Table Of Content

This is an electronic version of the print textbook. Due to electronic rights restrictions, some third party content may be suppressed. Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. The publisher reserves the right to remove content from this title at any time if subsequent rights restrictions require it. For valuable information on pricing, previous editions, changes to current editions, and alternate formats, please visit www.cengage.com/highered to search by ISBN#, author, title, or keyword for materials in your areas of interest. Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Student Solutions Manual Fundamentals of Analytical Chemistry NINTH EDITION Douglas A. Skoog Stanford University Donald M. West San Jose State University F. James Holler University of Kentucky Stanley R. Crouch University of Michigan Prepared by Stanley R. Crouch University of Michigan F. James Holler University of Kentucky Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. © 2014 Brooks/Cole, Cengage Learning ISBN-13: 978-0-495-55834-7 ISBN-10: 0-495-55834-6 ALL RIGHTS RESERVED. 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Table of Contents 3 Using Spreadsheets in Analytical Chemistry 1 4 Calculations Used in Analytical Chemistry 2 5 Errors in Chemical Analyses 12 6 Random Errors in Chemical Analysis 15 7 Statistical Data Treatment and Evaluation 23 8 Sampling, Standardization and Calibration 32 9 Aqueous Solutions and Chemical Equilibria 42 10 Effect of Electrolytes on Chemical Equilibria 52 11 Solving Equilibrium Problems for Complex Systems 63 12 Gravimetric Methods of Analysis 72 13 Titrations in Analytical Chemistry 79 14 Principles of Neutralization Titrations 85 15 Complex Acid/Base Systems 106 16 Applications of Neutralization Titrations 115 17 Complexation and Precipitation Reactions and Titrations 128 18 Introduction to Electrochemistry 141 19 Applications of Standard Electrode Potentials 150 20 Applications of Oxidation/Reduction Titrations 155 21 Potentiometry 159 22 Bulk Electrolysis: Electrogravimetry and Coulometry 165 23 Voltammetry 173 24 Introduction to Spectrochemical Methods 175 25 Instruments for Optical Spectrometry 178 26 Molecular Absorption Spectrometry 182 27 Molecular Fluorescence Spectroscopy 191 28 Atomic Spectroscopy 193 29 Mass Spectrometry 196 30 Kinetic Methods of Analysis 197 31 Introduction to Analytical Separations 202 32 Gas Chromatography 209 33 High-Performance Liquid Chromatography 212 34 Miscellaneous Separation Methods 215 iii Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Fundamentals of Analytical Chemistry: 9th ed. Chapter 3 Chapter 3 3-1. (a) SQRT returns the square root of a number or result of a calculation. (b) AVERAGE returns the arithmetic mean of a series of numbers. (c) PI returns the value of pi accurate to 15 digits (d) FACT returns the factorial of a number, equal to 1 × 2 × 3 × … × number. (e) EXP returns e raised to the value of a given number. (f) LOG returns the logarithm of a number to a base specified by the user. 1 Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Fundamentals of Analytical Chemistry: 9th ed. Chapter 4 Chapter 4 4-1. (a) The millimole is an amount of a chemical species, such as an atom, an ion, a molecule or an electron. There are particles mole particles 6.02×1023 × 10−3 = 6.02×1020 mole millimole millimole (c) The millimolar mass is the mass in grams of one millimole of a chemical species. 3 1000mL 1cm3 ⎛ 1 m ⎞ 4-3. The liter: 1 L = × ×⎜ ⎟ = 10−3m3 1 L 1 mL ⎝100cm⎠ 1mol 1 L 1mol Molar concentration: 1 M = × = 1 L 10−3m3 10−3m3 1 MHz 4-4. (a) 3.2×108Hz × = 320 MHz 106Hz 1 mol (c) 8.43×107μmol × = 84.3 mol 106 μmol 1 mm (e) 8.96×106nm × = 8.96 mm 106nm 4-5. For oxygen, for example 15.999 u/atom = 15.999 g/6.022 ×1023 atoms = 15.999 g/mol. So 1 u = 1 g/mol. Thus, 1g = 1mol u. 1mol Na PO 3mol Na+ 6.022×1023 Na+ 4-7. 2.92g Na PO × 3 4 × × =3.22 × 1022 Na+ 3 4 163.94 g 1 mol Na PO 1 mol Na+ 3 4 2 Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Fundamentals of Analytical Chemistry: 9th ed. Chapter 4 2 mol B 1 molB O 4-9. (a) 8.75gB O × × 2 3 = 0.251 molB 2 3 1 molB O 69.62gB O 2 3 2 3 1 g 7 mol O 167.2mgNa B O •10H O× × 2 4 7 2 1000mg 1 mol Na B O •10H O (b) 2 4 7 2 1 molNa B O •10H O × 2 4 7 2 =3.07 × 10−3molO = 3.07 mmol 381.37 g 1 molMn O 3 mol Mn (c) 4.96gMn O × 3 4 × =6.50 × 10−2molMn 3 4 228.81gMn O 1 molMn O 3 4 3 4 1 g molCaC O 2 mol C 333mgCaC O × × 2 4 × =5.20 × 10−3molC (d) 2 4 1000mg 128.10gCaC O 1 molCaC O 2 4 2 4 = 5.20 mmol 0.0555 molKMnO 1000mmol 4-11. (a) 4 × × 2.00L = 111mmolKMnO L 1 mol 4 3.25 × 10−3 MKSCN 1000mmol L × × ×750mL (b) L 1 mol 1000mL = 2.44mmolKSCN 3.33 mgCuSO 1 g 1 molCuSO 1000mmol 4× × 4 × × 3.50L (c) 1 L 1000mg 159.61gCuSO 1 mol 4 = 7.30×10−2mmolCuSO 4 0.414molKCl 1000mmol 1 L (d) × × × 250mL = 103.5 mmolKCl 1 L 1 mol 1000 mL 63.01gHNO 1000mg 4-13. (a) 0.367molHNO × 3× = 2.31 × 104mgHNO 3 1 molHNO 1 g 3 3 1 mol 40.30gMgO 1000mg (b) 245mmolMgO× × × = 9.87 × 103mgMgO 1000mmol 1 molMgO 1 g 3 Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Fundamentals of Analytical Chemistry: 9th ed. Chapter 4 80.04g NH NO 1000mg (c) 12.5molNH NO × 4 3× = 1.00 × 106mg NH NO 4 3 1 molNH NO 1 g 4 3 4 3 548.23g(NH ) Ce(NO ) 1000mg 4.95mol(NH ) Ce(NO ) × 4 2 3 6× (d) 4 2 3 6 1 mol(NH ) Ce(NO ) 1 g 4 2 3 6 = 2.71 × 106mg(NH ) Ce(NO ) 4 2 3 6 0.350molsucrose 1 L 342gsucrose 1000mg × × × 4-15. (a) 1 L 1000mL 1 molsucrose 1 g × 16.0mL = 1.92×103mgsucrose 3.76×10−3molH O 34.02gH O 1000mg 2 2× 2 2× (b) 1 L 1 molH O 1 g 2 2 × 1.92L = 246mgH O 2 2 0.264molH O 1 L 34.02gH O 2 2× × 2 2 × 250mL 4-16. (a) 1 L 1000mL 1 molH O 2 2 = 2.25gH O 2 2 5.75×10−4molbenzoicacid 1 L 122gbenzoicacid × × (b) 1 L 1000mL 1 molbenzoicacid × 37.0mL = 2.60×10−3gbenzoicacid 4-17. (a)pNa = −log(0.0635 + 0.0403) = −log(0.1038) = 0.9838 pCl = −log(0.0635) = 1.197 pOH = −log(0.0403) =1.395 (c) pH=−log(0.400) = 0.398 pCl=−log(0.400+2×0.100)=−log(0.600)=0.222 pZn =−log(0.100)= 1.00 4 Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.