Student Solutions Manual for Skoog / West Fundamentals of Analytical Chemistry, 9th Edition

224 Pages·2013·4.18 MB·English

by  Douglas A. Skoog, Donald M. West, F. James Holler

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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
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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
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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.

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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.
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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
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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
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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

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