SM fundamentals of analytical chemistry skoog 9th edition

Table of Contents 3 Using Spreadsheets in Analytical Chemistry 1 4 Calculations Used in Analytical Chemistry 2 6 Random Errors in Chemical Analysis 15 7 Statistical Data Treatment an

Trang 1

www.elsolucionario.org

Trang 3

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.

Trang 4

Student Solutions Manual

Prepared by

Stanley R Crouch University of Michigan

F James Holler University of Kentucky

Fundamentals of Analytical Chemistry

Trang 5

copyright herein may be reproduced, transmitted, stored, or

used in any form or by any means graphic, electronic, or

mechanical, including but not limited to photocopying,

recording, scanning, digitizing, taping, Web distribution,

information networks, or information storage and retrieval

systems, except as permitted under Section 107 or 108 of the

1976 United States Copyright Act, without the prior written

permission of the publisher

For product information and technology assistance, contact us at

Cengage Learning Customer & Sales Support,

1-800-354-9706

For permission to use material from this text or product, submit

all requests online at www.cengage.com/permissions

Further permissions questions can be emailed to

permissionrequest@cengage.com

ISBN-13: 978-0-495-55834-7 ISBN-10: 0-495-55834-6

Brooks/Cole

20 Davis Drive Belmont, CA 94002-3098 USA

Cengage Learning is a leading provider of customized learning solutions with office locations around the globe, including Singapore, the United Kingdom, Australia, Mexico, Brazil, and Japan Locate your local office at:

Trang 6

Table of Contents

3 Using Spreadsheets in Analytical Chemistry 1

4 Calculations Used in Analytical Chemistry 2

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

13 Titrations in Analytical Chemistry 79

14 Principles of Neutralization Titrations 85

16 Applications of Neutralization Titrations 115

17 Complexation and Precipitation Reactions and Titrations 128

19 Applications of Standard Electrode Potentials 150

20 Applications of Oxidation/Reduction Titrations 155

22 Bulk Electrolysis: Electrogravimetry and Coulometry 165

24 Introduction to Spectrochemical Methods 175

25 Instruments for Optical Spectrometry 178

26 Molecular Absorption Spectrometry 182

27 Molecular Fluorescence Spectroscopy 191

31 Introduction to Analytical Separations 202

33 High-Performance Liquid Chromatography 212

34 Miscellaneous Separation Methods 215

Trang 8

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

Trang 9

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

23 particles 3 mole 20 particles

×

Trang 10

4

3.33 mg CuSO 1 g 1 mol CuSO 1000 mmol

1 L 1000 mg 159.61 g CuSO 1 mol = 7.30×10 mmol CuSO−

(d) 0.414 mol KCl 1000 mmol× × 1 L 250 mL = 103.5 mmol KCl

1000 mmol 1 mol MgO 1 g

Trang 11

Trang 12

(c) pBa = −log(5.5 × 10–3) = 2.26; pOH = −log(2 × 5.5 × 10−3) = 1.96

(e) pCa = −log(8.7 × 10−3) = 2.06; pBa = −log(6.6 × 10–3) = 2.18

Trang 13

−

(b) There is 1 mole of Mg2+ per mole of KCliMgCl2, so the molar concentration of Mg2+

is the same as the molar concentration of KCliMgCl2 or 1.04 × 10−2M

Trang 14

100 g soln

500 g soln=23.8 g C H OH+x g water

x g water = 500 g soln 23.8 g C H OH = 476.2 g water

−Mix 23.8 g ethanol with 476.2 g water

100 mL soln4.75 mL C H OH

3

0.0750 mol AgNO0.0750 M AgNO

L0.0750 mol AgNO 169.87 g AgNO 1 L

Trang 15

(b)

2

0.285 mol HCl

×1 L = 0.285 mol HClL

1 L0.285 mol HCl × = 4.75×10 L HCl

71 g HClO 1.67 g reagent 1 g water 1000 mL mol HClO

Trang 16

(f)

+ +

4-33

3+

2 3

105.99 g0.0731 mol HCl 1 L

× × 100.0 mL = 7.31 × 10 mol HCl

−

Trang 17

Because one mole of CO2 is evolved for every mole Na2CO3 reacted, Na2CO3 is the limiting reagent Thus

1 mol CO 44.00 g CO2.094 10 mol Na CO × × = 9.214 × 10 g CO evolved

mol SO 64.06 g SO2.3 10 mol Na SO × × = 1.5 g SO evolved

2 4

Trang 18

4-39 A balanced chemical equation can be written as:

AgNO3 + KI → AgI(s) + KNO3

2 3

3 3

1 1 g 1 mol KI24.31 ppt KI × × × 200.0 mL × = 2.93×10 mol KI

10 ppt 1 mL 166.0 g

1 mol AgNO 1 L2.93 × 10 mol KI × × = 2.93 L AgNO

1 mol KI 0.0100 mol AgNO

−

2.93 L of 0.0100 M AgNO3 would be required to precipitate I− as AgI

Trang 19

Chapter 5

value while systematic error causes the mean of a data set to differ from the accepted value

(c) The absolute error of a measurement is the difference between the measured value and

the true value while the relative error is the absolute error divided by the true value

5-2 (1) Meter stick slightly longer or shorter than 1.0 m – systematic error

(2) Markings on the meter stick always read from a given angle – systematic error

(3) Variability in the sequential movement of the 1-m metal rule to measure the full 3-m table width – random error

(4) Variability in interpolation of the finest division of the meter stick – random error

5-4 (1) The analytical balance is miscalibrated

(2) After weighing an empty vial, fingerprints are placed on the vial while adding sample

Trang 20

5-7 Both constant and proportional systematic errors can be detected by varying the sample

size Constant errors do not change with the sample size while proportional errors

increase or decrease with increases or decreases in the samples size

Trang 21

3

0105.00104.00110

00013.001063.00105.0

00023.001063.00104.0

Trang 22

Chapter 6

6-1 (a) The standard error of the mean is the standard deviation of the mean and is given by

the standard deviation of the data set divided by the square root of the number of

measurements

(c) The variance is the square of the standard deviation

6-2 (a) The term parameter refers to quantities such as the mean and standard deviation of a

population or distribution of data The term statistic refers to an estimate of a parameter

that is made from a sample of data

(c) Random errors result from uncontrolled variables in an experiment while systematic

errors are those that can be ascribed to a particular cause and can usually be determined

6-3 (a) The sample standard deviation s is the standard deviation of a sample drawn from the

N

i

, where x is the sample mean

The population standard deviation σ is the standard deviation of an entire population

1

N i i

x N

μ

σ =

−

= ∑

, where μ is the population mean

6-5 Since the probability that a result lies between -1σ and +1σ is 0.683, the probability that a

result will lie between 0 and +1σ will be half this value or 0.342 The probability that a result will lie between +1σ and +2σ will be half the difference between the probability of the result being between -2σ and +2σ, and -1σ and +1σ, or ½ (0.954-0.683) = 0.136

Trang 23

6-7 Listing the data from Set A in order of increasing value:

9.5 90.25 8.5 72.25 9.1 82.81 9.3 86.49 9.1 82.81

(e) coefficient of variation: CV = (0.37/9.1) × 100% = 4.1%

Results for Sets A through F, obtained in a similar way, are given in the following table

Trang 24

71

1083.2

Trang 25

0.434 0.03 10

6.51 102.00 10

Trang 26

6-15 Since the titrant volume equals the final buret reading minus the initial buret reading, we

can introduce the values given into the equation for %A

Trang 27

6-17 We first calculate the mean transmittance and the standard deviation of the mean

T b

s c

Trang 28

6-19

(a) The standard deviations are s1 = 0.096, s2 = 0.077, s3 = 0.084, s4 = 0.090, s5 = 0.104, s6=

0.083

(b) spooled = 0.088 or 0.09

Trang 29

6-21

Trang 30

Chapter 7

7-1 The distribution of means is narrower than the distribution of single results Hence, the

standard error of the mean of 5 measurements is smaller than the standard deviation of a single result The mean is thus known with more confidence than is a single result

2.7 7.29 3.0 9.00 2.6 6.76 2.8 7.84 3.2 10.24

Since, for a small set of measurements we cannot be certain s is a good approximation of

σ, we should use the t statistic for confidence intervals From Table 7-3, at 95%

confidence t for 4 degrees of freedom is 2.78, therefore for set A,

CI for μ = 2.86 ± (2.78 0.24)( )

5 = 2.86 ± 0.30

Trang 31

Similarly, for the other data sets, we obtain the results shown in the following table:

The 95% confidence interval is the range within which the population mean is expected

to lie with a 95% probability

7-5 If s is a good estimate of σ then we can use z = 1.96 for the 95% confidence level For

set A, at the 95% confidence,

Trang 32

4 × = 3.22 ± 0.15 meq Ca/L

(b) 95% CI = 3.22 ±

3

056.096

1 × = 3.22 ± 0.06 meq Ca/L

7-13 (a) 0.3 = 2.58 0.38

N

× For the 99% CI, N = 10.7 ≅ 11

7-15 This is a two-tailed test where s → σ and from Table 7-1, zcrit = 2.58 for the 99%

confidence level

For As: 129 119 1.28 2.58

3 39.5

3 3

+

×

No significant difference exists at the 99% confidence level

Proceeding in a similar fashion for the other elements

For two of the elements there is a significant difference, but for three there are not Thus,

the defendant might have grounds for claiming reasonable doubt It would be prudent,

Trang 33

however, to analyze other windows and show that these elements are good diagnostics for the rare window

3.46

5

1.56.5

=

−

=

Q and Qcrit for 8 observations at 95% confidence = 0.526

Since Q < Qcrit the outlier value 5.6 cannot be rejected at the 95% confidence level

7-19. The null hypothesis is that for the pollutant the current level = the previous level (H0:

μcurrent = μprevious) The alternative hypothesis is Ha: μcurrent > μprevious This would be a one-tailed test The type I error for this situation would be that we reject the null

hypothesis when, in fact, it is true, i.e we decide the level of the pollutant is > the

previous level at some level of confidence when, in fact, it is not The type II error would

be that we accept the null hypothesis when, in fact, it is false, i.e we decide the level of the pollutant = the previous level when, in fact, it is > than the previous level

7-20 (a) H0: μΙ SE = μEDTA, Ha: μΙ SE ≠μEDTA This would be a two-tailed test The type I error

for this situation would be that we decide the methods agree when they do not The type

II error would be that we decide the methods do not agree when they do

7-21 (a) For the Top data set, x =26.338

For the bottom data set, x =26.254

spooled = 0.1199

degrees of freedom = 5 +5 – 2 = 8

Trang 34

For 8 degrees of freedom at 95% confidence tcrit = 2.31

26.338 26.254

1.11

5 50.1199

5 5

+

×

Since t < tcrit, we conclude that no significant difference

exists at 95% confidence level

(b) From the data, N = 5, d =0.084 and sd = 0.015166

For 4 degrees of freedom at 95% confidence t = 2.78

0.084 0

12.520.015 / 5

Since 12.52 > 2.78, a significant difference does exist at 95% confidence level

(c) The large sample to sample variability causes sTop and sBottom to be large and masks

the differences between the samples taken from the top and the bottom

7-23 For the first data set: x=2.2978

For the second data set: x =2.3106

spooled = 0.0027

Degrees of freedom = 4 + 3 – 2 = 5

207.634

340027

0

3106.22978

×+

−

=

t

For 5 degrees of freedom at the 99% confidence level, t = 4.03 and at the 99.9%

confidence level, t = 6.87 Thus, we can be between 99% and 99.9% confident that the

nitrogen prepared in the two ways is different The Excel TDIST(x,df,tails) function can

be used to calculate the probability of getting a t value of –6.207 In this case we find

TDIST(6.207,5,2) = 0.0016 Therefore, we can be 99.84% confident that the nitrogen

Trang 35

prepared in the two ways is different There is a 0.16% probability of this conclusion

(b) H0: μbrand1 = μbrand2 = μbrand3 = μbrand4 = μbrand5; Ha: at least two of the means differ

(c) The Excel FINV(prob,df1,df2) function can be used to calculate the F value for the

above problem In this case we find FINV(0.05,4,25) = 2.76 Since F calculated exceeds

F critical, we reject the null hypothesis and conclude that the average ascorbic acid

contents of the 5 brands of orange juice differ at the 95% confidence level

7-27

(a) H0: μAnalyst1 = μAnalyst2 = μAnalyst3 = μAnalyst4; Ha: at least two of the means differ

(b) See spreadsheet next page From Table 7-4 the F value for 3 degrees of freedom in

the numerator and 12 degrees of freedom in the denominator at 95% is 3.49 Since F

calculated exceeds F critical, we reject the null hypothesis and conclude that the analysts

differ at 95% confidence The F value calculated of 13.60 also exceeds the critical values

at the 99% and 99.9% confidence levels so that we can be certain that the analysts differ

at these confidence levels

(c) Based on the calculated LSD value there is a significant difference between analyst 2

and analysts 1 and 4, but not analyst 3 There is a significant difference between analyst

3 and analyst 1, but not analyst 4 There is a significant difference between analyst 1 and

analyst 4

Trang 36

Spreadsheet for Problem 7-27

Trang 37

7-29. (a) H0: μISE = μEDTA = μAA; Ha: at least two of the means differ

Trang 38

(c) Based on the calculated LSD value there is a significant difference between the

atomic absorption method and the EDTA titration There is no significant difference between the EDTA titration method and the ion-selective electrode method and there is

no significant difference between the atomic absorption method and the ion-selective electrode method

62.8410.85

70.8410.85

=

−

=

Since Q < Qcrit the outlier value 85.10 cannot be rejected with 95% confidence

62.8410.85

70.8410.85

=

−

=

Since Q > Qcrit the outlier value 85.10 can be rejected with 95% confidence

Trang 39

Chapter 8

8-1 The sample size is in the micro range and the analyte level is in the trace range Hence,

the analysis is a micro analysis of a trace constituent

8-3 Step 1: Identify the population from which the sample is to be drawn

Step 2: Collect the gross sample

Step 3: Reduce the gross sample to a laboratory sample, which is a small quantity of homogeneous material

Trang 40

8-9

2 2

2

)1

d d p p

N

r

B A B A

Định dạng
Số trang	225
Dung lượng	4,42 MB