Student Solution Manual Fundamentals of Analytical Chemistry 10e by Skoog

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  Molar concentration: 1 M = 2-4.

1 mol L 1 mol × −3 3 = −3 3 L 10 m 10 m

Simplify the following quantities using a unit with an appropriate prefix: Solutions: (a) 5.8 × 108 Hz.

5.8 × 108 Hz × (c)

MHz = 580 MHz 106 Hz

9.31 × 107 μmol. 9.31× 10 7 μmol ×

(e)

mol = 93.1 mol 10 μmol 6

3.96 × 106 nm.

3.96 × 106 nm ×

mm = 3.96 mm 106 nm

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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: 2.75 g Na3PO4 ×

2-9.

1 mol Na3PO4 3 mol Na+ 6.022 × 1023 Na+ × × = 3.03 × 1022 Na+ 163.94 g mol Na3PO4 mol Na+

Find the amount of the indicated element (in moles) in Solutions: (a) 5.32 g of B2O3. 5.32 g B2O3 ×

(b)

mol B2O3 2 mol B × = 0.153 mol B mol B2O3 69.62 g B2O3

195.7 mg of Na2B4O7 ⋅ 10H2O. 195.7 mg Na2B4O7 ⋅ 10H2O × ×

(c)

mol Na2B4O7 ⋅ 10H2O = 3.59 × 10−3 mol O = 3.59 mmol 381.37 g

4.96 g of Mn3O4. 4.96 g Mn3O4 ×

(d)

mol Mn3O4 3 mol Mn × = 6.50 × 10 −2 mol Mn 228.81 g Mn3O4 mol Mn3O4

333 mg of CaC2O4.

333 mg CaC2O4 × = 5.20 mmol 2-11.

g 7 mol O × 1000 mg mol Na2B4O7 ⋅ 10H2O

mol CaC2O4 g 2 mol C × × = 5.20 × 10−3 mol C 1000 mg 128.10 g CaC2O4 mol CaC2O4

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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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 L

(d)

×

mol CuSO4 g 1000 mmol × × × 3.50 L = 0.137 mmol CuSO4 mol 1000 mg 159.61 g CuSO4

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? 0.367 mol HNO3 ×

(b)

245 mmol of MgO? 245 mmol MgO ×

(c)

63.01 g HNO3 1000 mg × = 2.31× 104 mg HNO3 mol HNO3 g

40.30 g MgO 1000 mg mol × × = 9.87 × 10 3 mg MgO 1000 mmol mol MgO g

12.5 mol of NH4NO3 ? 12.5 m ol NH 4 NO 3 ×

(d)

80.04 g NH 4 NO 3 m ol NH 4 NO 3

×

1000 m g = 1.00 × 10 6 m g NH 4 NO 3 g

4.95 mol of (NH4 )2 e (NO3 )6 ( 548.23 g/mol) ? 4.95 mol (NH4 ) 2 Ce(NO 3 )6 ×

548.23 g (NH4 )2 Ce(NO 3 )6 mol (NH4 ) 2 Ce(NO 3 )6

×

1000 mg 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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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

(b)

×

34.02 g H2 O 2 L × × 250 mL = 2.25 g H2 O 2 1000 mL mol H2 O 2

37.0 mL of 5.75 × 10−4 M benzoic acid (122 g/mol)?

5.75 ×10−4 mol benzoicacid 122 g benzoicacid L × × 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

(e)

K+, OH−, and Fe ( CN )6

42

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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Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 2: Calculations Used in Analytical Chemistry

2-18.

Calculate the molar H3O+ ion concentration of a solution that has a pH of

Solutions: (a) 3.73. pH = 3.73, log[H3O+] = −3.73, [H3O+] = 1.9 × 10−4 M as in part (a)

(c)

0.59. [H3O+] = 0.26 M

(e)

7.62. [H3O+] = 2.4 × 10−8 M

(g)

20.76. [H3O+] = 5.8 M

2-19.

Calculate the p-functions for each ion in a solution that is

Solutions: (a) 0.0200 M in NaBr. pNa = pBr = −log(0.0200) = 1.699

(c)

4.5 × 1023 M in Ba ( OH) . 2 pBa = −log(4.5 × 10−3) = 2.35; pOH = −log(2 × 4.5 × 10−3) = 2.05

(e)

7.2 × 1023 M in CaCl and 8.2 × 1023 M in BaCl . 2 2 pCa = −log(7.2 × 10−3) = 2.14; pBa = −log(8.2 × 10−3) = 2.09 pCl = −log(2 × 7.2 × 10−3 + 2 × 8.2 × 10−3) = −log(0.0308) = 1.51

2-20.

Convert the following p-functions to molar concentrations:

Solutions: (a) pH = 1.102. pH = 1.102; log[H3O+] = −1.102; [H3O+] = 0.0791 M

(c)

pBr = 7.77.

pBr = 7.77; log[Br−] = −7.77; [Br−] = 1.70 × 10−8 M

(e)

pLi = 12.35.

pLi = 12.35; log[Li+] = −12.35; [Li+] = 4.5 × 10−13 M

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Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 2: Calculations Used in Analytical Chemistry

(g)

pMn = 0 .135.

pMn = 0.135; log[Mn2+] = −0.135; [Mn2+] = 0.733 M

2-21.

Seawater contains an average of 1.08 × 103 ppm of Na+ and 270 ppm of SO42−. Calculate

Solutions: (a) the molar concentrations of Na1 and SO4 22 given that the average density of seawater is 1.02 g/mL. 1.08 × 10 3 ppm Na + × 270 ppm SO 4 2 − ×

(b)

1 1.02 g 1000 mL mol Na + × × × = 4.79 × 10 −2 M Na + mL L 22.99 g 10 6 ppm

2− 1 1.02 g 1000 mL mol SO 4 × × × = 2.87 × 10 −3 M SO 4 2 − mL L 96.06 g 10 6 ppm

the pNa and pSO4 for seawater.

pNa = − log(4.79 × 10−2 ) = 1.320 pSO4 = − log(2.87 × 10−3 ) = 2.543

2-23.

A solution was prepared by dissolving 5.76 g of KCl ⋅ MgCl2 ⋅ 6H2O (277.85 g/mol) in sufficient water to give 2.000 L. Calculate

Solutions: (a) the molar analytical concentration of KCl ⋅ MgCl2 in this solution. 5.76 g KCl ⋅ MgCl2 ⋅ 6H2 O 2.00 L

(b)

×

mol KCl ⋅ MgCl2 ⋅ 6H2 O 277.85 g

= 1.04 × 10 −2 M KCl ⋅ MgCl2 ⋅ 6H2 O

the molar concentration of Mg21. There is 1 mole of Mg2+ per mole of KCl ⋅ MgCl2, so the molar concentration of Mg2+ is the same as the molar concentration of KCl ⋅ MgCl2 or 1.04 × 10−2 M.

(c)

the molar concentration of Cl2 . 1.04 × 10 − 2 M KCl ⋅ M gCl2 ⋅ 6H 2 O ×

(d)

3 m ol Cl− = 3.12 × 10 − 2 M Cl− m ol KCl ⋅ M gCl2 ⋅ 6H 2 O

the weight/volume percentage of KCl ⋅ MgCl2 ⋅ 6H2O. 5.76 g KCl ⋅ MgCl2 ⋅ 6H2 O L × × 100% = 0.288% (w/v) 2.00 L 1000 mL

(e)

the number of millimoles of Cl2 in 25.0 mL of this solution. 3.12 × 10 −2 mol Cl− L 1000 mmol × × × 25 mL = 7.8 × 10 −1 mmol Cl− L 1000 mL mol

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Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 2: Calculations Used in Analytical Chemistry

(f)

ppm K1. 1.04 × 10−2 M KCl ⋅ MgCl2 ⋅ 6H2O × =

(g)

407 mg = 407 ppm K + L

1 mol K + 39.10 g K + 1000 mg × × 1mol KCl ⋅ MgCl2 ⋅ 6H2O g mol K +

pMg for the solution. pMg = −log(1.04 × 10−2) = 1.983

(h)

pCl for the solution. pCl = −log(3.12 × 10−2) = 1.506

2-25.

A 5.85% (w/w) Fe(NO3)3 (241.86 g/mol) solution has a density of 1.059 g/mL. Calculate

Solutions: (a) the molar analytical concentration of Fe (NO3 )3 in this solution. 5.85% Fe(NO 3 )3 =

5.85 g Fe(NO 3 ) 3 g solution

×

1 1.059 g 1000 mL mol Fe(NO 3 ) 3 × × × 100 mL L 241.86 g

= 2.56 × 10 −1 M Fe(NO 3 ) 3 = 0.256 M

(b)

the molar NO32 concentration in the solution. 2.56 × 10 −1 M Fe(NO 3 ) 3 =

(c)

L

×

3 m ol NO 3 −

m ol Fe(NO 3 ) 3

= 7.68 × 10 − 1 M NO 3 −

the mass in grams of Fe(NO3 )3 contained in each liter of this solution. 2.56 × 10 −1 mol Fe(NO3 )3 L

2-27.

2.56 × 10 − 1 m ol Fe(NO 3 ) 3

×

241.86 g Fe(NO3 )3 mol

× L = 6.20 × 101 g Fe(NO3 )3 = 62.0 g

Describe the preparation of

Solutions: (a) 500 mL of 5.25% (w/v) aqueous ethanol ( C2H5OH, 46.1 g/mol). 5.25 g C2H5OH mL soln

×

1 × 500 mL soln = 26.3 g C2H5OH 100

Weigh 26.3 g ethanol and add enough water to give a final volume of 500 mL.

(b)

500 g of 5.25% (w/w) aqueous ethanol. 5.25% (w/w) C 2H5 OH =

5.25 g C 2H5 OH g soln

×

1 × 500 g soln = 2.63 × 101 g C 2H5 OH 100

500 g soln = 26.3 g C2H5OH + x g water

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Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 2: Calculations Used in Analytical Chemistry

x g water = 500 g soln − 26.3 g C2H5OH = 473.7 g water Mix 26.3 g ethanol with 473.7 g water.

(c)

500 mL of 5.25% (v/v) aqueous ethanol. 5.25% (v/v) C 2H5 OH = 5.25 mL C 2H5 OH 100 mL soln

5.25 mL C 2H5 OH 100 mL soln

× 500 mL soln = 26.3 mL C 2H5 OH

Dilute 26.3 mL ethanol with enough water to give a final volume of 500 mL.

2-29.

Describe the preparation of 500 mL of 3.00 M H3PO4 from the commercial reagent that is 86% H3PO4 (w/w) and has a specific gravity of 1.71.

Solution: 3.00 mol H3PO 4 L

86 g H3PO 4 g reagent

×

×

L × 500 mL = 1.50 mol H3PO 4 1000 mL

1 1 1.71 g reagent g water 1000 mL mol H3PO 4 1.50 × 10 mol H3PO 4 × × × × = 100 g water mL L 98.0 g L

volume 86% (w/w) H3PO 4 required = 1.50 mol H3PO 4 ×

L 1.50 × 10 mol H3PO 4 1

= 1.00 × 10 −1 L

Dilute 100 mL to 500 mL with water.

2-31.

Describe the preparation of

Solutions: (a) 500 mL of 0.1000 M AgNO3 from the solid reagent.

0.1000 M AgNO3 = =

0.1000 mol AgNO3

L 0.1000 mol AgNO3 L

×

169.87 g AgNO3 mol

×

L × 500 mL 1000 mL

= 8.49 g AgNO3 Dissolve 8.49 g AgNO3 in enough water to give a final volume of 500 mL.

(b)

1.00 L of 0.1000 M HCl, starting with a 6.00 M solution of the reagent. 0.1000 mol HCl × 1 L = 0.1000 mol HCl L L = 1.67 × 10−2 L HCl 0.1000 mol HCl × 6.00 mol HCl

Take 167 mL of the 6.00 M HCl and dilute to 1.00 L using water.

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Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 2: Calculations Used in Analytical Chemistry

(c)

250 mL of a solution that is 0.0810 M in K1 , starting with solid K 4Fe ( CN)6 .

0.0810 mol K + L × × 250 mL = 2.025 × 10−2 mol K + L 1000 mL 2.025 × 10 −2 mol K + ×

mol K 4Fe(CN)6 4 mol K

+

×

368.35 g K 4Fe(CN)6 mol

= 1.86 g K 4Fe(CN)6

Dissolve 1.86 g K4Fe(CN)6 in enough water to give a final volume of 250 mL.

(d)

500 mL of 3.00% (w/v) aqueous BaCl2 from a 0.400 M BaCl2 solution.

3.00 g BaCl2 1 × × 500 mL = 1.5 × 101 g BaCl2 mL soln 100 1.5 × 101 g BaCl2 ×

mol BaCl2 L × = 1.80 × 10 −1 L 208.23 g 0.400 mol BaCl2

Take 180 mL of the 0.400 M BaCl2 solution and dilute to 500 mL using water.

(e)

2.00 L of 0.120 M HClO4 from the commercial reagent [71.0% HClO4 (w/w), sp gr 1.67]. 0.120 mol HClO4 L

× 2.00 L = 0.240 mol HClO4

1 1.67 g reagent g water 1000 mL mol HClO4 1.18 × 10 mol HClO4 × × × = 100 g reagent g water mL L 100.46 g L

71 g HClO4

×

volume 71% (w/w) HClO4 required = 0.240 mol HClO4 =

L 1.18 × 101 mol HClO4

= 2.03 × 10 −2 L

Take 20.3 mL of the concentrated reagent and dilute to 2.00 L using water.

(f)

1.00 L of a solution that is 60.0 ppm in Na1 , starting with solid Na2 SO4 .

60 ppm Na+ =

60 mg Na+ L soln

60 mg Na+ × 1.00 L = 60 mg Na+ L soln 60 mg Na+ ×

g mol Na+ × = 2.61× 10 −3 mol Na+ 1000 mg 22.99 g

2.61× 10 −3 mol Na+ ×

mol Na2 SO 4 2 mol Na

+

×

142.04 g Na2 SO 4 mol

= 0.19 g Na2 SO4

Dissolve 0.19 g Na2SO4 in enough water to give a final volume of 1.00 L.

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Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 2: Calculations Used in Analytical Chemistry

2-33.

What mass of solid La(IO3)3 (663.6 g/mol) is formed when 50.0 mL of 0.250 M La3+ is mixed with 75.0 mL of 0.302 M IO3−?

Solution: 0.250 mol La3+ L × × 50.0 mL = 1.25 × 10 −2 mol La3+ L 1000 mL 0.302 M IO3 − =

0.302 mol IO3 − L

×

L × 75.0 mL = 2.27 × 10 −2 mol IO3 − 1000 mL

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

2-35.

mol La(IO3 )3 3 mol IO3

−

×

663.6 g La(IO3 )3 mol

= 5.01 g La(IO3 )3 formed

Exactly 0.118 g of pure Na2CO3 is dissolved in 100.0 mL of 0.0731 M HCl.

Solutions: (a) What mass in grams of CO2 is evolved? A balanced chemical equation can be written as: Na2CO3 + 2HCl → 2NaCl + H2O + CO2 (g) 0.1180 g Na2CO3 ×

mol Na2 CO3 105.99 g

= 1.113 × 10 −3 mol Na2CO3

0.0731 mol HCl L × × 100.0 mL = 7.31× 10−3 mol HCl L 1000 mL Because one mole of CO2 is evolved for every mole Na2CO3 reacted, Na2CO3 is the limiting reagent. Thus, 1.113 × 10 −3 mol Na2 CO3 ×

(b)

mol CO2

mol Na2 CO3

×

44.00 g CO2 mol

= 4.897 × 10 −2 g CO2 evolved

What is the molar concentration of the excess reactant (HCl or Na2CO3 ) ?

amnt HCl left = 7.31× 10−3 mol − (2 × 1.113 × 10−3 mol) = 5.08 × 10−3 mol 5.08 × 10−3 mol HCl 1000 mL × = 5.08 × 10−2 M HCl 100.0 mL L 2-37.

Exactly 75.00 mL of a 0.3132 M solution of Na2SO3 is treated with 150.0 mL of 0.4025 M HClO4 and boiled to remove the SO2 formed.

Solutions: (a) What is the mass in grams of SO2 that is evolved?

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Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 2: Calculations Used in Analytical Chemistry

A balanced chemical equation can be written as:

Na2 SO3 + 2HClO4 → 2NaClO4 + H2O + SO2 (g) 0.3132 M Na2 SO3 =

0.4025 M HClO4 = = 6.038 × 10−2

0.3132 mol Na2 SO3 L

×

L × 75 mL = 2.3 ×10 −2 mol Na2 SO3 1000 mL

0.4025 mol HClO4 L × × 150.0 mL L 1000 mL mol HClO4

Because one mole SO2 is evolved per mole Na2SO3, Na2SO3 is the limiting reagent. Thus, 2.3 × 10 −2 mol Na2 SO3 ×

(b)

mol SO2

mol Na2 SO3

×

64.06 g SO2 mol

= 1.5 g SO2 evolved

What is the concentration of the unreacted reagent (Na2 SO3 or HClO4 ) after the reaction is complete?

mol HClO4 unreacted = 6.038 × 10−2 mol − (2 × 2.3 × 10−2 ) = 1.4 × 10−2 mol 1.4 × 10 −2 mol HClO 4 225 mL

2-39.

×

1000 mL = 6.4 × 10 −2 M HClO 4 = 0.064 M L

What volume of 0.01000 M AgNO3 is required to precipitate all of the I− in 150 mL of a solution that contains 22.50 ppt KI?

Solution: A balanced chemical equation can be written as: AgNO3 + KI → AgI(s) + KNO3 22.50 ppt KI ×

1 g mol KI × × 150 mL × = 2.03 × 10−2 mol KI 166.0 g 10 ppt mL 3

2.03 × 10 −2 mol KI ×

mol AgNO3 mol KI

×

L = 2.03 L AgNO3 0.0100 mol AgNO3

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

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11

Student Solution Manual Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 3: Precision and Accuracy of Chemical Analyses 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 3 3-1.

Explain the difference between Answers: (a) random and systematic error. Random error causes data to be scattered more or less symmetrically around a mean value while systematic error causes the mean of a data set to differ from the accepted value. (c)

absolute and relative error. 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.

3-2.

Suggest two sources of systematic error and two sources of random error in measuring the length of a 3-m table with a 1-m metal rule. Answers: (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 length—random error. (4) Variability in interpolation of the finest division of the meter stick—random error.

3-4.

Describe at least three systematic errors that might occur while weighing a solid on an analytical balance. Answers: (1) The analytical balance is miscalibrated. (2) After weighing an empty vial, fingerprints are placed on the vial while adding sample to the vial. (3) A hygroscopic sample absorbs water from the atmosphere while placing it in a weighing vial.

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Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 3: Precision and Accuracy of Chemical Analyses

3-5.

Describe at least three ways in which a systematic error might occur while using a pipet to transfer a known volume of liquid. Answers: (1) The pipet is miscalibrated and holds a slightly different volume of liquid than the indicated volume. (2) The user repetitively reads the meiscus from an angle rather than at eye level. (3) The temperature differs from the calibration temperature.

3-7.

What kind of systematic errors are detected by varying the sample size? Answer: 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.

3-8.

A method of analysis yields masses of gold that are low by 0.4 mg. Calculate the percent relative error caused by this result if the mass of gold in the sample is Solutions: (a) 500 mg. (−0.4 mg/500 mg) × 100% = −0.08% As in part (a) (c)

125 mg. −0.32%

3-9.

The method described in Problem 3-8 is to be used for the analysis of ores that assay about 1.2% gold. What minimum sample mass should be taken if the relative error resulting from a 0.4-mg loss is not to exceed Solutions: (a) 20.1%? First determine how much gold is needed to achieve the desired relative error. (−0.4 mg/−0.1%) × 100% = 400 mg gold Then determine how much ore is needed to yield the required amount of gold. (400 mg/1.2%) × 100% = 33,000 mg ore or 33 g ore (c)

20.8% ? 4.2 g ore

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Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 3: Precision and Accuracy of Chemical Analyses

3-10.

The color change of a chemical indicator requires an overtitration of 0.03 mL. Calculate the percent relative error if the total volume of titrant is Solutions: (a) 50.00 mL. (0.03/50.00) × 100% = 0.060% As in part (a)  (b)

10.0 mL. 0.30%

(c)

25.0 mL. 0.12%

3-11.

A loss of 0.4 mg of Zn occurs in the course of an analysis for that element. Calculate the percent relative error due to this loss if the mass of Zn in the sample is Solutions: (a) 30 mg. (−0.4/30) × 100% = −1.3% (c)

300 mg. −0.13%

3-12.

Find the mean and median of each of the following sets of data. Determine the deviation from the mean for each data point within the sets, and find the mean deviation for each set. Use a spreadsheet if it is convenient. Solutions: (a) 0.0110

0.0104

0.0105

 0.0110 + 0.0104 + 0.0105  mean =   = 0.01063 ≈ 0.0106 3   Arranging the numbers in increasing value the median is: 0.0104 ← median 0.0105 0.0110

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Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 3: Precision and Accuracy of Chemical Analyses

The deviations from the mean are: 0.0104 − 0.01063 = 0.00023

0.0105 − 0.01063 = 0.00013 0.0110 − 0.01063 = 0.00037  0.00023 + 0.00013 + 0.00037  mean deviation =   = 0.00024 ≈ 0.0002 3   (c)

188

 190

mean = 190 (e)

39.83

  194 median = 189

39.61

39.25

  187 deviation from mean 2, 0, 4, 3

mean deviation = 2

39.68

mean = 39.59 median = 39.64 mean deviation = 0.17

deviation from mean 0.24, 0.02, 0.34, 0.09

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4

Student Solution Manual Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 4: Random Errors in Chemical Analysis 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 4 4-1.

Define Answers: (a) sample standard deviation. The sample standard deviation is the standard deviation of a sample of data. It applies to a small set of data from the population. It uses deviations from the experimental mean and the number of degrees or freedom N − 1 in place of the population mean μ and N. (c)

variance. The variance is the square of the standard deviation.

4-2.

Differentiate between Answers: (a) parameter and statistic. 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 and systematic error. 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.

4-3.

Distinguish between Answer: (a) the sample variance and the population variance. The sample variance s2 is the variance of a sample drawn from the population. It is N

given by s 2 =

(x i =1

i

− x)

N −1

2

, where x is the sample mean.

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1

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 4: Random Errors in Chemical Analysis

The population variance σ2 is the variance of an entire population given by N

σ2 = 4-5.

(x i =1

i

− μ)

2

N

, where μ is the population mean.

From the Gaussian (normal) error curve, what is the probability that a result from a population lies between 0 and +1σ of the mean? What is the probability of a result occurring that is between +1σ and +2σ of the mean? Answer: 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.

4-7.

Consider the following sets of replicate measurements: A

C

E

9.5

0.612

20.63

8.5

0.592

20.65

9.1

0.694

20.64

9.3

0.700

20.51

9.1 For each set, calculate the (a) mean, (b) median, (c) spread or range, (d) standard deviation, and (e) coefficient of variation. Solutions: For set A xi

9.5 8.5 9.1 9.3 9.1 Σxi = 45.5

xi2 90.25 72.25 82.81 86.49 82.81 Σxi2 = 414.61

mean: x = 45.5/5 = 9.1 median = 9.1 spread: w = 9.5 − 8.5 = 1.0

414.61− ( 45.5 ) / 5 2

standard deviation: s =

5 −1

= 0.37

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

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2

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 4: Random Errors in Chemical Analysis

Results for Sets A, C, and E, obtained in a similar way, are given in the following table.

(a)

x median w s CV, %

(b) (c) (d) (e) 4-9.

A 9.1 9.1 1.0 0.37 4.1

C 0.650 0.653 0.108 0.056 8.5

E 20.61 20.64 0.14 0.07 0.32

Estimate the absolute deviation and the coefficient of variation for the results of the following calculations. Round each result so that it contains only significant digits. The numbers in parentheses are absolute standard deviations. Solutions: (a)

y = 3.95 ( ±0.03) + 0.993 ( ±0.001) 27.025 ( ±0.001) = 22.082

( 0.03) + ( 0.001) + ( 0.001) 2

sy =

2

2

= 0.030

CV = (0.03/−2.082) × 100% = −1.4% y = −2.08(±0.03)

(c)

y = 29.2 ( ±0.3 ) × 2.03 ( ±0.02 ) ×10217 = 5.93928 × 10216 2

2

sy

 0.3   0.02 × 10 −17  =  = 0.01422  + −17  y  29.2   2.034 × 10 

CV = (0.0142) × 100% = 1.42% sy = (0.0142) × (5.93928 × 10−16) = 0.08446 × 10−16 y = 5.94(±0.08) × 10−16

(e)

y=

187 ( ±6 ) − 89 ( ±3 )

1240 ( ±1) + 57 ( ±8 )

snum =

(6 ) + ( 3)

sden =

(1) + ( 8 )

sy

2

2

2

2

2

= 7.5559 × 1022

= 6.71

= 8.06

ynum = 187 − 89 = 98 yden = 1240 + 57 = 1297

2

 6.71  8.06  =   +  = 0.0688 y  98   1297  CV = (0.0688) × 100% = 6.88% sy = (0.0688) × (0.075559) = 0.00520 y = 7.6(±0.5) × 10−2

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Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 4: Random Errors in Chemical Analysis

4-10.

Estimate the absolute standard deviation and the coefficient of variation for the results of the following calculations. Round each result to include only significant figures. The numbers in parentheses are absolute standard deviations.

Solutions: (a)

y = 1.02 ( ±0.02 ) × 1028 − 3.54 ( ±0.2 ) × 1029

( 0.02 × 10 ) + ( 0.2 × 10 )

sy =

−8

2

−9

2

= 2.83 × 10−10

y = 1.02×10−8 − 3.54×10−9 = 6.66×10−9 2.83 × 10−10 × 100% = 4.25% CV = 6.66 × 10−9 y = 6.7 ±0.3 × 10−9

(c)

y = 0.0040 ( ±0.0005) × 10.28 ( ±0.02 ) × 347 ( ±1)

2

sy

2

 0.0005   0.02   1  =   +  +  = 0.1250 y  0.0040   10.28   347  CV = (0.1250) × 100% = 12.5% y = 0.0040 × 10.28 × 347 = 14.27 sy = (0.125) × (14.27) = 1.78 y = 14(±2)

(e)

y=

100 ( ±1)

sy

2 ( ±1)

2

2

 1   1 =   +   = 0.500 y  100   2  CV = (0.500) × 100% = 50.0% y = 100 / 2 = 50.0 sy = (0.500) × (50.0) = 25 y = 50(±25)

4-11.

Calculate the absolute standard deviation and the coefficient of variation for the results of the following calculations. Round each result to include only significant figures. The numbers in parentheses are absolute standard deviations.

Solutions: (a)

y = log 2.00 ( ±0.03) × 1024 

( 0.434 ) ( 0.03 × 10 ) −4

y = log(2.00 × 10−4) = −3.6989

sy =

( 2.00 × 10 ) −4

= 6.51× 10 −3

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Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 4: Random Errors in Chemical Analysis

y = −3.699 ±0.006 CV = (0.006/3.699) × 100% = 0.16%

(c)

y = antilog 1.200 ( ±0.003)  sy

y = antilog(1.200) = 15.849

y

sy = (0.0069)(15.849) = 0.11 CV = (0.11/15.8) × 100% = 0.69%

4-12.

= ( 2.303 )( 0.003 ) = 0.0069

y = 15.8 ±0.1

Calculate the absolute standard deviation and the coefficient of variation for the results of the following calculations. Round each result to include only significant figures. The numbers in parentheses are absolute standard deviations.

Solution: (a)

y =  4.17 ( ±0.03 ) × 1024 

3

sy

 0.03 × 10−4  = 3 = 0.0216 −4  y  4.17 × 10  sy = (0.0216)(7.251 × 10−11) = 1.565 × 10−12 y = 7.3(±0.2) × 10−11 CV = (1.565 × 10−12/7.251 × 10−11) × 100% = 2.2% y = (4.17 × 10−4)3 = 7.251 × 10−11

4-13.

The standard deviation in measuring the diameter d of a sphere is ±0.02 cm. What is the standard deviation in the calculated volume V of the sphere if d = 2.35 cm?

Solution: From the equation for the volume of a sphere, we have 3

V=

3

4 3 4 d 4  2.35  3 πr = π   = π   = 6.80 cm 3 3 2 3  2 

Hence, we may write s sV 0.02 = 3× d = 3× = 0.0255 2.35 V d sV = 6.80 × 0.0255 = 0.173 V = 6.8(±0.2) cm3

4-15.

In a volumetric determination of an analyte A, the data obtained and their standard deviations are as follows: Initial buret reading

0.19 mL

0.02 mL

Final buret reading

9.26 mL

0.03 mL

Sample mass

45.0 mg

0.2 mg

From the data, find the coefficient of variation of the final result for the %A that is obtained by using the equation that follows and assuming there is no uncertainty in the equivalent mass.

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Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 4: Random Errors in Chemical Analysis

%A = titrant volume × equivalent mass × 100%/sample mass

Solution: 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. %A = [9.26(±0.03) − 0.19(±0.02)] × equivalent weight × 100/[45.0(±0.2)] Obtaining the value of the first factor and the error in the first factor sy =

( 0.03 ) + ( 0.02 ) 2

2

= 0.0361

y = 9.26 − 0.19 = 9.07

We can now obtain the relative error of the calculation s%A

2

2

 0.036   0.2  =   +  = 0.00596 %A  9.07   45.0 

The coefficient of variation is then CV = (0.00596) × 100% = 0.596% or 0.6%

4-17.

Chapter 22 shows that quantitative molecular absorption spectrometry is based on Beer’s law, which can be written as

2logT = ε bcX where T is the transmittance of a solution of an analyte X, b is the thickness of the absorbing solution, c  X is the molar concentration of X, and ε is an experimentally determined constant. By measuring a series of standard solutions of X, ε b was found to have a value of 3312(±12) M−1, where the number in parentheses is the absolute standard deviation. An unknown solution of X was measured in a cell identical to the one used to determine ε b. The replicate results were T = 0.213, 0.216, 0.208, and 0.214. Calculate (a) the molar concentration of the analyte cX, (b) the absolute standard deviation of the cX, and (c) the coefficient of variation of cX.

Solutions: We first calculate the mean transmittance and the standard deviation of the mean.  0.213 + 0.216 + 0.208 + 0.214  mean T =   = 0.2128 4   sT = 0.0034

(a) (b)

 − log T cX =   εb

 − log ( 0.2128 ) = 2.029 × 10−4 M = 3312 

For −logT, sy = (0.434)sT/T = 0.434 × (0.0034/0.2128) = 0.00693 −log(0.2128) = 0.672 − log T 0.672 ± 0.00693 cX = = εb 3312 ± 12

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Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 4: Random Errors in Chemical Analysis

sC

2

2

 0.00693   12  =   +  = 0.0109 cX  0.672   3312  X

(

)

sC X = ( 0.0109 ) 2.029 × 10 −4 = 2.22 × 10 −6

(c) 4-19.

CV = (2.22 × 10−6/2.029 × 10−4) × 100% = 1.1%

Six bottles of wine of the same variety were analyzed for residual sugar content with the following results:

Bottle

(a)

Percent (w/v) Residual Sugar

1

1.02, 0.84, 0.99

2

1.13, 1.02, 1.17, 1.02

3

1.12, 1.32, 1.13, 1.20, 1.25

4

0.77, 0.58, 0.61, 0.72

5

0.73, 0.92, 0.90

6

0.73, 0.88, 0.72, 0.70

Evaluate the standard deviation s for each set of data.

Solution:

The standard deviations are s1 = 0.096, s2 = 0.077, s3 = 0.084, s4 = 0.090, s5 = 0.104, s6 = 0.083

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Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 4: Random Errors in Chemical Analysis

(b)

Pool the data to obtain an absolute standard deviation for the method.

Solution: spooled = 0.088 or 0.09 4-21.

Calculate a pooled estimate of σ from the following spectrophotometric analysis for NTA (nitrilotriacetic acid) in water from the Ohio River:

Sample

NTA, ppb

1

13, 19, 12, 7

2

42, 40, 39

3

29, 25, 26, 23, 30

Solution:

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8

Student Solution Manual Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 5: Statistical Data Treatment and Evaluation 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 5 5-1.

Describe in your own words why the confidence interval for the mean of five measurements is smaller than that for a single result. Solution: 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.

5-4.

Consider the following sets of replicate measurements: A

C

E

0.514

70.24

0.812

0.503

70.22

0.792

0.486

70.10

0.794

0.497

0.900

0.472 Calculate the mean and the standard deviation for each of these six data sets. Calculate the 95% confidence interval for each set of data. What does this interval mean? Solutions: For Set A xi

xi2

0.514

0.2642

0.503

0.2530

0.486

0.2362

0.497

0.2470

0.472 Σxi = 2.472

0.2228 Σxi2 = 1.223

mean: x = 2.472/5 = 0.494

1.223 − ( 2.472 ) / 5 2

standard deviation: s =

5 −1

= 0.016

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1

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 5: Statistical Data Treatment and Evaluation

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 5-3, at 95% confidence t for 4 degrees of freedom is 2.78, therefore for set A, CI for μ = 0.494 ±

( 2.78 )( 0.016 ) 5

= 0.494 ± 0.020

Similarly, for sets A, C, and E, we obtain the results shown in the following table: A

C

E

x

0.494

70.19

0.824

s

0.016

0.08

0.051

95% CI 0.494 ± 0.020 70.19 ± 0.20 0.824 ± 0.081 The 95% confidence interval is the range within which the population mean is expected to lie with a 95% probability. 5-5.

Calculate the 95% confidence interval for each set of data in Problem 5-4 if s is a good estimate of σ and has a value of set A, 0.015; set C, 0.070; set E, 0.0090. Solutions: 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, CI for μ = 0.494 ±

CI 5-7.

(1.96 )( 0.015 ) 5

= 0.494 ± 0.013. Similarly for sets C and E, the limits are:

C

E

70.19 ± 0.079

0.824 ± 0.009

An atomic absorption method for the determination of the amount of iron present in used jet engine oil was found from pooling 30 triplicate analyses to have a standard deviation s = 2.9 μg Fe/mL. If s is a good estimate of σ , calculate the 95% and 99% confidence intervals for the result 17.2 μg Fe/mL if it was based on (a) a single analysis, (b) the mean of two analyses, and (c) the mean of four analyses.

Solutions: (a) 99% CI = 17.2 ± 2.58 × 2.9 = 17.2 ± 7.5 μg Fe/mL 95% CI = 17.2 ± 1.96 × 2.9 = 17.2 ± 5.7 μg Fe/mL

(b) 99% CI = 17.2 ± 95% CI = 17.2 ±

2.58 × 2.9 2 1.96 × 2.9 2

= 17.2 ± 5.3 μg Fe/mL = 17.2 ± 4.0 μg Fe/mL

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Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 5: Statistical Data Treatment and Evaluation

(c) 99% CI = 17.2 ± 95% CI = 17.2 ±

5-9.

4 1.96 × 2.9 4

= 17.2 ± 3.7 μg Fe/mL = 17.2 ± 2.8 μg Fe/mL

How many replicate measurements are needed to decrease the 95% and 99% confidence limits for the analysis described in Problem 5-7 to ±1.9 μg Fe/mL?

Solutions: 1.96 × 2.9 1.9 = N 1.9 =

5-11.

2.58 × 2.9

For a 95% CI, N = 8.9 ≅ 9

2.58 × 2.9

For a 99% CI, N = 15.6 ≅ 16

N

A volumetric calcium analysis on triplicate samples of the blood serum of a patient believed to be suffering from a hyperparathyroid condition produced the following data: mmol Ca/L = 3 .15, 3.25, 3.26. What is the 95% confidence interval for the mean of the data, assuming

Solutions: (a) no prior information about the precision of the analysis? For the data set, x = 3.22 and s = 0.06 95% CI = 3.22 ±

(b)

3

= 3.22 ± 0.15 meq Ca/L

s → σ = 0.056 mmol Ca/L? 95% CI = 3.22 ±

5-13.

4.30 × 0.06

1.96 × 0.056 3

= 3.22 ± 0.06 meq Ca/L

A standard method for the determination of glucose in serum is reported to have a standard deviation of 0.36 mg/dL. If s = 0.36 is a good estimate of σ , how many replicate determinations should be made in order for the mean for the analysis of a sample to be within

Solution: (a) 0.3 mg/dL of the true mean 99% of the time? 0.3 =

2.58 × 0.36 N

For the 99% CI, N = 9.6 ≅ 10

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Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 5: Statistical Data Treatment and Evaluation

5-15.

A prosecuting attorney in a criminal case presented as principal evidence small fragments of glass found imbedded in the coat of the accused. The attorney claimed that the fragments were identical in composition to a rare Belgian stained glass window broken during the crime. The average of triplicate analyses for five elements in the glass are in the table. On the basis of these data, does the defendant have grounds for claiming reasonable doubt as to guilt? Use the 99% confidence level as a criterion for doubt.

Concentration, ppm Element

From Clothes

As

Standard Deviation

From Window

129

s→σ

119

9.5

Co

0.53

0.60

0.025

La

3.92

3.52

0.20

Sb

2.75

2.71

0.25

Th

0.61

0.73

0.043

Solution: This is a two-tailed test where s → σ and from Table 5-1, zcrit = 2.58 for the 99% confidence level. For As: z =

129 − 119 3+3 9.5 3×3

= 1.28 ≤ 2.58

No significant difference exists at the 99% confidence level. Proceeding in a similar fashion for the other elements

Element

z

Significant Difference?

As

1.28

No

Co

−3.43

Yes

La

2.45

No

Sb

0.20

No

Th

−3.42

Yes

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

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Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 5: Statistical Data Treatment and Evaluation

5-17.

The week 3 measurement in the data set of Problem 5-16 is suspected of being an outlier. Use the Q test to determine if the value can be rejected at the 95% confidence level.

Solution: 5.6 − 5.1 Q= = 0.385 and Qcrit for 8 observations at 95% confidence = 0.526. 5.6 − 4.3 Since Q < Qcrit the outlier value 5.6 cannot be rejected at the 95% confidence level. 5-19.

The level of a pollutant in a river adjacent to a chemical plant is regularly monitored. Over a period of years, the normal level of the pollutant has been established by chemical analyses. Recently, the company has made several changes to the plant that appear to have increased the level of the pollutant. The Environmental Protection Agency (EPA) wants conclusive proof that the pollutant level has not increased. State the relevant null and alternative hypotheses, and describe the type I and type II errors that might occur in this situation.

Solution:

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.

5-20.

State quantitatively the null hypothesis H0 and the alternative hypothesis Ha for the following situations, and describe the type I and type II errors. If these hypotheses were to be tested statistically, comment on whether a one- or two-tailed test would be involved for each case.

Solutions: (a) The mean values for Ca determinations by an ion-selective electrode method and by an EDTA titration differ substantially. H0: μISE = μEDTA, Ha: μISE ≠ μEDTA. This would be a two-tailed test. The type I error for this situation would be that we decide the methods do not agree when they do The type II error would be that we decide the methods agree when they do not.

(c)

Results show that the batch-to-batch variation in the impurity content of Brand X acetonitrile is lower than Brand Y acetonitrile. H0: σ X2 = σ Y2 ; Ha: σ X2 < σ Y2 . This is a one-tailed test. The type I error would be that we decide that σ X2 < σ Y2 when it is not. The type II error would be that we decide that

σ X2 = σ Y2 when actually σ X2 < σ Y2 .

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5

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 5: Statistical Data Treatment and Evaluation

5-21.

The homogeneity of the chloride level in a water sample from a lake was tested by analyzing portions drawn from the top and from near the bottom of the lake, with the following results in ppm Cl:

Top

Bottom

26.30

26.22

26.43

26.32

26.28

26.20

26.19

26.11

26.49

26.42

Solutions: (a) Apply the t test at the 95% confidence level to determine if the chloride level from the top of the lake is different from that at the bottom. 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 For 8 degrees of freedom at 95% confidence tcrit = 2.31 t=

26.338 − 26.254

= 1.11 5+5 0.1199 5×5 Since t < tcrit, we conclude that no significant difference exists at 95% confidence level.

(b)

If each row in the table were samples analyzed in top-bottom pairs, use the paired t test and determine whether there is a significant difference between the top and bottom values at the 95% confidence level. From the data, N = 5, d = 0.084 and sd = 0.015166 For 4 degrees of freedom at 95% confidence t = 2.78 t=

0.084 − 0

= 12.52 0.015 / 5 Since 12.52 > 2.78, a significant difference does exist at 95% confidence level.

(c)

Why is a different conclusion drawn from using the paired t test than from just pooling the data and using the normal t test for differences in means? 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.

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6

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 5: Statistical Data Treatment and Evaluation

5-23.

Sir William Ramsey (Lord Rayleigh) prepared nitrogen samples by several different methods. The density of each sample was measured as the mass of gas required to fill a particular flask at a certain temperature and pressure. Masses of nitrogen samples prepared by decomposition of various nitrogen compounds were 2.29280, 2.29940, 2.29849, and 2.30054 g. Masses of “nitrogen” prepared by removing oxygen from air in various ways were 2.31001, 2.31163, and 2.31028 g. Is the density of nitrogen prepared from nitrogen compounds significantly different from that prepared from air? What are the chances of the conclusion being in error? (Study of this difference led to the discovery of the inert gases by Lord Rayleigh).

Solution: 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 t=

2.2978 − 2.3106 4+3 0.0027 4×3

= −6.207

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

5-25.

The ascorbic acid concentration in mmol/L of five different brands of orange juice was measured. Six replicate samples of each brand were analyzed. The following partial ANOVA table was obtained.

Variation Source

SS

df

MS

Between juices

______

______

______

Within juices

______

______

0.913

Total

______

______

F 8.45

Solutions: (a) Fill in the missing entries in the table. Source

SS

df

MS

Between juices

4 × 7.715 = 30.86

5−1=4

0.913 × 8.45 = 7.715

Within juices

25 × 0.913 = 22.825

30 − 5 = 25

0.913

Total

30.86 + 22.82 = 53.68

30 − 1 = 29

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F 8.45

7

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 5: Statistical Data Treatment and Evaluation

(b)

State the null and alternative hypotheses. H0: μbrand1 = μbrand2 = μbrand3 = μbrand4 = μbrand5; Ha: at least two of the means differ.

(c)

Is there a difference in the ascorbic acid content of the five juices at the 95% confidence level? 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.

5-27.

Four analysts perform replicate sets of Hg determinations on the same analytical sample. The results in ppb Hg are shown in the following table:

Determination

Analyst 1

Analyst 2

Analyst 3

Analyst 4

1

10.19

10.19

10.14

10.24

2

10.15

10.11

10.12

10.26

3

10.16

10.15

10.04

10.29

4

10.10

10.12

10.07

10.23

Solutions: (a) State the appropriate hypotheses. H0: μAnalyst1 = μAnalyst2 = μAnalyst3 = μAnalyst4; Ha: at least two of the means differ.

(b)

Do the analysts differ at the 95% confidence level? At the 99% confidence level ( Fcrit = 5.95 ) ? At the 99.9% confidence level ( Fcrit = 10.80 ) ? See spreadsheet next page. From Table 5-4 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)

Which analysts differ from each other at the 95% confidence level? Based on the calculated LSD value of 0.0573, there are significant differences between analysts 1 and 4, analysts 1 and 3, analysts 2 and 4, and analysts 3 and 4. There is no significant difference between analysts 1 and 2 and 2 and 3.

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8

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 5: Statistical Data Treatment and Evaluation

Spreadsheet for Problem 5-27.

5-29.

Three different analytical methods are compared for determining Ca in a biological sample. The laboratory is interested in knowing whether the methods differ. The results shown next represent Ca results in ppm determined by an ion-selective electrode (ISE) method, by EDTA titration, and by atomic absorption spectrometry:

Repetition No.

ISE

EDTA Titration

Atomic Absorption

1

39.2

29.9

44.0

2

32.8

28.7

49.2

3

41.8

21.7

35.1

4

35.3

34.0

39.7

5

33.5

39.1

45.9

Solutions: (a) State the null and alternative hypotheses. H0: μISE = μEDTA = μAA; Ha: at least two of the means differ.

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9

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 5: Statistical Data Treatment and Evaluation

(b)

Determine whether there are differences in the three methods at the 95% confidence levels. See spreadsheet.

From Table 5-4 the F value for 2 degrees of freedom in the numerator and 12 degrees of freedom in the denominator at 95% is 3.89. Since F calculated is greater than F critical, we reject the null hypothesis and conclude that the 3 methods give different results at the 95% confidence level.

(c)

If a difference was found at the 95% confidence level, determine which methods differ from each other. 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.

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10

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 5: Statistical Data Treatment and Evaluation

5-31.

Apply the Q test to the following data sets to determine whether the outlying result should be retained or rejected at the 95% confidence level.

Solutions: (a) 95.10, 94.62, 94.70

Q=

95.10 − 94.70

= 0.833 and Qcrit for 3 observations at 95% confidence = 0.970. 95.10 − 94.62 Since Q < Qcrit the outlier value 95.10 cannot be rejected with 95% confidence. (b)

95.10, 94.62, 94.65, 94.70

Q=

95.10 − 94.70

= 0.833 and Qcrit for 4 observations at 95% confidence = 0.829. 95.10 − 94.62 Since Q > Qcrit the outlier value 95.10 can be rejected with 95% confidence.

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11

Student Solution Manual Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 6: Sampling, Standardization, and Calibration 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 6 6-1.

A 0.005-g sample of a rock is to be analyzed, and iron is to be determined at the ppm level. Determine the type of analysis and type of constituent. Answer: 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.

6-3.

Describe the steps in a sampling operation. Answer: 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.

6-5.

The following results were obtained for the determination of calcium in a NIST limestone sample: %CaO = 51.33, 51.22, 51.36, 51.21, and 51.44. Five gross samples were then obtained for a carload of limestone. The average percent CaO values for the gross samples were found to be 48.53, 50.75, 48.60, 48.87, and 50.29. Calculate the relative standard deviation associated with the sampling step. Solution: so2 = ss2 + sm2 From the NIST sample:

sm2 = 0.00947

From the gross sample:

so2 = 1.073

ss = 1.07302 − 0.00947 = 1.0313 s   1.0313  The relative standard deviation =  s  × 100% =   × 100% = 2.01%  51.312  x

6-7.

Changes in the method used to coat the tablets in Problem 6-6 lowered the percentage of rejects from 8.0% to 3.0%. How many tablets should be taken for inspection if the permissible relative standard deviation in the measurement is to be

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Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 6: Sampling, Standardization, and Calibration

Solutions: (a) 15%? N= (b)

(1− p) = (1− 0.03) pσ

2 r

0.03(0.15)

2

=

32.333 = 1437 (0.15)2

10%?

N = 32.333/(0.10)2 = 3233 (c)

5%?

N = 32.333/(0.05)2 = 12,933 (d)

2%?

N = 32.333/(0.02)2 = 80,833 6-9.

Approximately 15% of the particles in a shipment of silver-bearing ore are judged to be argentite, Ag2S (d = 7.3 g cm−3, 87% Ag); the remainder are siliceous (d = 2.6 g cm−3) and contain essentially no silver.

Solutions: (a) Calculate the number of particles that should be taken for the gross sample if the relative standard deviation due to sampling is to be 2% or less. 2

d d  P −P N = p(1− p)  A 2 B   A B  d   σrP

  

2

d = 7.3 × 0.15 + 2.6 × 0.85 = 3.3 P = 0.15 × 7.3 × 0.87 × 100 / 3.3 = 29% 2

2

 7.3 × 2.6   87 − 0  N = 0.15(1 − 0.15)     = 8714 particles 2  (3.3)   0.020 × 29  (b)

Estimate the mass of the gross sample, assuming that the particles are spherical and have an average diameter of 3.5 mm. mass = (4/3)π(r)3 × d × N = (4/3) π (0.175 cm)3 × 3.3(g/cm3) × 8.714 × 103 = 650 g

(c)

The sample taken for analysis is to weigh 0.500 g and contain the same number of particles as the gross sample. To what diameter must the particles be ground to satisfy these criteria? 0.500 = (4/3)π(r)3 × 3.3(g/cm3) × 8.714 × 103

r = 0.016 cm (diameter = 0.32 mm)

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Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 6: Sampling, Standardization, and Calibration

6-11.

The data in the accompanying table represent the concentration of glucose in the blood serum of an adult patient. On four consecutive days, a blood sample was drawn from the patient and analyzed in triplicate. The variance for a given sample is an estimate of the measurement variance while the day-to-day variance reflects both the measurement variance and the sampling variance.

Day Glucose Concentration, mg/100 mL 1

62

60

63

2

58

57

57

3

51

47

48

4

54

59

57

Solutions: (a) Perform an analysis of variance, and see whether the mean concentrations vary significantly from day to day. The following single-factor ANOVA table was generated using Excel’s Data Analysis Tools:

The Between Groups SS value of 264.25, the large F value, and the small P value all indicate that the mean concentrations vary significantly from day to day.

(b)

Estimate the sum of squares due to sampling. SST is the total variance and is the sum of the within day variance, SSE, and the dayto-day variance, SSF; SST = SSE + SSF. The within day variance, SSE, reflects the method variance, SSM. The day-to-day variance, SSF, reflects the sum of the method variance, SSM, and the sampling variance, SSS; SSF = SSM + SSS. Thus, SST = SSM + SSM + SSS and SSS = SST – 2×SSM SSS = 290.92 – 2×26.67 = 237.58. The SSS could be used to extract the sampling variance (see D. C. Montgomery, Design and Analysis of Experiments 10th ed., Chap. 3, Hoboken, NJ, Wiley, 2020).

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3

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 6: Sampling, Standardization, and Calibration

(c)

What is the best way to lower the overall variance? The best approach to lowering the overall variance would be to reduce the sampling variance, since this is the major component of the total variance ( σ t2 = 88.08333).

6-13.

A method for the determination of the corticosteroid methylprednisolone acetate in solutions obtained from pharmaceutical preparations yielded a mean value of 3.7 mg mL−1 with a standard deviation of 0.3 mg mL−1. For quality control purposes, the relative uncertainty in the concentration should be no more than 3%. How many samples of each batch should be analyzed to ensure that the relative standard deviation does not exceed 7% at the 95% confidence level?

Solution: See Example 6-3 Using t = 1.96 for infinite samples

N=

(1.96)2 × (0.3)2 = 5.16 (3.7)2 × (0.07)2

Using t = 2.78 for 5 samples (4 df)

N=

(2.78)2 × (0.3)2 = 10.36 (3.7)2 × (0.07)2

Using t = 2.26 for 10 samples

N=

(2.26)2 × (0.3)2 = 6.85 (3.7)2 × (0.07)2

Using t = 2.45 for 7 samples

N=

(2.45)2 × (0.3)2 = 8.05 (3.7)2 × (0.07)2

Using t = 2.36 for 8 samples

N=

(2.36)2 × (0.3)2 = 7.47 (3.7)2 × (0.07)2

The iterations converge at between 7 and 8 samples, so 8 should be taken for safety.

6-15.

The following data were obtained in calibrating a calcium ion electrode for the determination of pCa. A linear relationship between the potential and pCa is known to exist.

pCa = −log [Ca2+]

E, mV

5.00

−53.8

4.00

−27.7

3.00

+2.7

2.00

+31.9

1.00

+65.1

Solutions: (b)

Find the least-squares expression for the best straight line among the points. Plot this line. Equation of the line: y = –29.74x + 92.86

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4

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 6: Sampling, Standardization, and Calibration

(d)

Calculate the pCa of a serum solution in which the electrode potential was 15.3 mV. Find the absolute and relative standard deviations for pCa if the result was from a single voltage measurement. pCaUnk = 2.608; SD in pCa = 0.079; RSD = 0.030 (CV = 3.0%)

6-17.

The data in the following table were obtained during a colorimetric determination of glucose in blood serum.

Glucose Concentration, mM

Absorbance, A

0.0

0.002

2.0

0.150

4.0

0.294

6.0

0.434

8.0

0.570

10.0

0.704

Solutions: (a) Assuming a linear relationship between the variables, find the least-squares estimates of the slope and intercept.

m = 0.07014 and b = 0.008286

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5

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 6: Sampling, Standardization, and Calibration

(b)

What are the standard deviations of the slope and intercept? What is the standard error of the estimate?

sm = 0.00067; sb = 0.004039; SE = 0.00558 (c)

Determine the 95% confidence intervals for the slope and intercept. 95% CIm = m ± t × sm = 0.07014 ± 0.0019 95% CIb = b ± t × sb = 0.0083 ± 0.0112

(d)

A serum sample gave an absorbance of 0.413. Find the 95% confidence interval for glucose in the sample.

cunk = 5.77 mM; sunk = 0.09; 6-19.

95% CIunk = cunk ± t × sUnk = 5.77 ± 0.24 mM

A study was made to determine the activation energy EA for a chemical reaction. The rate constant k was determined as a function of temperature T, and the data in the following table were obtained.

T, K

k, s−1

599

0.00054

629

0.0025

647

0.0052

666

0.014

683

0.025

700

0.064

Solutions: The data should fit a linear model of the form log k = log A − EA/(2.303RT ), where A is the preexponential factor, and R is the gas constant. (b)

Find the slope, intercept, and standard error of the estimate.

m = –8.456; b = 10.83 and SE = 0.0459 (c)

Noting that EA = −b × 2.303R × 1000, find the activation energy and its standard deviation (Use R = 1.987 cal mol−1 K−1).

EA = –m × 2.303 × R × 1000 (Note: m has units of mK) = –(–8.456 mK)×(2.303)×(1.987 cal mol–1 K–1)×(1000 K/mK) = 38697 cal/mol

sEA = sm × 2.303 × R × 1000 = 1069 cal/mol Thus, EA = 38,697 ± 1069 cal/mol or 38.7 ± 1.1 kcal/mol

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6

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 6: Sampling, Standardization, and Calibration

(d)

A theoretical prediction gave EA = 41.00 kcal mol−1 K−1. Test the null hypothesis that EA is this value at the 95% confidence level.

H0: EA = 41.00 kcal/mol; HA: EA ≠ 41.00 kcal/mol. t = (38.697 – 41.00)/1.069 = –2.15 t(0.025, 4) = 2.776 Since t > –tcrit we retain H0. There is no reason to doubt that EA is not 41.00 kcal/mol at the 95% confidence level.

6-21.

Potassium can be determined by flame emission spectrometry (flame photometry) using a lithium internal standard. The following data were obtained for standard solutions of KCl and an unknown containing a constant, known amount of LiCl as the internal standard. All the intensities were corrected for background by subtracting the intensity of a blank.

cK, ppm

Intensity of K Emission

Intensity of Li Emission

1.0

10.0

10.0

2.0

15.3

7.5

5.0

34.7

6.8

7.5

65.2

8.5

10.0

95.8

10.0

20.0

110.2

5.8

Unknown

47.3

9.1

Solution: (c)

Calculate the concentration of K in the unknown. 5.247 ppm rounded to 5.2 ppm

6-23.

The method of standard additions was used to determine nitrite in a soil sample. A 1.00mL portion of the sample was mixed with 24.00 mL of a colorimetric reagent, and the nitrite was converted to a colored product that produced a blank-corrected absorbance of 0.300. To 50.00 mL of the original sample, 1.00 mL of a standard solution of 1.00 × 10−3 M nitrite was added. The same color-forming procedure was followed, and the new absorbance was 0.530. What was the concentration of nitrite in the original undiluted sample?

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7

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 6: Sampling, Standardization, and Calibration

Solution: See Example 6-8. (0.300)(1.00 × 10 −3 )(1.00) = 2.4938 × 10–5 M (0.530)(51.00) − (0.300)(50.00) To obtain the concentration of the original sample, we need to multiply by 25.00/1.00.

cu =

cu = (2.4938×10–5 M)(25.00)/(1.00) = 6.23 × 10–4 M

6-25.

Atomic emission measurements were made to determine sodium in a blood serum sample. The following emission intensities were obtained for standards of 5.0 and 10.0 ng/mL and for the serum sample. All emission intensities were corrected for any blank emission. The mean value for the blank intensity (cNa = 0.0) was 0.000 with a standard deviation of 0.0071 (arbitrary units).

cNa, ng/mL

Emission Intensity

5.0

0.51

5.0

0.49

5.0

0.48

10.0

1.02

10.0

1.00

10.0

0.99

Serum

0.71

Serum

0.77

Serum

0.78

Solution: (c)

Find the detection limit for k values of 2 and 3. To what level of confidence do these correspond? For k = 2, DL = 0.14 ng/mL (92.1% confidence level) for k = 3, DL = 0.21 ng/mL (98.3% confidence level)

6-27.

The following table gives the sample means and standard deviations for six measurements each day of the purity of a polymer in a process. The purity is monitored for 24 days. Determine the overall mean and standard deviation of the measurements, and construct a control chart with upper and lower control limits. Do any of the means indicate a loss of statistical control?

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Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 6: Sampling, Standardization, and Calibration

Day Mean SD

Day Mean SD

1

96.50 0.80

13

96.64 1.59

2

97.38 0.88

14

96.87 1.52

3

96.85 1.43

15

95.52 1.27

4

96.64 1.59

16

96.08 1.16

5

96.87 1.52

17

96.48 0.79

6

95.52 1.27

18

96.63 1.48

7

96.08 1.16

19

95.47 1.30

8

96.48 0.79

20

96.43 0.75

9

96.63 1.48

21

97.06 1.34

10

95.47 1.30

22

98.34 1.60

11

97.38 0.88

23

96.42 1.22

12

96.85 1.43

24

95.99 1.18

Solution:

The process went out of control on Day 22.

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9

Student Solution Manual Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 7: Aqueous Solutions and Chemical Equilibria 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 7 7-1.

Briefly describe or define and give an example of Solutions: (a) a strong electrolyte. A strong electrolyte totally ionizes when dissolved in water. HCl is an example of a strong electrolyte. (c)

the conjugate acid of a Brønsted-Lowry base. The conjugate acid of a Brønsted-Lowry base is the potential proton donor formed when a Brønsted-Lowry base accepts a proton. For example, the NH4+ is a conjugate acid in the reaction, NH3 + proton  NH4 + .

(e)

an amphiprotic solute. An amphiprotic solute can act either as an acid or a base depending on the situation An amino acid is an amphiprotic solute.

(g)

autoprotolysis. Autoprotolysis is the act of self-ionization to produce both a conjugate acid and a conjugate base.

(i)

Le Châtelier’s principle. The Le Châtelier principle states that the position of an equilibrium always shifts in such a direction that it relieves the stress. A common ion like sulfate added to a solution containing sparingly soluble BaSO4 is an example.

7-2.

Briefly describe or define and give an example of Solutions: (a) an amphiprotic solvent. An amphiprotic solvent can act either as an acid or a base depending on the solute. Water is an example of an amphiprotic solvent since it can act as a proton donor or a proton acceptor.

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1

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 7: Aqueous Solutions and Chemical Equilibria

(c)

a leveling solvent. A leveling solvent shows no difference between strong acids. Perchloric acid and hydrochloric acid ionize completely in water; thus, water is a leveling solvent.

7-3.

Briefly explain why there is no term in an equilibrium-constant expression for water or for a pure solid, even though one (or both) appears in the balanced net ionic equation for the equilibrium. Solution: For dilute aqueous solutions, the concentration of water remains constant and is assumed to be independent of the equilibrium. Thus, its concentration is included within the equilibrium constant. For a pure solid, the concentration of the chemical species in the solid phase is constant. As long as some solid exists as a second phase, its effect on the equilibrium is constant and is included within the equilibrium constant.

7-4.

Identify the acid on the left and its conjugate base on the right in the following equations: (a)

PO4 3 − + H2PO4 −  2HPO4 2 −

(b)

NH+4 + H2O  NH3 + H3O+

(e)

HOCl + H2O  H3O+ + OCl−

Solutions: Acid

7-6.

Conjugate Base

(a)

H2PO4

(b)

NH+4

NH3

(e)

HOCl

OCl−

−

HPO42−

Write expressions for the autoprotolysis of Solutions: (a) H2 O.

2H2O  H3O+ + OH− (c)

CH3NH2 .

2CH3NH2  CH3NH3+ + CH3NH−

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Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 7: Aqueous Solutions and Chemical Equilibria

7-7.

Write the equilibrium-constant expressions and obtain numerical values for each constant in Solutions: (a) the basic dissociation of aniline, C 6H5NH2 .

C6H5NH2 +H2O  C6H5NH3+ +OH−

Kb = (c)

Kw Ka

=

+ − 1.00×10−14 [C6H5NH3 ][OH ] = = 3.98 × 10−4 [C6H5NH2 ] 2.51×10−11

the acidic dissociation of methyl ammonium hydrochloride, CH3NH3 Cl.

CH3NH3 + + H2O  CH3NH2 + H3O+ Ka =

(e)

K w [CH3NH2 ][H3 O+ ] = 2.3 × 10 −11 = + Kb [CH3NH3 ]

the dissociation of H3 AsO3 to H3O+ and AsO33 − . H3 AsO4 + H2O  H3O+ + H2 AsO4 − H2 AsO4 − + H2O  H3O+ + HAsO4 2 −   AsO4 2 − + H2O  H3O+ + AsO4 3 − H H3 AsO4 + 3H2O  3H3O+ + AsO4 3

K a1 =

[H3O+ ][H2 AsO4 − ]

K overall = 7-8.

K a2 =

[H3 AsO4 ]

[H3O+ ]3 [AsO4 3− ] [H3 AsO4 ]

[H3O+ ][HAsO4 2 − ] [H2 AsO4 − ]

K a3 =

[H3O+ ][AsO4 3− ] [HAsO4 2 − ]

= K a1K a2 K a3 = 5.8 × 10−3 × 1.1× 10−7 × 3.2 × 10−12 = 2.0 × 10−21

Generate the solubility-product expression for

Solutions: (a) CuBr. CuBr( s )  Cu+ + Br −

(b)

K sp = [Cu+ ][Br − ]

MgCO3. MgCO3 ( s)  Mg2 + + CO32 −

(c)

K sp = [Mg2 + ][CO32 − ]

PbCl2. PbCl2 ( s )  Pb2 + + 2Cl−

K sp = [Pb2 + ][Cl− ]2

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Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 7: Aqueous Solutions and Chemical Equilibria

7-10.

Calculate the solubility-product constant for each of the following substances, given that the molar concentrations of their saturated solutions are as indicated:

Solutions: (b)

RaSO4 (6.6 × 10−6 M). RaSO4  Ra2− + SO42− [Ra2+] = [SO42−] = 6.6 × 10−6 M

Ksp = [Ra2+][SO42−] = (6.6 × 10−6 M)2 = 4.4 × 10−11 (d)

Ce(IO3)3 (1.9 × 10–3 M). Ce(IO3)3  Ce3+ + 3IO3− [Ce3+] = 1.9 × 10−3 M

[IO3−] = 3 × 1.9 × 10−3 M = 5.7 × 10−3 M

Ksp = [Ce3+][IO3−]3 = 1.9 × 10−3 × (5.7 × 10−3)3 = 3.5 × 10−10 7-13.

What CrO4 2 − concentration is required to Ag2CrO4(s)  2Ag+ + CrO42−

Solutions: (a) initiate precipitation of Ag2CrO4 from a solution that is 5.24 × 10 − 3 M in Ag + ? [CrO4 2 − ] =

(b)

lower the concentration of Ag+ in a solution to 7.82 × 10−7 M? [CrO4 2 − ] =

7-15.

1.2 × 10 −12 = 4.37 × 10−8 M (5.24 × 10 −3 )2

1.2 × 10 −12 = 1.96 M (7.82 × 10 −7 )2

The solubility-product constant for Ce(IO3)3 is 3.2 × 10−10. What is the Ce3+ concentration in a solution prepared by mixing 50.00 mL of 0.0500 M Ce3+ with 50.00 mL of Ce3+ + 3IO3−  Ce(IO3)3(s)

Ksp = [Ce3+][IO3−]3 = 3.2 × 10−10 Solutions: (a) water? 50.00 mL × 0.0500 mmol/ml = 2.50 mmol Ce3+ [Ce 3 + ] =

2.50 mmol = 0.0250 M (50.00 + 50.00) mL

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4

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 7: Aqueous Solutions and Chemical Equilibria

(b)

0.0500 M IO3−? We mix 2.50 mmol Ce3+ with 50.00 mL × 0.050 mmol/mL = 2.50 mmol IO3−. Each mole of IO3− reacts with 1/3 mole of Ce3+ so 2.50 mmol would consume 1/3 × 2.50 mmol Ce3+ or the amount of unreacted Ce3+ = 2.50 − 2.50/3 = 1.666 mmol cCe3+ =

1.666 mmol = 0.0166 M 100 mL

[Ce3+] = 0.0166 + S (where S is the solubility). Assume S is small so [Ce3+] = 0.0166 [IO3−] = 3S

Ksp = [Ce3+][IO3−]3 = 0.0166 × (3S)3 = 3.2 × 10−10 1/3

 3.2 × 10−10  S = −2   27 × 1.66 × 10 

= 8.9 × 10−4

[Ce3+] = 1.66 × 10−2 + 8.9 × 10−4 = 1.7 × 10−2 M

(c)

0.250 M IO3−? Now we have 0.250 mmol IO3− × 50.00 mL = 12.5 mmol. Since 3 × 2.50 mmol = 7.50 mmol would be required to completely react with the Ce3+, we have excess IO3−. [IO3 − ] =

12.5 mmol − 7.50 mmol + 3S = 0.0500 + 3S 100 mL

[Ce3+] = S

Ksp = S(0.0500 + 3S)3 = 3.2 × 1010 Lets assume 3S BiI3 > CuI > AgI

(b)

0.20 M NaI. For CuI, S = 1 × 10−12/0.20 = 5 × 10−12 M For AgI, S = 8.3 × 10−17/0.20 = 4.2 × 10−16 M For PbI2, S = 7.1 × 10−9/(0.20)2 = 1.8 × 10−7 M For BiI3, S = 8.1 × 10−19/(0.20)3 = 1.0 × 10−16 M So, solubilities are in the order PbI2 > CuI > AgI > BiI3

(c)

a 0.020 M solution of the solute cation. For CuI, S = 1 × 10−12/0.020 = 5 × 10−11 M For AgI, S = 8.3 × 10−17/0.020 = 4.2 × 10−15 M For PbI2, S = For BiI3, S =

1 7.1× 10 −9 = 3.0 × 10 −4 M 2 0.020

1 3 8.1× 10 −19 = 1.1× 10 −6 M 3 0.020

So, solubilities are in the order, PbI2 > BiI3 > CuI > AgI

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6

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 7: Aqueous Solutions and Chemical Equilibria

7-20.

At 25ºC, what are the molar H3O+ and OH− concentrations in

Solutions: (a) 0.0300 M HCOOH? For formic acid, Ka = 180 × 10−5. Call formic acid HFm and the formate anion Fm− HFm + H2O  Fm− + H3O+ [Fm− ][H3O+ ]

Ka =

[HFm]

Mass balance cHFm = [HFm] + [Fm−] = 0.0300

= 1.80 × 10 −4

[Fm−] = [H3O+]

[H3O+ ]2 0.0300 − [H3O ] +

Thus, [HFm] = 0.0300 [Fm−] = 0.0300 − [H3O+]

= 1.80 × 10−4

Solving the quadratic or solving by iterations gives, [H3O+] = 2.24 × 10−3 M so [OH−] = 1.00 × 10−14/2.24 × 10−3 = 4.5 × 10−12 M

(c)

0.200 M ethylamine?

C2H5NH2 +H2O  C2H5NH3+ +OH− Kb =

[C2H5NH3+ ][OH− ] [C2H5NH2 ]

=

K w 1.0 × 10−14 = = 4.33 × 10 −4 K a 2.31× 10 −11

[OH ]=[C2H5NH3+ ]

[C2H5NH2 ] = 0.200 − [OH− ]

[OH− ]2 = 4.33 × 10−4 − (0.200 − [OH ])

[OH− ]2 = 4.33 × 10−4 (0.200 − [OH− ])

−

[OH− ]2 + 4.33 × 10 −4 [OH− ] − 8.66 × 10 −5 = 0 −4.33 × 10 −4 + (4.33 × 10 −4 )2 + 4(8.66 × 10 −5 ) = 9.09 × 10 −3 M 2 1.0 × 10 −14 [H3O+ ] = = 1.1× 10 −12 M −3 9.09 × 10 [OH− ] =

(e)

0.250 M C6H5COONa (sodium benzoate)? Bz + H2O  HBz + OH−

Kb = Kw/Ka = 1.00 × 10−14/6.28 × 10−5 = 1.60 × 10−10

[OH−] = [HBz]

[Bz] = 0.250 − [OH−]

[OH− ]2 = 1.60 × 10 −10 0.250 − [OH− ] [OH−] = 6.32 × 10−6 M

[H3O+] = 1.00 × 10−14/6.32 × 10−6 = 1.58 × 10−9 M

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7

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 7: Aqueous Solutions and Chemical Equilibria

(g)

0.250 M hydroxylamine hydrochloride?

Ka = 1.1 × 10−6

HONH3+ + H2O  HONH2 +H3O+ As in part (b)

[H3O+ ]2 0.250 − [H3O+ ]

= 1.1 × 10−6

[H3O+] = 5.24 × 10−4 M

7-21.

[OH−] = 1.91 × 10−11 M

At 25°C, what is the hydronium ion concentration in

Solutions: (a) 0.100 M chloroacetic acid? ClCH2 COOH+H2 O  ClCH2 COO− + H3O+ [H3O+ ] = [ClCH2COO− ] [H3O+ ]2 (0.100 − [H3O+ ])

Ka =

[ClCH2COO− ][H3O+ ] [ClCH2COOH]

= 1.36 × 10 −3

[ClCH2COOH] = 0.100 − [H3O+ ]

= 1.36 × 10 −3

[H3O+ ]2 =1.36×10 −3 (0.100 − [H3O+ ])

[H3O+ ]2 + 1.36 × 10 −3 [H3O+ ] − 1.36 × 10 −4 = 0 [H3O+ ] =

(b)

−1.36 × 10 −3 + (1.36 × 10 −3 )2 + 4(1.36 × 10 −4 ) = 1.10 × 10 −2 M 2

0.100 M sodium chloroacetate?

ClCH2COO− + H2O  ClCH2COOH + OH− Kb =

[ClCH2COOH][OH− ] [ClCH2COO− ]

=

Kw 1.0 × 10−14 = = 7.35 × 10−12 −3 Ka 1.36 × 10

[OH− ] = [ClCH2COOH] [OH− ]2 = 7.35 × 10−12 (0.100 − [OH− ])

[ClCH2COO− ] = 0.100 M − [OH− ] [OH− ]2 = 7.35 × 10−12 (0.100 − [OH− ])

[OH− ]2 + 7.35 × 10 −12 [OH− ] − 7.35 × 10−13 = 0 [OH− ] = 8.57 × 10−7 M [H3O+ ] = (e)

1.0 × 10−14 = 1.17 × 10−8 M 8.57 × 10−7

1.50 × 103 M aniline hydrochloride? C6H5NH3+ + H2O  C6H5NH2 + H3O+ +

[H3O ] = [C6H5NH2 ] [H3O+ ]2 (0.0015 − [H3O+ ])

Ka =

[C6H5NH2 ][H3O+ ] [C6H5NH3 ] +

= 2.51 × 10−5

[C6H5NH3+ ] = 0.0015 M − [H3O+ ]

= 2.51× 10 −5

Proceeding as in part (d), we find [H3O+] = 1.82 × 10−4 M

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8

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 7: Aqueous Solutions and Chemical Equilibria

7-23.

Define buffer capacity.

Answer: Buffer capacity of a solution is defined as the number of moles of a strong acid (or a strong base) that causes 1.00 L of a buffer to undergo a 1.00-unit change in pH. 7-25.

Consider solutions prepared by HOAc + H2O  OAc− + H3O+

OAc− + H2O  HOAc + OH−

Solutions: (a) dissolving 8.00 mmol of NaOAc in 200 mL of 0.100 M HOAc. 8.00 mmol = 4 × 10 −2 M 200 mL [HOAc] = 0.100 M [OAc − ] =

pH = −log(1.75 × 10 −5 ) + log

(b)

4 × 102 = 4.359 0.100

adding 100 mL of 0.0500 M NaOH to 100 mL of 0.175 M HOAc.

0.175 mmol × 100 mL = 17.5 mmol mL 0.0500 mmol 0.0500 M NaOH = × 100 mL = 5.00 mmol mL (17.5 − 5.00) mmol [HOAc] = = 6.25 × 10−2 M 200 mL 5 mmol [OAc− ] = = 2.50 × 10−2 M 200 mL 0.175 M HOAc =

pH = −log(1.75 × 10 −5 ) + log (c)

2.50 × 10−2 = 4.359 6.25 × 10−2

adding 40.0 mL of 0.1200 M HCl to 160.0 mL of 0.0420 M NaOAc. In what respects do these solutions resemble one another? How do they differ? 0.042 mmol 0.0420 M OAc − = × 160 mL = 6.72 mmol mL 0.1200 mmol 0.1200 M HCl = × 40.0 mL = 4.80 mmol mL (6.72 − 4.80) mmol [OAc − ] = = 9.6 × 10 −3 M 200 mL [HOAc] =

4.8 mmol = 2.4 × 10 −2 M 200 mL

pH = − log(1.75 × 10 −5 ) + log

9.6 × 10 −3 = 4.359 2.4 × 10 −2

The solutions all are buffers with the same pH, but they differ in buffer capacity with (a) having the greatest and (c) the least.

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9

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 7: Aqueous Solutions and Chemical Equilibria

7-26.

Consult Appendix 3, and pick out a suitable acid-base pair to prepare a buffer with a pH of

Solutions: (a) 10.3. The closest are C2H5NH3+/ C2H5NH2 (pKa = 10.64) and CH3NH3+/CH3NH2 (pKa = 10.64)

(c)

4.5. C6H5NH3+/C6H5NH2 (pKa = 4.60)

7-27.

What mass of sodium formate must be added to 500.0 mL of 1.00 M formic acid to produce a buffer solution that has a pH of 3.75?

Solution: pH = 3.75 = pK a + log

[HCOO− ] [HCOO− ] = − log(1.8 × 10−4 ) + log [HCOOH] [HCOOH]

[HCOO− ] [HCOO− ] = 100.01 = 1.02 [HCOOH] [HCOOH] mmol HCOOH 500 mL × 1.00 = 500 mmol HCOOH mL 3.75 = 3.74 + log

So amount of HCOO− needed = 1.02 × 500 mmol = 511.6 mmol 511.6 mmol × 10−3 mol/mmol = 0.5116 mol HCOO− Mass HCOONa = 0.5116 mol × 67.997 g/mol = 34.79 g

7-29.

What volume of 0.200 M HCl must be added to 500.0 mL of 0.300 M sodium mandelate to produce a buffer solution with a pH of 3.25?

Solution: Let HMn = mandelic acid, Mn− = mandelate anion. 500 mL × 0.300 M NaMn = 150 mmol Mn. For a pH of 3.25 need the ratio of Mn− to HMn to be pH = 3.25 = pK a + log

[Mn− ] [Mn− ] = 3.398 + log [HMn] [HMn]

log

[Mn− ] = 3.25 − 3.398 = −0.148 [HMn]

[Mn− ] = 0.711 [HMn] mmol Mn− − x mmol HCl = 0.711 x mmol HCl

0.711 × x mmol HCl = mmol Mn− x mmol HCl

x = mmol Mn−/1.711 = 150 mmol Mn−/1.711 = 87.66 mmol HCl Volume HCl = 87.66 mmol/(0.200 mmol/mL) = 438 mL

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10

Student Solution Manual Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 8: Effect of Electrolytes on Chemical Equilibria 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 8 8-1.

Make a distinction between Answers: (a) activity and activity coefficient. Activity, aA, is the effective concentration of a chemical species A in solution. The activity coefficient, γA, is the numerical factor necessary to convert the molar concentration of the chemical species A to activity as shown below: aA=γA[A] (b)

thermodynamic and concentration equilibrium constants. The thermodynamic equilibrium constant refers to an ideal system within which each chemical species is unaffected by any others. A concentration equilibrium constant takes into account the influence exerted by solute species upon one another. The thermodynamic equilibrium constant is numerically constant and independent of ionic strength; the concentration equilibrium constant depends on molar concentrations of reactants and products as well as other chemical species that may not participate in the equilibrium.

8-3.

Neglecting any effects caused by volume changes, would you expect the ionic strength to (1) increase, (2) decrease, or (3) remain essentially unchanged when NaOH is added to a dilute solution of Answers: (a) magnesium chloride [ Mg ( OH)2 ( s ) ] forms?

MgCl2 + 2NaOH  Mg(OH)2 (s) + 2NaCl Replacing divalent Mg2+ with Na+ causes the ionic strength to decrease. (b)

hydrochloric acid? HCl + NaOH  H2O + NaCl There is no change in the charge states of the ions present in the solution equilibria. The ionic strength is unchanged.

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1

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 8: Effect of Electrolytes on Chemical Equilibria

(c)

acetic acid? HOAc + NaOH  H2O + NaOAc The ionic strength will increase because NaOH and NaOAc are totally ionized wheras acetic acid is only partially ionized.

8-5.

Explain why the activity coefficient for dissolved ions in water is usually less than that for water itself. Answer: Water is a neutral molecule and its activity equals its concentration at all low to moderate ionic strengths. That is, its activity coefficient is unity. In solutions of low to moderate ionic strength, activity coefficients of ions decrease with increasing ionic strength because the ionic atmosphere surrounding the ion causes it to lose some of its chemical effectiveness and its activity is less than its concentration.

8-7.

Explain why the initial slope for Ca2+ in Figure 8-3 is steeper than that for K+? Answer: Multiply charged ions deviate from ideality more than singly charged ions because of the effect of the surrounding ionic atmosphere. The initial slope of the activity coefficient vs square root of ionic strength for Ca2+ is steeper than that for K+ because the activity coefficient of Ca2+ is more influenced by ionic strength than that for K+.

8-9.

Calculate the ionic strength of a solution that is Solutions: (a) 0.025 M in FeSO4 .

μ = ½[0.025 × 22 + 0.025 × 22] = 0.10 (c)

0.25 M in FeCl3 and 0.15 M in FeCl2 .

μ = ½[0.25 × 32 + 0.75 × 12 + 0.15 × 22 + 0.30 × 12] = 1.95 8-10.

Use Equation 8-5 to calculate the activity coefficient of − log γ X =

0.51Z X2 μ 1 + 3.3α X μ

This problem is easiest to work with a spreadsheet.

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2

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 8: Effect of Electrolytes on Chemical Equilibria

Rounding these results, gives for the γ X Solutions: (a) Fe31 at μ = 0.057. 0.23

(c)

Ce41 at μ = 0.065. 0.08

8-12.

′ for For a solution in which μ = 6.5 × 10−2 , calculate K sp

Solutions: (a) AgSCN. We must use − log γ X =

0.51Z X2 μ 1 + 3.3α X μ

For Ag , α Ag+ = 0.25. At μ = 0.065, γ Ag+ = 0.7809; For SCN−, α SCN − = 0.35 and +

γ SCN = 0.7935 retaining insignificant figures for later calculations. −

′ = K sp

K sp

γ Ag γ SCN +

(c)

=

−

1.1 × 10 −12 = 1.8 × 10−12 (0.7809)(0.7935)

La (IO3 )3 .

For La3+, γ La3+ = 0.2158. For IO3−, γ IO − = 0.7935 3

′ = K sp

K sp

γ La γ IO 3+

= 3

−

1.0 × 10 = 9.3 × 10 −11 (0.2158)(0.7935)3 −11

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3

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 8: Effect of Electrolytes on Chemical Equilibria

8-13.

Use activities to calculate the molar solubility of Zn ( OH)2 in

Solutions: (a) 0.0150 M KCl. Zn(OH)2(s)  Zn2+ + 2OH−

Ksp = 3.0 × 10−16

μ = ½[0.015 × 12 + 0.015 × 12] = 0.0150 Using Equation 8-5, γ Zn2+ = 0.6294 γ OH− = 0.8816 − 2 2 2 ′ = aZn2+ aOH [Zn2+ ] × γ OH K sp − = γ − [OH ] Zn2+

[Zn2 + ][OH− ]2 =

3.0 × 10 −16 2 γ Zn γ OH 2+

=

−

3.0 × 10 −16 = 6.133 × 10 −16 (0.6294)(0.8816)2

Solubility = S = [Zn2+] = ½[OH−] S(2S)2 = 6.133 × 10−16 1/3

 6.133 × 10 −16  S =  4  

(b)

= 5.35 × 10−6 M

0.0250 M K 2 SO4 .

μ = ½[2 × 0.0250 × 12 + 0.0250 × 22] = 0.15/2 = 0.075 From Equation 8-5, γ Zn2+ = 0.434 γ OH− = 0.783 2 2 − 2 ′ = aZn2+ aOH K sp [Zn2+ ] × γ OH − = γ − [OH ] Zn2+

[Zn2+ ][OH− ]2 =

3.0 × 10 −16 = 1.127 × 10 −15 (0.434)(0.783)2

Solubility = S = [Zn2+] = ½[OH−] S(2S)2 = 1.127 × 10−15 1

 1.127 × 10 −15  3 −6 S =  = 6.6 × 10 M 4  

(c)

the solution that results when you mix 40.0 mL of 0.250 M KOH with 60.0 mL of 0.0250 M ZnCl2 . 0.250 mmol × 40.0 mL = 10.0 mmol mL 0.0250 mmol amount of ZnCl2 = × 60.0 mL = 1.5 mmol mL

amount of KOH =

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4

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 8: Effect of Electrolytes on Chemical Equilibria

[K + ] =

10 mmol = 0.10 M 100.0 mL

[OH− ] =

(10 mmol − (2 × 1.5 mmol) ) = 0.07 M 100.0 mL

2 × 1.5 mmol [Cl− ] = = 0.03 M 100.0 mL [Zn2+ ] = 0

μ

= ½[0.10 × 12 + 0.07 × 12 + 0.03 × 12] = 0.10

From Table 8-2,

γ Zn = 0.40 2+

2 2 − 2 ′ = aZn2+ aOH [Zn2+ ] × γ OH K sp − = γ − [OH ] Zn2+

γ OH = 0.76 −

[Zn2+ ][OH− ]2 =

3.0 × 10−16 2

γ Zn γ OH 2+

=

−

2+

3.0 × 10−16 = 1.298 × 10 −15 2 (0.40)(0.76)

S (0.07)2 = 1.298 × 10−15

Solubility = S = [Zn ]

 1.298 × 10−15  −13 S =  = 2.6 × 10 M 2  (0.07) 

(d)

the solution that results when you mix 20.0 mL of 0.100 M KOH with 80.0 mL of 0.0250 M ZnCl2 . 0.100 mmol × 20.0 mL = 2.0 mmol mL 0.0250 mmol × 80.0 mL = 2.0 mmol amount ZnCl2 = mL 2 mmol = 0.02 M [K + ] = 100.0 mL amount KOH =

[OH− ] = 0 2 × 2.0 mmol = 0.04 M [Cl− ] = 100.0 mL 1

[Zn2 + ] =

2 mmol − (2 mmol) 2

100.0 mL

= 0.01 M

1 0.02 × 12 + 0.040 × 12 + 0.01× 22 = 0.05 2 From Table 10-2, γ Zn2+ = 0.48 γ OH− = 0.81

μ=

(

)

2 2 − 2 ′ = aZn2+ aOH K sp [Zn2+ ] × γ OH − = γ − [OH ] Zn2+

[Zn2+ ][OH− ]2 =

3.0 × 10 −16 2 γ Zn γ OH 2+

−

=

3.0 × 10−16 = 9.53 × 10 −16 (0.48)(0.81)2

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5

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 8: Effect of Electrolytes on Chemical Equilibria

Solubility = S = [OH− ] / 2 (0.01)[OH− ]2 = 9.53 × 10 −16 1

 9.53 × 10 −16  2 −7 [OH ] =   = 3.09 × 10 M 0.01   −7 S = 3.09 × 10 M /2 = 1.5 × 10 −7 M −

(

8-14.

)

Calculate the solubilities of the following compounds in a 0.0275 M solution of Mg ( ClO4 )2 using (1) activities and (2) molar concentrations:

Solutions: (a) AgSCN.

μ = ½[0.0275 × 22 + 2 × 0.0275 × 12] = 0.0825 Cannot use data in Table 8-2. Use Equation 8-5

AgSCN(s)  Ag+ + SCN− (1) For Ag+, γ Ag+ = 0.761; for SCN−, γ SCN − = 0.776 K sp = γ Ag+ [Ag + ]γ SCN− [SCN− ] = 1.1× 10 −12 [Ag + ][SCN− ] =

1.1× 10 −12 = 1.86 × 10 −12 0.761× 0.776

S = [Ag+] = [SCN−]

S = 1.86 × 10−12 = 1.4 × 10−6 M (2) S = 1.1× 10−12 = 1.0 × 10−6 M

(b)

PbI2 . PbI2 (s)  Pb2+ + 2I− ′ = aPb2+ aI2− = γ Pb2+ [Pb2+ ] × (γ I− [I− ])2 (1) γ Pb2+ = 0.388 γ I− = 0.769 K sp [Pb2+ ][I− ]2 =

7.9 × 10−9

γ Pb γ I 2+

2

=

−

Solubility = S = [Pb2 + ] = S (2S )2 = 3.44 × 10 −8

7.9 × 10 −9 = 3.44 × 10−8 (0.388)(0.769)2 1 − [I ] 2

1

 3.44 × 10−8  3 −3 S =  = 2.0 × 10 M 4   1

(2)

 7.9 × 10−9  3 −3 S =  = 1.3 × 10 M 4  

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6

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 8: Effect of Electrolytes on Chemical Equilibria

(c)

BaSO4 . BaSO4(s)  Ba2+ + SO42−

γ Ba = 0.40; γ SO 2+

[Ba2 + ][SO4 2 − ] =

4

2−

= 0.376

1.1× 10 −10

γ Ba γ SO 2+

4

2−

=

1.1× 10 −10 = 7.3 × 10−10 (0.40)(0.376)

Solubility = S = [Ba ] = [SO4 2 − ] 2+

S 2 = 7.3 × 10 −10 S = 7.3 × 10 −10 = 2.7 × 10 −5 M (2)

(d)

S = 1.1 × 10−10 = 1.0 × 10−5 M

Cd2Fe ( CN)6 .

Cd2Fe(CN)6 (s)  2Cd2+ + Fe(CN)6 4 − K sp = 3.2 × 10−17 Cd2Fe(CN)6(s)  2Cd2+ + Fe(CN)64−

(1) γ Cd2+ = 0.40 γ Fe(CN) 4− = 0.020 6

[Cd ] [Fe(CN)6 ] = 2+ 2

4−

3.2 × 10−17 2

γ Cd γ Fe(CN) 2+

6

=

4−

3.2 × 10−17 = 1.00 × 10 −14 2 (0.40) (0.020)

1 [Cd2 + ] = [Fe(CN)6 4 − ] 2 (2S )2 S = 1.00 × 10 −14 Solubility = S =

1

 1.00 × 10 −14  3 −5 S =  = 1.4 × 10 M 4   1

(2)

8-15.

 3.2 × 10 −17  3 −6 S =  = 2.0 × 10 M 4  

Calculate the solubilities of the following compounds in a 0.0167 M solution of Ba (NO3 )2 using (1) activities and (2) molar concentrations:

Solutions: AgIO3 . (a)

μ = ½[0.0167 × 22 + 2 × 0.0167 × 12] = 0.050 AgIO3(s)  Ag+ + IO3−

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7

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 8: Effect of Electrolytes on Chemical Equilibria

(1) γ Ag+ = 0.80 γ IO − = 0.82 3

[Ag + ][IO3 − ] =

3.1× 10 −8

γ Ag γ IO +

3

3.1× 10 −8 = 4.7 × 10 −8 (0.80)(0.82)

=

−

Solubility = S = [Ag ] = [IO3 − ] +

S 2 = 4.7 × 10 −8 S = 4.7 × 10−8 = 2.2 × 10 −4 M (2)

(b)

S = 3.1× 10−8 = 1.8 × 10 −4 M

Mg ( OH)2 .

Mg(OH)2(s)  Mg2+ + 2OH− (1)

γ Mg = 0.52 γ OH = 0.81 2+

−

[Mg2 + ][OH− ]2 =

7.1× 10 −12 2 γ Mg γ OH 2+

=

−

Solubility = S = [Mg2 + ] = S (2S )2 = 2.081× 10−11

7.1× 10 −12 = 2.081× 10−11 (0.52)(0.81)2

1 [OH− ] 2

1

 2.081× 10 −11  3 −4 S =  = 1.7 × 10 M 4   1

 7.1× 10 −12  3 −4 S =  = 1.2 × 10 M 4  

(2)

(c)

BaSO4 . BaSO4(s)  Ba2+ + SO42− (1) γ Ba2+ = 0.46 γ SO 2− = 0.44 4

2+

2−

[Ba ][SO4 ] =

1.1× 10 −10

γ Ba γ SO 2+

4

=

2−

1.1× 10 −10 = 5.435 × 10 −10 (0.46)(0.44)

2−

Solubility = S = [SO4 ] (0.0167) × S = 5.435 × 10−10  5.435 × 10 −10  −8 S =  = 3.3 × 10 M  0.0167   1.1× 10 −10  −9 (2) S =   = 6.6 × 10 M  0.0167 

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8

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 8: Effect of Electrolytes on Chemical Equilibria

(d)

La (IO3 )3 .

La(IO3)3(s)  La3+ + 3IO3

(1) γ La3+ = 0.24 [La3+ ][IO3 − ]3 =

γ IO = 0.82 K sp = aLa aIO3 = γ La [La3+ ] × (γ IO [IO3 − ])3 3

3+

−

1.0 × 10

−11

=

3

γ La γ IO 3+

3

−

3

−

3+

3

−

1.0 × 10 = 7.557 × 10 −11 (0.24)(0.82)3 −11

1

Solubility = S = [La3 + ] = [IO3 − ] 3

S (3S )3 = 7.557 × 10 −11 1

 7.557 × 10−11  4 −3 S =  = 1.3 × 10 M 27   1

(2)

8-16.

 1.0 × 10 −11  4 −4 S =  = 7.8 × 10 M 27  

Calculate the % relative error in solubility by using concentrations instead of activities for the following compounds in 0.0350 M KNO3 using the thermodynamic solubility products listed in Appendix 2.

Solutions: (a) CuCl (αCu+ = 0.3 nm) CuCl(s)  Cu+ + Cl− Since Cu+ has an effective diameter of 0.3, then

(1) γ Cu+ = 0.83 γ Cl− = 0.83 [Cu+ ][Cl− ] =

1.9 × 10 −7

γ Cu γ Cl +

=

−

′ = a + a − = γ + [Cu+ ] × γ − [Cl− ] K sp Cu Cl Cu Cl

1.9 × 10 −7 = 2.758 × 10−7 (0.83)(0.83)

+

Solubility = S = [Cu ] = [Cl− ] S 2 = 2.758 × 10−7 S = 2.758 × 10−7 = 5.25 × 10−4 M (2)

S = 1.9 × 10 −7 = 4.4 × 10 −4 M

relative error =

( 4.4 × 10

−4

− 5.25 × 10−4

5.25 × 10

−4

) × 100% = −16%

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9

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 8: Effect of Electrolytes on Chemical Equilibria

(c)

Fe ( OH)3

Fe(OH)3  Fe3+ + 3OH− (1) γ Fe3+ = 0.281 γ OH− = 0.835 [Fe3+ ][OH− ]3 =

2 × 10 −39 3

γ Fe γ OH 3+

=

−

3 ′ = aFe3+ aOH K sp [Fe3+ ] × (γ OH− [OH− ])3 − = γ Fe3+

2 × 10−39 = 1.223 × 10 −38 (0.281)(0.835)3

retaining figures until the end

1

Solubility = S = [Fe3 + ] = [OH− ] 3

S (3S )3 = 1.223 × 10 −38

 1.223 × 10 −38 S =  27 

1

4 −10  = 1.46 × 10 M  1

 2 × 10−39  4 −11 (2) S =   = 9.3 × 10 M 27   9.3 × 10−11 − 1.46 × 10−10 × 100% = −36% relative error = 1.46 × 10−10

(e)

Ag3 AsO4 (α AsO43− = 0.4 nm) Ag3(AsO4)(s)  3Ag+ + AsO43− (1) γ Ag+ = 0.827 [Ag + ]3 [AsO4 3 − ] =

γ AsO

4

3−

= 0.205

6 × 10 −23 3 γ Ag γ AsO +

4

3−

3 ′ = aAg = (γ Ag+ [Ag+ ])3 × γ AsO 3− [AsO4 3 − ] K sp + a AsO 3− 4

4

6 × 10 −23 = = 5.17 × 10 −22 (0.827)3 (0.205)

1

Solubility = S = [AsO3 4 − ] = [Ag + ] 3

(3S )3 S = 5.17 × 10−22 1

 5.17 × 10−22  4 −6 S =  = 2.1× 10 M 27   1

 6 × 10 −23  4 −6 (2) S =   = 1.2 × 10 M 27   1.2 × 10 −6 − 2.1 × 10 −6 relative error = × 100% = −43% 2.1× 10 −6

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10

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 8: Effect of Electrolytes on Chemical Equilibria

8-17.

Calculate the % relative error in hydronium ion concentration by using concentrations instead of activities in calculating the pH of the following buffer solutions using the thermodynamic constants found in Appendix 3.

Solution: (a) 0.175 M HOAc and 0.275 M NaOAc In this buffer solution, we assume [HOAc] = cHOAc and [OAC−] = cNaOAc. We also assume that the ionic strength is contributed solely by NaOAc, neglecting H3O+ and OH−.

μ = ½[0.275 × 12 + 0.275 × 12] = 0.275 (0.51)(1)2 0.275

− log γ H O+ =

1 + (3.3)(0.9) 0.275

3

− log γ OAc− = Ka = K a′ =

= 0.1046 γ H O+ = 0.786 3

(0.51)(1)2 0.275

1 + (3.3)(0.425) 0.275 γ H O+ [H3O+ ]γ OAc− [OAc − ]

= 0.1541 γ OAc− = 0.701

3

[HOAc] [H3O+ ][OAc − ]

[H3O+ ] =

[HOAc]

Ka

γ H O γ OAc 3

K a′ [HOAc]

pH = 4.69

=

[OAc ] −

=

+

= −

1.75 × 10 −5 = 3.176 × 10 −5 0.786 × 0.701

3.176 × 10 −5 × 0.175 = 2.0 × 10 −5 M 0.275

With no activity corrections 1.75 × 10 −5 × 0.175 = 1.1× 10 −5 M 0.275 pH = 4.96 [H3O+ ] =

relative error in [H3O+ ] =

1.1× 10 −5 − 2.0 × 10 −5 × 100% = −45% 2.0 × 10−5

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11

Student Solution Manual Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 9: Solving Equilibrium Problems for Complex Systems 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 9 9-2.

Why are simplifying assumptions in equilibrium problems restricted to relationships that are sums or differences? Answer: To simplify equilibrium calculations, we sometimes assume that the concentrations of one or more species are negligible and can be approximated as 0.00 M. In a sum or difference assuming a concentration is 0.00 M leads to an appropriate result. In contrast, if we were to simplify and equilibrium constant expression by assuming on or more concentrations are zero, we would be multiplying or dividing by 0.00, which would render the expression meaningless.

9-4.

Why do molar concentrations of some species appear as multiples in charge-balance equations? Answer: A charge-balance equation is derived by relating the concentration of cations and anions no. mol/L positive charge = no. mol/L negative charge For a doubly charged ion, such as Ba2+, the concentration of charge for each mole is twice the molar concentration of the Ba2+. That is, mol/L positive charge = 2[Ba2+] Thus, the molar concentration of all multiply charged species is always multiplied by the charge in a charge-balance equation.

9-5.

Write the mass-balance expressions for a solution that is Solutions: (a) 0.2 M in HF. 0.20 = [HF] + [F−] (c)

0.10 M in H3PO4 . 0.10 = [H3PO4] + [H2PO4−] + [HPO42−] + [PO43−]

(e)

0.0500 M in HClO2 and 0.100 M in NaClO2 . 0.0500 + 0.100 = [HClO2] + [ClO2−] [Na+ ] = cNaClO = 0.100 M 2

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1

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 9: Solving Equilibrium Problems for Complex Systems

(g)

0.100 M in NaOH and saturated with Zn(OH)2, which undergoes the reaction Zn(OH)2 + 2OH−  Zn(OH)42−. 0.100 = [Na+] = [OH−] + 2[Zn(OH)42−]

(i)

saturated with PbF2. [Pb2+] = ½([F−] + [HF])

9-7.

Calculate the molar solubility of ZnC2O4 in a solution that has a fixed H3O+ concentration of

Solutions: (a)

1.0 × 10−6 M.

(c)

1.0 × 10−9 M. Following the systematic procedure, using part (a) Step 1 ZnC2O4(s)  Zn2+ + C2O42− H2C2O4 + H2O  H3O+ + HC2O4− HC2O4 + H2O  H3O+ + C2O42−

Step 2

S = solubility = [Zn2+] = [C2O42−] + [HC2O4−] + [ H2C2O4]

Step 3

[Zn2+][C2O42−] = Ksp= 8 × 10−9

(1)

[H3O+ ][HC2O4 − ] = K 1 = 5.6 × 10−2 [H2C2O4 ] [H3O+ ][C2 O4 2 − ] [HC 2 O4 − ]

Step 4

= K 2 = 5.42 × 10 −5

(2) (3)

[Zn+] = [C2O42−] + [HC2O4−] + [ H2C2O4] (4) [H3O+] = 1.0 × 10−6 M

Step 5

No charge balance because an unknown buffer is maintaining the pH.

Step 6

Unknowns are [Zn2+], [C2O42−], [HC2O4−], [ H2C2O4]

Step 7

No approximations needed, because we have 4 equations and 4 unkowns.

Step 8

Substituting [H3O+] = 1.0 × 10−6 M into equation (3) and rearranging gives [HC2O4 − ] =

1 × 10−6 [C2 O4 2 − ] 5.42 × 10−5

= 1.845 × 10−2 [C2 O4 2 − ]

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2

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 9: Solving Equilibrium Problems for Complex Systems

Substituting this relationship and [H3O+] = 1.0 × 10−6 M into equation (2) and rearranging gives [H2C2O4 ] =

1 × 10 −6 × 1.845 × 10 −2 [C2O4 2 − ] 5.6 × 10

−2

= 3.295 × 10 −7 [C2O4 2 − ]

Substituting these last two relationships in to equation (4) gives [Zn2+] = [C2O42−] + 1.845 × 10−2[C2O42−] + 3.295 × 10−7[C2O42−] = 1.0185[C2O42−] Substituting this last relationship into equation (1) gives K sp =

[Zn2 + ][Zn2+ ] = 8 × 10−9 1.0185

[Zn2+] = (8 × 10−9 × 1.0185)1/2 = 9.03 × 10−5

S = [Zn2+] = 9.0× 10−5 M Substituting other values for [H3O+] gives the following:

[H3O+]

9-8.

S, M

(a)

−6

1.00 × 10

9.0 × 10−5

(c)

1.00 × 10−9

8.9 × 10−5

Calculate the molar solubility of BaSO4 in a solution in which H3O+  is

Solutions: (a)

3.5 M.

(c)

0.080 M. Proceeding as in Problem 9-7, we write BaSO4  Ba2+ + SO42−

Ksp = 1.1 × 10−10

HSO4 + H2O  H3O+ + SO42−

K2 = 1.02 × 10−2

S = [Ba2+] [Ba2+][ SO42−] = 1.1 × 10−10

(1)

[H3O+ ][SO4 2 − ]

(2)

[HSO4 − ]

= 1.02 × 10 −2

Mass balance requires that [Ba2+] = [SO42−] + [ HSO4−]

(3)

The unknowns are [Ba ], [ SO4 ], and [ HSO4 ] 2+

2−

−

We have 3 equations and 3 unknowns so no approximations are needed. Substituting equation (2) into (3) gives [Ba2 + ] = [SO4 2 − ] +

[H3O+ ][SO4 2 − ] 1.02 × 10

−2

 [H3O+ ]  = [SO4 2 − ]  1 +   1.02 × 10 −2  

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3

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 9: Solving Equilibrium Problems for Complex Systems

Substituting equation (1) to eliminate [SO42−], gives [Ba2 + ] =

[H3O+ ]  1.1 × 10−10 1.1 × 10 −10  × 1 + × 1 + 98.0[H3O+ ]   = −2  2+ × [Ba2 + ] 1.02 10 [Ba ]  

(

(

S = [Ba2+ ] = 1.1 × 10−10 1 + 98.0[H3O+ ]

)=

)

1.1 × 10−10 + 1.078 × 10−8 [H3O+ ]

Using the different values of [H3O+]

9-9.

[H3O+]

S, M

(a)

3.5

1.9 × 10−4

(c)

0.08

3.1 × 10−5

Calculate the molar solubility of PbS in a solution in which H3O+  is held constant at (a) 3.0 × 10 −1 M and (b) 3.0 × 10 −4 M.

Solution: The derivation that follows applies to problems 9-11. MS(s)  M2+ + S2−

Ksp

H2S + H2O  H3O+ + HS−

K1 = 9.6 × 10−8

HS + H2O  H3O+ + S2−

K2 = 1.3 × 10−14

Overall H2S + 2H2O  2H3O+ + S2−

K1K2 = 1.25 × 10−21

S = solubility = [M2+] [M2+][S2−] = Ksp

[H3O+ ][S2− ] [HS ] −

(1)

= K 2 = 1.3 × 10−14

[H3O+ ]2 [S2− ] = K1K 2 = 1.25 × 10−21 [H2S]

(2) (3)

Mass balance is: [M2+] = [S2−] + [HS−] + [H2S]

(4)

Substituting equation (2) and (3) into (4), gives: [M2 + ] = [S2 − ] +

 [H O+ ] [H3O+ ]2  [H3O+ ][S2 − ] [H3O+ ]2 [S2 − ] + + = [S2 − ]  1+ 3   K2 K 1K 2 K2 K 1K 2  

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(5)

4

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 9: Solving Equilibrium Problems for Complex Systems

Substituting equation (1) into (5), gives [M2 + ] =

K sp  [H3O+ ] [H3O+ ]2  +  1+  K2 K 1K 2  [M2 + ] 

 [H3O+ ] [H3O+ ]2  + [M2+ ] = K sp  1+   1.3 × 10−14 1.25 × 10−21   (a)

(6)

Substituting Ksp = 3 × 10−28 and [H3O+] = 3.0 × 10−1 into equation (6), gives   0.30 (0.30)2 + [M2 + ] = solubility = 3 × 10−28  1+ = 1.5 × 10−4 M −14 −21  × × 1.3 10 1.25 10  

(b)

Using the same Ksp, but [H3O+] = 3.0 × 10−4, gives [M2+] = solubility = 1.5 × 10−7 M

9-12.

Calculate the molar solubility of ZnCO3 in a solution buffered to a pH of 7.00.

Solution: Proceeding as in Problem 9-9, we find

 [H O+ ] [H3O+ ]2  + [Zn2 + ] = K sp  1+ 3   K2 K1K 2   For ZnCO3, Ksp = 1.0 × 10−10. For H2CO3, K1 = 4.45 × 10−7, and K2 = 4.69 × 10−11

  [H3O+ ] [H3O+ ]2 + [Zn2+ ] = 1 × 10−10  1 + −11 −7 −11   4.69 × 10 4.45 × 10 × 4.69 × 10   For pH = 7.00, [H3O+] = 1.00 × 10−7 [Zn2+] = 5.1 × 10−4 M

9-14.

Dilute NaOH is introduced into a solution that is 0.050 M in Cu2+ and 0.040 M in Mn2+ .

Solutions: [Cu2+][OH−]2 = 4.8 × 10−20 (a)

[Mn2+][ OH−]2 = 2 × 10−13

Which hydroxide precipitates first? Cu(OH)2 precipitates first

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5

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 9: Solving Equilibrium Problems for Complex Systems

(b)

What OH− concentration is needed to initiate precipitation of the first hydroxide? Cu2+ begins to precipitate when

[OH− ] = (c)

4.8 × 10−20 = 9.8 × 10−10 M 0.05

What is the concentration of the cation forming the less soluble hydroxide when the more soluble hydroxide begins to form? Mn2+ begins to precipitate when

2 × 10−13 = 2.24 × 10−6 M 0.04 [Cu2+] = 4.8 × 10−20/(2.24 × 10−6)2 = 9.6 × 10−9 M [OH− ] =

9-16.

Silver ion is being considered for separating I− from SCN− in a solution that is 0.040 M in KI and 0.080 M in NaSCN.

Solutions: (a)

What Ag + concentration is needed to lower the I− concentration to 1.0 × 10 −6 M? [Ag+] = Ksp/[I−] = 8.3 × 10−17/(1.0 × 10−6) = 8.3 × 10−11 M

(b)

What is the Ag + concentration of the solution when AgSCN begins to precipitate? [Ag+] = Ksp/[SCN−] = 1.1 × 10−12/(0.080) = 1.375 × 10−11 M ≈ 1.4 × 10−11 M

(c)

What is the ratio of SCN− to I− when AgSCN begins to precipitate? [I−] when [Ag+] = 1.375 × 10−11 M [I−] = 8.3 × 10−17/(1.375 × 10−11) = 6.0 × 10−6 M [SCN−]/[I−] = 0.080/(6.0 × 10−6) = 1.3 × 104

(d)

What is the ratio of SCN− to I− when the Ag + concentration is 1 .0 × 10 −3 M? [I] = 8.3 × 10−17/(1.0 × 10−3) = 8.3 × 10−14 M [SCN−] = 1.1 × 10−12/(1.0 × 10−3) = 1.1 × 10−9 M [SCN−]/[I−] = 1.1 × 10−9/(8.3 × 10−14) = 1.3 × 104 Note that this ratio is independent of [Ag+] as long as some AgSCN(s) is present.

9-18.

What mass of AgBr dissolves in 200 mL of 0.200 M NaCN?

Ag+ + 2CN−  Ag(CN)2 −

β2 = 1.3 × 1021

Solution: AgBr  Ag+ + Br−

Ksp = 5.0 × 10−13 = [Ag+][Br]

Ag+ + 2CN−  Ag(CN)2−

β2 = 1.3 × 1021 =

[Ag(CN)2 ] −

[Ag + ][CN− ]2

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(1) (2)

6

Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e, © 2022, 978-0-357-45055-0, Chapter 9: Solving Equilibrium Problems for Complex Systems

It is readily shown that CN− + H2O  HCN + OH− can be neglected in this problem so that only the two equilibria shown above need to be considered. Solubility = [Br−] Mass balance requires that [Br−] = [Ag+] + [Ag(CN)2−]

(3)

0.200 = [CN ] + 2[Ag(CN)2 ] −

(4)

−

We now have 4 equations and 4 unknowns. Because β2 is very large, let us assume that [CN−]