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Skoog
ManualetSkoog
al.,
1 ofFundamentals
233
et al., Fundamentals
of Analytical
of Analytical
Chemistry,
Chemistry,
10e, © 2022,
10e, ©
978-0-357-45055-0
2022, 978-0-357-45055-0
Student Solution Manual
Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e,
© 2022, 978-0-357-45055-0, Chapter 2: Calculations Used in Analytical Chemistry
Some of the answers below may differ in format but have the same value as your result. Please
check with your instructor if a specific format is desired.
Chapter 2
2-1. Define
Answers:
(a) molar mass.
The molar mass is the mass in grams of one mole of a chemical species.
(c) millimolar mass.
The millimolar mass is the mass in grams of one millimole of a chemical species.
2-3. Give two examples of units derived from the fundamental base SI units.
Solution:
3
1000 mL 1 cm3 m −3 3
The liter: 1 L = × × = 10 m
1L mL 100 cm
1 mol L 1 mol
Molar concentration: 1 M = × −3 3 = −3 3
L 10 m 10 m
2-4. Simplify the following quantities using a unit with an appropriate prefix:
Solutions:
(a) 5.8 × 108 Hz.
MHz
5.8 × 108 Hz × = 580 MHz
106 Hz
(c) 9.31 × 107 µmol.
mol
9.31× 10 7 µmol × 6
= 93.1 mol
10 µmol
(e) 3.96 × 106 nm.
mm
3.96 × 106 nm × = 3.96 mm
106 nm
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publicly accessible website, in whole or in part.
1
StudentStudent
SolutionSolution
Manual
StudentManual
Student
Solution
Student
Solution
Manual
Solution
Manual
StudentManual
Skoog
Solution
et
Skoog
al.,
Manual
Fundamentals
et al.,
Skoog
Fundamentals
et al.,
of Analytical
Fundamentals
of Analytical
Chemistry,
of Chemistry,
Analytical
10e, ©Chemistry,
10e,
2022,
©978-0-357-45055-0
2022,
10e,978-0-357-45055-0.pdf
© 2022, 978-0-357-45055-0
,2026/2027:
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Skoog
ManualetSkoog
al.,
2 ofFundamentals
233
et al., Fundamentals
of Analytical
of Analytical
Chemistry,
Chemistry,
10e, © 2022,
10e, ©
978-0-357-45055-0
2022, 978-0-357-45055-0
Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e,
© 2022, 978-0-357-45055-0, Chapter 2: Calculations Used in Analytical Chemistry
2-5. Why is 1 g no longer exactly 1 mole of unified atomic mass units?
Answer:
The dalton is defined as 1/12 the mass of a neutral 12C atom. With the redefinition of SI
base units in 2019, the definition of the dalton remained the same. However, the
definition of the mole and the kilogram changed in such a way that the molar mass unit
is no longer exactly 1 g/mol.
2-7. Find the number of Na+ ions in 2.75 g of Na3PO4?
Solution:
1 mol Na3PO4 3 mol Na+ 6.022 × 1023 Na+
2.75 g Na3PO4 × × × = 3.03 × 1022 Na+
163.94 g mol Na3PO4 mol Na+
2-9. Find the amount of the indicated element (in moles) in
Solutions:
(a) 5.32 g of B2O3.
2 mol B mol B2O3
5.32 g B2O3 × × = 0.153 mol B
mol B2O3 69.62 g B2O3
(b) 195.7 mg of Na2B4O7 ⋅ 10H2O.
g 7 mol O
195.7 mg Na2B4O7 ⋅ 10H2O × ×
1000 mg mol Na2B4O7 ⋅ 10H2O
mol Na2B4O7 ⋅ 10H2O
× = 3.59 × 10−3 mol O = 3.59 mmol
381.37 g
(c) 4.96 g of Mn3O4.
mol Mn3O4 3 mol Mn
4.96 g Mn3O4 × × = 6.50 × 10 −2 mol Mn
228.81 g Mn3O4 mol Mn3O4
(d) 333 mg of CaC2O4.
g mol CaC2O4 2 mol C
333 mg CaC2O4 × × × = 5.20 × 10−3 mol C
1000 mg 128.10 g CaC2O4 mol CaC2O4
= 5.20 mmol
2-11. Find the number of millimoles of solute in
Solutions:
(a) 2.00 L of 0.0449 MKMnO .
4
0.0449 mol KMnO4 1000 mmol
× × 2.00 L = 89.8 mmol KMnO4
L mol
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publicly accessible website, in whole or in part.
2
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SolutionSolution
Manual
StudentManual
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Solution
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Solution
Manual
Solution
Manual
StudentManual
Skoog
Solution
et
Skoog
al.,
Manual
Fundamentals
et al.,
Skoog
Fundamentals
et al.,
of Analytical
Fundamentals
of Analytical
Chemistry,
of Chemistry,
Analytical
10e, ©Chemistry,
10e,
2022,
©978-0-357-45055-0
2022,
10e,978-0-357-45055-0.pdf
© 2022, 978-0-357-45055-0
,2026/2027:
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Skoog
ManualetSkoog
al.,
3 ofFundamentals
233
et al., Fundamentals
of Analytical
of Analytical
Chemistry,
Chemistry,
10e, © 2022,
10e, ©
978-0-357-45055-0
2022, 978-0-357-45055-0
Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e,
© 2022, 978-0-357-45055-0, Chapter 2: Calculations Used in Analytical Chemistry
(b) 750 mL of 5 .35 × 1023 M KSCN.
5.35 × 10 −3 M KSCN 1000 mmol L
× × × 750 mL = 4.01 mmol KSCN
L mol 1000 mL
(c) 3.50 L of a solution that contains 6.23 ppm of CuSO4 .
6.23 mg CuSO4 g mol CuSO4 1000 mmol
× × × × 3.50 L = 0.137 mmol CuSO4
L 1000 mg 159.61 g CuSO4 mol
(d) 250 mL of 0.414 mM KCl.
0.414 mmol KCl 1L
× × 250 mL = 0.104 mmol KCl
L 1000 mL
2-13. What is the mass in milligrams of
Solutions:
(a) 0.367 mol of HNO3?
63.01 g HNO3 1000 mg
0.367 mol HNO3 × × = 2.31× 104 mg HNO3
mol HNO3 g
(b) 245 mmol of MgO?
mol 40.30 g MgO 1000 mg
245 mmol MgO × × × = 9.87 × 10 3 mg MgO
1000 mmol mol MgO g
(c) 12.5 mol of NH4NO3 ?
80.04 g NH 4 NO 3 1000 m g
12.5 m ol NH 4 NO 3 × × = 1.00 × 10 6 m g NH 4 NO 3
m ol NH 4 NO 3 g
(d) 4.95 mol of (NH4 )2 e (NO3 )6 ( 548.23 g/mol) ?
548.23 g (NH4 )2 Ce(NO 3 )6 1000 mg
4.95 mol (NH4 ) 2 Ce(NO 3 )6 × ×
mol (NH4 ) 2 Ce(NO 3 )6 g
= 2.71× 10 6 mg (NH4 ) 2 Ce(NO 3 )6
2-15. What is the mass in milligrams of solute in
Solutions:
(a) 16.0 mL of 0.350 M sucrose (342 g/mol)?
0.350 mol sucrose L 342 g sucrose 1000 mg
× × ×
L 1000 mL mol sucrose g
× 16.0 mL = 1.92 × 10 3 mg sucrose
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publicly accessible website, in whole or in part.
3
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SolutionSolution
Manual
StudentManual
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Solution
Student
Solution
Manual
Solution
Manual
StudentManual
Skoog
Solution
et
Skoog
al.,
Manual
Fundamentals
et al.,
Skoog
Fundamentals
et al.,
of Analytical
Fundamentals
of Analytical
Chemistry,
of Chemistry,
Analytical
10e, ©Chemistry,
10e,
2022,
©978-0-357-45055-0
2022,
10e,978-0-357-45055-0.pdf
© 2022, 978-0-357-45055-0
, 2026/2027:
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4 ofFundamentals
233
et al., Fundamentals
of Analytical
of Analytical
Chemistry,
Chemistry,
10e, © 2022,
10e, ©
978-0-357-45055-0
2022, 978-0-357-45055-0
Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e,
© 2022, 978-0-357-45055-0, Chapter 2: Calculations Used in Analytical Chemistry
(b) 1.92 L of 3.76 × 1023 M H2O2 ?
3.76 × 10 − 3 m ol H 2 O 2 34.02 g H 2 O 2 1000 m g
× × × 1.92 L = 246 m g H 2 O 2
L m ol H 2 O 2 g
2-16. What is the mass in grams of solute in
Solutions:
(a) 250 mL of 0.264 M H2O2 ?
0.264 mol H2 O 2 L 34.02 g H2 O 2
× × × 250 mL = 2.25 g H2 O 2
L 1000 mL mol H2 O 2
(b) 37.0 mL of 5.75 × 10−4 M benzoic acid (122 g/mol)?
5.75 ×10−4 mol benzoicacid L 122 g benzoicacid
× ×
L 1000 mL mol benzoicacid
× 37.0 mL = 2.60 ×10−3 g benzoicacid
2-17. Calculate the p-value for each of the listed ions in the following:
Solutions:
− −
(a) Na1 , Cl , and OH in a solution that is 0.0635 M in NaCl and 0.0403 M in NaOH.
pNa = − log (0.0635 + 0.0403) = −log(0.1038) = 0.9838
pCl = −log(0.0635) = 1.197
pOH = − log (0.0403) = 1.395
(c) H+, Cl−, and Zn21 in a solution that is 0.400 M in HCl and 0.100 M in ZnCl2 .
pH = − log(0.400) = 0.398
pCl = − log(0.400 + 2 × 0.100) = − log(0.600) = 0.222
pZn = − log(0.100) = 1.00
42
(e) K+, OH−, and Fe ( CN )6 in a solution that is 1.62 × 10−7 M in K 4Fe ( CN)6 and 5.12 × 10−7 M
in KOH.
pK = − log(4 × 1.62 × 10−7 + 5.12 × 10−7 ) = − log(1.16 × 10−6 ) = 5.94
pOH = − log(5.12 × 10−7 ) = 6.291
pFe(CN)6 = − log(1.62 × 10−7 ) = 6.790
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publicly accessible website, in whole or in part.
4
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SolutionSolution
Manual
StudentManual
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Solution
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Solution
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Solution
Manual
StudentManual
Skoog
Solution
et
Skoog
al.,
Manual
Fundamentals
et al.,
Skoog
Fundamentals
et al.,
of Analytical
Fundamentals
of Analytical
Chemistry,
of Chemistry,
Analytical
10e, ©Chemistry,
10e,
2022,
©978-0-357-45055-0
2022,
10e,978-0-357-45055-0.pdf
© 2022, 978-0-357-45055-0
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Skoog
ManualetSkoog
al.,
1 ofFundamentals
233
et al., Fundamentals
of Analytical
of Analytical
Chemistry,
Chemistry,
10e, © 2022,
10e, ©
978-0-357-45055-0
2022, 978-0-357-45055-0
Student Solution Manual
Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e,
© 2022, 978-0-357-45055-0, Chapter 2: Calculations Used in Analytical Chemistry
Some of the answers below may differ in format but have the same value as your result. Please
check with your instructor if a specific format is desired.
Chapter 2
2-1. Define
Answers:
(a) molar mass.
The molar mass is the mass in grams of one mole of a chemical species.
(c) millimolar mass.
The millimolar mass is the mass in grams of one millimole of a chemical species.
2-3. Give two examples of units derived from the fundamental base SI units.
Solution:
3
1000 mL 1 cm3 m −3 3
The liter: 1 L = × × = 10 m
1L mL 100 cm
1 mol L 1 mol
Molar concentration: 1 M = × −3 3 = −3 3
L 10 m 10 m
2-4. Simplify the following quantities using a unit with an appropriate prefix:
Solutions:
(a) 5.8 × 108 Hz.
MHz
5.8 × 108 Hz × = 580 MHz
106 Hz
(c) 9.31 × 107 µmol.
mol
9.31× 10 7 µmol × 6
= 93.1 mol
10 µmol
(e) 3.96 × 106 nm.
mm
3.96 × 106 nm × = 3.96 mm
106 nm
© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
publicly accessible website, in whole or in part.
1
StudentStudent
SolutionSolution
Manual
StudentManual
Student
Solution
Student
Solution
Manual
Solution
Manual
StudentManual
Skoog
Solution
et
Skoog
al.,
Manual
Fundamentals
et al.,
Skoog
Fundamentals
et al.,
of Analytical
Fundamentals
of Analytical
Chemistry,
of Chemistry,
Analytical
10e, ©Chemistry,
10e,
2022,
©978-0-357-45055-0
2022,
10e,978-0-357-45055-0.pdf
© 2022, 978-0-357-45055-0
,2026/2027:
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Skoog
ManualetSkoog
al.,
2 ofFundamentals
233
et al., Fundamentals
of Analytical
of Analytical
Chemistry,
Chemistry,
10e, © 2022,
10e, ©
978-0-357-45055-0
2022, 978-0-357-45055-0
Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e,
© 2022, 978-0-357-45055-0, Chapter 2: Calculations Used in Analytical Chemistry
2-5. Why is 1 g no longer exactly 1 mole of unified atomic mass units?
Answer:
The dalton is defined as 1/12 the mass of a neutral 12C atom. With the redefinition of SI
base units in 2019, the definition of the dalton remained the same. However, the
definition of the mole and the kilogram changed in such a way that the molar mass unit
is no longer exactly 1 g/mol.
2-7. Find the number of Na+ ions in 2.75 g of Na3PO4?
Solution:
1 mol Na3PO4 3 mol Na+ 6.022 × 1023 Na+
2.75 g Na3PO4 × × × = 3.03 × 1022 Na+
163.94 g mol Na3PO4 mol Na+
2-9. Find the amount of the indicated element (in moles) in
Solutions:
(a) 5.32 g of B2O3.
2 mol B mol B2O3
5.32 g B2O3 × × = 0.153 mol B
mol B2O3 69.62 g B2O3
(b) 195.7 mg of Na2B4O7 ⋅ 10H2O.
g 7 mol O
195.7 mg Na2B4O7 ⋅ 10H2O × ×
1000 mg mol Na2B4O7 ⋅ 10H2O
mol Na2B4O7 ⋅ 10H2O
× = 3.59 × 10−3 mol O = 3.59 mmol
381.37 g
(c) 4.96 g of Mn3O4.
mol Mn3O4 3 mol Mn
4.96 g Mn3O4 × × = 6.50 × 10 −2 mol Mn
228.81 g Mn3O4 mol Mn3O4
(d) 333 mg of CaC2O4.
g mol CaC2O4 2 mol C
333 mg CaC2O4 × × × = 5.20 × 10−3 mol C
1000 mg 128.10 g CaC2O4 mol CaC2O4
= 5.20 mmol
2-11. Find the number of millimoles of solute in
Solutions:
(a) 2.00 L of 0.0449 MKMnO .
4
0.0449 mol KMnO4 1000 mmol
× × 2.00 L = 89.8 mmol KMnO4
L mol
© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
publicly accessible website, in whole or in part.
2
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StudentManual
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Solution
Manual
StudentManual
Skoog
Solution
et
Skoog
al.,
Manual
Fundamentals
et al.,
Skoog
Fundamentals
et al.,
of Analytical
Fundamentals
of Analytical
Chemistry,
of Chemistry,
Analytical
10e, ©Chemistry,
10e,
2022,
©978-0-357-45055-0
2022,
10e,978-0-357-45055-0.pdf
© 2022, 978-0-357-45055-0
,2026/2027:
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3 ofFundamentals
233
et al., Fundamentals
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of Analytical
Chemistry,
Chemistry,
10e, © 2022,
10e, ©
978-0-357-45055-0
2022, 978-0-357-45055-0
Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e,
© 2022, 978-0-357-45055-0, Chapter 2: Calculations Used in Analytical Chemistry
(b) 750 mL of 5 .35 × 1023 M KSCN.
5.35 × 10 −3 M KSCN 1000 mmol L
× × × 750 mL = 4.01 mmol KSCN
L mol 1000 mL
(c) 3.50 L of a solution that contains 6.23 ppm of CuSO4 .
6.23 mg CuSO4 g mol CuSO4 1000 mmol
× × × × 3.50 L = 0.137 mmol CuSO4
L 1000 mg 159.61 g CuSO4 mol
(d) 250 mL of 0.414 mM KCl.
0.414 mmol KCl 1L
× × 250 mL = 0.104 mmol KCl
L 1000 mL
2-13. What is the mass in milligrams of
Solutions:
(a) 0.367 mol of HNO3?
63.01 g HNO3 1000 mg
0.367 mol HNO3 × × = 2.31× 104 mg HNO3
mol HNO3 g
(b) 245 mmol of MgO?
mol 40.30 g MgO 1000 mg
245 mmol MgO × × × = 9.87 × 10 3 mg MgO
1000 mmol mol MgO g
(c) 12.5 mol of NH4NO3 ?
80.04 g NH 4 NO 3 1000 m g
12.5 m ol NH 4 NO 3 × × = 1.00 × 10 6 m g NH 4 NO 3
m ol NH 4 NO 3 g
(d) 4.95 mol of (NH4 )2 e (NO3 )6 ( 548.23 g/mol) ?
548.23 g (NH4 )2 Ce(NO 3 )6 1000 mg
4.95 mol (NH4 ) 2 Ce(NO 3 )6 × ×
mol (NH4 ) 2 Ce(NO 3 )6 g
= 2.71× 10 6 mg (NH4 ) 2 Ce(NO 3 )6
2-15. What is the mass in milligrams of solute in
Solutions:
(a) 16.0 mL of 0.350 M sucrose (342 g/mol)?
0.350 mol sucrose L 342 g sucrose 1000 mg
× × ×
L 1000 mL mol sucrose g
× 16.0 mL = 1.92 × 10 3 mg sucrose
© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
publicly accessible website, in whole or in part.
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SolutionSolution
Manual
StudentManual
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Student
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Solution
Manual
StudentManual
Skoog
Solution
et
Skoog
al.,
Manual
Fundamentals
et al.,
Skoog
Fundamentals
et al.,
of Analytical
Fundamentals
of Analytical
Chemistry,
of Chemistry,
Analytical
10e, ©Chemistry,
10e,
2022,
©978-0-357-45055-0
2022,
10e,978-0-357-45055-0.pdf
© 2022, 978-0-357-45055-0
, 2026/2027:
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233
et al., Fundamentals
of Analytical
of Analytical
Chemistry,
Chemistry,
10e, © 2022,
10e, ©
978-0-357-45055-0
2022, 978-0-357-45055-0
Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e,
© 2022, 978-0-357-45055-0, Chapter 2: Calculations Used in Analytical Chemistry
(b) 1.92 L of 3.76 × 1023 M H2O2 ?
3.76 × 10 − 3 m ol H 2 O 2 34.02 g H 2 O 2 1000 m g
× × × 1.92 L = 246 m g H 2 O 2
L m ol H 2 O 2 g
2-16. What is the mass in grams of solute in
Solutions:
(a) 250 mL of 0.264 M H2O2 ?
0.264 mol H2 O 2 L 34.02 g H2 O 2
× × × 250 mL = 2.25 g H2 O 2
L 1000 mL mol H2 O 2
(b) 37.0 mL of 5.75 × 10−4 M benzoic acid (122 g/mol)?
5.75 ×10−4 mol benzoicacid L 122 g benzoicacid
× ×
L 1000 mL mol benzoicacid
× 37.0 mL = 2.60 ×10−3 g benzoicacid
2-17. Calculate the p-value for each of the listed ions in the following:
Solutions:
− −
(a) Na1 , Cl , and OH in a solution that is 0.0635 M in NaCl and 0.0403 M in NaOH.
pNa = − log (0.0635 + 0.0403) = −log(0.1038) = 0.9838
pCl = −log(0.0635) = 1.197
pOH = − log (0.0403) = 1.395
(c) H+, Cl−, and Zn21 in a solution that is 0.400 M in HCl and 0.100 M in ZnCl2 .
pH = − log(0.400) = 0.398
pCl = − log(0.400 + 2 × 0.100) = − log(0.600) = 0.222
pZn = − log(0.100) = 1.00
42
(e) K+, OH−, and Fe ( CN )6 in a solution that is 1.62 × 10−7 M in K 4Fe ( CN)6 and 5.12 × 10−7 M
in KOH.
pK = − log(4 × 1.62 × 10−7 + 5.12 × 10−7 ) = − log(1.16 × 10−6 ) = 5.94
pOH = − log(5.12 × 10−7 ) = 6.291
pFe(CN)6 = − log(1.62 × 10−7 ) = 6.790
© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
publicly accessible website, in whole or in part.
4
StudentStudent
SolutionSolution
Manual
StudentManual
Student
Solution
Student
Solution
Manual
Solution
Manual
StudentManual
Skoog
Solution
et
Skoog
al.,
Manual
Fundamentals
et al.,
Skoog
Fundamentals
et al.,
of Analytical
Fundamentals
of Analytical
Chemistry,
of Chemistry,
Analytical
10e, ©Chemistry,
10e,
2022,
©978-0-357-45055-0
2022,
10e,978-0-357-45055-0.pdf
© 2022, 978-0-357-45055-0
