050 M H 3 BO 3 with 0.50 M NaOH obtainedby monitoring (a) pH, and (b) temperature. The red arrows show the end points for the titrations.2SH SH + + S−2+ −K s= [ SH ][ S ]2You should recognize that K w is just specificform of K s when the solvent is water.and the solvent’s pH and pOH are+pH =−log[ SH ]−pOH=−log[ S ]2The most important limitation imposed by K s is the change in pH duringa titration. To understand why this is true, let’s consider the titrationof 50.0 mL of 1.010 –4 M HCl using 1.010 –4 M NaOH. Before theequivalence point, the pH is determined by the untitrated strong acid. Forexample, when the volume of NaOH is 90% of V eq , the concentration ofH 3 O + is+MV − MV[ HO ] =3V + Va a b bab−(. 10×10 4 −M)(50.0 mL) − (. 10×104 M)(45.0 mL)=50.0 mL + 45.0 mL= 53 . × 10 −6Mand the pH is 5.3. When the volume of NaOH is 110% of V eq , the concentrationof OH – is−MV − MVb b a a[ OH ] =V + Vab−(. 10× 10 4 M)(55.0 mL) −(.10×10=50.0 mL + 55.0 mL= 48 . × 10 −6M− 4M)(50.0 mL)The titration’s equivalence point requires50.0 mL of NaOH; thus, 90% of V eq is45.0 mL of NaOH.The titration’s equivalence point requires50.0 mL of NaOH; thus, 110% of V eq is55.0 mL of NaOH.

436 Analytical Chemistry 2.0and the pOH is 5.3. The titrand’s pH ispH = pK − pOH = 14 − 53 . = 87 .wand the change in the titrand’s pH as the titration goes from 90% to 110%of V eq is∆pH = 87 . − 53 . = 34 .If we carry out the same titration in a nonaqueous solvent with a K s of1.010 –20 , the pH after adding 45.0 mL of NaOH is still 5.3. However,the pH after adding 55.0 mL of NaOH isIn this case the change in pHpH = pK − pOH = 20 − 53 . = 14.7s∆pH = 14. 7− 5. 3=9.4is significantly greater than that obtained when the titration is carried outin water. Figure 9.17 shows the titration curves in both the aqueous andthe nonaqueous solvents.Another parameter affecting the feasibility of an acid–base titration isthe titrand’s dissociation constant. Here, too, the solvent plays an importantrole. The strength of an acid or a base is a relative measure of the easetransferring a proton from the acid to the solvent, or from the solvent tothe base. For example, HF, with a K a of 6.8 10 –4 , is a better proton donorthan CH 3 COOH, for which K a is 1.75 10 –5 .The strongest acid that can exist in water is the hydronium ion, H 3 O + .HCl and HNO 3 are strong acids because they are better proton donors thanH 3 O + and essentially donate all their protons to H 2 O, leveling their acid2015(b)pH10(a)5Figure 9.17 Titration curves for 50.0 mL of 1.0 10 –4M HCl using 1.0 10 –4 M NaOH in (a) water,K w = 1.0 10 –14 , and (b) a nonaqueous solvent,K s = 1.0 10 – 20 .00 20 40 60 80 100Volume of NaOH(mL)

Chapter 9 Titrimetric Methods437strength to that of H 3 O + . In a different solvent HCl and HNO 3 may notbehave as strong acids.If we place acetic acid in water the dissociation reaction+ −CH COOH( aq) + HO() l HO ( aq) + CH COO ( aq )3 2 3 3does not proceed to a significant extent because CH 3 COO – is a strongerbase than H 2 O, and H 3 O + is a stronger acid than CH 3 COOH. If we placeacetic acid in a solvent, such as ammonia, that is a stronger base than water,then the reactionCH COOH + NH NH + CH COO+ −3 3 4 3proceeds to a greater extent. In fact, both HCl and CH 3 COOH are strongacids in ammonia.All other things being equal, the strength of a weak acid increases ifwe place it in a solvent that is more basic than water, and the strength of aweak base increases if we place it in a solvent that is more acidic than water.In some cases, however, the opposite effect is observed. For example, thepK b for NH 3 is 4.75 in water and it is 6.40 in the more acidic glacial aceticacid. In contradiction to our expectations, NH 3 is a weaker base in themore acidic solvent. A full description of the solvent’s effect on the pK a ofweak acid or the pK b of a weak base is beyond the scope of this text. Youshould be aware, however, that a titration that is not feasible in water maybe feasible in a different solvent.Representative Method 9.1Description o f t h e Me t h o dDetermination of Protein in BreadThis method is based on a determination of %w/w nitrogen using theKjeldahl method. The protein in a sample of bread is oxidized to NH 4+using hot concentrated H 2 SO 4 . After making the solution alkaline, whichconverts the NH 4 + to NH 3 , the ammonia is distilled into a flask containinga known amount of HCl. The amount of unreacted HCl is determinedby a back titration with standard strong base titrant. Becausedifferent cereal proteins contain similar amounts of nitrogen, multiplyingthe experimentally determined %w/w N by a factor of 5.7 gives the %w/wprotein in the sample (on average there are 5.7 g protein for every gramof nitrogen).The best way to appreciate the theoreticaland practical details discussed in this sectionis to carefully examine a typical acid–base titrimetric method. Although eachmethod is unique, the following descriptionof the determination of protein inbread provides an instructive example ofa typical procedure. The description hereis based on Method 13.86 as published inOfficial Methods of Analysis, 8th Ed., Associationof Official Agricultural Chemists:Washington, D. C., 1955.Pr o c e d u r eTransfer a 2.0-g sample of bread, which has previously been air-dried andground into a powder, to a suitable digestion flask, along with 0.7 g of aHgO catalyst, 10 g of K 2 SO 4 , and 25 mL of concentrated H 2 SO 4 . Bringthe solution to a boil. Continue boiling until the solution turns clear andthen boil for at least an additional 30 minutes. After cooling the solution

438 Analytical Chemistry 2.0below room temperature, remove the Hg 2+ catalyst by adding 200 mL ofH 2 O and 25 mL of 4% w/v K 2 S. Add a few Zn granules to serve as boilingstones and 25 g of NaOH. Quickly connect the flask to a distillationapparatus and distill the NH 3 into a collecting flask containing a knownamount of standardized HCl. The tip of the condenser must be placedbelow the surface of the strong acid. After the distillation is complete,titrate the excess strong acid with a standard solution of NaOH usingmethyl red as an indicator (Figure 9.18).Qu e s t i o n s1. Oxidizing the protein converts all of its nitrogen to NH 4 + . Why isthe amount of nitrogen not determined by titrating the NH 4 + witha strong base?There are two reasons for not directly titrating the ammonium ion.First, because NH 4 + is a very weak acid (its K a is 5.6 10 –10 ), itstitration with NaOH yields a poorly defined end point. Second, evenif the end point can be determined with acceptable accuracy and precision,the solution also contains a substantial larger concentrationof unreacted H 2 SO 4 . The presence of two acids that differ greatly inconcentration makes for a difficult analysis. If the titrant’s concentrationis similar to that of H 2 SO 4 , then the equivalence point volumefor the titration of NH 4 + is too small to measure reliably. On theother hand, if the titrant’s concentration is similar to that of NH 4 + ,the volume needed to neutralize the H 2 SO 4 is unreasonably large.2. Ammonia is a volatile compound as evidenced by the strong smell ofeven dilute solutions. This volatility is a potential source of determinateerror. Is this determinate error negative or positive?Any loss of NH 3 is loss of nitrogen and, therefore, a loss of protein.The result is a negative determinate error.3. Discuss the steps in this procedure that minimize this determinateerror.Three specific steps minimize the loss of ammonia: (1) the solution iscooled below room temperature before adding NaOH; (2) after add-Figure 9.18 Methyl red’s endpoint for the titration of a strong acidwith a strong base; the indicator is: (a) red prior to the end point;(b) orange at the end point; and (c) yellow after the end point.

Chapter 9 Titrimetric Methods439ing NaOH, the digestion flask is quickly connected to the distillationapparatus; and (3) the condenser’s tip is placed below the surface ofthe HCl to ensure that the NH 3 reacts with the HCl before it can belost through volatilization.4. How does K 2 S remove Hg 2+ , and why is its removal important?Adding sulfide precipitates Hg 2+ as HgS. This is important becauseNH 3 forms stable complexes with many metal ions, including Hg 2+ .Any NH 3 that reacts with Hg 2+ is not collected during distillation,providing another source of determinate error.9B.4 Qu a n t i t a t i ve Ap p l i c a t i o n sAlthough many quantitative applications of acid–base titrimetry have beenreplaced by other analytical methods, a few important applications continueto be relevant. In this section we review the general application ofacid–base titrimetry to the analysis of inorganic and organic compounds,with an emphasis on applications in environmental and clinical analysis.First, however, we discuss the selection and standardization of acidic andbasic titrants.Se l e c t i n g a n d St a n d a r d i z i n g a Ti t r a n tThe most common strong acid titrants are HCl, HClO 4 , and H 2 SO 4 . Solutionsof these titrants are usually prepared by diluting a commerciallyavailable concentrated stock solution. Because the concentrations of concentratedacids are known only approximately, the titrant’s concentrationis determined by standardizing against one of the primary standard weakbases listed in Table 9.5.The most common strong base titrant is NaOH. Sodium hydroxide isavailable both as an impure solid and as an approximately 50% w/v solution.Solutions of NaOH may be standardized against any of the primaryweak acid standards listed in Table 9.5.Using NaOH as a titrant is complicated by potential contaminationfrom the following reaction between CO 2 and OH – .−−CO ( aq) + 2OH ( aq) → CO ( aq) + HO()l2 32During the titration, NaOH reacts with both the titrand and CO 2 , increasingthe volume of NaOH needed to reach the titration’s end point. This isnot a problem if end point pH is less than 6. Below this pH the CO 32–fromreaction 9.7 reacts with H 3 O + to form carbonic acid.29.72− +CO ( aq) + 2HO ( aq) → HCO ( aq) + 2H O()l 9.833 2 3 2Combining reaction 9.7 and reaction 9.8 gives an overall reaction that doesnot include OH – .CO ( aq ) + HO () l →HCO( aq )2 2 2 3The nominal concentrations of the concentratedstock solutions are 12.1 M HCl,11.7 M HClO 4 , and 18.0 M H 2 SO 4 .Any solution in contact with the atmospherecontains a small amount ofCO 2 (aq) from the equilibriumCO( g) CO ( aq )2 2

440 Analytical Chemistry 2.0Table 9.5 Selected Primary Standards for Standardizing Strong Acid andStrong Base TitrantsStandardization of Acidic TitrantsPrimary Standard Titration Reaction CommentNa 2 CO 3Na CO + 2HO → HCO + 2Na + 2H O+ + a2 3 3 2 32(HOCH 2 ) 3 CNH 2( HOCH ) CNH + H O → ( HOCH ) CNH + HO+ + b2 3 2 3 2 3 3 2+ +Na 2 B 4 O 7Na BO + 2HO + 3HO→ 2Na + 4HBO2 4 7 3 2 3 3Standardization of Basic TitrantsPrimary Standard Titration Reaction CommentKHC 8 H 4 O 4 KHC HO + OH → K + CHO + H O− + − c8 4 4 8 4 4 2C 6 H 5 COOH CHCOOH + OH → CHCOO + H O− − d6 5 6 5 2− + −KH(IO 3 ) 2KH(IO ) 2+ OH → K + 2IO + HO3 3 2a The end point for this titration is improved by titrating to the second equivalence point, boiling the solution to expelCO 2 , and retitrating to the second equivalence point. The reaction in this case isNa CO + 2HO → CO + 2Na + 3K2 3 3 2Under these conditions the presence of CO 2 does not affect the quantity ofOH – used in the titration and is not a source of determinate error.If the end point pH is between 6 and 10, however, the neutralizationof CO 3 2– requires one proton2− + −CO ( aq) + HO ( aq) → HCO ( aq) + H O()l3+ + +b Tris-(hydroxymethyl)aminomethane often goes by the shorter name of TRIS or THAM.c Potassium hydrogen phthalate often goes by the shorter name of KHP.d Because it is not very soluble in water, dissolve benzoic acid in a small amount of ethanol before diluting with water.3 32and the net reaction between CO 2 and OH – is−−CO 2( aq) + OH ( aq) →HCO 3( aq)Under these conditions some OH – is consumed in neutralizing CO 2 , resultingin a determinate error. We can avoid the determinate error if weuse the same end point pH in both the standardization of NaOH and theanalysis of our analyte, although this often is not practical.Solid NaOH is always contaminated with carbonate due to its contactwith the atmosphere, and can not be used to prepare a carbonate-free solutionof NaOH. Solutions of carbonate-free NaOH can be prepared from50% w/v NaOH because Na 2 CO 3 is insoluble in concentrated NaOH.

Chapter 9 Titrimetric Methods441When CO 2 is absorbed, Na 2 CO 3 precipitates and settles to the bottom ofthe container, allowing access to the carbonate-free NaOH. When preparinga solution of NaOH, be sure to use water that is free from dissolvedCO 2 . Briefly boiling the water expels CO 2 , and after cooling, it may be usedto prepare carbonate-free solutions of NaOH. A solution of carbonate-freeNaOH is relatively stable f we limit its contact with the atmosphere. Standardsolutions of sodium hydroxide should not be stored in glass bottlesas NaOH reacts with glass to form silicate; instead, store such solutions inpolyethylene bottles.In o r g a n i c An a l y s i sAcid –base titrimetry is a standard method for the quantitative analysis ofmany inorganic acids and bases. A standard solution of NaOH can beused to determine the concentration of inorganic acids, such as H 3 PO 4 orH 3 AsO 4 , and inorganic bases, such as Na 2 CO 3 can be analyzed using astandard solution of HCl.An inorganic acid or base that is too weak to be analyzed by an aqueousacid–base titration can be analyzed by adjusting the solvent, or by an indirectanalysis. For example, when analyzing boric acid, H 3 BO 3 , by titratingwith NaOH, accuracy is limited by boric acid’s small acid dissociationconstant of 5.8 10 –10 . Boric acid’s K a value increases to 1.5 10 –4 in thepresence of mannitol, because it forms a complex with the borate ion. Theresult is a sharper end point and a more accurate titration. Similarly, theanalysis of ammonium salts is limited by the small acid dissociation constantof 5.7 10 –10 for NH 4 + . In this case, we can convert NH 4 + to NH 3by neutralizing with strong base. The NH 3 , for which K b is 1.58 10 –5 , isthen removed by distillation and titrated with HCl.We can analyze a neutral inorganic analyte if we can first convert it intoan acid or base. For example, we can determine the concentration of NO 3–by reducing it to NH 3 in a strongly alkaline solution using Devarda’s alloy,a mixture of 50% w/w Cu, 45% w/w Al, and 5% w/w Zn.Figure 9.16a shows a typical result for thetitration of H 3 BO 3 with NaOH.− − −3NO ( aq) + 8Al() s + 5OH ( aq) + 2HO() l →8AlO( aq) + 3NH ( aq)3 223The NH 3 is removed by distillation and titrated with HCl. Alternatively,we can titrate NO 3 – as a weak base by placing it in an acidic nonaqueoussolvent such as anhydrous acetic acid and using HClO 4 as a titrant.Acid–base titrimetry continues to be listed as a standard method for thedetermination of alkalinity, acidity, and free CO 2 in waters and wastewaters.Alkalinity is a measure of a sample’s capacity to neutralize acids. The mostimportant sources of alkalinity are OH – , HCO 3 – , and CO 3 2– , althoughother weak bases, such as phosphate, may contribute to the overall alkalinity.Total alkalinity is determined by titrating to a fixed end point pH of 4.5(or to the bromocresol green end point) using a standard solution of HClor H 2 SO 4 . Results are reported as mg CaCO 3 /L.Although a variety of strong bases andweak bases may contribute to a sample’salkalinity, a single titration cannot distinguishbetween the possible sources. Reportingthe total alkalinity as if CaCO 3 isthe only source provides a means for comparingthe acid-neutralizing capacities ofdifferent samples.

442 Analytical Chemistry 2.01412(a) 14 (b) 14 (c)1212101010pH86pH86pH8644422200 20 40 60 80 100Volume of NaOH (mL)Figure 9.19 Titration curves for 50.0 mL of (a) 0.10 M NaOH, (b) 0.050 M Na 2 CO 3 , and (c) 0.10 M NaHCO 3 using 0.10M HCl. The dashed lines indicate the fixed pH end points of 8.3 and 4.5. The color gradients show the phenolphthalein(red colorless) and bromocresol green (blue green) endpoints. When titrating to the phenolphthalein endpoint, the titrationcontinues until the last trace of red is lost.Solutions containing OH – and HCO 3–alkalinities are unstable with respect to theformation of CO 32– . Problem 9.15 in theend of chapter problems asks you to explainwhy this is true.00 20 40 60 80 100Volume of NaOH (mL)0 20 40 60 80 100Volume of NaOH (mL)When the sources of alkalinity are limited to OH – , HCO 3 – , and CO 3 2– ,separate titrations to a pH of 4.5 (or the bromocresol green end point)and a pH of 8.3 (or the phenolphthalein end point) allow us to determinewhich species are present and their respective concentrations. Titrationcurves for OH – , HCO 3 – , and CO 3 2– are shown in Figure 9.19. For a solutioncontaining only OH – alkalinity, the volumes of strong acid needed toreach the two end points are identical (Figure 9.19a). When the only sourceof alkalinity is CO 3 2– , the volume of strong acid needed to reach the endpoint at a pH of 4.5 is exactly twice that needed to reach the end point at apH of 8.3 (Figure 9.19b). If a solution contains only HCO 3 – alkalinity, thevolume of strong acid needed to reach the end point at a pH of 8.3 is zero,but that for the pH 4.5 end point is greater than zero (Figure 9.19c).Mixtures of OH – and CO 3 2– , or of HCO 3 – and CO 3 2– also are possible.Consider, for example, a mixture of OH – and CO 3 2– . The volumeof strong acid to titrate OH – is the same whether we titrate to a pH of 8.3or a pH of 4.5. Titrating CO 3 2– to a pH of 4.5, however, requires twice asmuch strong acid as titrating to a pH of 8.3. Consequently, when titratinga mixture of these two ions, the volume of strong acid to reach a pH of 4.5is less than twice that to reach a pH of 8.3. For a mixture of HCO 3 – andCO 3 2– the volume of strong acid to reach a pH of 4.5 is more than twicethat to reach a pH of 8.3. Table 9.6 summarizes the relationship betweenthe sources of alkalinity and the volumes of titrant needed to reach the twoend points.Acidity is a measure of a water sample’s capacity for neutralizing base,and is conveniently divided into strong acid and weak acid acidity. Strongacid acidity, from inorganic acids such as HCl, HNO 3 , and H 2 SO 4 , iscommon in industrial effluents and acid mine drainage. Weak acid acidityis usually dominated by the formation of H 2 CO 3 from dissolved CO 2 , butalso includes contributions from hydrolyzable metal ions such as Fe 3+ , Al 3+ ,0

Chapter 9 Titrimetric Methods443Table 9.6 Relationship Between End Point Volumes andSources of AlkalinitySource of Alkalinity Relationship Between End Point VolumesOH – V pH 4.5 = V pH 8.32–CO 3 V pH 4.5 = 2 V pH 8.3–HCO 3 V pH 4.5 > 0; V pH 8.3 = 0OH – 2–and CO 3 V pH 4.5 < 2 V pH 8.3CO 2– –3 and HCO 3 V pH 4.5 > 2 V pH 8.3and Mn 2+ . In addition, weak acid acidity may include a contribution fromorganic acids.Acidity is determined by titrating with a standard solution of NaOHto fixed pH of 3.7 (or the bromothymol blue end point) and a fixed pH8.3 (or the phenolphthalein end point). Titrating to a pH of 3.7 provides ameasure of strong acid acidity, and titrating to a pH of 8.3 provides a measureof total acidity. Weak acid acidity is the difference between the totaland strong acid acidities. Results are expressed as the amount of CaCO 3that can be neutralized by the sample’s acidity. An alternative approach fordetermining strong acid and weak acid acidity is to obtain a potentiometrictitration curve and use a Gran plot to determine the two equivalence points.This approach has been used, for example, to determine the forms of acidityin atmospheric aerosols. 4Water in contact with either the atmosphere, or with carbonate-bearingsediments contains free CO 2 that exists in equilibrium with CO 2 (g) andaqueous H 2 CO 3 , HCO – 3 , and CO 2– 3 . The concentration of free CO 2 isdetermined by titrating with a standard solution of NaOH to the phenolphthaleinend point, or to a pH of 8.3, with results reported as mg CO 2 /L.This analysis is essentially the same as that for the determination of totalacidity, and can only be applied to water samples that do not contain strongacid acidity.As is the case with alkalinity, acidity is reportedas mg CaCO 3 /L.Free CO 2 is the same thing as CO 2 (aq).Or g a n i c An a l y s i sAcid–base titrimetry continues to have a small, but important role for theanalysis of organic compounds in pharmaceutical, biochemical, agricultural,and environmental laboratories. Perhaps the most widely employed acid–base titration is the Kjeldahl analysis for organic nitrogen. Examples ofanalytes determined by a Kjeldahl analysis include caffeine and saccharin inpharmaceutical products, proteins in foods, and the analysis of nitrogen infertilizers, sludges, and sediments. Any nitrogen present in a –3 oxidationstate is quantitatively oxidized to NH 4 + . Because some aromatic heterocycliccompounds, such as pyridine, are difficult to oxidize, a catalyst is usedto ensure a quantitative oxidation. Nitrogen in other oxidation states, suchSee Representative Method 9.1 for oneapplication of a Kjeldahl analysis.4 Ferek, R. J.; Lazrus, A. L.; Haagenson, P. L.; Winchester, J. W. Environ. Sci. Technol. 1983, 17,315–324.

444 Analytical Chemistry 2.0Table 9.7 Selected Elemental Analyses Based on an Acid–Base TitrationElement Convert to... Reaction Producing Titratable Acid or Base a Titration Details+ − add HCl in excess and backN NH 3 (g) NH ( g) + HCl( aq) → NH ( aq) + Cl ( aq )3 4titrate with NaOHS SO 2 (g) SO ( 2g ) + HO ( ) ( )2 2aq →HSO aq2 4titrate H 2 SO 4 with NaOHC CO 2 (g) CO ( 2g ) + Ba(OH) ( aq ) → BaCO () 3s + HO () add excess Ba(OH)22l2 and backtitrate with HClCl HCl(g) — titrate HCl with NaOHF SiF 4 (g) 3SiF ( aq) + 2HO() l → 2HSiF( aq) + SiO () s4 22 6 2titrate H 2 SiF 4 with NaOHa The species that is titrated is shown in bold.as nitro and azo nitrogens, may be oxidized to N 2 , resulting in a negativedeterminate error. Including a reducing agent, such as salicylic acid, convertsthis nitrogen to a –3 oxidation state, eliminating this source of error.Table 9.7 provides additional examples in which an element is quantitativeconverted into a titratable acid or base.Several organic functional groups are weak acids or weak bases. Carboxylic(–COOH), sulfonic (–SO 3 H) and phenolic (–C 6 H 5 OH) functionalgroups are weak acids that can be successfully titrated in either aqueousor nonaqueous solvents. Sodium hydroxide is the titrant of choice foraqueous solutions. Nonaqueous titrations are often carried out in a basicsolvent, such as ethylenediamine, using tetrabutylammonium hydroxide,(C 4 H 9 ) 4 NOH, as the titrant. Aliphatic and aromatic amines are weak basesthat can be titrated using HCl in aqueous solution, or HClO 4 in glacialacetic acid. Other functional groups can be analyzed indirectly following areaction that produces or consumes an acid or base. Typical examples areshown in Table 9.8.Many pharmaceutical compounds are weak acids or bases that canbe analyzed by an aqueous or nonaqueous acid–base titration; examplesinclude salicylic acid, phenobarbital, caffeine, and sulfanilamide. Aminoacids and proteins can be analyzed in glacial acetic acid using HClO 4 as thetitrant. For example, a procedure for determining the amount of nutritionallyavailable protein uses an acid–base titration of lysine residues. 5Qu a n t i t a t i ve Ca l c u l at i o n sThe quantitative relationship between the titrand and the titrant is determinedby the stoichiometry of the titration reaction. If the titrand is polyprotic,then we must know to which equivalence point we are titrating. Thefollowing example illustrates how we can use a ladder diagram to determinea titration reaction’s stoichiometry.5 (a) Molnár-Perl, I.; Pintée-Szakács, M. Anal. Chim. Acta 1987, 202, 159–166; (b) Barbosa, J.;Bosch, E.; Cortina, J. L.; Rosés, M. Anal. Chim. Acta 1992, 256, 177–181.

Chapter 9 Titrimetric Methods445Table 9.8 Selected Acid–Base Titrimetric Procedures for Organic Functional GroupsBased on the Production or Consumption of Acid or BaseFunctionalGroup Reaction Producing Titratable Acid or Base a Titration Details−ester RCOOR′ ( aq) + OH − ( aq) → RCOO ( aq) + HOR′( aq)titrate OH – with HClcarbonylRC= O( aq) + NHOHHCl i ( aq)→2 2RC= NOH( aq) + HCl ( aq)+ HO 2() lalcohol b [] 1 (CH CO) O+ ROH → CH COOR + CH COOH3 2 33[ 2](CH CO) O+ H O3 2 2→2CH COOH3a The species that is titrated is shown in bold.2titrate HCl with NaOHtitrate CH 3 COOH withNaOH; a blank titration ofacetic anhydride, (CH 3 CO) 2 O,corrects for the contribution ofreaction [2]b The acetylation reaction [1] is carried out in pyridine to prevent the hydrolysis of acetic by water. After the acetylation reactionis complete, water is added to covert any unreacted acetic anhydride to acetic acid [2].Example 9.2A 50.00 mL sample of a citrus drink requires 17.62 mL of 0.04166 MNaOH to reach the phenolphthalein end point. Express the sample’s acidityas grams of citric acid, C 6 H 8 O 7 , per 100 mL.(a)more basicCit 3–pK a3 = 6.40So l u t i o nBecause citric acid is a triprotic weak acid, we must first determine if thephenolphthalein end point corresponds to the first, second, or third equivalencepoint. Citric acid’s ladder diagram is shown in Figure 9.20a. Basedon this ladder diagram, the first equivalence point is between a pH of 3.13and a pH of 4.76, the second equivalence point is between a pH of 4.76and a pH of 6.40, and the third equivalence point is greater than a pHof 6.40. Because phenolphthalein’s end point pH is 8.3–10.0 (see Table9.4), the titration proceeds to the third equivalence point and the titrationreaction is−3−CHO ( aq) + 3OH ( aq) → C HO ( aq) + 3HO()l6 8 7 6 5 7In reaching the equivalence point, each mole of citric acid consumes threemoles of NaOH; thus0. 04166 MNaOH× 0.01762 LNaOH= 7.3405× 10 −5moles NaOH1 molC HO57. 3405× 10−6 8 7molNaOH×3mol NaOH= 2. 4468× 10 −4mol CHO26 8 7pH14121086420(b)pHmore acidicHCit 2–pK a2 = 4.76H 2 Cit –pK a1 = 3.13H 3 Cit0 5 10 15 20 25 30Volume of NaOH (mL)Figure 9.20 (a) Ladder diagram forcitric acid; (b) Titration curve for thesample in Example 9.2 showing phenolphthalein’spH transition region.

446 Analytical Chemistry 2.0192.13 gC HO42.4468× 10− molC HO ×6 8 7molC HO6 8 76 8 7= 0.04701 gC HO6 8 7Because this is the amount of citric acid in a 50.00 mL sample, the concentrationof citric acid in the citrus drink is 0.09402 g/100 mL. The completetitration curve is shown in Figure 9.20b.Practice Exercise 9.6Your company recently received a shipment of salicylic acid, C 7 H 6 O 3 , tobe used in the production of acetylsalicylic acid (aspirin). The shipmentcan be accepted only if the salicylic acid is more than 99% pure. To evaluatethe shipment’s purity, a 0.4208-g sample is dissolved in water andtitrated to the phenolphthalein end point, requiring 21.92 mL of 0.1354M NaOH. Report the shipment’s purity as %w/w C 7 H 6 O 3 . Salicylic acidis a diprotic weak acid with pK a values of 2.97 and 13.74.Click here to review your answer to this exercise.In an indirect analysis the analyte participates in one or more preliminaryreactions, one of which produces or consumes acid or base. Despitethe additional complexity, the calculations are straightforward.Example 9.3The purity of a pharmaceutical preparation of sulfanilamide, C 6 H 4 N 2 O 2 S,is determined by oxidizing sulfur to SO 2 and bubbling it through H 2 O 2 toproduce H 2 SO 4 . The acid is titrated to the bromothymol blue end pointwith a standard solution of NaOH. Calculate the purity of the preparationgiven that a 0.5136-g sample requires 48.13 mL of 0.1251 M NaOH.So l u t i o nThe bromothymol blue end point has a pH range of 6.0–7.6. Sulfuric acidis a diprotic acid, with a pK a2 of 1.99 (the first K a value is very large andthe acid dissociation reaction goes to completion, which is why H 2 SO 4 isa strong acid). The titration, therefore, proceeds to the second equivalencepoint and the titration reaction is−2−HSO ( aq) + 2OH ( aq) → 2H O() l + SO ( aq )2 4 24Using the titration results, there are0. 1251 MNaOH× 0. 04813 LNaOH = 6.021× 10 −3mol NaOH31 molH SO6. 021× 10−2 4molNaOH× = 3.010× 10 −3molH SO2mol NaOH2 4

Chapter 9 Titrimetric Methods447produced by bubbling SO 2 through H 2 O 2 . Because all the sulfur in H 2 SO 4comes from the sulfanilamide, we can use a conservation of mass to determinethe amount of sulfanilamide in the sample.31 molS3. 010× 10− molH SO × ×2 4molH SO2 41 mol CHNOS 168 gC HNOS6 4 2 2 6 4 2 2× . 18molS molC H NOS6 42 2= 0.5062 gC HNOS6 4 2 20.5062 gC HNOS6 4 2 2× 100 = 98.56%w/w CHNOS6 4 2 20.5136 gsamplePractice Exercise 9.7The concentration of NO 2 in air can be determined by passing the samplethrough a solution of H 2 O 2 , which oxidizes NO 2 to HNO 3 , andtitrating the HNO 3 with NaOH. What is the concentration of NO 2 , inmg/L, if a 5.0 L sample of air requires 9.14 mL of 0.01012 M NaOH toreach the methyl red end pointClick here to review your answer to this exercise.Example 9.4The amount of protein in a sample of cheese is determined by a Kjeldahlanalysis for nitrogen. After digesting a 0.9814-g sample of cheese, the nitrogenis oxidized to NH 4 + , converted to NH 3 with NaOH, and distilledinto a collection flask containing 50.00 mL of 0.1047 M HCl. The excessHCl is back titrated with 0.1183 M NaOH, requiring 22.84 mL to reachthe bromothymol blue end point. Report the %w/w protein in the cheeseassuming that there are 6.38 grams of protein for every gram of nitrogenin most dairy products.For a back titration we must consider twoacid–base reactions. Again, the calculationsare straightforward.So l u t i o nThe HCl in the collection flask reacts with two bases+ −HCl( aq) + NH ( aq) → NH ( aq) + Cl ( aq)3−−HCl( aq) + OH ( aq) → H O() l + Cl ( aq )The collection flask originally contains0. 1047 MHCl× 0. 05000 LHCl = 5.235× 10 −3molHClof which24

448 Analytical Chemistry 2.00. 1183 MNaOH× 0.02284 LNaOH×1 molHClmolNaOH= 2.702× 10 −3molHClreact with NaOH. The difference between the total moles of HCl and themoles of HCl reacting with NaOH5. 235× 10 −3−molHCl − 2. 702× 10 3 molHCl = 2.533× 10 −3molHClis the moles of HCl reacting with NH 3 . Because all the nitrogen in NH 3comes from the sample of cheese, we use a conservation of mass to determinethe grams of nitrogen in the sample.31 molNH 14.01 g2.533× 10− molHCl× ×molHCl molNHThe mass of protein, therefore, is3N3= 0.03549638 . gprotein0.03549 gN× = 0.2264 gproteingNand the % w/w protein is0.2264 gprotein× 100 = 23.%1 w/w protein0.9814 gsamplegNPractice Exercise 9.8Limestone consists mainly of CaCO 3 , with traces of iron oxides andother metal oxides. To determine the purity of a limestone, a 0.5413-gsample is dissolved using 10.00 mL of 1.396 M HCl. After heating toexpel CO 2 , the excess HCl was titrated to the phenolphthalein end point,requiring 39.96 mL of 0.1004 M NaOH. Report the sample’s purity as%w/w CaCO 3 .Click here to review your answer to this exercise.Earlier we noted that we can use an acid–base titration to analyze amixture of acids or bases by titrating to more than one equivalence point.The concentration of each analyte is determined by accounting for its contributionto each equivalence point.Example 9.5The alkalinity of natural waters is usually controlled by OH – , HCO 3 – , andCO 3 2– , which may be present singularly or in combination. Titrating a100.0-mL sample to a pH of 8.3 requires 18.67 mL of a 0.02812 M HCl.A second 100.0-mL aliquot requires 48.12 mL of the same titrant to reach

Chapter 9 Titrimetric Methods449a pH of 4.5. Identify the sources of alkalinity and their concentrations inmilligrams per liter.So l u t i o nBecause the volume of titrant to reach a pH of 4.5 is more than twice thatneeded to reach a pH of 8.3, we know, from Table 9.6, that the sample’salkalinity is controlled by CO 3 2– and HCO 3 – . Titrating to a pH of 8.3neutralizes CO 3 2– to HCO 3–2− − −CO ( aq) + HCl( aq) → HCO ( aq) + Cl ( aq)3but there is no reaction between the titrant and HCO 3 – (see Figure 9.19).The concentration of CO 3 2– in the sample, therefore, is0. 02812 MHCl× 0.01867 LHCl×5.250×100.1000 LmolCO1 molCOmol HCl2−33= 5.250× 10 −4 molCO 2−360.01 gCO 1000 mg×× = 315.1 mg/L2− mol CO g− 4 2 − 2 −3 3Titrating to a pH of 4.5 neutralizes CO 3 2– to H 2 CO 3 , and HCO 3 – toH 2 CO 3 (see Figures 9.19).2−−CO ( aq) + 2HCl( aq) → H CO ( aq) + 2Cl( aq)332 3−−HCO ( aq) + HCl( aq) → HCO ( aq) + Cl ( aq)23 2 3Because we know how many moles of CO 3 2– are in the sample, we cancalculate the volume of HCl it consumes.−4 2−2 molHCl5. 250× 10 molCO × ×32−molCO1 L HCl 1000 mL× =0.02812 molHCl L337.34This leaves 48.12 mL - 37.34 mL, or 10.78 mL of HCl to react withHCO 3 – . The amount of HCO 3 – in the sample is3.031×100. 02812 MHCl× 0.01078 LHCl×0.1000 LmolHCO1 molHCOmol HCl−3mL= 3.031× 10 −4molHCO −361.02 gHCO mg×mol HCO× 1000− g=−4− −3 33185.0mg/L

450 Analytical Chemistry 2.0Practice Exercise 9.9Samples containing the monoprotic weak acids 2–methylaniliniumchloride (C 7 H 10 NCl, pK a = 4.447) and 3–nitrophenol (C 6 H 5 NO 3 ,pK a = 8.39) can be analyzed by titrating with NaOH. A 2.006-g samplerequires 19.65 mL of 0.200 M NaOH to reach the bromocresol purpleend point and 48.41 mL of 0.200 M NaOH to reach the phenolphthaleinend point. Report the %w/w of each compound in the sample.Click here to review your answer to this exercise.9B.5 Qualitative ApplicationsExample 9.5 shows how we can use an acid–base titration to assign theforms of alkalinity in waters. We can easily extend this approach to othersystems. For example, by titrating with either a strong acid or a strong baseto the methyl orange and phenolphthalein end points we can determine thecomposition of solutions containing one or two of the following species:H 3 PO 4 , H 2 PO 4 – , HPO 4 2– , PO 4 3– , HCl, and NaOH. As outlined in Table9.9, each species or mixture of species has a unique relationship betweenthe volumes of titrant needed to reach these two end points.9B.6 Characterization ApplicationsAn acid–base titration can be use to characterize the chemical and physicalproperties of matter. Two useful characterization applications are theTable 9.9 Relationship Between End Point Volumes for Mixtures of PhosphateSpecies with HCl and NaOHRelationship Between End Point Relationship Between End PointSolution CompositionVolumes with Strong Base Titrant a Volumes With Strong Acid Titrant aH 3 PO 4 V PH = 2 V MO— bH 2 PO 4–HPO 42–PO 43–V PH > 0; V MO = 0 —— V MO > 0; V PH = 0— V MO = 2 V PHHCl V PH = V MO—NaOH — V MO = V PHHCl and H 3 PO 4 V PH < 2 V MO—H 3 PO 4 and H 2 PO 4–H 2 PO 4–and HPO 42–HPO 42–and PO 43–V PH > 2 V MO—V PH > 0; V MO = 0 V MO > 0; V PH = 0— V MO > 2 V PHPO 43–and NaOH — V MO < 2 V PHa VPH and V MO are, respectively, the volume of titrant at the phenolphthalein and methyl orange end points.b When no information is provided, the volume of titrant to each end point is zero.

Chapter 9 Titrimetric Methods451determination of a compound’s equivalent weight and its acid or its basedissociation constant.Eq u i v a l e nt We i g h t sSuppose we titrate a sample containing an impure weak acid to a welldefinedend point using a monoprotic strong base as the titrant. If we assumethat the titration involves the transfer of n protons, then the moles oftitrant needed to reach the end point isn molestitrantmolestitrant = × molesanalytemolesanalyteIf we know the analyte’s identity, we can use this equation to determine theamount of analyte in the sample1moleanalytegramsanalyte = molestitrant× × FW analyten molestitrantwhere FW is the analyte’s formula weight.But what if we do not know the analyte’s identify? If we can titrate apure sample of the analyte, we can obtain some useful information that mayhelp in establishing its identity. Because we do not know the number ofprotons being titrated, we let n = 1 and replace the analyte’s formula weightwith its equivalent weight (EW)1equivalentanalytegramsanalyte = molestitrant××EW analyte1moles titrantwhereFW= n×EWExample 9.6A 0.2521-g sample of an unknown weak acid is titrated with 0.1005 MNaOH, requiring 42.68 mL to reach the phenolphthalein end point. Determinethe compound’s equivalent weight. Which of the following compoundsis most likely to be the unknown weak acid?So l u t i o nascorbic acid C 8 H 8 O 6 FW = 176.1 monoproticmalonic acid C 3 H 4 O 4 FW = 104.1 diproticsuccinic acid C 4 H 6 O 4 FW = 118.1 diproticcitric acid C 6 H 8 O 7 FW = 192.1 triproticThe moles of NaOH needed to reach the end point is

452 Analytical Chemistry 2.00. 1005 MNaOH× 0. 04268 LNaOH = 4.289× 10 −3mol NaOHwhich gives the analyte’s equivalent weight asganalyte 0.2521 gEW = == 58.78 g/equivalent− 3equiv. analyte 4.289×10 equiv.The possible formula weights for the weak acid aren = 1: FW = EW = 5878 . g/equivalentn = 2:FW = 2× EW = 117.6 g/equivalentn = 3: FW = 3× EW = 176.3 g/equivalentIf the analyte is a monoprotic weak acid, then its formula weight is 58.78g/mol, eliminating ascorbic acid as a possibility. If it is a diprotic weak acid,then the analyte’s formula weight is either 58.78 g/mol or 117.6 g/mol, dependingon whether the titration is to the first or second equivalence point.Succinic acid, with a formula weight of 118.1 g/mole is a possibility, butmalonic acid is not. If the analyte is a triprotic weak acid, then its formulaweight is 58.78 g/mol, 117.6 g/mol, or 176.3 g/mol. None of these valuesis close to the formula weight for citric acid, eliminating it as a possibility.Only succinic acid provides a possible match.Practice Exercise 9.10Figure 9.21 shows the potentiometric titration curve for the titration of a0.500-g sample an unknown weak acid. The titrant is 0.1032 M NaOH.What is the weak acid’s equivalent weight?Click here to review your answer to this exercise.1412108pH6Figure 9.21 Titration curve for Practice Exercise 9.10.4200 20 40 60 80 100 120Volume of NaOH (mL)

Chapter 9 Titrimetric Methods4531412V eqpK apH108642½×V eq0Equilibrium Co n s t a n t s0 10 20 30 40 50Volume of NaOH (mL)Figure 9.22 Estimating acetic acid’s pK a using its potentiometrictitration curve.Another application of acid–base titrimetry is the determination of equilibriumconstants. Consider, for example, a solution of acetic acid, CH 3 COOH,for which the dissociation constant is[ HO ][ CH COO ]+ −3 3K a=[ CH COOH]3When the concentrations of CH 3 COOH and CH 3 COO – are equal, the K aexpression reduces to K a = [H 3 O + ], or pH = pK a . If we titrate a solution ofacetic acid with NaOH, the pH equals the pK a when the volume of NaOHis approximately ½V eq . As shown in Figure 9.22, a potentiometric titrationcurve provides a reasonable estimate of acetic acid’s pK a .This method provides a reasonable estimate of a weak acid’s pK a if theacid is neither too strong nor too weak. These limitations are easily to appreciateif we consider two limiting cases. For the first case let’s assumethat the weak acid, HA, is more than 50% dissociated before the titrationbegins (a relatively large K a value). The concentration of HA before theequivalence point is always less than the concentration of A – , and there isno point on the titration curve where [HA] = [A – ]. At the other extreme, ifthe acid is too weak, less than 50% of the weak acid reacts with the titrantat the equivalence point. In this case the concentration of HA before theequivalence point is always greater than that of A – . Determining the pK aby the half-equivalence point method overestimates its value if the acid istoo strong and underestimates its value if the acid is too weak.Practice Exercise 9.11Use the potentiometric titration curve in Figure 9.21 to estimate the pK avalues for the weak acid in Practice Exercise 9.10.Click here to review your answer to this exercise.

454 Analytical Chemistry 2.0A second approach for determining a weak acid’s pK a is to use a Granplot. For example, earlier in this chapter we derived the following equationfor the titration of a weak acid with a strong base.[ HO+ ] × V = K V − K V3 b a eq a bA plot of [H 3 O + ] V b versus V b , for volumes less than the equivalencepoint, yields a straight line with a slope of –K a . Other linearizations havebeen developed which use the entire titration curve, or that require no assumptions.6 This approach to determining acidity constants has been usedto study the acid–base properties of humic acids, which are naturally occurring,large molecular weight organic acids with multiple acidic sites. In onestudy a humic acid was found to have six titratable sites, three of which wereidentified as carboxylic acids, two of which were believed to be secondary ortertiary amines, and one of which was identified as a phenolic group. 79B.7 Evaluation of Acid–Base TitrimetrySc a l e o f Op e r a t i o nAcid–base titrimetry is an example of a totalanalysis technique in which the signalis proportional to the absolute amount ofanalyte. See Chapter 3 for a discussion ofthe difference between total analysis techniquesand concentration techniques.In an acid–base titration the volume of titrant needed to reach the equivalencepoint is proportional to the moles of titrand. Because the pH of thetitrand or the titrant is a function of its concentration, however, the changein pH at the equivalence point—and thus the feasibility of an acid–basetitration—depends on their respective concentrations. Figure 9.23, for example,shows a series of titration curves for the titration of several concentrationsof HCl with equimolar solutions NaOH. For titrand and titrantconcentrations smaller than 10 –3 M, the change in pH at the end pointmay be too small to provide accurate and precise results.6 (a) Gonzalez, A. G.; Asuero, A. G. Anal. Chim. Acta 1992, 256, 29–33; (b) Papanastasiou, G.;Ziogas, I.; Kokkindis, G. Anal. Chim. Acta 1993, 277, 119–135.7 Alexio, L. M.; Godinho, O. E. S.; da Costa, W. F. Anal. Chim. Acta 1992, 257, 35–39.pH14121086(a)(b)(c)(d)(e)4Figure 9.23 Titration curves for 25.0 mL of (a) 10 –1 MHCl, (b) 10 –2 M HCl, (c) 10 –3 M HCl, (d) 10 –4 MHCl, and (e) 10 –5 M HCl. In each case the titrant is anequimolar solution of NaOH.200 10 20 30 40 50Volume of NaOH (mL)

Chapter 9 Titrimetric Methods455piezoelectric ceramicpH electroderotatingsample stagesampletitrantA minimum concentration of 10 –3 M places limits on the smallestamount of analyte that we can successfully analyze. For example, supposeour analyte has a formula weight of 120 g/mol. To successfully monitorthe titration’s end point using an indicator or with a pH probe, the titrandneeds an initial volume of approximately 25 mL. If we assume that the analyte’sformula weight is 120 g/mol, then each sample must contain at least3 mg of analyte. For this reason, acid–base titrations are generally limitedto major and minor analytes (see Figure 3.6 in Chapter 3). We can extendthe analysis of gases to trace analytes by pulling a large volume of the gasthrough a suitable collection solution.One goal of analytical chemistry is to extend analyses to smaller samples.Here we describe two interesting approaches to titrating mL and pLsamples. In one experimental design (Figure 9.24), samples of 20–100 mLwere held by capillary action between a flat-surface pH electrode and astainless steel sample stage. 8 The titrant was added by using the oscillationsof a piezoelectric ceramic device to move an angled glass rod in and out ofa tube connected to a reservoir containing the titrant. Each time the glasstube was withdrawn an approximately 2 nL microdroplet of titrant wasreleased. The microdroplets were allowed to fall onto the sample, with mixingaccomplished by spinning the sample stage at 120 rpm. A total of 450microdroplets, with a combined volume of 0.81 –0.84 mL, was dispensedbetween each pH measurement. In this fashion a titration curve was constructed.This method was used to titrate solutions of 0.1 M HCl and 0.1 MCH 3 COOH with 0.1 M NaOH. Absolute errors ranged from a minimumof +0.1% to a maximum of –4.1%, with relative standard deviations from0.15% to 4.7%. Sample as small as 20 mL were successfully titrated.Figure 9.24 Experimental design for a microdroplettitration apparatus.8 Steele, A.; Hieftje, G. M. Anal. Chem. 1984, 56, 2884–2888.

456 Analytical Chemistry 2.0microburetagar gel membranesample dropheptaneFigure 9.25 Experimental set-up for a diffusionalmicrotitration. The indicator is amixture of bromothymol blue and bromocresolpurple.Another approach carries out the acid–base titration in a single drop ofsolution. 9 The titrant is delivered using a microburet fashioned from a glasscapillary micropipet (Figure 9.25). The microburet has a 1-2 mm tip filledwith an agar gel membrane. The tip of the microburet is placed within adrop of the sample solution, which is suspended in heptane, and the titrantis allowed to diffuse into the sample. The titration’s progress is monitoredusing an acid–base indicator, and the time needed to reach the end pointis measured. The rate of the titrant’s diffusion from the microburet is determinedby a prior calibration. Once calibrated the end point time can beconverted to an end point volume. Samples usually consisted of picolitervolumes (10 –12 liters), with the smallest sample being 0.7 pL. The precisionof the titrations was usually about 2%.Titrations conducted with microliter or picoliter sample volumes requirea smaller absolute amount of analyte. For example, diffusional titrationshave been successfully conducted on as little as 29 femtomoles (10 –15moles) of nitric acid. Nevertheless, the analyte must still be present in thesample at a major or minor level for the titration to be performed accuratelyand precisely.Ac c u r a c yindicator’s color changeSee Figure 3.5 in Chapter 3 to review thecharacteristics of macro–major and macro–minorsamples.When working with a macro–major or a macro–minor sample, an acid–base titration can achieve a relative error of 0.1–0.2%. The principal limitationto accuracy is the difference between the end point and the equivalencepoint.9 (a) Gratzl, M.; Yi, C. Anal. Chem. 1993, 65, 2085–2088; (b) Yi, C.; Gratzl, M. Anal. Chem.1994, 66, 1976–1982; (c) Hui, K. Y.; Gratzl, M. Anal. Chem. 1997, 69, 695–698; (d) Yi, C.;Huang, D.; Gratzl, M. Anal. Chem. 1996, 68, 1580–1584; (e) Xie, H.; Gratzl, M. Anal. Chem.1996, 68, 3665–3669.

Chapter 9 Titrimetric Methods457PrecisionAn acid–base titration’s relative precision depends primarily on the precisionwith which we can measure the end point volume and the precision indetecting the end point. Under optimum conditions, an acid–base titrationhas a relative precision of 0.1–0.2%. We can improve the relative precisionby using the largest possible buret and ensuring that we use most of its capacityin reaching the end point. A smaller volume buret is a better choicewhen using costly reagents, when waste disposal is a concern, or when thetitration must be completed quickly to avoid competing chemical reactions.Automatic titrators are particularly useful for titrations requiring small volumesof titrant because they provide significantly better precision (typicallyabout ±0.05% of the buret’s volume).The precision of detecting the end point depends on how it is measuredand the slope of the titration curve at the end point. With an indicator theprecision of the end point signal is usually ±0.03–0.10 mL. Potentiometricend points usually are more precise.SensitivityFor an acid–base titration we can write the following general analyticalequation relating the titrant’s volume to the absolute amount of titrandvolume of titrant= k × moles of titrandwhere k, the sensitivity, is determined by the stoichiometry between thetitrand and the titrant. Consider, for example, the determination of sulfurousacid, H 2 SO 3 , by titrating with NaOH to the first equivalence point−−HSO ( aq) + OH ( aq) → H O() l + HSO ( aq )2 3 2At the equivalence point the relationship between the moles of NaOH andthe moles of H 2 SO 3 ismolNaOH=mol HSO 2 3Substituting the titrant’s molarity and volume for the moles of NaOH andrearrangingM× V = molH SONaOH NaOH 2 33VNaOH1= ×molH SOMNaOH2 3we find that k isk= 1M NaOH

458 Analytical Chemistry 2.0There are two ways in which we can improve a titration’s sensitivity. Thefirst, and most obvious, is to decrease the titrant’s concentration because itis inversely proportional to the sensitivity, k.The second approach, which only applies if the titrand is multiprotic,is to titrate to a later equivalence point. If we titrate H 2 SO 3 to the secondequivalence point−2−HSO ( aq) + 2OH ( aq) → 2H O() l + SO ( aq )2 3 23then each mole of H 2 SO 3 consumes two moles of NaOHmolNaOH = 2×molH 2SO 3and the sensitivity becomesk= 2M NaOHIn practice, however, any improvement in sensitivity is offset by a decreasein the end point’s precision if the larger volume of titrant requiresus to refill the buret. Consequently, standard acid–base titrimetric proceduresare written to ensure that titrations require 60–100% of the buret’svolume.Se l e c t i v i t yAcid–base titrants are not selective. A strong base titrant, for example, reactswith all acids in a sample, regardless of their individual strengths. If thetitrand contains an analyte and an interferent, then selectivity depends ontheir relative acid strengths. Two limiting situations must be considered.If the analyte is a stronger acid than the interferent, then the titrantwill react with the analyte before it begins reacting with the interferent. Thefeasibility of the analysis depends on whether the titrant’s reaction with theinterferent affects the accurate location of the analyte’s equivalence point. Ifthe acid dissociation constants are substantially different, the end point forthe analyte can be accurately determined. Conversely, if the acid dissociationconstants for the analyte and interferent are similar, then an accurateend point for the analyte may not be found. In the latter case a quantitativeanalysis for the analyte is not possible.In the second limiting situation the analyte is a weaker acid than theinterferent. In this case the volume of titrant needed to reach the analyte’sequivalence point is determined by the concentration of both the analyteand the interferent. To account for the interferent’s contribution to theend point, an end point for the interferent must be present. Again, if theacid dissociation constants for the analyte and interferent are significantlydifferent, then the analyte’s determination is possible. If the acid dissociationconstants are similar, however, there is only a single equivalence point

Chapter 9 Titrimetric Methods459and the analyte’s and interferent’s contributions to the equivalence pointvolume can not be separated.Tim e , Co s t, a n d Eq u i pm e n tAcid–base titrations require less time than most gravimetric procedures, butmore time than many instrumental methods of analysis, particularly whenanalyzing many samples. With an automatic titrator, however, concernsabout analysis time are less significant. When performing a titration manuallyour equipment needs—a buret and, perhaps, a pH meter—are few innumber, inexpensive, routinely available, and easy to maintain. Automatictitrators are available for between $3000 and $10 000.9CComplexation TitrationsThe earliest examples of metal–ligand complexation titrations are Liebig’sdeterminations, in the 1850s, of cyanide and chloride using, respectively,Ag + and Hg 2+ as the titrant. Practical analytical applications of complexationtitrimetry were slow to develop because many metals and ligandsform a series of metal–ligand complexes. Liebig’s titration of CN – withAg + was successful because they form a single, stable complex of Ag(CN) 2 – ,giving a single, easily identified end point. Other metal–ligand complexes,such as CdI 4 2– , are not analytically useful because they form a series ofmetal–ligand complexes (CdI + , CdI 2 (aq), CdI 3 – and CdI 4 2– ) that producea sequence of poorly defined end points.In 1945, Schwarzenbach introduced aminocarboxylic acids as multidentateligands. The most widely used of these new ligands—ethylenediaminetetraaceticacid, or EDTA—forms strong 1:1 complexes with manymetal ions. The availability of a ligand that gives a single, easily identifiedend point made complexation titrimetry a practical analytical method.9C.1 Chemistry and Properties of EDTAEthylenediaminetetraacetic acid, or EDTA, is an aminocarboxylic acid.EDTA, which is shown in Figure 9.26a in its fully deprotonated form, isa Lewis acid with six binding sites—four negatively charged carboxylategroups and two tertiary amino groups—that can donate six pairs of electronsto a metal ion. The resulting metal–ligand complex, in which EDTAforms a cage-like structure around the metal ion (Figure 9.26b), is verystable. The actual number of coordination sites depends on the size of themetal ion, however, all metal–EDTA complexes have a 1:1 stoichiometry.Recall that an acid–base titration curve fora diprotic weak acid has a single end pointif its two K a values are not sufficiently different.See Figure 9.11 for an example.(a)− OONONO O − O(b)− OO – OO –OO −NNM 2+O –Figure 9.26 Structures of (a) EDTA, in its fully deprotonatedform, and (b) in a six-coordinate metal–EDTAcomplex with a divalent metal ion.OO –O

460 Analytical Chemistry 2.0pH10.246.162.662.01.50.0Y 4–HY 3–H 2 Y 2–H 3 Y –H 4 YH 5 Y +H 6 Y 2+Figure 9.27 Ladder diagram for EDTA.Me t a l–EDTA Fo r m a t i o n Co n s t a n t sTo illustrate the formation of a metal–EDTA complex, let’s consider thereaction between Cd 2+ and EDTACd 2 +Y 4 −CdY2 −( aq) + ( aq) ( aq)9.9where Y 4– is a shorthand notation for the fully deprotonated form of EDTAshown in Figure 9.26a. Because the reaction’s formation constantK f2−= [ CdY ]Cd Y= × 1629 . 10 9.102+ 4−[ ][ ]is large, its equilibrium position lies far to the right. Formation constantsfor other metal–EDTA complexes are found in Appendix 12.EDTA is a We a k Ac i dIn addition to its properties as a ligand, EDTA is also a weak acid. The fullyprotonated form of EDTA, H 6 Y 2+ , is a hexaprotic weak acid with successivepK a values ofpK = 00 . pK = 15 . pK = 20 .a1 a2 a3pK = 266 . pK = 6. 16 pK a6= 10.24a4a5The first four values are for the carboxylic acid protons and the last twovalues are for the ammonium protons. Figure 9.27 shows a ladder diagramfor EDTA. The specific form of EDTA in reaction 9.9 is the predominatespecies only at pH levels greater than 10.17.Co n d i t i o n a l Me t a l–Li g a n d Fo r m a t i o n Co n s t a n t sThe formation constant for CdY 2– in equation 9.10 assumes that EDTA ispresent as Y 4– . Because EDTA has many forms, when we prepare a solutionof EDTA we know it total concentration, C EDTA , not the concentration ofa specific form, such as Y 4– . To use equation 9.10, we need to rewrite it interms of C EDTA .At any pH a mass balance on EDTA requires that its total concentrationequal the combined concentrations of each of its forms.2+ +C EDTA 6 5 4= [ HY ] + [ HY ] + [ HY]+2−[ HY ] + [ HY ] + [ HY−3 2−] + [ Y ]3−4To correct the formation constant for EDTA’s acid–base properties we needto calculate the fraction, a Y 4–, of EDTA present as Y 4– .α Y4− =4−[ Y ]CEDTA9.11

Chapter 9 Titrimetric Methods461Table 9.10 Values of a Y 4– for Selected pH LevelspH a Y 4– pH a Y 4–1 1.9 10 –18 8 5.6 10 –32 3.4 10 –14 9 5.4 10 –23 2.6 10 –11 10 0.374 3.8 10 –9 11 0.855 3.7 10 –7 12 0.986 2.4 10 –5 13 1.007 5.0 10 –4 14 1.00Table 9.10 provides values of a Y 4– for selected pH levels. Solving equation9.11 for [Y 4– ] and substituting into equation 9.10 for the CdY 2– formationconstantKf2−[ CdY ]=2+[ Cd ] α C4−YEDTAProblem 9.42 from the end of chapterproblems asks you to verify the values inTable 9.10 by deriving an equation fora Y 4-.and rearranging givesK2−′[ CdY ]= K × α 4−=2+[ Cd ] Cf f YEDTA9.12where K f´ is a pH-dependent conditional formation constant. Asshown in Table 9.11, the conditional formation constant for CdY 2– becomessmaller and the complex becomes less stable at more acidic pHs.EDTA Co m p e t e s Wi t h Ot h e r Li g a n d sTo maintain a constant pH during a complexation titration we usually adda buffering agent. If one of the buffer’s components is a ligand that bindsCd 2+ , then EDTA must compete with the ligand for Cd 2+ . For example, anTable 9.11 Conditional Formation Constants for CdY 2–pH K f´ pH K f´1 5.5 10 –2 8 1.6 10 142 1.0 10 3 9 1.6 10 153 7.7 10 5 10 1.1 10 164 1.1 10 8 11 2.5 10 165 1.1 10 10 12 2.9 10 166 6.8 10 11 13 2.9 10 167 1.5 10 13 14 2.9 10 16

462 Analytical Chemistry 2.0NH 4 + /NH 3 buffer includes NH 3 , which forms several stable Cd 2+ –NH 3complexes. Because EDTA forms a stronger complex with Cd 2+ it will displaceNH 3 , but the stability of the Cd 2+ –EDTA complex decreases.We can account for the effect of an auxiliary complexing agent, suchas NH 3 , in the same way we accounted for the effect of pH. Before addingEDTA, the mass balance on Cd 2+ , C Cd , is2+ 2+C = [ Cd ] + [ Cd( NH ) ] +Cd 32+[ Cd( NH ) ] + [ Cd(NH ) 2+ ] [ ( ) 2++ Cd NH ]3 23 33 4The value of a Cd 2+ depends on the concentrationof NH 3 . Contrast this witha Y 4-, which depends on pH.and the fraction of uncomplexed Cd 2+ , a Cd 2+, isα Cd2+2+[ Cd ]=CCd9.13Solving equation 9.13 for [Cd 2+ ] and substituting into equation 9.12givesK′ = K × α 4 − =αf f Y2−[ CdY ]C C2+Cd Cd EDTABecause the concentration of NH 3 in a buffer is essentially constant, wecan rewrite this equationK2−′′[ CdY ]= K × α 4−× α2+=C Cf f Y CdCdEDTA9.14to give a conditional formation constant, K f´´, that accounts for both pHand the auxiliary complexing agent’s concentration. Table 9.12 providesvalues of a M 2+ for several metal ion when NH 3 is the complexing agent.9C.2 Complexometric EDTA Titration CurvesNow that we know something about EDTA’s chemical properties, we areready to evaluate its usefulness as a titrant. To do so we need to know theTable 9.12 Values of a M 2+ for Selected Concentrations of Ammonia[NH 3 ] (M) a Ca 2+ a Cd 2+ a Co 2+ a Cu 2+ a Mg 2+ a Ni 2+ a Zn 2+1 5.50 10 –1 6.09 10 –8 1.00 10 –6 3.79 10 –14 1.76 10 –1 9.20 10 –10 3.95 10 –100.5 7.36 10 –1 1.05 10 –6 2.22 10 –5 6.86 10 –13 4.13 10 –1 3.44 10 –8 6.27 10 –90.1 9.39 10 –1 3.51 10 –4 6.64 10 –3 4.63 10 –10 8.48 10 –1 5.12 10 –5 3.68 10 –60.05 9.69 10 –1 2.72 10 –3 3.54 10 –2 7.17 10 –9 9.22 10 –1 6.37 10 –4 5.45 10 –50.01 9.94 10 –1 8.81 10 –2 3.55 10 –1 3.22 10 –6 9.84 10 –1 4.32 10 –2 1.82 10 –20.005 9.97 10 –1 2.27 10 –1 5.68 10 –1 3.62 10 –5 9.92 10 –1 1.36 10 –1 1.27 10 –10.001 9.99 10 –1 6.09 10 –1 8.84 10 –1 4.15 10 –3 9.98 10 –1 5.76 10 –1 7.48 10 –1

Chapter 9 Titrimetric Methods463shape of a complexometric EDTA titration curve. In section 9B we learnedthat an acid–base titration curve shows how the titrand’s pH changes as weadd titrant. The analogous result for a complexation titration shows thechange in pM, where M is the metal ion, as a function of the volume ofEDTA. In this section we will learn how to calculate a titration curve usingthe equilibrium calculations from Chapter 6. We also will learn how toquickly sketch a good approximation of any complexation titration curveusing a limited number of simple calculations.Ca l c u l at i n g t h e Ti t r a t i o n Cu r v eLet’s calculate the titration curve for 50.0 mL of 5.00 10 –3 M Cd 2+ usinga titrant of 0.0100 M EDTA. Furthermore, let’s assume that the titrand isbuffered to a pH of 10 with a buffer that is 0.0100 M in NH 3 .Because the pH is 10, some of the EDTA is present in forms other thanY 4– . In addition, EDTA must compete with NH 3 for the Cd 2+ . To evaluatethe titration curve, therefore, we first need to calculate the conditionalformation constant for CdY 2– . From Table 9.10 and Table 9.11 we find thata Y 4– is 0.35 at a pH of 10, and that a Cd 2+ is 0.0881 when the concentrationof NH 3 is 0.0100 M. Using these values, the conditional formationconstant isStep 1: Calculate the conditional formationconstant for the metal–EDTA complex.K′′16= K × α 4−× α = ( 29 . × 10 )( 037 . )( 0. 0881) = 95 . × 10 14f f Y Cd 2+Because K f´´ is so large, we can treat the titration reactionCd 2 +Y 4 −CdY2 −( aq) + ( aq) → ( aq)as if it proceeds to completion.The next task in calculating the titration curve is to determine the volumeof EDTA needed to reach the equivalence point. At the equivalencepoint we know thatStep 2: Calculate the volume of EDTAneeded to reach the equivalence point.molesEDTA = moles Cd 2+M × V = M × VEDTA EDTA Cd CdSubstituting in known values, we find that it requiresVeqM VCd Cd( 500 . × 10 −3M)(50.0 mL)= V = =EDTAM0.0100 MEDTA= 25.0mLof EDTA to reach the equivalence point.Before the equivalence point, Cd 2+ is present in excess and pCd isdetermined by the concentration of unreacted Cd 2+ . Because not all theunreacted Cd 2+ is free—some is complexed with NH 3 —we must accountfor the presence of NH 3 . For example, after adding 5.0 mL of EDTA, thetotal concentration of Cd 2+ isStep 3: Calculate pM values before theequivalence point by determining theconcentration of unreacted metal ions.

464 Analytical Chemistry 2.0initialmoles Cd − molesEDTAadded M V − M V=total volumeV + V2+C Cd=( 500 . × 10 −3=M)(50.0 mL) −(0.0100 M)(5.0 mL)50.0 mL + 50 . mLCd Cd EDTA EDTACdEDTA= 364 . × 10 −3MTo calculate the concentration of free Cd 2+ we use equation 9.132+ −4[ Cd ] = α × C = ( 0. 0881)( 364 . × 10 M) = 321 . × 10 −4M2+CdCdwhich gives a pCd of2+ −4pCd=− log[ Cd ] =− log( 321 . × 10 ) = 3.49Step 4: Calculate pM at the equivalencepoint using the conditional formationconstant.At the equivalence point all the Cd 2+ initially in the titrand is now presentas CdY 2– . The concentration of Cd 2+ , therefore, is determined by thedissociation of the CdY 2– complex. First, we calculate the concentrationof CdY 2– .2+initialmoles Cd[ CdY2− Cd] = = M V Cdtotalvolume V + V× −CdEDTA3( 500 . 10 M)(50.0 mL)== 333 . × 10 −3M50.0 mL + 25.0 mLNext, we solve for the concentration of Cd 2+ in equilibrium with CdY 2– .At the equivalence point the initial molesof Cd 2+ and the moles of EDTA addedare equal. The total concentrations ofCd 2+ , C Cd , and the total concentrationof EDTA, C EDTA , are equal.Kf2−−3′′[ CdY ] 333 . × 10 − x= == 95 . × 10 14C C ( x)( x)CdEDTAx = C = 19 . × 10 −9MCdOnce again, to find the concentration of uncomplexed Cd 2+ we must accountfor the presence of NH 3 ; thus2+ −9[ Cd ] = α × C = ( 0. 0881)( 19 . × 10 M) = 170 . × 10 −102+CdCdMStep 5: Calculate pM after the equivalencepoint using the conditional formationconstant.and pCd is 9.77 at the equivalence point.After the equivalence point, EDTA is in excess and the concentration ofCd 2+ is determined by the dissociation of the CdY 2– complex. First, we calculatethe concentrations of CdY 2– and of unreacted EDTA. For example,after adding 30.0 mL of EDTA2+initialmoles Cd[ CdY2− Cd] = = M V Cdtotalvolume V + V× −CdEDTA3( 500 . 10 M)(50.0 mL)== 3.13× 10 −350.0 mL + 30.0mLM

Chapter 9 Titrimetric Methods465CEDTAM V − M V=V + VEDTA EDTA Cd CdCdEDTAM)= ( 0 .−30100 (30.0 mL) − (5.00×10 M)(50.0 mL)50. 0 mL + 30.0mL= 625 . × 10 −4MSubstituting into equation 9.14 and solving for [Cd 2+ ] gives2−−3[ CdY ] 313 . × 10 M=−4C C C ( 625 . × 10 M)Cd EDTA Cd2+CdCdC Cd= 54 . × 10 −15M= 95 . × 10 142+ −15[ Cd ] = α × C = ( 0. 0881)( 54 . × 10 M) = 48 . × 10 −16Ma pCd of 15.32. Table 9.13 and Figure 9.28 show additional results for thistitration.Table 9.13 Titration of 50.0 mL of 5.00x10 -3 M Cd 2+with 0.0100 M EDTA at a pH of 10 and in thePresence of 0.0100 M NH3Volume ofEDTA (mL) pCdVolume ofEDTA (mL) pCd0.00 3.36 27.0 14.955.00 3.49 30.0 15.3310.0 3.66 35.0 15.6115.0 3.87 40.0 15.7620.0 4.20 45.0 15.8623.0 4.62 50.0 15.9425.0 9.7720After the equilibrium point we know the equilibriumconcentrations of CdY 2- and EDTA. Wecan solve for the equilibrium concentration ofC Cd using K f´´ and then calculate [Cd 2+ ] usinga Cd 2+. Because we use the same conditional formationconstant, K f´´, for all calculations, this isthe approach shown here.There is a second method for calculating [Cd 2+ ]after the equivalence point. Because the calculationuses only [CdY 2- ] and C EDTA , we can useK f´ instead of K f´´; thus[ CdY ]2+[ Cd ] C−32−EDTA4− ×Y313 . × 10 M16= ( 037 . )( 2. 9×10 )2+−4[ Cd ]( 625 . × 10 M)Solving gives [Cd 2+ ] = 4.710 –16 M and a pCdof 15.33. We will use this approach when learninghow to sketch a complexometric titration curve.=αKf15pCd10500 10 20 30 40 50Volume of EDTA (mL)Figure 9.28 Titration curve for the titration of 50.0 mL of5.0010 –3 M Cd 2+ with 0.0100 M EDTA at a pH of 10 and inthe presence of 0.0100 M NH 3 . The red points correspond tothe data in Table 9.13. The blue line shows the complete titrationcurve.

466 Analytical Chemistry 2.0Practice Exercise 9.12Calculate titration curves for the titration of 50.0 mL of 5.0010 –3M Cd 2+ with 0.0100 M EDTA (a) at a pH of 10 and (b) at a pH of 7.Neither titration includes an auxiliary complexing agent. Compare yourresults with Figure 9.28 and comment on the effect of pH and of NH 3on the titration of Cd 2+ with EDTA.Click here to review your answer to this exercise.Sk e t c h i n g a n EDTA Ti t r a t i o n Cu r v eThis is the same example that we used indeveloping the calculations for a complexationtitration curve. You can review theresults of that calculation in Table 9.13and Figure 9.28.See Table 9.13 for the values.See the side bar comment on the previouspage for an explanation of why we areignoring the effect of NH 3 on the concentrationof Cd 2+ .To evaluate the relationship between a titration’s equivalence point and itsend point, we need to construct only a reasonable approximation of theexact titration curve. In this section we demonstrate a simple method forsketching a complexation titration curve. Our goal is to sketch the titrationcurve quickly, using as few calculations as possible. Let’s use the titration of50.0 mL of 5.0010 –3 M Cd 2+ with 0.0100 M EDTA in the presence of0.0100 M NH 3 to illustrate our approach.We begin by calculating the titration’s equivalence point volume, which,as we determined earlier, is 25.0 mL. Next, we draw our axes, placing pCdon the y-axis and the titrant’s volume on the x-axis. To indicate the equivalencepoint’s volume, we draw a vertical line corresponding to 25.0 mL ofEDTA. Figure 9.29a shows the result of the first step in our sketch.Before the equivalence point, Cd 2+ is present in excess and pCd isdetermined by the concentration of unreacted Cd 2+ . Because not all theunreacted Cd 2+ is free—some is complexed with NH 3 —we must accountfor the presence of NH 3 . The calculations are straightforward, as we sawearlier. Figure 9.29b shows the pCd after adding 5.00 mL and 10.0 mL ofEDTA.The third step in sketching our titration curve is to add two points afterthe equivalence point. Here the concentration of Cd 2+ is controlled by thedissociation of the Cd 2+ –EDTA complex. Beginning with the conditionalformation constantK′ 2−= [ CdY ]KY fCd C= α × = ×4−( 037 . )( 2.9 10 ) = 1.1×10[ ]f 2+EDTAwe take the log of each side and rearrange, arriving atlog log[ ] log [ 2−CdYK ′ =− Cd +]2+fC EDTA16 16C EDTA2pCd = logK′ + logf[−CdY ]

Chapter 9 Titrimetric Methods46720(a)20(b)1515pCd10pCd1055000 10 20 30 40 50Volume of EDTA (mL)0 10 20 30 40 50Volume of EDTA (mL)20(c)20(d)1515pCd10pCd105500 10 20 30 40 50Volume of EDTA (mL)00 10 20 30 40 50Volume of EDTA (mL)20(e)20(f)1515pCd10pCd105500 10 20 30 40 50Volume of EDTA (mL)00 10 20 30 40 50Volume of EDTA (mL)Figure 9.29 Illustrations showing the steps in sketching an approximate titration curve for the titration of50.0 mL of 5.00 10 –3 M Cd 2+ with 0.0100 M EDTA in the presence of 0.0100 M NH 3 : (a) locating theequivalence point volume; (b) plotting two points before the equivalence point; (c) plotting two points after theequivalence point; (d) preliminary approximation of titration curve using straight-lines; (e) final approximationof titration curve using a smooth curve; (f) comparison of approximate titration curve (solid black line) andexact titration curve (dashed red line). See the text for additional details.

468 Analytical Chemistry 2.0Note that after the equivalence point, the titrand’s solution is a metal–ligand complexation buffer, with pCd determined by C EDTA and [CdY 2– ].The buffer is at its lower limit of pCd = logK f´ – 1 whenC EDTA2+molesEDTAadded initial moles2−[ CdY ]= − Cd= 12+initialmoles Cd10Making appropriate substitutions and solving, we find thatM V − M VEDTA EDTA Cd CdM VCdCd= 110M V − M V = 01 . × M VEDTA EDTA Cd Cd Cd CdVEDTA11 . M VCd Cd= × = 11 . × VMEDTAeqOur derivation here is general and appliesto any complexation titration usingEDTA as a titrant.Thus, when the titration reaches 110% of the equivalence point volume, pCdis logK f´ – 1. A similar calculation should convince you that pCd = logK f´when the volume of EDTA is 2V eq .Figure 9.29c shows the third step in our sketch. First, we add a ladderdiagram for the CdY 2– complex, including its buffer range, using its logK f´value of 16.04. Next, we add points representing pCd at 110% of V eq (apCd of 15.04 at 27.5 mL) and at 200% of V eq (a pCd of 16.04 at 50.0mL).Next, we draw a straight line through each pair of points, extending theline through the vertical line representing the equivalence point’s volume(Figure 9.29d). Finally, we complete our sketch by drawing a smooth curvethat connects the three straight-line segments (Figure 9.29e). A comparisonof our sketch to the exact titration curve (Figure 9.29f) shows that they arein close agreement.Practice Exercise 9.13Sketch titration curves for the titration of 50.0 mL of 5.0010 –3 M Cd 2+with 0.0100 M EDTA (a) at a pH of 10 and (b) at a pH of 7. Compareyour sketches to the calculated titration curves from Practice Exercise9.12.Click here to review your answer to this exercise.9C.3 Selecting and Evaluating the End pointThe equivalence point of a complexation titration occurs when we reactstoichiometrically equivalent amounts of titrand and titrant. As is the casewith acid–base titrations, we estimate the equivalence point of a complexationtitration using an experimental end point. A variety of methods are

Chapter 9 Titrimetric Methods469available for locating the end point, including indicators and sensors thatrespond to a change in the solution conditions.Fi n d i n g t h e En d p o i n t w it h a n In d i c a t o rMost indicators for complexation titrations are organic dyes—known asmetallochromic indicators—that form stable complexes with metalions. The indicator, In m– , is added to the titrand’s solution where it forms astable complex with the metal ion, MIn n– . As we add EDTA it reacts firstwith free metal ions, and then displaces the indicator from MIn n– .n 4 2MIn + Y → MY + In− − − m−If MIn n– and In m– have different colors, then the change in color signalsthe end point.The accuracy of an indicator’s end point depends on the strength of themetal–indicator complex relative to that of the metal–EDTA complex. Ifthe metal–indicator complex is too strong, the change in color occurs afterthe equivalence point. If the metal–indicator complex is too weak, however,the end point occurs before we reach the equivalence point.Most metallochromic indicators also are weak acids. One consequenceof this is that the conditional formation constant for the metal–indicatorcomplex depends on the titrand’s pH. This provides some control over anindicator’s titration error because we can adjust the strength of a metal–indicator complex by adjusted the pH at which we carry out the titration.Unfortunately, because the indicator is a weak acid, the color of the uncomplexedindicator also changes with pH. Figure 9.30, for example, showsthe color of the indicator calmagite as a function of pH and pMg, whereH 2 In – , HIn 2– , and In 3– are different forms of the uncomplexed indicator,and MgIn – is the Mg 2+ –calmagite complex. Because the color of calmagite’smetal–indicator complex is red, it use as a metallochromic indicatorhas a practical pH range of approximately 8.5–11 where the uncomplexedindicator, HIn 2– , has a blue color.Table 9.14 provides examples of metallochromic indicators and themetal ions and pH conditions for which they are useful. Even if a suitableindicator does not exist, it is often possible to complete an EDTA titrationTable 9.14 Selected Metallochromic IndicatorsIndicator pH Range Metal Ions acalmagite 8.5–11 Ba, Ca, Mg, Zneriochrome Black T 7.5–10.5 Ba, Ca, Mg, Zneriochrome Blue Black R 8–12 Ca, Mg, Zn, Cumurexide 6–13 Ca, Ni, CuPAN 2–11 Cd, Cu, Znsalicylic acid 2–3 Fea metal ions in italic font have poor end pointsFigure 9.30 is essentially a two-variableladder diagram. The solid lines are equivalentto a step on a conventional ladder diagram,indicating conditions where two (orthree) species are equal in concentration.

470 Analytical Chemistry 2.0(a) 10(b)pMgpMg8642H 2 In – HIn 2– In 3–MgIn –Mg(OH) 2 (s)HIn 2–pMg = logK f,MgIn –MgIn –indicatoris color of HIn 2–indicator’scolor transitionrangeindicatoris color of MgIn –7 8 9 10 11 12 13 14pHFigure 9.30 (a) Predominance diagram for the metallochromic indicator calmagite showing the most important formand color of calmagite as a function of pH and pMg, where H 2 In – , HIn 2– , and In 3– are uncomplexed forms of calmagite,and MgIn – is its complex with Mg 2+ . Conditions to the right of the dashed line, where Mg 2+ precipitates as Mg(OH) 2 ,are not analytically useful for a complexation titration. A red to blue end point is possible if we maintain the titrand’spH in the range 8.5–11. (b) Diagram showing the relationship between the concentration of Mg 2+ (as pMg) and theindicator’s color. The ladder diagram defines pMg values where MgIn – and HIn – are predominate species. The indicatorchanges color when pMg is between logK f – 1 and logK f + 1.by introducing a small amount of a secondary metal–EDTA complex, ifthe secondary metal ion forms a stronger complex with the indicator and aweaker complex with EDTA than the analyte. For example, calmagite givespoor end points when titrating Ca 2+ with EDTA. Adding a small amount ofMg 2+ –EDTA to the titrand gives a sharper end point. Because Ca 2+ formsa stronger complex with EDTA, it displaces Mg 2+ , which then forms thered-colored Mg 2+ –calmagite complex. At the titration’s end point, EDTAdisplaces Mg 2+ from the Mg 2+ –calmagite complex, signaling the end pointby the presence of the uncomplexed indicator’s blue form.Two other methods for finding the endpoint of a complexation titration are athermometric titration, in which we monitorthe titrand’s temperature as we add thetitrant, and a potentiometric titration inwhich we use an ion selective electrode tomonitor the metal ion’s concentration aswe add the titrant. The experimental approachis essentially identical to that describedearlier for an acid–base titration,to which you may refer.See Chapter 11 for more details about ionselective electrodes.Fi n d i n g t h e En d p o i n t b y Mo n i t o r i n g Ab s o r b a n c eAn important limitation when using an indicator is that we must be ableto see the indicator’s change in color at the end point. This may be difficultif the solution is already colored. For example, when titrating Cu 2+ withEDTA, ammonia is used to adjust the titrand’s pH. The intensely coloredCu(NH 3 ) 4 2+ complex obscures the indicator’s color, making an accuratedetermination of the end point difficult. Other absorbing species presentwithin the sample matrix may also interfere. This is often a problem whenanalyzing clinical samples, such as blood, or environmental samples, suchas natural waters.

Chapter 9 Titrimetric Methods471If at least one species in a complexation titration absorbs electromagneticradiation, we can identify the end point by monitoring the titrand’sabsorbance at a carefully selected wavelength. For example, we can identifythe end point for a titration of Cu 2+ with EDTA, in the presence of NH 3 bymonitoring the titrand’s absorbance at a wavelength of 745 nm, where theCu(NH 3 ) 4 2+ complex absorbs strongly. At the beginning of the titrationthe absorbance is at a maximum. As we add EDTA, however, the reaction2+ 4− 2−Cu(NH ) ( aq) + Y ( aq) → CuY ( aq) + 4NH( aq)3 4decreases the concentration of Cu(NH 3 ) 4 2+ and decreases the absorbanceuntil we reach the equivalence point. After the equivalence point the absorbanceremains essentially unchanged. The resulting spectrophotometrictitration curve is shown in Figure 9.31a. Note that the titration curve’sy-axis is not the actual absorbance, A, but a corrected absorbance, A corrA A V + VEDTA= ×corrVwhere V EDTA and V Cu are, respectively, the volumes of EDTA and Cu. Correctingthe absorbance for the titrand’s dilution ensures that the spectrophotometrictitration curve consists of linear segments that we can extrapolateto find the end point. Other common spectrophotometric titration curvesare shown in Figures 9.31b-f.CuCu3(a)(b)A corrA corrVolume of TitrantVolume of Titrant(c)(d)A corrA corrVolume of TitrantVolume of Titrant(e)A corrVolume of Titrant(f)A corrVolume of TitrantFigure 9.31 Examples of spectrophotometric titrationcurves: (a) only the titrand absorbs; (b) only the titrantabsorbs; (c) only the product of the titration reactionabsorbs; (d) both the titrand and the titrant absorb;(e) both the titration reaction’s product and the titrantabsorb; (f) only the indicator absorbs. The red arrowsindicate the end points for each titration curve.

472 Analytical Chemistry 2.0The best way to appreciate the theoreticaland practical details discussed in this sectionis to carefully examine a typical complexationtitrimetric method. Althougheach method is unique, the followingdescription of the determination of thehardness of water provides an instructiveexample of a typical procedure. The descriptionhere is based on Method 2340Cas published in Standard Methods for theExamination of Water and Wastewater,20th Ed., American Public Health Association:Washington, D. C., 1998.Representative Method 9.2Determination of Hardness of Water and WastewaterDescription o f t h e Me t h o dThe operational definition of water hardness is the total concentrationof cations in a sample capable of forming insoluble complexes with soap.Although most divalent and trivalent metal ions contribute to hardness,the most important are Ca 2+ and Mg 2+ . Hardness is determined by titratingwith EDTA at a buffered pH of 10. Calmagite is used as an indicator.Hardness is reported as mg CaCO 3 /L.Pr o c e d u r eSelect a volume of sample requiring less than 15 mL of titrant to keepthe analysis time under 5 minutes and, if necessary, dilute the sample to50 mL with distilled water. Adjust the sample’s pH by adding 1–2 mLof a pH 10 buffer containing a small amount of Mg 2+ –EDTA. Add 1–2drops of indicator and titrate with a standard solution of EDTA until thered-to-blue end point is reached (Figure 9.32).Qu e s t i o n s1. Why is the sample buffered to a pH of 10? What problems might youexpect at a higher pH or a lower pH?Of the cations contributing to hardness, Mg 2+ forms the weakestcomplex with EDTA and is the last cation to be titrated. Calmagiteis a useful indicator because it gives a distinct end point when titratingMg 2+ . Because of calmagite’s acid–base properties, the range ofpMg values over which the indicator changes color is pH–dependent(Figure 9.30). Figure 9.33 shows the titration curve for a 50-mL solutionof 10 –3 M Mg 2+ with 10 –2 M EDTA at pHs of 9, 10, and 11.Superimposed on each titration curve is the range of conditions forwhich the average analyst will observe the end point. At a pH of 9 anearly end point is possible, leading to a negative determinate error. AFigure 9.32 End point for the titration of hardness withEDTA using calmagite as an indicator; the indicator is:(a) red prior to the end point due to the presence ofthe Mg 2+ –indicator complex; (b) purple at the titration’send point; and (c) blue after the end point due to thepresence of uncomplexed indicator.

Chapter 9 Titrimetric Methods473late end point and a positive determinate error are possible if we usea pH of 11.2. Why is a small amount of the Mg 2+ –EDTA complex added to thebuffer?The titration’s end point is signaled by the indicator calmagite. Theindicator’s end point with Mg 2+ is distinct, but its change in colorwhen titrating Ca 2+ does not provide a good end point. If the sampledoes not contain any Mg 2+ as a source of hardness, then the titration’send point is poorly defined, leading to inaccurate and impreciseresults.Adding a small amount of Mg 2+ –EDTA to the buffer ensures thatthe titrand includes at least some Mg 2+ . Because Ca 2+ forms a strongercomplex with EDTA, it displaces Mg 2+ from the Mg 2+ –EDTAcomplex, freeing the Mg 2+ to bind with the indicator. This displacementis stoichiometric, so the total concentration of hardness cationsremains unchanged. The displacement by EDTA of Mg 2+ from theMg 2+ –indicator complex signals the titration’s end point.3. Why does the procedure specify that the titration take no longer than5 minutes?A time limitation suggests that there is a kinetically controlled interference,possibly arising from a competing chemical reaction. In thiscase the interference is the possible precipitation of CaCO 3 at a pHof 10.9C.4 Quantitative ApplicationsAlthough many quantitative applications of complexation titrimetry havebeen replaced by other analytical methods, a few important applicationscontinue to be relevant. In the section we review the general application ofcomplexation titrimetry with an emphasis on applications from the analysisof water and wastewater. First, however, we discuss the selection and standardizationof complexation titrants.Se l e c t i o n a n d St a n d a r d i z a t i o n o f Ti t r a n t sEDTA is a versatile titrant that can be used to analyze virtually all metalions. Although EDTA is the usual titrant when the titrand is a metal ion, itcannot be used to titrate anions. In the later case, Ag + or Hg 2+ are suitabletitrants.Solutions of EDTA are prepared from its soluble disodium salt,Na 2 H 2 Y•2H 2 O and standardized by titrating against a solution made fromthe primary standard CaCO 3 . Solutions of Ag + and Hg 2+ are prepared usingAgNO 3 and Hg(NO 3 ) 2 , both of which are secondary standards. Standardizationis accomplished by titrating against a solution prepared fromprimary standard grade NaCl.pMgpMgpMg1086420108642086420pH 90 2 4 6 8 10Volume of EDTA (mL)0 2 4 6 8 10Volume of EDTA (mL)10 pH 11early end pointpH 10late end point0 2 4 6 8 10Volume of EDTA (mL)Figure 9.33 Titration curves for 50 mLof 10 –3 M Mg 2+ with 10 –3 M EDTAat pHs 9, 10, and 11 using calmagiteas an indicator. The range of pMg andvolume of EDTA over which the indicatorchanges color is shown for eachtitration curve.

474 Analytical Chemistry 2.0In o r g a n i c An a l y s i sComplexation titrimetry continues to be listed as a standard method forthe determination of hardness, Ca 2+ , CN – , and Cl – in waters and wastewaters.The evaluation of hardness was described earlier in RepresentativeMethod 9.2. The determination of Ca 2+ is complicated by the presence ofMg 2+ , which also reacts with EDTA. To prevent an interference the pH isadjusted to 12–13, precipitating Mg 2+ as Mg(OH) 2 . Titrating with EDTAusing murexide or Eriochrome Blue Black R as the indicator gives theconcentration of Ca 2+ .Cyanide is determined at concentrations greater than 1 mg/L by makingthe sample alkaline with NaOH and titrating with a standard solutionof AgNO 3 , forming the soluble Ag(CN) 2 – complex. The end point is determinedusing p-dimethylaminobenzalrhodamine as an indicator, with thesolution turning from a yellow to a salmon color in the presence of excessAg + .Chloride is determined by titrating with Hg(NO 3 ) 2 , forming HgCl 2 (aq).The sample is acidified to a pH of 2.3–3.8 and diphenylcarbazone, whichforms a colored complex with excess Hg 2+ , serves as the indicator. A pHindicator—xylene cyanol FF—is added to ensure that the pH is within thedesired range. The initial solution is a greenish blue, and the titration iscarried out to a purple end point.Qu a n t i t a t i ve Ca l c u l at i o n sNote that in this example, the analyte isthe titrant.The quantitative relationship between the titrand and the titrant is determinedby the stoichiometry of the titration reaction. For a titration usingEDTA, the stoichiometry is always 1:1.Example 9.7The concentration of a solution of EDTA was determined by standardizingagainst a solution of Ca 2+ prepared using a primary standard of CaCO 3 . A0.4071-g sample of CaCO 3 was transferred to a 500-mL volumetric flask,dissolved using a minimum of 6 M HCl, and diluted to volume. Aftertransferring a 50.00-mL portion of this solution to a 250-mL Erlenmeyerflask, the pH was adjusted by adding 5 mL of a pH 10 NH 3 –NH 4 Cl buffercontaining a small amount of Mg 2+ –EDTA. After adding calmagite asan indicator, the solution was titrated with the EDTA, requiring 42.63mL to reach the end point. Report the molar concentration of EDTA inthe titrant.So l u t i o nThe primary standard of Ca 2+ has a concentration of2+0.4071 gCaCO31 molCa×0.5000 L 100.09 gCaCO3= 8.135× 10 −3MCa2+

Chapter 9 Titrimetric Methods475The moles of Ca 2+ in the titrand is−2+ 2+8. 135× 10 MCa × 0. 05000 LCa = 4.068×103 − 4molCa 2+which means that 4.068×10 –4 moles of EDTA are used in the titration.The molarity of EDTA in the titrant is−44.068×10 molEDTA−= 9.543×103 MEDTA0.04263 LEDTAPractice Exercise 9.14A 100.0-mL sample is analyzed for hardness using the procedure outlinedin Representative Method 9.2, requiring 23.63 mL of 0.0109 MEDTA. Report the sample’s hardness as mg CaCO 3 /L.Click here to review your answer to this exercise.As shown in the following example, we can easily extended this calculationto complexation reactions using other titrants.Example 9.8The concentration of Cl – in a 100.0-mL sample of water from a freshwateraquifer was tested for the encroachment of sea water by titratingwith 0.0516 M Hg(NO 3 ) 2 . The sample was acidified and titrated to thediphenylcarbazone end point, requiring 6.18 mL of the titrant. Report theconcentration of Cl – , in mg/L, in the aquifer.So l u t i o nThe reaction between Cl – and Hg 2+ produces a metal–ligand complex ofHgCl 2 (aq). Each mole of Hg 2+ reacts with 2 moles of Cl – ; thus0.0516 molHg(NO )L3 2× 0.00618 LHg(NO )−2 mol Cl×molHg(NO )3 23 2−35.453 gCl× = 00 .226 gCl −−molClare in the sample. The concentration of Cl – in the sample is0.0226 gCl0.1000 L−1000 mg× = 226 mg / LgPractice Exercise 9.15A 0.4482-g sample of impure NaCN is titrated with 0.1018 M AgNO 3 , requi i iring39.68 mL to reach the end point. Report the purity of the sample as %w/w NaCN.Click here to review your answer to this exercise.

476 Analytical Chemistry 2.0Finally, complex titrations involving multiple analytes or back titrationsare possible.Example 9.9An alloy of chromel containing Ni, Fe, and Cr was analyzed by a complexationtitration using EDTA as the titrant. A 0.7176-g sample of the alloywas dissolved in HNO 3 and diluted to 250 mL in a volumetric flask. A50.00-mL aliquot of the sample, treated with pyrophosphate to mask theFe and Cr, required 26.14 mL of 0.05831 M EDTA to reach the murexideend point. A second 50.00-mL aliquot was treated with hexamethylenetetramineto mask the Cr. Titrating with 0.05831 M EDTA required 35.43mL to reach the murexide end point. Finally, a third 50.00-mL aliquotwas treated with 50.00 mL of 0.05831 M EDTA, and back titrated to themurexide end point with 6.21 mL of 0.06316 M Cu 2+ . Report the weightpercents of Ni, Fe, and Cr in the alloy.So l u t i o nThe stoichiometry between EDTA and each metal ion is 1:1. For each ofthe three titrations, therefore, we can easily equate the moles of EDTA tothe moles of metal ions that are titrated.Titration1:moles Ni = molesEDTATitration2: molesNi+ moles Fe = molesEDTATitration3:molesNi+ moles Fe + molesCr+ moles Cu = moles EDTAWe can use the first titration to determine the moles of Ni in our 50.00-mLportion of the dissolved alloy. The titration uses0.05831molEDTAL× 0. 02614 LEDTA = 1.524× 10 −3molEDTAwhich means the sample contains 1.524×10 –3 mol Ni.Having determined the moles of EDTA reacting with Ni, we can use thesecond titration to determine the amount of Fe in the sample. The secondtitration uses0.05831molEDTAL× 0. 03543 LEDTA = 2.066× 10 −3molEDTAof which 1.524×10 –3 mol are used to titrate Ni. This leaves 5.42×10 –4mol of EDTA to react with Fe; thus, the sample contains 5.42×10 –4 molof Fe.Finally, we can use the third titration to determine the amount of Cr inthe alloy. The third titration uses0.05831molEDTAL× 0. 05000 LEDTA = 2.916× 10 −3molEDTA

Chapter 9 Titrimetric Methods477of which 1.524×10 –3 mol are used to titrate Ni and 5.42×10 –4 mol areused to titrate Fe. This leaves 8.50×10 –4 mol of EDTA to react with Cuand Cr. The amount of EDTA reacting with Cu is2+0.06316 molCu2+× 0.00621 LCuL1 molEDTA−× = 392 . × 10 4 molEDTAmolCu2+leaving 4.58×10 –4 mol of EDTA to react with Cr. The sample, therefore,contains 4.58×10 –4 mol of Cr.Having determined the moles of Ni, Fe, and Cr in a 50.00-mL portionof the dissolved alloy, we can calculate the %w/w of each analyte in thealloy.–31.524×10 molNi58. 69 g Ni× 250.0 mL×50.00 mLmolNi= 0.44720.4472 gNi× 100 = 62. 32%w/wNi0.7176 gsample–45.42×10 molFe55. 847 g Fe× 250.0 mL×50.00 mLmolFe0.151 gFe× 100 = 21.%0 w/wFe0.7176 gsample–44.58×10 molCr51. 996 g Cr× 250.0 mL×50.00 mLmolCr0.119 gCr× 100 = 16.%6 w/wFe0.7176 gsamplegNi= 0.151 gFe= 0.119gCrPractice Exercise 9.16A indirect complexation titration with EDTA can be used to determinethe concentration of sulfate, SO 2– 4 , in a sample. A 0.1557-g sample isdissolved in water, any sulfate present is precipitated as BaSO 4 by addingBa(NO 3 ) 2 . After filtering and rinsing the precipitate, it is dissolvedin 25.00 mL of 0.02011 M EDTA. The excess EDTA is then titratedw iith 0.01113 M Mg 2+ , requiring 4.23 mL to reach the end point.Calculate the %w/w Na 2 SO 4 in the sample.Click here to review your answer to this exercise.

478 Analytical Chemistry 2.0108(a)pH 9108(b) pH 3 Ni 2+pH 9 Ca 2+pCa64pNi or pCa642pH 32A corrfirst end pointFigure 9.35 Spectrophotometric titrationcurve for the complexation titration of amixture of two analytes. The red arrows indicatethe end points for each analyte.00 20 40 60 80 100Volume of EDTA (mL)Figure 9.34 Titration curves illustrating how we can use the titrand’s pH to control EDTA’s selectivity.(a) Titration of 50.0 mL of 0.010 M Ca 2+ at a pH of 3 and a pH of 9 using 0.010 M EDTA.At a pH of 3 the CaY 2– complex is too weak to successfully titrate. (b) Titration of a 50.0 mLmixture of 0.010 M Ca 2+ and 0.010 M Ni 2+ at a pH of 3 and a pH of 9 using 0.010 M EDTA.At a pH of 3 EDTA reacts only with Ni 2+ . When the titration is complete, raising the pH to 9allows for the titration of Ca 2+ .second end pointVolume of Titrant9C.5 Evaluation of Complexation TitrimetryThe scale of operations, accuracy, precision, sensitivity, time, and cost of acomplexation titration are similar to those described earlier for acid–basetitrations. Complexation titrations, however, are more selective. AlthoughEDTA forms strong complexes with most metal ion, by carefully controllingthe titrand’s pH we can analyze samples containing two or more analytes.The reason we can use pH to provide selectivity is shown in Figure9.34a. A titration of Ca 2+ at a pH of 9 gives a distinct break in the titrationcurve because the conditional formation constant for CaY 2– of 2.6 × 10 9 islarge enough to ensure that the reaction of Ca 2+ and EDTA goes to completion.At a pH of 3, however, the conditional formation constant of 1.23 isso small that very little Ca 2+ reacts with the EDTA.Suppose we need to analyze a mixture of Ni 2+ and Ca 2+ . Both analytesreact with EDTA, but their conditional formation constants differ significantly.If we adjust the pH to 3 we can titrate Ni 2+ with EDTA withouttitrating Ca 2+ (Figure 9.34b). When the titration is complete, we adjustthe titrand’s pH to 9 and titrate the Ca 2+ with EDTA.A spectrophotometric titration is a particularly useful approach for analyzinga mixture of analytes. For example, as shown in Figure 9.35, we candetermine the concentration of a two metal ions if there is a differencebetween the absorbance of the two metal-ligand complexes.9DRedox Titrations00 20 40 60 80 100Volume of EDTA (mL)Analytical titrations using redox reactions were introduced shortly after thedevelopment of acid–base titrimetry. The earliest Redox titration tookadvantage of the oxidizing power of chlorine. In 1787, Claude Berthollet

Chapter 9 Titrimetric Methods479introduced a method for the quantitative analysis of chlorine water (a mixtureof Cl 2 , HCl, and HOCl) based on its ability to oxidize indigo, a dyethat is colorless in its oxidized state. In 1814, Joseph Gay-Lussac developeda similar method for determining chlorine in bleaching powder. In bothmethods the end point is a change in color. Before the equivalence pointthe solution is colorless due to the oxidation of indigo. After the equivalencepoint, however, unreacted indigo imparts a permanent color to thesolution.The number of redox titrimetric methods increased in the mid-1800swith the introduction of MnO 4 – , Cr 2 O 7 2– , and I 2 as oxidizing titrants,and of Fe 2+ and S 2 O 3 2– as reducing titrants. Even with the availability ofthese new titrants, redox titrimetry was slow to develop due to the lack ofsuitable indicators. A titrant can serve as its own indicator if its oxidizedand reduced forms differ significantly in color. For example, the intenselypurple MnO 4 – ion serves as its own indicator since its reduced form, Mn 2+ ,is almost colorless. Other titrants require a separate indicator. The first suchindicator, diphenylamine, was introduced in the 1920s. Other redox indicatorssoon followed, increasing the applicability of redox titrimetry.9D.1 Redox Titration CurvesTo evaluate a redox titration we need to know the shape of its titration curve.In an acid–base titration or a complexation titration, the titration curveshows how the concentration of H 3 O + (as pH) or M n+ (as pM) changes aswe add titrant. For a redox titration it is convenient to monitor the titrationreaction’s potential instead of the concentration of one species.You may recall from Chapter 6 that the Nernst equation relates a solution’spotential to the concentrations of reactants and products participatingin the redox reaction. Consider, for example, a titration in which a titrandin a reduced state, A red , reacts with a titrant in an oxidized state, B ox .A + B B + Ared ox red oxwhere A ox is the titrand’s oxidized form, and B red is the titrant’s reducedform. The reaction’s potential, E rxn , is the difference between the reductionpotentials for each half-reaction.E = E −Erxn Box / BredAox / AredAfter each addition of titrant the reaction between the titrand and thetitrant reaches a state of equilibrium. Because the potential at equilibriumis zero, the titrand’s and the titrant’s reduction potentials are identical.E= EBox / BredAox / AredThis is an important observation because we can use either half-reaction tomonitor the titration’s progress.

480 Analytical Chemistry 2.0Although the Nernst equation is writtenin terms of the half-reaction’s standardstate potential, a matrix-dependent formalpotential often is used in its place.See Appendix 13 for the standard state potentialsand formal potentials for selectedhalf-reactions.Before the equivalence point the titration mixture consists of appreciablequantities of the titrand’s oxidized and reduced forms. The concentration ofunreacted titrant, however, is very small. The potential, therefore, is easierto calculate if we use the Nernst equation for the titrand’s half-reactionErxn= o RT AE −redln [ ]Aox/ Ared nF [ A ]After the equivalence point it is easier to calculate the potential using theNernst equation for the titrant’s half-reaction.Erxnox= o RT BE −redln [ ]Box/ Bred nF [ B ]oxCa l c u l at i n g t h e Ti t r a t i o n Cu r v eIn 1 M HClO 4 , the formal potential forthe reduction of Fe 3+ to Fe 2+ is +0.767 V,and the formal potential for the reductionof Ce 4+ to Ce 3+ is +1.70 V.Step 1: Calculate the volume of titrantneeded to reach the equivalence point.Let’s calculate the titration curve for the titration of 50.0 mL of 0.100 MFe 2+ with 0.100 M Ce 4+ in a matrix of 1 M HClO 4 . The reaction in thiscase isFe 2 +Ce 4 +Ce 3 +Fe3 +( aq) + ( aq) ( aq) + ( aq)9.15Because the equilibrium constant for reaction 9.15 is very large—it is approximately6 × 10 15 —we may assume that the analyte and titrant reactcompletely.The first task is to calculate the volume of Ce 4+ needed to reach the titration’sequivalence point. From the reaction’s stoichiometry we know thatmolesFe= moles Ce2+ 4+M × V = M × VFe Fe Ce CeStep 2: Calculate the potential before theequivalence point by determining theconcentrations of the titrand’s oxidizedand reduced forms, and using the Nernstequation for the titrand’s reduction halfreaction.Solving for the volume of Ce 4+ gives the equivalence point volume asVeqM VFe FeM)(50.0 mL)= V = = ( 0.100 = 50.0 mLCeM (0.100 M)CeBefore the equivalence point, the concentration of unreacted Fe 2+ andthe concentration of Fe 3+ are easy to calculate. For this reason we find thepotential using the Nernst equation for the Fe 3+ /Fe 2+ half-reaction.2+o RT FeE = E3+ 2+− log [ ] =Fe / Fe3+nF [ Fe ]+ 0. 767V−0. log [ 2+Fe05916]3+[ Fe ]9.16For example, the concentrations of Fe 2+ and Fe 3+ after adding 10.0 mL oftitrant are

Chapter 9 Titrimetric Methods4812+ 4+2+initialmoles Fe − molesCe added M V[ Fe ] ==total volumeV( 0.100 M) (50.0 mL) −( 0.100 M)(10.0 mL)=50.0 mL + 10.0 mL4+3+molesCe addedCe[ Fe ] = = M Vtotalvolume + VV FeCeCeFe Fe Ce CeFe− M V+ V( 0.100 M)(10.0 mL)== 1.67× 10 −2M50.0 mL + 10.0 mLCe= 667 . × 10 −2Substituting these concentrations into equation 9.16 gives a potential of⎛667 . × 10E =+ 0. 767 V −0. 05916log⎝⎜167 . × 10−2−2M⎞M⎠⎟ =+0 . 731 VMAfter the equivalence point, the concentration of Ce 3+ and the concentrationof excess Ce 4+ are easy to calculate. For this reason we find thepotential using the Nernst equation for the Ce 4+ /Ce 3+ half-reaction.3+RTE = E − Ce CnF=4+ 3+log [ ]oCe/ e4+[ Ce ]+ 170 . V −0. log [ 3+Ce05916]4+[ Ce ]9.17For example, after adding 60.0 mL of titrant, the concentrations of Ce 3+and Ce 4+ are2+3+initialmoles FeFe F[ Ce ] = = M V etotalvolume V + V( 0.100 M)(50.0 mL)=50.0 mL + 60.0mLFeCe= 455 . × 10 −3MStep 3: Calculate the potential after theequivalence point by determining theconcentrations of the titrant’s oxidizedand reduced forms, and using the Nernstequation for the titrant’s reduction halfreaction.4+molesCe added−initial molesFe[ Ce ] =total volume4+ 2+M V=V( 0.100 M) (60.0 mL) −( 0.100 M)(50.0 mL)=50.0 mL + 60.0 mLCe Ce Fe FeFe− M V+ VCe= 909 . × 10 −3Substituting these concentrations into equation 9.17 gives a potential of⎛455 . × 10E =+ 170 . V −0. 05916log⎝⎜909 . × 10−2−3M⎞M⎠⎟ =+166 . VMAt the titration’s equivalence point, the potential, E eq , in equation 9.16and equation 9.17 are identical. Adding the equations together to givesStep 4: Calculate the potential at theequivalence point.

482 Analytical Chemistry 2.0Table 9.15 Data for the Titration of 50.0 mL of 0.100 MFe 2+ with 0.100 M Ce 4+Volume of Ce 4+ (mL) E (V) Volume Ce 4+ (mL) E (V)10.0 0.731 60.0 1.6620.0 0.757 70.0 1.6830.0 0.777 80.0 1.6940.0 0.803 90.0 1.6950.0 1.23 100.0 1.7022E E ooFe=3 2E Ce4 C30 05916eq + +++ +− . log [ + 3+][ Ce ]Fe / Fe/ e3+ 4+[ Fe ][ Ce ]Because [Fe 2+ ] = [Ce 4+ ] and [Ce 3+ ] = [Fe 3+ ] at the equivalence point, thelog term has a value of zero and the equivalence point’s potential isEeqooE3+ 2+ + EFe Fe Ce4+ C3+/ / e0. 767 V+170 . V==22= 123 . VAdditional results for this titration curve are shown in Table 9.15 and Figure9.36.Practice Exercise 9.17Calculate the titration curve for the titration of 50.0 mL of 0.0500 MSn 2+ with 0.100 M Tl 3+ . Both the titrand and the titrant are 1.0 M inHCl. The titration reaction is3+2+ 4+ +Sn ( aq) + Tl ( aq) → Sn ( aq) + Tl ( aq)Click here to review your answer to this exercise.1.61.4E (V)1.21.0Figure 9.36 Titration curve for the titration of 50.0 mLof 0.100 M Fe 2+ with 0.100 M Ce 4+ . The red pointscorrespond to the data in Table 9.15. The blue lineshows the complete titration curve.0.80.60 20 40 60 80 100Volume of Ce 4+ (mL)

Chapter 9 Titrimetric Methods483Sk e t c h i n g a Re d o x Ti t r a t i o n Cu r v eTo evaluate the relationship between a titration’s equivalence point and itsend point we need to construct only a reasonable approximation of theexact titration curve. In this section we demonstrate a simple method forsketching a redox titration curve. Our goal is to sketch the titration curvequickly, using as few calculations as possible. Let’s use the titration of 50.0mL of 0.100 M Fe 2+ with 0.100 M Ce 4+ in a matrix of 1 M HClO 4 .We begin by calculating the titration’s equivalence point volume, which,as we determined earlier, is 50.0 mL. Next, we draw our axes, placing thepotential, E, on the y-axis and the titrant’s volume on the x-axis. To indicatethe equivalence point’s volume, we draw a vertical line correspondingto 50.0 mL of Ce 4+ . Figure 9.37a shows the result of the first step in oursketch.Before the equivalence point, the potential is determined by a redoxbuffer of Fe 2+ and Fe 3+ . Although we can easily calculate the potential usingthe Nernst equation, we can avoid this calculation by making a simpleassumption. You may recall from Chapter 6 that a redox buffer operatesover a range of potentials that extends approximately ±(0.05916/n) unitoon either side of E Fe3+ Fe2+. The potential is at the buffer’s lower limit/oE = E Fe3+ Fe2+– 0.05916/when the concentration of Fe 2+ is 10 greater than that of Fe 3+ . The bufferreaches its upper potentialoE = E Fe3+ Fe2++ 0.05916/when the concentration of Fe 2+ is 10 smaller than that of Fe 3+ . The redoxbuffer spans a range of volumes from approximately 10% of the equivalencepoint volume to approximately 90% of the equivalence point volume.Figure 9.37b shows the second step in our sketch. First, we superimposea ladder diagram for Fe 2+ on the y-axis, using its E Fe3+ Fe2+value of 0.767/V and including the buffer’s range of potentials. Next, we add points representingthe pH at 10% of the equivalence point volume (a potential of0.708 V at 5.0 mL) and at 90% of the equivalence point volume (a potentialof 0.826 V at 45.0 mL).The third step in sketching our titration curve is to add two points after theequivalence point. Here the potential is controlled by a redox buffer of Ce 3+and Ce 4+ . The redox buffer is at its lower limit of E = E Ce4+ Ce3+– 0.05916/when the titrant reaches 110% of the equivalence point volume and theopotential is E Ce4+ Ce3+when the volume of Ce 4+ is 2V/eq .Figure 9.37c shows the third step in our sketch. First, we add a ladderdiagram for Ce 4+ o, including its buffer range, using its E Ce4+ Ce3+value of/1.70 V. Next, we add points representing the potential at 110% of V eq (avalue of 1.66 V at 55.0 mL) and at 200% of V eq (a value of 1.70 V at 100.0mL).ooThis is the same example that we used indeveloping the calculations for a redoxtitration curve. You can review the resultsof that calculation in Table 9.15 and Figure9.36.We used a similar approach when sketchingthe acid–base titration curve for thetitration of acetic acid with NaOH.We used a similar approach when sketchingthe complexation titration curve forthe titration of Mg 2+ with EDTA.

484 Analytical Chemistry 2.01.6(a)(b)1.61.41.4E (V)1.21.0E (V)1.21.00.80.80.60 20 40 60 80 100Volume of Ce 4+ (mL)0.60 20 40 60 80 100Volume of Ce 4+ (mL)(c)1.6(d)1.61.41.4E (V)1.2E (V)1.21.01.00.80.80.60 20 40 60 80 100Volume of Ce 4+ (mL)0.60 20 40 60 80 100Volume of Ce 4+ (mL)(e)1.6(f)1.61.41.4E (V)1.21.0E (V)1.21.00.80.80.60 20 40 60 80 100Volume of Ce 4+ (mL)0.60 20 40 60 80 100Volume of Ce 4+ (mL)Figure 9.37 Illustrations showing the steps in sketching an approximate titration curve for the titration of 50.0mL of 0.100 M Fe 2+ with 0.100 M Ce 4+ in 1 M HClO 4 : (a) locating the equivalence point volume; (b) plottingtwo points before the equivalence point; (c) plotting two points after the equivalence point; (d) preliminaryapproximation of titration curve using straight-lines; (e) final approximation of titration curve using a smoothcurve; (f) comparison of approximate titration curve (solid black line) and exact titration curve (dashed red line).See the text for additional details.

Chapter 9 Titrimetric Methods485Next, we draw a straight line through each pair of points, extending theline through the vertical line representing the equivalence point’s volume(Figure 9.37d). Finally, we complete our sketch by drawing a smooth curvethat connects the three straight-line segments (Figure 9.37e). A comparisonof our sketch to the exact titration curve (Figure 9.37f) shows that they arein close agreement.Practice Exercise 9.18Sketch the titration curve for the titration of 50.0 mL of 0.0500 M Sn 4+with 0.100 M Tl + . Both the titrand and the titrant are 1.0 M in HCl.The titration reaction is3+2+ 4+ +Sn ( aq) + Tl ( aq) → Sn ( aq) + Tl ( aq)Compare your sketch to your calculated titration curve from PracticeExercise 9.17.Click here to review your answer to this exercise.9D.2 Selecting and Evaluating the End pointA redox titration’s equivalence point occurs when we react stoichiometricallyequivalent amounts of titrand and titrant. As is the case with acid–baseand complexation titrations, we estimate the equivalence point of a complexationtitration using an experimental end point. A variety of methodsare available for locating the end point, including indicators and sensorsthat respond to a change in the solution conditions.Wh e r e is t h e Eq u i v a l e n c e Po i nt ?For an acid–base titration or a complexometric titration the equivalencepoint is almost identical to the inflection point on the steeping rising part ofthe titration curve. If you look back at Figure 9.7 and Figure 9.28, you willsee that the inflection point is in the middle of this steep rise in the titrationcurve, which makes it relatively easy to find the equivalence point when yousketch these titration curves. We call this a symmetric equivalence point.If the stoichiometry of a redox titration is symmetric—one mole of titrantreacts with each mole of titrand—then the equivalence point is symmetric.If the titration reaction’s stoichiometry is not 1:1, then the equivalencepoint is closer to the top or to bottom of the titration curve’s sharp rise. Inthis case we have an asymmetric equivalence point.Example 9.10Derive a general equation for the equivalence point’s potential when titratingFe 2+ with MnO 4 – .2+ − + 3+ 2+5Fe ( aq) + MnO ( aq) + 8H ( aq) → 5Fe ( aq) + Mn ( aq) + 4H O42We often use H + instead of H 3 O + whenwriting a redox reaction.

486 Analytical Chemistry 2.0So l u t i o nThe half-reactions for Fe 2+ and MnO 4 – areFe( aq) → Fe ( aq)+ e2+ 3+ −− + −2+MnO ( aq) + 8H ( aq)+ 5e → Mn ( aq) + 4H O()l42for which the Nernst equations are2+oFeE = E3+ 2+−0. 05916log [ ]Fe / Fe3+[ Fe ]2+o 0.05916 [ Mn ]E = E− 2+− logMnO Mn− +4 /5 [ MnO ][ H ]Before adding these two equations together we must multiply the secondequation by 5 so that we can combine the log terms; thus2+ 2+oo[ Fe ][ Mn ]6E = E3+ 2+ + 5E− 2+−0. 05916logFe / FeMnO4/ Mn3+ − +[ Fe ][ MnO ][ H ]At the equivalence point we know that2+ −[ Fe ] = 5×[ MnO ]3+ 2+[ Fe ] = 5×[ Mn ]Substituting these equalities into the previous equation and rearranginggives us a general equation for the potential at the equivalence point.− 2+56E = oE + o3 25E−[MnO ][ Mn ]420 05916eq + + − +. logFe / FeMnO4/ Mn2+ − +5[ Mn ][ MnO ][ H ]EEeqeqE=E=Eeq+ 5Eoo3+ 2+ − 2+Fe / FeMnO4/ Mn .6+ 5E−oo3+ 2+ − 2+Fe / FeMnO4/ Mn .E=6+ 5Eoo3+ 2+ − 2+Fe / FeMnO / Mn6440 0591660 05916×8+6841log[ H + ]84−0.07888pH4log[ H + ]88Instead of standard state potentials, youcan use formal potentials.Our equation for the equivalence point has two terms. The first term isa weighted average of the titrand’s and the titrant’s standard state potentials,in which the weighting factors are the number of electrons in theirrespective half-reactions. The second term shows that E eq for this titration

Chapter 9 Titrimetric Methods4871.4E (V)1.21.0equivalence point0.80.60.40 20 40 60 80 100Volume of MnO – 4 (mL)Figure 9.38 Titration curve for the titration of 50.0 mLof 0.100 M Fe 2+ with 0.0200 M MnO 4– at a fixed pH of1 (using H 2 SO 4 ). The equivalence point is shown by thered dot.is pH-dependent. At a pH of 1 (in H 2 SO 4 ), for example, the equivalencepoint has a potential ofE eq0. 768+ 5×1.51=− 0. 07888× 1=1.31 V6Figure 9.38 shows a typical titration curve for titration of Fe 2+ with MnO 4 – .Note that the titration’s equivalence point is asymmetrical.Practice Exercise 9.19Derive a general equation for the equivalence point’s potential for thetitration of U 4+ with Ce 4+ . The unbalanced reaction is4+ 4+ 2+ 3+Ce ( aq) + U ( aq) → UO ( aq) + Ce ( aq)What is the equivalence point’s potential if the pH is 1?Click here to review your answer to this exercise.Fi n d i n g t h e En d p o i n t w it h a n In d i c a t o rThree types of indicators are used to signal a redox titration’s end point. Theoxidized and reduced forms of some titrants, such as MnO 4 – , have differentcolors. A solution of MnO 4 – is intensely purple. In an acidic solution,however, permanganate’s reduced form, Mn 2+ , is nearly colorless. Whenusing MnO 4 – as a titrant, the titrand’s solution remains colorless until theequivalence point. The first drop of excess MnO 4 – produces a permanenttinge of purple, signaling the end point.Some indicators form a colored compound with a specific oxidized orreduced form of the titrant or the titrand. Starch, for example, forms a darkblue complex with I 3 – . We can use this distinct color to signal the presenceof excess I 3 – as a titrant—a change in color from colorless to blue—or the2

488 Analytical Chemistry 2.0For simplicity, In ox and In red are shownwithout specific charges. Because there isa change in oxidation state, In ox and In redcannot both be neutral.This is the same approach we took in consideringacid–base indicators and complexationindicators.completion of a reaction consuming I 3 – as the titrand—a change in colorfrom blue to colorless. Another example of a specific indicator is thiocyanate,SCN – , which forms a soluble red-colored complex of Fe(SCN) 2+ withFe 3+ .The most important class of indicators are substances that do not participatein the redox titration, but whose oxidized and reduced forms differin color. When we add a redox indicator to the titrand, the indicatorimparts a color that depends on the solution’s potential. As the solution’spotential changes with the addition of titrant, the indicator changes oxidationstate and changes color, signaling the end point.To understand the relationship between potential and an indicator’scolor, consider its reduction half-reactionInox+ ne− Inwhere In ox and In red are, respectively, the indicator’s oxidized and reducedforms. The Nernst equation for this half-reaction isoE = E −In ox /Inred0.05916nred[In ]redlog[In ]As shown in Figure 9.39, if we assume that the indicator’s color changesfrom that of In ox to that of In red when the ratio [In red ]/[In ox ] changes from0.1 to 10, then the end point occurs when the solution’s potential is withinthe rangeoE = E ±In ox /Inred0.05916noxEIn oxindicatoris color of In oxFigure 9.39 Diagram showing the relationship between Eand an indicator’s color. The ladder diagram defines potentialswhere In red and In ox are the predominate species. Theindicator changes color when E is within the rangeoE = E ±In ox /Inred0.05916noE = E In ox/InredIn redindicator’scolor transitionrangeindicatoris color of In red

Chapter 9 Titrimetric Methods489Table 9.16Selected Examples of Redox IndicatorsIndicator Color of In ox Color of In redoE In ox /In redindigo tetrasulfate blue colorless 0.36methylene blue blue colorless 0.53diphenylamine violet colorless 0.75diphenylamine sulfonic acid red-violet colorless 0.85tris(2,2´-bipyridine)iron pale blue red 1.120ferroin pale blue red 1.147tris(5-nitro-1,10-phenanthroline)iron pale blue red-violet 1.25A partial list of redox indicators is shown in Table 9.16. Examples of appropriateand inappropriate indicators for the titration of Fe 2+ with Ce 4+are shown in Figure 9.40.Ot h e r Me t h o d s f o r Fi n d i n g t h e En d p o i n tAnother method for locating a redox titration’s end point is a potentiometrictitration in which we monitor the change in potential while addingthe titrant to the titrand. The end point is found by visually examiningthe titration curve. The simplest experimental design for a potentiometrictitration consists of a Pt indicator electrode whose potential is governed bythe titrand’s or titrant’s redox half-reaction, and a reference electrode thathas a fixed potential. A further discussion of potentiometry is found inChapter 11. Other methods for locating the titration’s end point includethermometric titrations and spectrophotometric titrations.1.61.4E (V)1.21.00.80.6ferroindiphenylaminesulfonic acid0 20 40 60 80 100Volume of Ce 4+ (mL)Figure 9.40 Titration curve for the titration of 50.0mL of 0.100 M Fe 2+ with 0.100 M Ce 4+ . The endpoint transitions for the indicators diphenylaminesulfonic acid and ferroin are superimposed on thetitration curve. Because the transition for ferroinis too small to see on the scale of the x-axis—it requiresonly 1–2 drops of titrant—the color changeis expanded to the right.

490 Analytical Chemistry 2.0The best way to appreciate the theoreticaland practical details discussed in this sectionis to carefully examine a typical redoxtitrimetric method. Although each methodis unique, the following description ofthe determination of the total chlorineresidual in water provides an instructiveexample of a typical procedure. The descriptionhere is based on Method 4500-Cl B as published in Standard Methods forthe Examination of Water and Wastewater,20th Ed., American Public Health Association:Washington, D. C., 1998.Representative Method 9.3Determination of Total Chlorine ResidualDescription o f t h e Me t h o dThe chlorination of public water supplies produces several chlorine-containingspecies, the combined concentration of which is called the totalchlorine residual. Chlorine may be present in a variety of states, includingthe free residual chlorine, consisting of Cl 2 , HOCl and OCl – , and thecombined chlorine residual, consisting of NH 2 Cl, NHCl 2 , and NCl 3 .The total chlorine residual is determined by using the oxidizing powerof chlorine to convert I – to I 3 – . The amount of I 3 – formed is then determinedby titrating with Na 2 S 2 O 3 using starch as an indicator. Regardlessof its form, the total chlorine residual is reported as if Cl 2 is the onlysource of chlorine, and is reported as mg Cl/L.Pr o c e d u r eSelect a volume of sample requiring less than 20 mL of Na 2 S 2 O 3 to reachthe end point. Using glacial acetic acid, acidify the sample to a pH of 3–4,and add about 1 gram of KI. Titrate with Na 2 S 2 O 3 until the yellow colorof I 3 – begins to disappear. Add 1 mL of a starch indicator solution andcontinue titrating until the blue color of the starch–I 3 – complex disappears(Figure 9.41). Use a blank titration to correct the volume of titrantneeded to reach the end point for reagent impurities.Qu e s t i o n s1. Is this an example of a direct or an indirect analysis?This is an indirect analysis because the chlorine-containing speciesdo not react with the titrant. Instead, the total chlorine residual oxidizesI – to I 3 – , and the amount of I 3 – is determined by titrating withNa 2 S 2 O 3 .2. Why does the procedure rely on an indirect analysis instead of directlytitrating the chlorine-containing species using KI as a titrant?Because the total chlorine residual consists of six different species,a titration with I – does not have a single, well-defined equivalenceFigure 9.41 Endpoint for the determination of the total chlorine residual.(a) Acidifying the sample and adding KI forms a brown solutionof I 3– . (b) Titrating with Na2 S 2 O 3 converts I 3– to I– with thesolution fading to a pale yellow color as we approach the end point.(c) Adding starch forms the deep purple starch–I 3– complex. (d) As thetitration continues, the end point is a sharp transition from a purpleto a colorless solution. The change in color from (c) to (d) typicallytakes 1–2 drops of titrant.

Chapter 9 Titrimetric Methods491point. By converting the chlorine residual to an equivalent amountof I 3 – , the indirect titration with Na 2 S 2 O 3 has a single, useful equivalencepoint.Even if the total chlorine residual is from a single species, such asHOCl, a direct titration with KI is impractical. Because the productof the titration, I 3 – , imparts a yellow color, the titrand’s color wouldchange with each addition of titrant, making it difficult to find a suitableindicator.3. Both oxidizing and reducing agents can interfere with this analysis.Explain the effect of each type of interferent has on the total chlorineresidual.An interferent that is an oxidizing agent converts additional I – to I 3 – .Because this extra I 3 – requires an additional volume of Na 2 S 2 O 3 toreach the end point, we overestimate the total chlorine residual. Ifthe interferent is a reducing agent, it reduces back to I – some of theI 3 – produced by the reaction between the total chlorine residual andiodide. As a result. we underestimate the total chlorine residual.9D.3 Quantitative ApplicationsAlthough many quantitative applications of redox titrimetry have been replacedby other analytical methods, a few important applications continueto be relevant. In this section we review the general application of redoxtitrimetry with an emphasis on environmental, pharmaceutical, and industrialapplications. We begin, however, with a brief discussion of selectingand characterizing redox titrants, and methods for controlling the titrand’soxidation state.Ad j u s t i n g t h e Ti t r a n d ’s Oxidation St a t eIf a redox titration is to be used in a quantitative analysis, the titrand mustinitially be present in a single oxidation state. For example, iron can be determinedby a redox titration in which Ce 4+ oxidizes Fe 2+ to Fe 3+ . Dependingon the sample and the method of sample preparation, iron may initiallybe present in both the +2 and +3 oxidation states. Before titrating, we mustreduce any Fe 3+ to Fe 2+ . This type of pretreatment can be accomplishedusing an auxiliary reducing agent or oxidizing agent.A metal that is easy to oxidize—such as Zn, Al, and Ag—can serve asan auxiliary reducing agent. The metal, as a coiled wire or powder, isadded to the sample where it reduces the titrand. Because any unreactedauxiliary reducing agent will react with the titrant, it must be removed beforebeginning the titration. This can be accomplished by simply removingthe coiled wire, or by filtering.An alternative method for using an auxiliary reducing agent is to immobilizeit in a column. To prepare a reduction column an aqueous slurry

492 Analytical Chemistry 2.0of the finally divided metal is packed in a glass tube equipped with a porousplug at the bottom. The sample is placed at the top of the column andmoves through the column under the influence of gravity or vacuum suction.The length of the reduction column and the flow rate are selected toensure the analyte’s complete reduction.Two common reduction columns are used. In the Jones reductor thecolumn is filled with amalgamated zinc, Zn(Hg), prepared by briefly placingZn granules in a solution of HgCl 2 . Oxidation of zincZn(Hg) () s → Zn ( aq) + Hg()l + 2e2+ −provides the electrons for reducing the titrand. In the Walden reductorthe column is filled with granular Ag metal. The solution containing thetitrand is acidified with HCl and passed through the column where theoxidation of silverAg() s + Cl − ( aq) → AgCl()s + eprovides the necessary electrons for reducing the titrand. Table 9.17 providesa summary of several applications of reduction columns.Several reagents are commonly used as auxiliary oxidizing agents,including ammonium peroxydisulfate, (NH 4 ) 2 S 2 O 8 , and hydrogen peroxide,H 2 O 2 . Peroxydisulfate is a powerful oxidizing agentSO2( aq) + 2e→2SO( aq)2− − −2 84Table 9.17 Examples of Reactions For Reducing a Titrand’s Oxidation StateUsing a Reduction ColumnOxidizedTitrand Walden Reductor Jones ReductorCr 3+ — Cr 3 + −Cr2 +( aq) + e → ( aq)Cu 2+ 2+ − +2+ −Cu ( aq) + e →Cu( aq)Cu ( aq) + 2e→Cr()sFe 3+ Fe 3 + −Fe2 +( aq) + e → ( aq)Fe 3 + −Fe2 +( aq) + e → ( aq)−TiO 2+ —2+ 2+ − +MoO 2 MoO ( aq) + e →MoO( aq)22TiOMoO( aq) + 2H( aq)+ e2 + + −2+23+→ Ti ( aq) + HO()l( aq) + 4H( aq)+ 3e+ −3+→ Mo ( aq) + 2HO()l22VO 2+VO+( aq) + H ( aq)+ e VO ( aq) + 4H( aq)+ 3e+ 2+ −22+→ VO ( aq) + HO()l22+ −2+→ V ( aq) + 2H O()l2

Chapter 9 Titrimetric Methods493capable of oxidizing Mn 2+ to MnO 4 – , Cr 3+ to Cr 2 O 7 2– , and Ce 3+ to Ce 4+ .Excess peroxydisulfate is easily destroyed by briefly boiling the solution.The reduction of hydrogen peroxide in acidic solutionHO( aq) + 2H + ( aq) + 2e− → 2H O()l2 2 2provides another method for oxidizing a titrand. Excess H 2 O 2 is destroyedby briefly boiling the solution.Se l e c t i n g a n d St a n d a r d i z i n g a Ti t r a n tIf it is to be used quantitatively, the titrant’s concentration must remainstable during the analysis. Because a titrant in a reduced state is susceptibleto air oxidation, most redox titrations use an oxidizing agent as thetitrant. There are several common oxidizing titrants, including MnO 4 – ,Ce 4+ , Cr 2 O 7 2– , and I 3 – . Which titrant is used often depends on how easyit is to oxidize the titrand. A titrand that is a weak reducing agent needsa strong oxidizing titrant if the titration reaction is to have a suitable endpoint.The two strongest oxidizing titrants are MnO 4 – and Ce 4+ , for which thereduction half-reactions are− + − +MnO ( aq) + 8H ( aq) + 5e Mn ( aq) + 4HO()l42Ce( aq) + e Ce ( aq)4 + − 3 +2Solutions of Ce 4+ usually are prepared from the primary standard ceriumammonium nitrate, Ce(NO 3 ) 4 •2NH 4 NO 3 , in 1 M H 2 SO 4 . When preparedusing a reagent grade material, such as Ce(OH) 4 , the solution isstandardized against a primary standard reducing agent such as Na 2 C 2 O 4or Fe 2+ (prepared using iron wire) using ferroin as an indicator. Despite itsavailability as a primary standard and its ease of preparation, Ce 4+ is not asfrequently used as MnO 4 – because it is more expensive.Solutions of MnO 4 – are prepared from KMnO 4 , which is not availableas a primary standard. Aqueous solutions of permanganate are thermodynamicallyunstable due to its ability to oxidize water.The standardization reactions areCe2Ce4+ 2+( aq ) + Fe ( aq ) →Ce3+ 3+( aq ) + Fe ( aq )4+( aq ) HCO ( )2 2 4 aq+ →3+ +2Ce ( aq ) + 2CO ( g) + 2H( aq )2−−4MnO ( aq) + 2HO() l 4MnO () s + 3O ( g) + 4OH( aq )4 22 2This reaction is catalyzed by the presence of MnO 2 , Mn 2+ , heat, light, andthe presence of acids and bases. A moderately stable solution of permanganatecan be prepared by boiling it for an hour and filtering through a sinteredglass filter to remove any solid MnO 2 that precipitates. Standardization isaccomplished against a primary standard reducing agent such as Na 2 C 2 O 4or Fe 2+ (prepared using iron wire), with the pink color of excess MnO 4–signaling the end point. A solution of MnO 4 – prepared in this fashion isstable for 1–2 weeks, although the standardization should be recheckedperiodically.The standardization reactions are− 2+ +MnO4 ( aq ) + 5Fe ( aq ) + 8H( aq ) →−2+ 3+Mn ( aq ) + 5Fe ( aq ) + 4HO( l )2MnO 4 ( aq ) + 5HCO H2 2 4 ( aq ) + 6 ( aq ) →2+2Mn( aq) + 10CO( g) + 8HO( l )2 2+2

494 Analytical Chemistry 2.0Potassium dichromate is a relatively strong oxidizing agent whose principaladvantages are its availability as a primary standard and the long termstability of its solutions. It is not, however, as strong an oxidizing agentas MnO 4 – or Ce 4+ , which makes it less useful when the titrand is a weakreducing agent. Its reduction half-reaction is2− + − 3+Cr O ( aq) + 14H ( aq) + 6e 2Cr ( aq) + 7HO()l2 72Although a solution of Cr 2 O 7 2– is orange and a solution of Cr 3+ is green,neither color is intense enough to serve as a useful indicator. Diphenylaminesulfonic acid, whose oxidized form is red-violet and reduced form iscolorless, gives a very distinct end point signal with Cr 2 O 7 2– .Iodine is another important oxidizing titrant. Because it is a weakeroxidizing agent than MnO 4 – , Ce 4+ , and Cr 2 O 7 2– , it is useful only whenthe titrand is a stronger reducing agent. This apparent limitation, however,makes I 2 a more selective titrant for the analysis of a strong reducing agentin the presence of a weaker reducing agent. The reduction half-reaction forI 2 isI ( aq) + 2e− 2I− ( aq)2Because iodine is not very soluble in water, solutions are prepared byadding an excess of I – . The complexation reactionI ( aq) + I − ( aq) I− ( aq)2 3increases the solubility of I 2 by forming the more soluble triiodide ion, I 3 – .Even though iodine is present as I 3 – instead of I 2 , the number of electronsin the reduction half-reaction is unaffected.The standardization reaction is−2−3 2 3I( aq ) + 2S O ( aq ) →−3I( aq ) + 2S O ( aq )42−6A freshly prepared solution of KI is clear,but after a few days it may show a faint yellowcoloring due to the presence of I 3– .I− ( aq) + 2e− 3I− ( aq)3Solutions of I 3 – are normally standardized against Na 2 S 2 O 3 using starch asa specific indicator for I 3 – .An oxidizing titrant such as MnO 4 – , Ce 4+ , Cr 2 O 7 2– , and I 3 – , is usedwhen the titrand is in a reduced state. If the titrand is in an oxidized state,we can first reduce it with an auxiliary reducing agent and then completethe titration using an oxidizing titrant. Alternatively, we can titrate it usinga reducing titrant. Iodide is a relatively strong reducing agent that couldserve as a reducing titrant except that a solution of I – is susceptible to theair-oxidation of I – to I 3 – .3I ( aq) I 3( aq)+2e− − −Instead, adding an excess of KI reduces the titrand, releasing a stoichiometricamount of I 3 – . The amount of I 3 – produced is then determined by a backtitration using thiosulfate, S 2 O 3 2– , as a reducing titrant.

Chapter 9 Titrimetric Methods4952S O2( aq) 2S O ( aq)+2e2 − − −2 34 6Solutions of S 2 O 3 2– are prepared using Na 2 S 2 O 3 •5H 2 O, and must bestandardized before use. Standardization is accomplished by dissolving acarefully weighed portion of the primary standard KIO 3 in an acidic solutioncontaining an excess of KI. The reaction between IO 3 – and I –− − + −IO ( aq) + 8I ( aq) + 6H ( aq) → 3I ( aq) + 3H O()l3 3liberates a stoichiometric amount of I 3 – . By titrating this I 3 – with thiosulfate,using starch as a visual indicator, we can determine the concentrationof S 2 O 3 2– in the titrant.Although thiosulfate is one of the few reducing titrants that is not readilyoxidized by contact with air, it is subject to a slow decomposition tobisulfite and elemental sulfur. If used over a period of several weeks, a solutionof thiosulfate should be restandardized periodically. Several forms ofbacteria are able to metabolize thiosulfate, which also can lead to a changein its concentration. This problem can be minimized by adding a preservativesuch as HgI 2 to the solution.Another useful reducing titrant is ferrous ammonium sulfate,Fe(NH 4 ) 2 (SO4) 2 •6H 2 O, in which iron is present in the +2 oxidation state.A solution of Fe 2+ is susceptible to air-oxidation, but when prepared in 0.5M H 2 SO 4 it remains stable for as long as a month. Periodic restandardizationwith K 2 Cr 2 O 7 is advisable. The titrant can be used to directly titratethe titrand by oxidizing Fe 2+ to Fe 3+ . Alternatively, ferrous ammoniumsulfate is added to the titrand in excess and the quantity of Fe 3+ produceddetermined by back titrating with a standard solution of Ce 4+ or Cr 2 O 7 2– .2The standardization titration is−2−3 2 3I( aq ) + 2S O ( aq ) →−3I( aq ) + 2S O ( aq )42−6which is the same reaction used to standardizesolutions of I 3– . This approachto standardizing solutions of S 2 O 32– issimilar to the determination of the totalchlorine residual outlined in RepresentativeMethod 9.3.In o r g a n i c An a l y s i sOne of the most important applications of redox titrimetry is evaluatingthe chlorination of public water supplies. Representative Method 9.3,for example, describes an approach for determining the total chlorine residualby using the oxidizing power of chlorine to oxidize I – to I 3 – . Theamount of I 3 – is determined by back titrating with S 2 O 3 2– .The efficiency of chlorination depends on the form of the chlorinatingspecies. There are two contributions to the total chlorine residual—the freechlorine residual and the combined chlorine residual. The free chlorineresidual includes forms of chlorine that are available for disinfecting thewater supply. Examples of species contributing to the free chlorine residualinclude Cl 2 , HOCl and OCl – . The combined chlorine residual includesthose species in which chlorine is in its reduced form and, therefore, nolonger capable of providing disinfection. Species contributing to the combinedchlorine residual are NH 2 Cl, NHCl 2 and NCl 3 .When a sample of iodide-free chlorinated water is mixed with an excessof the indicator N,N-diethyl-p-phenylenediamine (DPD), the free chlorine

496 Analytical Chemistry 2.0oxidizes a stoichiometric portion of DPD to its red-colored form. Theoxidized DPD is then back titrated to its colorless form using ferrous ammoniumsulfate as the titrant. The volume of titrant is proportional to thefree residual chlorine.Having determined the free chlorine residual in the water sample, a smallamount of KI is added, catalyzing the reduction monochloramine, NH 2 Cl,and oxidizing a portion of the DPD back to its red-colored form. Titratingthe oxidized DPD with ferrous ammonium sulfate yields the amount ofNH 2 Cl in the sample. The amount of dichloramine and trichloramine aredetermined in a similar fashion.The methods described above for determining the total, free, or combinedchlorine residual also are used to establish a water supply’s chlorinedemand. Chlorine demand is defined as the quantity of chlorine needed tocompletely react with any substance that can be oxidized by chlorine, whilealso maintaining the desired chlorine residual. It is determined by addingprogressively greater amounts of chlorine to a set of samples drawn fromthe water supply and determining the total, free, or combined chlorineresidual.Another important example of redox titrimetry, which finds applicationsin both public health and environmental analyses is the determinationof dissolved oxygen. In natural waters, such as lakes and rivers, the level ofdissolved O 2 is important for two reasons: it is the most readily availableoxidant for the biological oxidation of inorganic and organic pollutants;and it is necessary for the support of aquatic life. In a wastewater treatmentplant dissolved O 2 is essential for the aerobic oxidation of waste materials.If the concentration of dissolved O 2 falls below a critical value, aerobicbacteria are replaced by anaerobic bacteria, and the oxidation of organicwaste produces undesirable gases, such as CH 4 and H 2 S.One standard method for determining the dissolved O 2 content ofnatural waters and wastewaters is the Winkler method. A sample of wateris collected without exposing it to the atmosphere, which might change theconcentration of dissolved O 2 . The sample is first treated with a solution ofMnSO 4 , and then with a solution of NaOH and KI. Under these alkalineconditions the dissolved oxygen oxidizes Mn 2+ to MnO 2 .2+ −2Mn ( aq) + 4OH ( aq) + O ( g) → 2MnO () s + 2HO()l2 2After the reaction is complete, the solution is acidified with H 2 SO 4 . Underthe now acidic conditions I – is oxidized to I 3 – by MnO 2 .− + + −MnO () s + 3I ( aq) + 4H ( aq) → Mn + I ( aq) + 2HO(l)22The amount of I 3 – formed is determined by titrating with S 2 O 3 2– usingstarch as an indicator. The Winkler method is subject to a variety of interferences,and several modifications to the original procedure have beenproposed. For example, NO 2 – interferes because it can reduce I 3 – to I – un-322

Chapter 9 Titrimetric Methods497der acidic conditions. This interference is eliminated by adding sodiumazide, NaN 3 , reducing NO 2 – to N 2 . Other reducing agents, such as Fe 2+ ,are eliminated by pretreating the sample with KMnO 4 , and destroying theexcess permanganate with K 2 C 2 O 4 .Another important example of redox titrimetry is the determination ofwater in nonaqueous solvents. The titrant for this analysis is known as theKarl Fischer reagent and consists of a mixture of iodine, sulfur dioxide, pyridine,and methanol. Because the concentration of pyridine is sufficientlylarge, I 2 and SO 2 react with pyridine (py) to form the complexes py•I 2 andpy•SO 2 . When added to a sample containing water, I 2 is reduced to I – andSO 2 is oxidized to SO 3 .pyiI + pySO i + py+ H O→ 2pyHI i + pySO i2 2 23Methanol is included to prevent the further reaction of py•SO 3 with water.The titration’s end point is signaled when the solution changes from theproduct’s yellow color to the brown color of the Karl Fischer reagent.Or g a n i c An a l y s i sRedox titrimetry also is used for the analysis of organic analytes. Oneimportant example is the determination of the chemical oxygen demand(COD) of natural waters and wastewaters. The COD provides a measureof the quantity of oxygen necessary to completely oxidize all the organicmatter in a sample to CO 2 and H 2 O. Because no attempt is made to correctfor organic matter that can not be decomposed biologically, or forslow decomposition kinetics, the COD always overestimates a sample’strue oxygen demand. The determination of COD is particularly importantin managing industrial wastewater treatment facilities where it is used tomonitor the release of organic-rich wastes into municipal sewer systems orthe environment.A sample’s COD is determined by refluxing it in the presence of excessK 2 Cr 2 O 7 , which serves as the oxidizing agent. The solution is acidified withH 2 SO 4 using Ag 2 SO 4 to catalyze the oxidation of low molecular weightfatty acids. Mercuric sulfate, HgSO 4 , is added to complex any chloride thatis present, preventing the precipitation of the Ag + catalyst as AgCl. Underthese conditions, the efficiency for oxidizing organic matter is 95–100%.After refluxing for two hours, the solution is cooled to room temperatureand the excess Cr 2 O 7 2– is determined by back titrating using ferrous ammoniumsulfate as the titrant and ferroin as the indicator. Because it is difficultto completely remove all traces of organic matter from the reagents, ablank titration must be performed. The difference in the amount of ferrousammonium sulfate needed to titrate the sample and the blank is proportionalto the COD.Iodine has been used as an oxidizing titrant for a number of compoundsof pharmaceutical interest. Earlier we noted that the reaction of S 2 O 32–

498 Analytical Chemistry 2.0with I 3 – produces the tetrathionate ion, S 4 O 6 2– . The tetrathionate ion isactually a dimer consisting of two thiosulfate ions connected through adisulfide (–S–S–) linkage. In the same fashion, I 3 – can be used to titratemercaptans of the general formula RSH, forming the dimer RSSR as aproduct. The amino acid cysteine also can be titrated with I 3 – . The productof this titration is cystine, which is a dimer of cysteine. Triiodide also can beused for the analysis of ascorbic acid (vitamin C) by oxidizing the enediolfunctional group to an alpha diketoneHOOHOOHOOHOO+ 2H + + 2eHOOHOOand for the analysis of reducing sugars, such as glucose, by oxidizing thealdehyde functional group to a carboxylate ion in a basic solution.HHOHHCHOOHHOHOHCH 2 OH+ 3OHHHOHHCO 2OHHOHOHCH 2 OHAn organic compound containing a hydroxyl, a carbonyl, or an aminefunctional group adjacent to an hydoxyl or a carbonyl group can be oxidizedusing metaperiodate, IO 4 – , as an oxidizing titrant.− − − −IO ( aq) + HO() l + 2e IO ( aq) + 2OH( aq )4 23A two-electron oxidation cleaves the C–C bond between the two functionalgroups, with hydroxyl groups being oxidized to aldehydes or ketones,carbonyl functional groups being oxidized to carboxylic acids, andamines being oxidized to an aldehyde and an amine (ammonia if a primaryamine). The analysis is conducted by adding a known excess of IO 4 – to thesolution containing the analyte, and allowing the oxidation to take placefor approximately one hour at room temperature. When the oxidation iscomplete, an excess of KI is added, which converts any unreacted IO 4 – toIO 3 – and I 3 – .IO − ( aq) + 3I − ( aq) + H O() l → IO − ( aq) + I − ( aq ) + 2OH − ( aq)4 23 3The I 3 – is then determined by titrating with S 2 O 3 2– using starch as anindicator.

Chapter 9 Titrimetric Methods499Qu a n t i t a t i ve Ca l c u l at i o n sThe quantitative relationship between the titrand and the titrant is determinedby the stoichiometry of the titration reaction. If you are unsure ofthe balanced reaction, you can deduce the stoichiometry by rememberingthat the electrons in a redox reaction must be conserved.Example 9.11The amount of Fe in a 0.4891-g sample of an ore was determined bytitrating with K 2 Cr 2 O 7 . After dissolving the sample in HCl, the iron wasbrought into the +2 oxidation state using a Jones reductor. Titration to thediphenylamine sulfonic acid end point required 36.92 mL of 0.02153 MK 2 Cr 2 O 7 . Report the ore’s iron content as %w/w Fe 2 O 3 .So l u t i o nBecause we have not been provided with the titration reaction, let’s use aconservation of electrons to deduce the stoichiometry. During the titrationthe analyte is oxidized from Fe 2+ to Fe 3+ , and the titrant is reduced fromCr 2 O 7 2– to Cr 3+ . Oxidizing Fe 2+ to Fe 3+ requires only a single electron.Reducing Cr 2 O 7 2– , in which each chromium is in the +6 oxidation state,to Cr 3+ requires three electrons per chromium, for a total of six electrons.A conservation of electrons for the titration, therefore, requires that eachmole of K 2 Cr 2 O 7 reacts with six moles of Fe 2+ .The moles of K 2 Cr 2 O 7 used in reaching the end point is( 0. 02153 MK Cr O) × ( 0.03692 LK Cr O)2 2 7 2 2 7which means that the sample contains2+46 molFe7. 949× 10− molK Cr O ×2 2 7molK Cr O= 79 . 49× 10 −4molK Cr O2 2 7Thus, the %w/w Fe 2 O 3 in the sample of ore isAlthough we can deduce the stoichiometry betweenthe titrant and the titrand without balancing thetitration reaction, the balanced reactionK CrO Fe 14H2 2 7 ( ) 2+ ( ) +6( ) 3+ + 3+2Cr ( aq ) + 2K ( aq ) + 6Fe ( aq ) + 7H O( l )2does provide useful information. For example, thepresence of H + reminds us that the reaction’s feasibilityis pH-dependent.2 2 7molFe2 3= 4.769× 10 − 3 2+3 21 molFeO4. 769× 10− +2 3molFe × ×2+2mol Fe159.69 gFeO2 3= 0.3808 gFeOmolFeO2 30.3808 gFeO2 3× 100 = 77. 86%w/ wFeO0.4891 gsample

500 Analytical Chemistry 2.0Practice Exercise 9.20The purity of a sample of sodium oxalate, Na 2 C 2 O 4 , is determined bytitrating with a standard solution of KMnO 4 . If a 0.5116-g sample requires35.62 mL of 0.0400 M KMnO 4 to reach the titration’s end point,what is the %w/w Na 2 C 2 O 4 in the sample.Click here to review your answer to this exercise.As shown in the following two examples, we can easily extend this approachto an analysis that requires an indirect analysis or a back titration.Example 9.12A 25.00-mL sample of a liquid bleach was diluted to 1000 mL in a volumetricflask. A 25-mL portion of the diluted sample was transferred bypipet into an Erlenmeyer flask containing an excess of KI, reducing theOCl – to Cl – , and producing I 3 – . The liberated I 3 – was determined bytitrating with 0.09892 M Na 2 S 2 O 3 , requiring 8.96 mL to reach the starchindicator end point. Report the %w/v NaOCl in the sample of bleach.The balanced reactions for this analysis are:− − +OCl ( aq ) + 3I ( aq ) + 2H( aq ) →− −( aq) + Cl ( aq) + HO( l)32I−2−2−−I ( aq ) + 2S O ( aq ) → S O ( aq ) + 3I ( aq )3 2 34 6So l u t i o nTo determine the stoichiometry between the analyte, NaOCl, and thetitrant, Na 2 S 2 O 3 , we need to consider both the reaction between OCl –and I – , and the titration of I 3 – with Na 2 S 2 O 3 .First, in reducing OCl – to Cl – , the oxidation state of chlorine changesfrom +1 to –1, requiring two electrons. The oxidation of three I – to formI 3 – releases two electrons as the oxidation state of each iodine changes from–1 in I – to –⅓ in I 3 – . A conservation of electrons, therefore, requires thateach mole of OCl – produces one mole of I 3 – .Second, in the titration reaction, I 3 – is reduced to I – and S 2 O 3 2– is oxidizedto S 4 O 6 2– . Reducing I 3 – to 3I – requires two elections as each iodine changesfrom an oxidation state of –⅓ to –1. In oxidizing S 2 O 32–to S 4 O 6 2– , eachsulfur changes its oxidation state from +2 to +2.5, releasing one electronfor each S 2 O 3 2– . A conservation of electrons, therefore, requires that eachmole of I 3 – reacts with two moles of S 2 O 3 2– .Finally, because each mole of OCl – produces one mole of I 3 – , and eachmole of I 3 – reacts with two moles of S 2 O 3 2– , we know that every mole ofNaOCl in the sample ultimately results in the consumption of two molesof Na 2 S 2 O 3 .The moles of Na 2 S 2 O 3 used in reaching the titration’s end point is( 0. 09892 MNaSO) × ( 0.00896 LNaSO)2 2 3 2 2 3= 88 . 6× 10 −4molNaSO2 2 3which means the sample contains

Chapter 9 Titrimetric Methods50141 molNaOCl886 . × 10− molNaSO ××2 2 32mol Na S O74.44 gNaOClmolNaOCl2 2 3Thus, the %w/v NaOCl in the diluted sample is= 0.03299 gNaOCl0.03299 gNaOCl× 100 = 132 . % w/vNaOCl25.00 mLBecause the bleach was diluted by a factor of 40 (25 mL to 1000 mL), theconcentration of NaOCl in the bleach is 5.28% (w/v).Example 9.13The amount of ascorbic acid, C 6 H 8 O 6 , in orange juice was determinedby oxidizing the ascorbic acid to dehydroascorbic acid, C 6 H 6 O 6 , witha known amount of I 3 – , and back titrating the excess I 3 – with Na 2 S 2 O 3 .A 5.00-mL sample of filtered orange juice was treated with 50.00 mL of0.01023 M I 3 – . After the oxidation was complete, 13.82 mL of 0.07203M Na 2 S 2 O 3 was needed to reach the starch indicator end point. Reportthe concentration ascorbic acid in mg/100 mL.So l u t i o nFor a back titration we need to determine the stoichiometry between I 3–and the analyte, C 6 H 8 O 6 , and between I 3 – and the titrant, Na 2 S 2 O 3 . Thelater is easy because we know from Example 9.12 that each mole of I 3 – reactswith two moles of Na 2 S 2 O 3 .In oxidizing ascorbic acid to dehydroascorbic acid, the oxidation stateof carbon changes from +⅔ in C 6 H 8 O 6 to +1 in C 6 H 6 O 6 . Each carbonreleases ⅓ of an electron, or a total of two electrons per ascorbic acid. Aswe learned in Example 9.12, reducing I 3 – requires two electrons; thus, aconservation of electrons requires that each mole of ascorbic acid consumesone mole of I 3 – .The total moles of I 3 – reacting with C 6 H 8 O 6 and with Na 2 S 2 O 3 isThe balanced reactions for this analysis are:CHO6 8−6 3( aq ) + I ( aq ) →3I−( aq ) + C HO ( aq ) + 2H6 6 6− 2− 2− −3 2 34 6+( aq )I ( aq ) + 2S O ( aq ) → S O ( aq ) + 3I( aq )− − −( 0. 01023 MI ) × ( 0. 05000 LI ) = 5.115× 10 4 −molI 3The back titration consumes0.01382 LNaSO3 32 2 30.07203 molNaSO2 2 3××LNaSO21 molI−32 32 molNaSO2 2 3−= 4 977×10 4 −. mol I 3

502 Analytical Chemistry 2.0Subtracting the moles of I 3 – reacting with Na 2 S 2 O 3 from the total molesof I 3 – gives the moles reacting with ascorbic acid.4 45. 115× 10 − molI− − 4. 977× 10 − molI− = 138 . ×10 −53 3molI −3The grams of ascorbic acid in the 5.00-mL sample of orange juice is−5−1 molC HO6 8 6138 . × 10 molI × ×3−molI176.13gC HOmolC HO36 8 66 8 6= 243 . × 10 −3gC HO6 8 6There are 2.43 mg of ascorbic acid in the 5.00-mL sample, or 48.6 mg per100 mL of orange juice.E (V)1.21.00.80.60.40.2end point for Fe 2+end point for Sn 2+0 10 20 30 40 50 60 70Volume of Ce 4+ (mL)Figure 9.42 Titration curve for the titrationof 50.0 mL of 0.0125 M Sn 2+ and0.0250 M Fe 2+ with 0.050 M Ce 4+ . Boththe titrand and the titrant are 1M in HCl.Practice Exercise 9.21A quantitative analysis for ethanol, C 2 H 6 O, can be accomplished by aredox back titration. Ethanol is oxidized to acetic acid, C 2 H 4 O 2 , usingexcess dichromate, Cr 2 O 7 2– , which is reduced to Cr 3+ . The excess dichromateis titrated with Fe 2+ , giving Cr 3+ and Fe 3+ as products. In a typicalanalysis, a 5.00-mL sample of a brandy is diluted to 500 mL in a volumetricflask. A 10.00-mL sample is taken and the ethanol is removed bydistillation and collected in 50.00 mL of an acidified solution of 0.0200M K 2 Cr 2 O 7 . A back titration of the unreacted Cr 2 O 7 2– requires 21.48mL of 0.1014 M Fe 2+ . Calculate the %w/v ethanol in the brandy.Click here to review your answer to this exercise.9D.4 Evaluation of Redox TitrimetryThe scale of operations, accuracy, precision, sensitivity, time, and cost ofa redox titration are similar to those described earlier in this chapter foracid–base or a complexation titration. As with acid–base titrations, we canextend a redox titration to the analysis of a mixture of analytes if there is asignificant difference in their oxidation or reduction potentials. Figure 9.42shows an example of the titration curve for a mixture of Fe 2+ and Sn 2+ usingCe 4+ as the titrant. A titration of a mixture of analytes is possible if theirstandard state potentials or formal potentials differ by at least 200 mV.9EPrecipitation TitrationsThus far we have examined titrimetric methods based on acid–base, complexation,and redox reactions. A reaction in which the analyte and titrantform an insoluble precipitate also can serve as the basis for a titration. Wecall this type of titration a precipitation titration.One of the earliest precipitation titrations—developed at the end ofthe eighteenth century—was the analysis of K 2 CO 3 and K 2 SO 4 in potash.

Chapter 9 Titrimetric Methods503Calcium nitrate, Ca(NO 3 ) 2 , was used as the titrant, forming a precipitateof CaCO 3 and CaSO 4 . The titration’s end point was signaled by notingwhen the addition of titrant ceased to generate additional precipitate. Theimportance of precipitation titrimetry as an analytical method reached itszenith in the nineteenth century when several methods were developed fordetermining Ag + and halide ions.9E.1 Titration CurvesA precipitation titration curve follows the change in either the titrand’s orthe titrant’s concentration as a function of the titrant’s volume. As we havedone with other titrations, we first show how to calculate the titration curveand then demonstrate how we can quickly sketch a reasonable approximationof the titration curve.Ca l c u l at i n g t h e Ti t r a t i o n Cu r v eLet’s calculate the titration curve for the titration of 50.0 mL of 0.0500 MNaCl with 0.100 M AgNO 3 . The reaction in this case is+ −Ag ( aq) + Cl ( aq) AgCl()sBecause the reaction’s equilibrium constant is so largeK− − −= ( K ) = (. 18× 10 ) = 56 . × 10sp1 10 1 9we may assume that Ag + and Cl – react completely.By now you are familiar with our approach to calculating a titrationcurve. The first task is to calculate the volume of Ag + needed to reach theequivalence point. The stoichiometry of the reaction requires thatStep 1: Calculate the volume of AgNO 3needed to reach the equivalence point.molesAgSolving for the volume of Ag +Veq= moles Cl+ −M × V = M × VAg Ag Cl ClM VCl ClM)(50.0 mL)= V = = ( 0.0500 AgM (0.100 M)Ag= 25.0mLshows that we need 25.0 mL of Ag + to reach the equivalence point.Before the equivalence point the titrand, Cl – , is in excess. The concentrationof unreacted Cl – after adding 10.0 mL of Ag + , for example, is− +− initialmoles Cl − molesAg added M V[ Cl ] ==total volumeV( 0. 0500 M)(50. 0mL) −( 0. 100 M)( 10. 0 mL)=50. 0 mL + 100 . mLCl Cl Ag AgCl− M V+ VAg= 250 . × 10 −2MStep 2: Calculate pCl before the equivalencepoint by determining the concentrationof unreacted NaCl.

504 Analytical Chemistry 2.0Step 3: Calculate pCl at the equivalencepoint using the K sp for AgCl to calculatethe concentration of Cl – .Table 9.18 Titration of 50.0 mL of 0.0500 M NaCl with 0.100 M AgNO 3Volume of AgNO 3 (mL) pCl Volume of AgNO 3 (mL) pCl0.00 1.30 30.0 7.545.00 1.44 35.0 7.8210.0 1.60 40.0 7.9715.0 1.81 45.0 8.0720.0 2.15 50.0 8.1425.0 4.89which corresponds to a pCl of 1.60.At the titration’s equivalence point, we know that the concentrationsof Ag + and Cl – are equal. To calculate the concentration of Cl – we use theK sp expression for AgCl; thus+ − −K = [ Ag ][ Cl ] = ( x)( x) = 18 . × 10 10spStep 4: Calculate pCl after the equivalencepoint by first calculating the concentrationof excess AgNO 3 and then calculatingthe concentration of Cl – using the K spfor AgCl.Solving for x gives [Cl – ] as 1.3 × 10 –5 M, or a pCl of 4.89.After the equivalence point, the titrant is in excess. We first calculate theconcentration of excess Ag + and then use the K sp expression to calculate theconcentration of Cl – . For example, after adding 35.0 mL of titrant+ molesAg added−initial molesCl[ Ag ] =total volume+ −M V=V( 0. 100 M)(35. 0 mL) −( 0. 0500 M)( 50. 0 mL)=50. 0 mL + 350 . mLAg Ag Cl ClCl− M V+ VAg= 118 . × 10 −2MK−− sp 18 . × 10[ Cl ] = =+[ Ag ] 118 . × 1010−2= 15 . × 10−8Mor a pCl of 7.81. Additional results for the titration curve are shown inTable 9.18 and Figure 9.43.1086pCl4Figure 9.43 Titration curve for the titration of 50.0mL of 0.0500 M NaCl with 0.100 M AgNO 3 . The redpoints corresponds to the data in Table 9.18. The blueline shows the complete titration curve.200 10 20 30 40 50Volume of AgNO 3

Chapter 9 Titrimetric Methods505Practice Exercise 9.22When calculating a precipitation titration curve, you can choose to followthe change in the titrant’s concentration or the change in the titrand’sconcentration. Calculate the titration curve for the titration of 50.0 mLof 0.0500 M AgNO 3 with 0.100 M NaCl as pAg versus V NaCl , and aspCl versus V NaCl .Click here to review your answer to this exercise.Sk e t c h i n g t h e Ti t r a t i o n Cu r v eTo evaluate the relationship between a titration’s equivalence point and itsend point we need to construct only a reasonable approximation of theexact titration curve. In this section we demonstrate a simple method forsketching a precipitation titration curve. Our goal is to sketch the titrationcurve quickly, using as few calculations as possible. Let’s use the titration of50.0 mL of 0.0500 M NaCl with 0.100 M AgNO 3 .We begin by calculating the titration’s equivalence point volume, which,as we determined earlier, is 25.0 mL. Next we draw our axes, placing pClon the y-axis and the titrant’s volume on the x-axis. To indicate the equivalencepoint’s volume, we draw a vertical line corresponding to 25.0 mL ofAgNO3. Figure 9.44a shows the result of this first step in our sketch.Before the equivalence point, Cl – is present in excess and pCl is determinedby the concentration of unreacted Cl – . As we learned earlier, thecalculations are straightforward. Figure 9.44b shows pCl after adding 10.0mL and 20.0 mL of AgNO 3 .After the equivalence point, Ag + is in excess and the concentration ofCl – is determined by the solubility of AgCl. Again, the calculations arestraightforward. Figure 4.43c shows pCl after adding 30.0 mL and 40.0mL of AgNO 3 .Next, we draw a straight line through each pair of points, extendingthem through the vertical line representing the equivalence point’s volume(Figure 9.44d). Finally, we complete our sketch by drawing a smooth curvethat connects the three straight-line segments (Figure 9.44e). A comparisonof our sketch to the exact titration curve (Figure 9.44f) shows that they arein close agreement.This is the same example that we used indeveloping the calculations for a precipitationtitration curve. You can review theresults of that calculation in Table 9.18and Figure 9.43.See Table 9.18 for the values.See Table 9.18 for the values.9E.2 Selecting and Evaluating the End pointAt the beginning of this section we noted that the first precipitation titrationused the cessation of precipitation to signal the end point. At best,this is a cumbersome method for detecting a titration’s end point. Beforeprecipitation titrimetry became practical, better methods for identifyingthe end point were necessary.

506 Analytical Chemistry 2.010(a)10(b)8866pClpCl4422000 10 20 30 40 500 10 20 30 40 50Volume of AgNO 3Volume of AgNO 310(c)10(d)8866pClpCl4422000 10 20 30 40 500 10 20 30 40 50Volume of AgNO 3Volume of AgNO 310(e)10(f)8866pClpCl442200 10 20 30 40 50Volume of AgNO 30 10 20 30 40 50Volume of AgNO 3Figure 9.44 Illustrations showing the steps in sketching an approximate titration curve for the titration of 50.0mL of 0.0500 M NaCl with 0.100 M AgNO 3 : (a) locating the equivalence point volume; (b) plotting two pointsbefore the equivalence point; (c) plotting two points after the equivalence point; (d) preliminary approximationof titration curve using straight-lines; (e) final approximation of titration curve using a smooth curve; (f) comparisonof approximate titration curve (solid black line) and exact titration curve (dashed red line). See the textfor additional details. A better fit is possible if the two points before the equivalence point are further apart—forexample, 0 mL and 20 mL— and the two points after the equivalence point are further apart.0

Chapter 9 Titrimetric Methods507Fi n d i n g t h e En d p o i n t Wi t h a n In d i c a t o rThere are three general types of indicators for precipitation titrations, eachof which changes color at or near the titration’s equivalence point. The firsttype of indicator is a species that forms a precipitate with the titrant. In theMohr method for Cl – using Ag + as a titrant, for example, a small amountof K 2 CrO 4 is added to the titrand’s solution. The titration’s end point is theformation of a reddish-brown precipitate of Ag 2 CrO 4 .Because CrO 4 2– imparts a yellow color to the solution, which mightobscure the end point, only a small amount of K 2 CrO 4 is added. As a result,the end point is always later than the equivalence point. To compensate forthis positive determinate error, an analyte-free reagent blank is analyzed todetermine the volume of titrant needed to affect a change in the indicator’scolor. Subtracting the end point for the reagent blank from the titrand’send point gives the titration’s end point. Because CrO 4 2– is a weak base, thetitrand’s solution is made slightly alkaline. If the pH is too acidic, chromateis present as HCrO 4 – instead of CrO 4 2– , and the Ag 2 CrO 4 end point isdelayed. The pH also must be less than 10 to avoid the precipitation ofsilver hydroxide.A second type of indicator uses a species that forms a colored complexwith the titrant or the titrand. In the Volhard method for Ag + usingKSCN as the titrant, for example, a small amount of Fe 3+ is added to thetitrand’s solution. The titration’s end point is the formation of the reddishcoloredFe(SCN) 2+ complex. The titration must be carried out in an acidicsolution to prevent the precipitation of Fe 3+ as Fe(OH) 3 .The third type of end point uses a species that changes color when itadsorbs to the precipitate. In the Fajans method for Cl – using Ag + as atitrant, for example, the anionic dye dichlorofluoroscein is added to thetitrand’s solution. Before the end point, the precipitate of AgCl has a negativesurface charge due to the adsorption of excess Cl – . Because dichlorofluorosceinalso carries a negative charge, it is repelled by the precipitateand remains in solution where it has a greenish-yellow color. After the endpoint, the surface of the precipitate carries a positive surface charge dueto the adsorption of excess Ag + . Dichlorofluoroscein now adsorbs to theprecipitate’s surface where its color is pink. This change in the indicator’scolor signals the end point.The Mohr method was first published in1855 by Karl Friedrich Mohr.The Volhard method was first published in1874 by Jacob Volhard.The Fajans method was first published inthe 1920s by Kasimir Fajans.Fi n d i n g t h e En d p o i n t Po t e n t i o m e t r i c a l l yAnother method for locating the end point is a potentiometric titration inwhich we monitor the change in the titrant’s or the titrand’s concentrationusing an ion-selective electrode. The end point is found by visually examiningthe titration curve. A further discussion of potentiometry is found inChapter 11.

508 Analytical Chemistry 2.0Table 9.19 Representative Examples of PrecipitationTitrationsTitrand Titrant a End Point b3–AsO 4 AgNO 3 , KSCN VolhardBr – AgNO 3Mohr or FajansAgNO 3 , KSCN VolhardCl – AgNO 3Mohr or FajansAgNO 3 , KSCN Volhard*2–CO 3 AgNO 3 , KSCN Volhard*2–C 2 O 4 AgNO 3 , KSCN Volhard*2–CrO 4 AgNO 3 , KSCN Volhard*I – AgNO 3FajansAgNO 3 , KSCN Volhard3–PO 4 AgNO 3 , KSCN Volhard*S 2– AgNO 3 , KSCN Volhard*SCN – AgNO 3 , KSCN Volhard*a When two reagents are listed, the analysis is by a back titration. The first reagentis added in excess and the second reagent used to back titrate the excess.b For those Volhard methods identified with an asterisk ( *) the precipitated silversalt must be removed before carrying out the back titration.9E.3 Quantitative ApplicationsAlthough precipitation titrimetry is rarely listed as a standard method ofanalysis, it may still be useful as a secondary analytical method for verifyingother analytical methods. Most precipitation titrations use Ag + as eitherthe titrand or the titration. A titration in which Ag + is the titrant is calledan argentometric titration. Table 9.19 provides a list of several typicalprecipitation titrations.Qu a n t i t a t i ve Ca l c u l at i o n sThe quantitative relationship between the titrand and the titrant is determinedby the stoichiometry of the titration reaction. If you are unsure ofthe balanced reaction, you can deduce the stoichiometry from the precipitate’sformula. For example, in forming a precipitate of Ag 2 CrO 4 , eachmole of CrO 4 2– reacts with two moles of Ag + .Example 9.14A mixture containing only KCl and NaBr is analyzed by the Mohr method.A 0.3172-g sample is dissolved in 50 mL of water and titrated to theAg 2 CrO 4 end point, requiring 36.85 mL of 0.1120 M AgNO 3 . A blanktitration requires 0.71 mL of titrant to reach the same end point. Reportthe %w/w KCl in the sample.

Chapter 9 Titrimetric Methods509So l u t i o nTo find the moles of titrant reacting with the sample, we first need to correctfor the reagent blank; thusV Ag= 36. 85 mL − 071 . mL = 3614 . mL(0.1120 MAgNO ) × ( 0. 03614 LAgNO ) = 4.048×10 −3 molAgNO3 33Titrating with AgNO 3 produces a precipitate of AgCl and AgBr. In formingthe precipitates, each mole of KCl consumes one mole of AgNO 3 andeach mole of NaBr consumes one mole of AgNO 3 ; thusmolesKCl + molesNaBr = 4.048× 10 −3We are interested in finding the mass of KCl, so let’s rewrite this equationin terms of mass. We know thatgKClmolesKCl =74.551 gKCl/mol KClgNaBrmolesNaBr=102.89 gNaBr/molNaBrwhich we substitute back into the previous equationgKCl74.551 gKCl/mol KClgNaBr+102.89 gNaBr/mol NaBr= 4.048× 10 −3Because this equation has two unknowns—g KCl and g NaBr—we needanother equation that includes both unknowns. A simple equation takesadvantage of the fact that the sample contains only KCl and NaBr; thus,gNaBr= 0.3172 g−gKClgKCl74.551 gKCl/mol KCl0.3172 g−gKCl+102.89 gNaBr/molNaBr= 4.048× 10 −31. 341× 10 −2 (g KCl) + 3. 083× 10 −3 − 9. 719×10−3( gKCl) = 4.048× 10 −3−369 . × 10 (g KCl) = 9.65×103 − 4The sample contains 0.262 g of KCl and the %w/w KCl in the sample is0.262 gKCl× 100 = 82.%6 w/wKCl0.3172 gsampleThe analysis for I – using the Volhard method requires a back titration.A typical calculation is shown in the following example.

510 Analytical Chemistry 2.0Example 9.15The %w/w I – in a 0.6712-g sample was determined by a Volhard titration.After adding 50.00 mL of 0.05619 M AgNO 3 and allowing the precipitateto form, the remaining silver was back titrated with 0.05322 M KSCN,requiring 35.14 mL to reach the end point. Report the %w/w I – in thesample.So l u t i o nThere are two precipitates in this analysis: AgNO 3 and I – form a precipitateof AgI, and AgNO 3 and KSCN form a precipitate of AgSCN. Each moleof I – consumes one mole of AgNO 3 , and each mole of KSCN consumesone mole of AgNO 3 ; thusmolesAgNO = moles I − + moles KSCN3Solving for the moles of I – we findmolesI − = molesAgNO −moles KSCNmolesI − = M × V − M × VAg Ag KSCN KSCN3molesI − = ( 0. 05619 MAgNO ) × ( 0.05000LAgNO) −3 3( 0. 05322 MKSCN) × ( 0.03514 LKSCN)that there are 9.393 × 10 –4 moles of I – in the sample. The %w/w I – in thesample is−4− 126.9 gI( 9. 393× 10 molI ) ×0.6712 gsample−molI−× 100 = 17. 76%w/wI−Practice Exercise 9.23A 1.963-g sample of an alloy is dissolved in HNO3 and diluted to volumein a 100-mL volumetric flask. Titrating a 25.00-mL portion with0.1078 M KSCN requires 27.19 mL to reach the end point. Calculatethe %w/w Ag in the alloy.Click here to review your answer to this exercise.9E.4 Evaluation of Precipitation TitrimetryThe scale of operations, accuracy, precision, sensitivity, time, and cost of aprecipitation titration is similar to those described elsewhere in this chapterfor acid–base, complexation, and redox titrations. Precipitation titrationsalso can be extended to the analysis of mixtures provided that there is a sig-

Chapter 9 Titrimetric Methods5111510end point for I –pAg50end point for Cl –0 20 40 60 80Volume of AgNO 3 (mL)Figure 9.45 Titration curve for the titration of a 50.0 mLmixture of 0.0500 M I – and 0.0500 M Cl – using 0.100 MAg + as a titrant. The red arrows show the end points. Notethat the end point for I – is earlier than the end point for Cl –because AgI is less soluble than AgCl.nificant difference in the solubilities of the precipitates. Figure 9.45 showsan example of a titration curve for a mixture of I – and Cl – using Ag + as atitrant.9FKey Termsacid–base titration acidity alkalinityargentometric titrationasymmetric equivalencepointauxiliary complexing agentauxiliary oxidizing agent auxiliary reducing agent back titrationburet complexation titration conditional formationconstantdirect titration displacement titration end pointequivalence point Fajans method formal potentialGran plot indicator Jones reductorKjeldahl analysis leveling metallochromic indicatorMohr method potentiometric titration precipitation titrationredox indicator redox titration spectrophotometrictitrationsymmetric equivalencepointthermometric titrationtitrandtitrant titration curve titration errortitrimetry Volhard method Walden reductor9G Chapter SummaryIn a titrimetric method of analysis, the volume of titrant reacting stoichiometricallywith a titrand provides quantitative information about theamount of analyte in a sample. The volume of titrant corresponding to thisstoichiometric reaction is called the equivalence point. Experimentally wedetermine the titration’s end point using an indicator that changes colornear the equivalence point. Alternatively, we can locate the end point byAs you review this chapter, try to define akey term in your own words. Check youranswer by clicking on the key term, whichwill take you to the page where it was firstintroduced. Clicking on the key termthere, will bring you back to this page sothat you can continue with another keyterm.

512 Analytical Chemistry 2.0continuously monitoring a property of the titrand’s solution—absorbance,potential, and temperature are typical examples—that changes as the titrationprogresses. In either case, an accurate result requires that the end pointclosely match the equivalence point. Knowing the shape of a titration curveis critical to evaluating the feasibility of a titrimetric method.Many titrations are direct, in which the analyte participates in the titrationas the titrand or the titrant. Other titration strategies may be usedwhen a direct reaction between the analyte and titrant is not feasible. Ina back titration a reagent is added in excess to a solution containing theanalyte. When the reaction between the reagent and the analyte is complete,the amount of excess reagent is determined by a titration. In a displacementtitration the analyte displaces a reagent, usually from a complex, and theamount of displaced reagent is determined by an appropriate titration.Titrimetric methods have been developed using acid–base, complexation,redox, and precipitation reactions. Acid–base titrations use a strongacid or a strong base as a titrant. The most common titrant for a complexationtitration is EDTA. Because of their stability against air oxidation,most redox titrations use an oxidizing agent as a titrant. Titrations withreducing agents also are possible. Precipitation titrations often involve Ag +as either the analyte or titrant.9HProblemsAnswers, but not worked solutions, tomost end-of-chapter problems are availablehere.Some of the problems that follow require one ormore equilibrium constants or standard state potentials.For your convenience, here are hyperlinksto the appendices containing these constantsAppendix 10: Solubility ProductsAppendix 11: Acid Dissociation ConstantsAppendix 12: Metal-Ligand Formation ConstantsAppendix 13: Standard State Reduction Potentials1. Calculate or sketch titration curves for the following acid–base titrations.a. 25.0 mL of 0.100 M NaOH with 0.0500 M HClb. 50.0 mL of 0.0500 M HCOOH with 0.100 M NaOHc. 50.0 mL of 0.100 M NH 3 with 0.100 M HCld. 50.0 mL of 0.0500 M ethylenediamine with 0.100 M HCle. 50.0 mL of 0.0400 M citric acid with 0.120 M NaOHf. 50.0 mL of 0.0400 M H 3 PO 4 with 0.120 M NaOH2. Locate the equivalence point for each titration curve in problem 1.What is the stoichiometric relationship between the moles of acid andthe moles of base at each of these equivalence points?3. Suggest an appropriate visual indicator for each of the titrations inproblem 1.4. In sketching the titration curve for a weak acid we approximate the pHat 10% of the equivalence point volume as pK a – 1, and the pH at 90%of the equivalence point volume as pK a + 1. Show that these assumptionsare reasonable.

Chapter 9 Titrimetric Methods5135. Tartaric acid, H 2 C 4 H 4 O 6 , is a diprotic weak acid with a pK a1 of 3.0and a pK a2 of 4.4. Suppose you have a sample of impure tartaric acid(purity > 80%), and that you plan to determine its purity by titratingwith a solution of 0.1 M NaOH using an indicator to signal the endpoint. Describe how you will carry out the analysis, paying particularattention to how much sample to use, the desired pH range for theindicator, and how you will calculate the %w/w tartaric acid.6. The following data for the titration of a monoprotic weak acid with astrong base were collected using an automatic titrator. Prepare normal,first derivative, second derivative, and Gran plot titration curves for thisdata, and locate the equivalence point for each.Volume of NaOH (ml) pH Volume of NaOH (mL) pH0.25 3.0 49.95 7.80.86 3.2 49.97 8.01.63 3.4 49.98 8.22.72 3.6 49.99 8.44.29 3.8 50.00 8.76.54 4.0 50.01 9.19.67 4.2 50.02 9.413.79 4.4 50.04 9.618.83 4.6 50.06 9.824.47 4.8 50.10 10.030.15 5.0 50.16 10.235.33 5.2 50.25 10.439.62 5.4 50.40 10.642.91 5.6 50.63 10.845.28 5.8 51.01 11.046.91 6.0 51.61 11.248.01 6.2 52.58 11.448.72 6.4 54.15 11.649.19 6.6 56.73 11.849.48 6.8 61.11 12.049.67 7.0 68.83 12.249.79 7.2 83.54 12.449.78 7.4 116.14 12.649.92 7.6Some of the problems that follow require one ormore equilibrium constants or standard state potentials.For your convenience, here are hyperlinksto the appendices containing these constantsAppendix 10: Solubility ProductsAppendix 11: Acid Dissociation ConstantsAppendix 12: Metal-Ligand Formation ConstantsAppendix 13: Standard State Reduction Potentials7. Schwartz published the following simulated data for the titration of a1.02 × 10 –4 M solution of a monoprotic weak acid (pK a = 8.16) with

514 Analytical Chemistry 2.0Some of the problems that follow require one ormore equilibrium constants or standard state potentials.For your convenience, here are hyperlinksto the appendices containing these constantsAppendix 10: Solubility ProductsAppendix 11: Acid Dissociation ConstantsAppendix 12: Metal-Ligand Formation ConstantsAppendix 13: Standard State Reduction Potentials1.004 × 10 –3 M NaOH. 10 The simulation assumes that a 50-mL pipetis used to transfer a portion of the weak acid solution to the titrationvessel. A calibration of the pipet shows that it delivers a volume of only49.94 mL. Prepare normal, first derivative, second derivative, and Granplot titration curves for this data, and determine the equivalence pointfor each. How do these equivalence points compare to the expectedequivalence point? Comment on the utility of each titration curve forthe analysis of very dilute solutions of very weak acids.mL of NaOH pH mL of NaOH pH0.03 6.212 4.79 8.8580.09 6.504 4.99 8.9260.29 6.936 5.21 8.9940.72 7.367 5.41 9.0561.06 7.567 5.61 9.1181.32 7.685 5.85 9.1801.53 7.776 6.05 9.2311.76 7.863 6.28 9.2831.97 7.938 6.47 9.3272.18 8.009 6.71 9.3742.38 8.077 6.92 9.4142.60 8.146 7.15 9.4512.79 8.208 7.36 9.4843.01 8.273 7.56 9.5143.41 8.332 7.79 9.5453.60 8.458 8.21 9.5723.80 8.521 8.44 9.5993.99 8.584 8.64 9.6454.18 8.650 8.84 9.6664.40 8.720 9.07 9.6884.57 8.784 9.27 9.7068. Calculate or sketch the titration curve for a 50.0 mL solution of a0.100 M monoprotic weak acid (pK a = 8) with 0.1 M strong base in anonaqueous solvent with K s = 10 –20 . You may assume that the changein solvent does not affect the weak acid’s pK a . Compare your titrationcurve to the titration curve when water is the solvent.9. The titration of a mixture of p-nitrophenol (pK a = 7.0) and m-nitrophenol(pK a = 8.3) can be followed spectrophotometrically. Neither acidabsorbs at a wavelength of 545 nm, but their respective conjugate bases10 Schwartz, L. M. J. Chem. Educ. 1992, 69, 879–883.

Chapter 9 Titrimetric Methods515do absorb at this wavelength. The m-nitrophenolate ion has a greaterabsorbance than an equimolar solution of the p-nitrophenolate ion.Sketch the spectrophotometric titration curve for a 50.00-mL mixtureconsisting of 0.0500 M p-nitrophenol and 0.0500 M m-nitrophenolwith 0.100 M NaOH. Compare your result to the expected potentiometrictitration curves.10. The quantitative analysis for aniline (C 6 H 5 NH 2 , K b = 3.94 × 10 –10 )can be carried out by an acid–base titration using glacial acetic acidas the solvent and HClO 4 as the titrant. A known volume of samplecontaining 3–4 mmol of aniline is transferred to a 250-mL Erlenmeyerflask and diluted to approximately 75 mL with glacial acetic acid. Twodrops of a methyl violet indicator are added, and the solution is titratedwith previously standardized 0.1000 M HClO 4 (prepared in glacialacetic acid using anhydrous HClO 4 ) until the end point is reached.Results are reported as parts per million aniline.(a) Explain why this titration is conducted using glacial acetic acid asthe solvent instead of water.(b) One problem with using glacial acetic acid as solvent is its relativelyhigh coefficient of thermal expansion of 0.11%/ o C. For example,100.00 mL of glacial acetic acid at 25 o C occupies 100.22 mL at27 o C. What is the effect on the reported concentration of anilineif the standardization of HClO 4 is conducted at a temperature thatis lower than that for the analysis of the unknown?(c) The procedure calls for a sample containing 3–4 mmoles of aniline.Why is this requirement necessary?Some of the problems that follow require one ormore equilibrium constants or standard state potentials.For your convenience, here are hyperlinksto the appendices containing these constantsAppendix 10: Solubility ProductsAppendix 11: Acid Dissociation ConstantsAppendix 12: Metal-Ligand Formation ConstantsAppendix 13: Standard State Reduction Potentials11. Using a ladder diagram, explain why the presence of dissolved CO 2leads to a determinate error for the standardization of NaOH if theend point’s pH falls between 6–10, but no determinate error if the endpoint’s pH is less than 6.12. A water sample’s acidity is determined by titrating to fixed end pointpHs of 3.7 and 8.3, with the former providing a measure of the concentrationof strong acid, and the later a measure of the combinedconcentrations of strong acid and weak acid. Sketch a titration curvefor a mixture of 0.10 M HCl and 0.10 M H 2 CO 3 with 0.20 M strongbase, and use it to justify the choice of these end points.13. Ethylenediaminetetraacetic acid, H 4 Y, is a weak acid with successiveacid dissociation constants of 0.010, 2.19 × 10 –3 , 6.92 × 10 –7 , and5.75 × 10 –11 . Figure 9.46 shows a titration curve for H 4 Y with NaOH.What is the stoichiometric relationship between H 4 Y and NaOH at theequivalence point marked with the red arrow?pH141210864200 10 20 30 40Volume of NaOH (mL)Figure 9.46 Titration curve for Problem9.13.

516 Analytical Chemistry 2.0Some of the problems that follow require one ormore equilibrium constants or standard state potentials.For your convenience, here are hyperlinksto the appendices containing these constantsAppendix 10: Solubility ProductsAppendix 11: Acid Dissociation ConstantsAppendix 12: Metal-Ligand Formation ConstantsAppendix 13: Standard State Reduction Potentials14. A Gran plot method has been described for the quantitative analysisof a mixture consisting of a strong acid and a monoprotic weak acid. 11A 50.00-mL mixture of HCl and CH 3 COOH is transferred to an Erlenmeyerflask and titrated by using a digital pipet to add successive1.00-mL aliquots of 0.09186 M NaOH. The progress of the titrationis monitored by recording the pH after each addition of titrant. Usingthe two papers listed in the footnote as a reference, prepare a Gran plotfor the following data, and determine the concentrations of HCl andCH 3 COOH.Volume ofNaOH (ml) pHVolume ofNaOH (mL) pHVolume ofNaOH (ml) pH1.00 1.83 24.00 4.45 47.00 12.142.00 1.86 25.00 4.53 48.00 12.173.00 1.89 26.00 4.61 49.00 12.204.00 1.92 27.00 4.69 50.00 12.235.00 1.95 28.00 4.76 51.00 12.266.00 1.99 29.00 4.84 52.00 12.287.00 2.03 30.00 4.93 53.00 12.308.00 2.10 31.00 5.02 54.00 12.329.00 2.18 32.00 5.13 55.00 12.3410.00 2.31 33.00 5.23 56.00 12.3611.00 2.51 34.00 5.37 57.00 12.3812.00 2.81 35.00 5.52 58.00 12.3913.00 3.16 36.00 5.75 59.00 12.4014.00 3.36 37.00 6.14 60.00 12.4215.00 3.54 38.00 10.30 61.00 12.4316.00 3.69 39.00 11.31 62.00 12.4417.00 3.81 40.00 11.58 63.00 12.4518.00 3.93 41.00 11.74 64.00 12.4719.00 4.02 42.00 11.85 65.00 12.4820.00 4.14 43.00 11.93 66.00 12.4921.00 4.22 44.00 12.00 67.00 12.5022.00 4.30 45.00 12.05 68.00 12.5123.00 4.38 46.00 12.10 69.00 12.5215. Explain why it is not possible for a sample of water to simultaneouslyhave OH – and HCO 3 – as sources of alkalinity.16. For each of the following, determine the sources of alkalinity (OH – ,HCO 3 – , CO 3 2– ) and their respective concentrations in parts per mil-11 (a) Boiani, J. A. J. Chem. Educ. 1986, 63, 724–726; (b) Castillo, C. A.; Jaramillo, A. J. Chem.Educ. 1989, 66, 341.

Chapter 9 Titrimetric Methods517lion In each case a 25.00-mL sample is titrated with 0.1198 M HCl tothe bromocresol green and the phenolphthalein end points.Volume of HCl (mL) to thephenolphthalein end pointVolume of HCl (mL) to thebromocresol green end pointa 21.36 21.38b 5.67 21.13c 0.00 14.28d 17.12 34.26e 21.36 25.6917. A sample may contain any of the following: HCl, NaOH, H 3 PO 4 ,H 2 PO 4 – , HPO 4 2– , or PO 4 3– . The composition of a sample is determinedby titrating a 25.00-mL portion with 0.1198 M HCl or 0.1198M NaOH to the phenolphthalein and the methyl orange end points.For each of the following, determine which species are present in thesample, and their respective molar concentrations.TitrantPhenolphthalein endpoint volume (mL)methyl orange end pointvolume (mL)a HCl 11.54 35.29b NaOH 19.79 9.89c HCl 22.76 22.78d NaOH 39.42 17.4818. The protein in a 1.2846-g sample of an oat cereal is determined by aKjeldahl analysis. The sample is digested with H 2 SO 4 , the resultingsolution made basic with NaOH, and the NH 3 distilled into 50.00 mLof 0.09552 M HCl. The excess HCl is back titrated using 37.84 mL of0.05992 M NaOH. Given that the proteins in grains average 17.54%w/w N, report the %w/w protein in the sample.Some of the problems that follow require one ormore equilibrium constants or standard state potentials.For your convenience, here are hyperlinksto the appendices containing these constantsAppendix 10: Solubility ProductsAppendix 11: Acid Dissociation ConstantsAppendix 12: Metal-Ligand Formation ConstantsAppendix 13: Standard State Reduction Potentials19. The concentration of SO 2 in air is determined by bubbling a sample ofair through a trap containing H 2 O 2 . Oxidation of SO 2 by H 2 O 2 resultsin the formation of H 2 SO 4 , which is then determined by titrating withNaOH. In a typical analysis, a sample of air was passed through theperoxide trap at a rate of 12.5 L/min for 60 min and required 10.08 mLof 0.0244 M NaOH to reach the phenolphthalein end point. Calculatethe mL/L SO 2 in the sample of air. The density of SO 2 at the temperatureof the air sample is 2.86 mg/mL.20. The concentration of CO 2 in air is determined by an indirect acid–base titration. A sample of air is bubbled through a solution containingan excess of Ba(OH) 2 , precipitating BaCO 3 . The excess Ba(OH) 2 isback titrated with HCl. In a typical analysis a 3.5-L sample of air wasbubbled through 50.00 mL of 0.0200 M Ba(OH) 2 . Back titrating with

518 Analytical Chemistry 2.00.0316 M HCl required 38.58 mL to reach the end point. Determinethe ppm CO 2 in the sample of air given that the density of CO 2 at thetemperature of the sample is 1.98 g/L.21. The purity of a synthetic preparation of methylethyl ketone, C 3 H 8 O, isdetermined by reacting it with hydroxylamine hydrochloride, liberatingHCl (see reaction in Table 9.8). In a typical analysis a 3.00-mL samplewas diluted to 50.00 mL and treated with an excess of hydroxylaminehydrochloride. The liberated HCl was titrated with 0.9989 M NaOH,requiring 32.68 mL to reach the end point. Report the percent purityof the sample given that the density of methylethyl ketone is 0.805 g/mL.Some of the problems that follow require one ormore equilibrium constants or standard state potentials.For your convenience, here are hyperlinksto the appendices containing these constantsAppendix 10: Solubility ProductsAppendix 11: Acid Dissociation ConstantsAppendix 12: Metal-Ligand Formation ConstantsAppendix 13: Standard State Reduction PotentialspH1412108640 10 20 30 40 50Volume of NaOH (mL)Figure 9.47 Titration curve for Problem9.24.22. Animal fats and vegetable oils are triesters formed from the reactionbetween glycerol (1,2,3-propanetriol) and three long-chain fatty acids.One of the methods used to characterize a fat or an oil is a determinationof its saponification number. When treated with boiling aqueousKOH, an ester saponifies into the parent alcohol and fatty acids (ascarboxylate ions). The saponification number is the number of milligramsof KOH required to saponify 1.000 gram of the fat or the oil.In a typical analysis a 2.085-g sample of butter is added to 25.00 mLof 0.5131 M KOH. After saponification is complete the excess KOH isback titrated with 10.26 mL of 0.5000 M HCl. What is the saponificationnumber for this sample of butter?23. A 250.0-mg sample of an organic weak acid is dissolved in an appropriatesolvent and titrated with 0.0556 M NaOH, requiring 32.58 mL toreach the end point. Determine the compound’s equivalent weight.24. Figure 9.47 shows a potentiometric titration curve for a 0.4300-g sampleof a purified amino acid that was dissolved in 50.00 mL of waterand titrated with 0.1036 M NaOH. Identify the amino acid from thepossibilities listed in the following table.amino acid formula weight (g/mol) K aalanine 89.1 1.36 × 10 –10glycine 75.1 1.67 × 10 –10methionine 149.2 8.9 × 10 –10taurine 125.2 1.8 × 10 –9asparagine 150 1.9 × 10 –9leucine 131.2 1.79 × 10 –10phenylalanine 166.2 4.9 × 10 –10valine 117.2 1.91 × 10 –10

Chapter 9 Titrimetric Methods51925. Using its titration curve, determine the acid dissociation constant forthe weak acid in problem 9.6.26. Where in the scale of operations do the microtitration techniques discussedin section 9B.7 belong?27. An acid–base titration may be used to determine an analyte’s gramequivalent weight, but it can not be used to determine its gram formulaweight. Explain why.28. Commercial washing soda is approximately 30–40% w/w Na 2 CO 3 .One procedure for the quantitative analysis of washing soda containsthe following instructions:Transfer an approximately 4-g sample of the washing soda to a250-mL volumetric flask. Dissolve the sample in about 100 mL ofH 2 O and then dilute to the mark. Using a pipet, transfer a 25-mLaliquot of this solution to a 125-mL Erlenmeyer flask, and add 25-mL of H 2 O and 2 drops of bromocresol green indicator. Titratethe sample with 0.1 M HCl to the indicator’s end point.What modifications, if any, are necessary if you want to adapt thisprocedure to evaluate the purity of commercial Na 2 CO 3 that is >98%pure?29. A variety of systematic and random errors are possible when standardizinga solution of NaOH against the primary weak acid standard potassiumhydrogen phthalate (KHP). Identify, with justification, whetherthe following are systematic or random sources of error, or if they haveno effect. If the error is systematic, then indicate whether the experimentallydetermined molarity for NaOH is too high or too low. Thestandardization reaction isSome of the problems that follow require one ormore equilibrium constants or standard state potentials.For your convenience, here are hyperlinksto the appendices containing these constantsAppendix 10: Solubility ProductsAppendix 11: Acid Dissociation ConstantsAppendix 12: Metal-Ligand Formation ConstantsAppendix 13: Standard State Reduction Potentials− − 2−CHO ( aq) + OH ( aq) → C HO ( aq) + HO()l8 5 4 8 4 42(a) The balance used to weigh KHP is not properly calibrated andalways reads 0.15 g too low.(b) The indicator for the titration changes color between a pH of 3–4.(c) An air bubble, which is lodged in the buret’s tip at the beginning ofthe analysis, dislodges during the titration.(d) Samples of KHP are weighed into separate Erlenmeyer flasks, butthe balance is only tarred with the first flask.(e) The KHP is not dried before it was used.(f) The NaOH is not dried before it was used.

520 Analytical Chemistry 2.0(g) The procedure states that the sample of KHP should be dissolved in25 mL of water, but it is accidentally dissolved in 35 mL of water.30. The concentration of o-phthalic acid in an organic solvent, such as n-butanol, is determined by an acid–base titration using aqueous NaOHas the titrant. As the titrant is added, the o-phthalic acid is extractedinto the aqueous solution where it reacts with the titrant. The titrantmust be added slowly to allow sufficient time for the extraction to takeplace.(a) What type of error do you expect if the titration is carried out tooquickly?(b) Propose an alternative acid–base titrimetric method that allows fora more rapid determination of the concentration of o-phthalic acidin n‐butanol.31. Calculate or sketch titration curves for 50.00 mL of 0.0500 Mg 2+ with0.0500 M EDTA at a pH of 7 and 10. Locate the equivalence point foreach titration curve.Some of the problems that follow require one ormore equilibrium constants or standard state potentials.For your convenience, here are hyperlinksto the appendices containing these constantsAppendix 10: Solubility ProductsAppendix 11: Acid Dissociation ConstantsAppendix 12: Metal-Ligand Formation ConstantsAppendix 13: Standard State Reduction Potentials32. Calculate or sketch titration curves for 25.0 mL of 0.0500 M Cu 2+ with0.025 M EDTA at a pH of 10, and in the presence of 10 –3 M and 10 –1M NH 3 . Locate the equivalence point for each titration curve.33. Sketch the spectrophotometric titration curve for the titration of amixture of 5.00 × 10 –3 M Bi 3+ and 5.00 × 10 –3 M Cu 2+ with 0.0100M EDTA. Assume that only the Cu 2+ –EDTA complex absorbs at theselected wavelength.34. The EDTA titration of mixtures of Ca 2+ and Mg 2+ can be followedthermometrically because the formation of the Ca 2+ –EDTA complexis exothermic and the formation of the Mg 2+ –EDTA complex is endothermic.Sketch the thermometric titration curve for a mixture of5.00 × 10 –3 M Ca 2+ and 5.00 × 10 –3 M Mg 2+ with 0.0100 M EDTA.The heats of formation for CaY 2– and MgY 2– are, respectively, –23.9kJ/mole and 23.0 kJ/mole.35. EDTA is one member of a class of aminocarboxylate ligands that formvery stable 1:1 complexes with metal ions. The following table provideslogK f values for the complexes of six such ligands with Ca 2+ and Mg 2+ .Which ligand is the best choice for the direct titration of Ca 2+ in thepresence of Mg 2+ ?Mg 2+ Ca 2+EDTA ethylenediaminetetraacetic acid 8.7 10.7HEDTA N-hydroxyethylenediaminetriacetic acid 7.0 8.0

Chapter 9 Titrimetric Methods521Mg 2+ Ca 2+EEDTA ethyletherdiaminetetraacetic acid 8.3 10.0DGTA ethyleneglycol-bis(b-aminoethylether)- 5.4 10.9N,N´-tetraacetic acidDTPA diethylenetriaminepentaacetic acid 9.0 10.7CyDTA cyclohexanediaminetetraacetic acid 10.3 12.336. The amount of calcium in physiological fluids can be determined by acomplexometric titration with EDTA. In one such analysis a 0.100-mLsample of a blood serum was made basic by adding 2 drops of NaOHand titrated with 0.00119 M EDTA, requiring 0.268 mL to reach theend point. Report the concentration of calcium in the sample as milligramsCa per 100 mL.37. After removing the membranes from an eggshell, the shell is dried andits mass recorded as 5.613 g. The eggshell is transferred to a 250-mLbeaker and dissolved in 25 mL of 6 M HCl. After filtering, the solutioncontaining the dissolved eggshell is diluted to 250 mL in a volumetricflask. A 10.00-mL aliquot is placed in a 125-mL Erlenmeyer flask andbuffered to a pH of 10. Titrating with 0.04988 M EDTA requires 44.11mL to reach the end point. Determine the amount of calcium in theeggshell as %w/w CaCO 3 .38. The concentration of cyanide, CN – , in a copper electroplating bathcan be determined by a complexometric titration with Ag + , formingthe soluble Ag(CN) 2 – complex. In a typical analysis a 5.00-mL samplefrom an electroplating bath is transferred to a 250-mL Erlenmeyer flask,and treated with 100 mL of H 2 O, 5 mL of 20% w/v NaOH and 5 mLof 10% w/v KI. The sample is titrated with 0.1012 M AgNO 3 , requiring27.36 mL to reach the end point as signaled by the formation of ayellow precipitate of AgI. Report the concentration of cyanide as partsper million of NaCN.39. Before the introduction of EDTA most complexation titrations usedAg + or CN – as the titrant. The analysis for Cd 2+ , for example, was accomplishedindirectly by adding an excess of KCN to form Cd(CN) 4 2– ,and back titrating the excess CN – with Ag + , forming Ag(CN) 2 – . In onesuch analysis a 0.3000-g sample of an ore was dissolved and treatedwith 20.00 mL of 0.5000 M KCN. The excess CN – required 13.98 mLof 0.1518 M AgNO 3 to reach the end point. Determine the %w/w Cdin the ore.40. Solutions containing both Fe 3+ and Al 3+ can be selectively analyzed forFe 3+ by buffering to a pH of 2 and titrating with EDTA. The pH of thesolution is then raised to 5 and an excess of EDTA added, resulting inSome of the problems that follow require one ormore equilibrium constants or standard state potentials.For your convenience, here are hyperlinksto the appendices containing these constantsAppendix 10: Solubility ProductsAppendix 11: Acid Dissociation ConstantsAppendix 12: Metal-Ligand Formation ConstantsAppendix 13: Standard State Reduction Potentials

522 Analytical Chemistry 2.0Some of the problems that follow require one ormore equilibrium constants or standard state potentials.For your convenience, here are hyperlinksto the appendices containing these constantsAppendix 10: Solubility ProductsAppendix 11: Acid Dissociation ConstantsAppendix 12: Metal-Ligand Formation ConstantsAppendix 13: Standard State Reduction Potentialsthe formation of the Al 3+ –EDTA complex. The excess EDTA is backtitratedusing a standard solution of Fe 3+ , providing an indirect analysisfor Al 3+ .(a) At a pH of 2, verify that the formation of the Fe 3+ –EDTA complexis favorable, and that the formation of the Al 3+ –EDTA complex isnot favorable.(b) A 50.00-mL aliquot of a sample containing Fe 3+ and Al 3+ is transferredto a 250-mL Erlenmeyer flask and buffered to a pH of 2. Asmall amount of salicylic acid is added, forming the soluble redcoloredFe 3+ –salicylic acid complex. The solution is titrated with0.05002 M EDTA, requiring 24.82 mL to reach the end point assignaled by the disappearance of the Fe 3+ –salicylic acid complex’sred color. The solution is buffered to a pH of 5 and 50.00 mL of0.05002 M EDTA is added. After ensuring that the formation ofthe Al 3+ –EDTA complex is complete, the excess EDTA was backtitrated with 0.04109 M Fe 3+ , requiring 17.84 mL to reach theend point as signaled by the reappearance of the red-colored Fe 3+ –salicylic acid complex. Report the molar concentrations of Fe 3+ andAl 3+ in the sample.41. Prada and colleagues described an indirect method for determiningsulfate in natural samples, such as seawater and industrial effluents. 12The method consists of three steps: precipitating the sulfate as PbSO 4 ;dissolving the PbSO 4 in an ammonical solution of excess EDTA toform the soluble PbY 2– complex; and titrating the excess EDTA witha standard solution of Mg 2+ . The following reactions and equilibriumconstants are known22PbSO () s Pb + ( aq) + SO− ( aq ) K sp = 1.6 × 10 –84Pb 2 +Y 4 −PbY2 −( aq) + ( aq) ( aq)K f = 1.1 × 10 18Mg 2 +Y 4 −MgY2 −( aq) + ( aq) ( aq)K f = 4.9 × 10 8Zn 2 +Y 4 −ZnY2 −( aq) + ( aq) ( aq)K f = 3.2 × 10 16(a) Verify that a precipitate of PbSO 4 dissolves in a solution of Y 4– .4(b) Sporek proposed a similar method using Zn 2+ as a titrant and foundthat the accuracy was frequently poor. 13 One explanation is thatZn 2+ might react with the PbY 2– complex, forming ZnY 2– . Showthat this might be a problem when using Zn 2+ as a titrant, but thatit is not a problem when using Mg 2+ as a titrant. Would such a12 Prada, S.; Guekezian, M.; Suarez-Iha, M. E. V. Anal. Chim. Acta 1996, 329, 197–202.13 Sporek, K. F. Anal. Chem. 1958, 30, 1030–1032.

Chapter 9 Titrimetric Methods523displacement of Pb 2+ by Zn 2+ lead to the reporting of too much ortoo little sulfate?(c) In a typical analysis, a 25.00-mL sample of an industrial effluentwas carried through the procedure using 50.00 mL of 0.05000 MEDTA. Titrating the excess EDTA required 12.42 mL of 0.1000M Mg 2+ . Report the molar concentration of SO 4 2– in the sampleof effluent.42. Table 9.10 provides values for the fraction of EDTA present as Y 4- , a Y 4–.Values of a Y 4– are calculated using the equationα Y4− =4−[ Y ]CEDTAwhere [Y 4- ] is the concentration of the fully deprotonated EDTA andC EDTA is the total concentration of EDTA in all of its forms2+ +C EDTA 6 5 4= [ HY ] + [ HY ] + [ HY]+2−[ HY ] + [ HY ] + [ HY−3 2−] + [ Y ]3−4Using the following equilibria2+ + +HY ( aq) + HO() l HO ( aq) + HY ( aq )6 2 3 5+ +HY ( aq) + HO() l HO ( aq) + HY( aq )5 2 3 4+ −HY( aq) + HO() l HO ( aq) + HY ( aq )4 2 3 3− + 2−HY ( aq) + HO() l HO ( aq) + HY ( aq )3 2 3 22− + 3−HY ( aq) + HO() l HO ( aq) + HY ( aq )2 2 33− + 4−HY ( aq) + HO() l HO ( aq) + Y ( aq )2 3K a1K a2K a3K a4K a5K a6Some of the problems that follow require one ormore equilibrium constants or standard state potentials.For your convenience, here are hyperlinksto the appendices containing these constantsAppendix 10: Solubility ProductsAppendix 11: Acid Dissociation ConstantsAppendix 12: Metal-Ligand Formation ConstantsAppendix 13: Standard State Reduction Potentialsshow thatα Y4−=K K K K K Ka1 a2 a3 a4 a5 a6dwhered = + 6+ + 5K + + 4[ H ] [ H ] [ H ] K K ++ 3[ H ] K K Ka1 a2 a3a1 a1 a2+ 2+ [ H ] K K K K +a1 a2 a3 a4+ 1[ H ] K K K K K + Ka1 a2 a3 a4 a5 a 1K a2K a3K a4K a5K a6

524 Analytical Chemistry 2.043. Calculate or sketch titration curves for the following (unbalanced) redoxtitration reactions at 25 o C. Assume the analyte is initially presentat a concentration of 0.0100 M and that a 25.0-mL sample is taken foranalysis. The titrant, which is the underlined species in each reaction,is 0.0100 M.(a) V 2 +Ce 4 +V 3 +Ce3 +( aq) + ( aq) → ( aq) + ( aq)(b) Ti 2 +Fe 3 +Ti 3 +Fe2 +( aq) + ( aq) → ( aq) + ( aq)2+ − 3+ 2+(c) Fe ( aq) + MnO ( aq) → Fe ( aq) + Mn ( aq ) (atpH = 1)444. What is the equivalence point for each titration in problem 43?45. Suggest an appropriate indicator for each titration in problem 43.Some of the problems that follow require one ormore equilibrium constants or standard state potentials.For your convenience, here are hyperlinksto the appendices containing these constantsAppendix 10: Solubility ProductsAppendix 11: Acid Dissociation ConstantsAppendix 12: Metal-Ligand Formation ConstantsAppendix 13: Standard State Reduction Potentials46. The iron content of an ore can be determined by a redox titration usingK 2 Cr 2 O 7 as the titrant. A sample of the ore is dissolved in concentratedHCl using Sn 2+ to speed its dissolution by reducing Fe 3+ to Fe 2+ . Afterthe sample is dissolved, Fe 2+ and any excess Sn 2+ are oxidized to Fe 3+and Sn 4+ using MnO 4 – . The iron is then carefully reduced to Fe 2+ byadding a 2–3 drop excess of Sn 2+ . A solution of HgCl 2 is added and, ifa white precipitate of Hg 2 Cl 2 forms, the analysis is continued by titratingwith K 2 Cr 2 O 7 . The sample is discarded without completing theanalysis if a precipitate of Hg 2 Cl 2 does not form, or if a gray precipitate(due to Hg) forms.(a) Explain why the analysis is not completed if a white precipitate ofHg 2 Cl 2 forms, or if a gray precipitate forms.(b) Is a determinate error introduced if the analyst forgets to add Sn 2+in the step where the iron ore is dissolved?(c) Is a determinate error introduced if the iron is not quantitativelyoxidized back to Fe 3+ by the MnO 4 – ?47. The amount of Cr 3+ in an inorganic salt can be determined by a redoxtitration. A portion of sample containing approximately 0.25 g of Cr 3+is accurately weighed and dissolved in 50 mL of H 2 O. The Cr 3+ isoxidized to Cr 2 O 7 2– by adding 20 mL of 0.1 M AgNO 3 , which servesas a catalyst, and 50 mL of 10%w/v (NH 4 ) 2 S 2 O 8 , which serves as theoxidizing agent. After the reaction is complete the resulting solutionis boiled for 20 minutes to destroy the excess S 2 O 8 2– , cooled to roomtemperature, and diluted to 250 mL in a volumetric flask. A 50-mLportion of the resulting solution is transferred to an Erlenmeyer flask,treated with 50 mL of a standard solution of Fe 2+ , and acidified with200 mL of 1 M H 2 SO 4 , reducing the Cr 2 O 7 2– to Cr 3+ . The excessFe 2+ is then determined by a back titration with a standard solution

Chapter 9 Titrimetric Methods525of K 2 Cr 2 O 7 using an appropriate indicator. The results are reported as%w/w Cr 3+ .(a) There are several places in the procedure where a reagent’s volume isspecified (see underlined text). Which of these measurements mustbe made using a volumetric pipet.(b) Excess peroxydisulfate, S 2 O 8 2– is destroyed by boiling the solution.What is the effect on the reported %w/w Cr 3+ if some of the S 2 O 82–is not destroyed during this step?(c) Solutions of Fe 2+ undergo slow air oxidation to Fe 3+ . What is theeffect on the reported %w/w Cr 3+ if the standard solution of Fe 2+is inadvertently allowed to be partially oxidized?48. The exact concentration of H 2 O 2 in a solution that is nominally 6%w/v H 2 O 2 can be determined by a redox titration with MnO 4 – . A 25-mL aliquot of the sample is transferred to a 250-mL volumetric flaskand diluted to volume with distilled water. A 25-mL aliquot of thediluted sample is added to an Erlenmeyer flask, diluted with 200 mLof distilled water, and acidified with 20 mL of 25% v/v H 2 SO 4 . Theresulting solution is titrated with a standard solution of KMnO 4 untila faint pink color persists for 30 s. The results are reported as %w/vH 2 O 2 .(a) Many commercially available solutions of H 2 O 2 contain an inorganicor organic stabilizer to prevent the autodecomposition of theperoxide to H 2 O and O 2 . What effect does the presence of thisstabilizer have on the reported %w/v H 2 O 2 if it also reacts withMnO 4 – ?(b) Laboratory distilled water often contains traces of dissolved organicmaterial that may react with MnO 4 – . Describe a simple method tocorrect for this potential interference.(c) What modifications to the procedure, if any, are need if the samplehas a nominal concentration of 30% w/v H 2 O 2 .Some of the problems that follow require one ormore equilibrium constants or standard state potentials.For your convenience, here are hyperlinksto the appendices containing these constantsAppendix 10: Solubility ProductsAppendix 11: Acid Dissociation ConstantsAppendix 12: Metal-Ligand Formation ConstantsAppendix 13: Standard State Reduction Potentials49. The amount of iron in a meteorite was determined by a redox titrationusing KMnO 4 as the titrant. A 0.4185-g sample was dissolved in acidand the liberated Fe 3+ quantitatively reduced to Fe 2+ using a Waldenreductor. Titrating with 0.02500 M KMnO 4 requires 41.27 mL toreach the end point. Determine the %w/w Fe 2 O 3 in the sample ofmeteorite.50. Under basic conditions, MnO 4 – can be used as a titrant for the analysisof Mn 2+ , with both the analyte and the titrant forming MnO 2 . In theanalysis of a mineral sample for manganese, a 0.5165-g sample is dissolvedand the manganese reduced to Mn 2+ . The solution is made basic

526 Analytical Chemistry 2.0and titrated with 0.03358 M KMnO 4 , requiring 34.88 mL to reach theend point. Calculate the %w/w Mn in the mineral sample.51. The amount of uranium in an ore can be determined by a redox backtitration. The analysis is accomplished by dissolving the ore in sulfuricacid and reducing the resulting UO 2 2+ to U 4+ with a Walden reductor.The resulting solution is treated with an excess of Fe 3+ , forming Fe 2+and U 6+ . The Fe 2+ is titrated with a standard solution of K 2 Cr 2 O 7 . Ina typical analysis a 0.315-g sample of ore is passed through the Waldenreductor and treated with 50.00 mL of 0.0125 M Fe 3+ . Back titratingwith 0.00987 M K 2 Cr 2 O 7 requires 10.52 mL. What is the %w/w U inthe sample?Some of the problems that follow require one ormore equilibrium constants or standard state potentials.For your convenience, here are hyperlinksto the appendices containing these constantsAppendix 10: Solubility ProductsAppendix 11: Acid Dissociation ConstantsAppendix 12: Metal-Ligand Formation ConstantsAppendix 13: Standard State Reduction Potentials52. The thickness of the chromium plate on an auto fender was determinedby dissolving a 30.0-cm 2 section in acid, and oxidizing the liberatedCr 3+ to Cr 2 O 7 2– with peroxydisulfate. After removing the excessperoxydisulfate by boiling, 500.0 mg of Fe(NH 4 ) 2 (SO 4 ) 2 •6H 2 O wasadded, reducing the Cr 2 O 7 2– to Cr 3+ . The excess Fe 2+ was back titrated,requiring 18.29 mL of 0.00389 M K 2 Cr 2 O 7 to reach the end point.Determine the average thickness of the chromium plate given that thedensity of Cr is 7.20 g/cm 3 .53. The concentration of CO in air can be determined by passing a knownvolume of air through a tube containing I 2 O 5 , forming CO 2 and I 2 .The I 2 is removed from the tube by distilling it into a solution containingan excess of KI, producing I 3 – . The I 3 – is titrated with a standardsolution of Na 2 S 2 O 3 . In a typical analysis a 4.79-L sample of air wassampled as described here, requiring 7.17 mL of 0.00329 M Na 2 S 2 O 3to reach the end point. If the air has a density of 1.23 × 10 –3 g/mL,determine the parts per million CO in the air.54. The level of dissolved oxygen in a water sample can be determined bythe Winkler method. In a typical analysis a 100.0-mL sample is madebasic and treated with a solution of MnSO 4 , resulting in the formationof MnO 2 . An excess of KI is added and the solution is acidified, resultingin the formation of Mn 2+ and I 2 . The liberated I 2 is titrated with asolution of 0.00870 M Na 2 S 2 O 3 , requiring 8.90 mL to reach the starchindicator end point. Calculate the concentration of dissolved oxygen asparts per million O 2 .55. The analysis for Cl – using the Volhard method requires a back titration.A known amount of AgNO 3 is added, precipitating AgCl. Theunreacted Ag + is determined by back titrating with KSCN. There is acomplication, however, because AgCl is more soluble than AgSCN.

Chapter 9 Titrimetric Methods527(a) Why do the relative solubilities of AgCl and AgSCN lead to a titrationerror?(b) Is the resulting titration error a positive or a negative determinateerror?(c) How might you modify the procedure to prevent this eliminate thissource of determinate error?(d) Will this source of determinate error be of concern when using theVolhard method to determine Br – ?56. Voncina and co-workers suggest that a precipitation titration can bemonitored by measuring pH as a function of the volume of titrantif the titrant is a weak base. 14 For example, when titrating Pb 2+ withCrO 4 2– the solution containing the analyte is initially acidified to a pHof 3.50 using HNO 3 . Before the equivalence point the concentrationof CrO 4 2– is controlled by the solubility product of PbCrO 4 . After theequivalence point the concentration of CrO 4 2– is determined by theamount of excess titrant. Considering the reactions controlling the concentrationof CrO 4 2– , sketch the expected titration curve of pH versusvolume of titrant.57. Calculate or sketch the titration curve for the titration of 50.0 mL of0.0250 M KI with 0.0500 M AgNO 3 . Prepare separate titration curveusing pAg and pI on the y-axis.58. Calculate or sketch the titration curve for the titration of 25.0 mL mixtureof 0.0500 M KI and 0.0500 M KSCN with 0.0500 M AgNO 3 .59. A 0.5131-g sample containing KBr is dissolved in 50 mL of distilledwater. Titrating with 0.04614 M AgNO 3 requires 25.13 mL to reachthe Mohr end point. A blank titration requires 0.65 mL to reach thesame end point. Report the %w/w KBr in the sample.Some of the problems that follow require one ormore equilibrium constants or standard state potentials.For your convenience, here are hyperlinksto the appendices containing these constantsAppendix 10: Solubility ProductsAppendix 11: Acid Dissociation ConstantsAppendix 12: Metal-Ligand Formation ConstantsAppendix 13: Standard State Reduction Potentials60. A 0.1093-g sample of impure Na 2 CO 3 was analyzed by the Volhardmethod. After adding 50.00 mL of 0.06911 M AgNO 3 , the sample wasback titrated with 0.05781 M KSCN, requiring 27.36 mL to reach theend point. Report the purity of the Na 2 CO 3 sample.61. A 0.1036-g sample containing only BaCl 2 and NaCl is dissolved in50 mL of distilled water. Titrating with 0.07916 M AgNO 3 requires19.46 mL to reach the Fajans end point. Report the %w/w BaCl 2 in thesample.14 VonČina, D. B.; DobČnik, D.; GomiŠČek, S. Anal. Chim. Acta 1992, 263, 147–153.

528 Analytical Chemistry 2.09ISolutions to Practice ExercisesPractice Exercise 9.1The volume of HCl needed to reach the equivalence point isVeq= MVV = b bM)(25.0 mL)aM= ( 0.1250.0625 M= 50.0 mLaBefore the equivalence point, NaOH is present in excess and the pH isdetermined by the concentration of unreacted OH – . For example, afteradding 10.0 mL of HCl[ OH ]− =( 0. 125 M)(25.0 mL) −(0.0625 M)(10.0 mL)25.0 mL + 10.0 mL= 0.0714Mthe pH is 12.85.−+Kw100 . × 10[ HO ] = =3−[ OH ] 0.0714 M14= 140 . × 10− 13 MFor the titration of a strong base with a strong acid the pH at the equivalencepoint is 7.00.For volumes of HCl greater than the equivalence point, the pH is determinedby the concentration of excess HCl. For example, after adding 70.0mL of titrant the concentration of HCl is( 0. 0625 M)(70.0 mL) −(0.125 M)(25.0 mL)[ HCl]=70.0 mL + 25.0 mLgiving a pH of 1.88. Some additional results are shown here.= 0.0132Volume of HCl (mL) pH Volume of HCl (mL) pH0 13.10 60 2.1310 12.85 70 1.8820 12.62 80 1.7530 12.36 90 1.6640 11.98 100 1.6050 7.00Click here to return to the chapter.Practice Exercise 9.2The volume of HCl needed to reach the equivalence point isVeq= MVV = b bM)(25.0 mL)aM= ( 0.1250.0625 M= 50.0 mLaBefore adding HCl the pH is that for a solution of 0.100 M NH 3 .M

Chapter 9 Titrimetric Methods529Kb− +[ OH ][ NH ]4( x)( x)= =NH 0.125− x= × −175 . 10 5[ ]3x = −−[ OH ] = 147 . × 10 3 MThe pH at the beginning of the titration, therefore, is 11.17.Before the equivalence point the pH is determined by an NH 3 /NH 4+buffer. For example, after adding 10.0 mL of HCl(0.125 M)(25.0 mL) (0.0625 M)(10.0 mL[ NH ] = − )= 0.0714 M325.0 mL + 10.0 mL[ NH ]+ =4( 0.0625 M)(10.0 mL)=25.0 mL + 10.0 mL0. 0179 M0 0714 MpH = 9. 244 + log . = 984 .0.0179 MAt the equivalence point the predominate ion in solution is NH 4 + . Tocalculate the pH we first determine the concentration of NH 4+[ NH ]+ =4and then calculate the pHKaobtaining a value of 5.31.( 0.125 M)(25.0 mL)= 0.0417 M25.0 mL + 50.0 mL+[ HO ][ NH ]3 3( x)( x)= =NH 0.0417 − x= 570 . × 10 −10+[ ]4x = [ HO+ ] = 488 . × 10 −6M3After the equivalence point, the pH is determined by the excess HCl. Forexample, after adding 70.0 mL of HCl( 0. 0625 M)(70.0 mL) −(0.125 M)(25.0 mL)[ HCl]=25.0 mL + 70.0 mLand the pH is 1.88. Some additional results are shown here.= 0.0132Volume of HCl (mL) pH Volume of HCl (mL) pH0 11.17 60 2.1310 9.84 70 1.8820 9.42 80 1.7530 9.07 90 1.6640 8.64 100 1.6050 5.31Click here to return to the chapter.M

530 Analytical Chemistry 2.0pHFigure 9.48 Titration curve for Practice Exercise9.3. The black dots and curve are theapproximate sketch of the titration curve.The points in red are the calculations fromPractice Exercise 9.2.pH14121086420141210864200 20 40 60 80 100Volume of HCl (mL)0 20 40 60 80 100Volume of NaOH (mL)Figure 9.49 Titration curve for Practice Exercise9.4. The black points and curve arethe approximate titration curve, and the redcurve is the exact titration curve.Practice Exercise 9.3Figure 9.48 shows a sketch of the titration curve. The two points beforethe equivalence point (V HCl = 5 mL, pH = 10.24 and V HCl = 45 mL,pH = 8.24) are plotted using the pK a of 9.244 for NH 4 + . The two pointsafter the equivalence point (V HCl = 60 mL, pH = 2.13 and V HCl = 80 mL,pH = 1.75 ) are from the answer to Practice Exercise 9.2.Click here to return to the chapter.Practice Exercise 9.4Figure 9.49 shows a sketch of the titration curve. The titration curve hastwo equivalence points, one at 25.0 mL (H 2 A HA – ) and one at 50.0mL (HA – A 2– ). In sketching the curve, we plot two points before thefirst equivalence point using the pK a of 3 for H 2 AV HCl = 2.5 mL, pH = 2 and V HCl = 22.5 mL, pH = 4two points between the equivalence points using the pK a of 5 for HA –V HCl = 27.5 mL, pH = 3, and V HCl = 47.5 mL, pH = 5and two points after the second equivalence pointV HCl = 70 mL, pH = 12.22 and V HCl = 90 mL, pH = 12.46)Drawing a smooth curve through these points presents us with the followingdilemma—the pH appears to increase as the titrant’s volume approachesthe first equivalence point and then appears to decrease as itpasses through the first equivalence point. This is, of course, absurd; aswe add NaOH the pH cannot decrease. Instead, we model the titrationcurve before the second equivalence point by drawing a straight line fromthe first point (V HCl = 2.5 mL, pH = 2) to the fourth (V HCl = 47.5 mL,pH = 5), ignoring the second and third points. The results is a reasonableapproximation of the exact titration curve.Click here to return to the chapter.Practice Exercise 9.5The pH at the equivalence point is 5.31 (see Practice Exercise 9.2) andthe sharp part of the titration curve extends from a pH of approximately7 to a pH of approximately 4. Of the indicators in Table 9.4, methyl redis the best choice because it pK a value of 5.0 is closest to the equivalencepoint’s pH and because the pH range of 4.2–6.3 for its change in colorwill not produce a significant titration error.Click here to return to the chapter.Practice Exercise 9.6Because salicylic acid is a diprotic weak acid, we must first determineto which equivalence point it is being titrated. Using salicylic acid’s pK a

Chapter 9 Titrimetric Methods531values as a guide, the pH at the first equivalence point is between a pHof 2.97 and 13.74, and the second equivalence points is at a pH greaterthan 13.74. From Table 9.4, phenolphthalein’s end point falls in the pHrange 8.3–10.0. The titration, therefore, is to the first equivalence pointfor which the moles of NaOH equal the moles of salicylic acid; thus0.1354molNaOHL× 0. 02192 L= 2.968× 10 −3mol NaOH1 molC HO32. 968× 10−7 6 3molNaOH× ×molNaOH13812. gC HO7 6 3= 0.4099 gC HOmolC HO7 6 30.4099 gC HO7 6 3 × 100 = 97. 41%w/w CHO0.4208 gsample7 6 37 6 3Because the purity of the sample is less than 99%, we reject the shipment.Click here to return to the chapter.Practice Exercise 9.7The moles of HNO 3 produced by pulling the air sample through thesolution of H 2 O 2 is0.01012 molNaOH× 0.00914 L×L1 molHNO3= 925 . × 10 −5molHNOmolNaOH3A conservation of mass on nitrogen requires that each mole of NO 2 inthe sample of air produces one mole of HNO 3 ; thus, the mass of NO 2in the sample is51 molNO925 . × 10−2molHNO × ×3molHNO46.01 gNOmolNOand the concentration of NO 2 is426 . × 10− 35 LairgNOClick here to return to the chapter.2223= 426 . × 10 −3gNO1000 mg× = 0.852 mg NO /L air2g2

532 Analytical Chemistry 2.0Practice Exercise 9.8The total moles of HCl used in this analysis isOf this,1.396molNaOHL× 0. 01000 L= 1.396× 10 −2mol HCl0.1004 molNaOH1 molHCl× 0.03996 L×= 4.012× 10 −3molHClLmolNaOHare consumed in the back titration with NaOH, which means that−2 −31. 396× 10 molHCl − 4. 012× 10 molHCl = 995 . × 10 −3molHClreact with the CaCO 3 . Because CO 3 2– is dibasic, each mole of CaCO 3consumes two moles of HCl; thus31 molCaCO995 . × 10−3molHCl× ×2mol HCl100.09 gCaCO3= 0.498 gCaCOmolCaCO0.498 gCaCO 3× 100 = 96.%8 w/w CaCO0.5143 gsample3Click here to return to the chapter.Practice Exercise 9.9Of the two analytes, 2-methylanilinium is the stronger acid and is the firstto react with the titrant. Titrating to the bromocresol purple end point,therefore, provides information about the amount of 2-methylaniliniumin the sample.0.200molNaOHL31 molC H NCl7 10× 0.01965 L×mol NaOH143.61 gC H NCl7 10× = 0. 564 gC H NCl7 10molC H NCl7 100.564 gC H NCl7 10× 100 = 28.%1 w/ wC H NCl7 102.006 gsampleTitrating from the bromocresol purple end point to the phenolphthaleinend point, a total of 48.41 mL – 19.65 mL, or 28.76 mL, gives theamount of NaOH reacting with 3-nitrophenol. The amount of 3-nitrophenolin the sample, therefore, is3

Chapter 9 Titrimetric Methods5330.200molNaOHL1 molC HNO6 5 3× 0.02876 L×mol NaOH139.11 gC HNO6 5 3× = 0 800 gC HNOmolC HNO6 5 3.6 5 30.800 gC HNO6 5 3× 100 = 38.%8 w/w CHNO2.006 gsampleClick here to return to the chapter.Practice Exercise 9.106 5 3The first of the two visible end points is approximately 37 mL of NaOH.The analyte’s equivalent weight, therefore, is0.1032molNaOHL1 equivalent× 0.037 L×molNaOH0.5000 gEW == 13 . × 10−338 . × 10 equivalentsClick here to return to the chapter.Practice Exercise 9.11= 38 . × 10 −32g/equivalentequivalentsAt ½V eq , or approximately 18.5 mL, the pH is approximately 2.2; thus,we estimate that the analyte’s pK a is 2.2.Click here to return to the chapter.Practice Exercise 9.12Let’s begin with the calculations at a pH of 10. Ata pH of 10 some of theEDTA is present in forms other than Y 4– . To evaluate the titration curve,therefore, we need the conditional formation constant for CdY 2– , which,from Table 9.11 is K f´ = 1.1 × 10 16 . Note that the conditional formationconstant is larger in the absence of an auxiliary complexing agent.The titration’s equivalence point requiresVeqof EDTA.M VCd Cd( 500 . × 10 −3M)(50.0 mL)= V = =EDTAM0.0100 MEDTA= 25.0Before the equivalence point, Cd 2+ is present in excess and pCd is determinedby the concentration of unreacted Cd 2+ . For example, after adding5.00 mL of EDTA, the total concentration of Cd 2+ ismL

534 Analytical Chemistry 2.0−32+( 500 . × 10 M)(50.0 mL) −(0.0100 M)(5. 00 mL)[ Cd ] =50.0 mL + 500 . mL= 364 . × 10 −3 Mwhich gives a pCd of 2.43.At the equivalence point all the Cd 2+ initially in the titrand is now presentas CdY 2– . The concentration of Cd 2+ , therefore, is determined by thedissociation of the CdY 2– complex. First, we calculate the concentrationof CdY 2– .−32−( 500 . × 10 M)(50.0 mL)[ CdY ] =50. 0 mL + 250 . mL= 333 . × 10 −3Next, we solve for the concentration of Cd 2+ in equilibrium with CdY 2– .Kf2−−3′[ CdY ] 333 . × 10 − x= =+= 11 . × 10 162[ Cd ] C ( x)( x)EDTAMSolving gives [Cd 2+ ] as 5.50 × 10 –10 M, or a pCd of 9.26 at the equivalencepoint.After the equivalence point, EDTA is in excess and the concentration ofCd 2+ is determined by the dissociation of the CdY 2– complex. First, wecalculate the concentrations of CdY 2– and of unreacted EDTA. For example,after adding 30.0 mL of EDTA−32−( 500 . × 10 M)(50.0 mL)[ CdY ] =50. 0 mL + 300 . mL= 313 . × 10 −3( 0.0100 M)(30.0 mL) − (5.00× 10 −3M)(50. 0mL)C EDTA=50.0 mL + 30.0 mL= 625 . × 10 −4MSubstituting into the equation for the conditional formation constant andsolving for [Cd 2+ ] gives−3313 . × 10 M= 11 . × 102+ −4[ Cd ]( 625 . × 10 M)[Cd 2+ ] as 4.55× 10 –16 M, or a pCd of 15.34.The calculations at a pH of 7 are identical, except the conditional formationconstant for CdY 2– is 1.5 × 10 13 instead of 1.1 × 10 16 . The followingtable summarizes results for these two titrations as well as the results fromTable 9.13 for the titration of Cd 2+ at a pH of 10 in the presence of 0.0100M NH 3 as an auxiliary complexing agent.16M

Chapter 9 Titrimetric Methods535Volume ofEDTA (mL)pCdat pH 10pCdat pH 10 w/0.0100 M NH 3pCdat pH 70 2.30 3.36 2.305.00 2.43 3.49 2.4310.0 2.60 3.66 2.6015.0 2.81 3.87 2.8120.0 3.15 4.20 3.1523.0 3.56 4.62 3.5625.0 9.26 9.77 7.8327.0 14.94 14.95 12.0830.0 15.34 15.33 12.4835.0 15.61 15.61 12.7840.0 15.76 15.76 12.9545.0 15.86 15.86 13.0850.0 15.94 15.94 13.18Examining these results allows us to draw several conclusions. First, inthe absence of an auxiliary complexing agent the titration curve beforethe equivalence point is independent of pH (compare columns 2 and 4).Second, for any pH, the titration curve after the equivalence point is thesame regardless of whether or not an auxiliary complexing agent is present(compare columns 2 and 3). Third, the largest change in pH through theequivalence point occurs at higher pHs and in the absence of an auxiliarycomplexing agent. For example, from 23.0 mL to 27.0 mL of EDTA thechange in pCd is 11.38 at a pH of 10, 10.33 at a pH of 10 and in thepresence of 0.0100 M NH 3 , and 8.52 at a pH of 7.Click here to return to the chapter.Practice Exercise 9.13Figure 9.50 shows a sketch of the titration curves. The two points beforethe equivalence point (V EDTA = 5 mL, pCd = 2.43 and V EDTA = 15 mL,pCd = 2.81) are the same for both pHs and are taken from the results ofPractice Exercise 9.12. The two points after the equivalence point for a pHof 7 (V EDTA = 27.5 mL, pCd = 12.2 and V EDTA = 50 mL, pCd = 13.2 )are plotted using the logK f´ of 13.2 for CdY 2- . The two points after theequivalence point for a pH of 10 (V EDTA = 27.5 mL, pCd = 15.0 andV EDTA = 50 mL, pCd = 16.0 ) are plotted using the logK f´ of 16.0 forCdY 2- .Click here to return to the chapter.pCd201510500 10 20 30 40 50Volume of EDTA (mL)Figure 9.50 Titration curve for Practice Exercise9.13. The black dots and curve arethe approximate sketches of the two titrationcurves. The points in red are the calculationsfrom Practice Exercise 9.12 for apH of 10, and the points in green are thecalculations from Practice Exercise 9.12 fora pH of 7.

536 Analytical Chemistry 2.0Practice Exercise 9.14In an analysis for hardness we treat the sample as if Ca 2+ is the only metalion reacting with EDTA. The grams of Ca 2+ in the sample, therefore, is0.0109molEDTAL2+1 molCa× 0.02363 L×molEDTA= 258 . × 10 −441258 . × 10−2+molCaCO3molCa × ×2+molCa100.09 gCaCO3= 0.0258 gCaCOmolCaCOand the sample’s hardness is0.0258 gCaCO 31000 mg× = 258 mg CaCO3 0.1000 L g/L33molCa 2+Click here to return to the chapter.Practice Exercise 9.15The titration of CN – with Ag + produces a metal-ligand complex ofAg(CN) 2 2– ; thus, each mole of AgNO 3 reacts with two moles of NaCN.The grams of NaCN in the sample is0.1018 molAgNO 32 molNaCN× 0.03968 L×LmolAgNO49.01 gNaCN× = 0.3959 gNaCNmolNaCNand the purity of the sample is0.3959 gNaCN× 100 = 88. 33%w/w NaCN0.4482 gsampleClick here to return to the chapter.Practice Exercise 9.16The total moles of EDTA used in this analysis is0.02011 molEDTALOf this,× 0. 02500 L= 5. 028× 10 −4mol EDTA2+0.01113 molMg× 0.00423 L×L1 molEDTA= 4.708× 10 −5molEDTA2+molMg3

Chapter 9 Titrimetric Methods537are consumed in the back titration with Mg 2+ , which means that−5. 028× 10 molEDTA− 4.708×104 −5mol EDTA= 45 . 57× 10 −4molEDTAreact with the BaSO 4 . Each mole of BaSO 4 reacts with one mole ofEDTA; thus41 molBaSO4. 557× 10−4molEDTA × ×molEDTA1 mol Na SO 142 gNaSO2 4× . 04molBaSO molNaSO422 44= 0.064730.06473 gNaSO2 4× 100 = 41. 23%w/wNaSO0.1557 gsampleClick here to return to the chapter.Practice Exercise 9.172 4The volume of Tl 3+ needed to reach the equivalence point isVeqM VSn SnM)(50.0 mL)= V = = ( 0.050 = 25.0 mLTlM (0.100 M)TlgNaSO2 4Before the equivalence point, the concentration of unreacted Sn 2+ andthe concentration of Sn 4+ are easy to calculate. For this reason we findthe potential using the Nernst equation for the Sn 4+ /Sn 2+ half-reaction.For example, the concentrations of Sn 2+ and Sn 4+ after adding 10.0 mLof titrant are2 ( 0. 050 M)(50.0 mL) − ( 0.100 M)(10.0 mL)[ Sn+ ] ==50.0 mL + 10.0 mLand the potential is4 ( 0.100 M)(10.0 mL)[ Sn+ ] == 0.0167 M50.0 mL + 10.0 mLE =+ 0.139 V −0.059162⎛0.0250 M⎞log⎝⎜0.0167 M⎠⎟ 0.0250=+0 . 134 VAfter the equivalence point, the concentration of Tl + and the concentrationof excess Tl 3+ are easy to calculate. For this reason we find the potentialusing the Nernst equation for the Tl 3+ /Tl + half-reaction. For example,after adding 40.0 mL of titrant, the concentrations of Tl + and Tl 3+ are[ Tl ]+ =( 0.0500 M)(50.0 mL)= 00278 . M50.0 mL + 40.0mLM

538 Analytical Chemistry 2.0E (V)0.80.60.40.20.00 10 20 30 40 50Volume of Tl 3+ (mL)Figure 9.51 Titration curve for Practice Exercise9.18. The black dots and curve are theapproximate sketch of the titration curve.The points in red are the calculations fromPractice Exercise 9.17.3 ( 0. 100 M)(40.0 mL) −( 0.0500 M)(50.0 mL)[ Tl+ ] =50.0 mL + 40.0 mLand the potential isE =+ 077 . V −0.059162= 0.0167⎛0.0278 M⎞log⎝⎜0.0167 M⎠⎟ =+076 . VAt the titration’s equivalence point, the potential, E eq , potential is0. 139 V+077 . VE eq== 045 . V2Some additional results are shown here.Volume of Tl 3+ (mL) E (V) Volume of Tl 3+ (mL) E (V)5 0.121 30 0.7510 0.134 35 0.7515 0.144 40 0.7620 0.157 45 0.7625 0.45 50 0.76Click here to return to the chapter.Practice Exercise 9.18Figure 9.51 shows a sketch of the titration curve. The two points beforethe equivalence pointV Tl = 2.5 mL, E = +0.109 V and V Tl = 22.5 mL, E = +0.169 Vare plotted using the redox buffer for Sn 4+ /Sn 2+ , which spans a potentialrange of +0.139 ± 0.5916/2. The two points after the equivalence pointV Tl = 27.5 mL, E = +0.74 V and V EDTA = 50 mL, E = +0.77 Vare plotted using the redox buffer for Tl 3+ /Tl + , which spans the potentialrange of +0.139 ± 0.5916/2.Click here to return to the chapter.MPractice Exercise 9.19The two half reactions areCe( aq) + e →Ce( aq)4+ − 3+4+2+U ( aq) + 2H O→ UO ( aq)+ 4H + ( aq)+2e2for which the Nernst equations are3+o 0.05916 CeE = E4+ 3+− log [ ]Ce / Ce4+1 [ Ce ]2−

Chapter 9 Titrimetric Methods5394+o 0.05916 [ U ]E = E+ +− logUO2 U4+ +2 /2 [ UO ][ H ]2 42Before adding these two equations together we must multiply the secondequation by 2 so that we can combine the log terms; thus3+4+oo[ Ce ][ U ]3E = E4+ 3+ + 2E2+ 4+−0. 05916logCe / CeUO2/ U4+ + +[ Ce ][ UO ][ H ]At the equivalence point we know that3+ 2+[ Ce ] = 2×[ UO ]4+ 4+[ Ce ] = 2×[ U ]22 42Substituting these equalities into the previous equation and rearranginggives us a general equation for the potential at the equivalence point.2+ 4+oo2[UO ][ U ]23E = E4+ 3+ + 2E2+ 4+−0. 05916logCe / CeUO2/ U4+ + +2[ U ][ UO ][ H ]EE =EE =+ 2Eoo4+ 3+ + +Ce / CeUO2 2 / U4 .3+ 2E−oo4+ 3+ + +Ce / CeUO2 2 / U4 .EE =3+ 2Eoo4+ 3+ + +Ce / CeUO2 2 / U430 059163log[+ ] 4H10 05916×4+log[ H3−0.07888pHAt a pH of 1 the equivalence point has a potential ofE eq172 . + 2×0.327=− 0. 07888× 1=0.712 V3Click here to return to the chapter.Practice Exercise 9.202 42Because we have not been provided with a balanced reaction, let’s use aconservation of electrons to deduce the stoichiometry. Oxidizing C 2 O 4 2– ,in which each carbon has a +3 oxidation state, to CO 2 , in which carbonhas an oxidation state of +4, requires one electron per carbon, or a total oftwo electrons for each mole of C 2 O 4 2– . Reducing MnO 4 – , in which eachmanganese is in the +7 oxidation state, to Mn 2+ requires five electrons.A conservation of electrons for the titration, therefore, requires that twomoles of KMnO 4 (10 moles of e - ) reacts with five moles of Na 2 C 2 O 4 (10moles of e - ).The moles of KMnO 4 used in reaching the end point is+ ]

540 Analytical Chemistry 2.0( 0. 0400 MKMnO ) × ( 0.03562 LKMnO )which means that the sample contains4 435 molNaCO142 . × 10− molKMnO ×42mol KMnO= 142 . × 10−3 molKMnO 42 2 4Thus, the %w/w Na 2 C 2 O 4 in the sample of ore is3134.00 gNaCO355 . × 10− molNaCO ×2 2 4molNaCO4= 355 . × 10 −32 2 42 2 4= 0.476molNaCO2 2 4gNaCO2 2 40.476 gNaCO2 2 4× 100 = 93.%0 w/ wNaCO0.5116 gsampleClick here to return to the chapter.Practice Exercise 9.212 2 4For a back titration we need to determine the stoichiometry betweenCr 2 O 72–and the analyte, C 2 H 6 O, and between Cr 2 O 7 2– and the titrant,Fe 2+ . In oxidizing ethanol to acetic acid, the oxidation state of carbonchanges from –2 in C 2 H 6 O to 0 in C 2 H 4 O 2 . Each carbon releases twoelectrons, or a total of four electrons per C 2 H 6 O. In reducing Cr 2 O 7 2– ,in which each chromium has an oxidation state of +6, to Cr 3+ , each chromiumloses three electrons, for a total of six electrons per Cr 2 O 7 2– . Oxidationof Fe 2+ to Fe 3+ requires one electron. A conservation of electronsrequires that each mole of K 2 Cr 2 O 7 (6 moles of e - ) reacts with six molesof Fe 2+ (6 moles of e - ), and that four moles of K 2 Cr 2 O 7 (24 moles of e - )react with six moles of C 2 H 6 O (24 moles of e - ).The total moles of K 2 Cr 2 O 7 reacting with C 2 H 6 O and with Fe 2+ is−( 0. 0200 MK Cr O ) × ( 0. 05000 LI ) = 100 . × 10−3 molK Cr O2 2 7 3The back titration with Fe 2+ consumes2 2 72+2+0.1014 molFe0.02148 LFe × ×2+LFe1 mol KCrO2 2 7−= 363 . × 10 4 molK Cr O2+2 2 76 molFeSubtracting the moles of K 2 Cr 2 O 7 reacting with Fe 2+ from the total molesof K 2 Cr 2 O 7 gives the moles reacting with the analyte.−100 . × 10 KCrO − 3.63×10molK Cr O3 −42 2 7 2 2 7= 63 . 7× 10 −4molK Cr O2 2 7The grams of ethanol in the 10.00-mL sample of diluted brandy is

Chapter 9 Titrimetric Methods5416mol CHO637 . × 10 −42 6molK Cr O ××2 2 74mol KCr OThe %w/v C 2 H 6 O in the brandy is0.0444 gC HO2 6×10.00 mL dilute brandyClick here to return to the chapter.Practice Exercise 9.222 2 746.50 gC HO2 6= 0.0444 gC HO2 6molC HO2 6500.0 mLdilutebrandy5.00 mL brandy× 100 = 44.% 4 w/ vC HO2 6The first task is to calculate the volume of NaCl needed to reach theequivalence point; thusVeqM VAg Ag M)(50.0 mL)= V = = ( 0.0500 NaClM(0.100 M)NaCl= 25.0Before the equivalence point the titrand, Ag + , is in excess. The concentrationof unreacted Ag + after adding 10.0 mL of NaCl, for example, is[ Ag ]+ =mL( 0. 0500 M)( 50. 0 mL) −( 0. 100 M)( 10.0 mL)= 250 . × 10 −2M50. 0 mL + 100 . mLwhich corresponds to a pAg of 1.60. To find the concentration of Cl – weuse the K sp for AgCl; thusor a pCl of 8.14.K−− sp 18 . × 10[ Cl ] = =+[ Ag ] 250 . × 1010−2= 72 . × 10At the titration’s equivalence point, we know that the concentrations ofAg + and Cl – are equal. To calculate their concentrations we use the K spexpression for AgCl; thus−9+ − −K = [ Ag ][ Cl ] = ( x)( x) = 18 . × 10 10spSolving for x gives a concentration of Ag + and the concentration of Cl – as1.3 × 10 –5 M, or a pAg and a pCl of 4.89.After the equivalence point, the titrant is in excess. For example, afteradding 35.0 mL of titrant[ Cl ]− =( 0. 100 M)( 35. 0 mL) −( 0. 0500 M)( 50.0 mL)= 118 . × 10 −2M50. 0 mL + 350 . mLM

542 Analytical Chemistry 2.0or a pCl of 1.93. To find the concentration of Ag + we use the K sp forAgCl; thusK−+ sp 18 . × 10[ Ag ] = =−[ Cl ] 118 . × 1010−2= 15 . × 10or a pAg of 7.82. The following table summarizes additional results forthis titration.−8Volume ofNaCl (mL) pAg pCl0 1.30 –5.00 1.44 8.3110.0 1.60 8.1415.0 1.81 7.9320.0 2.15 7.6025.0 4.89 4.8930.0 7.54 2.2035.0 7.82 1.9340.0 7.97 1.7845.0 8.07 1.6850.0 8.14 1.60Click here to return to the chapter.Practice Exercise 9.23The titration uses0.1078MKSCNLM× 0. 02719 L= 2.931× 10 −3mol KSCNThe stoichiometry between SCN – and Ag + is 1:1; thus, there are− 3+ 107.87 gAg2.931× 10 molAg × = 0.3162 gAgmolAgin the 25.00 mL sample. Because this represents ¼ of the total solution,there are 0.3162 × 4 or 1.265 g Ag in the alloy. The %w/w Ag in the alloyis1.265 gAg× 100 = 64. 44%w/wAg1.963 gsampleClick here to return to the chapter.

DRAFTChapter 10Spectroscopic MethodsChapter OverviewSection 10A Overview of SpectroscopySection 10B Spectroscopy Based on AbsorptionSection 10C UV/Vis and IR SpectroscopySection 10D Atomic Absorption SpectroscopySection 10E Emission SpectroscopySection 10F Photoluminescent SpectroscopySection 10G Atomic Emission SpectroscopySection 10H Spectroscopy Based on ScatteringSection 10I Key TermsSection 10J Chapter SummarySection 10K ProblemsSection 10L Solutions to Practice ExercisesAn early example of a colorimetric analysis is Nessler’s method for ammonia, which wasintroduced in 1856. Nessler found that adding an alkaline solution of HgI 2 and KI to a dilutesolution of ammonia produces a yellow to reddish brown colloid, with the colloid’s colordepending on the concentration of ammonia. By visually comparing the color of a sample to thecolors of a series of standards, Nessler was able to determine the concentration of ammonia.Colorimetry, in which a sample absorbs visible light, is one example of a spectroscopicmethod of analysis. At the end of the nineteenth century, spectroscopy was limited to theabsorption, emission, and scattering of visible, ultraviolet, and infrared electromagnetic radiation.Since its introduction, spectroscopy has expanded to include other forms of electromagneticradiation—such as X-rays, microwaves, and radio waves—and other energetic particles—suchas electrons and ions.Copyright: David Harvey, 2009543

544 Analytical Chemistry 2.0Figure 10.1 The Golden Gate bridge asseen through rain drops. Refraction of lightby the rain drops produces the distortedimages. Source: Mila Zinkova (commons.wikipedia.org).10AOverview of SpectroscopyThe focus of this chapter is on the interaction of ultraviolet, visible, andinfrared radiation with matter. Because these techniques use optical materialsto disperse and focus the radiation, they often are identified as opticalspectroscopies. For convenience we will use the simpler term spectroscopyin place of optical spectroscopy; however, you should understand that weare considering only a limited part of a much broader area of analyticaltechniques.Despite the difference in instrumentation, all spectroscopic techniquesshare several common features. Before we consider individual examples ingreater detail, let’s take a moment to consider some of these similarities.As you work through the chapter, this overview will help you focus onsimilarities between different spectroscopic methods of analysis. You willfind it easier to understand a new analytical method when you can see itsrelationship to other similar methods.10A.1 What is Electromagnetic RadiationElectromagnetic radiation—light—is a form of energy whose behavioris described by the properties of both waves and particles. Some propertiesof electromagnetic radiation, such as its refraction when it passes from onemedium to another (Figure 10.1), are explained best by describing light asa wave. Other properties, such as absorption and emission, are better describedby treating light as a particle. The exact nature of electromagneticradiation remains unclear, as it has since the development of quantummechanics in the first quarter of the 20 th century. 1 Nevertheless, the dualmodels of wave and particle behavior provide a useful description for electromagneticradiation.Wa v e Pr o p e r t i e s o f El e c t r o m a g n e t i c RadiationElectromagnetic radiation consists of oscillating electric and magnetic fieldsthat propagate through space along a linear path and with a constant velocity.In a vacuum electromagnetic radiation travels at the speed of light,c, which is 2.997 92 10 8 m/s. When electromagnetic radiation movesthrough a medium other than a vacuum its velocity, v, is less than the speedof light in a vacuum. The difference between v and c is sufficiently small(

Chapter 10 Spectroscopic Methods545electric fieldmagnetic fieldAλdirectionofpropagationFigure 10.2 Plane-polarized electromagnetic radiation showing the oscillating electric field in red and the oscillatingmagnetic field in blue. The radiation’s amplitude, A, and its wavelength, l, are shown. Normally, electromagneticradiation is unpolarized, with oscillating electric and magnetic fields present in all possible planesperpendicular to the direction of propagation.tion, and direction of propagation. 2 For example, the amplitude of theoscillating electric field at any point along the propagating wave isA = A sin( 2πνt+ Φ)tewhere A t is the magnitude of the electric field at time t, A e is the electricfield’s maximum amplitude, n is the wave’s frequency—the number ofoscillations in the electric field per unit time—and F is a phase angle,which accounts for the fact that A t need not have a value of zero at t = 0.The identical equation for the magnetic field isA = A sin( 2πνt+ Φ)tmwhere A m is the magnetic field’s maximum amplitude.Other properties also are useful for characterizing the wave behavior ofelectromagnetic radiation. The wavelength, l, is defined as the distancebetween successive maxima (see Figure 10.2). For ultraviolet and visibleelectromagnetic radiation the wavelength is usually expressed in nanometers(1 nm = 10 –9 m), and for infrared radiation it is given in microns (1mm = 10 –6 m). The relationship between wavelength and frequency isλ= cνAnother unit useful unit is the wavenumber, ν , which is the reciprocal ofwavelengthν= 1λWavenumbers are frequently used to characterize infrared radiation, withthe units given in cm –1 .2 Ball, D. W. Spectroscopy 1994, 9(5), 24–25.When electromagnetic radiation movesbetween different media—for example,when it moves from air into water—itsfrequency, n, remains constant. Becauseits velocity depends upon the medium inwhich it is traveling, the electromagneticradiation’s wavelength, l, changes. If wereplace the speed of light in a vacuum, c,with its speed in the medium, v, then thewavelength isλ = vνThis change in wavelength as light passesbetween two media explains the refractionof electromagnetic radiation shownin Figure 10.1.

546 Analytical Chemistry 2.0Example 10.1In 1817, Josef Fraunhofer studied the spectrum of solar radiation, observinga continuous spectrum with numerous dark lines. Fraunhofer labeledthe most prominent of the dark lines with letters. In 1859, Gustav Kirchhoffshowed that the D line in the sun’s spectrum was due to the absorptionof solar radiation by sodium atoms. The wavelength of the sodium D lineis 589 nm. What are the frequency and the wavenumber for this line?So l u t i o nThe frequency and wavenumber of the sodium D line are8c 300 . × 10 m/sν = == 509 . × 10−9λ 589×10 ms14 −11 1ν = =λ 589×10−91 m× = 170 . × 10m 100 cmcm4 −1Practice Exercise 10.1Another historically important series of spectral lines is the Balmer seriesof emission lines form hydrogen. One of the lines has a wavelength of656.3 nm. What are the frequency and the wavenumber for this line?Click here to review your answer to this exercise.Pa r t i c l e Pr o p e r t i e s o f El e c t r o m a g n e t i c RadiationWhen matter absorbs electromagnetic radiation it undergoes a change inenergy. The interaction between matter and electromagnetic radiation iseasiest to understand if we assume that radiation consists of a beam of energeticparticles called photons. When a photon is absorbed by a sampleit is “destroyed,” and its energy acquired by the sample. 3 The energy of aphoton, in joules, is related to its frequency, wavelength, and wavenumberby the following equalitieshcE = hν= = hcνλwhere h is Planck’s constant, which has a value of 6.626 10 –34 J . s.Example 10.2What is the energy of a photon from the sodium D line at 589 nm?So l u t i o nThe photon’s energy is3 Ball, D. W. Spectroscopy 1994, 9(6) 20–21.

Chapter 10 Spectroscopic Methods547−hc ( 6.626× 10 J⋅ s)(3.00×10E = =−9λ589×10 m34 8m/s)= 337 . × 10−19JPractice Exercise 10.2What is the energy of a photon for the Balmer line at a wavelength of656.3 nm?Click here to review your answer to this exercise.Th e El e c t r o m a g n e t i c Sp e c t r u mThe frequency and wavelength of electromagnetic radiation vary over manyorders of magnitude. For convenience, we divide electromagnetic radiationinto different regions—the electromagnetic spectrum—based on thetype of atomic or molecular transition that gives rise to the absorption oremission of photons (Figure 10.3). The boundaries between the regions ofthe electromagnetic spectrum are not rigid, and overlap between spectralregions is possible.10A.2 Photons as a Signal SourceIn the previous section we defined several characteristic properties of electromagneticradiation, including its energy, velocity, amplitude, frequency,phase angle, polarization, and direction of propagation. A spectroscopicmeasurement is possible only if the photon’s interaction with the sampleleads to a change in one or more of these characteristic properties.Increasing Frequency (ν)10 24 10 22 10 20 10 18 10 16 10 14 10 12 10 10 10 8 10 6 10 4 10 2 10 0 ν (s -1 )10 -16 10 -14 10 -12 10 -10 10 -8 10 -6 10 -4 10 -2 10 0 10 2 10 4 10 6 10 8γ-rays X-rays UV IR Microwave FM AM Long radio wavesRadio wavesVisible Spectrum400 500600 700Increasing Wavelength (λ) in nmIncreasing Wavelength (λ)λ (nm)Types of Atomic & Molecular Transitionsγ-rays: nuclearX-rays: core-level electronsUltraviolet (UV): valence electronsVisible (Vis): valence electronsInfrared (IR): molecular vibrationsMicrowave: molecular roations; electron spinRadio waves: nuclear spinFigure 10.3 The electromagnetic spectrum showing the boundaries between different regions and the typeof atomic or molecular transition responsible for the change in energy. The colored inset shows the visiblespectrum. Source: modified from Zedh (www.commons.wikipedia.org).

548 Analytical Chemistry 2.0Table 10.1 Examples of Spectroscopic Techniques Involving an Exchange of EnergyBetween a Photon and the SampleRegion ofType of Energy Transfer Electromagnetic Spectrum Spectroscopic Technique aabsorption g-ray Mossbauer spectroscopyX-rayX-ray absorption spectroscopyUV/VisUV/Vis spectroscopyatomic absorption spectroscopyIRinfrared spectroscopyraman spectroscopyMicrowavemicrowave spectroscopyRadio waveelectron spin resonance spectroscopynuclear magnetic resonance spectroscopyemission (thermal excitation) UV/Visatomic emission spectroscopyphotoluminescence X-ray X-ray fluorescenceUV/Visfluorescence spectroscopyphosphorescence spectroscopyatomic fluorescence spectroscopychemiluminescence UV/Vis chemiluminescence spectroscopya Techniques discussed in this text are shown in italics.We can divide spectroscopy into two broad classes of techniques. In oneclass of techniques there is a transfer of energy between the photon and thesample. Table 10.1 provides a list of several representative examples.In absorption spectroscopy a photon is absorbed by an atom or molecule,which undergoes a transition from a lower-energy state to a higherenergy,or excited state (Figure 10.4). The type of transition depends on thephoton’s energy. The electromagnetic spectrum in Figure 10.3, for example,shows that absorbing a photon of visible light promotes one of the atom’sor molecule’s valence electrons to a higher-energy level. When an moleculeabsorbs infrared radiation, on the other hand, one of its chemical bondsexperiences a change in vibrational energy.Figure 10.4 Simplified energy diagram showing theabsorption and emission of a photon by an atom ora molecule. When a photon of energy hn strikes theatom or molecule, absorption may occur if the differencein energy, DE, between the ground state and theexcited state is equal to the photon’s energy. An atomor molecule in an excited state may emit a photonand return to the ground state. The photon’s energy,hn, equals the difference in energy, DE, between thetwo states.Energyabsorptionemissionexicted stateshν ΔE = hν ΔE = hνground statehν

Chapter 10 Spectroscopic Methods5490.80.6absorbance0.40.20.0400 450 500 550 600 650 700 750wavelength (nm)Figure 10.5 Visible absorbance spectrum for cranberry juice. Theanthocyanin dyes in cranberry juice absorb visible light with blue,green, and yellow wavelengths (see Figure 10.3). As a result, thejuice appears red.When it absorbs electromagnetic radiation the number of photons passingthrough a sample decreases. The measurement of this decrease in photons,which we call absorbance, is a useful analytical signal. Note that theeach of the energy levels in Figure 10.4 has a well-defined value becausethey are quantized. Absorption occurs only when the photon’s energy, hn,matches the difference in energy, DE, between two energy levels. A plot ofabsorbance as a function of the photon’s energy is called an absorbancespectrum. Figure 10.5, for example, shows the absorbance spectrum ofcranberry juice.When an atom or molecule in an excited state returns to a lower energystate, the excess energy often is released as a photon, a process we call emission(Figure 10.4). There are several ways in which an atom or moleculemay end up in an excited state, including thermal energy, absorption of aphoton, or by a chemical reaction. Emission following the absorption ofa photon is also called photoluminescence, and that following a chemicalreaction is called chemiluminescence. A typical emission spectrum isshown in Figure 10.6.Molecules also can release energy in theform of heat. We will return to this pointlater in the chapter.emission intensity (arbitrary units)400 450 500 550 600 650wavelength (nm)Figure 10.6 Photoluminescence spectrum of the dye coumarin343, which is incorporated in a reverse micelle suspended in cyclohexanol.The dye’s absorbance spectrum (not shown) has abroad peak around 400 nm. The sharp peak at 409 nm is fromthe laser source used to excite coumarin 343. The broad bandcentered at approximately 500 nm is the dye’s emission band.Because the dye absorbs blue light, a solution of coumarin 343appears yellow in the absence of photoluminescence. Its photoluminescentemission is blue-green. Source: data from BridgetGourley, Department of Chemistry & Biochemistry, DePauwUniversity).

550 Analytical Chemistry 2.0Table 10.2 Examples of Spectroscopic Techniques That Do Not Involve anExchange of Energy Between a Photon and the SampleRegion ofElectromagnetic Spectrum Type of Interaction Spectroscopic Technique aX-ray diffraction X-ray diffractionUV/Vis refraction refractometryscatteringnephelometryturbidimetrydispersionoptical rotary dispersiona Techniques discussed in this text are shown in italics.In the second broad class of spectroscopic techniques, the electromagneticradiation undergoes a change in amplitude, phase angle, polarization,or direction of propagation as a result of its refraction, reflection, scattering,diffraction, or dispersion by the sample. Several representative spectroscopictechniques are listed in Table 10.2.You will find a more detailed treatmentof these components in the additional resourcesfor this chapter.10A.3 Basic Components of Spectroscopic InstrumentsThe spectroscopic techniques in Table 10.1 and Table 10.2 use instrumentsthat share several common basic components, including a source of energy,a means for isolating a narrow range of wavelengths, a detector for measuringthe signal, and a signal processor that displays the signal in a form convenientfor the analyst. In this section we introduce these basic components.Specific instrument designs are considered in later sections.Emission Intensity (arbitrary units)500 550 600 650wavelength (nm)Figure 10.7 Spectrum showing the emissionfrom a green LED, which providescontinuous emission over a wavelengthrange of approximately 530–640 nm.So u r c e s o f En e r g yAll forms of spectroscopy require a source of energy. In absorption andscattering spectroscopy this energy is supplied by photons. Emission andphotoluminescence spectroscopy use thermal, radiant (photon), or chemicalenergy to promote the analyte to a suitable excited state.Sources of Electromagnetic Radiation. A source of electromagnetic radiationmust provide an output that is both intense and stable. Sources ofelectromagnetic radiation are classified as either continuum or line sources.A continuum source emits radiation over a broad range of wavelengths,with a relatively smooth variation in intensity (Figure 10.7). A line source,on the other hand, emits radiation at selected wavelengths (Figure 10.8).Table 10.3 provides a list of the most common sources of electromagneticradiation.Sources of Thermal Energy. The most common sources of thermal energyare flames and plasmas. Flames sources use the combustion of a fuel and anoxidant to achieve temperatures of 2000–3400 K. Plasmas, which are hot,ionized gases, provide temperatures of 6000–10 000 K.

Chapter 10 Spectroscopic Methods551Table 10.3 Common Sources of Electromagnetic RadiationSource Wavelength Region Useful for...H 2 and D 2 lamp continuum source from 160–380 nm molecular absorptiontungsten lamp continuum source from 320–2400 nm molecular absorptionXe arc lamp continuum source from 200–1000 nm molecular fluorescencenernst glower continuum source from 0.4–20 mm molecular absorptionglobar continuum source from 1–40 mm molecular absorptionnichrome wire continuum source from 0.75–20 mm molecular absorptionhollow cathode lamp line source in UV/Visible atomic absorptionHg vapor lamp line source in UV/Visible molecular fluorescencelaser line source in UV/Visible/IR atomic and molecular absorption,fluorescence, and scatteringChemical Sources of Energy Exothermic reactions also may serve as asource of energy. In chemiluminescence the analyte is raised to a higherenergystate by means of a chemical reaction, emitting characteristic radiationwhen it returns to a lower-energy state. When the chemical reactionresults from a biological or enzymatic reaction, the emission of radiationis called bioluminescence. Commercially available “light sticks” and theflash of light from a firefly are examples of chemiluminescence and bioluminescence.Wave l e n g t h Se l e c t i o nIn Nessler’s original colorimetric method for ammonia, described at thebeginning of the chapter, the sample and several standard solutions of ammoniaare placed in separate tall, flat-bottomed tubes. As shown in Figure10.9, after adding the reagents and allowing the color to develop, the analystevaluates the color by passing natural, ambient light through the bottomof the tubes and looking down through the solutions. By matching thesample’s color to that of a standard, the analyst is able to determine theconcentration of ammonia in the sample.In Figure 10.9 every wavelength of light from the source passes throughthe sample. If there is only one absorbing species, this is not a problem.If two components in the sample absorbs different wavelengths of light,then a quantitative analysis using Nessler’s original method becomes impossible.Ideally we want to select a wavelength that only the analyte absorbs.Unfortunately, we can not isolate a single wavelength of radiation from acontinuum source. As shown in Figure 10.10, a wavelength selector passesa narrow band of radiation characterized by a nominal wavelength, aneffective bandwidth and a maximum throughput of radiation. The effectivebandwidth is defined as the width of the radiation at half of itsmaximum throughput.Emission Intensity (arbitrary units)200 250 300 350 400wavelength (nm)Figure 10.8 Emission spectrum from a Cuhollow cathode lamp. This spectrum consistsof seven distinct emission lines (thefirst two differ by only 0.4 nm and are notresolved in this spectrum). Each emissionline has a width of approximately 0.01 nmat ½ of its maximum intensity.

552 Analytical Chemistry 2.0radiant powermaximumthroughputFigure 10.9 Nessler’s original method for comparing the color of twosolutions. Natural light passes upwards through the samples and standardsand the analyst views the solutions by looking down toward thelight source. The top view, shown on the right, is what the analyst sees.To determine the analyte’s concentration, the analyst exchanges standardsuntil the two colors match.nominal wavelengthwavelengtheffectivebandwidthFigure 10.10 Radiation exiting a wavelengthselector showing the band’s nominalwavelength and its effective bandwidth.The ideal wavelength selector has a high throughput of radiation and anarrow effective bandwidth. A high throughput is desirable because morephotons pass through the wavelength selector, giving a stronger signal withless background noise. A narrow effective bandwidth provides a higher resolution,with spectral features separated by more than twice the effectivebandwidth being resolved. As shown in Figure 10.11, these two featuresof a wavelength selector generally are in opposition. Conditions favoring ahigher throughput of radiation usually provide less resolution. Decreasingthe effective bandwidth improves resolution, but at the cost of a noisiersignal. 4 For a qualitative analysis, resolution is usually more important thannoise, and a smaller effective bandwidth is desirable. In a quantitative analysisless noise is usually desirable.Wavelength Selection Using Filters. The simplest method for isolating anarrow band of radiation is to use an absorption or interference filter.Absorption filters work by selectively absorbing radiation from a narrowregion of the electromagnetic spectrum. Interference filters use constructiveand destructive interference to isolate a narrow range of wavelengths. Asimple example of an absorption filter is a piece of colored glass. A purplefilter, for example, removes the complementary color green from 500–560nm. Commercially available absorption filters provide effective bandwidthsof 30–250 nm, although the throughput may be only 10% of the source’semission intensity at the low end of this range. Interference filters are moreexpensive than absorption filters, but have narrower effective bandwidths,typically 10–20 nm, with maximum throughputs of at least 40%.Wavelength Selection Using Monochromators. A filter has one significantlimitation—because a filter has a fixed nominal wavelength, if you need tomake measurements at two different wavelengths, then you need to use two4 Jiang, S.; Parker, G. A. Am. Lab. 1981, October, 38–43.side viewtop view

Chapter 10 Spectroscopic Methods553larger effective bandwidthsmaller effective bandwidthsignalsignalwavelengthwavelengthFigure 10.11 Example showing the effect of the wavelength selector’s effective bandwidth on resolutionand noise. The spectrum with the smaller effective bandwidth has a better resolution, allowing us to seethe presence of three peaks, but at the expense of a noisier signal. The spectrum with the larger effectivebandwidth has less noise, but at the expense of less resolution between the three peaks.different filters. A monochromator is an alternative method for selectinga narrow band of radiation that also allows us to continuously adjust theband’s nominal wavelength.The construction of a typical monochromator is shown in Figure 10.12.Radiation from the source enters the monochromator through an entranceslit. The radiation is collected by a collimating mirror, which reflects a parallelbeam of radiation to a diffraction grating. The diffraction grating isan optically reflecting surface with a large number of parallel grooves (seeinsert to Figure 10.12). The diffraction grating disperses the radiation anda second mirror focuses the radiation onto a planar surface containing anexit slit. In some monochromators a prism is used in place of the diffractiongrating.Radiation exits the monochromator and passes to the detector. As shownin Figure 10.12, a monochromator converts a polychromatic source ofcollimating mirrorfocusing mirrorPolychromatic means many colored. Polychromaticradiation contains many differentwavelengths of light.diffraction gratingλ 1 λ2λ 3entrance slitlight sourceλ 1 λ 3λ exit slit2detectorFigure 10.12 Schematic diagram of amonochromator that uses a diffractiongrating to disperse the radiation.

554 Analytical Chemistry 2.0Monochromatic means one color, or onewavelength. Although the light exiting amonochromator is not strictly of a singlewavelength, its narrow effective bandwidthallows us to think of it as monochromatic.radiation at the entrance slit to a monochromatic source of finite effectivebandwidth at the exit slit. The choice of which wavelength exits the monochromatoris determined by rotating the diffraction grating. A narrower exitslit provides a smaller effective bandwidth and better resolution, but allowsa smaller throughput of radiation.Monochromators are classified as either fixed-wavelength or scanning.In a fixed-wavelength monochromator we select the wavelength by manuallyrotating the grating. Normally a fixed-wavelength monochromator isused for a quantitative analysis where measurements are made at one or twowavelengths. A scanning monochromator includes a drive mechanism thatcontinuously rotates the grating, allowing successive wavelengths to exitfrom the monochromator. Scanning monochromators are used to acquirespectra, and, when operated in a fixed-wavelength mode, for a quantitativeanalysis.Interferometers. An interferometer provides an alternative approachfor wavelength selection. Instead of filtering or dispersing the electromagneticradiation, an interferometer allows source radiation of all wavelengthsto reach the detector simultaneously (Figure 10.13). Radiation from thesource is focused on a beam splitter that reflects half of the radiation to afixed mirror and transmits the other half to a movable mirror. The radiationrecombines at the beam splitter, where constructive and destructive interferencedetermines, for each wavelength, the intensity of light reaching thedetector. As the moving mirror changes position, the wavelengths of lightexperiencing maximum constructive interference and maximum destructiveinterference also changes. The signal at the detector shows intensity asa function of the moving mirror’s position, expressed in units of distance ortime. The result is called an interferogram, or a time domain spectrum.fixed mirrorcollimating mirrorlight sourcebeam splittermoving mirrordetectorFigure 10.13 Schematic diagram of an interferometers.focusing mirror

Chapter 10 Spectroscopic Methods555The time domain spectrum is converted mathematically, by a process calleda Fourier transform, to a spectrum (also called a frequency domain spectrum)showing intensity as a function of the radiation’s energy.In comparison to a monochromator, an interferometer has two significantadvantages. The first advantage, which is termed Jacquinot’s advantage,is the higher throughput of source radiation. Because an interferometerdoes not use slits and has fewer optical components from whichradiation can be scattered and lost, the throughput of radiation reachingthe detector is 80–200 greater than that for a monochromator. The resultis less noise. The second advantage, which is called Fellgett’s advantage,is a savings in the time needed to obtain a spectrum. Because the detectormonitors all frequencies simultaneously, an entire spectrum takes approximatelyone second to record, as compared to 10–15 minutes with a scanningmonochromator.The mathematical details of the Fouriertransform are beyond the level of thistextbook. You can consult the chapter’sadditional resources for additional information.De t e c t o r sIn Nessler’s original method for determining ammonia (Figure 10.9) theanalyst’s eye serves as the detector, matching the sample’s color to that of astandard. The human eye, of course, has a poor range—responding only tovisible light—nor is it particularly sensitive or accurate. Modern detectorsuse a sensitive transducer to convert a signal consisting of photons intoan easily measured electrical signal. Ideally the detector’s signal, S, is a linearfunction of the electromagnetic radiation’s power, P,S = kP+DTransducer is a general term that refersto any device that converts a chemical orphysical property into an easily measuredelectrical signal. The retina in your eye,for example, is a transducer that convertsphotons into an electrical nerve impulse.where k is the detector’s sensitivity, and D is the detector’s dark current,or the background current when we prevent the source’s radiation fromreaching the detector.There are two broad classes of spectroscopic transducers: thermal transducersand photon transducers. Table 10.4 provides several representativeexamples of each class of transducers.Table 10.4 Examples of Transducers for SpectroscopyTransducer Class Wavelength Range Output Signalphototube photon 200–1000 nm currentphotomultiplier photon 110–1000 nm currentSi photodiode photon 250–1100 nm currentphotoconductor photon 750–6000 nm change in resistancephotovoltaic cell photon 400–5000 nm current or voltagethermocouple thermal 0.8–40 mm voltagethermistor thermal 0.8–40 mm change in resistancepneumatic thermal 0.8–1000 mm membrane displacementpyroelectric thermal 0.3–1000 mm current

556 Analytical Chemistry 2.0dynodehνelectronsopticalwindowanodephotoemissive cathodeFigure 10.14 Schematic of a photomultiplier.A photon strikes the photoemissivecathode producing electrons, which acceleratetoward a positively charged dynode.Collision of these electrons with thedynode generates additional electrons,which accelerate toward the next dynode.A total of 10 6 –10 7 electrons per photoneventually reach the anode, generating anelectrical current.If the retina in your eye is a transducer,then your brain is a signal processor.Photon Transducers. Phototubes and photomultipliers contain a photosensitivesurface that absorbs radiation in the ultraviolet, visible, or nearIR, producing an electrical current proportional to the number of photonsreaching the transducer (Figure 10.14). Other photon detectors use a semiconductoras the photosensitive surface. When the semiconductor absorbsphotons, valence electrons move to the semiconductor’s conduction band,producing a measurable current. One advantage of the Si photodiode isthat it is easy to miniaturize. Groups of photodiodes may be gathered togetherin a linear array containing from 64–4096 individual photodiodes.With a width of 25 mm per diode, for example, a linear array of 2048 photodiodesrequires only 51.2 mm of linear space. By placing a photodiodearray along the monochromator’s focal plane, it is possible to monitorsimultaneously an entire range of wavelengths.Thermal Transducers. Infrared photons do not have enough energy to producea measurable current with a photon transducer. A thermal transducer,therefore, is used for infrared spectroscopy. The absorption of infrared photonsby a thermal transducer increases its temperature, changing one ormore of its characteristic properties. A pneumatic transducer, for example,is a small tube of xenon gas with an IR transparent window at one endand a flexible membrane at the other end. Photons enter the tube and areabsorbed by a blackened surface, increasing the temperature of the gas. Asthe temperature inside the tube fluctuates, the gas expands and contractsand the flexible membrane moves in and out. Monitoring the membrane’sdisplacement produces an electrical signal.Signal ProcessorsA transducer’s electrical signal is sent to a signal processor where it isdisplayed in a form that is more convenient for the analyst. Examples ofsignal processors include analog or digital meters, recorders, and computersequipped with digital acquisition boards. A signal processor also is usedto calibrate the detector’s response, to amplify the transducer’s signal, toremove noise by filtering, or to mathematically transform the signal.10BSpectroscopy Based on AbsorptionIn absorption spectroscopy a beam of electromagnetic radiation passesthrough a sample. Much of the radiation passes through the sample withouta loss in intensity. At selected wavelengths, however, the radiation’s intensityis attenuated. This process of attenuation is called absorption.10B.1 Absorbance SpectraThere are two general requirements for an analyte’s absorption of electromagneticradiation. First, there must be a mechanism by which the radiation’selectric field or magnetic field interacts with the analyte. For ultra-

Chapter 10 Spectroscopic Methods557violet and visible radiation, absorption of a photon changes the energy ofthe analyte’s valence electrons. A bond’s vibrational energy is altered by theabsorption of infrared radiation.The second requirement is that the photon’s energy, hn, must exactlyequal the difference in energy, DE, between two of the analyte’s quantizedenergy states. Figure 10.4 shows a simplified view of a photon’s absorption,which is useful because it emphasizes that the photon’s energy must matchthe difference in energy between a lower-energy state and a higher-energystate. What is missing, however, is information about what types of energystates are involved, which transitions between energy states are likely tooccur, and the appearance of the resulting spectrum.We can use the energy level diagram in Figure 10.15 to explain an absorbancespectrum. The lines labeled E 0 and E 1 represent the analyte’s ground(lowest) electronic state and its first electronic excited state. Superimposedon each electronic energy level is a series of lines representing vibrationalenergy levels.Figure 10.3 provides a list of the types ofatomic and molecular transitions associatedwith different types of electromagneticradiation.In f r a r e d Sp e c t r a f o r Mo l e c u l e s a n d Po l y a t o m i c Io n sThe energy of infrared radiation produces a change in a molecule’s or apolyatomic ion’s vibrational energy, but is not sufficient to effect a changein its electronic energy. As shown in Figure 10.15, vibrational energy levelsare quantized; that is, a molecule may have only certain, discrete vibrationalenergies. The energy for an allowed vibrational mode, E n , is⎛E = 1⎞v+hνν⎝⎜2⎠⎟0ν 3ννν 421UV/Vis IRν 0E 1energyν 4ν 3ν 2ν 1ν 0E 0Figure 10.15 Diagram showing two electronic energy levels(E 0 and E 1 ), each with five vibrational energy levels (n 0 –n 4 ).Absorption of ultraviolet and visible radiation leads to a changein the analyte’s electronic energy levels and, possibly, a changein vibrational energy as well. A change in vibrational energywithout a change in electronic energy levels occurs with theabsorption of infrared radiation.

558 Analytical Chemistry 2.0Why does a non-linear molecule have3N – 6 vibrational modes? Consider amolecule of methane, CH 4 . Each of thefive atoms in methane can move in oneof three directions (x, y, and z) for a totalof 5 3 = 15 different ways in which themolecule can move. A molecule can movein three ways: it can move from one placeto another, what we call translational motion;it can rotate around an axis; and itsbonds can stretch and bend, what we callvibrational motion.Because the entire molecule can move inthe x, y, and z directions, three of methane’s15 different motions are translational. Inaddition, the molecule can rotate aboutits x, y, and z axes, accounting for threeadditional forms of motion. This leaves15 – 3 – 3 = 9 vibrational modes.A linear molecule, such as CO 2 , has3N – 5 vibrational modes because it canrotate around only two axes.where n is the vibrational quantum number, which has values of 0, 1, 2,…, and n 0 is the bond’s fundamental vibrational frequency. The value of n 0 ,which is determined by the bond’s strength and by the mass at each endof the bond, is a characteristic property of a bond. For example, a carboncarbonsingle bond (C–C) absorbs infrared radiation at a lower energy thana carbon-carbon double bond (C=C) because a single bond is weaker thana double bond.At room temperature most molecules are in their ground vibrationalstate (n = 0). A transition from the ground vibrational state to the firstvibrational excited state (n = 1) requires absorption of a photon with anenergy of hn 0 . Transitions in which Dn is ±1 give rise to the fundamentalabsorption lines. Weaker absorption lines, called overtones, result fromtransitions in which Dn is ±2 or ±3. The number of possible normal vibrationalmodes for a linear molecule is 3N – 5, and for a non-linear moleculeis 3N – 6, where N is the number of atoms in the molecule. Not surprisingly,infrared spectra often show a considerable number of absorption bands.Even a relatively simple molecule, such as ethanol (C 2 H 6 O), for example,has 3 9 – 6, or 21 possible normal modes of vibration, although not allof these vibrational modes give rise to an absorption. The IR spectrum forethanol is shown in Figure 10.16.UV/Vis Sp e c t r a f o r Mo l e c u l e s a n d Io n sThe valence electrons in organic molecules and polyatomic ions, such asCO 3 2– , occupy quantized sigma bonding, s, pi bonding, p, and non-bonding,n, molecular orbitals (MOs). Unoccupied sigma antibonding, s*, andpi antibonding, p*, molecular orbitals are slightly higher in energy. Becausethe difference in energy between the highest-energy occupied MOs and100percent transmittance80604020Figure 10.16 Infrared spectrum of ethanol.04000 3000 2000 1000wavenumber (cm-1)

Chapter 10 Spectroscopic Methods559Table 10.5 Electronic Transitions Involving n, s, and pMolecular OrbitalsTransition Wavelength Range Examplesσ→ σ

560 Analytical Chemistry 2.01.21.0absorbance0.80.60.4Figure 10.18 UV/Vis spectrum for the metal–ligandcomplex Fe(phen) 32+ , where phen is the ligand o-phenanthroline.why this is true. When a species absorbs UV/Vis radiation, the transitionbetween electronic energy levels may also include a transition between vibrationalenergy levels. The result is a number of closely spaced absorptionbands that merge together to form a single broad absorption band.UV/Vis Sp e c t r a f o r At o m s0.20.0400 450 500 550 600 650 700wavelength (nm)The valence shell energy level diagramin Figure 10.19 might strike you as oddbecause it shows that the 3p orbitals aresplit into two groups of slightly differentenergy. The reasons for this splitting areunimportant in the context of our treatmentof atomic absorption. For furtherinformation about the reasons for thissplitting, consult the chapter’s additionalresources.The energy of ultraviolet and visible electromagnetic radiation is sufficientto cause a change in an atom’s valence electron configuration. Sodium, forexample, has a single valence electron in its 3s atomic orbital. As shown inFigure 10.19, unoccupied, higher energy atomic orbitals also exist.Absorption of a photon is accompanied by the excitation of an electronfrom a lower-energy atomic orbital to an orbital of higher energy. Not allpossible transitions between atomic orbitals are allowed. For sodium theonly allowed transitions are those in which there is a change of ±1 in theorbital quantum number (l); thus transitions from s p orbitals are allowed,and transitions from s d orbitals are forbidden.5p4d5sFigure 10.19 Valence shell energy level diagram for sodium.The wavelengths (in wavenumbers) correspondingto several transitions are shown.Energy3s4s1138.34p330.3330.21140.4818.3819.5589.0 3p3d

Chapter 10 Spectroscopic Methods5611.0absorbance0.80.60.43s3p0.23s4p0330.0 330.2 330.4588.5 589.0 589.5wavelength (nm)The atomic absorption spectrum for Na is shown in Figure 10.20, andis typical of that found for most atoms. The most obvious feature of thisspectrum is that it consists of a small number of discrete absorption linescorresponding to transitions between the ground state (the 3s atomic orbital)and the 3p and 4p atomic orbitals. Absorption from excited states,such as the 3p 4s and the 3p 3d transitions included in Figure 10.19,are too weak to detect. Because an excited state’s lifetime is short—typicallyan excited state atom takes 10 –7 to 10 –8 s to return to a lower energy state—an atom in the exited state is likely to return to the ground state before ithas an opportunity to absorb a photon.Another feature of the atomic absorption spectrum in Figure 10.20 isthe narrow width of the absorption lines, which is a consequence of thefixed difference in energy between the ground and excited states. Naturalline widths for atomic absorption, which are governed by the uncertaintyprinciple, are approximately 10 –5 nm. Other contributions to broadeningincrease this line width to approximately 10 –3 nm.10B.2 Transmittance and AbsorbanceAs light passes through a sample, its power decreases as some of it is absorbed.This attenuation of radiation is described quantitatively by twoseparate, but related terms: transmittance and absorbance. As shown inFigure 10.21a, transmittance is the ratio of the source radiation’s powerexiting the sample, P T , to that incident on the sample, P 0 .TP= T P0590.010.1Multiplying the transmittance by 100 gives the percent transmittance, %T,which varies between 100% (no absorption) and 0% (complete absorption).All methods of detecting photons—including the human eye andmodern photoelectric transducers—measure the transmittance of electromagneticradiation.Figure 10.20 Atomic absorption spectrum for sodium.Note that the scale on the x-axis includes a break.

562 Analytical Chemistry 2.0(a)P 0sampleP T(b)blankP 0We will show that this is true in Section10B.3.Figure 10.21 (a) Schematic diagram showing the attenuation of radiation passingthrough a sample; P 0 is the radiant power from the source and P T is the radiantpower transmitted by the sample. (b) Schematic diagram showing how we redefineP 0 as the radiant power transmitted by the blank. Redefining P 0 in this way correctsthe transmittance in (a) for the loss of radiation due to scattering, reflection,or absorption by the sample’s container and absorption by the sample’s matrix.Equation 10.1 does not distinguish between different mechanisms thatprevent a photon emitted by the source from reaching the detector. In additionto absorption by the analyte, several additional phenomena contributeto the attenuation of radiation, including reflection and absorption by thesample’s container, absorption by other components in the sample’s matrix,and the scattering of radiation. To compensate for this loss of the radiation’spower, we use a method blank. As shown in Figure 10.21b, we redefine P 0as the power exiting the method blank.An alternative method for expressing the attenuation of electromagneticradiation is absorbance, A, which we define asPA=− log T =− log T P010.2Absorbance is the more common unit for expressing the attenuation ofradiation because it is a linear function of the analyte’s concentration.Example 10.3A sample has a percent transmittance of 50%. What is its absorbance?So l u t i o nA percent transmittance of 50.0% is the same as a transmittance of 0.500Substituting into equation 10.2 givesA=− logT=− log( 0. 500) = 0.301Practice Exercise 10.3What is the %T for a sample if its absorbance is 1.27?Click here to review your answer to this exercise.

Chapter 10 Spectroscopic Methods563Equation 10.1 has an important consequence for atomic absorption. Aswe saw in Figure 10.20, atomic absorption lines are very narrow. Even witha high quality monochromator, the effective bandwidth for a continuumsource is 100–1000 greater than the width of an atomic absorption line.As a result, little of the radiation from a continuum source is absorbed (P 0 ≈P T ), and the measured absorbance is effectively zero. For this reason, atomicabsorption requires a line source instead of a continuum source.10B.3 Absorbance and Concentration: Beer’s LawWhen monochromatic electromagnetic radiation passes through an infinitesimallythin layer of sample of thickness dx, it experiences a decrease inits power of dP (Figure 10.22). The fractional decrease in power is proportionalto the sample’s thickness and the analyte’s concentration, C; thusdP− = αCdx10.3Pwhere P is the power incident on the thin layer of sample, and a is a proportionalityconstant. Integrating the left side of equation 10.3 over theentire samplePTdP− ∫ = αCP0P⎛ln P ⎝⎜P0T∫o⎞bC⎠⎟ =αconverting from ln to log, and substituting in equation 10.2, givesA= abC10.4where a is the analyte’s absorptivity with units of cm –1 conc –1 . If we expressthe concentration using molarity, then we replace a with the molarabsorptivity, e, which has units of cm –1 M –1 .A=εbC10.5The absorptivity and molar absorptivity are proportional to the probabilitythat the analyte absorbs a photon of a given energy. As a result, values forboth a and e depend on the wavelength of the absorbed photon.bdxP 0 P P – dP P Tx = 0dxx = bFigure 10.22 Factors used in derivingthe Beer-Lambert law.

564 Analytical Chemistry 2.0Example 10.4A 5.00 10 –4 M solution of an analyte is placed in a sample cell with apathlength of 1.00 cm. When measured at a wavelength of 490 nm, thesolution’s absorbance is 0.338. What is the analyte’s molar absorptivity atthis wavelength?So l u t i o nSolving equation 10.5 for e and making appropriate substitutions givesA0.338ε= == 676 cm−4bC (. 100 cm)(5.00×10 M)−1M −1Practice Exercise 10.4A solution of the analyte from Example 10.4 has an absorbance of 0.228in a 1.00-cm sample cell. What is the analyte’s concentration?Click here to review your answer to this exercise.Equation 10.4 and equation 10.5, which establish the linear relationshipbetween absorbance and concentration, are known as the Beer-Lambertlaw, or more commonly, as Beer’s law. Calibration curves based onBeer’s law are common in quantitative analyses.10B.4 Beer’s Law and Multicomponent SamplesWe can extend Beer’s law to a sample containing several absorbing components.If there are no interactions between the components, the individualabsorbances, A i , are additive. For a two-component mixture of analyte’s Xand Y, the total absorbance, A tot , isA = A + A = ε bC + ε bCtotX Y X X Y YabsorbancepositivedeviationconcentrationidealnegativedeviationGeneralizing, the absorbance for a mixture of n components, A mix , isn∑10B.5 Limitations to Beer’s Lawn∑A = A = ε bCmix ii i10.6i= 1 i=1Beer’s law suggests that a calibration curve is a straight line with a y-interceptof zero and a slope of ab or eb. In many cases a calibration curve deviatesfrom this ideal behavior (Figure 10.23). Deviations from linearity aredivided into three categories: fundamental, chemical, and instrumental.Figure 10.23 Calibration curves showingpositive and negative deviationsfrom the ideal Beer’s law calibrationcurve, which is a straight line.Fu n d a m e n t a l Limitations t o Be e r’s La wBeer’s law is a limiting law that is valid only for low concentrations of analyte.There are two contributions to this fundamental limitation to Beer’s law. At

Chapter 10 Spectroscopic Methods565higher concentrations the individual particles of analyte no longer behaveindependently of each other. The resulting interaction between particles ofanalyte may change the analyte’s absorptivity. A second contribution is thatthe analyte’s absorptivity depends on the sample’s refractive index. Becausethe refractive index varies with the analyte’s concentration, the values of aand e may change. For sufficiently low concentrations of analyte, the refractiveindex is essentially constant and the calibration curve is linear.Ch e m ic a l Limitations t o Be e r’s La wA chemical deviation from Beer’s law may occur if the analyte is involved inan equilibrium reaction. Consider, as an example, an analysis for the weakacid, HA. To construct a Beer’s law calibration curve we prepare a seriesof standards—each containing a known total concentration of HA—andmeasure each standard’s absorbance at the same wavelength. Because HA isa weak acid, it is in equilibrium with its conjugate weak base, A – .+ −HA( aq) + HO() l HO ( aq) + A ( aq )2 3If both HA and A – absorb at the chosen wavelength, then Beer’s law isA= ε bC + ε bCHA HA A A10.7where C HA and C A are the equilibrium concentrations of HA and A – . Becausethe weak acid’s total concentration, C total , isC = C + Ctotal HA Athe concentrations of HA and A – can be written asCC=α C10.8HA HA total= ( 1−α ) C10.9A HA totalwhere a HA is the fraction of weak acid present as HA. Substituting equation10.8 and equation 10.9 into equation 10.7 and rearranging, givesA= ( ε α + ε − ε α ) bCHA HA A A HA total10.10To obtain a linear Beer’s law calibration curve, one of two conditions mustbe met. If a HA and a A have the same value at the chosen wavelength, thenequation 10.10 simplifies toA=ε AbC totalAlternatively, if a HA has the same value for all standards, then each termwithin the parentheses of equation 10.10 is constant—which we replacewith k—and a linear calibration curve is obtained at any wavelength.A= kbCtotal

566 Analytical Chemistry 2.0For a monoprotic weak acid, the equationfor a HA is+[ HO ]3α HA=+[ HO ] + K3 aProblem 10.6 in the end of chapter problemsasks you to explore this chemicallimitation to Beer’s law.Problem 10.7 in the end of chapter problemsask you to explore the effect of polychromaticradiation on the linearity ofBeer’s law.Another reason for measuring absorbanceat the top of an absorbance peak is thatit provides for a more sensitive analysis.Note that the green calibration curvein Figure 10.24 has a steeper slope—agreater sensitivity—than the red calibrationcurve.Because HA is a weak acid, the value of a HA varies with pH. To holda HA constant we buffer each standard to the same pH. Depending on therelative values of a HA and a A , the calibration curve has a positive or anegative deviation from Beer’s law if we do not buffer the standards to thesame pH.In s t r u m e n t a l Limitations t o Be e r’s La wThere are two principal instrumental limitations to Beer’s law. The first limitationis that Beer’s law assumes that the radiation reaching the sample is ofa single wavelength—that is, that the radiation is purely monochromatic.As shown in Figure 10.10, however, even the best wavelength selector passesradiation with a small, but finite effective bandwidth. Polychromatic radiationalways gives a negative deviation from Beer’s law, but the effect issmaller if the value of e is essentially constant over the wavelength rangepassed by the wavelength selector. For this reason, as shown in Figure 10.24,it is better to make absorbance measurements at the top of a broad absorptionpeak. In addition, the deviation from Beer’s law is less serious if thesource’s effective bandwidth is less than one-tenth of the natural bandwidthof the absorbing species. 5 When measurements must be made on a slope,linearity is improved by using a narrower effective bandwidth.Stray radiation is the second contribution to instrumental deviationsfrom Beer’s law. Stray radiation arises from imperfections in the wavelengthselector that allow light to enter the instrument and reach the detectorwithout passing through the sample. Stray radiation adds an additionalcontribution, P stray , to the radiant power reaching the detector; thusP + PA =−log TP + Pstray0 strayFor a small concentration of analyte, P stray is significantly smaller than P 0and P T , and the absorbance is unaffected by the stray radiation. At a higherconcentration of analyte, however, less light passes through the sample and5 (a) Strong, F. C., III Anal. Chem. 1984, 56, 16A–34A; Gilbert, D. D. J. Chem. Educ. 1991, 68,A278–A281.1.21.0absorbance0.80.60.4absorbanceFigure 10.24 Effect of the choice ofwavelength on the linearity of a Beer’slaw calibration curve.0.20.0400 450 500 550 600 650 700wavelength (nm)concentration

Chapter 10 Spectroscopic Methods567P T and P stray may be similar in magnitude. This results is an absorbance thatis smaller than expected, and a negative deviation from Beer’s law.Problem 10.8 in the end of chapter problemsask you to explore the effect of strayradiation on the linearity of Beer’s law.10CUV/Vis and IR SpectroscopyIn Figure 10.9 we examined Nessler’s original method for matching thecolor of a sample to the color of a standard. Matching the colors was a laborintensive process for the analyst. Not surprisingly, spectroscopic methods ofanalysis were slow to develop. The 1930s and 1940s saw the introductionof photoelectric transducers for ultraviolet and visible radiation, and thermocouplesfor infrared radiation. As a result, modern instrumentation forabsorption spectroscopy became routinely available in the 1940s—progresshas been rapid ever since.10C.1 InstrumentationFrequently an analyst must select—from among several instruments of differentdesign—the one instrument best suited for a particular analysis. Inthis section we examine several different instruments for molecular absorptionspectroscopy, emphasizing their advantages and limitations. Methodsof sample introduction are also covered in this section.In s t r u m e n t De s i g n s f o r Mo l e c u l a r UV/Vis Ab s o r p t i o nFilter Photometer. The simplest instrument for molecular UV/Vis absorptionis a filter photometer (Figure 10.25), which uses an absorption orinterference filter to isolate a band of radiation. The filter is placed betweenthe source and the sample to prevent the sample from decomposing whenexposed to higher energy radiation. A filter photometer has a single opticalpath between the source and detector, and is called a single-beam instrument.The instrument is calibrated to 0% T while using a shutter to blockthe source radiation from the detector. After opening the shutter, the in-sourcefiltershutteropenclosedsampledetectorThe percent transmittance varies between0% and 100%. As we learned in Figure10.21, we use a blank to determine P 0 ,which corresponds to 100% T. Even in theabsence of light the detector records a signal.Closing the shutter allows us to assign0% T to this signal. Together, setting 0%T and 100% T calibrates the instrument.The amount of light passing through asample produces a signal that is greaterthan or equal to that for 0% T and smallerthan or equal to that for 100%T.blanksignalprocessorFigure 10.25 Schematic diagram of a filter photometer.The analyst either inserts a removable filter or thefilters are placed in a carousel, an example of which isshown in the photographic inset. The analyst selects afilter by rotating it into place.

568 Analytical Chemistry 2.0sourceshutteropenP Tdetectorλ 1 λ2λ 3sampleblankmonochromatorclosedsamplecompartmentP 0Figure 10.26 Schematic diagram of a fixed-wavelengthsingle-beam spectrophotometer. The photographic insetshows a typical instrument. The shutter remainsclosed until the sample or blank is placed in the samplecompartment. The analyst manually selects the wavelengthby adjusting the wavelength dial. Photo modifiedfrom: Adi (www.commons.wikipedia.org).0% T and 100% Tadjustmentwavelengthdialsignalprocessorstrument is calibrated to 100% T using an appropriate blank. The blank isthen replaced with the sample and its transmittance measured. Because thesource’s incident power and the sensitivity of the detector vary with wavelength,the photometer must be recalibrated whenever the filter is changed.Photometers have the advantage of being relatively inexpensive, rugged,and easy to maintain. Another advantage of a photometer is its portability,making it easy to take into the field. Disadvantages of a photometer includethe inability to record an absorption spectrum and the source’s relativelylarge effective bandwidth, which limits the calibration curve’s linearity.Single-Beam Spectrophotometer. An instrument that uses a monochromatorfor wavelength selection is called a spectrophotometer. Thesimplest spectrophotometer is a single-beam instrument equipped witha fixed-wavelength monochromator (Figure 10.26). Single-beam spectrophotometersare calibrated and used in the same manner as a photometer.One example of a single-beam spectrophotometer is Thermo Scientific’sSpectronic 20D+, which is shown in the photographic insert to Figure10.26. The Spectronic 20D+ has a range of 340–625 nm (950 nm whenusing a red-sensitive detector), and a fixed effective bandwidth of 20 nm.Battery-operated, hand-held single-beam spectrophotometers are available,which are easy to transport into the field. Other single-beam spectrophotometersalso are available with effective bandwidths of 2–8 nm. Fixedwavelength single-beam spectrophotometers are not practical for recordingspectra because manually adjusting the wavelength and recalibrating thespectrophotometer is awkward and time-consuming. The accuracy of asingle-beam spectrophotometer is limited by the stability of its source anddetector over time.Double-Beam Spectrophotometer. The limitations of fixed-wavelength,single-beam spectrophotometers are minimized by using a double-beamspectrophotometer (Figure 10.27). A chopper controls the radiation’s path,

Chapter 10 Spectroscopic Methods569blanksampleshuttersourceλ 1 λ2λ 3choppersampleP TmirrorsignalprocessormonochromatortemperaturemirrorsamplecontrolcompartmentFigure 10.27 Schematic diagram of a scanning, double-beam spectrophotometer. A chopper directs the source’sradiation, using a transparent window to pass radiation to the sample and a mirror to reflect radiation to theblank. The chopper’s opaque surface serves as a shutter, which allows for a constant adjustment of the spectrophotometer’s0% T. The photographic insert shows a typical instrument. The unit in the middle of the photois a temperature control unit that allows the sample to be heated or cooled.alternating it between the sample, the blank, and a shutter. The signalprocessor uses the chopper’s known speed of rotation to resolve the signalreaching the detector into the transmission of the blank, P 0 , and the sample,P T . By including an opaque surface as a shutter, it is possible to continuouslyadjust 0% T. The effective bandwidth of a double-beam spectrophotometeris controlled by adjusting the monochromator’s entrance and exitslits. Effective bandwidths of 0.2–3.0 nm are common. A scanning monochromatorallows for the automated recording of spectra. Double-beaminstruments are more versatile than single-beam instruments, being usefulfor both quantitative and qualitative analyses, but also are more expensive.Diode Array Spectrometer. An instrument with a single detector canmonitor only one wavelength at a time. If we replace a single photomultiplierwith many photodiodes, we can use the resulting array of detectorsto record an entire spectrum simultaneously in as little as 0.1 s. In a diodearray spectrometer the source radiation passes through the sample and isdispersed by a grating (Figure 10.28). The photodiode array is situated atthe grating’s focal plane, with each diode recording the radiant power overa narrow range of wavelengths. Because we replace a full monochromatorwith just a grating, a diode array spectrometer is small and compact.blankP 0semitransparentmirrordetector

570 Analytical Chemistry 2.0Figure 10.28 Schematic diagram of a diodearray spectrophotometer. The photographic insertshows a typical instrument. Note that the50-mL beaker provides a sense of scale.sourceshutteropenclosedsampleP Tgratingλ 1 λ2λ 3detectorsabsorbanceabsorbanceabsorbance1.00.80.60.40.201.00.80.60.40.201.00.80.60.40.20500 550 600 650 700wavelength1 scan500 550 600 650 700wavelength4 scans500 550 600 650 700wavelength16 scansFigure 10.29 Effect of signal averagingon a spectrum’s signal-to-noise ratio.From top to bottom: spectrum for asingle scan; average spectrum after fourscans; and average spectrum after adding16 scans.samplecompartmentOne advantage of a diode array spectrometer is the speed of data acquisition,which allows to collect several spectra for a single sample. Individualspectra are added and averaged to obtain the final spectrum. This signalaveraging improves a spectrum’s signal-to-noise ratio. If we add togethern spectra, the sum of the signal at any point, x, increases as nS x , where S x isthe signal. The noise at any point, N x , is a random event, which increases asnN xwhen we add together n spectra. The signal-to-noise ratio (S/N)after n scans isSNblanknSx= =nNxn S xNwhere S x /N x is the signal-to-noise ratio for a single scan. The impact of signalaveraging is shown in Figure 10.29. The first spectrum shows the signalfor a single scan, which consists of a single, noisy peak. Signal averagingusing 4 scans and 16 scans decreases the noise and improves the signal-tonoiseratio. One disadvantage of a photodiode array is that the effectivebandwidth per diode is roughly an order of magnitude larger than that fora high quality monochromator.Sample Cells. The sample compartment provides a light-tight environmentthat limits the addition of stray radiation. Samples are normally in theliquid or solution state, and are placed in cells constructed with UV/Vistransparent materials, such as quartz, glass, and plastic (Figure 10.30). Aquartz or fused-silica cell is required when working at a wavelength

Chapter 10 Spectroscopic Methods571Figure 10.30 Examples of sample cells for UV/Vis spectroscopy. From left to right (with path lengths in parentheses):rectangular plastic cuvette (10.0 mm), rectangular quartz cuvette (5.000 mm), rectangular quartzcuvette (1.000 mm), cylindrical quartz cuvette (10.00 mm), cylindrical quartz cuvette (100.0 mm). Cells oftenare available as a matched pair, which is important when using a double-beam instrument.are useful when analyzing a very dilute solution, or for gas samples. Thehighest quality cells allow the radiation to strike a flat surface at a 90 o angle,minimizing the loss of radiation to reflection. A test tube is often used asa sample cell with simple, single-beam instruments, although differencesin the cell’s pathlength and optical properties add an additional source oferror to the analysis.If we need to monitor an analyte’s concentration over time, it may notbe possible to physically remove samples for analysis. This is often the case,for example, when monitoring industrial production lines or waste lines,when monitoring a patient’s blood, or when monitoring environmentalsystems. With a fiber-optic probe we can analyze samples in situ. An exampleof a remote sensing fiber-optic probe is shown in Figure 10.31. Theprobe consists of two bundles of fiber-optic cable. One bundle transmitsradiation from the source to the probe’s tip, which is designed to allow thesample to flow through the sample cell. Radiation from the source passesthrough the solution and is reflected back by a mirror. The second bundlereflectingmirrorlight inlight outflow cellprobe’s tipFigure 10.31 Example of a fiber-optic probe. Theinset photographs provide a close-up look at theprobe’s flow cell and the reflecting mirror

572 Analytical Chemistry 2.0of fiber-optic cable transmits the nonabsorbed radiation to the wavelengthselector. Another design replaces the flow cell shown in Figure 10.31 witha membrane containing a reagent that reacts with the analyte. When theanalyte diffuses across the membrane it reacts with the reagent, producinga product that absorbs UV or visible radiation. The nonabsorbed radiationfrom the source is reflected or scattered back to the detector. Fiber opticprobes that show chemical selectivity are called optrodes. 6In s t r u m e n t De s i g n s f o r In f r a r e d Ab s o r p t i o nFilter Photometer. The simplest instrument for IR absorption spectroscopyis a filter photometer similar to that shown in Figure 10.25 for UV/Vis absorption.These instruments have the advantage of portability, and typicallyare used as dedicated analyzers for gases such as HCN and CO.Double-beam spectrophotometer. Infrared instruments using a monochromatorfor wavelength selection use double-beam optics similar to thatshown in Figure 10.27. Double-beam optics are preferred over single-beamoptics because the sources and detectors for infrared radiation are less stablethan those for UV/Vis radiation. In addition, it is easier to correct for theabsorption of infrared radiation by atmospheric CO 2 and H 2 O vapor whenusing double-beam optics. Resolutions of 1–3 cm –1 are typical for mostinstruments.Fourier transform spectrometer. In a Fourier transform infrared spectrometer,or FT–IR, the monochromator is replaced with an interferometer(Figure 10.13). Because an FT-IR includes only a single optical path, itis necessary to collect a separate spectrum to compensate for the absorbanceof atmospheric CO 2 and H 2 O vapor. This is done by collecting abackground spectrum without the sample and storing the result in the instrument’scomputer memory. The background spectrum is removed fromthe sample’s spectrum by ratioing the two signals. In comparison to otherinstrument designs, an FT–IR provides for rapid data acquisition, allowingan enhancement in signal-to-noise ratio through signal-averaging.Sample Cells. Infrared spectroscopy is routinely used to analyze gas, liquid,and solid samples. Sample cells are made from materials, such as NaCl andKBr, that are transparent to infrared radiation. Gases are analyzed usinga cell with a pathlength of approximately 10 cm. Longer pathlengths areobtained by using mirrors to pass the beam of radiation through the sampleseveral times.A liquid samples may be analyzed using a variety of different samplecells (Figure 10.32). For non-volatile liquids a suitable sample can be preparedby placing a drop of the liquid between two NaCl plates, forming athin film that typically is less than 0.01 mm thick. Volatile liquids must beplaced in a sealed cell to prevent their evaporation.6 (a) Seitz, W. R. Anal. Chem. 1984, 56, 16A–34A; (b) Angel, S. M. Spectroscopy 1987, 2(2),38–48.

Chapter 10 Spectroscopic Methods573(a) (b) (c)Figure 10.32 Three examples of IR sample cells: (a) NaCl salts plates; (b) fixed pathlength (0.5 mm) sample cell withNaCl windows; (c) disposable card with a polyethylene window that is IR transparent with the exception of strong absorptionbands at 2918 cm –1 and 2849 cm –1 .The analysis of solution samples is limited by the solvent’s IR absorbingproperties, with CCl 4 , CS 2 , and CHCl 3 being the most common solvents.Solutions are placed in cells containing two NaCl windows separated by aTeflon spacer. By changing the Teflon spacer, pathlengths from 0.015–1.0mm can be obtained.Transparent solid samples can be analyzed directly by placing them inthe IR beam. Most solid samples, however, are opaque, and must be dispersedin a more transparent medium before recording the IR spectrum. Ifa suitable solvent is available, then the solid can be analyzed by preparinga solution and analyzing as described above. When a suitable solvent is notavailable, solid samples may be analyzed by preparing a mull of the finelypowdered sample with a suitable oil. Alternatively, the powdered samplecan be mixed with KBr and pressed into an optically transparent pellet.The analysis of an aqueous sample is complicated by the solubility ofthe NaCl cell window in water. One approach to obtaining infrared spectraon aqueous solutions is to use attenuated total reflectance instead oftransmission. Figure 10.33 shows a diagram of a typical attenuated totalreflectance (ATR) FT–IR instrument. The ATR cell consists of a high refractiveindex material, such as ZnSe or diamond, sandwiched between alow refractive index substrate and a lower refractive index sample. Radiationfrom the source enters the ATR crystal where it undergoes a series oftotal internal reflections before exiting the crystal. During each reflectionthe radiation penetrates into the sample to a depth of a few microns. Theresult is a selective attenuation of the radiation at those wavelengths where

574 Analytical Chemistry 2.0fromsourceATRcrystalsampleglass substratetodetectorpressuretower andcompression tipsampleplatformsampleslotFigure 10.33 FT-IR spectrometer equipped with a diamond ATR sample cell. The inserts show a close-upphoto of the sample platform, a sketch of the ATR’s sample slot, and a schematic showing how the source’sradiation interacts with the sample. The pressure tower is used to ensure the contact of solid samples with theATR crystal.Further details about these, and othermethods for preparing solids for infraredanalysis can be found in this chapter’s additionalresources.the sample absorbs. ATR spectra are similar, but not identical, to thoseobtained by measuring the transmission of radiation.Solid samples also can be analyzed using an ATR sample cell. Afterplacing the solid in the sample slot, a compression tip ensures that it is incontact with the ATR crystal. Examples of solids that have been analyzedby ATR include polymers, fibers, fabrics, powders, and biological tissuesamples. Another reflectance method is diffuse reflectance, in which radiationis reflected from a rough surface, such as a powder. Powdered samplesare mixed with a non-absorbing material, such as powdered KBr, and thereflected light is collected and analyzed. As with ATR, the resulting spectrumis similar to that obtained by conventional transmission methods.10C.2 Quantitative ApplicationssampleslotFigure 10.18 shows the visible spectrumfor Fe(phen) 3 2+ .The determination of an analyte’s concentration based on its absorption ofultraviolet or visible radiation is one of the most frequently encounteredquantitative analytical methods. One reason for its popularity is that manyorganic and inorganic compounds have strong absorption bands in the UV/Vis region of the electromagnetic spectrum. In addition, if an analyte doesnot absorb UV/Vis radiation—or if its absorbance is too weak—we oftencan react it with another species that is strongly absorbing. For example, adilute solution of Fe 2+ does not absorb visible light. Reacting Fe 2+ with o-phenanthroline, however, forms an orange–red complex of Fe(phen) 3 2+ thathas a strong, broad absorbance band near 500 nm. An additional advantageto UV/Vis absorption is that in most cases it is relatively easy to adjust experimentaland instrumental conditions so that Beer’s law is obeyed.

Chapter 10 Spectroscopic Methods575A quantitative analysis based on the absorption of infrared radiation,although important, is less frequently encountered than those for UV/Visabsorption. One reason is the greater tendency for instrumental deviationsfrom Beer’s law when using infrared radiation. Because an infrared absorptionband is relatively narrow, any deviation due to the lack of monochromaticradiation is more pronounced. In addition, infrared sources are lessintense than UV/Vis sources, making stray radiation more of a problem.Differences in pathlength for samples and standards when using thin liquidfilms or KBr pellets are a problem, although an internal standard canbe used to correct for any difference in pathlength. Finally, establishing a100% T (A = 0) baseline is often difficult because the optical properties ofNaCl sample cells may change significantly with wavelength due to contaminationand degradation. We can minimize this problem by measuringabsorbance relative to a baseline established for the absorption band. Figure10.34 shows how this is accomplished.Another approach is to use a cell with afixed pathlength, such as that shown inFigure 10.32b.En v i r o nm e n t a l Ap p l i c a t i o n sThe analysis of waters and wastewaters often relies on the absorption of ultravioletand visible radiation. Many of these methods are outlined in Table10.6. Several of these methods are described here in more detail.Although the quantitative analysis of metals in waters and wastewatersis accomplished primarily by atomic absorption or atomic emissionspectroscopy, many metals also can be analyzed following the formationof a colored metal–ligand complex. One advantage to these spectroscopicmethods is that they are easily adapted to the analysis of samples in the fieldusing a filter photometer. One ligand that is used in the analysis of severalmetals is diphenylthiocarbazone, also known as dithizone. Dithizone isnot soluble in water, but when a solution of dithizone in CHCl 3 is shaken100Atomic absorption is the subject of Section10D and atomic emission is the subjectof Section 10G.The structure of dithizoneis shown to the right. SeeChapter 7 for a discussion ofextracting metal ions usingdithizone.SHNNNHN80% transmittance604020P 0P TA = –log P TP 00800 850 900 950 1000wavenumberFigure 10.34 Method for determining absorbancefrom an IR spectrum.

576 Analytical Chemistry 2.0Table 10.6 Examples of the Molecular UV/Vis Analysis of Waters andWastewatersAnalyte Method l (nm)Trace Metalsaluminumreact with Eriochrome cyanide R dye at pH 6; forms red topink complex535arsenicreduce to AsH 3 using Zn and react with silver diethyldithiocarbamate;forms red complex535cadmiumextract into CHCl 3 containing dithizone from a samplemade basic with NaOH; forms pink to red complex518chromiumoxidize to Cr(VI) and react with diphenylcarbazide; formsred-violet product540copperreact with neocuprine in neutral to slightly acid solutionand extract into CHCl 3 /CH 3 OH; forms yellow complex457ironreduce to Fe 2+ and react with o-phenanthroline; formsorange-red complex510extract into CHCl 3 containing dithizone from sampleleadmade basic with NH 3 /NH + 4 buffer; forms cherry red 510complexmanganese oxidize to MnO – 4 with persulfate; forms purple solution 525mercuryextract into CHCl 3 containing dithizone from acidicsample; forms orange complex492zinc react with zincon at pH 9; forms blue complex 620Inorganic Nonmetalsammoniareaction with hypochlorite and phenol using a manganoussalt catalyst; forms blue indophenol as product630cyanidereact with chloroamine-T to form CNCl and then with apyridine-barbituric acid; forms a red-blue dye578fluoridereact with red Zr-SPADNS lake; formation of ZrF 2– 6 decreasescolor of the red lake570chlorine (residual) react with leuco crystal violet; forms blue product 592nitratereact with Cd to form NO – 2 and then react with sulfanilamideand N-(1-napthyl)-ethylenediamine; forms red azo 543dyephosphatereact with ammonium molybdate and then reduce withSnCl 2 ; forms molybdenum blue690Organicsphenolreact with 4-aminoantipyrine and K 3 Fe(CN) 6 ; forms yellowantipyrine dye460anionic surfactantreact with cationic methylene blue dye and extract intoCHCl 3 ; forms blue ion pair652

Chapter 10 Spectroscopic Methods577with an aqueous solution containing an appropriate metal ion, a coloredmetal–dithizonate complex forms that is soluble in CHCl 3 . The selectivityof dithizone is controlled by adjusting the sample’s pH. For example, Cd 2+is extracted from solutions that are made strongly basic with NaOH, Pb 2+from solutions that are made basic with an NH 3 /NH 4 + buffer, and Hg 2+from solutions that are slightly acidic.When chlorine is added to water the portion available for disinfectionis called the chlorine residual. There are two forms of chlorine residual. Thefree chlorine residual includes Cl 2 , HOCl, and OCl – . The combined chlorineresidual, which forms from the reaction of NH 3 with HOCl, consistsof monochloramine, NH 2 Cl, dichloramine, NHCl 2 , and trichloramine,NCl 3 . Because the free chlorine residual is more efficient at disinfection,there is an interest in methods that can distinguish between the differentforms of the total chlorine residual. One such method is the leucocrystal violet method. The free residual chlorine is determined by addingleuco crystal violet to the sample, which instantaneously oxidizes to givea blue colored compound that is monitored at 592 nm. Completing theanalysis in less than five minutes prevents a possible interference from thecombined chlorine residual. The total chlorine residual (free + combined)is determined by reacting a separate sample with iodide, which reacts withboth chlorine residuals to form HOI. When the reaction is complete, leucocrystal violet is added and oxidized by HOI, giving the same blue coloredproduct. The combined chlorine residual is determined by difference.The concentration of fluoride in drinking water may be determinedindirectly by its ability to form a complex with zirconium. In the presenceof the dye SPADNS, solutions of zirconium form a red colored compound,called a lake, that absorbs at 570 nm. When fluoride is added, the formationof the stable ZrF 6 2– complex causes a portion of the lake to dissociate,decreasing the absorbance. A plot of absorbance versus the concentrationof fluoride, therefore, has a negative slope.Spectroscopic methods also are used to determine organic constituentsin water. For example, the combined concentrations of phenol, and orthoandmeta- substituted phenols are determined by using steam distillation toseparate the phenols from nonvolatile impurities. The distillate reacts with4-aminoantipyrine at pH 7.9 ± 0.1 in the presence of K 3 Fe(CN) 6 , forminga yellow colored antipyrine dye. After extracting the dye into CHCl 3 , itsabsorbance is monitored at 460 nm. A calibration curve is prepared usingonly the unsubstituted phenol, C 6 H 5 OH. Because the molar absorptivityof substituted phenols are generally less than that for phenol, the reportedconcentration represents the minimum concentration of phenoliccompounds.Molecular absorption also can be used for the analysis of environmentallysignificant airborne pollutants. In many cases the analysis is carriedout by collecting the sample in water, converting the analyte to an aqueousform that can be analyzed by methods such as those described in TableIn Chapter 9 we explored how the totalchlorine residual can be determined by aredox titration; see Representative Method9.3 for further details. The methoddescribed here allows us to divide the totalchlorine residual into its componentparts.SPADNS, which is shown below, isan abbreviation for the sodium salt of2-(4-sulfophenylazo)-1,8-dihydroxy-3,6-napthalenedisulfonic acid, which is amouthful to say.NaOO 3 SOHOHNSO 3 NaONH 24-aminoantipyreneSO 3 Na

578 Analytical Chemistry 2.0H 2 NO 3 SN NNH 2red azo dyeC 2 H 4 NH 310.6. For example, the concentration of NO 2 can be determined by oxidizingNO 2 to NO 3 – . The concentration of NO 3 – is then determined by firstreducing it to NO 2 – with Cd, and then reacting NO 2 – with sulfanilamideand N-(1-naphthyl)-ethylenediamine to form a red azo dye. Another importantapplication is the analysis for SO 2 , which is determined by collectingthe sample in an aqueous solution of HgCl 4 2– where it reacts to formHg(SO 3 ) 2 2– . Addition of p-rosaniline and formaldehyde produces a purplecomplex that is monitored at 569 nm. Infrared absorption is useful for theanalysis of organic vapors, including HCN, SO 2 , nitrobenzene, methylmercaptan, and vinyl chloride. Frequently, these analyses are accomplishedusing portable, dedicated infrared photometers.Clinical Ap p l i c a t i o n sThe analysis of clinical samples is often complicated by the complexity ofthe sample matrix, which may contribute a significant background absorptionat the desired wavelength. The determination of serum barbituratesprovides one example of how this problem is overcome. The barbituratesare first extracted from a sample of serum with CHCl 3 and then extractedfrom the CHCl 3 into 0.45 M NaOH (pH ≈ 13). The absorbance of theaqueous extract is measured at 260 nm, and includes contributions fromthe barbiturates as well as other components extracted from the serumsample. The pH of the sample is then lowered to approximately 10 byadding NH 4 Cl and the absorbance remeasured. Because the barbituratesdo not absorb at this pH, we can use the absorbance at pH 10, A pH 10 , tocorrect the absorbance at pH 13, A pH 13AV= A −barb pH 13samp+ VVsampNH4Cl× ApH 10where A barb is the absorbance due to the serum barbiturates, and V samp andV NH 4Cl are the volumes of sample and NH 4Cl, respectively. Table 10.7 providesa summary of several other methods for analyzing clinical samples.In d u s t r i a l An a l y s i sUV/Vis molecular absorption is used for the analysis of a diverse array ofindustrial samples including pharmaceuticals, food, paint, glass, and metals.In many cases the methods are similar to those described in Table 10.6 andTable 10.7. For example, the amount of iron in food can be determined bybringing the iron into solution and analyzing using the o-phenanthrolinemethod listed in Table 10.6.Many pharmaceutical compounds contain chromophores that makethem suitable for analysis by UV/Vis absorption. Products that have beenanalyzed in this fashion include antibiotics, hormones, vitamins, and analgesics.One example of the use of UV absorption is in determining the

Chapter 10 Spectroscopic Methods579Table 10.7 Examples of the Molecular UV/Vis Analysis of Clinical SamplesAnalyte Method l (nm)total serum protein react with NaOH and Cu 2+ ; forms blue-violet complex 540serum cholesterolreact with Fe 3+ in presence of isopropanol, acetic acid, andH 2 SO 4 ; forms blue-violet complex540uric acid react with phosphotungstic acid; forms tungsten blue 710serum barbituratesextract into CHCl 3 to isolate from interferents and thenextract into 0.45 M NaOH260glucose react with o-toludine at 100 o C; forms blue-green complex 630protein-bound iodinedecompose protein to release iodide, which catalyzes redoxreaction between Ce 3+ and As 3+ ; forms yellow colored Ce 4+ 420purity of aspirin tablets, for which the active ingredient is acetylsalicylicacid. Salicylic acid, which is produced by the hydrolysis of acetylsalicylicacid, is an undesirable impurity in aspirin tablets, and should not be presentat more than 0.01% w/w. Samples can be screened for unacceptable levelsof salicylic acid by monitoring the absorbance at a wavelength of 312 nm.Acetylsalicylic acid absorbs at 280 nm, but absorbs poorly at 312 nm. Conditionsfor preparing the sample are chosen such that an absorbance ofgreater than 0.02 signifies an unacceptable level of salicylic acid.Fo r e n s i c Ap p l i c a t i o n sUV/Vis molecular absorption is routinely used for the analysis of narcoticsand for drug testing. One interesting forensic application is the determinationof blood alcohol using the Breathalyzer test. In this test a 52.5-mLbreath sample is bubbled through an acidified solution of K 2 Cr 2 O 7 , whichoxidizes ethanol to acetic acid. The concentration of ethanol in the breathsample is determined by the decrease in absorbance at 440 nm where thedichromate ion absorbs. A blood alcohol content of 0.10%, which is abovethe legal limit, corresponds to 0.025 mg of ethanol in the breath sample.De v e l o p i n g a Qu a n t i t a t i ve Me t h o d f o r a Si n g l e Co m p o n e n tIn developing a quantitative analytical method, the conditions under whichBeer’s law is obeyed must be established. First, the most appropriate wavelengthfor the analysis is determined from an absorption spectrum. In mostcases the best wavelength corresponds to an absorption maximum becauseit provides greater sensitivity and is less susceptible to instrumental limitations.Second, if an instrument with adjustable slits is being used, then anappropriate slit width needs to be chosen. The absorption spectrum alsoaids in selecting a slit width. Usually we set the slits to be as wide as possiblebecause this increases the throughput of source radiation, while also beingnarrow enough to avoid instrumental limitations to Beer’s law. Finally, acalibration curve is constructed to determine the range of concentrationsfor which Beer’s law is valid. Additional considerations that are important

580 Analytical Chemistry 2.0in any quantitative method are the effect of potential interferents and establishingan appropriate blank.The best way to appreciate the theoreticaland practical details discussed in this sectionis to carefully examine a typical analyticalmethod. Although each method isunique, the following description of thedetermination of iron in water and wastewaterprovides an instructive example of atypical procedure. The description here isbased on Method 3500- Fe B as publishedin Standard Methods for the Examination ofWater and Wastewater, 20th Ed., AmericanPublic Health Association: Washington,D. C., 1998.Figure 10.18 shows the visible spectrumfor Fe(phen) 3 2+ .Representative Method 10.1Determination of Iron in Water and WastewaterDescription o f Me t h o dIron in the +2 oxidation state reacts with o-phenanthroline to form theorange-red Fe(phen) 3 2+ complex. The intensity of the complex’s color isindependent of solution acidity between a pH of 3 and 9. Because thecomplex forms more rapidly at lower pH levels, the reaction is usuallycarried out within a pH range of 3.0–3.5. Any iron present in the +3oxidation state is reduced with hydroxylamine before adding o-phenanthroline.The most important interferents are strong oxidizing agents,polyphosphates, and metal ions such as Cu 2+ , Zn 2+ , Ni 2+ , and Cd 2+ . Aninterference from oxidizing agents is minimized by adding an excess ofhydroxylamine, and an interference from polyphosphate is minimizedby boiling the sample in the presence of acid. The absorbance of samplesand standards are measured at a wavelength of 510 nm using a 1-cm cell(longer pathlength cells also may be used). Beer’s law is obeyed for concentrationsof within the range of 0.2–4.0 mg Fe/LNNo-phenanthrolinePr o c e d u r eFor samples containing less than 2 mg Fe/L, directly transfer a 50-mLportion to a 125-mL Erlenmeyer flask. Samples containing more than 2mg Fe/L must be diluted before acquiring the 50-mL portion. Add 2 mLof concentrated HCl and 1 mL of hydroxylamine to the sample. Heatthe solution to boiling and continue boiling until the solution’s volumeis reduced to between 15 and 20 mL. After cooling to room temperature,transfer the solution to a 50-mL volumetric flask, add 10 mL of an ammoniumacetate buffer, 2 mL of a 1000 ppm solution of o-phenanthroline,and dilute to volume. Allow 10–15 minutes for color development beforemeasuring the absorbance, using distilled water to set 100% T. Calibrationstandards, including a blank, are prepared by the same procedureusing a stock solution containing a known concentration of Fe 2+ .Qu e s t i o n s1. Explain why strong oxidizing agents are interferents, and why anexcess of hydroxylamine prevents the interference.

Chapter 10 Spectroscopic Methods581A strong oxidizing agent oxidizes some Fe 2+ to Fe 3+ . BecauseFe(phen) 3+ 3 does not absorb as strongly as Fe(phen) 2+ 3 , the absorbancedecreases, producing a negative determinate error. The excesshydroxylamine reacts with the oxidizing agents, removing them fromthe solution.2. The color of the complex is stable between pH levels of 3 and 9. Whatare some possible complications at more acidic or more basic pH’s?Because o-phenanthroline is a weak base, its conditional formationconstant for Fe(phen) 3 2+ is less favorable at more acidic pH levels,where o-phenanthroline is protonated. The result is a decrease inabsorbance and a less sensitive analytical method. When the pH isgreater than 9, competition between OH – and o-phenanthroline forFe 2+ also decreased the absorbance. In addition, if the pH is sufficientlybasic there is a risk that the iron will precipitate as Fe(OH) 2 .3. Cadmium is an interferent because it forms a precipitate with o-phenanthroline. What effect would the formation of precipitate haveon the determination of iron?Because o-phenanthroline is present in large excess (2000 mg of o-phenanthroline for 100 mg of Fe 2+ ), it is not likely that the interferenceis due to an insufficient amount of o-phenanthroline being availableto react with the Fe 2+ . The presence of a precipitate in the samplecell results in the scattering of radiation, which causes an apparentincrease in absorbance. Because the measured absorbance increases,the reported concentration is too high.4. Even high quality ammonium acetate contains a significant amountof iron. Why is this source of iron not a problem?Because all samples and standards are prepared using the same volumeof ammonium acetate buffer, the contribution of this source ofiron is accounted for by the calibration curve’s reagent blank.In Chapter 9 we saw the same effect of pHon the complexation reactions betweenEDTA and metal ions.Although scattering is a problem here, itcan serve as the basis of a useful analyticalmethod. See Section 10H for furtherdetails.Qu a n t i t a t i ve An a l y s i s f o r a Si n g l e An a l y t eTo determine the concentration of a an analyte we measure its absorbanceand apply Beer’s law using any of the standardization methods described inChapter 5. The most common methods are a normal calibration curve usingexternal standards and the method of standard additions. A single pointstandardization is also possible, although we must first verify that Beer’s lawholds for the concentration of analyte in the samples and the standard.Example 10.5The determination of Fe in an industrial waste stream was carried out bythe o‐phenanthroline described in Representative Method 10.1. Using thedata in the following table, determine the mg Fe/L in the waste stream.

582 Analytical Chemistry 2.0mg Fe/L absorbance0.00 0.0001.00 0.1832.00 0.3643.00 0.5464.00 0.727sample 0.269absorbance0.80.60.4So l u t i o nLinear regression of absorbance versus the concentration of Fe in the standardsgives a calibration curve with the following equation.A = 0. 0006 + 0. 1817×(mgFe/L)0.20.00 1 2 3 4mg Fe/LSubstituting the sample’s absorbance into the calibration expression givesthe concentration of Fe in the waste stream as 1.48 mg Fe/LPractice Exercise 10.5The concentration of Cu 2+ in a sample can be determined by reacting itwith the ligand cuprizone and measuring its absorbance at 606 nm ina 1.00-cm cell. When a 5.00-mL sample is treated with cuprizone anddiluted to 10.00 mL, the resulting solution has an absorbance of 0.118.A second 5.00-mL sample is mixed with 1.00 mL of a 20.00 mg/L standardof Cu 2+ , treated with cuprizone and diluted to 10.00 mL, giving anabsorbance of 0.162. Report the mg Cu 2+ /L in the sample.Click here to review your answer to this exercise.Qu a n t i t a t i ve An a l y s i s o f Mi x t u r e sSuppose we need to determine the concentration of two analytes, X andY, in a sample. If each analyte has a wavelength where the other analytedoes not absorb, then we can proceed using the approach in Example 10.5.Unfortunately, UV/Vis absorption bands are so broad that it frequently isnot possible to find suitable wavelengths. Because Beer’s law is additive themixture’s absorbance, A mix , is( A ) = ( ε ) bC + ( ε ) bC10.11mix λ1 X λ1 X Y λ1Ywhere l1 is the wavelength at which we measure the absorbance. Becauseequation 10.11 includes terms for the concentration of both X and Y, theabsorbance at one wavelength does not provide enough information todetermine either C X or C Y . If we measure the absorbance at a second wavelength

Chapter 10 Spectroscopic Methods583( A ) = ( ε ) bC + ( ε ) bC10.12mix λ2 X λ2 X Y λ2Ythen C X and C Y can be determined by solving simultaneously equation10.11and equation 10.12. Of course, we also must determine the value for e X ande Y at each wavelength. For a mixture of n components, we must measurethe absorbance at n different wavelengths.Example 10.6The concentrations of Fe 3+ and Cu 2+ in a mixture can be determined followingtheir reaction with hexacyanoruthenate (II), Ru(CN) 6 4– , whichforms a purple-blue complex with Fe 3+ (l max = 550 nm) and a pale-greencomplex with Cu 2+ (l max = 396 nm). 7 The molar absorptivities (M –1cm –1 ) for the metal complexes at the two wavelengths are summarized inthe following table.e 550 e 396Fe 3+ 9970 84Cu 2+ 34 856When a sample containing Fe 3+ and Cu 2+ is analyzed in a cell with a pathlengthof 1.00 cm, the absorbance at 550 nm is 0.183 and the absorbanceat 396 nm is 0.109. What are the molar concentrations of Fe 3+ and Cu 2+in the sample?So l u t i o nSubstituting known values into equations 10.11 and 10.12 givesA = 0 183 = 9970C + 34C550.Fe CuA = 0 109 = 84C + 856C396.Fe CuTo determine C Fe and C Cu we solve the first equation for C CuCCu0.183−9970C=34and substitute the result into the second equation.0.183−9970C0.109 = 84C+ 856×Fe34FeFe= 4. 607 −( 251 . ×10 5 )C FeAnother approach is to multiply the firstequation by 856/34 giving4. 607 = 251009C+ 856CFeCuSubtracting the second equation from thisequation4.607 = 251009C+ 856CFeFe− 0.109 = 84C+ 856C4. 498 = 250925CFeCuCuwe find that C Fe is 1.7910 –5 . Havingdetermined C Fe we can substitute backinto one of the other equations to solvefor C Cu , which is 1.2610 –5 .Solving for C Fe gives the concentration of Fe 3+ as 1.79 10 –5 M. Substitutingthis concentration back into the equation for the mixture’s absorbanceat 396 nm gives the concentration of Cu 2+ as 1.26 10 –4 M.7 DiTusa, M. R.; Schlit, A. A. J. Chem. Educ. 1985, 62, 541–542.

584 Analytical Chemistry 2.0Practice Exercise 10.6The absorbance spectra for Cr 3+ and Co 2+ overlap significantly. To determinethe concentration of these analytes in a mixture, its absorbance wasmeasured at 400 nm and at 505 nm, yielding values of 0.336 and 0.187,respectively. The individual molar absorptivities (M –1 cm –1 ) aree 400 e 505Cr 3+ 15.2 0.533Co 2+ 5.60 5.07Click here to review your answer to this exercise.For example, in Example 10.6 the molarabsorptivity for Fe 3+ at 550 nm is 119that for Cu 2+ , and the molar absorptivityfor Cu 2+ at 396 nm is 10.2 that forFe 3+ .To obtain results with good accuracy and precision the two wavelengthsshould be selected so that e X > e Y at one wavelength and e X < e Y at theother wavelength. It is easy to appreciate why this is true. Because the absorbanceat each wavelength is dominated by one analyte, any uncertaintyin the concentration of the other analyte has less of an impact. Figure10.35 shows that the choice of wavelengths for Practice Exercise 10.6 arereasonable. When the choice of wavelengths is not obvious, one methodfor locating the optimum wavelengths is to plot e X /e Y as function of wavelength,and determine the wavelengths where e X /e Y reaches maximum andminimum values. 8When the analyte’s spectra overlap severely, such that e X ≈ e Y at allwavelength, other computational methods may provide better accuracy andprecision. In a multiwavelength linear regression analysis, for example, amixture’s absorbance is compared to that for a set of standard solutions atseveral wavelengths. 9 If A SX and A SY are the absorbance values for standardsolutions of components X and Y at any wavelength, then8 Mehra, M. C.; Rioux, J. J. Chem. Educ. 1982, 59, 688–689.9 Blanco, M.; Iturriaga, H.; Maspoch, S.; Tarin, P. J. Chem. Educ. 1989, 66, 178–180.1.00.8mixtureabsorbance0.60.4Figure 10.35 Visible absorption spectra for 0.0250 MCr 3+ , 0.0750 M Co 2+ , and for a mixture of Cr 3+ andCo 2+ . The two wavelengths used for analyzing the mixtureof Cr 3+ and Co 2+ are shown by the dashed lines. Thedata for the two standard solutions are from reference 7.0.20.0Cr 3+Co 2+400 450 500 550 600wavelength

Chapter 10 Spectroscopic Methods585A=ε bC10.13SX X SXA=ε bC10.14SY Y SYwhere C SX and C SY are the known concentrations of X and Y in the standardsolutions. Solving equation 10.13 and equation 10.14 for e X and e Y ,substituting into equation 10.11, and rearranging, givesAAmixSXC C AX Y= + ×C C ASXTo determine C X and C Y the mixture’s absorbance and the absorbancesof the standard solutions are measured at several wavelengths. GraphingA mix /A SX versus A SY /A SX gives a straight line with a slope of C Y /C SY and ay-intercept of C X /C SX . This approach is particularly helpful when it is notpossible to find wavelengths where e X > e Y and e X < e Y .Example 10.7Figure 10.35 shows visible absorbance spectra for a standard solution of0.0250 M Cr 3+ , a standard solution of 0.0750 M Co 2+ , and a mixturecontaining unknown concentrations of each ion. The data for these spectraare shown here. 10SYSYSXThe approach outlined here for a multiwavelengthlinear regression uses a singlestandard solution for each analyte. A morerigorous approach uses multiple standardsfor each analyte. The math behindthe analysis of this data—what we call amultiple linear regression—is beyond thelevel of this text. For more details aboutmultiple linear regression see Brereton,R. G. Chemometrics: Data Analysis for theLaboratory and Chemical Plant, Wiley:Chichester, England, 2003.l (nm) A Cr A Co A mix l (nm) A Cr A Co A mix375 0.26 0.01 0.53 520 0.19 0.38 0.63400 0.43 0.03 0.88 530 0.24 0.33 0.70425 0.39 0.07 0.83 540 0.28 0.26 0.73440 0.29 0.13 0.67 550 0.32 0.18 0.76455 0.20 0.21 0.54 570 0.38 0.08 0.81470 0.14 0.28 0.47 575 0.39 0.06 0.82480 0.12 0.30 0.44 580 0.38 0.05 0.79490 0.11 0.34 0.45 600 0.34 0.03 0.70500 0.13 0.38 0.51 625 0.24 0.02 0.49Use a multiwavelength regression analysis to determine the compositionof the unknown.So l u t i o nFirst we need to calculate values for A mix /A SX and for A SY /A SX . Let’s defineX as Co 2+ and Y as Cr 3+ . For example, at a wavelength of 375 nmA mix /A SX is 0.53/0.01, or 53 and A SY /A SX is 0.26/0.01, or 26. Completingthe calculation for all wavelengths and graphing A mix /A SX versus A SY /A SX10 The data for the two standards are from Brewer, S. Solving Problems in Analytical Chemistry, JohnWiley & Sons: New York, 1980.

586 Analytical Chemistry 2.0A mix /A SX504030201000 5 10 15 20 25A SY /A SXFigure 10.36 Multiwavelength linear regressionanalysis for the data in Example10.7.gives the result shown in Figure 10.36. Fitting a straight-line to the datagives a regression model ofAAmixSX= 0. 636 + 201 . ×Using the y-intercept, the concentration of Co 2+ isCCXSXAASYSXCCo= = 0.6360.0750 Mor C Co = 0.048 M, and using the slope the concentration of Cr 3+ isCCYSY= CCr0 0250 M= 201 ..or C Cr = 0.050 M.Practice Exercise 10.7There are many additional ways to analyzemixtures spectrophotometrically, includinggeneralized standard additions,H-point standard additions, principalcomponent regression to name a few.Consult the chapter’s additional resourcesfor further information.A mixture of MnO 4 – and Cr 2 O 7 2– , and standards of 0.10 mM KMnO 4and of 0.10 mM K 2 Cr 2 O 7 gives the results shown in the following table.Determine the composition of the mixture. The data for this problem isfrom Blanco, M. C.; Iturriaga, H.; Maspoch, S.; Tarin, P. J. Chem. Educ.1989, 66, 178–180.l (nm) A Mn A Cr A mix266 0.042 0.410 0.766288 0.082 0.283 0.571320 0.168 0.158 0.422350 0.125 0.318 0.672360 0.056 0.181 0.366Click here to review your answer to this exercise.10C.3 Qualitative ApplicationsAs discussed earlier in Section 10B.1, ultraviolet, visible, and infrared absorptionbands result from the absorption of electromagnetic radiation byspecific valence electrons or bonds. The energy at which the absorptionoccurs, and its intensity is determined by the chemical environment of theabsorbing moiety. For example, benzene has several ultraviolet absorptionbands due to p p* transitions. The position and intensity of two of thesebands, 203.5 nm (e = 7400 M –1 cm –1 ) and 254 nm (e = 204 M –1 cm –1 ),are sensitive to substitution. For benzoic acid, in which a carboxylic acidgroup replaces one of the aromatic hydrogens, the two bands shift to 230nm (e = 11 600 M –1 cm –1 ) and 273 nm (e = 970 M –1 cm –1 ). A variety of

Chapter 10 Spectroscopic Methods587rules have been developed to aid in correlating UV/Vis absorption bandsto chemical structure. Similar correlations have been developed for infraredabsorption bands. For example a carbonyl’s C=O stretch is sensitive toadjacent functional groups, occurring at 1650 cm –1 for acids, 1700 cm –1for ketones, and 1800 cm –1 for acid chlorides. The interpretation of UV/Vis and IR spectra receives adequate coverage elsewhere in the chemistrycurriculum, notably in organic chemistry, and is not considered further inthis text.With the availability of computerized data acquisition and storage it ispossible to build digital libraries of standard reference spectra. The identityof an a unknown compound can often be determined by comparing itsspectrum against a library of reference spectra, a process is known as spectralsearching. Comparisons are made using an algorithm that calculatesthe cumulative difference between the sample’s spectrum and a referencespectrum. For example, one simple algorithm uses the following equationnD= ∑ ( A ) −( A )sample i reference ii=1where D is the cumulative difference, A sample is the sample’s absorbance atwavelength or wavenumber i, A reference is the absorbance of the referencecompound at the same wavelength or wavenumber, and n is the number ofdigitized points in the spectra. The cumulative difference is calculated foreach reference spectrum. The reference compound with the smallest valueof D provides the closest match to the unknown compound. The accuracyof spectral searching is limited by the number and type of compoundsincluded in the library, and by the effect of the sample’s matrix on thespectrum.Another advantage of computerized data acquisition is the ability tosubtract one spectrum from another. When coupled with spectral searchingit may be possible, by repeatedly searching and subtracting reference spectra,to determine the identity of several components in a sample withoutthe need of a prior separation step. An example is shown in Figure 10.37in which the composition of a two-component mixture is determined bysuccessive searching and subtraction. Figure 10.37a shows the spectrumof the mixture. A search of the spectral library selects cocaine . HCl (Figure10.37b) as a likely component of the mixture. Subtracting the referencespectrum for cocaine . HCl from the mixture’s spectrum leaves a result(Figure 10.37c) that closely matches mannitol’s reference spectrum (Figure10.37d). Subtracting the reference spectrum for leaves only a small residualsignal (Figure 10.37e).10C.4 Characterization ApplicationsMolecular absorption, particularly in the UV/Vis range, has been used fora variety of different characterization studies, including determining the

588 Analytical Chemistry 2.0IR spectra traditionally are displayed usingpercent transmittance, %T, along they-axis ( for example, see Figure 10.16).Because absorbance—not percent transmittance—isa linear function of concentration,spectral searching and spectralsubtraction, is easier to do when displayingabsorbance on the y-axis.0.80.4absorbance1.20.0A: mixture4000 3500 3000 2500 2000 1500 1000 500wavenumber (cm -1 )0.80.4absorbance1.20.0B: cocaine hydrochloride4000 3500 3000 2500 2000 1500 1000 500wavenumber (cm -1 )0.80.4absorbance1.20.0C: first subtraction (A–B)4000 3500 3000 2500 2000 1500 1000 500wavenumber (cm -1 )Figure 10.37 Identifying the componentsof a mixture by spectral searchingand subtracting. (a) IR spectrum of themixture; (b) Reference IR spectrum ofcocaine . HCl; (c) Result of subtractingthe spectrum of cocaine . HCl from themixture’s spectrum; (d) Reference IRspectrum of mannitol; and (e) The residualspectrum after removing mannitol’scontribution to the mixture’sspectrum.0.80.4absorbance1.20.00.80.4absorbance1.20.0D: mannitol4000 3500 3000 2500 2000 1500 1000 500wavenumber (cm -1 )E: second subtraction (C–D)4000 3500 3000 2500 2000 1500 1000 500wavenumber (cm -1 )stoichiometry of metal–ligand complexes and determining equilibriumconstants. Both of these examples are examined in this section.St o i c h i o m e t r y o f a Me t a l-Li g a n d Co m p l e xWe can determine the stoichiometry of a metal–ligand complexation reactionM+ yL MLyusing one of three methods: the method of continuous variations, themole-ratio method, and the slope-ratio method. Of these approaches, the

Chapter 10 Spectroscopic Methods589method of continuous variations, also called Job’s method, is the mostpopular. In this method a series of solutions is prepared such that the totalmoles of metal and ligand, n total , in each solution is the same. If (n M ) i and(n L ) i are, respectively, the moles of metal and ligand in solution i, thenn = ( n ) + ( n )total M i L iThe relative amount of ligand and metal in each solution is expressed as themole fraction of ligand, (X L ) i , and the mole fraction of metal, (X M ) i ,( X )Mi( X )Li( n )L i=ntotal( n ) ( n )L i M i= 1− =n ntotaltotalThe concentration of the metal–ligand complex in any solution is determinedby the limiting reagent, with the greatest concentration occurringwhen the metal and the ligand are mixed stoichiometrically. If we monitorthe complexation reaction at a wavelength where only the metal–ligandcomplex absorbs, a graph of absorbance versus the mole fraction of ligandwill have two linear branches—one when the ligand is the limiting reagentand a second when the metal is the limiting reagent. The intersection ofthese two branches represents a stoichiometric mixing of the metal and theligand. We can use the mole fraction of ligand at the intersection to determinethe value of y for the metal–ligand complex ML y .Example 10.8ny = X XLn= LX= L1 − XMTo determine the formula for the complex between Fe 2+ and o-phenanthroline,a series of solutions is prepared in which the total concentrationof metal and ligand is held constant at 3.15 10 –4 M. The absorbance ofeach solution is measured at a wavelength of 510 nm. Using the followingdata, determine the formula for the complex.X L absorbance X L absorbance0.000 0.000 0.600 0.6930.100 0.116 0.700 0.8090.200 0.231 0.800 0.6930.300 0.347 0.900 0.3470.400 0.462 1.000 0.0000.500 0.578MLYou also can plot the data as absorbanceversus the mole fraction of metal. In thiscase, y is equal to (1–X M )/X M .To prepare the solutions for this exampleI first prepared a solution of 3.15 10 -4M Fe 2+ and a solution of 3.15 10 -4M o-phenanthroline. Because the twostock solutions are of equal concentration,diluting a portion of one solutionwith the other solution gives a mixturein which the combined concentration ofo-phenanthroline and Fe 2+ is 3.15 10 -4M. If each solution has the same volume,then each solution contains the same totalmoles of metal and ligand.

590 Analytical Chemistry 2.01.0moles metal = moles ligand0.8absorbance0.60.4metal in excessligand in excess0.2Figure 10.38 Continuous variations plotfor Example 10.8. The photo shows thesolutions used in gathering the data. Eachsolution is displayed directly below its correspondingpoint on the continuous variationsplot.0.00.0 0.2 0.4 0.6 0.8 1.0X LPractice Exercise 10.8So l u t i o nA plot of absorbance versus the mole fraction of ligand is shown in Figure10.38. To find the maximum absorbance, we extrapolate the two linearportions of the plot. The two lines intersect at a mole fraction of ligand of0.75. Solving for y givesXL075 .y = = = 31 − X 1 − 0.75LThe formula for the metal–ligand complex is Fe(o-phenanthroline) 3 2+ .Use the continuous variations data in the following table to determine the formula for the complexbetween Fe 2+ and SCN – . The data for this problem is adapted from Meloun, M.; Havel, J.; Högfeldt,E. Computation of Solution Equilibria, Ellis Horwood: Chichester, England, 1988, p. 236.X L absorbance X L absorbance X L absorbance X L absorbance0.0200 0.068 0.2951 0.670 0.5811 0.790 0.8923 0.3250.0870 0.262 0.3887 0.767 0.6860 0.701 0.9787 0.0710.1792 0.471 0.4964 0.807 0.7885 0.540Click here to review your answer to this exercise.

Chapter 10 Spectroscopic Methods591Several precautions are necessary when using the method of continuousvariations. First, the metal and the ligand must form only one metal–ligandcomplex. To determine if this condition is true, plots of absorbance versusX L are constructed at several different wavelengths and for several differentvalues of n total . If the maximum absorbance does not occur at the samevalue of X L for each set of conditions, then more than one metal–ligandcomplex must be present. A second precaution is that the metal–ligandcomplex’s absorbance must obey Beer’s law. Third, if the metal–ligand complex’sformation constant is relatively small, a plot of absorbance versus X Lmay show significant curvature. In this case it is often difficult to determinethe stoichiometry by extrapolation. Finally, because the stability of a metal–ligand complex may be influenced by solution conditions, the compositionof the solutions must be carefully controlled. When the ligand is a weakbase, for example, the solutions must be buffered to the same pH.In the mole-ratio method the amount of one reactant, usually themoles of metal, is held constant, while the amount of the other reactant isvaried. The absorbance is monitored at a wavelength where the metal–ligandcomplex absorbs. A plot of absorbance as a function of the ligand-to-metalmole ratio, n L /n M , has two linear branches, which intersect at a mole–ratiocorresponding to the complex’s formula. Figure 10.39a shows a mole-ratioplot for the formation of a 1:1 complex in which the absorbance is monitoredat a wavelength where only the complex absorbs. Figure 10.39b showsa mole-ratio plot for a 1:2 complex in which all three species—the metal,the ligand, and the complex—absorb at the selected wavelength. Unlike themethod of continuous variations, the mole-ratio method can be used forcomplexation reactions that occur in a stepwise fashion if there is a differencein the molar absorptivities of the metal–ligand complexes, and if theformation constants are sufficiently different. A typical mole-ratio plot forthe step-wise formation of ML and ML 2 is shown in Figure 10.39c.In both the method of continuous variations and the mole-ratio methodwe determine the complex’s stoichiometry by extrapolating absorbance data(a) (b) (c)MLML 2ML 2MLabsorbanceabsorbanceincreasing absorbancedue to excess ligandabsorbanceinitial absorbance of metal0 1 2 3n L /n M0 1 2 3n L /n M0 1 2 3n L /n MFigure 10.39 Mole-ratio plots for: (a) a 1:1 metal–ligand complex in which only the complex absorbs; (b) a 1:2 metal–ligand complex in which the metal, the ligand, and the complex absorb; and (c) the stepwise formation of a 1:1 and a1:2 metal–ligand complex.

592 Analytical Chemistry 2.0from conditions in which there is a linear relationship between absorbanceand the relative amounts of metal and ligand. If a metal–ligand complex isvery weak, a plot of absorbance versus X L or n L /n M may be so curved that itis impossible to determine the stoichiometry by extrapolation. In this casethe slope-ratio may be used.In the slope-ratio method two sets of solutions are prepared. The firstset of solutions contains a constant amount of metal and a variable amountof ligand, chosen such that the total concentration of metal, C M , is muchlarger than the total concentration of ligand, C L . Under these conditionswe may assume that essentially all the ligand reacts in forming the metal–ligand complex. The concentration of the complex, which has the generalform M x L y , isC[ ML]=x yyLIf we monitor the absorbance at a wavelength where only M x L y absorbs,thenεbCA= εb[ ML]=x yyand a plot of absorbance versus C L is linear with a slope, s L , ofbs = εLyA second set of solutions is prepared with a fixed concentration of ligandthat is much greater than a variable concentration of metal; thusC[ ML]=x yxMεbCA= εb[ ML]=x yxbs = εMxA ratio of the slopes provides the relative values of x and y.ssMLεb/x= =εb/yAn important assumption in the slope-ratio method is that the complexationreaction continues to completion in the presence of a sufficientlylarge excess of metal or ligand. The slope-ratio method also is limited toyxLM

Chapter 10 Spectroscopic Methods593systems in which only a single complex is formed and for which Beer’s lawis obeyed.De t e r m i n a t i o n o f Equilibrium Co n s t a n t sAnother important application of molecular absorption spectroscopy is thedetermination of equilibrium constants. Let’s consider, as a simple example,an acid–base reaction of the general form+ −HIn( aq) + H O() l H O ( aq) + In ( aq )2 3where HIn and In – are the conjugate weak acid and weak base forms of anacid–base indicator. The equilibrium constant for this reaction is+ −[ HO ][ In ]=[ HIn]K a3To determine the equilibrium constant’s value, we prepare a solution inwhich the reaction is in a state of equilibrium and determine the equilibriumconcentration of H 3 O + , HIn, and In – . The concentration of H 3 O + iseasy to determine by simply measuring the solution’s pH. To determine theconcentration of HIn and In – we can measure the solution’s absorbance.If both HIn and In – absorb at the selected wavelength, then, from equation10.6, we know that−A= ε b[ HIn] + ε b[ In ] 10.15HInwhere e HIn and e In are the molar absorptivities for HIn and In – . The totalconcentration of indicator, C, is given by a mass balance equationInC = [ HIn] + [ In − ]10.16Solving equation 10.16 for [HIn] and substituting into equation 10.15giveswhich we simplify toA= b C − −−ε ( [ In ]) + ε b[ In ]HInA= bC − b −−ε ε [ In ] + ε b[ In ]InHIn HIn In−A= A + b[ In ]( ε − ε )10.17HIn In HInwhere A HIn , which is equal to e HIn bC, is the absorbance when the pH isacidic enough that essentially all the indicator is present as HIn. Solvingequation 10.17 for the concentration of In – gives[ In ]− =A−AHIn=b( ε − ε )InHIn10.18

594 Analytical Chemistry 2.0Proceeding in the same fashion, we can derive a similar equation for theconcentration of HInA − AIn[ HIn]=b( ε − ε )InHIn10.19where A In , which is equal to e In bC, is the absorbance when the pH is basicenough that only In – contributes to the absorbance. Substituting equation10.18 and equation 10.19 into the equilibrium constant expression for HIngivesKa =+A−AHIn[ HO 3]A − AIn10.20We can use equation 10.20 to determine the value of K a in one of twoways. The simplest approach is to prepare three solutions, each of whichcontains the same amount, C, of indicator. The pH of one solution ismade sufficiently acidic such that [HIn] >> [In – ]. The absorbance of thissolution gives A HIn . The value of A In is determined by adjusting the pH ofthe second solution such that [In-] >> [HIn]. Finally, the pH of the thirdsolution is adjusted to an intermediate value, and the pH and absorbance,A, recorded. The value of K a is calculated using equation 10.20.Example 10.9The acidity constant for an acid–base indicator is determined by preparingthree solutions, each of which has a total indicator concentration of5.00 10 –5 M. The first solution is made strongly acidic with HCl and hasan absorbance of 0.250. The second solution was made strongly basic andhas an absorbance of 1.40. The pH of the third solution is 2.91 and has anabsorbance of 0.662. What is the value of K a for the indicator?So l u t i o nThe value of K a is determined by making appropriate substitutions into10.20; thus−30. 662−0.250K a= (. 123× 10 ) ×= 687 . × 1140 . −0.6620 −4Practice Exercise 10.9To determine the K a of a merocyanine dye, the absorbance of a solution of3.510 –4 M dye was measured at a pH of 2.00, a pH of 6.00, and a pHof 12.00, yielding absorbances of 0.000, 0.225, and 0.680, respectively.What is the value of K a for this dye? The data for this problem is adaptedfrom Lu, H.; Rutan, S. C. Anal. Chem., 1996, 68, 1381–1386.Click here to review your answer to this exercise.

Chapter 10 Spectroscopic Methods595A second approach for determining K a is to prepare a series of solutions,each containing the same amount of indicator. Two solutions are used todetermine values for A HIn and A In . Taking the log of both sides of equation10.20 and rearranging leave us with the following equation.⎛log A − A ⎞HInK⎝⎜A − A ⎠⎟ = pH − pa10.21InA plot of log[(A – A HIn )/(A In – A)] versus pH is a straight-line with a slopeof +1 and a y-intercept of –pK a .Practice Exercise 10.10To determine the K a of the indicator bromothymol blue, the absorbanceof a series of solutions containing the same concentration of the indicatorwas measured at pH levels of 3.35, 3.65, 3.94, 4.30, and 4.64, yieldingabsorbances of 0.170, 0.287, 0.411, 0.562, and 0.670, respectively.Acidifying the first solution to a pH of 2 changes its absorbance to 0.006,and adjusting the pH of the last solution to 12 changes its absorbance to0.818. What is the value of K a for this day? The data for this problem isfrom Patterson, G. S. J. Chem. Educ., 1999, 76, 395–398.Click here to review your answer to this exercise.In developing these approaches for determining K a we considered arelatively simple system in which the absorbance of HIn and In – are easy tomeasure and for which it is easy to determine the concentration of H 3 O + .In addition to acid–base reactions, we can adapt these approaches to anyreaction of the general formX( aq) + Y( aq) Z( aq)including metal–ligand complexation reactions and redox reactions, providedthat we can determine spectrophotometrically the concentration ofthe product, Z, and one of the reactants, and that the concentration ofthe other reactant can be measured by another method. With appropriatemodifications, more complicated systems, in which one or more of theseparameters can not be measured, also can be treated. 1110C.5 Evaluation of UV/Vis and IR SpectroscopySc a l e o f Op e r a t i o nMolecular UV/Vis absorption is routinely used for the analysis of trace analytesin macro and meso samples. Major and minor analytes can be determinedby diluting the sample before analysis, while concentrating a samplemay allow for the analysis of ultratrace analytes. The scale of operations forinfrared absorption is generally poorer than that for UV/Vis absorption.11 Ramette, R. W. Chemical Equilibrium and Analysis, Addison-Wesley: Reading, MA, 1981,Chapter 13.See Figure 3.5 to review the meaning ofmacro and meso for describing samples,and the meaning of major, minor, and ultratracefor describing analytes.

596 Analytical Chemistry 2.0Ac c u r a c yUnder normal conditions a relative error of 1–5% is easy to obtained withUV/Vis absorption. Accuracy is usually limited by the quality of the blank.Examples of the type of problems that may be encountered include thepresence of particulates in a sample that scatter radiation and interferentsthat react with analytical reagents. In the latter case the interferent mayreact to form an absorbing species, giving rise to a positive determinateerror. Interferents also may prevent the analyte from reacting, leading toa negative determinate error. With care, it may be possible to improve theaccuracy of an analysis by as much as an order of magnitude.PrecisionIn absorption spectroscopy, precision is limited by indeterminate errors—primarily instrumental noise—introduced when measuring absorbance.Precision is generally worse for low absorbances where P 0 ≈ P T , and forhigh absorbances when P T approaches 0. We might expect, therefore, thatprecision will vary with transmittance.We can derive an expression between precision and transmittance byapplying the propagation of uncertainty as described in Chapter 4. To doso we rewrite Beer’s law asC =− 1 logT10.22εbTable 4.10 in Chapter 4 helps us in completing the propagation of uncertaintyfor equation 10.22, giving the absolute uncertainty in the concentration,s C , as0.4343 sTs =− ×10.23CεbTwhere s T is the absolute uncertainty in the transmittance. Dividing equation10.23 by equation 10.22 gives the relative uncertainty in concentration,s C /C, assCCs= 0. 4343T logTIf we know the absolute uncertainty in transmittance, we can determine therelative uncertainty in concentration for any transmittance.Determining the relative uncertainty in concentration is complicatedbecause s T may be a function of the transmittance. As shown in Table 10.8,three categories of indeterminate instrumental error have been observed. 12A constant s T is observed for the uncertainty associated with reading %Ton a meter’s analog or digital scale. Typical values are ±0.2–0.3% (a k 1 of±0.002–0.003) for an analog scale, and ±0.001% a (k 1 of ±0.000 01) for12 Rothman, L. D.; Crouch, S. R.; Ingle, J. D. Jr. Anal. Chem. 1975, 47, 1226–1233.T

Chapter 10 Spectroscopic Methods597Table 10.8 Effect of Indeterminate Errors on Relative Uncertainty in ConcentrationCategory Sources of Indeterminate Error Relative Uncertainty in Concentrations = k%T readout resolutionsT 1Cknoise in thermal detectors= 0 . 4343 1C T logTs = 2k T + noise in photon detectorsTT 2skTT = 3positioning of sample cellfluctuations in source intensitya digital scale. A constant s T also is observed for the thermal transducersused in infrared spectrophotometers. The effect of a constant s T on therelative uncertainty in concentration is shown by curve A in Figure 10.40.Note that the relative uncertainty is very large for both high and low absorbances,reaching a minimum when the absorbance is 0.4343. This sourceof indeterminate error is important for infrared spectrophotometers and forinexpensive UV/Vis spectrophotometers. To obtain a relative uncertaintyin concentration of ±1–2%, the absorbance must be kept within the range0.1–1.Values of s T are a complex function of transmittance when indeterminateerrors are dominated by the noise associated with photon detectors.Curve B in Figure 10.40 shows that the relative uncertainty in concentrationis very large for low absorbances, but is less at higher absorbances.Although the relative uncertainty reaches a minimum when the absorbanceis 0.963, there is little change in the relative uncertainty for absorbanceswithin the range 0.5–2. This source of indeterminate error generally limitss kCC= 0.434321T+ 1log Ts Ck= 0 . 4343 3C logT5A4s C /C×1003210BC0.0 0.5 1.0 1.5 2.0absorbanceFigure 10.40 Percent relative uncertainty in concentration asa function of absorbance for the categories of indeterminateerrors in Table 10.8. A: k 1 = ±0.0030; B: k 2 = ±0.0030; C:k 3 = ±0.0130. The dashed lines correspond to the minimumuncertainty for curve A (absorbance of 0.4343) and for curveB (absorbance of 0.963).

598 Analytical Chemistry 2.00% Tsample standard100% T(a)0% T 100% T0% Tstandard sample 100% T(b)0% T 100% T0% Tstandard100% Tsample standard(c)0% T 100% TFigure 10.41 Methods for improving theprecision of absorption methods: (a) highabsorbancemethod; (b) low-absorbancemethod; (c) maximum precision method.the precision of high quality UV/Vis spectrophotometers for mid-to-highabsorbances.Finally, the value of s T is directly proportional to transmittance for indeterminateerrors resulting from fluctuations in the source’s intensity andfrom uncertainty in positioning the sample within the spectrometer. Thelatter is particularly important because the optical properties of any samplecell are not uniform. As a result, repositioning the sample cell may lead toa change in the intensity of transmitted radiation. As shown by curve C inFigure 10.40, the effect is only important at low absorbances. This sourceof indeterminate errors is usually the limiting factor for high quality UV/Vis spectrophotometers when the absorbance is relatively small.When the relative uncertainty in concentration is limited by the %Treadout resolution, the precision of the analysis can be improved by redefining100% T and 0% T. Normally 100% T is established using a blank and0% T is established while preventing the source’s radiation from reachingthe detector. If the absorbance is too high, precision can be improved byresetting 100% T using a standard solution of the analyte whose concentrationis less than that of the sample (Figure 10.41a). For a sample whoseabsorbance is too low, precision can be improved by redefining 0% T usinga standard solution of the analyte whose concentration is greater than thatof the analyte (Figure 10.41b). In this case a calibration curve is requiredbecause a linear relationship between absorbance and concentration nolonger exists. Precision can be further increased by combining these twomethods (Figure 10.41c). Again, a calibration curve is necessary since therelationship between absorbance and concentration is no longer linear.SensitivitySee Figure 10.24 for an example of howthe choice of wavelength affects a calibrationcurve’s sensitivity.The sensitivity of a molecular absorption method, which is the slope of aBeer’s law calibration curve, is the product of the analyte’s absorptivity andthe pathlength of the sample cell (eb). You can improve a method’s sensitivityby selecting a wavelength where absorbance is at a maximum or byincreasing the pathlength.Se l e c t i v i t ySelectivity is rarely a problem in molecular absorption spectrophotometry.In many cases it is possible to find a wavelength where only the analyte absorbs.When two or more species do contribute to the measured absorbance,a multicomponent analysis is still possible, as shown in Example 10.6 andExample 10.7.Tim e , Co s t, a n d Eq u i pm e n tThe analysis of a sample by molecular absorption spectroscopy is relativelyrapid, although additional time may be required if we need to chemicallyconvert a nonabsorbing analyte into an absorbing form. The cost of UV/

Chapter 10 Spectroscopic Methods599Vis instrumentation ranges from several hundred dollars for a simple filterphotometer, to more than $50,000 for a computer controlled high resolution,double-beam instrument equipped with variable slits, and operatingover an extended range of wavelengths. Fourier transform infrared spectrometerscan be obtained for as little as $15,000–$20,000, although moreexpensive models are available.10DAtomic Absorption SpectroscopyGuystav Kirchoff and Robert Bunsen first used atomic absorption spectroscopy—alongwith atomic emission—in 1859 and 1860 as a means foridentify atoms in flames and hot gases. Although atomic emission continuedto develop as an analytical technique, progress in atomic absorptionlanguished for almost a century. Modern atomic absorption spectroscopyhas its beginnings in 1955 as a result of the independent work of A. C.Walsh and C. T. J. Alkemade. 13 Commercial instruments were in place bythe early 1960s, and the importance of atomic absorption as an analyticaltechnique was soon evident.10D.1 InstrumentationAtomic absorption spectrophotometers use the same single-beam or double-beamoptics described earlier for molecular absorption spectrophotometers(see Figure 10.26 and Figure 10.27). There is, however, an importantadditional need in atomic absorption spectroscopy—we must covert theanalyte into free atoms. In most cases our analyte is in solution form. If oursample is a solid, then we must bring it into solution before the analysis.When analyzing a lake sediment for Cu, Zn, and Fe, for example, we bringthe analytes into solution as Cu 2+ , Zn 2+ , and Fe 3+ by extracting them with asuitable reagent. For this reason, only the introduction of solution samplesis considered in this text.At o m i z at i o nThe process of converting an analyte to a free gaseous atom is called atomization.Converting an aqueous analyte into a free atom requires that westrip away the solvent, volatilize the analytes, and, if necessary, dissociatethe analyte into free atoms. Desolvating an aqueous solution of CuCl 2 , forexample, leaves us with solid particulates of CuCl 2 . Converting the particulateCuCl 2 to gas phases atoms of Cu and Cl requires thermal energy.CuCl CuCl Cu Cl2 ( aq) → () s → ( g) + 22 ( g )What reagent we choose to use dependson our research goals. If we need to knowthe total amount of metal in the sediment,then we might use a microwave digestionusing a mixture of concentrated acids,such as HNO 3 , HCl, and HF. This destroysthe sediment’s matrix and bringseverything into solution. On the otherhand, if our interest is biologically availablemetals, we might extract the sampleunder milder conditions, such as a dilutesolution of HCl or CH 3 COOH at roomtemperature.There are two common atomization methods: flame atomization and electrothermalatomization, although a few elements are atomized using othermethods.13 (a) Walsh, A. Anal. Chem. 1991, 63, 933A–941A; (b) Koirtyohann, S. R. Anal. Chem. 1991, 63,1024A–1031A; (c) Slavin, W. Anal. Chem. 1991, 63, 1033A–1038A.

600 Analytical Chemistry 2.0(a)burner slotburner headnebulizerspraychambercapillarytubeFigure 10.42 Flame atomization assemblywith expanded views of (a) the burnerhead showing the burner slot where theflame is located; (b) the nebulizer’s impactbead; and (c) the interior of the spraychamber. Although the unit shown here isfrom an older instrument, the basic componentsof a modern flame AA spectrometerare the same.(b)air inletimpactbeadwaste lineacetylene inlet(c)interior of spray chamberFl a m e At o m i z e rCompressed air is one of the two gaseswhose combustion produces the flame.Figure 10.42 shows a typical flame atomization assembly with close-upviews of several key components. In the unit shown here, the aqueoussample is drawn into the assembly by passing a high-pressure stream ofcompressed air past the end of a capillary tube immersed in the sample.When the sample exits the nebulizer it strikes a glass impact bead, convertingit into a fine aerosol mist within the spray chamber. The aerosol mist isswept through the spray chamber by the combustion gases—compressedair and acetylene in this case—to the burner head where the flame’s thermalenergy desolvates the aerosol mist to a dry aerosol of small, solid particles.The flame’s thermal energy then volatilizes the particles, producing a vaporconsisting of molecular species, ionic species, and free atoms.Burner. The slot burner in Figure 10.42a provides a long optical pathlengthand a stable flame. Because absorbance increases linearly with thepath length, a long path length provides greater sensitivity. A stable flameminimizes uncertainty due to fluctuations in the flame.The burner is mounted on an adjustable stage that allows the entireassembly to move horizontally and vertically. Horizontal adjustments ensurethat the flame is aligned with the instrument’s optical path. Vertical

Chapter 10 Spectroscopic Methods601adjustments adjust the height within the flame from which absorbance ismonitored. This is important because two competing processes affect theconcentration of free atoms in the flame. The more time the analyte spendsin the flame the greater the atomization efficiency; thus, the production offree atoms increases with height. On the other hand, a longer residencetime allows more opportunity for the free atoms to combine with oxygento form a molecular oxide. For an easily oxidized metal, such as Cr, the concentrationof free atoms is greatest just above the burner head. For metals,such as Ag, which are difficult to oxidize, the concentration of free atomsincreases steadily with height (Figure 10.43). Other atoms show concentrationprofiles that maximize at a characteristic height.Flame. The flame’s temperature, which affects the efficiency of atomization,depends on the fuel–oxidant mixture, several examples of which are listedin Table 10.9. Of these, the air–acetylene and the nitrous oxide–acetyleneflames are the most popular. Normally the fuel and oxidant are mixed inan approximately stoichiometric ratio; however, a fuel-rich mixture may benecessary for easily oxidized analytes.Figure 10.44 shows a cross-section through the flame, looking downthe source radiation’s optical path. The primary combustion zone is usuallyrich in gas combustion products that emit radiation, limiting is usefulnessfor atomic absorption. The interzonal region generally is rich in free atomsand provides the best location for measuring atomic absorption. The hottestpart of the flame is typically 2–3 cm above the primary combustion zone.As atoms approach the flame’s secondary combustion zone, the decrease intemperature allows for formation of stable molecular species.Sample Introduction. The most common means for introducing samplesinto a flame atomizer is a continuous aspiration in which the sample flowsthrough the burner while we monitor the absorbance. Continuous aspirationis sample intensive, typically requiring from 2–5 mL of sample.Flame microsampling allows us to introduce a discrete sample of fixedvolume, and is useful when we have a limited amount of sample or whenthe sample’s matrix is incompatible with the flame atomizer. For example,continuously aspirating a sample that has a high concentration of dissolvedsolids—sea water, for example, comes to mind—may build-up a solid depositon the burner head that obstructs the flame and that lowers the absorbance.Flame microsampling is accomplished using a micropipet to placeTable 10.9 Fuels and Oxidants Used for Flame Combustionfuel oxidant temperature range ( o C)natural gas air 1700–1900hydrogen air 2000–2100acetylene air 2100–2400acetylene nitrous oxide 2600–2800acetylene oxygen 3050–3150absorbanceFigure 10.43 Absorbance versus heightprofiles for Ag and Cr in flame atomicabsorption spectroscopy.secondarycombustion zonetypicaloptical pathheight above burner headburner headAgFigure 10.44 Profile of typical flameusing a slot burner. The relative size ofeach zone depends on many factors,including the choice of fuel and oxidant,and their relative proportions.Crinterzonalregionprimarycombustion zone

602 Analytical Chemistry 2.0This is the reason for the waste line shownat the bottom of the spray chamber in Figure10.42.50–250 mL of sample in a Teflon funnel connected to the nebulizer, or bydipping the nebulizer tubing into the sample for a short time. Dip samplingis usually accomplished with an automatic sampler. The signal for flamemicrosampling is a transitory peak whose height or area is proportional tothe amount of analyte that is injected.Advantages and Disadvantages of Flame Atomization. The principal advantageof flame atomization is the reproducibility with which the sample isintroduced into the spectrophotometer. A significant disadvantage to flameatomizers is that the efficiency of atomization may be quite poor. Thereare two reasons for poor atomization efficiency. First, the majority of theaerosol droplets produced during nebulization are too large to be carriedto the flame by the combustion gases. Consequently, as much as 95% ofthe sample never reaches the flame. A second reason for poor atomizationefficiency is that the large volume of combustion gases significantly dilutesthe sample. Together, these contributions to the efficiency of atomizationreduce sensitivity because the analyte’s concentration in the flame may bea factor of 2.5 10 –6 less than that in solution. 14El e c t r o t h e r m a l At o m i z e rsA significant improvement in sensitivity is achieved by using the resistiveheating of a graphite tube in place of a flame. A typical electrothermal atomizer,also known as a graphite furnace, consists of a cylindrical graphitetube approximately 1–3 cm in length and 3–8 mm in diameter. As shownin Figure 10.45, the graphite tube is housed in an sealed assembly that hasoptically transparent windows at each end. A continuous stream of an inertgas is passed through the furnace, protecting the graphite tube from oxidationand removing the gaseous products produced during atomization. Apower supply is used to pass a current through the graphite tube, resultingin resistive heating.Samples of between 5–50 mL are injected into the graphite tube througha small hole at the top of the tube. Atomization is achieved in three stages.In the first stage the sample is dried to a solid residue using a current thatraises the temperature of the graphite tube to about 110 o C. In the secondstage, which is called ashing, the temperature is increased to between14 Ingle, J. D.; Crouch, S. R. Spectrochemical Analysis, Prentice-Hall: Englewood Cliffs, NJ, 1988;p. 275.opticalwindowsampleopticalwindowoptical pathoptical pathFigure 10.45 Diagram showing a crosssectionof an electrothermal analyzer.graphite tubeinert gas inlets

Chapter 10 Spectroscopic Methods603350–1200 o C. At these temperatures any organic material in the sample isconverted to CO 2 and H 2 O, and volatile inorganic materials are vaporized.These gases are removed by the inert gas flow. In the final stage the sample isatomized by rapidly increasing the temperature to between 2000–3000 o C.The result is a transient absorbance peak whose height or area is proportionalto the absolute amount of analyte injected into the graphite tube.Together, the three stages take approximately 45–90 s, with most of thistime used for drying and ashing the sample.Electrothermal atomization provides a significant improvement in sensitivityby trapping the gaseous analyte in the small volume within thegraphite tube. The analyte’s concentration in the resulting vapor phase maybe as much as 1000 greater than in a flame atomization. 15 This improvementin sensitivity—and the resulting improvement in detection limits—is offset by a significant decrease in precision. Atomization efficiency isstrongly influenced by the sample’s contact with the graphite tube, whichis difficult to control reproducibly.Mi s ce l l a n e o u s At o m i z at i o n Me t h o d sA few elements may be atomized by a chemical reaction that produces avolatile product. Elements such as As, Se, Sb, Bi, Ge, Sn, Te, and Pb, forexample, form volatile hydrides when reacted with NaBH 4 in acid. Aninert gas carries the volatile hydrides to either a flame or to a heated quartzobservation tube situated in the optical path. Mercury is determined by thecold-vapor method in which it is reduced to elemental mercury with SnCl 2 .The volatile Hg is carried by an inert gas to an unheated observation tubesituated in the instrument’s optical path.10D.2 Quantitative ApplicationsAtomic absorption is widely used for the analysis of trace metals in a varietyof sample matrices. Using Zn as an example, atomic absorption methodshave been developed for its determination in samples as diverse as waterand wastewater, air, blood, urine, muscle tissue, hair, milk, breakfast cereals,shampoos, alloys, industrial plating baths, gasoline, oil, sediments, androcks.Developing a quantitative atomic absorption method requires severalconsiderations, including choosing a method of atomization, selecting thewavelength and slit width, preparing the sample for analysis, minimizingspectral and chemical interferences, and selecting a method of standardization.Each of these topics is considered in this section.De v e l o p i n g a Qu a n t i t a t i ve Me t h o dFlame or Electrothermal Atomization? The most important factor inchoosing a method of atomization is the analyte’s concentration. Because15 Parsons, M. L.; Major, S.; Forster, A. R. Appl. Spectrosc. 1983, 37, 411–418.

604 Analytical Chemistry 2.0Because atomic absorption lines are narrow,we need to use a line source instead ofa continuum source (compare, for example,Figure 10.18 with Figure 10.20). Theeffective bandwidth when using a continuumsource is roughly 1000 larger thanan atomic absorption line; thus, P T ≈ P 0 ,%T ≈ 100, and A ≈ 0. Because a hollowcathode lamp is a line source, P T and P 0have different values giving a %T < 100and A > 0.Table 10.10 Concentration of Analyte Yielding an Absorbance of 0.20Concentration (mg/L) aelement flame atomization electrothermal atomizationAg 1.5 0.0035Al 40 0.015As 40 b 0.050Ca 0.8 0.003Cd 0.6 0.001Co 2.5 0.021Cr 2.5 0.0075Cu 1.5 0.012Fe 2.5 0.006Hg 70 b 0.52Mg 0.15 0.00075Mn 1 0.003Na 0.3 0.00023Ni 2 0.024Pb 5 0.080Pt 70 0.29Sn 50 b 0.023Zn 0.3 0.00071a Source: Varian Cookbook, SpectraAA Software Version 4.00 Pro.b As: 10 mg/L by hydride vaporization; Hg: 11.5 mg/L by cold-vapor; and Sn:18 mg/L by hydride vaporizationof its greater sensitivity, it takes less analyte to achieve a given absorbancewhen using electrothermal atomization. Table 10.10, which compares theamount of analyte needed to achieve an absorbance of 0.20 when usingflame atomization and electrothermal atomization, is useful when selectingan atomization method. For example, flame atomization is the method ofchoice if our samples contain 1–10 mg Zn 2+ /L, but electrothermal atomizationis the best choice for samples containing 1–10 mg Zn 2+ /L.Selecting the Wavelength and Slit Width. The source for atomic absorptionis a hollow cathode lamp consisting of a cathode and anode enclosedwithin a glass tube filled with a low pressure of Ne or Ar (Figure 10.46). Applyinga potential across the electrodes ionizes the filler gas. The positivelycharged gas ions collide with the negatively charged cathode, sputteringatoms from the cathode’s surface. Some of the sputtered atoms are in theexcited state and emit radiation characteristic of the metal(s) from whichthe cathode was manufactured. By fashioning the cathode from the metallicanalyte, a hollow cathode lamp provides emission lines that correspondto the analyte’s absorption spectrum.

Chapter 10 Spectroscopic Methods605hν + MM*opticalwindowMNe +anodecathodeFigure 10.46 Photo of a typical multielemental hollow cathode lamp. The cathode in this lampis fashioned from an alloy containing Co, Cr, Cu, Fe, Mn, and Ni, and is surrounded by a glassshield to isolate it from the anode. The lamp is filled with Ne gas. Also shown is the processleading to atomic emission. See the text for an explanation.Each element in a hollow cathode lamp provides several atomic emissionlines that we can use for atomic absorption. Usually the wavelengththat provides the best sensitivity is the one we choose to use, although a lesssensitive wavelength may be more appropriate for a larger concentration ofanalyte. For the Cr hollow cathode lamp in Table 10.11, for example, thebest sensitivity is obtained using a wavelength of 357.9 nm.Another consideration is the intensity of the emission line. If severalemission lines meet our need for sensitivity, we may wish to use the emissionline with the largest relative P 0 because there is less uncertainty in measuringP 0 and P T . When analyzing samples containing ≈10 mg Cr/L, forexample, the first three wavelengths in Table 10.11 provide an appropriatesensitivity. The wavelengths of 425.5 nm and 429.0 nm, however, have agreater P 0 and will provide less uncertainty in the measured absorbance.The emission spectrum from a hollow cathode lamp includes, besidesemission lines for the analyte, additional emission lines for impurities presentin the metallic cathode and from the filler gas. These additional linesTable 10.11 Atomic Emission Lines for a Cr Hollow Cathode Lampwavelength (nm) slit width (nm) mg Cr/L giving A = 0.20 P 0 (relative)357.9 0.2 2.5 40425.4 0.2 12 85429.0 0.5 20 100520.5 0.2 1500 15520.8 0.2 500 20

606 Analytical Chemistry 2.0See Chapter 7 to review different methodsfor preparing samples for analysis.are a source of stray radiation that leads to an instrumental deviation fromBeer’s law. The monochromator’s slit width is set as wide as possible, improvingthe throughput of radiation, while, at the same time, being narrowenough to eliminate the stray radiation.Preparing the Sample. Flame and electrothermal atomization require thatthe sample be in solution. Solid samples are brought into solution by dissolvingin an appropriate solvent. If the sample is not soluble it may bedigested, either on a hot-plate or by microwave, using HNO 3 , H 2 SO 4 , orHClO 4 . Alternatively, we can extract the analyte using a Soxhlet extractor.Liquid samples may be analyzed directly or extracted if the matrix is incompatiblewith the method of atomization. A serum sample, for instance, isdifficult to aspirate when using flame atomization and may produce an unacceptablyhigh background absorbance when using electrothermal atomization.A liquid–liquid extraction using an organic solvent and a chelatingagent is frequently used to concentrate analytes. Dilute solutions of Cd 2+ ,Co 2+ , Cu 2+ , Fe 3+ , Pb 2+ , Ni 2+ , and Zn 2+ , for example, can be concentratedby extracting with a solution of ammonium pyrrolidine dithiocarbamatein methyl isobutyl ketone.Minimizing Spectral Interference. A spectral interference occurs whenan analyte’s absorption line overlaps with an interferent’s absorption lineor band. Because they are so narrow, the overlap of two atomic absorptionlines is seldom a problem. On the other hand, a molecule’s broad absorptionband or the scattering of source radiation is a potentially serious spectralinterference.An important consideration when using a flame as an atomizationsource is its effect on the measured absorbance. Among the products ofcombustion are molecular species that exhibit broad absorption bands andparticulates that scatter radiation from the source. If we fail to compensatefor these spectral interference, then the intensity of transmitted radiationdecreases. The result is an apparent increase in the sample’s absorbance. Fortunately,absorption and scattering of radiation by the flame are correctedby analyzing a blank.Spectral interferences also occur when components of the sample’s matrixother than the analyte react to form molecular species, such as oxidesand hydroxides. The resulting absorption and scattering constitutes thesample’s background and may present a significant problem, particularly atwavelengths below 300 nm where the scattering of radiation becomes moreimportant. If we know the composition of the sample’s matrix, then we canprepare our samples using an identical matrix. In this case the backgroundabsorption is the same for both the samples and standards. Alternatively,if the background is due to a known matrix component, then we can addthat component in excess to all samples and standards so that the contributionof the naturally occurring interferent is insignificant. Finally, manyinterferences due to the sample’s matrix can be eliminated by increasing the

Chapter 10 Spectroscopic Methods607atomization temperature. For example, by switching to a higher temperatureflame it may be possible to prevent the formation of interfering oxidesand hydroxides.If the identity of the matrix interference is unknown, or if it is not possibleto adjust the flame or furnace conditions to eliminate the interference,then we must find another method to compensate for the background interference.Several methods have been developed to compensate for matrixinterferences, and most atomic absorption spectrophotometers include oneor more of these methods.One of the most common methods for background correction is touse a continuum source, such as a D 2 lamp. Because a D 2 lamp is a continuumsource, absorbance of its radiation by the analyte’s narrow absorptionline is negligible. Only the background, therefore, absorbs radiation fromthe D 2 lamp. Both the analyte and the background, on the other hand, absorbthe hollow cathode’s radiation. Subtracting the absorbance for the D 2lamp from that for the hollow cathode lamp gives a corrected absorbancethat compensates for the background interference. Although this methodof background correction may be quite effective, it does assume that thebackground absorbance is constant over the range of wavelengths passedby the monochromator. If this is not true, subtracting the two absorbancesmay underestimate or overestimate the background.Minimizing Chemical Interferences. The quantitative analysis of someelements is complicated by chemical interferences occurring during atomization.The two most common chemical interferences are the formationof nonvolatile compounds containing the analyte and ionization of theanalyte.One example of the formation of nonvolatile compounds is the effect ofPO 4 3– or Al 3+ on the flame atomic absorption analysis of Ca 2+ . In one study,for example, adding 100 ppm Al 3+ to a solution of 5 ppm Ca 2+ decreasedthe calcium ion’s absorbance from 0.50 to 0.14, while adding 500 ppmPO 4 3– to a similar solution of Ca 2+ decreased the absorbance from 0.50 to0.38. These interferences were attributed to the formation of nonvolatileparticles of Ca 3 (PO 4 ) 2 and an Al–Ca–O oxide. 16When using flame atomization, we can minimize the formation ofnonvolatile compounds by increasing the flame’s temperature, either bychanging the fuel-to-oxidant ratio or by switching to a different combinationof fuel and oxidant. Another approach is to add a releasing agentor a protecting agent to the samples. A releasing agent is a species thatreacts with the interferent, releasing the analyte during atomization. AddingSr 2+ or La 3+ to solutions of Ca 2+ , for example, minimizes the effect ofPO 43–and Al 3+ by reacting in place of the analyte. Thus, adding 2000 ppmSrCl 2 to the Ca 2+ /PO 4 3– and Ca 2+ /Al 3+ mixtures described in the previousparagraph increased the absorbance to 0.48. A protecting agent reactswith the analyte to form a stable volatile complex. Adding 1% w/w EDTA16 Hosking, J. W.; Snell, N. B.; Sturman, B. T. J. Chem. Educ. 1977, 54, 128–130.Other methods of background correctionhave been developed, including Zeemaneffect background correction and Smith–Hieftje background correction, both ofwhich are included in some commerciallyavailable atomic absorption spectrophotometers.Consult the chapter’s additionalresources for additional information.

608 Analytical Chemistry 2.0Most instruments include several differentalgorithms for computing the calibrationcurve. The instrument in my lab, forexample, includes five algorithms. Threeof the algorithms fit absorbance data usinglinear, quadratic, or cubic polynomialfunctions of the analyte’s concentration. Italso includes two algorithms that fit theconcentrations of the standards to quadraticfunctions of the absorbance.The best way to appreciate the theoreticaland practical details discussed in this sectionis to carefully examine a typical analyticalmethod. Although each method isunique, the following description of thedetermination of Cu and Zn in biologicaltissues provides an instructive example ofa typical procedure. The description hereis based on Bhattacharya, S. K.; Goodwin,T. G.; Crawford, A. J. Anal. Lett. 1984,17, 1567–1593, and Crawford, A. J.;Bhattacharya, S. K. Varian Instruments atWork, Number AA–46, April 1985.to the Ca 2+ /PO 4 3– solution described in the previous paragraph increasedthe absorbance to 0.52.Ionization interferences occur when thermal energy from the flame orthe electrothermal atomizer is sufficient to ionize the analyteM( g) M+ ( g)+ e−10.24where M is the analyte. Because the absorption spectra for M and M + aredifferent, the position of the equilibrium in reaction 10.24 affects absorbanceat wavelengths where M absorbs. To limit ionization we add a highconcentration of an ionization suppressor, which is simply a species thationizes more easily than the analyte. If the concentration of the ionizationsuppressor is sufficient, then the increased concentration of electrons inthe flame pushes reaction 10.24 to the left, preventing the analyte’s ionization.Potassium and cesium are frequently used as an ionization suppressorbecause of their low ionization energy.Standardizing the Method. Because Beer’s law also applies to atomic absorption,we might expect atomic absorption calibration curves to be linear.In practice, however, most atomic absorption calibration curves are nonlinear,or linear for only a limited range of concentrations. Nonlinearity inatomic absorption is a consequence of instrumental limitations, includingstray radiation from the hollow cathode lamp and the variation in molarabsorptivity across the absorption line. Accurate quantitative work, therefore,often requires a suitable means for computing the calibration curvefrom a set of standards.When possible, a quantitative analysis is best conducted using externalstandards. Unfortunately, matrix interferences are a frequent problem,particularly when using electrothermal atomization. For this reason themethod of standard additions is often used. One limitation to this methodof standardization, however, is the requirement that there be a linear relationshipbetween absorbance and concentration.Representative Method 10.2Determination of Cu and Zn in Tissue SamplesDescription o f Me t h o dCopper and zinc are isolated from tissue samples by digesting the samplewith HNO 3 after first removing any fatty tissue. The concentration ofcopper and zinc in the supernatant are determined by atomic absorptionusing an air-acetylene flame.Pr o c e d u r eTissue samples are obtained by a muscle needle biopsy and dried for 24–30 h at 105 o C to remove all traces of moisture. The fatty tissue in thedried samples is removed by extracting overnight with anhydrous ether.After removing the ether, the sample is dried to obtain the fat-free dry

Chapter 10 Spectroscopic Methods609tissue weight (FFDT). The sample is digested at 68 o C for 20–24 h using3 mL of 0.75 M HNO 3 . After centrifuging at 2500 rpm for 10 minutes,the supernatant is transferred to a 5-mL volumetric flask. The digestionis repeated two more times, for 2–4 hours each, using 0.9-mL aliquotsof 0.75 M HNO 3 . These supernatants are added to the 5-mL volumetricflask, which is diluted to volume with 0.75 M HNO 3 . The concentrationsof Cu and Zn in the diluted supernatant are determined by flame atomicabsorption spectroscopy using an air-acetylene flame and external standards.Copper is analyzed at a wavelength of 324.8 nm with a slit widthof 0.5 nm, and zinc is analyzed at 213.9 nm with a slit width of 1.0 nm.Background correction using a D 2 lamp is necessary for zinc. Results arereported as mg of Cu or Zn per gram of FFDT.Qu e s t i o n s1. Describe the appropriate matrix for the external standards and for theblank?The matrix for the standards and the blank should match the matrixof the samples; thus, an appropriate matrix is 0.75 M HNO 3 . Anyinterferences from other components of the sample matrix are minimizedby background correction.2. Why is a background correction necessary for the analysis of Zn, butnot for the analysis of Cu?Background correction compensates for background absorption andscattering due to interferents in the sample. Such interferences aremost severe when using a wavelength less than 300 nm. This is thecase for Zn, but not for Cu.3. A Cu hollow cathode lamp has several emission lines. Explain whythis method uses the line at 324.8 nm.wavelength(nm)slit width(nm)mg Cu/L forA = 0.20 P 0 (relative)217.9 0.2 15 3218.2 0.2 15 3222.6 0.2 60 5244.2 0.2 400 15249.2 0.5 200 24324.8 0.5 1.5 100327.4 0.5 3 87With 1.5 mg Cu/L giving an absorbance of 0.20, the emissionline at 324.8 nm has the best sensitivity. In addition, it is themost intense emission line, which decreases the uncertainty inthe measured absorbance.

610 Analytical Chemistry 2.0Example 10.10To evaluate the method described in Representative Method 10.2, a seriesof external standard is prepared and analyzed, providing the results shownhere. 17 mg Cu/mL absorbance mg Cu/mL absorbance0.000 0.000 0.500 0.0330.100 0.006 0.600 0.0390.200 0.013 0.700 0.0460.300 0.020 1.000 0.0660.400 0.026A bovine liver standard reference material was used to evaluate the method’saccuracy. After drying and extracting the sample, a 11.23-mg FFDTtissue sample gives an absorbance of 0.023. Report the amount of copperin the sample as mg Cu/g FFDT.absorbance0.080.060.040.020.000.0 0.2 0.4 0.6 0.8 1.0μg Cu/mLSo l u t i o nLinear regression of absorbance versus the concentration of Cu in the standardsgives a calibration curve with the following equation.µgCuA = − 0. 0002 + 0.0661×mLSubstituting the sample’s absorbance into the calibration equation givesthe concentration of copper as 0.351 mg/mL. The concentration of copperin the tissue sample, therefore, is0.351 µ gCu× 5.000 mLmL= 156 µgCu/g FFDT0.01123 gsample10D.3 - Evaluation of Atomic Absorption SpectroscopySc a l e o f Op e r a t i o nSee Figure 3.5 to review the meaning ofmacro and meso for describing samples,and the meaning of major, minor, and ultratracefor describing analytes.Atomic absorption spectroscopy is ideally suited for the analysis of trace andultratrace analytes, particularly when using electrothermal atomization. Forminor and major analyte, sample can be diluted before the analysis. Mostanalyses use a macro or a meso sample. The small volume requirement forelectrothermal atomization or flame microsampling, however, makes practicalthe analysis micro and ultramicro samples.17 Crawford, A. J.; Bhattacharya, S. K. “Microanalysis of Copper and Zinc in Biopsy-Sized TissueSpecimens by Atomic Absorption Spectroscopy Using a Stoichiometric Air-Acetylene Flame,”Varian Instruments at Work, Number AA–46, April 1985.

Chapter 10 Spectroscopic Methods611Ac c u r a c yIf spectral and chemical interferences are minimized, an accuracy of 0.5–5%is routinely attainable. When the calibration curve is nonlinear, accuracymay be improved by using a pair of standards whose absorbances closelybracket the sample’s absorbance and assuming that the change in absorbanceis linear over this limited concentration range. Determinate errors forelectrothermal atomization are often greater than that obtained with flameatomization due to more serious matrix interferences.PrecisionFor absorbance values greater than 0.1–0.2, the relative standard deviationfor atomic absorption is 0.3–1% for flame atomization and 1–5% forelectrothermal atomization. The principle limitation is the variation in theconcentration of free analyte atoms resulting from variations in the rateof aspiration, nebulization, and atomization when using a flame atomizer,and the consistency of injecting samples when using electrothermalatomization.SensitivityThe sensitivity of a flame atomic absorption analysis is influenced stronglyby the flame’s composition and by the position in the flame from whichwe monitor the absorbance. Normally the sensitivity of an analysis is optimizedby aspirating a standard solution of the analyte and adjusting operatingconditions, such as the fuel-to-oxidant ratio, the nebulizer flow rate,and the height of the burner, to give the greatest absorbance. With electrothermalatomization, sensitivity is influenced by the drying and ashingstages that precede atomization. The temperature and time used for eachstage must be optimized for each type of sample.Sensitivity is also influenced by the sample’s matrix. We have alreadynoted, for example, that sensitivity can be decreased by chemical interferences.An increase in sensitivity may be realized by adding a low molecularweight alcohol, ester, or ketone to the solution, or by using an organicsolvent.See Chapter 14 for several strategies foroptimizing experiments.Se l e c t i v i t yDue to the narrow width of absorption lines, atomic absorption providesexcellent selectivity. Atomic absorption can be used for the analysis of over60 elements at concentrations at or below the level of mg/L.Tim e , Co s t, a n d Eq u i pm e n tThe analysis time when using flame atomization is short, with samplethroughputs of 250–350 determinations per hour when using a fully automatedsystem. Electrothermal atomization requires substantially more time

612 Analytical Chemistry 2.0per analysis, with maximum sample throughputs of 20–30 determinationsper hour. The cost of a new instrument ranges from between $10,000–$50,000 for flame atomization, and from $18,000–$70,000 for electrothermalatomization. The more expensive instruments in each price rangeinclude double-beam optics, automatic samplers, and can be programmedfor multielemental analysis by allowing the wavelength and hollow cathodelamp to be changed automatically.10EEmission SpectroscopyAn analyte in an excited state possesses an energy, E 2 , that is greater than itsenergy when it is in a lower energy state, E 1 . When the analyte returns to itslower energy state—a process we call relaxation—the excess energy, DE∆E = E 2−E1must be released. Figure 10.4 shows a simplified picture of this process.The amount of time the analyte spends in the excited state—its lifetime—isshort, typically 10 –5 –10 –9 s for electronic excited states and 10 –15 sfor vibrational excited states. Relaxation of an analyte’s excited-state, A*, occursthrough several mechanisms, including collisions with other species inthe sample, by photochemical reactions, and by the emission of photons.In the first process, which is called vibrational relaxation, or nonradiativerelaxation, the excess energy is released as heat.A* → A+heatRelaxation by a photochemical reaction may involve a decomposition reactionA* → X+Yor a reaction between A* and another speciesA * + Z→ X+YIn both cases the excess energy is used up in the chemical reaction or releasedas heat.In the third mechanism, the excess energy is released as a photon ofelectromagnetic radiation.A* → A+hνThe release of a photon following thermal excitation is called emission andthat following the absorption of a photon is called photoluminescence. Inchemiluminescence and bioluminescence, excitation results from a chemicalor biochemical reaction, respectively. Spectroscopic methods based onphotoluminescence are the subject of the next section and atomic emissionis covered in Section 10G.

Chapter 10 Spectroscopic Methods613(a) (b) (c)10Fsingletground statesingletexcited statePhotoluminescence Spectroscopytripletexcited statePhotoluminescence is divided into two categories: fluorescence and phosphorescence.A pair of electrons occupying the same electronic ground statehave opposite spins and are said to be in a singlet spin state (Figure 10.47a).When an analyte absorbs an ultraviolet or visible photon, one of its valenceelectrons moves from the ground state to an excited state with a conservationof the electron’s spin (Figure 10.47b). Emission of a photon from thesinglet excited state to the singlet ground state—or between any twoenergy levels with the same spin—is called fluorescence. The probabilityof fluorescence is very high and the average lifetime of an electron in theexcited state is only 10 –5 –10 –8 s. Fluorescence, therefore, decays rapidlyonce the source of excitation is removed.In some cases an electron in a singlet excited state is transformed to atriplet excited state (Figure 10.47c) in which its spin is no longer pairedwith the ground state. Emission between a triplet excited state and a singletground state—or between any two energy levels that differ in their respectivespin states–is called phosphorescence. Because the average lifetimefor phosphorescence ranges from 10 –4 –10 4 s, phosphorescence may continuefor some time after removing the excitation source.The use of molecular fluorescence for qualitative analysis and semiquantitativeanalysis can be traced to the early to mid 1800s, with moreaccurate quantitative methods appearing in the 1920s. Instrumentationfor fluorescence spectroscopy using a filter or a monochromator for wavelengthselection appeared in, respectively, the 1930s and 1950s. Althoughthe discovery of phosphorescence preceded that of fluorescence by almost200 years, qualitative and quantitative applications of molecular phosphorescencedid not receive much attention until after the development offluorescence instrumentation.10F.1 Fluorescence and Phosphorescence SpectraTo appreciate the origin of fluorescence and phosphorescence we must considerwhat happens to a molecule following the absorption of a photon.Let’s assume that the molecule initially occupies the lowest vibrational energylevel of its electronic ground state, which is a singlet state labeled S 0 inFigure 10.47 Electron configurations for(a) a singlet ground state; (b) a singlet excitedstate; and (c) a triplet excited state.

614 Analytical Chemistry 2.0S 2vrvrvrS 1vricvrvrvrT 1vriscic ecic ecvrvrvrisc ecvrvrvrvrvrvrvrvrvrvrvrvrvrvrvrvrvrvrS vr0vrvrvrvrvrvrvrvrabsorptionfluorescenceabsorptionFigure 10.48 Energy level diagram for a molecule showing pathways for the deactivation of an excited state: vr is vibrationalrelaxation; ic is internal conversion; ec is external conversion; and isc is an intersystem crossing. The lowestvibrational energy for each electronic state is indicated by the thicker line. The electronic ground state is shown in blackand the three electronic excited states are shown in green. The absorption, fluorescence, and phosphorescence of photonsalso are shown.fluorescenceFigure 10.48. Absorption of a photon excites the molecule to one of severalvibrational energy levels in the first excited electronic state, S 1 , or the secondelectronic excited state, S 2 , both of which are singlet states. Relaxationto the ground state occurs by a number of mechanisms, some involving theemission of photons and others occurring without emitting photons. Theserelaxation mechanisms are shown in Figure 10.48. The most likely relaxationpathway is the one with the shortest lifetime for the excited state.phosphorescenceRadiationless De a c t i v a t i o nWhen a molecule relaxes without emitting a photon we call the processradiationless deactivation. One example of radiationless deactivationis vibrational relaxation, in which a molecule in an excited vibrationalenergy level loses energy by moving to a lower vibrational energy level inthe same electronic state. Vibrational relaxation is very rapid, with an averagelifetime of

Chapter 10 Spectroscopic Methods615to the ground electronic state without emitting a photon. A related formof radiationless deactivation is an external conversion in which excessenergy is transferred to the solvent or to another component of the sample’smatrix.A final form of radiationless deactivation is an intersystem crossingin which a molecule in the ground vibrational energy level of an excitedelectronic state passes into a higher vibrational energy level of a lower energyelectronic state with a different spin state. For example, an intersystemcrossing is shown in Figure 10.48 between a singlet excited state, S 1 , and atriplet excited state, T 1 .Re l a x a t i o n b y Fl u o r e s c e n c eFluorescence occurs when a molecule in an excited state’s lowest vibrationalenergy level returns to a lower energy electronic state by emitting a photon.Because molecules return to their ground state by the fastest mechanism,fluorescence is observed only if it is a more efficient means of relaxationthan a combination of internal conversions and vibrational relaxations.A quantitative expression of fluorescence efficiency is the fluorescentquantum yield, F f , which is the fraction of excited state molecules returningto the ground state by fluorescence. Fluorescent quantum yields rangefrom 1, when every molecule in an excited state undergoes fluorescence, to0 when fluorescence does not occur.The intensity of fluorescence, I f , is proportional to the amount of radiationabsorbed by the sample, P 0 – P T , and the fluorescent quantum yieldI = kΦ ( P −P) 10.25f f 0 Twhere k is a constant accounting for the efficiency of collecting and detectingthe fluorescent emission. From Beer’s law we know thatPTbC= 10−ε 10.26P0where C is the concentration of the fluorescing species. Solving equation10.26 for P T and substituting into equation 10.25 gives, after simplifyingbCI = kΦ P − − ε0( 1 10 )ff10.27When ebC < 0.01, which often is the case when concentration is small,equation 10.27 simplifies toI = 2. 303kΦεbCP = kP ′ff 0 010.28where k′ is a collection of constants. The intensity of fluorescent emission,therefore, increases with an increase in the quantum efficiency, the source’sincident power, and the molar absorptivity and the concentration of thefluorescing species.Fluorescence is generally observed when the molecule’s lowest energyabsorption is a p p* transition, although some n p* transitions showLet’s use Figure 10.48 to illustrate howa molecule can relax back to its groundstate without emitting a photon. Supposeour molecule is in the highest vibrationalenergy level of the second electronic excitedstate. After a series of vibrationalrelaxations brings the molecule to thelowest vibrational energy level of S 2 , itundergoes an internal conversion into ahigher vibrational energy level of the firstexcited electronic state. Vibrational relaxationsbring the molecule to the lowestvibrational energy level of S 1 . Followingan internal conversion into a higher vibrationalenergy level of the ground state,the molecule continues to undergo vibrationalrelaxation until it reaches the lowestvibrational energy level of S 0 .

616 Analytical Chemistry 2.0quinineOHOFigure 10.49 Tonic water, which containsquinine, is fluorescent when placedunder a UV lamp. Source: Splarka (commons.wikipedia.org).NNweak fluorescence. Most unsubstituted, nonheterocyclic aromatic compoundshave favorable fluorescence quantum yields, although substitutionson the aromatic ring can significantly effect F f . For example, the presenceof an electron-withdrawing group, such as –NO 2 , decreases F f , while addingan electron-donating group, such as –OH, increases F f . Fluorescencealso increases for aromatic ring systems and for aromatic molecules withrigid planar structures. Figure 10.49 shows the fluorescence of quinineunder a UV lamp.A molecule’s fluorescent quantum yield is also influenced by externalvariables, such as temperature and solvent. Increasing the temperature generallydecreases F f because more frequent collisions between the moleculeand the solvent increases external conversion. A decrease in the solvent’sviscosity decreases F f for similar reasons. For an analyte with acidic or basicfunctional groups, a change in pH may change the analyte’s structure andits fluorescent properties.As shown in Figure 10.48, fluorescence may return the molecule to anyof several vibrational energy levels in the ground electronic state. Fluorescence,therefore, occurs over a range of wavelengths. Because the change inenergy for fluorescent emission is generally less than that for absorption, amolecule’s fluorescence spectrum is shifted to higher wavelengths than itsabsorption spectrum.Re l a x a t i o n b y Ph o s p h o r e s c e n c eA molecule in a triplet electronic excited state’s lowest vibrational energylevel normally relaxes to the ground state by an intersystem crossing to a singletstate or by an external conversion. Phosphorescence occurs when themolecule relaxes by emitting a photon. As shown in Figure 10.48, phosphorescenceoccurs over a range of wavelengths, all of which are at lower energiesthan the molecule’s absorption band. The intensity of phosphorescence,I p , is given by an equation similar to equation 10.28 for fluorescenceI = 2. 303kΦεbCP = kP ′pp 0 010.29where F p is the phosphorescent quantum yield.Phosphorescence is most favorable for molecules with n p* transitions,which have a higher probability for an intersystem crossing thanp p* transitions. For example, phosphorescence is observed with aromaticmolecules containing carbonyl groups or heteroatoms. Aromaticcompounds containing halide atoms also have a higher efficiency for phosphorescence.In general, an increase in phosphorescence corresponds to adecrease in fluorescence.Because the average lifetime for phosphorescence is very long, rangingfrom 10 –4 –10 4 s, the phosphorescent quantum yield is usually quite small.An improvement in F p is realized by decreasing the efficiency of externalconversion. This may be accomplished in several ways, including loweringthe temperature, using a more viscous solvent, depositing the sample on a

Chapter 10 Spectroscopic Methods617(a) (b) (c)Figure 10.50 An europium doped strontium silicate-aluminum oxide powder under (a) naturallight, (b) a long-wave UV lamp, and (c) in total darkness. The photo taken in total darkness showsthe phosphorescent emission. Source: modified from Splarka (commons.wikipedia.org).solid substrate, or trapping the molecule in solution. Figure 10.50 showsan example of phosphorescence.Exc i t a t i o n v e r s u s Em i s s i o n Sp e c t r aPhotoluminescence spectra are recorded by measuring the intensity ofemitted radiation as a function of either the excitation wavelength or theemission wavelength. An excitation spectrum is obtained by monitoringemission at a fixed wavelength while varying the excitation wavelength.When corrected for variations in the source’s intensity and the detector’sresponse, a sample’s excitation spectrum is nearly identical to its absorbancespectrum. The excitation spectrum provides a convenient means for selectingthe best excitation wavelength for a quantitative or qualitative analysis.In an emission spectrum a fixed wavelength is used to excite thesample and the intensity of emitted radiation is monitored as function ofwavelength. Although a molecule has only a single excitation spectrum, ithas two emission spectra, one for fluorescence and one for phosphorescence.Figure 10.51 shows the UV absorption spectrum and the UV fluorescenceemission spectrum for tyrosine.absorbanceabsorbancespectrumfluorescenceemissionspectrum220 260 300 340 380wavelength (nm)emission intensityFigure 10.51 Absorbance spectrum and fluorescenceemission spectrum for tyrosine in apH 7, 0.1 M phosphate buffer. The emissionspectrum uses an excitation wavelength of260 nm. Source: modified from Mark Somoza(commons.wikipedia.org).

618 Analytical Chemistry 2.0z-axisy-axisP 0sampleP TI f or I px-axisFigure 10.52 Schematic diagram showingthe orientation of the source and the detectorwhen measuring fluorescence and phosphorescence.Contrast this to Figure 10.21,which shows the orientation for absorptionspectroscopy.10F.2 InstrumentationThe basic instrumental needs for monitoring fluorescence and phosphorescence—asource of radiation, a means of selecting a narrow band ofradiation, and a detector—are the same as those for absorption spectroscopy.The unique demands of both techniques, however, require somemodifications to the instrument designs seen earlier in Figure 10.25 (filterphotometer), Figure 10.26 (single-beam spectrophotometer), Figure 10.27(double-beam spectrophotometer), and Figure 10.28 (diode array spectrometer).The most important difference is the detector cannot be placeddirectly across from the source. Figure 10.52 shows why this is the case. Ifwe place the detector along the source’s axis it will receive both the transmittedsource radiation, P T , and the fluorescent, I f , or phosphorescent, I p ,radiation. Instead, we rotate the director and place it at 90 o to the source.In s t r u m e n t s f o r Me a s u r i n g Fl u o r e s c e n c eFigure 10.53 shows the basic design of an instrument for measuring fluorescence,which includes two wavelength selectors, one for selecting anexcitation wavelength from the source and one for selecting the emissionwavelength from the sample. In a fluorimeter the excitation and emissionwavelengths are selected using absorption or interference filters. Theexcitation source for a fluorimeter is usually a low-pressure Hg vapor lampthat provides intense emission lines distributed throughout the ultravioletand visible region (254, 312, 365, 405, 436, 546, 577, 691, and 773nm). When a monochromator is used to select the excitation and emissionwavelengths, the instrument is called a spectrofluorimeter. With amonochromator the excitation source is usually high-pressure Xe arc lamp,which has a continuous emission spectrum. Either instrumental design is

Chapter 10 Spectroscopic Methods619sourceexcitation wavelength selectorλ 1 λ2λ 3ormonochromator filtersamplesignal processormonochromator filterλ 1 λ2λ 3appropriate for quantitative work, although only a spectrofluorimeter canbe used to record an excitation or emission spectrum.The sample cells for molecular fluorescence are similar to those formolecular absorption. Remote sensing with fiber optic probes also can beadapted for use with either a fluorimeter or spectrofluorimeter. An analytethat is fluorescent can be monitored directly. For analytes that are not fluorescent,a suitable fluorescent probe molecule can be incorporated into thetip of the fiber optic probe. The analyte’s reaction with the probe moleculeleads to an increase or decrease in fluorescence.In s t r u m e n t s f o r Me a s u r i n g Ph o s p h o r e s c e n c eInstrumentation for molecular phosphorescence must discriminate betweenphosphorescence and fluorescence. Because the lifetime for fluorescence isshorter than that for phosphorescence, discrimination is easily achieved byincorporating a delay between exciting the sample and measuring phosphorescentemission. Figure 10.54 shows how two out-of-phase chopperscan be use to block emission from reaching the detector when the sampleis being excited, and to prevent source radiation from reaching the samplewhile we are measuring the phosphorescent emission.Because phosphorescence is such a slow process, we must prevent theexcited state from relaxing by external conversion. Traditionally, this hasbeen accomplished by dissolving the sample in a suitable organic solvent,usually a mixture of ethanol, isopentane, and diethylether. The resultingsolution is frozen at liquid-N 2 temperatures, forming an optically clearsolid. The solid matrix minimizes external conversion due to collisions betweenthe analyte and the solvent. External conversion also is minimizedoremission wavelength selectordetectorFigure 10.53 Schematic diagram for measuring fluorescenceshowing the placement of the wavelength selectorsfor excitation and emission. When a filter is used the instrumentis called a fluorimeter, and when a monochromator isused the instrument is called a spectrofluorimeter.fromsourceExcitating SampleopenblockedsampletodetectorMeasuring EmissionfromsourceopenblockedchoppersblockedsampleopentodetectorFigure 10.54 Schematic diagram showinghow choppers are used to prevent fluorescentemission from interfering with themeasurement of phosphorescent emission.

620 Analytical Chemistry 2.0by immobilizing the sample on a solid substrate, making possible roomtemperature measurements. One approach is to place a drop of the solutioncontaining the analyte on a small disc of filter paper. After drying thesample under a heat lamp, the sample is placed in the spectrofluorimeter foranalysis. Other solid surfaces that have been used include silica gel, alumina,sodium acetate, and sucrose. This approach is particularly useful for theanalysis of thin layer chromatography plates.10F.3 Quantitative ApplicationsMolecular fluorescence and, to a lesser extent, phosphorescence have beenused for the direct or indirect quantitative analysis of analytes in a variety ofmatrices. A direct quantitative analysis is possible when the analyte’s fluorescentor phosphorescent quantum yield is favorable. When the analyte isnot fluorescent or phosphorescent, or if the quantum yield is unfavorable,then an indirect analysis may be feasible. One approach is to react theanalyte with a reagent to form a product with fluorescent or phosphorescentproperties. Another approach is to measure a decrease in fluorescenceor phosphorescence when the analyte is added to a solution containing afluorescent or phosphorescent probe molecule. A decrease in emission isobserved when the reaction between the analyte and the probe molecule enhancesradiationless deactivation, or produces a nonemittng product. Theapplication of fluorescence and phosphorescence to inorganic and organicanalytes are considered in this section.In o r g a n i c An a l y t e sHOHOOHNFigure 10.55 Structure of alizarin garnetR and its metal–ligand complexwith Al 3+ .NHOalizarin garnet RONAl 3+NOfluorescent complexOHOHExcept for a few metal ions, most notably UO 2 + , most inorganic ions arenot sufficiently fluorescent for a direct analysis. Many metal ions may be determinedindirectly by reacting with an organic ligand to form a fluorescent,or less commonly, a phosphorescent metal–ligand complex. One example isthe reaction of Al 3+ with the sodium salt of 2, 4, 3′-trihydroxyazobenzene-5′-sulfonic acid—also known as alizarin garnet R—which forms a fluorescentmetal–ligand complex (Figure 10.55). The analysis is carried outusing an excitation wavelength of 470 nm, monitoring fluorescence at 500nm. Table 10.12 provides additional examples of chelating reagents thatform fluorescent metal–ligand complexes with metal ions. A few inorganicnonmetals are determined by their ability to decrease, or quench, the fluorescenceof another species. One example is the analysis for F – based on itsability to quench the fluorescence of the Al 3+ –alizarin garnet R complex.Or g a n i c An a l y t e sAs noted earlier, organic compounds containing aromatic rings generally arefluorescent and aromatic heterocycles are often phosphorescent. As shownin Table 10.13, several important biochemical, pharmaceutical, and environmentalcompounds may be analyzed quantitatively by fluorimetry or

Chapter 10 Spectroscopic Methods621Table 10.12 Chelating Agents for the Fluorescence Analysis of Metal Ionschelating agentmetal ions8-hydroxyquinolineAl 3+ , Be 2+ , Zn 2+ , Li + , Mg 2+ (and others)flavonal Zr 2+ , Sn 4+benzoin B 4 O 2– 7 , Zn 2+2′, 3, 4′, 5, 7-pentahydroxyflavone Be 2+2-(o-hydroxyphenyl) benzoxazole Cd 2+Table 10.13 Examples of Naturally Photoluminescent Organic Analytesclasscompounds (F = fluorescence; P = phosphorescence)aromatic amino acids phenylalanine (F)tyrosine (F)tryptophan (F, P)vitaminsvitamin A (F)vitamin B2 (F)vitamin B6 (F)vitamin B12 (F)vitamin E (F)folic acid (F)catecholaminesdopamine (F)norepinephrine (F)pharmaceuticals and drugs quinine (F)salicylic acid (F, P)morphine (F)barbiturates (F)LSD (F)codeine (P)caffeine (P)sulfanilamide (P)environmental pollutants pyrene (F)benzo[a]pyrene (F)organothiophosphorous pesticides (F)carbamate insecticides (F)DDT (P)phosphorimetry. If an organic analyte is not naturally fluorescent or phosphorescent,it may be possible to incorporate it into a chemical reactionthat produces a fluorescent or phosphorescent product. For example, theenzyme creatine phosphokinase can be determined by using it to catalyzethe formation of creatine from phosphocreatine. Reacting the creatine withninhydrin produces a fluorescent product of unknown structure.

622 Analytical Chemistry 2.0The best way to appreciate the theoreticaland practical details discussed in this sectionis to carefully examine a typical analyticalmethod. Although each method isunique, the following description of thedetermination of quinine in urine providesan instructive example of a typicalprocedure. The description here is basedon Mule, S. J.; Hushin, P. L. Anal. Chem.1971, 43, 708–711, and O’Reilly, J. E.; J.Chem. Educ. 1975, 52, 610–612.Figure 10.49 shows the fluorescence of thequinine in tonic water.St a n d a r d i z i n g t h e Me t h o dFrom equation 10.28 and equation 10.29 we know that the intensity offluorescent or phosphorescent emission is a linear function of the analyte’sconcentration provided that the sample’s absorbance of source radiation(A = ebC) is less than approximately 0.01. Calibration curves often arelinear over four to six orders of magnitude for fluorescence and over two tofour orders of magnitude for phosphorescence. For higher concentrationsof analyte the calibration curve becomes nonlinear because the assumptionsleading to equation 10.28 and equation 10.29 no longer apply. Nonlinearitymay be observed for small concentrations of analyte due to the presenceof fluorescent or phosphorescent contaminants. As discussed earlier, quantumefficiency is sensitive to temperature and sample matrix, both of whichmust be controlled when using external standards. In addition, emissionintensity depends on the molar absorptivity of the photoluminescent species,which is sensitive to the sample matrix.Representative Method 10.3Description o f Me t h o dDetermination of Quinine in UrineQuinine is an alkaloid used in treating malaria. It is a strongly fluorescentcompound in dilute solutions of H 2 SO 4 (F f = 0.55). Quinine’s excitationspectrum has absorption bands at 250 nm and 350 nm and its emissionspectrum has a single emission band at 450 nm. Quinine is rapidly excretedfrom the body in urine and is easily determined by measuring itsfluorescence following its extraction from the urine sample.Pr o c e d u r eTransfer a 2.00-mL sample of urine to a 15-mL test tube and adjust its pHto between 9 and 10 using 3.7 M NaOH. Add 4 mL of a 3:1 (v/v) mixtureof chloroform and isopropanol and shake the contents of the test tube forone minute. Allow the organic and the aqueous (urine) layers to separateand transfer the organic phase to a clean test tube. Add 2.00 mL of 0.05M H 2 SO 4 to the organic phase and shake the contents for one minute.Allow the organic and the aqueous layers to separate and transfer the aqueousphase to the sample cell. Measure the fluorescent emission at 450 nmusing an excitation wavelength of 350 nm. Determine the concentrationof quinine in the urine sample using a calibration curve prepared with aset of external standards in 0.05 M H 2 SO 4 , prepared from a 100.0 ppmsolution of quinine in 0.05 M H 2 SO 4 . Use distilled water as a blank.Qu e s t i o n s1. Chloride ion quenches the intensity of quinine’s fluorescent emission.For example, in the presence of 100 ppm NaCl (61 ppm Cl – ) quinine’semission intensity is only 83% of its emission intensity in the

Chapter 10 Spectroscopic Methods623absence of chloride. The presence of 1000 ppm NaCl (610 ppm Cl – )further reduces quinine’s fluorescent emission to less than 30% of itsemission intensity in the absence of chloride. The concentration ofchloride in urine typically ranges from 4600–6700 ppm Cl – . Explainhow this procedure prevents an interference from chloride.The procedure uses two extractions. In the first of these extractions,quinine is separated from urine by extracting it into a mixture of chloroformand isopropanol, leaving the chloride behind in the originalsample.2. Samples of urine may contain small amounts of other fluorescentcompounds, which interfere with the analysis if they are carriedthrough the two extractions. Explain how you can modify the procedureto take this into account?One approach is to prepare a blank using a sample of urine known tobe free of quinine. Subtracting the blank’s fluorescent signal from themeasured fluorescence from urine samples corrects for the interferingcompounds.3. The fluorescent emission for quinine at 450 nm can be induced usingan excitation frequency of either 250 nm or 350 nm. The fluorescentquantum efficiency is the same for either excitation wavelength. Quinine’sabsorption spectrum shows that e 250 is greater than e 350 . Giventhat quinine has a stronger absorbance at 250 nm, explain why itsfluorescent emission intensity is greater when using 350 nm as theexcitation wavelength.From equation 10.28 we know that I f is a function of the followingterms: k, F f , P 0 , e, b, and C. We know that F f , b, and C are the samefor both excitation wavelengths and that e is larger for a wavelengthof 250 nm; we can, therefore, ignore these terms. The greater emissionintensity when using an excitation wavelength of 350 nm mustbe due to a larger value for P 0 or k . In fact, P 0 at 350 nm for a highpressureXe arc lamp is about 170% of that at 250 nm. In addition,the sensitivity of a typical photomultiplier detector (which contributesto the value of k) at 350 nm is about 140% of that at 250 nm.Example 10.11To evaluate the method described in Representative Method 10.3, a seriesof external standard was prepared and analyzed, providing the resultsshown in the following table. All fluorescent intensities were correctedusing a blank prepared from a quinine-free sample of urine. The fluorescentintensities are normalized by setting I f for the highest concentrationstandard to 100.

624 Analytical Chemistry 2.0[quinine] (mg/mL) I f1.00 10.113.00 30.205.00 49.847.00 69.8910.00 100.0After ingesting 10.0 mg of quinine, a volunteer provided a urine sample24-h later. Analysis of the urine sample gives an relative emission intensityof 28.16. Report the concentration of quinine in the sample in mg/L andthe percent recovery for the ingested quinine.relative emission intensity1008060402000 2 4 6 8 10μg quinine/mLIt can take up 10–11 days for the body tocompletely excrete quinine.So l u t i o nLinear regression of the relative emission intensity versus the concentrationof quinine in the standards gives a calibration curve with the followingequation.I f= 0. 124 + 9.978×µgquininemLSubstituting the sample’s relative emission intensity into the calibrationequation gives the concentration of quinine as 2.81 mg/mL. Because thevolume of urine taken, 2.00 mL, is the same as the volume of 0.05 MH 2 SO 4 used in extracting quinine, the concentration of quinine in theurine also is 2.81 mg/mL. The recovery of the ingested quinine is281 . µ g1 mg× 200 . mL urine×ml urine1000 µ g× 100 = 0. 0562%10.0 mg quinineingested10F.4 Evaluation of Photoluminescence SpectroscopySc a l e o f Op e r a t i o nSee Figure 3.5 to review the meaning ofmacro and meso for describing samples,and the meaning of major, minor, and ultratracefor describing analytes.Photoluminescence spectroscopy is used for the routine analysis of traceand ultratrace analytes in macro and meso samples. Detection limits forfluorescence spectroscopy are strongly influenced by the analyte’s quantumyield. For an analyte with F f > 0.5, a picomolar detection limit is possiblewhen using a high quality spectrofluorimeter. For example, the detectionlimit for quinine sulfate, for which F f is 0.55, is generally between 1 partper billion and 1 part per trillion. Detection limits for phosphorescenceare somewhat higher, with typical values in the nanomolar range for lowtemperaturephosphorimetry, and in the micromolar range for room-temperaturephosphorimetry using a solid substrate.

Chapter 10 Spectroscopic Methods625(a)(b)fromsourcefromsourcetodetectortodetectorFigure 10.56 Use of slit orientation to change the volume from which fluorescence is measured: (a) verticalslit orientation; (b) horizontal slit orientation. Suppose the slit’s dimensions are 0.1 mm 3 mm. In (a) thedimensions of the sampling volume are 0.1 mm 0.1mm 3 mm, or 0.03 mm 3 . For (b) the dimensions ofthe sampling volume are 0.1 mm 3 mm 3 mm, or 0.9 mm 3 , a 30-fold increase in the sampling volume.Ac c u r a c yThe accuracy of a fluorescence method is generally between 1–5% whenspectral and chemical interferences are insignificant. Accuracy is limited bythe same types of problems affecting other optical spectroscopic methods.In addition, accuracy is affected by interferences influencing the fluorescentquantum yield. The accuracy of phosphorescence is somewhat greater thanthat for fluorescence.PrecisionThe relative standard deviation for fluorescence is usually between 0.5–2%when the analyte’s concentration is well above its detection limit. Precisionis usually limited by the stability of the excitation source. The precision forphosphorescence is often limited by reproducibility in preparing samplesfor analysis, with relative standard deviations of 5–10% being common.SensitivityFrom equation 10.28 and equation 10.29 we know that the sensitivity ofa fluorescent or phosphorescent method is influenced by a number of parameters.The importance of quantum yield and the effect of temperatureand solution composition on F f and F p already have been considered.Besides quantum yield, the sensitivity of an analysis can be improved byusing an excitation source that has a greater emission intensity, P 0 , at thedesired wavelength, and by selecting an excitation wavelength that has agreater absorbance. Another approach for improving sensitivity is to increasethe volume in the sample from which emission is monitored. Figure10.56 shows how rotating a monochromator’s slits from their usual verticalorientation to a horizontal orientation increases the sampling volume. Theresult can increase the emission from the sample by 5–30.

626 Analytical Chemistry 2.0Se l e c t i v i t yThe selectivity of fluorescence and phosphorescence is superior to that ofabsorption spectrophotometry for two reasons: first, not every compoundthat absorbs radiation is fluorescent or phosphorescent; and, second, selectivitybetween an analyte and an interferent is possible if there is a differencein either their excitation or their emission spectra. The total emission intensityis a linear sum of that from each fluorescent or phosphorescent species.The analysis of a sample containing n components, therefore, can beaccomplished by measuring the total emission intensity at n wavelengths.Tim e , Co s t, a n d Eq u i pm e n tAs with other optical spectroscopic methods, fluorescent and phosphorescentmethods provide a rapid means for analyzing samples and are capableof automation. Fluorimeters are relatively inexpensive, ranging from severalhundred to several thousand dollars, and often are satisfactory for quantitativework. Spectrofluorimeters are more expensive, with models oftenexceeding $50,000.Energy3s5s4s1138.35p4p330.3330.21140.4818.3819.5589.0 3pFigure 10.57 Valence shell energy leveldiagram for sodium. The wavelengthscorresponding to several transitions areshown. Note that this is the same energylevel diagram as Figure 10.19.4d3d10GAtomic Emission SpectroscopyThe focus of this section is on the emission of ultraviolet and visible radiationfollowing the thermal excitation of atoms. Atomic emission spectroscopyhas a long history. Qualitative applications based on the color offlames were used in the smelting of ores as early as 1550 and were more fullydeveloped around 1830 with the observation of atomic spectra generatedby flame emission and spark emission. 18 Quantitative applications basedon the atomic emission from electric sparks were developed by Lockyer inthe early 1870 and quantitative applications based on flame emission werepioneered by Lundegardh in 1930. Atomic emission based on emissionfrom a plasma was introduced in 1964.10G.1 Atomic Emission SpectraAtomic emission occurs when a valence electron in a higher energy atomicorbital returns to a lower energy atomic orbital. Figure 10.57 shows a portionof the energy level diagram for sodium, which consists of a series ofdiscrete lines at wavelengths corresponding to the difference in energy betweentwo atomic orbitals.The intensity of an atomic emission line, I e , is proportional to the numberof atoms, N*, populating the excited state,I = kN *10.30ewhere k is a constant accounting for the efficiency of the transition. If asystem of atoms is in thermal equilibrium, the population of excited state iis related to the total concentration of atoms, N, by the Boltzmann distribu-18 Dawson, J. B. J. Anal. At. Spectrosc. 1991, 6, 93–98.

Chapter 10 Spectroscopic Methods627tion. For many elements at temperatures of less than 5000 K the Boltzmanndistribution is approximated asN* = N ⎛ g ⎝⎜g⎞e⎠⎟i −Ei/ kT010.31where g i and g 0 are statistical factors that account for the number of equivalentenergy levels for the excited state and the ground state, E i is the energyof the excited state relative to a ground state energy, E 0 , of 0, k is Boltzmann’sconstant (1.3807 10 –23 J/K), and T is the temperature in kelvin.From equation 10.31 we expect that excited states with lower energies havelarger populations and more intense emission lines. We also expect emissionintensity to increase with temperature.10G.2 EquipmentAn atomic emission spectrometer is similar in design to the instrumentationfor atomic absorption. In fact, it is easy to adapt most flame atomicabsorption spectrometers for atomic emission by turning off the hollowcathode lamp and monitoring the difference in the emission intensity whenaspirating the sample and when aspirating a blank. Many atomic emissionspectrometers, however, are dedicated instruments designed to take advantageof features unique to atomic emission, including the use of plasmas,arcs, sparks, and lasers as atomization and excitation sources, and an enhancedcapability for multielemental analysis.At o m i z at i o n a n d Exc i t a t i o nAtomic emission requires a means for converting a solid, liquid, or solutionanalyte into a free gaseous atom. The same source of thermal energy usuallyserves as the excitation source. The most common methods are flamesand plasmas, both of which are useful for liquid or solution samples. Solidsamples may be analyzed by dissolving in a solvent and using a flame orplasma atomizer.Fl a m e So u r c e sAtomization and excitation in flame atomic emission is accomplished usingthe same nebulization and spray chamber assembly used in atomic absorption(Figure 10.42). The burner head consists of single or multiple slots, ora Meker style burner. Older atomic emission instruments often used a totalconsumption burner in which the sample is drawn through a capillary tubeand injected directly into the flame.Pl a s m a So u r c e sA plasma is a hot, partially ionized gas that contains an abundant concentrationof cations and electrons. The plasmas used in atomic emission areformed by ionizing a flowing stream of argon gas, producing argon ions and

628 Analytical Chemistry 2.06000 K8000 K10 000 Kplasma tubecoolent tubesample aerosol inletquartz bonnetRF induction coilcapillary injection tubetangential Ar flowAr plasma gas inletFigure 10.58 Schematic diagram of aninductively coupled plasma torch. Source:modified from Xvlun (commons.wikipedia.org).electrons. A plasma’s high temperature results from resistive heating as theelectrons and argon ions move through the gas. Because plasmas operateat much higher temperatures than flames, they provide better atomizationand a higher population of excited states.A schematic diagram of the inductively coupled plasma source (ICP) isshown in Figure 10.58. The ICP torch consists of three concentric quartztubes, surrounded at the top by a radio-frequency induction coil. Thesample is mixed with a stream of Ar using a nebulizer, and is carried tothe plasma through the torch’s central capillary tube. Plasma formation isinitiated by a spark from a Tesla coil. An alternating radio-frequency currentin the induction coils creates a fluctuating magnetic field that inducesthe argon ions and the electrons to move in a circular path. The resultingcollisions with the abundant unionized gas give rise to resistive heating,providing temperatures as high as 10 000 K at the base of the plasma, andbetween 6000 and 8000 K at a height of 15–20 mm above the coil, whereemission is usually measured. At these high temperatures the outer quartztube must be thermally isolated from the plasma. This is accomplished bythe tangential flow of argon shown in the schematic diagram.Mu l t i e l e m e n t a l An a l y s i sAtomic emission spectroscopy is ideally suited for multielemental analysisbecause all analytes in a sample are excited simultaneously. If the instrumentincludes a scanning monochromator, we can program it to move rapidly toan analyte’s desired wavelength, pause to record its emission intensity, andthen move to the next analyte’s wavelength. This sequential analysis allowsfor a sampling rate of 3–4 analytes per minute.Another approach to a multielemental analysis is to use a multichannelinstrument that allows us to simultaneously monitor many analytes. Asimple design for a multichannel spectrometer couples a monochromatorwith multiple detectors that can be positioned in a semicircular arrayaround the monochromator at positions corresponding to the wavelengthsfor the analytes (Figure 10.59).10G.3 Quantitative ApplicationsAtomic emission is widely used for the analysis of trace metals in a varietyof sample matrices. The development of a quantitative atomic emissionmethod requires several considerations, including choosing a source foratomization and excitation, selecting a wavelength and slit width, preparingthe sample for analysis, minimizing spectral and chemical interferences,and selecting a method of standardization.Ch o i c e o f At o m i z at i o n a n d Exc i t a t i o n So u r c eExcept for the alkali metals, detection limits when using an ICP are significantlybetter than those obtained with flame emission (Table 10.14).

Chapter 10 Spectroscopic Methods629detector 3detector 4detector 2detector 5detector 1detector 6λ 1 λ2λ 3ICP torchmonochromatorFigure 10.59 Schematic diagram of a multichannel atomicemission spectrometer for the simultaneous analysis ofseveral elements. The ICP torch is modified from Xvlun(commons.wikipedia.org). Instruments may contain asmany as 48–60 detectors.Table 10.14 Detection Limits for Atomic Emission adetection limit in mg/mLelement flame emission ICPAg 2 0.2Al 3 0.2As 2000 2Ca 0.1 0.0001Cd 300 0.07Co 5 0.1Cr 1 0.08Fe 10 0.09Hg 150 1K 0.01 30Li 0.001 0.02Mg 1 0.003Mn 1 0.01Na 0.01 0.1Ni 10 0.2Pb 0.2 1Pt 2000 0.9Sn 100 3Zn 1000 0.1a Source: Parsons, M. L.; Major, S.; Forster, A. R.; App. Spectrosc. 1983, 37, 411–418.

630 Analytical Chemistry 2.0Plasmas also are subject to fewer spectral and chemical interferences. Forthese reasons a plasma emission source is usually the better choice.Se l e c t i n g t h e Wave l e n g t h a n d Sl i t Wi d t hThe choice of wavelength is dictated by the need for sensitivity and the needto avoid interferences from the emission lines of other constituents in thesample. Because an analyte’s atomic emission spectrum has an abundanceof emission lines—particularly when using a high temperature plasmasource—it is inevitable that there will be some overlap between emissionlines. For example, an analysis for Ni using the atomic emission line at349.30 nm is complicated by the atomic emission line for Fe at 349.06 nm.Narrower slit widths provide better resolution, but at the cost of less radiationreaching the detector. The easiest approach to selecting a wavelength isto record the sample’s emission spectrum and look for an emission line thatprovides an intense signal and is resolved from other emission lines.Pr e p a r i n g t h e Sa m p l eFlame and plasma sources are best suited for samples in solution and liquidform. Although a solid sample can be analyzed by directly inserting it intothe flame or plasma, they usually are first brought into solution by digestionor extraction.Minimizing Sp e c t r a l In t e r fe r e n ce semission intensityFigure 10.60 Method for correctingan analyte’s emission for the flame’sbackground emission.I ewavelengthThe most important spectral interference is broad, background emissionfrom the flame or plasma and emission bands from molecular species. Thisbackground emission is particularly severe for flames because the temperatureis insufficient to break down refractory compounds, such as oxidesand hydroxides. Background corrections for flame emission are made byscanning over the emission line and drawing a baseline (Figure 10.60).Because a plasma’s temperature is much higher, a background interferencedue to molecular emission is less of a problem. Although emission from theplasma’s core is strong, it is insignificant at a height of 10–30 mm above thecore where measurements normally are made.Minimizing Ch e m ic a l In t e r fe r e n ce sFlame emission is subject to the same types of chemical interferences asatomic absorption. These interferences are minimized by adjusting theflame’s composition and adding protecting agents, releasing agents, or ionizationsuppressors. An additional chemical interference results from selfabsorption.Because the flame’s temperature is greatest at its center, theconcentration of analyte atoms in an excited state is greater at the flame’scenter than at its outer edges. If an excited state atom in the flame’s centeremits a photon while returning to its ground state, then a ground state atomin the cooler, outer regions of the flame may absorb the photon, decreasing

Chapter 10 Spectroscopic Methods631the emission intensity. For higher concentrations of analyte self-absorptionmay invert the center of the emission band (Figure 10.61).Chemical interferences with plasma sources generally are not significantbecause the plasma’s higher temperature limits the formation of nonvolatilespecies. For example, PO 4 3– is a significant interferent when analyzingsamples for Ca 2+ by flame emission, but has a negligible effect when usinga plasma source. In addition, the high concentration of electrons from theionization of argon minimizes ionization interferences.St a n d a r d i z i n g t h e Me t h o dFrom equation 10.30 we know that emission intensity is proportional tothe population of the analyte’s excited state, N*. If the flame or plasma isin thermal equilibrium, then the excited state population is proportionalto the analyte’s total population, N, through the Boltzmann distribution(equation 10.31).A calibration curve for flame emission is usually linear over two to threeorders of magnitude, with ionization limiting linearity when the analyte’sconcentrations is small and self-absorption limiting linearity for higherconcentrations of analyte. When using a plasma, which suffers from fewerchemical interferences, the calibration curve often is linear over four tofive orders of magnitude and is not affected significantly by changes in thematrix of the standards.Emission intensity may be affected significantly by many parameters,including the temperature of the excitation source and the efficiency ofatomization. An increase in temperature of 10 K, for example, producesa 4% increase in the fraction of Na atoms occupying the 3p excited state.This is potentially significant uncertainty that may limit the use of externalstandards. The method of internal standards can be used when variationsin source parameters are difficult to control. To compensate for changes inthe temperature of the excitation source, the internal standard is selectedso that its emission line is close to the analyte’s emission line. In addition,the internal standard should be subject to the same chemical interferencesto compensate for changes in atomization efficiency. To accurately compensatefor these errors the analyte and internal standard emission lines mustbe monitored simultaneously.Representative Method 10.4Determination of Sodium in a Salt SubstituteDescription o f Me t h o dSalt substitutes, which are used in place of table salt for individuals onlow–sodium diets, replaces NaCl with KCl. Depending on the brand,fumaric acid, calcium hydrogen phosphate, or potassium tartrate also maybe present. Although intended to be sodium-free, salt substitutes containsmall amounts of NaCl as an impurity. Typically, the concentration of(a)(b)Figure 10.61 Atomic emission linesfor (a) a low concentration of analyte,and (b) a high concentration of analyteshowing the effect of self-absorption.The best way to appreciate the theoreticaland practical details discussed in this sectionis to carefully examine a typical analyticalmethod. Although each method isunique, the following description of thedetermination of sodium in salt substitutesprovides an instructive example of atypical procedure. The description here isbased on Goodney, D. E. J. Chem. Educ.1982, 59, 875–876.

632 Analytical Chemistry 2.0sodium in a salt substitute is about 100 mg/g The exact concentrationof sodium is easily determined by flame atomic emission. Because it isdifficult to match the matrix of the standards to that of the sample, theanalysis is accomplished by the method of standard additions.Pr o c e d u r eA sample is prepared by placing an approximately 10-g portion of the saltsubstitute in 10 mL of 3 M HCl and 100 mL of distilled water. After thesample has dissolved, it is transferred to a 250-mL volumetric flask anddiluted to volume with distilled water. A series of standard additions isprepared by placing 25-mL portions of the diluted sample into separate50-mL volumetric flasks, spiking each with a known amount of an approximately10 mg/L standard solution of Na + , and diluting to volume.After zeroing the instrument with an appropriate blank, the instrument isoptimized at a wavelength of 589.0 nm while aspirating a standard solutionof Na + . The emission intensity is measured for each of the standardaddition samples and the concentration of sodium in the salt substituteis reported in mg/g.Qu e s t i o n s1. Potassium ionizes more easily than sodium. What problem might thispresent if you use external standards prepared from a stock solutionof 10 mg Na/L instead of using a set of standard additions?Because potassium is present at a much higher concentration thansodium, its ionization suppresses the ionization of sodium. Normallysuppressing ionization is a good thing because it increases emissionintensity. In this case, however, the difference between the matrix ofthe standards and the sample’s matrix means that the sodium in astandard experiences more ionization than an equivalent amount ofsodium in a sample. The result is a determinate error.2. One way to avoid a determinate error when using external standardsis to match the matrix of the standards to that of the sample. Wecould, for example, prepare external standards using reagent gradeKCl to match the matrix to that of the sample. Why is this not a goodidea for this analysis?Sodium is a common contaminant, which is found in many chemicals.Reagent grade KCl, for example, may contain 40–50 mg Na/g.This is a significant source of sodium, given that the salt substitutecontains approximately 100 mg Na/g..3. Suppose you decide to use an external standardization. Given theanswer to the previous questions, is the result of your analysis likelyto underestimate or overestimate the amount of sodium in the saltsubstitute?

Chapter 10 Spectroscopic Methods633emissionidealcalibration curvenot accounting forextra sodiumsample’semissiondecrease in emissiondue to ionizationactualconcentrationreportedconcentrationsmg Na/LFigure 10.62 External standards calibration curves for the flame atomic emissionanalysis of Na in a salt substitute. The solid black line shows the ideal calibrationcurve assuming matrix matching of samples and standards with pure KCl. Thelower of the two dashed red lines shows the effect of failing to add KCl to theexternal standards, which decreases emission. The other dashed red line shows theeffect of using KCl that is contaminated with NaCl, which causes us to underestimatethe concentration of Na in the standards. In both cases, the result is a positivedeterminate error in the analysis of samples.The solid black line in Figure 10.62 shows the ideal calibration curveassuming that we match the matrix of the standards to the sample’smatrix, and that we do so without adding an additional sodium. Ifwe prepare the external standards without adding KCl, the emissionfor each standard decreases due to increased ionization. This is shownby the lower of the two dashed red lines. Preparing the standards byadding reagent grade KCl increases the concentration of sodium dueto its contamination. Because we underestimate the actual concentrationof sodium in the standards, the resulting calibration curve isshown by the other dashed red line. In both cases, the sample’s emissionresults in our overestimating the concentration of sodium in thesample.4. One problem with analyzing salt samples is their tendency to clogthe aspirator and burner assembly. What effect does this have on theanalysis?Clogging the aspirator and burner assembly decreases the rate of aspiration,which decreases the analyte’s concentration in the flame. Theresult is a decrease in the emission intensity and a negative determinateerror.

634 Analytical Chemistry 2.0See Section 5C.3 in Chapter 5 to reviewthe method of standard additions.Example 10.12To evaluate the method described in Representative Method 10.4, a seriesof standard additions is prepared using a 10.0077-g sample of a salt substitute.The results of a flame atomic emission analysis of the standards isshown here. 19 added Na (mg/mL) I e (arb. units)emission intensity (arbitrary units)76543210-2 -1 0 1 2 3 4μg Na/mL0.000 1.790.420 2.631.051 3.542.102 4.943.153 6.18What is the concentration of sodium, in mg/g, in the salt substitute.So l u t i o nLinear regression of emission intensity versus the concentration of addedNa gives a standard additions calibration curve with the following equation.I e= 197 . + 1.37×µgNamLThe concentration of sodium in the sample is equal to the absolute valueof the calibration curve’s x-intercept. Substituting zero for the emission intensityand solving for sodium’s concentration gives a result of 1.44 mg Na/mL. The concentration of sodium in the salt substitute is144 . µ gNa 50.00 mL× × 250.0 mLmL 25.00 mL= 71. 9 µ gNa/g10.0077 gsample10G.4 Evaluation of Atomic Emission SpectroscopySc a l e o f Op e r a t i o nSee Figure 3.5 to review the meaning ofmacro and meso for describing samples,and the meaning of major, minor, and ultratracefor describing analytes.The scale of operations for atomic emission is ideal for the direct analysisof trace and ultratrace analytes in macro and meso samples. With appropriatedilutions, atomic emission also can be applied to major and minoranalytes.Ac c u r a c yWhen spectral and chemical interferences are insignificant, atomic emissionis capable of producing quantitative results with accuracies of between19 Goodney, D. E. J. Chem. Educ. 1982, 59, 875–876

Chapter 10 Spectroscopic Methods6351–5%. Accuracy frequently is limited by chemical interferences. Becausethe higher temperature of a plasma source gives rise to more emission lines,the accuracy of using plasma emission often is limited by stray radiationfrom overlapping emission lines.PrecisionFor samples and standards in which the analyte’s concentration exceeds thedetection limit by at least a factor of 50, the relative standard deviation forboth flame and plasma emission is about 1–5%. Perhaps the most importantfactor affecting precision is the stability of the flame’s or the plasma’stemperature. For example, in a 2500 K flame a temperature fluctuationof ±2.5 K gives a relative standard deviation of 1% in emission intensity.Significant improvements in precision may be realized when using internalstandards.SensitivitySensitivity is strongly influenced by the temperature of the excitation sourceand the composition of the sample matrix. Sensitivity is optimized by aspiratinga standard solution of analyte and maximizing the emission byadjusting the flame’s composition and the height from which we monitorthe emission. Chemical interferences, when present, decrease the sensitivityof the analysis. The sensitivity of plasma emission is less affected by thesample matrix. In some cases a calibration curve prepared using standardsin a matrix of distilled water can be used for samples with more complexmatrices.Se l e c t i v i t yThe selectivity of atomic emission is similar to that of atomic absorption.Atomic emission has the further advantage of rapid sequential or simultaneousanalysis.Tim e , Co s t, a n d Eq u i pm e n tSample throughput with atomic emission is very rapid when using automatedsystems capable of multielemental analysis. For example, samplingrates of 3000 determinations per hour have been achieved using a multichannelICP, and 300 determinations per hour with a sequential ICP. Flameemission is often accomplished using an atomic absorption spectrometer,which typically costs between $10,000–$50,000. Sequential ICP’s rangein price from $55,000–$150,000, while an ICP capable of simultaneousmultielemental analysis costs between $80,000–$200,000. CombinationICP’s that are capable of both sequential and simultaneous analysis rangein price from $150,000–$300,000. The cost of Ar, which is consumed insignificant quantities, can not be overlooked when considering the expenseof operating an ICP.

636 Analytical Chemistry 2.0(a)source(b)sourcesampleFigure 10.63 Distribution of radiationfor (a) Rayleigh, or small-particlescattering, and (b) large-particle scattering.10HSpectroscopy Based on ScatteringThe blue color of the sky during the day and the red color of the sun at sunsetresult from the scattering of light by small particles of dust, moleculesof water, and other gases in the atmosphere. The efficiency of a photon’sscattering depends on its wavelength. We see the sky as blue during the daybecause violet and blue light scatter to a greater extent than other, longerwavelengths of light. For the same reason, the sun appears red at sunset becausered light is less efficiently scattered and is more likely to pass throughthe atmosphere than other wavelengths of light. The scattering of radiationhas been studied since the late 1800s, with applications beginning soonthereafter. The earliest quantitative applications of scattering, which datefrom the early 1900s, used the elastic scattering of light by colloidal suspensionsto determine the concentration of colloidal particles.10H.1 Origin of ScatteringIf we send a focused, monochromatic beam of radiation with a wavelengthl through a medium of particles with dimensions

Chapter 10 Spectroscopic Methods637(a)λ 1 λ2λ 3shutteropendetectorsourcemonochromatorclosedsample or blank(b)λ 1 λ2λ 3shutteropensample or blanksignalprocessorsourcemonochromatorclosedsignalprocessordetectorFigure 10.64 Schematic diagrams for (a) a turbimeter,and (b) a nephelometer.Tu r b i d i m e t r y o r Ne p h e l o m e t r y?When developing a scattering method the choice of using turbidimetry ornephelometry is determined by two factors. The most important considerationis the intensity of the scattered radiation relative to the intensity of thesource’s radiation. If the solution contains a small concentration of scatteringparticles the intensity of the transmitted radiation, I T , is approximatelythe same as the intensity of the source’s radiation, I 0 . As we learned earlierin the section on molecular absorption, there is substantial uncertainty indetermining a small difference between two intense signals. For this reason,nephelometry is a more appropriate choice for samples containing few scatteringparticles. Turbidimetry is a better choice when the sample containsa high concentration of scattering particles.A second consideration in choosing between turbidimetry and nephelometryis the size of the scattering particles. For nephelometry, the intensityof scattered radiation at 90 o increases when the particles are small andRayleigh scattering is in effect. For larger particles, as shown in Figure 10.63,the intensity of scattering is diminished at 90 o . When using an ultravioletor visible source of radiation, the optimum particle size is 0.1–1 mm. Thesize of the scattering particles is less important for turbidimetry where thesignal is the relative decrease in transmitted radiation. In fact, turbidimetricmeasurements are still feasible even when the size of the scattering particlesresults in an increase in reflection and refraction, although a linear relationshipbetween the signal and the concentration of scattering particles mayno longer hold.De t e r m i n i n g Co n c e n t r a t i o n b y Tu r b i d i m e t r yIn turbidimetry the measured transmittance, T, is the ratio of the intensityof source radiation transmitted by the sample, I T , to the intensity of sourceradiation transmitted by a blank, I 0 .

638 Analytical Chemistry 2.0TI=IT0The relationship between transmittance and the concentration of the scatteringparticles is similar to that given by Beer’s law− logT = kbC 10.32where C is the concentration of the scattering particles in mass per unitvolume (w/v), b is the pathlength, and k is a constant that depends on severalfactors, including the size and shape of the scattering particles and thewavelength of the source radiation. As with Beer’s law, equation 10.32 mayshow appreciable deviations from linearity. The exact relationship is establishedby a calibration curve prepared using a series of standards containingknown concentrations of analyte.De t e r m i n i n g Co n c e n t r a t i o n b y Ne p h e l o m e t r yIn nephelometry the relationship between the intensity of scattered radiation,I S , and the concentration of scattering particles isIS= k IC 010.33where k S is an empirical constant for the system and I 0 is the source radiation’sintensity. The value of k S is determined from a calibration curveprepared using a series of standards containing known concentrations ofanalyte.Se l e c t i n g a Wave l e n g t h f o r t h e In c i d e n t RadiationThe choice of wavelength is based primarily on the need to minimize potentialinterferences. For turbidimetry, where the incident radiation is transmittedthrough the sample, a monochromator or filter allow us to avoid wavelengthsthat are absorbed by the sample. For nephelometry, the absorptionof incident radiation is not a problem unless it induces fluorescence fromthe sample. With a nonfluorescent sample there is no need for wavelengthselection, and a source of white light may be used as the incident radiation.For both techniques, other considerations in choosing a wavelengthincluding the effect of wavelength on scattering intensity, the transducer’ssensitivity, and the source’s intensity. For example, many common photontransducers are more sensitive to radiation at 400 nm than at 600 nm.SPr e p a r i n g t h e Sa m p l e f o r An a l y s i sAlthough equation 10.32 and equation 10.33 relate scattering to the concentrationof scattering particles, the intensity of scattered radiation is alsoinfluenced by the particle’s size and shape. Samples containing the samenumber of scattering particles may show significantly different values for-logT or I S , depending on the average diameter of the particles. For a quan-

Chapter 10 Spectroscopic Methods639titative analysis, therefore, it is necessary to maintain a uniform distributionof particle sizes throughout the sample and between samples and standards.Most turbidimetric and nephelometric methods rely on precipitationto form the scattering particles. As we learned in Chapter 8, the propertiesof a precipitate are determined by the conditions under which it forms. Tomaintain a reproducible distribution of particle sizes between samples andstandards, it is necessary to control parameters such as the concentrationof reagents, the order of adding reagents, the pH and temperature, theagitation or stirring rate, the ionic strength, and the time between theprecipitate’s initial formation and the measurement of transmittance orscattering. In many cases a surface-active agent—such as glycerol, gelatin,or dextrin—is added to stabilize the precipitate in a colloidal state and toprevent the coagulation of the particles.Ap p l i c a t i o n sTurbidimetry and nephelometry are widely used to determine the clarityof water, beverages, and food products. For example, a nephelometric determinationof the turbidity of water compares the sample’s scattering tothe scattering of a set of standards. The primary standard for measuringturbidity is formazin, which is an easily prepared, stable polymer suspension(Figure 10.65). 20 Formazin prepared by mixing a 1 g/100 mL solutionof hydrazine sulfate, N 2 H 4. H2 SO 4 , with a 10 g/100 mL solution of hexamethylenetetramineproduces a suspension that is defined as 4000 nephelometricturbidity units (NTU). A set of standards with NTUs between 0and 40 is prepared and used to construct a calibration curve. This methodis readily adapted to the analysis of the clarity of orange juice, beer, andmaple syrup.A number of inorganic cations and anions can be determined by precipitatingthem under well-defined conditions and measuring the transmittanceor scattering of radiation from the precipitated particles. The trans-20 Hach, C. C.; Bryant, M. “Turbidity Standards,” Technical Information Series, Booklet No. 12,Hach Company: Loveland, CO, 1995.NNNNhexamethylenetetramine+ 6H 2 O + 2H 2 SO 4 6H 2 CO + 2(NH 4 )SO 4nH 2 CO + (n/2)H 2 NNH 2NNNNformazinn/4+ nH 2 OFigure 10.65 Scheme for preparing formazinfor use as a turbidity standard.

640 Analytical Chemistry 2.0The best way to appreciate the theoreticaland practical details discussed in this sectionis to carefully examine a typical analyticalmethod. Although each method isunique, the following description of thedetermination of sulfate in water providesan instructive example of a typicalprocedure. The description here is basedon Method 4500–SO 4 2– –C in StandardMethods for the Analysis of Water andWastewater, American Public Health Association:Washington, D. C. 20 th Ed.,1998.Table 10.15 Examples of Analytes Determined byTurbidimetry or Nephelometryanalyte precipitant precipitateAg + NaCl AgClCa 2+ Na 2 C 2 O 4 CaC 2 O 4Cl – AgNO 3 AgClCN – AgNO 3 AgCN2–CO 3 BaCl 2 BaCO 3F – CaCl 2 CaF 22–SO 4 BaCl 2 BaSO 4mittance or scattering, as given by equation 10.32 or equation 10.33 isproportional to the concentration of the scattering particles, which, in turn,is related by the stoichiometry of the precipitation reaction to the analyte’sconcentration. Examples of analytes that have been determined in this wayare listed in Table 10.15.Representative Method 10.5Turbidimetric Determination of Sulfate in WaterDescription o f Me t h o dAdding BaCl 2 to an acidified sample precipitates SO 4 2– as BaSO 4 . Theconcentration of SO 4 2– may be determined either by turbidimetry or bynephelometry using an incident source of radiation of 420 nm. Externalstandards containing know concentrations of SO 4 2– are used to standardizethe method.Pr o c e d u r eTransfer a 100-mL sample to a 250-mL Erlenmeyer flask along with 20.00mL of a buffer. For samples containing >10 mg SO 4 2– /L, the buffer contains30 g of MgCl 2. 6H2 O, 5 g of CH 3 COONa.3H 2 O, 1.0 g of KNO 3 ,and 20 mL of glacial CH 3 COOH per liter. The buffer for samples containing

Chapter 10 Spectroscopic Methods641Qu e s t i o n s1. What is the purpose of the buffer?If the precipitate’s particles are too small, I T may be too small tomeasure reliably. Because rapid precipitation favors the formationof microcrystalline particles of BaSO 4 , we use conditions that favorprecipitate growth over nucleation. The buffer’s high ionic strengthand its acidity favors precipitate growth and prevents the formationof microcrystalline BaSO 4 .2. Why is it important to use the same stirring rate and time for thesamples and standards?How fast and how long we stir the sample after adding BaCl 2 influencethe size of the precipitate’s particles.3. Many natural waters have a slight color due to the presence of humicand fulvic acids, and may contain suspended matter (Figure 10.66).Explain why these might interfere with the analysis for sulfate. Foreach interferent, suggest a method for minimizing its effect on theanalysis.Suspended matter in the sample contributes to the scattering, resultingin a positive determinate error. We can eliminate this interferenceby filtering the sample prior to its analysis. A sample that is coloredmay absorb some of the source’s radiation, leading to a positive determinateerror. We can compensate for this interference by taking asample through the analysis without adding BaCl 2 . Because no precipitateforms, we can use the transmittance of this sample blank tocorrect for the interference.4. Why is Na 2 SO 4 added to the buffer for samples containing

642 Analytical Chemistry 2.0Example 10.13To evaluate the method described in Representative Method 10.5, a seriesof external standard was prepared and analyzed, providing the resultsshown in the following table.mg SO 2– 4 /L transmittance0.00 1.0010.00 0.64620.00 0.41730.00 0.26940.00 0.174–logT1.00.80.60.40.20.00 10 20 30 40mg sulfate/LAnalysis of a 100.0-mL sample of a surface water gives a transmittance of0.538. What is the concentration of sulfate in the sample?So l u t i o nLinear regression of -logT versus concentration of SO 4 2– gives a standardizationequation of− =− × − 5mg SOlog T 104 . 10 + 0.0190×LSubstituting the sample’s transmittance into the calibration curve’s equationgives the concentration of sulfate in sample as 14.2 mg SO 4 2– /L.2−4As you review this chapter, try to define akey term in your own words. Check youranswer by clicking on the key term, whichwill take you to the page where it was firstintroduced. Clicking on the key termthere, will bring you back to this page sothat you can continue with another keyterm.10IKey Termsabsorbance absorbance spectrum absorptivityamplitude attenuated total reflectance atomizationbackground correction Beer’s law chemiluminescencechromophore continuum source dark currentdouble-beam effective bandwidth electromagnetic radiationelectromagnetic spectrum emission emission spectrumexcitation spectrum external conversion Fellgett’s advantagefiber-optic probe filter filter photometerfluorescence fluorescent quantum yield fluorimeterfrequency graphite furnace interferograminterferometer internal conversion intersystem crossingionization suppressor Jacquinot’s advantage lifetimeline sourcemethod of continuousvariationsmolar absorptivitymole-ratio method monochromatic monochromatornephelometry nominal wavelength phase angle

Chapter 10 Spectroscopic Methods643phosphorescencephosphorescent quantumyieldphotoluminescence photon plasmaphotodiode arraypolychromatic protecting agent radiationless deactivationrelaxation releasing agent resolutionself-absorption signal averaging signal processorsignal-to-noise ratio single-beam singlet excited stateslope-ratio method spectral searching spectrofluorimeterspectrophotometer spectroscopy stray radiationtransducer transmittance triplet excited stateturbidimetry vibrational relaxation wavelengthwavenumber10JChapter SummaryThe spectrophotometric methods of analysis covered in this chapter includethose based on the absorption, emission, or scattering of electromagneticradiation. When a molecule absorbs UV/Vis radiation it undergoesa change in its valence shell configuration. A change in vibrational energyresults from the absorption of IR radiation. Experimentally we measurethe fraction of radiation transmitted, T, by the sample. Instrumentationfor measuring absorption requires a source of electromagnetic radiation, ameans for selecting a wavelength, and a detector for measuring transmittance.Beer’s law relates absorbance to both transmittance and to the concentrationof the absorbing species (A = –logT = ebC).In atomic absorption we measure the absorption of radiation by gasphase atoms. Samples are atomized using thermal energy from either aflame or a graphite furnace. Because the width of an atom’s absorptionband is so narrow, the continuum sources common for molecular absorptioncan not be used. Instead, a hollow cathode lamp provides the necessaryline source of radiation. Atomic absorption suffers from a number ofspectral and chemical interferences. The absorption or scattering of radiationfrom the sample’s matrix are important spectral interferences that maybe minimized by background correction. Chemical interferences includethe formation of nonvolatile forms of the analyte and ionization of theanalyte. The former interference is minimized by using a releasing agent ora protecting agent, and an ionization suppressor helps minimize the latterinterference.When a molecule absorbs radiation it moves from a lower energy stateto a higher energy state. In returning to the lower energy state the moleculemay emit radiation. This process is called photoluminescence. One formof photoluminescence is fluorescence in which the analyte emits a photonwithout undergoing a change in its spin state. In phosphorescence, emissionoccurs with a change in the analyte’s spin state. For low concentrations

644 Analytical Chemistry 2.0of analyte, both fluorescent and phosphorescent emission intensities are alinear function of the analyte’s concentration. Thermally excited atoms alsoemit radiation, forming the basis for atomic emission spectroscopy. Thermalexcitation is achieved using either a flame or a plasma.Spectroscopic measurements may also involve the scattering of lightby a particulate form of the analyte. In turbidimetry, the decrease in theradiation’s transmission through the sample is measured and related to theanalyte’s concentration through an equation similar to Beer’s law. In nephelometrywe measure the intensity of scattered radiation, which varies linearlywith the analyte’s concentration.10KProblemsAnswers, but not worked solutions, tomost end-of-chapter problems are availablehere.1. Provide the missing information in the following table.wavelength(m)frequency(s –1 )wavenumber(cm –1 )energy(J)4.50 10 –9 1.33 10 15 32157.20 10 –192. Provide the missing information in the following table.[analyte](M) absorbance %Tmolarabsorptivity(M –1 cm –1 )pathlength(cm)1.40 10 –4 1120 1.000.563 750 1.002.56 10 –4 0.225 4401.55 10 –3 0.167 5.0033.3 565 1.004.35 10 –3 21.2 15501.20 10 –4 81.3 10.003. A solution’s transmittance is 35.0%. What is the transmittance if youdilute 25.0 mL of the solution to 50.0 mL?4. A solution’s transmittance is 85.0% when measured in a cell with apathlength of 1.00 cm. What is the %T if you increase the pathlengthto 10.00 cm?

Chapter 10 Spectroscopic Methods6455. The accuracy of a spectrophotometer can be evaluated by preparing asolution of 60.06 ppm K 2 Cr 2 O 7 in 0.0050 M H 2 SO 4 , and measuringits absorbance at a wavelength of 350 nm in a cell with a pathlength of1.00 cm. The absorbance should be 0.640. What is the molar absorptivityof K 2 Cr 2 O 7 at this wavelength?6. A chemical deviation to Beer’s law may occur if the concentration of anabsorbing species is affected by the position of an equilibrium reaction.Consider a weak acid, HA, for which K a is 2 10 –5 . Construct Beer’slaw calibration curves of absorbance versus the total concentration ofweak acid (C total = [HA] + [A – ]), using values for C total of 1 10 –5 ,3 10 –5 , 5 10 –5 , 9 10 –5 , 11 10 –5 , and 13 10 –5 M for the followingsets of conditions:(a) e HA = e A – = 2000 M –1 cm –1 ; unbuffered solution.(b) e HA = 2000 M –1 cm –1 and e A – = 500 M –1 cm –1 ; unbuffered solution.(c) e HA = 2000 M –1 cm –1 and e A – = 500 M –1 cm –1 ; solution is bufferedto a pH of 4.5.Assume a constant pathlength of 1.00 cm for all samples.7. One instrumental limitation to Beer’s law is the effect of polychromaticradiation. Consider a line source that emits radiation at two wavelengths,l′ and l″. When treated separately, the absorbances at thesewavelengths, A′ and A″, areP ′A′ =−log T ′ = ε′bCP0P ′′A′′ =−log T ′′ = ε′′bCP0If both wavelengths are measured simultaneously the absorbance isP ′ + P ′′A =−log ( )T T( P′ + P′′)0 0(a) Show that if the molar absorptivity at l′ and l″ are the same (e′ =e″ = e), the absorbance is equivalent toA=εbC(b) Construct Beer’s law calibration curves over the concentrationrange of zero to 1 10 –4 M using e′ = 1000 and e″ = 1000, ande′ = 1000 and e″ = 100. Assume a value of 1.00 cm for the pathlength.Explain the difference between the two curves.

646 Analytical Chemistry 2.08. A second instrumental limitation to Beer’s law is stray radiation. Thefollowing data were obtained using a cell with a pathlength of 1.00 cmwhen stray light is insignificant (P stray = 0).[analyte] (mM) absorbance0.00 0.002.00 0.404.00 0.806.00 1.208.00 1.6010.00 2.00Calculate the absorbance of each solution when P stray is 5% of P 0 , andplot Beer’s law calibration curves for both sets of data. Explain any differencesbetween the two curves. (Hint: Assume that P 0 is 100).9. In the process of performing a spectrophotometric determination ofFe, an analyst prepares a calibration curve using a single-beam spectrophotometersimilar to that shown in Figure 10.26. After preparing thecalibration curve, the analyst drops and breaks the cuvette. The analystacquires a new cuvette, measures the absorbance of his sample, anddetermines the %w/w Fe in the sample. Will the change in cuvette leadto a determinate error in the analysis? Explain.10. The spectrophotometric methods for determining Mn in steel and fordetermining glucose use a chemical reaction to produce a colored specieswhose absorbance we can monitor. In the analysis of Mn in steel,colorless Mn2+ is oxidized to give the purple MnO4- ion. To analyzefor glucose, which is colorless, we react it with a yellow colored solutionof the Fe(CN)63-, forming the colorless Fe(CN)64- ion. The directionsfor the analysis of Mn do not specify precise reaction conditions, andsamples and standards may be treated separately. The conditions for theanalysis of glucose, however, require that the samples and standards betreated simultaneously at exactly the same temperature and for exactlythe same length of time. Explain why these two experimental proceduresare so different.11. One method for the analysis of Fe 3+ , which can be used with a varietyof sample matrices, is to form the highly colored Fe 3+ –thioglycolic acidcomplex. The complex absorbs strongly at 535 nm. Standardizing themethod is accomplished using external standards. A 10.00 ppm Fe 3+working standard is prepared by transferring a 10-mL aliquot of a 100.0ppm stock solution of Fe 3+ to a 100-mL volumetric flask and dilutingto volume. Calibration standards of 1.00, 2.00, 3.00, 4.00, and 5.00ppm are prepared by transferring appropriate amounts of the 10.0 ppm

Chapter 10 Spectroscopic Methods647working solution into separate 50-mL volumetric flasks, each containing5 mL of thioglycolic acid, 2 mL of 20% w/v ammonium citrate, and5 mL of 0.22 M NH 3 . After diluting to volume and mixing, the absorbancesof the external standards are measured against an appropriateblank. Samples are prepared for analysis by taking a portion known tocontain approximately 0.1 g of Fe 3+ , dissolving in a minimum amountof HNO 3 , and diluting to volume in a 1-L volumetric flask. A 1.00-mLaliquot of this solution is transferred to a 50-mL volumetric flask, alongwith 5 mL of thioglycolic acid, 2 mL of 20% w/v ammonium citrate,and 5 mL of 0.22 M NH 3 and diluted to volume. The absorbanceof this solution is used to determine the concentration of Fe 3+ in thesample.(a) What is an appropriate blank for this procedure?(b) Ammonium citrate is added to prevent the precipitation of Al 3+ .What is the effect on the reported concentration of iron in thesample if there is a trace impurity of Fe 3+ in the ammonium citrate?(c) Why does the procedure specify that the sample contain approximately0.1 g of Fe 3+ ?(d) Unbeknownst to the analyst, the 100-mL volumetric flask used toprepare the 10.00 ppm working standard of Fe 3+ has a volume thatis significantly smaller than 100.0 mL. What effect will this haveon the reported concentration of iron in the sample?12. A spectrophotometric method for the analysis of iron has a linear calibrationcurve for standards of 0.00, 5.00, 10.00, 15.00, and 20.00mg Fe/L. An iron ore sample that is 40–60% w/w is to be analyzedby this method. An approximately 0.5-g sample is taken, dissolved ina minimum of concentrated HCl, and diluted to 1 L in a volumetricflask using distilled water. A 5.00 mL aliquot is removed with a pipet.To what volume—10, 25, 50, 100, 250, 500, or 1000 mL—should itbe diluted to minimize the uncertainty in the analysis? Explain.13. Lozano-Calero and colleagues describe a method for the quantitativeanalysis of phosphorous in cola beverages based on the formationof the intensely blue-colored phosphomolybdate complex,(NH 4 ) 3 [PO 4 (MoO 3 ) 12 ]. 21 The complex is formed by adding(NH 4 ) 6 Mo 7 O 24 to the sample in the presence of a reducing agent, suchas ascorbic acid. The concentration of the complex is determined spectrophotometricallyat a wavelength of 830 nm, using a normal calibrationcurve as a method of standardization.21 Lozano-Calero, D.; Martín-Palomeque, P.; Madueño-Loriguillo, S. J. Chem. Educ. 1996, 73,1173–1174.

648 Analytical Chemistry 2.0Crystal Pepsi was a colorless, caffeine-freesoda produced by PepsiCo. It was availablein the United States from 1992 to1993.In a typical analysis, a set of standard solutions containing knownamounts of phosphorous was prepared by placing appropriate volumesof a 4.00 ppm solution of P 2 O 5 in a 5-mL volumetric flask, adding 2mL of an ascorbic acid reducing solution, and diluting to volume withdistilled water. Cola beverages were prepared for analysis by pouring asample into a beaker and allowing it to stand for 24 h to expel the dissolvedCO 2 . A 2.50-mL sample of the degassed sample was transferredto a 50-mL volumetric flask and diluted to volume. A 250-mL aliquotof the diluted sample was then transferred to a 5-mL volumetric flask,treated with 2 mL of the ascorbic acid reducing solution, and dilutedto volume with distilled water.(a) The authors note that this method can be applied only to noncoloredcola beverages. Explain why this is true.(b) How might you modify this method so that it could be applied toany cola beverage?(c) Why is it necessary to remove the dissolved gases?(d) Suggest an appropriate blank for this method?(e) The author’s report a calibration curve ofA =− 002 . + 0. 72×ppm PO 2 5A sample of Crystal Pepsi, analyzed as described above, yields anabsorbance of 0.565. What is the concentration of phosphorous,reported as ppm P, in the original sample of Crystal Pepsi?14. EDTA forms colored complexes with a variety of metal ions that mayserve as the basis for a quantitative spectrophotometric method of analysis.The molar absorptivities of the EDTA complexes of Cu 2+ , Co 2+ ,and Ni 2+ at three wavelengths are summarized in the following table(all values of e are in M –1 cm –1 ).metal e 462.9 e 732.0 e 378.7Co 2+ 15.8 2.11 3.11Cu 2+ 2.32 95.2 7.73Ni 2+ 1.79 3.03 13.5Using this information determine the following:(a) The concentration of Cu 2+ in a solution that has an absorbance of0.338 at a wavelength of 732.0 nm.(b) The concentrations of Cu 2+ and Co 2+ in a solution that has anabsorbance of 0.453 at a wavelength of 732.0 nm and 0.107 at awavelength of 462.9 nm.

Chapter 10 Spectroscopic Methods649(c) The concentrations of Cu 2+ , Co 2+ , and Ni 2+ in a sample that hasan absorbance of 0.423 at a wavelength of 732.0 nm, 0.184 at awavelength of 462.9 nm, and 0.291 at a wavelength of 378.7 nm.The pathlength, b, is 1.00 cm for all measurements.15. The concentration of phenol in a water sample is determined by separatingthe phenol from non-volatile impurities by steam distillation,followed by reacting with 4-aminoantipyrine and K 3 Fe(CN) 6 at pH7.9 to form a colored antipyrine dye. A phenol standard with a concentrationof 4.00 ppm has an absorbance of 0.424 at a wavelengthof 460 nm using a 1.00 cm cell. A water sample is steam distilled anda 50.00-mL aliquot of the distillate is placed in a 100-mL volumetricflask and diluted to volume with distilled water. The absorbance of thissolution is found to be 0.394. What is the concentration of phenol (inparts per million) in the water sample?16. Saito describes a quantitative spectrophotometric procedure for ironbased on a solid-phase extraction using bathophenanthroline in apoly(vinyl chloride) membrane. 22 In the absence of Fe 2+ the membraneis colorless, but when immersed in a solution of Fe 2+ and I – , themembrane develops a red color as a result of the formation of an Fe 2+ –bathophenanthroline complex. A calibration curve determined usinga set of external standards with known concentrations of Fe 2+ gave astandardization relationship ofA = (. 860× 10 3 M− 1 ) × [ Fe2+]What is the concentration of iron, in mg Fe/L, for a sample with anabsorbance of 0.100?17. In the DPD colorimetric method for the free chlorine residual, which isreported as mg Cl 2 /L, the oxidizing power of free chlorine converts thecolorless amine N,N-diethyl-p-phenylenediamine to a colored dye thatabsorbs strongly over the wavelength range of 440–580 nm. Analysis ofa set of calibration standards gave the following results.mg Cl 2 /L absorbance0.00 0.0000.50 0.2701.00 0.5431.50 0.8132.00 1.08422 Saito, T. Anal. Chim. Acta 1992, 268, 351–355.

650 Analytical Chemistry 2.0A sample from a public water supply is analyzed to determine the freechlorine residual, giving an absorbance of 0.113. What is the free chlorineresidual for the sample in mg Cl 2 /L?18. Lin and Brown described a quantitative method for methanol basedon its effect on the visible spectrum of methylene blue. 23 In the absenceof methanol, the visible spectrum for methylene blue shows twoprominent absorption bands centered at approximately 610 nm and660 nm, corresponding to the monomer and dimer, respectively. Inthe presence of methanol, the intensity of the dimer’s absorption banddecreases, while that of the monomer increases. For concentrations ofmethanol between 0 and 30% v/v, the ratio of the absorbance at 663nm, A 663 , to that at 610 nm, A 610 , is a linear function of the amountof methanol. Using the following standardization data, determine the%v/v methanol in a sample for which A 610 is 0.75 and A 663 is 1.07.%v/v methanol A 663 / A 6100.0 1.215.0 1.2910.0 1.4215.0 1.5220.0 1.6225.0 1.7430.0 1.8419. The concentration of the barbiturate barbital in a blood sample wasdetermined by extracting 3.00 mL of blood with 15 mL of CHCl 3 . Thechloroform, which now contains the barbital, is extracted with 10.0mL of 0.45 M NaOH (pH ≈ 13). A 3.00-mL sample of the aqueousextract is placed in a 1.00-cm cell and an absorbance of 0.115 is measured.The pH of the sample in the absorption cell is then adjusted toapproximately 10 by adding 0.5 mL of 16% w/v NH 4 Cl, giving an absorbanceof 0.023. When 3.00 mL of a standard barbital solution witha concentration of 3 mg/100 mL is taken through the same procedure,the absorbance at pH 13 is 0.295 and the absorbance at a pH of 10 is0.002. Report the mg barbital/100 mL in the sample.20. Jones and Thatcher developed a spectrophotometric method for analyzinganalgesic tablets containing aspirin, phenacetin, and caffeine. 24 Thesample is dissolved in CHCl 3 and extracted with an aqueous solutionof NaHCO 3 to remove the aspirin. After the extraction is complete, thechloroform is transferred to a 250-mL volumetric flask and diluted tovolume with CHCl 3 . A 2.00-mL portion of this solution is diluted tovolume in a 200-mL volumetric flask with CHCl 3 . The absorbance of23 Lin, J.; Brown, C. W. Spectroscopy 1995, 10(5), 48–51.24 Jones, M.; Thatcher, R. L. Anal. Chem. 1951, 23, 957–960.

Chapter 10 Spectroscopic Methods651the final solution is measured at wavelengths of 250 nm and 275 nm,at which the absorptivities, in ppm –1 cm –1 , for caffeine and phenacetinarea 250 a 275caffeine 0.0131 0.0485phenacetin 0.0702 0.0159Aspirin is determined by neutralizing the NaHCO 3 in the aqueoussolution and extracting the aspirin into CHCl 3 . The combined extractsare diluted to 500 mL in a volumetric flask. A 20.00-mL portion of thesolution is placed in a 100-mL volumetric flask and diluted to volumewith CHCl 3 . The absorbance of this solution is measured at 277 nm,where the absorptivity of aspirin is 0.00682 ppm –1 cm –1 . An analgesictablet treated by this procedure is found to have absorbances of 0.466at 250 nm, 0.164 at 275 nm, and 0.600 at 277 nm when using a cellwith a 1.00 cm pathlength. Report the milligrams of aspirin, caffeine,and phenacetin in the analgesic tablet.21. The concentration of SO 2 in a sample of air was determined by thep-rosaniline method. The SO 2 was collected in a 10.00-mL solutionof HgCl 4 2– , where it reacts to form Hg(SO 3 ) 2 2– , by pulling the airthrough the solution for 75 min at a rate of 1.6 L/min. After addingp-rosaniline and formaldehyde, the colored solution was diluted to 25mL in a volumetric flask. The absorbance was measured at 569 nm ina 1‐cm cell, yielding a value of 0.485. A standard sample was preparedby substituting a 1.00-mL sample of a standard solution containing theequivalent of 15.00 ppm SO 2 for the air sample. The absorbance of thestandard was found to be 0.181. Report the concentration of SO 2 inthe air in mg SO 2 /L. The density of air is 1.18 g/liter.22. Seaholtz and colleagues described a method for the quantitative analysisof CO in automobile exhaust based on the measurement of infraredradiation at 2170 cm –1 . 25 A calibration curve was prepared by filling a10-cm IR gas cell with a known pressure of CO and measuring the absorbanceusing an FT-IR. The standardization relationship was foundto beA=− × − 4 −411 . 10 + ( 99 . × 10 ) × PCOSamples were prepared by using a vacuum manifold to fill the gas cell.After measuring the total pressure, the absorbance of the sample at2170 cm –1 was measured. Results are reported as %CO (P CO /P total ).The analysis of five exhaust samples from a 1973 coupe give the followingresults.25 Seaholtz, M. B.; Pence, L. E.; Moe, O. A. Jr. J. Chem. Educ. 1988, 65, 820–823.

652 Analytical Chemistry 2.0P total (torr) absorbance595 0.1146354 0.0642332 0.0591233 0.0412143 0.0254Determine the %CO for each sample, and report the mean value andthe 95% confidence interval.23. Figure 10.32 shows an example of a disposable IR sample card madeusing a thin sheet of polyethylene. To prepare an analyte for analysis, itis dissolved in a suitable solvent and a portion of the sample placed onthe IR card. After the solvent evaporates, leaving the analyte behind asa thin film, the sample’s IR spectrum is obtained. Because the thicknessof the polyethylene film is not uniform, the primary application of IRcards is for a qualitative analysis. Zhao and Malinowski reported howan internal standardization with KSCN can be used for a quantitativeIR analysis of polystyrene. 26 Polystyrene was monitored at 1494 cm –1and KSCN at 2064 cm –1 . Standard solutions were prepared by placingweighed portions of polystyrene in a 10‐mL volumetric flask anddiluting to volume with a solution of 10 g/L KSCN in methyl isobutylketone. A typical set of results is shown here.g polystyrene 0.1609 0.3290 0.4842 0.6402 0.8006A 1494 0.0452 0.1138 0.1820 0.3275 0.3195A 2064 0.1948 0.2274 0.2525 0.3580 0.2703When a 0.8006-g sample of a poly(styrene/maleic anhydride) copolymerwas analyzed, the following results were obtained.replicate A 1494 A 20641 0.2729 0.35822 0.2074 0.28203 0.2785 0.3642What is the %w/w polystyrene in the copolymer? Given that the reported%w/w polystyrene is 67%, is there any evidence for a determinateerror at a = 0.05?24. The following table lists molar absorptivities for the Arsenazo complexesof copper and barium. 27 Suggest appropriate wavelengths for analyzingmixtures of copper and barium using their Arsenzao complexes.26 Zhao, Z.; Malinowski, E. R. Spectroscopy 1996, 11(7), 44–49.27 Grossman, O.; Turanov, A. N. Anal. Chim. Acta 1992, 257, 195–202.

Chapter 10 Spectroscopic Methods653wavelength (nm) e Cu (M –1 cm –1 ) e Ba (M –1 cm –1 )595 11900 7100600 15500 7200607 18300 7400611 19300 6900614 19300 7000620 17800 7100626 16300 8400635 10900 9900641 7500 10500645 5300 10000650 3500 8600655 2200 6600658 1900 6500665 1500 3900670 1500 2800680 1800 150025. Blanco and colleagues report several applications of multiwavelengthlinear regression analysis for the simultaneous determination of twocomponentmixtures. 9 For each of the following, determine the molarconcentration of each analyte in the mixture.(a) Titanium and vanadium were determined by forming complexeswith H 2 O 2 . Results for a mixture of Ti(IV) and V(V) and for standardsof 63.1 ppm Ti(IV) and 96.4 ppm V(V) are listed in thefollowing table.absorbancewavelength (nm) Ti(V) standard V(V) standard mixture390 0.895 0.326 0.651430 0.884 0.497 0.743450 0.694 0.528 0.665470 0.481 0.512 0.547510 0.173 0.374 0.314(b) Copper and zinc were determined by forming colored complexeswith 2-pyridyl-azo-resorcinol (PAR). The absorbances for PAR, amixture of Cu 2+ and Zn 2+ , and standards of 1.00 ppm Cu 2+ and1.00 ppm Zn 2+ are listed in the following table. Note that you mustcorrect the absorbances for the metal for the contribution fromPAR.

654 Analytical Chemistry 2.0absorbancewavelength (nm) PAR Cu standard Zn standard mixture480 0.211 0.698 0.971 0.656496 0.137 0.732 1.018 0.668510 0.100 0.732 0.891 0.627526 0.072 0.602 0.672 0.498540 0.056 0.387 0.306 0.29026. The stoichiometry of a metal–ligand complex, ML n , was determined bythe method of continuous variations. A series of solutions was preparedin which the combined concentrations of M and L were held constantat 5.15 10 –4 M. The absorbances of these solutions were measured ata wavelength where only the metal–ligand complex absorbs. Using thefollowing data, determine the formula of the metal–ligand complex.mole fraction M mole fraction L absorbance1.0 0.0 0.0010.9 0.1 0.1260.8 0.2 0.2600.7 0.3 0.3890.6 0.4 0.5150.5 0.5 0.6420.4 0.6 0.7750.3 0.7 0.7710.2 0.8 0.5130.1 0.9 0.2530.0 1.0 0.00027. The stoichiometry of a metal–ligand complex, ML n , was determinedby the mole-ratio method. A series of solutions was prepared in whichthe concentration of metal was held constant at 3.65 10 –4 M, and theligand’s concentration was varied from 1 10 –4 M to 1 10 –3 M. Usingthe following data, determine the stoichiometry of the metal-ligandcomplex.[ligand] (M) absorbance1 .0 10 –4 0.1222 .0 10 –4 0.2513 .0 10 –4 0.3764 .0 10 –4 0.4965 .0 10 –4 0.6256 .0 10 –4 0.752

Chapter 10 Spectroscopic Methods6557 .0 10 –4 0.8738 .0 10 –4 0.9379 .0 10 –4 0.9621 .0 10 –3 1.00228. The stoichiometry of a metal–ligand complex, ML n , was determinedby the slope-ratio method. Two sets of solutions were prepared. For thefirst set of solutions the concentration of the metal was held constantat 0.010 M and the concentration of the ligand was varied. When theabsorbance of these solutions was measured at a wavelength where onlythe metal–ligand complex absorbs, the following data were obtained.[ligand] (M) absorbance1 .0 10 –5 0.0122 .0 10 –5 0.0293 .0 10 –5 0.0424 .0 10 –5 0.0555 .0 10 –5 0.069For the second set of solutions the concentration of the ligand was heldconstant at 0.010 M, and the concentration of the metal was varied,yielding the following absorbances.[metal] (M) absorbance1 .0 10 –5 0.0402 .0 10 –5 0.0853 .0 10 –5 0.1254 .0 10 –5 0.1625 .0 10 –5 0.206Using this data, determine the stoichiometry of the metal-ligand complex.29. Kawakami and Igarashi developed a spectrophotometric method fornitrite based on its reaction with 5, 10, 15, 20-tetrakis(4-aminophenyl)porphrine (TAPP). As part of their study they investigated the stoichiometryof the reaction between TAPP and NO 2 – . The following dataare derived from a figure in their paper. 28[TAPP] (M) [NO – 2 ] (M) absorbance8 .0 10 –7 1.6 10 –7 0.2278 .0 10 –7 3.2 10 –7 0.19228 Kawakami, T.; Igarashi, S. Anal. Chim. Acta 1996, 33, 175–180.

656 Analytical Chemistry 2.08 .0 10 –7 4.8 10 –7 0.1588 .0 10 –7 8 .0 10 –7 0.1268 .0 10 –7 1.6 10 –6 0.0658 .0 10 –7 2.4 10 –6 0.0478 .0 10 –7 3.2 10 –6 0.0428 .0 10 –7 4.0 10 –6 0.042What is the stoichiometry of the reaction?30. The equilibrium constant for an acid–base indicator is determined bypreparing three solutions, each of which has a total indicator concentrationof 1.35 10 –5 M. The pH of the first solution is adjusted untilit is acidic enough to ensure that only the acid form of the indicator ispresent, yielding an absorbance of 0.673. The absorbance of the secondsolution, whose pH is adjusted to give only the base form of the indicator,is 0.118. The pH of the third solution is adjusted to 4.17 and hasan absorbance of 0.439. What is the acidity constant for the acid–baseindicator?31. The acidity constant for an organic weak acid was determined by measuringits absorbance as a function of pH while maintaining a constanttotal concentration of the acid. Using the data in the following table,determine the acidity constant for the organic weak acid.pH absorbance1.53 0.0102.20 0.0103.66 0.0354.11 0.0724.35 0.1034.75 0.1694.88 0.1935.09 0.2275.69 0.2887.20 0.3177.78 0.31732. Suppose you need to prepare a set of calibration standards for the spectrophotometricanalysis of an analyte that has a molar absorptivity of1138 M –1 cm –1 at a wavelength of 625 nm. To maintain an acceptableprecision for the analysis, the %T for the standards should be between15% and 85%.

Chapter 10 Spectroscopic Methods657(a) What is the concentration of the most concentrated and the leastconcentrated standard you should prepare, assuming a 1.00-cmsample cell.(b) Explain how you will analyze samples with concentrations that are10 mM, 0.1 mM, and 1.0 mM in the analyte.33. When using a spectrophotometer whose precision is limited by theuncertainty of reading %T, the analysis of highly absorbing solutionscan lead to an unacceptable level of indeterminate errors. Consider theanalysis of a sample for which the molar absorptivity is 1.0 10 4 M –1cm –1 and the pathlength is 1.00 cm.(a) What is the relative uncertainty in concentration for an analytewhose concentration is 2.0 10 –4 M if s T is ±0.002?(b) What is the relative uncertainty in the concentration if the spectrophotometeris calibrated using a blank consisting of a 1.0 10 –4 Msolution of the analyte?34. Hobbins reported the following calibration data for the flame atomicabsorption analysis for phosphorous. 29mg P/L absorbance2130 0.0484260 0.1106400 0.1738530 0.230To determine the purity of a sample of Na 2 HPO 4 , a 2.469-g sample isdissolved and diluted to volume in a 100-mL volumetric flask. Analysisof the resulting solution gives an absorbance of 0.135. What is thepurity of the Na 2 HPO 4 ?35. Bonert and Pohl reported results for the atomic absorption analysis ofseveral metals in the caustic suspensions produced during the manufactureof soda by the ammonia-soda process. 30(a) The concentration of Cu was determined by acidifying a 200-mLsample of the caustic solution with 20 mL of concentrated HNO 3 ,adding 1 mL of 27% w/v H 2 O 2 , and boiling for 30 min. The resultingsolution was diluted to 500 mL, filtered, and analyzed by flameatomic absorption using matrix matched standards. The results fora typical analysis are shown in the following table.29 Hobbins, W. B. “Direct Determination of Phosphorous in Aqueous Matricies by Atomic Absorption,”Varian Instruments at Work, Number AA-19, February 1982.30 Bonert, K.; Pohl, B. “The Determination of Cd, Cr, Cu, Ni, and Pb in Concentrated CaCl 2 /NaCl solutions by AAS,” AA Instruments at Work (Varian) Number 98, November, 1990.

658 Analytical Chemistry 2.0solution mg Cu/L absorbanceblank 0.000 0.007standard 1 0.200 0.014standard 2 0.500 0.036standard 3 1.000 0.072standard 4 2.000 0.146sample 0.027Determine the concentration of Cu in the caustic suspension.(b) The determination of Cr was accomplished by acidifying a 200-mL sample of the caustic solution with 20 mL of concentratedHNO 3 , adding 0.2 g of Na 2 SO 3 and boiling for 30 min. The Crwas isolated from the sample by adding 20 mL of NH 3 , producinga precipitate that includes the chromium as well as other oxides.The precipitate was isolated by filtration, washed, and transferredto a beaker. After acidifying with 10 mL of HNO 3 , the solutionwas evaporated to dryness. The residue was redissolved in a combinationof HNO 3 and HCl and evaporated to dryness. Finally, theresidue was dissolved in 5 mL of HCl, filtered, diluted to volume ina 50-mL volumetric flask, and analyzed by atomic absorption usingthe method of standard additions. The atomic absorption resultsare summarized in the following table.sample mg Cr added /L absorbanceblank 0.001sample 0.045standard addition 1 0.200 0.083standard addition 2 0.500 0.118standard addition 3 1.000 0.192Report the concentration of Cr in the caustic suspension.36. Quigley and Vernon report results for the determination of trace metalsin seawater using a graphite furnace atomic absorption spectrophotometerand the method of standard additions. 31 The trace metals were firstseparated from their complex, high-salt matrix by coprecipitating withFe 3+ . In a typical analysis a 5.00-mL portion of 2000 ppm Fe 3+ wasadded to 1.00 L of seawater. The pH was adjusted to 9 using NH 4 OH,and the precipitate of Fe(OH) 3 allowed to stand overnight. After isolatingand rinsing the precipitate, the Fe(OH) 3 and coprecipitated metalswere dissolved in 2 mL of concentrated HNO 3 and diluted to volumein a 50-mL volumetric flask. To analyze for Mn 2+ , a 1.00-mL sample ofthis solution was diluted to 100 mL in a volumetric flask. The followingsamples were injected into the graphite furnace and analyzed.31 Quigley, M. N.; Vernon, F. J. Chem. Educ. 1996, 73, 671–673.

Chapter 10 Spectroscopic Methods659sampleabsorbance2.5-mL sample + 2.5 mL of 0 ppb Mn 2+ 0.2232.5-mL sample + 2.5 mL of 2.5 ppb Mn 2+ 0.2942.5-mL sample + 2.5 mL of 5.0 ppb Mn 2+ 0.361Report the parts per billion Mn 2+ in the sample of seawater.37. The concentration of Na in plant materials may be determined by flameatomic emission. The material to be analyzed is prepared by grinding,homogenizing, and drying at 103 o C. A sample of approximately 4 gis transferred to a quartz crucible and heated on a hot plate to char theorganic material. The sample is heated in a muffle furnace at 550 o Cfor several hours. After cooling to room temperature the residue is dissolvedby adding 2 mL of 1:1 HNO 3 and evaporated to dryness. Theresidue is redissolved in 10 mL of 1:9 HNO 3 , filtered and diluted to50 mL in a volumetric flask. The following data were obtained duringa typical analysis for the concentration of Na in a 4.0264-g sample ofoat bran.sample mg Na/L emission (arbitrary units)blank 0.00 0.0standard 1 2.00 90.3standard 2 4.00 181standard 3 6.00 272standard 4 8.00 363standard 5 10.00 448sample 238Determine the mg Na/L in the sample of oat bran.38. Gluodenis describes the use of ICP atomic emission to analyze samplesof brass for Pb and Ni. 32 The analysis for Pb uses external standards preparedfrom brass samples containing known amounts of lead. Resultsare shown in the following table.%w/w Pb emission intensity0.000 4.29 10 40.0100 1.87 10 50.0200 3.20 10 50.0650 1.28 10 60.350 6.22 10 60.700 1.26 10 732 Gluodenis, T. J. Jr. Am. Lab. November 1998, 245–275.

660 Analytical Chemistry 2.01.04 1.77 10 72.24 3.88 10 73.15 5.61 10 79.25 1.64 10 8What is the %w/w Pb in a sample of brass that gives an emission intensityof 9.25 10 4 ?The analysis for Ni uses an internal standard. Results for a typical calibrationare show in the following table.% w/w Ni emission intensity ratio0.000 0.002670.0140 0.001540.0330 0.003120.130 0.1200.280 0.2460.280 0.2470.560 0.5331.30 1.204.82 4.44What is the %w/w Ni in a sample for which the ratio of emission intensityis 1.10 10 –3 ?39. Yan and colleagues developed a method for the analysis of iron basedits formation of a fluorescent metal–ligand complex with the ligand5-(4-methylphenylazo)-8-aminoquinoline. 33 In the presence of the surfactantcetyltrimethyl ammonium bromide the analysis is carried outusing an excitation wavelength of 316 nm with emission monitored at528 nm. Standardization with external standards gives the followingcalibration curve.I f=− 003 . + 1.594×mg FeL3+A 0.5113-g sample of dry dog food was ashed to remove organic materials,and the residue dissolved in a small amount of HCl and diluted tovolume in a 50-mL volumetric flask. Analysis of the resulting solutiongave a fluorescent emission intensity of 5.72. Determine the mg Fe/Lin the sample of dog food.40. A solution of 5.00 10 –5 M 1,3-dihydroxynaphthelene in 2 M NaOHhas a fluorescence intensity of 4.85 at a wavelength of 459 nm. What33 Yan, G.; Shi, G.; Liu, Y. Anal. Chim. Acta 1992, 264, 121–124.

Chapter 10 Spectroscopic Methods661is the concentration of 1,3-dihydroxynaphthelene in a solution with afluorescence intensity of 3.74 under identical conditions?41. The following data was recorded for the phosphorescence intensity forseveral standard solutions of benzo[a]pyrene.[benzo[a]pyrene] (M) emission intensity0 0.001.00 10 –5 0.983.00 10 –5 3.226.00 10 –5 6.251.00 10 –4 10.21What is the concentration of benzo[a]pyrene in a sample yielding aphosphorescent emission intensity of 4.97?42. The concentration of acetylsalicylic acid, C 9 H 8 O 4 , in aspirin tabletscan be determined by hydrolyzing to the salicylate ion, C 7 H 5 O 2 – , anddetermining the concentration of the salicylate ion spectrofluorometrically.A stock standard solution is prepared by weighing 0.0774 g ofsalicylic acid, C 7 H 6 O 2 , into a 1-L volumetric flask and diluting tovolume with distilled water. A set of calibration standards is preparedby pipeting 0, 2.00, 4.00, 6.00, 8.00, and 10.00 mL of the stock solutioninto separate 100-mL volumetric flasks containing 2.00 mL of 4M NaOH and diluting to volume with distilled water. The fluorescenceof the calibration standards was measured at an emission wavelength of400 nm using an excitation wavelength of 310 nm; results are listed inthe following table.mL of stock solution emission intensity0.00 0.002.00 3.024.00 5.986.00 9.188.00 12.1310.00 14.96Several aspirin tablets are ground to a fine powder in a mortar andpestle. A 0.1013-g portion of the powder is placed in a 1-L volumetricflask and diluted to volume with distilled water. A portion of this solutionis filtered to remove insoluble binders and a 10.00-mL aliquottransferred to a 100-mL volumetric flask containing 2.00 mL of 4 MNaOH. After diluting to volume the fluorescence of the resulting solutionis found to be 8.69. What is the %w/w acetylsalicylic acid in theaspirin tablets?

662 Analytical Chemistry 2.043. Selenium (IV) in natural waters can be determined by complexing withammonium pyrrolidine dithiocarbamate and extracting into CHCl 3 .This step serves to concentrate the Se(IV) and to separate it from Se(VI).The Se(IV) is then extracted back into an aqueous matrix using HNO 3 .After complexing with 2,3-diaminonaphthalene, the complex is extractedinto cyclohexane. Fluorescence is measured at 520 nm followingits excitation at 380 nm. Calibration is achieved by adding knownamounts of Se(IV) to the water sample before beginning the analysis.Given the following results what is the concentration of Se(IV) in thesample.[Se (IV)] added (nM) emission intensity0.00 3232.00 5974.00 8626.00 112344. Fibrinogen is a protein that is produced by the liver and found in humanplasma. Its concentration in plasma is clinically important. Many of theanalytical methods used to determine the concentration of fibrinogenin plasma are based on light scattering following its precipitation. Forexample, da Silva and colleagues describe a method in which fibrinogenprecipitates in the presence of ammonium sulfate in a guanidinehydrochloride buffer. 34 Light scattering is measured nephelometricallyat a wavelength of 340 nm. Analysis of a set of external calibrationstandards gives the following calibration equationI =− 466 . + 9907 . 63 × Cswhere I s is the intensity of scattered light and C is the concentration offibrinogen in g/L. A 9.00-mL sample of plasma was collected from apatient and mixed with 1.00 mL of an anticoagulating agent. A 1.00-mL aliquot of this solution was then diluted to 250 mL in a volumetricflask. Analysis of the resulting solution gave a scattering intensity of44.70. What is the concentration of fibrinogen, in gram per liter, in theplasma sample?10LSolutions to Practice ExercisesPractice Exercise 10.1The frequency and wavenumber for the line are8c 300 . × 10 m/sν = == 457 . × 10−9λ 656.3×10 ms14 −134 da Silva, M. P.; Fernandez-Romero, J. M.; Luque de Castro, M. D. Anal. Chim. Acta 1996, 327,101–106.

Chapter 10 Spectroscopic Methods6631 1ν = =λ 656.3×10−91 m× = 1.524×10m 100 cmcm4 −1Click here to return to the chapter.Practice Exercise 10.2The photon’s energy is−hc ( 6.626× 10 J⋅ s)(3.00×10E = =−9λ656.3×10 mClick here to return to the chapter.34 8m/s)= 303 . × 10−19JPractice Exercise 10.3To find the transmittance, T, we begin by noting thatSolving for TA = 1.27 = –logT–1.27 = logT10 –1.27 = Tgives a transmittance of 0.054, or a %T of 5.4%.Click here to return to the chapter.Practice Exercise 10.4Making appropriate substitutions into Beer’s lawA = 0.228 = ebC = (676 M –1 cm –1 )(1 cm)Cand solving for C gives a concentration of 3.3710 -4 M.Click here to return to the chapter.Practice Exercise 10.5For this standard addition we can write the following equations relatingabsorbance to the concentration of Cu 2+ in the sample. First, for thesample, we have0. 118 =εbC Cuand for the standard addition we have⎛ 20. 00 100 .0.162 = mg Cu mL ⎞εb C Cu+ ×⎝⎜L 10.00 mL⎠⎟The value of eb is the same in both equation. Solving each equation foreb and equating

664 Analytical Chemistry 2.0C Cu0.16220. 00 mg Cu 100 . mL+ ×L 10.00 mL0.118=CCuleaves us with an equation in which C Cu is the only variable. Solving forC Cu gives its value asCCu0.162 0.118=+ 200 . mg Cu/L CCu0. 162C= 0. 118C+ 0.236 mg Cu/LCu Cu0. 044 = 0.236 mg Cu/LC CuC Cu= 54 .mg Cu/LClick here to return to the chapter.Practice Exercise 10.6Substituting into equation 10.11 and equation 10.12 givesA = 0. 336 = 15. 2C + 5.60C400 CrA = 0. 187 = 0. 533C + 507 . C505 CrTo determine C Cr and C Co we solve the first equation for C CoCCo0. 336−15.2C=560 .and substitute the result into the second equation.0. 336−15.2CCr0. 187 = 0. 533C+ 507 . ×= 0.3042− 1323 . CCr560 .CrSolving for C Cr gives the concentration of Cr 3+ as 8.86 10 –3 M. Substitutingthis concentration back into the equation for the mixture’s absorbanceat 400 nm gives the concentration of Co 2+ as 3.60 10 –2 M.Click here to return to the chapter.Practice Exercise 10.7Letting X represent MnO 4 – and Y represent Cr 2 O 7 2– , we plot the equationAAmixSXCrC C AX Y= + ×C C ASXSYSYSXCoCo

Chapter 10 Spectroscopic Methods665placing A mix /A SX on the y-axis and A SY /A SX on the x-axis. For example,at a wavelength of 266 nm the value A mix /A SX of is 0.766/0.042, or 18.2,and the value of A SY /A SX is 0.410/0.042, or 9.76. Completing the calculationsfor all wavelengths and plotting the data gives the result shown inFigure 10.67. Fitting a straight-line to the data gives a regression modelofAAmixSX= 0. 8147 + 1.7839×Using the y-intercept, the concentration of MnO 4 – isCC= [ MnO ]= 0.8147×−X4−4−SX10 . 10 MMnO4AASYSXA mix /A SX201510500 2 4 6 8 10A SY /A SXFigure 10.67 Multiwavelength linearregression analysis for the data in PracticeExercise 10.7.or 8.15 10 –5 M MnO 4 – , and using the slope, the concentration ofCr 2 O 7 2– isCCYSY[ Cr O ]2−= 2 7× MCrO=−42−10 . 102 71.78391.0or 1.7810 –4 M Cr 2 O 7 2– .Click here to return to the chapter.Practice Exercise 10.8Figure 10.68 shows a continuous variations plot for the data in this exercise.Although the individual data points show substantial curvature—enoughcurvature that there is little point in trying to draw linear branches forexcess metal and excess ligand—the maximum absorbance clearly occursat an X L of approximately 0.5. The complex’s stoichiometry, therefore, isFe(SCN) 2+ .Click here to return to the chapter.Practice Exercise 10.9The value of K a is−60. 225−0.000K a= (. 100× 10 ) ×= 495 . × 10 −70. 680 −0.225Click here to return to the chapter.Practice Exercise 10.10To determine K a we use equation 10.21, plotting log[(A – A HIn )/(A In – A)]versus pH, as shown in Figure 10.69. Fitting a straight-line to the datagives a regression model ofabsorbance0.80.60.40.20.00.0 0.2 0.4 0.6 0.8 1.0X LFigure 10.68 Continuous variationsplot for the data in Practice Exercise10.8.log A–A HInAIn–A1.00.50.0-0.5-1.03.0 3.5 4.0 4.5 5.0pHFigure 10.69 Determining the pKa ofbromothymol blue using the data inPractice Exercise 10.10.

666 Analytical Chemistry 2.0A−AHInlog . .A − A=− 380 + 0 962pHInThe y-intercept is –pK a ; thus, the pK a is 3.80 and the K a is 1.5810 –4 .Click here to return to the chapter.

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