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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

www.elsolucionario.org 

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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 learning solutions with office locations aro including Singapore, the United Kingdom, Mexico, Brazil, and Japan. Locate your loc www.cengage.com/global www.cengage.com/global Cengage Learning products are represent Canada by Nelson Education, Ltd. To learn more about Brooks/Cole, visit www.cengage.com/brookscole Purchase any of our products at your loca store or at our preferred online store www.cengagebrain.com

Table of Contents 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34

Using Spreadsheets in Analytical Chemistry Calculations Used in Analytical Chemistry Errors in Chemical Analyses Random Errors in Chemical Analysis Statistical Data Treatment and Evaluation Sampling, Standardization and Calibration

1 2 12 15 23 32

Aqueous Solutions and Chemical Equilibria Effect of Electrolytes on Chemical Equilibria Solving Equilibrium Problems for Complex Systems

42 52 63

Gravimetric Methods of Analysis Titrations in Analytical Chemistry Principles of Neutralization Titrations Complex Acid/Base Systems Applications of Neutralization Titrations Complexation and Precipitation Reactions and Titrations

72 79 85 106 115 128

Introduction to Electrochemistry Applications of Standard Electrode Potentials Applications of Oxidation/Reduction Titrations Potentiometry Bulk Electrolysis: Electrogravimetry and Coulometry Voltammetry

141 150 155 159 165 173

Introduction to Spectrochemical Methods Instruments for Optical Spectrometry Molecular Absorption Spectrometry

175 178 182

Molecular Fluorescence Spectroscopy Atomic Spectroscopy Mass Spectrometry

191 193 196

Kinetic Methods of Analysis Introduction to Analytical Separations Gas Chromatography High-Performance Liquid Chromatography Miscellaneous Separation Methods

197 202 209 212 215

Chapter 3 3-1.

(a) SQRT SQRT returns returns the the square square root root of of aa number number or or result result of of aa calculation. calculation.  (b) AVERAGE returns the arit arithmetic hmetic mean of a series of numbers. (c) PI returns the value of pi accurate to 15 digits (d) FACT returns tthe he factorial of a number, equal to 1 ×2 ×3 ×… ×number. (e) EXP EXP returns returns ee raised to the the value value of a given number. number. (f) LOG LOG returns returns the the logarithm logarithm of of aa number number to to aa base base specified specified by by the the user. user.

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 6.02 ×10

23

particles mole

(c) The millimolar

4-3.

×

The liter: 1 L

10

massis

10 8Hz (a) 3.2 × 10

6 (e) 8.96× 1 10 0 nm

1 mol

10 6 Hz

×

×

particles

20

millimole



106 nm

×

=

=

1L

=

10 3 m 3 −

1L

1 mol 6 10 μmol 1 mm

3

⎛ 1m ⎞ × ⎜ ⎟ 1 mL ⎝ 100 cm ⎠

1 cm 3

×

1 MHz

×

(c) 8.43× 107 μmol

4-5.

millimole

1L

Molar concentration: 1 M =

4-4.

6.02 ×10

=

the mass in grams of one millimole of a chemical speci

1000 mL

=

mole

3

−

=

10 3 m3  −

1 mol 10 3 m 3 −



320 MHz 

=

84.3 mol

8.96 mm  23

For For oxygen, oxygen, for for example example 15.999 15.999 u/atom u/atom == 15.999 15.999 g/6.022 g/6.022 ×10 ×10 atoms = 15.999 g So 1 u = 1 g/mol. Thus, 1g = 1mol u. +

4-7.

mo ol Na 3 PO4 × 3 mol Na Na 2.92 g Na 3PO 4 × 1 m 16 3 . 9 4 g 1 mol Na Na3 PO4

2

23

+

× Na × 6. 0 2 2 1 0 1 mol Na Na +

=

3 .2 2

×

1022

16 7 . 2 m mg gN Naa2 B4 O7 • 10H2 O× (b) ×

1 mol Na Na 2 B4 O7 • 10H2 O

=

381.37 g (c) 4.96 g Mn 3O 4 ×

1g × 7 mol O 1000 mg mg 1 mol Na2 B4 O7 • 10H2 O 

3.07 3.07

1 mol Mn 3O 4

×

10 3 molO = 3. 3.07 07 mm mmol ol −

3 mol Mn

×

=

228.81 g Mn 3O 4 1 mol Mn 3O 4

1g mol CaC2 O4 × 333 mg CaC 2O 4 × (d)  1000 mg 128.10 g CaC2 O4

×

6.50

×

10 2mol Mn  −

2 mol C 1 mol C CaaC2 O4

=

5.20

×

3

−

10 m

= 5.20 mmol 0.0555 mol KMnO 4 1000 mmol × × 2.00 2.00 L = 11 111 1 mm mmol ol KM KMnO nO4  4-11. (a)  L 1 mol 3. 2 5

×

(b)  

10 3 M K KS SC N 1 0 0 0 m mm mol L × 1 mol ×1000 mL ×750 mL  L −

= 2 2.44 .44 mm mmol ol KS KSCN CN 3.33 mg CuSO4 1L

(c)

×

1g

×

1 mol CuSO4

1000 mg 159.61 g CuSO4

×

1000 mmol × 3. 5 0 L 1 mol 

= 7. 7.30 30×1 ×10 0 2 mmol Cu CuSO SO4 −

(d) 

0.414 mo m ol K KC Cl 1000 m mm mol 1L × × 1L 1 mol 1000 mL

63.0 63.01 1 g HN HNO O

×

250 mL = 103.5 mmol KCl

1000 1000 mg 4 = 2 . 3 1 × 1 0 m g HNO 3  1g

3 4-13. (a) 0.367 mol HNO × 1 mol HNO 3 × 3

(b)  245 mmol MgO×

1 mol

40.30 g MgO 1000 mg mg × = 9.87 × 10 3mg MgO  1000 mm mmol 1 mol M Mg gO 1g ×

3

(d) 

4 2

4.95 mol (NH4 )2 Ce(NO3 )6 ×

3 6

1 mol (NH4 )2 Ce(NO3 )6 6 = 2.71 × 10 mg (NH4 )2 Ce(NO3 )6

×

0.350 mo m ol su sucrose 1L 342 g sucrose 1000 m mg g × 1 000 mL × 1 mol sucrose× 1 g 1L  4-15. (a)  3

× 16 16.0 .0 mL = 1 1.9 .92× 2×10 10 mg su sucr cros osee 3

−

3.76×10 mol H2 O2 34.02 g H2 O2 1000 m mg g × × 1L 1 mol H 2O 2 1g  (b)  × 1.92 L = 246 mg H2 O2

0.264 mol H 2 O 2 1L

4-16. (a) 

×

1L

34.02 g H2 O2 × × 250 mL 1000 mL 1 mol H2 O2 

= 2.25 g H 2 O 2 4

−

5.75×10 mol benzoic aaccid 1L 122 g benzoic aaccid × × 1L 10 00 mL 1 mol benzoic acid  (b) × 37 37.0 .0 mL = 2. 2.60 60×1 ×10 0 3 g benz benzoi oicc acid acid −

4-17. (a) (a)pNa 8 pNa = − log(0.0 log(0.0635 635 + 0 0.0403) .0403) = − log(0.10 log(0.1038) 38) = 0.983 0.9838

pCl =

−

pOH =

log(0 log(0.0635) .0635) = 1.19 1.197 7

−

log(0.0403 log(0.0403))

=

1.395

(c)

pH = − log(0.4 log(0.400) 00) = 0.398 pCl

=−

log(0.400 + 2 × 0.100) = − log(0.600) log(0.600) = 0.222 

pZn = − log(0.100) = 1.00 

4

1g



pFe(C pFe(CN) N)6

7

−

log( log(1.62 1.62× 10 ) = 6.790 

= −

–5 4-18. (a)  pH = 4.31, log[H3O+] = −4.31, [H 3O+] = 4.9 × 10 M

as in part (a) (c) [H3O+]= 0.26 M  +

−8

(e) [H3O ] = 2.4 × 10 M +

(g)[H3O ] = 5.8 M 4-19. (a) pNa = pBr = −log(0.0300) = = 1.523 1.523 (c)pBa = −log(5.5 × 10–3) = 2.26; pOH = −log(2 × 5.5 × 10 −3) = 1.96 –3 (e) pCa = −log(8.7 × 10−3) = 2.06; pBa = −log(6.6 × 10 ) = 2.18

pCl = −log(2 × 8.7 × 10 3+ 2 × 6.6 × 10 −3) = −log(0.0306) log(0.0306) = 1.51 −

+

+

4-20. (a) pH = 1.020 ;log[H3O ] = −1.020; [H 3O ] = 0.0955 M – (c)pBr = 7.77; ] = 1.70 × 10 8M 7.77; [Br [Br −

+

−13

(e)pLi = 12.35; [Li ] = 4.5 × 10

M

2+

(g)pMn = 0.135; [Mn ] = 0.733 M +

4-21. (a) 1.08 × 103 ppm Na+

270 ppm SO 4 2

−

×

1 mL L × 1 mol Na Na × 1 .0 2 g × 1 0 0 0 m 6 10 ppp pm 1 mL 1L 2 2 .9 9 g

×

1

×

6

1.02 g 1000 mL

10 pp ppm

×

1 mL −

=−

×

1L

96.06 g

2

log(2.8 log(2.87 7 × 10 3 ) = 2.542 −

 5



4 . 7 9 ×1 0 2 M N −

−

(b) pNa = − log(4.79 × 10 ) = 1.320  pSO 4

1 mol mol SO 4 3

=

=

2.87 ×10 3M SO 4 2  −

−

is the same as the molar concentration of KCl MgCl2or 1.04 × 10 2M −

i

−

3 mol Cl

6H2 O O× × =3.12×1 0 2 M Cl  (c) 1.04×10 M KCl•MgC l2 ••6 1m mo ol KCl Cl••MgCl2 •6H2 O −

(d)

2

5.76 g KCl•MgCl 2•6H 2O 2.00 L

3.12× 10 2 mol Cl −

(e)

−

1L

×

×

1L 1000 mL

1L

1000 mmol ×25 mL 1000 mL 1 mol ×

=

1

−

 39.10 g K+

1.04 10 M KCl•MgCl •6H O 1 mol K KC Cl•MgC2l •6 H2 O =

2

40 407 7 mg 1L

2

−

7.8 ×10 mmol Cl

1 mol K+

2

×

−

×100%=0.288% ((w w/v) 

−

(f)

−

×

×

+

1 mol K

10 0 0 m ×

1g

= 407 ppm K+ 

− (g)pMg = –log(1.04 × × 10 2) = 1.983

−2

(h) pCl = –log(3.12 × 10 ) = 1.506

4-25. (a)

6.42% Fe(NO3 )3 = =

1

−

2 2..81× 10

6.42 g Fe(NO 3 ) 3 1.059 g 1000 mL 1 mol Fe(NO3 )3 × × × 100 g solution mL 1L 2 41 . 8 6 g 

M Fe(NO3 )3 = 0.281 M

(b) -1

1

−

2.81 ×10 M Fe(NO3 )3

=

2.81×10 mol Fe F e(NO3 )3

×

L

3m mo ol NO3

−

=

1 mol Fe(NO3 )3

2.81×10 1mol Fe(NO 3) 3 241.86 g Fe(NO 3) 3

8. 4 3 × 1 0 1 M −

−

(c)

×

L

1 mol

×

1L

=

6



www.elsolucionario.org

1

6.80 ×10 g Fe(NO 3) 3 = 68

(b) 500 g soln=23.8 g C 2 H5 OH+x g water x g water = 500 g soln − 23.8 g C2 H5 OH = 476. 6.2 2 g wate ater



Mix Mix 23.8 23.8 gg ethanol ethanol with with 476.2 476.2 gg water water 4.75% (v/v) C2 H 5OH= (c)

4.75 mL C 2H 5OH 100 mL soln soln

4.75 mL C2 H 5OH 100 mL soln soln

× 500 mL sso oln

=

 1

2.38 ×10 mL C 2H 5OH

Dilute Dilute 23.8 23.8 mL mL ethanol ethanol with with enough enough water water to to give give aa final final volume volume of of 500 500 mL. mL. 4-29.

6.00 mol H3 PO4 L 86 g H3 PO4

1L × ×75 0 mL = 4.50 mo mol H3 PO4 1000 mL

1.71 g reagent g water 1000 mL mol H3 PO4 × × × 100 g reagent g water mL L 9 8 .0 g =

×

1.5 1.50×101mol H3 PO4 L

volume 86% (w/w) H3 PO4 requ quiired = 4.50 mol H3 PO4 ×

L 1.5 1.50×10 mol H3 PO4 1

=

3 .0

 0.0 0.075 750 0 mo moll Ag AgNO NO3 L 0.0750 mol Ag NO3 169.87 g AgNO3 1L 4-31. (a) = × × ×500 mL  L 1 mol 1000 mL 0. 0.07 0750 50 M AgNO gNO3

=

=

6. 6.37 37 g AgN gNO O3

Dissolve 6.37 g AgNO 3in in enough enough water water to to give give aa final final volume volume of of 500 500 mL. mL.

7



 F

d

t l

fA

l ti l Ch

i t

9th d

Ch

(c)

2 + 4 6 4 6 3.24× 10 mol K ×1 mol K Fe(+CN) × 368.43 g K Fe(CN) = 2.98 g K4 Fe Fe(C 4 mol K mol −

in enough enough water water to to give give aa final final volume volume of of 400 400 mL. mL. Dissolve 2.98 g K4Fe(CN)6in 3. 3.00 00 g BaCl aCl2 (d)

100 m mL L soln soln

× 6 600 00 mL = 1 1.8 .8×1 ×10 01 g BaCl BaCl2

1 mol BaCl2 L 1.8 ×101 g BaCl2 × × = 2 . 16 × 1 0 1 L 20 8 . 2 3 g 0 . 40 0 m mo ol B BaaC2l



−

Take Take 216 216 mL mL of of the the 0.400 0.400 M M BaCl BaCl2solution and dilute to 6600 00 mL with water (e)

0. 0.120 120 mo moll HC HClO lO4 L 71 g HClO4

×2 ×2.0 .00 0L= 0 0.2 .240 40 mo moll HC HClO lO4

1.67 g reagent 1 g water 1000 mL mol HClO4 × × × 100 g reagent 1 g water 1 mL 1L 1 0 0 . 46 g =

×

1.1 1.18× 8×10 101mo moll HC HClO lO4 L

volum olumee 71 71% % (w/w) w/w) HClO ClO4 requi equirred = 0. 0.24 240 0 mol HC HCllO4 ×

L

1. 1.18 18×1 ×10 01 mo moll HC HClO lO4 Take Take 20.3 20.3 mL mL of of the the concentrated concentrated reagent reagent and and dilute dilute to to 2.00 2.00 L L with with water. water.

=2

8



 Fundamentals of Analytical Chemistry: 9thed. 

Ch

Dissolve 1.7 g Na 2SO4in in enough enough water water to to give give aa final final volume volume of of 9.00 9.00 L. 4-33. 3+

0.250 mol La 1L 2 3+ × × 5 0 .0 m mL L = 1.25×10 mol La La L 1 000 mL −

0.302 M IO3

−

=

0. 0.30 302 2 mol IO3

−

1L

×

1L

 2

−

1000 mL

× 75.0 mL = 2.27×10 −

mol IO3

−

−

Because each mole of La(IO 3)3requires three moles IO 3 , IO3 is the limiting reage Thus, 2.27 × 10 2 mol IO 3 −

4-35.

−

×

1 mol La(IO 3 ) 3 3 mol IO3

−

×

663.6 g La(IO 3) 3 1 mol

= 5.01 g La(IO 3) 3fo formed 

A balanced chemical equation can be written as: Na 2 CO3 + 2HCl

→

2NaCl + H2 O + CO2 (g)

(a)

1 mol Na2 CO3 0. 2 2 2 0 g N Naa2 CO3 × = 2.094 × 10 3 mol Na2 CO3 10 105.9 5.99 9g  0. 0 7 3 1 m mo ol HC H Cl 1L × × 100.0 mL mL = 7.31 × 10 3 mol H HC Cl L 1000 mL −

−

9



www.elsolucionario.org

 th

Fundamentals of Analytical Chemistry: 9 ed. Ch  Because one mole of CO 2is evolved for every mole Na 2CO3reacted, Na2CO3is th

100.0 mL 4-37

1L

A balanced chemical equation can be written as: Na 2SO3 + 2HClO4

→

2NaClO4 + H2 O+ SO2 (g) 

(a)

0.3132 M Na2 SO3

=

0.31 .3132 mol Na2 SO3 L

L × × 75 mL = 2.3 × 10 2 mol Na2 S 1000 mL −

0.4025 mo mol HC HClO4 L 2 ×1000 mL × 150.0 mL = 6.038 × 10 mol H L −

0.4025 M HClO

4 =

Because one mole SO2is evolved per mole Na 2SO3, Na2SO3is the limiting reagent Thus, 2.3 × 10 2 mol Na 2SO 3×

mol S SO O2

−

mol Na 2S SO O3

×

64.06 g S SO O2 mol

= 1.5 g SO 2 evolved 

(b) 

mol HClO4 unreacted = (6.038 × 10 2 mol− (2 × 2.3× 10 2 ) = 1.4× 10 2 mol −

1. 4

×

10 2 mol H HC ClO4 −

225 mL

×

1000 mL mL L

−

−

= 6.4 × 10 2 M HClO4 = 0.064 M −



10



 th Fundamentals of Analytical Chemistry: 9 ed.  4-39. A balanced chemical equation can be written as: AgNO3 + KI → AgI(s) + KNO 3

Ch

11



 th Fundamentals of Analytical Chemistry: 9 ed.

Ch

Chapter 5 5-1.

(a) Random error causes data to bbee scattered more or less symmetrically around a m

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

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

5-4.

(1) The analytical balance is miscalibrated. (2) (2) After After weighing weighing an an empty empty vial, vial, fingerprints fingerprints are are placed placed on on the the vial vial while while adding adding ss to the vial. (3) (3) A A hygroscop...