CHAPTER 19 IONIC EQUILIBRIA IN AQUEOUS SYSTEMS...will have more conjugate base particles than weak acid particles. The buffer in scene 3 has 6 conjugate base particles to 3 weak acid

19-1

CHAPTER 19 IONIC EQUILIBRIA IN AQUEOUS SYSTEMS FOLLOW–UP PROBLEMS 19.1A Plan: The problems are both equilibria with the initial concentration of reactant and product given. For part (a), set

up a reaction table for the dissociation of HF. Set up an equilibrium expression and solve for [H3O+], assuming that the change in [HF] and [F−] is negligible. Check this assumption after finding [H3O+]. Convert [H3O+] to pH. For part (b), first find the concentration of OH− added. Then, use the neutralization reaction to find the change in initial [HF] and [F−]. Repeat the solution in part (a) to find pH.

a) Solution: Concentration (M) HF(aq) + H2O(l) F−(aq) + H3O+(aq) Initial 0.50 — 0.45 0 Change – x — +x +x Equilibrium 0.50 – x — 0.45 + x x Assumed that x is negligible with respect to 0.50 M and 0.45 M.

[ ]

+3

a

H O F =

HFK

−

[H3O+] = [ ]

aHF

FK

−

= ( ) [ ][ ]

4 0.506.8x10

0.45− = 7.5556x10–4 = 7.6x10–4

Check assumption: Percent error in assuming x is negligible: ( )47.5556x10 100

0.45

−

= 0.17%.

The error is less than 5%, so the assumption is valid. Solve for pH: pH = –log (7.5556 x 10–4) = 3.12173 = 3.12 The other method to calculate the pH of a buffer is to use the Henderson-Hasselbalch equation:

a[base]pH = p + log[acid]

K

a[F ]pH = p + log[HF]

K−

pKa = –log Ka = –log(6.8x10–4) = 3.16749

[0.45]pH = 3.16749 + log[0.50]

pH = 3.12173 = 3.12 Since [HF] and [F−] are similar, the pH should be close to pKa, which equals log (6.8x10–4) = 3.17. The pH should be slightly less (more acidic) than pKa because [HF] > [F–]. The calculated pH of 3.12 is slightly less than pKa of 3.17.

b) Solution: What is the initial molarity of the OH– ion?

Molarity =

−

NaOHmol1OHmol1

NaOHg00.40NaOHmol1

LNaOHg40.0

= 0.010 M OH−

Set up reaction table for neutralization of 0.010 M OH− (note the quantity of water is irrelevant). Concentration (M) HF(aq) + OH−(aq) → F−(aq) + H2O(l) Before addition 0.50 — 0.45 — Addition — 0.010 — — Change – 0.010 – 0.010 + 0.010________________ After addition 0.49 0 0.46 —

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Following the same solution path with the same assumptions as part (a):

[H3O+] = [ ]

aHF

FK

−

= ( ) [ ][ ]

4 0.496.8x10

0.46− = 7.2434782x10–4 = 7.2x10–4 M

Check assumption:

( )47.2434782x10 100

0.46

−

= 0.16%, which is less than the 5% maximum so the assumption is valid.

Solve for pH: pH = –log (7.2434782x10–4) = 3.14005 = 3.14 or using the Henderson-Hasselbalch equation:


K−

= [0.46]3.16749 + log[0.49]

pH = 3.14005 = 3.14 With addition of base, the pH should increase and it does, from 3.12 to 3.14. However, the pH should still be slightly less than pKa: 3.14 is still less than 3.17.

19.1B Plan: The problems are both equilibria with the initial concentration of reactant and product given. Take the

inverse log of –pKb to solve for the Kb. Then use Kb to solve for Ka: Ka x Kb = 1.0x10-14. For part (a), set up a reaction table for the dissociation of (CH3)2NH2

+ (the acid component of the buffer; Cl– is a spectator ion and does not participate in the buffer reaction). Set up an equilibrium expression and solve for [H3O+], assuming that the change in [(CH3)2NH2

+] and [(CH3)2NH] is negligible. Check this assumption after finding [H3O+]. Convert [H3O+] to pH. For part (b), first find the concentration of H3O+ added. Then, use the neutralization reaction to find the change in initial [(CH3)2NH2

+] and [(CH3)2NH]. Repeat the solution in part (a) to find pH. a) Solution: Kb = 10−pKb = 10−3.23 = 5.8884x10–4 = 5.9x10–4

Ka = Kw

Kb =

1.0x10–14

5.9x10–4 = 1.7x10–11

Concentration (M) (CH3)2NH2+(aq) + H2O(l) (CH3)2NH(aq) + H3O+(aq)

Initial 0.25 — 0.30 0 Change – x — +x +x Equilibrium 0.25 – x — 0.30 + x x Assumed that x is negligible with respect to 0.25 M and 0.30 M.

+

3 2 3a +

3 2 2

(CH ) NH H O =

(CH ) NHK

[H3O+] = 3 2 2

a3 2

(CH ) NH

(CH ) NHK

+

= (1.7x10–11) (0.25)(0.30)

= 1.4167x10–11 = 1.4x10–11

Check assumption: Percent error in assuming x is negligible: 1.4x10–11

0.25 (100) = 5.6x10–9%

The error is less than 5%, so the assumption is valid. Solve for pH: pH = –log (1.4x10–11) = 10.8539 = 10.85 The other method to calculate the pH of a buffer is to use the Henderson-Hasselbalch equation:


K

pH = pKa + log 3 2

3 2 2

(CH ) NH

(CH ) NH +

pKa = –log Ka = –log(1.7x10–11) = 10.7696

pH = 10.7696 + log (0.30)(0.25)

pH = 10.8488 = 10.85

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Since [(CH3)2NH2+] and [(CH3)2NH] are similar, the pH should be close to pKa, which equals

log(1.7x10–11) = 10.7696. The pH should be slightly greater (more basic) than the pKa because [(CH3)2NH] > [(CH3)2NH2

+]. The calculated pH of 10.85 is slightly greater than the pKa of 10.77. b) Solution:

What is the initial molarity of the H3O+ ion?

Molarity = �0.73 g HCl1 L

� � 1 mol HCl36.46 g HCl

� �1 mol H3O+

1 mol HCl� = 0.020 M H3O+

Set up reaction table for neutralization of 0.020 M H3O+ (note the quantity of water is irrelevant). Concentration (M) (CH3)2NH(aq) + H3O+ (aq) → (CH3)2NH2

+ (aq) + H2O(l) Before addition 0.30 — 0.25 — Addition — 0.020 — — Change – 0.020 – 0.020 + 0.020________________________ After addition 0.28 0 0.27 —

Following the same solution path with the same assumptions as part (a):

[H3O+] = 3 2 2

a3 2

(CH ) NH

(CH ) NHK

+

= (1.7x10–11) (0.27)(0.28)

= 1.6393x10–11 = 1.6x10–11

Check assumption: Percent error in assuming x is negligible: 1.6x10–11

0.25 (100) = 6.4x10–9%

The error is less than 5%, so the assumption is valid. Solve for pH: pH = –log (1.6x10–11) = 10.7959 = 10.80 or using the Henderson-Hasselbalch equation:

pH = pKa + log 3 2

3 2 2

(CH ) NH

(CH ) NH +

pH = 10.7696 + log (0.28)(0.27)

pH = 10.7854 = 10.78

(The slight differences in the pH calculated using the reaction table and the pH calculated from the Henderson-Hasselbalch equation are due to differences in rounding.) With addition of acid, the pH should decrease and it does, from 10.85 to 10.80.

19.2A Plan: For high buffer capacity, the components of a buffer should be concentrated and the concentrations of the

base and acid components should be similar. Solution:

a) The buffer has a much larger amount of weak acid than of the conjugate weak base. Addition of strong base would convert some HB into B− to make the ration [B−]/[HB] closer to 1 according to the reaction:

HB(aq) + OH−(aq) → B−(aq) + H2O(l) b) The buffer with the highest possible buffer capacity would have 4 HB particles and 4 B− particles. Addition of strong base would convert 3 HB particles to 3 B− particles so that there are 4 of each of the weak acid and weak base.

19.2B Plan: The pH of a given buffer solution depends on the relative concentrations of conjugate base and acid. If [A−] > [HA], the pH of the buffer solution will be greater than its pKa. If [A−] < [HA], the pH of the buffer solution will be less than its pKa. Buffers are able to lessen the effects of added strong acid or base by neutralizing the strong acid or base. Added strong base reacts with the acid component of the buffer, and added strong acid reacts with the conjugate base component of the buffer. The most effective buffers consist of a weak acid with a pKa close to (within one pH unit of) the buffering pH and its conjugate base.

19-4

Solution: a) The buffer with pH > pKa will have more conjugate base particles than weak acid particles. The buffer in scene 3 has 6 conjugate base particles to 3 weak acid particles and, thus, would have pH > pKa. b) Strong base reacts with the weak acid component of the buffer, so the more weak acid there is, the more strong base the buffer can react with. There are more weak acid particles in the buffer in scene 2 than in the other scenes. c) Buffers are most effective at buffering pH values within one unit of their pKa. The pKa of this buffer, 4.2, is more than one unit away from the buffering pH, 6.1, so this buffer would not be effective at buffering a pH of 6.1.

19.3A Plan: Sodium benzoate is a salt so it dissolves in water to form Na+ ions and C6H5COO− ions. Only the benzoate ion is involved in the buffer system represented by the equilibrium:

C6H5COOH(aq) + H2O(l) C6H5COO−(aq) + H3O+(aq) Given in the problem are the volume and pH of the buffer and the concentration of the base, benzoate ion. The question asks for the mass of benzoic acid to add to the sodium benzoate solution. First, find the concentration of C6H5COOH needed to make a buffer with a pH of 4.25. Multiply the volume by the concentration to find moles of C6H5COOH and use the molar mass to find grams of benzoic acid.

Solution: The concentration of benzoic acid is calculated from the Henderson-Hasselbalch equation:

6 5a

6 5

[C H COO ]pH = p + log

[C H COOH]K

−

pKa = –log Ka = –log 6.3x10–5 = 4.20066

[0.050]4.25 = 4.20066 + log[x]

0.04934 = [0.050]log[x]

Raise each side to 10x .

1.1203146 = [0.050][x]

x = 4.46303x10–2 M The number of moles of benzoic acid is determined from the concentration and volume: (4.46303x10–2 M C6H5COOH) x (5.0 L) = 0.2231515 mol C6H5COOH Mass (g) of C6H5COOH =

( ) 6 56 5

6 5

122.12 g C H COOH0.2231515 mol C H COOH

1 mol C H COOH

= 27.2513 = 27 g C6H5COOH

Prepare a benzoic acid/benzoate buffer by dissolving 27 g of C6H5COOH into 5.0 L of 0.050 M C6H5COONa. Using a pH meter, adjust the pH to 4.25 with strong acid or base.

19.3B Plan: Ammonium chloride is a salt so it dissolves in water to form NH4+ ions and Cl− ions. Only the ammonium

ion is involved in the buffer system represented by the equilibrium: NH4

+(aq) + H2O(l) NH3(aq) + H3O+(aq) Given in the problem are the volume and pH of the buffer and the concentration of the conjugate base, ammonia. The question asks for the mass of ammonium chloride to add to the ammonia solution. First, find the ratio of ammonia to ammonium ion needed to make a buffer with a pH of 9.18. Multiply the volume of the solution by the concentration of the ammonia to find moles of ammonia in the buffer. Then use the ratio of ammonia to ammonium as a conversion factor to calculate the moles of ammonium ion. Then use the molar mass to find grams of ammonium chloride.

Solution: The ratio of ammonia to ammonium ion is calculated from the Henderson-Hasselbalch equation:

pH = pKa + log[ ]3

4

NH

NH +

19-5

Ka of NH4+ =

Kw

Kb of NH3 = 1.0x10–14

1.76x10–5 = 5.68x10–10

pKa = –log Ka = –log (5.68x10–10) = 9.25

9.18 = 9.25 + log [ ]3

4

NH

NH +

–0.07 = log[ ]3

4

NH

NH +

Raise each side to 10x .

0.85 = [ ]3

4

NH

NH +

There are 0.85 moles of NH3 for every 1 mole of NH4+, or 0.85 mol NH3 = 1 mol NH4

+.

Amount (mol) of NH3 in buffer = (0.75 L buffer) �0.15 mol NH3

1 L� = 0.1125 mol NH3

Mass (g) of NH4Cl = ( ) 4 4 43

3 44

1 mol NH 1 mol NH Cl 53.49 g NH Cl0.1125 mol NH

0.85 mol NH 1 mol NH Cl1 mol NH

+

+

= 7.1 g NH4Cl

19.4A Plan: The titration is of a weak acid, HBrO, with a strong base, NaOH. The reactions involved are: 1) Neutralization of weak acid with strong base: HBrO(aq) + OH−(aq) → H2O(l) + BrO−(aq) Note that the reaction goes to completion and produces the conjugate base, BrO−. 2) The weak acid and its conjugate base are in equilibrium based on the acid dissociation reaction: a) HBrO(aq) + H2O(l) BrO−(aq) + H3O+(aq) or b) the base dissociation reaction: BrO–(aq) + H2O(l) HBrO(aq) + OH–(aq) The pH of the solution is controlled by the relative concentrations of HBrO and BrO−.

For each step in the titration, first think about what is present initially in the solution. Then use the two reactions to determine solution pH. It is useful in a titration problem to first determine at what volume of titrant the equivalence point occurs. For part (e), use the pH values from (a) – (d) to plot the titration curve. Solution:

a) Before any base is added, the solution contains only HBrO and water. Equilibrium reaction 2a applies and pH can be found in the same way as for a weak acid solution: Concentration (M) HBrO(aq) + H2O(aq) BrO−(aq) + H3O+(aq) Initial 0.2000 — 0 0 Change –x — +x +x Equilibrium 0.2000 – x — x x

Ka = [ ]

3H O BrO

HBrO

+ − = 2.3x10–9 =

[ ][ ][ ]

x x0.2000 x−

Assume 0.2000 – x = 0.2000 M

2.3x10–9 = [ ][ ][ ]

x x0.2000

x = [H3O+] = 2.144761x10–5 M pH = –log [H3O+] = –log (2.144761x10–5) = 4.66862 = 4.67 Check the assumption by calculating the percent error in [HBrO]eq.

Percent error = ( )52.144761x10 100

0.2000

−

= 0.01%; this is well below the 5% maximum.

b) When [HBrO] = [BrO−], the solution contains significant concentrations of both the weak acid and its conjugate base. Use the equilibrium expression for reaction 2a to find pH. Since [HBrO] = [BrO−], their ratio equals 1.

19-6

Ka = [ ]

3H O BrO

HBrO

+ −

[H3O+] = [ ]

aHBrO

BrOK

−

= ( )[ ]92.3x10 1− = 2.3x10–9 M

pH = –log [H3O+] = –log (2.3x10–9) = 8.63827 = 8.64 Note that when [HBrO] = [BrO−], the titration is at the midpoint (half the volume to the equivalence point) and pH = pKa. c) At the equivalence point, the total number of moles of HBrO present initially in solution equals the number of moles of base added. Therefore, reaction 1 goes to completion to produce that number of moles of BrO−. The solution consists of BrO− and water. Calculate the concentration of BrO−, and then find the pH using the base dissociation equilibrium, reaction 2b. First, find equivalence point volume of NaOH.

Volume (mL) of NaOH = ( )3

30.2000 mol HBrO 10 L 1 mol NaOH 1 L 1 mL20.00 mL

L 1 mL 1 mol HBrO 0.1000 mol NaOH 10 L

−

−

= 40.00 mL NaOH added All of the HBrO present at the beginning of the titration is neutralized and converted to BrO− at the equivalence point. Calculate the concentration of BrO−.

Initial moles of HBrO: (0.2000 M)(0.02000 L) = 0.004000 mol Moles of added NaOH: (0.1000 M)(0.04000 L) = 0.004000 mol Amount (mol) HBrO(aq) + OH−(aq) → H2O(l) + BrO−(aq) Before addition 0.004000 mol — — 0 Addition — 0.004000 mol — — Change – 0.004000 mol – 0.004000 mol — +0.004000 mol After addition 0 0 — 0.004000 mol

At the equivalence point, 40.00 mL of NaOH solution has been added (see calculation above) to make the total volume of the solution (20.00 + 40.00) mL = 60.00 mL.

[BrO–] = 30.004000mol BrO 1 mL

60.00 mL 10 L

−

−

= 0.06666667 M

Set up reaction table with reaction 2b, since only BrO− and water are present initially: Concentration (M) BrO−(aq) + H2O(l) HBrO(aq) + OH−(aq) Initial 0.06666667 — 0 0 Change – x — +x +x Equilibrium 0.06666667 – x — x x Kb = Kw/Ka = (1.0x10–14/2.3x10–9) = 4.347826x10–6

Kb = [ ]OH HBrO

BrO

−

−

= [ ][ ]

[ ]x x

0.06666667 x− = 4.347826x10–6

Assume that x is negligible, since [BrO−] >> Kb.

4.347826x10–6 = [ ][ ]

[ ]x x

0.06666667

x = [OH−] = 5.3838191x10–4 = 5.4 x 10–4 M Check the assumption by calculating the percent error in [BrO−]eq.

Percent error = ( )45.3838191x10 100

0.06666667

−

= 0.8%, which is well below the 5% maximum.

pOH = –log (5.3838191x10–4) = 3.26891 pH = 14 – pOH = 14 – 3.26891 = 10.73109 = 10.73

d) After the equivalence point, the concentration of excess strong base determines the pH. Find the concentration of excess base and use it to calculate the pH.

19-7

Initial moles of HBrO: (0.2000 M)(0.02000 L) = 0.004000 mol Moles of added NaOH: 2 x 0.004000 mol = 0.008000 mol NaOH

Amount (mol) HBrO(aq) + OH−(aq) → H2O(l) + BrO−(aq) Before addition 0.004000 mol — — 0 Addition — 0.008000 mol — — Change – 0.004000 mol – 0.004000 mol — +0.004000 mol After addition 0 0.004000 mol — 0.004000 mol Excess NaOH: 0.004000 mol

Volume (mL) of added NaOH = ( ) 31 L 1 mL0.0080000 mol NaOH

0.1000 mol NaOH 10 L−

= 80.00 mL Total volume: 20.00 mL + 80.00 mL = 100.0 mL

[NaOH] = 0.004000 mol NaOH/0.1000 L = 0.0400 M pOH = –log (0.0400) = 1.3979 pH = 14 – pOH = 14 – 1.3979 = 12.60 e) Plot the pH values calculated in the preceding parts of this problem as a function of the volume of titrant. The plot and pH values follow the pattern for a weak acid vs. strong base titration. The pH at the midpoint of the titration does equal pKa. The equivalence point should be, and is, greater than 7.

19.4B Plan: The titration is of a weak acid, C6H5COOH, with a strong base, NaOH. The reactions involved are: 1) Neutralization of weak acid with strong base: C6H5COOH(aq) + OH−(aq) → H2O(l) + C6H5COO−(aq) Note that the reaction goes to completion and produces the conjugate base, C6H5COO−. 2) The weak acid and its conjugate base are in equilibrium based on the acid dissociation reaction: a) C6H5COOH(aq) + H2O(l) C6H5COO−(aq) + H3O+(aq) or b) the base dissociation reaction: C6H5COO–(aq) + H2O(l) C6H5COOH(aq) + OH–(aq) The pH of the solution is controlled by the relative concentrations of C6H5COOH and C6H5COO−.

For each step in the titration, first think about what is present initially in the solution. Then use the two reactions to determine solution pH.

3456789

1011121314

0 10 20 30 40 50 60 70 80 90 100

pH

Volume of 0.1000 M NaOH, mL

Titration of HBrO with NaOH

19-8

Solution: a) Before any base is added, the solution contains only C6H5COOH and water. Equilibrium reaction 2a

applies and pH can be found in the same way as for a weak acid solution:

Concentration (M) C6H5COOH(aq) + H2O(aq) C6H5COO−(aq) + H3O+(aq) Initial 0.1000 — 0 0 Change –x — +x +x Equilibrium 0.1000 – x — x x

Ka = 6.3x10–5 = �C6H5HCOO–�[H3O+]

[C6H5COOH] =

(x)(x)(0.10 – x)

Assume 0.1000 – x = 0.1000 M

Ka = 6.3x10–5 = (x)(x)(0.10)

x = [H3O+] = 2.50998x10–3 = 2.5x10–3M pH = –log [H3O+] = –log (2.5x10–3) = 2.6021 = 2.60 Check the assumption by calculating the percent error in [C6H5COOH]eq.

2.5x103

0.10 (100) = 2.5% which is smaller than 5%, so the assumption is valid.

b) Any NaOH added to the buffer reacts with the benzoic acid. Calculate the amount (mol) of NaOH (OH−) added.

Amount (mol) of OH− added = (12.00 mL OH−) � 1 L1000 mL

� �0.1500 mol OH–

1 L� = 0.001800 mol OH−

The added OH− will react with the benzoic acid: Initial moles of C6H5COOH: (0.1000 M)(0.03000 L) = 0.003000 mol Amount (mol) C6H5COOH(aq) + OH−(aq) → H2O(l) + C6H5COO−(aq) Before addition 0.003000 mol — — 0 Addition — 0.001800 mol — — Change – 0.001800 mol – 0.001800 mol — +0.001800 mol After addition 0.001200 mol 0 — 0.001800 mol

At this point in the titration, 12.00 mL of NaOH solution has been added (see calculation above) to make the total volume of the solution (12.00 + 30.00) mL = 42.00 mL (0.04200 L). Calculate the new concentrations of [C6H5COOH] and [C6H5COO−]:

Molarity of C6H5COOH = 0.001200 mol

0.04200 L = 0.02857 M

Molarity of C6H5COO− = 0.001800 mol

0.04200 L = 0.04286 M

Ka = 6.3x10–5 = �C6H5HCOO–�[H3O+]

[C6H5COOH] , so [H3O+] = Ka

[C6H5COOH]

�C6H5HCOO–�

[H3O+] = (6.3x10–5) (0.02857)(0.04286)

= 4.1995x10–5 = 4.2x10–5

pH = –log [H3O+] = –log (4.2x10–5) = 4.3768 = 4.38 c) At the equivalence point, the total number of moles of C6H5COOH present initially in solution equals the number of moles of base added. Therefore, reaction 1 goes to completion to produce that number of moles of C6H5COO−. The solution consists of C6H5COO− and water. Calculate the concentration of C6H5COO−, and then find the pH using the base dissociation equilibrium, reaction 2b. First, find equivalence point volume of NaOH. Volume (mL) of NaOH =

19-9

(30.00 mL C6H5COOH) � 1 L1000 mL

� �0.1000 mol C6H5COOH 1 L

� � 1 mol NaOH1 mol C6H5COOH

� � 1 L0.1500 mol NaOH

�

= 0.02000 L (20.00 mL) NaOH added All of the C6H5COOH present at the beginning of the titration is neutralized and converted to C6H5COO− at the equivalence point. Calculate the concentration of C6H5COO−.

Initial moles of C6H5COOH: (0.1000 M)(0.03000 L) = 0.003000 mol Moles of added NaOH: (0.1500 M)(0.02000 L) = 0.003000 mol Amount (mol) C6H5COOH(aq) + OH−(aq) → H2O(l) + C6H5COO−(aq) Before addition 0.003000 mol — — 0 Addition — 0.003000 mol — — Change – 0.003000 mol – 0.003000 mol — +0.003000 mol After addition 0 0 — 0.003000 mol

At this point in the titration, 20.00 mL of NaOH solution has been added (see calculation above) to make the total volume of the solution (20.00 + 30.00) mL = 50.00 mL (0.05000 L). Calculate the new concentration [C6H5COO−]:

Molarity of C6H5COO− = 0.003000 mol

0.05000 L = 0.06000 M

Set up reaction table with reaction 2b, since only C6H5COO− and water are present initially: Concentration (M) C6H5COO−(aq) + H2O(l) C6H5COOH (aq) + OH−(aq) Initial 0.06000 — 0 0 Change – x — +x +x Equilibrium 0.06000 – x — x x Kb = Kw/Ka = (1.0x10–14/6.3x10–5) = 1.6x10–10

Kb = [C6H5COOH][OH−]

�C6H5HCOO–� =

[x][x]�0.06000 – x�

= 1.6x10–10

Assume that x is negligible, since [C6H5COO−] >> Kb.

1.6x10–10 = [x][x]

[0.06000]

x = [OH−] = 3.0984x10–6 = 3.1x10–6 M Check the assumption by calculating the percent error in [C6H5COO−]eq.

3.1x10–6

0.06000 (100) = 0.0052% which is smaller than 5%, so the assumption is valid.

pOH = –log (3.1x10–6) = 5.5086 pH = 14 – pOH = 14 – 5.50864 = 8.4914 = 8.49

d) After the equivalence point, the concentration of excess strong base determines the pH. Find the concentration of excess base and use it to calculate the pH. Initial moles of C6H5COOH: (0.1000 M)(0.03000 L) = 0.003000 mol Moles of added NaOH: (0.1500 M)(0.02200 L) = 0.003300 mol Amount (mol) C6H5COOH(aq) + OH−(aq) → H2O(l) + C6H5COO−(aq)

Before addition 0.003000 mol — — 0 Addition — 0.003300 mol — — Change – 0.003000 mol – 0.003000 mol — +0.003000 mol After addition 0 0.000300 mol — 0.003000 mol Excess NaOH: 0.000300 mol Total volume: 22.00 mL + 30.00 mL = 52.00 mL (0.05200 L)

[NaOH] = 0.000300 mol NaOH/0.05200 L = 0.00577 M pOH = –log (0.00577) = 2.2388 pH = 14 – pOH = 14 – 2.2388 = 11.76 (pH is shown to two decimal places)

19-10

19.5A Plan: Write the formula of the salt and the reaction showing the equilibrium of a saturated solution. The ion-product expression can be written from the stoichiometry of the solution reaction as the coefficients in the reaction become exponents in the ion-product expression.

Solution: a) The formula of calcium sulfate is CaSO4. The equilibrium reaction is: CaSO4(s) Ca2+(aq) + SO4

2−(aq) Ion–product expression: Ksp = [Ca2+][SO4

2−] b) Chromium(III) carbonate is Cr2(CO3)3. Cr2(CO3)3(s) 2Cr3+(aq) + 3CO3

2−(aq) Ion–product expression: Ksp = [Cr3+]2[CO3

2−]3 c) Magnesium hydroxide is Mg(OH)2. Mg(OH)2(s) Mg2+(aq) + 2OH−(aq) Ion–product expression: Ksp = [Mg2+][OH−]2 d) Arsenic(III) sulfide is As2S3. As2S3(s) 2As3+(aq) + 3S2−(aq) 3{S2−(aq) + H2O(l) HS−(aq) + OH−(aq)} As2S3(s) + 3H2O(l) 2As3+(aq) + 3HS−(aq) + 3OH−(aq)

The second equilibrium must be considered in this case because its equilibrium constant is large, so essentially all the sulfide ion is converted to HS− and OH−.

Ion-product expression: Ksp = [As3+]2[HS−]3[OH−]3 19.5B Plan: Examine the ion-product expressions. The exponents in the ion-product expression are the subscripts of

those ions in the chemical formula. Solution: a) The compound is lead(II) chromate. Its formula is PbCrO4.

b) The compound is iron(II) sulfide. Its formula is FeS. (Note: metal sulfides have a special form of the ion-product expression. The sulfide ion, S2−, is so basic that it reacts completely with water to form the hydrogen sulfide ion, HS− and hydroxide. Thus, the presence of the term [HS−][OH−] is an indication that the anion in the compound is the sulfide ion, S2−. Because the term [HS−][OH−] has an exponent of 1 in this ion-product expression, there is one sulfide ion in the chemical formula for this compound.) c) The compound is strontium fluoride. Its formula is SrF2.

d) The compound is copper(II) phosphate. Its formula is Cu3(PO4)2. 19.6A Plan: Calculate the solubility of CaF2 as molarity and use molar ratios to find the molarity of

Ca2+ and F− dissolved in solution. Calculate Ksp from [Ca2+] and [F−] using the ion-product expression. Solution: Convert the solubility to molar solubility:

Molarity = 4

2 23

2

1.5x10 g CaF 1 mol CaF1 mL10.0 mL 78.08 g CaF10 L

−

−

= 1.9211x10–4 M CaF2

[Ca2+] = [CaF2] = 1.9211x10–4 M because there is 1 mol of calcium ions in each mol of CaF2. [F−] = 2[CaF2] = 3.8422x10–4 M because there are 2 mol of fluoride ions in each mol of CaF2. The solubility equilibrium is: CaF2(s) Ca2+(aq) + 2F−(aq) Ksp = [Ca2+][F−]2

Calculate Ksp using the solubility product expression from above and the saturated concentrations of calcium and fluoride ions.

Ksp = [Ca2+][F−]2 = (1.9211x10–4)( 3.8422x10–4)2 = 2.836024x10–11 = 2.8x10–11 The Ksp for CaF2 is 2.8x10–11 at 18°C. 19.6B Plan: Calculate the solubility of Ag3PO4 as molarity and use molar ratios to find the molarity of Ag+ and PO4

3− dissolved in solution. Calculate Ksp from [Ag+] and [PO4

3−] using the ion-product expression. Solution: Convert the solubility to molar solubility:

Molarity = �3.2x10–4 g Ag3PO4

50. mL� �1000 mL

1 L� � 1 mol Ag3PO4

418.7 g Ag3PO4� = 1.5x10–5 M Ag3PO4

19-11

[Ag+] = 3[Ag3PO4] = 4.5x10–5 M because there are 3 mol of silver ions in each mol of Ag3PO4. [PO4

3−] = [Ag3PO4] = 1.5x10–5 M because there is 1 mol of phosphate ions in each mol of Ag3PO4. The solubility equilibrium is: Ag3PO4 (s) 3 Ag+(aq) + PO4

3−(aq) Ksp = [Ag+]3[ PO43−]

Calculate Ksp using the solubility product expression from above and the saturated concentrations of silver and phosphate ions.

Ksp = [Ag+]3[ PO43−] = (4.5x10–5)3(1.5x10–5) = 1.3669x10–18 = 1.4x10–18

The Ksp for Ag3PO4 is 1.4x10–18 at 20°C. 19.7A Plan: Write the solubility reaction for Mg(OH)2 and set up a reaction table, where S is the unknown molar

solubility of the Mg2+ ion. Use the ion-product expression to solve for the concentration of Mg(OH)2 in a saturated solution (also called the solubility of Mg(OH)2).

Solution: Concentration (M) Mg(OH)2(s) Mg2+(aq) + 2OH−(aq) Initial — 0 0 Change — +S +2S Equilibrium — S 2S Ksp = [Mg2+][OH−]2 = (S)(2S)2 = 4S3 = 6.3x10–10 S = 5.4004114x10–4 = 5.4x10–4 M Mg(OH)2

The solubility of Mg(OH)2 is equal to S, the concentration of magnesium ions at equilibrium, so the molar solubility of magnesium hydroxide in pure water is 5.4x10–4 M.

19.7B Plan: Write the solubility reaction for Ca3(PO4)2 and set up a reaction table, where S is the unknown molar

solubility of Ca3(PO4)2. Use the ion-product expression to solve for the concentration of Ca3(PO4)2 in a saturated solution (also called the solubility of Ca3(PO4)2).

Solution: Concentration (M) Ca3(PO4)2(s) 3Ca2+(aq) + 2PO4

3−(aq) Initial — 0 0 Change — +3S +2S Equilibrium — 3S 2S Ksp = [Ca2+]3[PO4

3−]2 = (3S)3(2S)2 = 108S5 = 1.2x10–29 S = 6.4439x10–7 = 6.4x10–7 M Ca3(PO4)2

The solubility of Ca3(PO4)2 in pure water, S, is 6.4x10–7 M. 19.8A Plan: Write the solubility reaction of BaSO4. For part (a) set up a reaction table in which

[Ba2+] = [SO42−] = S, which also equals the solubility of BaSO4. Then, solve for S using the ion-product

expression. For part (b), there is an initial concentration of sulfate, so set up the reaction table including this initial [SO4

2−]. Solve for the solubility, S, which equals [Ba2+] at equilibrium. Solution: a) Set up reaction table. Concentration (M) BaSO4(s) Ba2+(aq) + SO4

2−(aq) Initial — 0 0 Change — +S +S Equilibrium — S S Ksp = [Ba2+][SO4

2–] = S2 = 1.1x10–10 S = 1.0488x10–5 = 1.0x10–5 M The molar solubility of BaSO4 in pure water is 1.0x10–5 M. b) Set up another reaction table with initial [SO4

2−] = 0.10 M (from the Na2SO4). Concentration (M) BaSO4(s) Ba2+(aq) + SO4

2−(aq) Initial — 0 0.10 Change — +S +S Equilibrium — S 0.10 + S

Ksp = [Ba2+][SO42–] = S(0.10 + S) = 1.1x10–10

19-12

Assume that 0.10 + S is approximately equal to 0.10, which appears to be a good assumption based on the fact that 0.10 > 1x10–10, Ksp.

Ksp = S(0.10) = 1.1x10–10 S = 1.1x10–9 M Molar solubility of BaSO4 in 0.10 M Na2SO4 is 1.1x10–9 M.

The solubility of BaSO4 decreases when sulfate ions are already present in the solution. The calculated decrease is from 10–5 M to 10–9 M, for a 10,000-fold decrease. This decrease is expected to be large because of the high concentration of sulfate ions.

19.8B Plan: Write the solubility reaction of CaF2. For part (a) set up a reaction table in which [Ca2+] = S, which also equals the solubility of CaF2. Then, solve for S using the ion-product expression. For part (b), there is an initial concentration of calcium, so set up the reaction table including this initial [Ca2+]. Solve for the solubility, S, which equals [Ca2+] at equilibrium. For part (c), there is an initial concentration of fluoride, so set up the reaction table including this initial [F−]. Solve for the solubility, S, which equals [Ca2+] at equilibrium.

Solution: a) Set up reaction table. Concentration (M) CaF2(s) Ca2+(aq) + 2F−(aq) Initial — 0 0 Change — +S +2S Equilibrium — S 2S Ksp = [Ca2+][ F−]2 = (S)(2S)2 = 4S3 = 3.2x10–11 S = 2.0x10–4 M The molar solubility of CaF2 in pure water is 2.0x10–4 M. b) Set up another reaction table with initial [Ca2+] = 0.20 M (from the CaCl2). Concentration (M) CaF2(s) Ca2+(aq) + 2F−(aq) Initial — 0.20 0 Change — +S +2S Equilibrium — 0.20 + S 2S

Ksp = [Ca2+][ F−]2 = (0.20 + S)(2S)2 = 1.1x10–10 Assume that 0.20 + S is approximately equal to 0.20, which appears to be a good assumption based on the fact that 0.20 > 3.2x10–11, Ksp.

Ksp = (0.20)(2S)2 = 3.2x10–11 S = 6.3x10–6 M Molar solubility of CaF2 in 0.20 M CaCl2 is 6.3x10–6 M. b) Set up another reaction table with initial [F−] = 0.40 M (from the NiF2; there are two fluoride ions per NiF2

unit, so an NiF2 concentration of 0.20 M gives a fluoride ion concentration of 0.40 M.). Concentration (M) CaF2(s) Ca2+(aq) + 2F−(aq) Initial — 0 0.40 Change — +S +2S Equilibrium — S 0.40 + 2S

Ksp = [Ca2+][ F−]2 = (S)(0.40 + 2S)2 = 1.1x10–10 Assume that 0.40 + 2S is approximately equal to 0.40, which appears to be a good assumption based on the fact that 0.40 > 3.2x10–11, Ksp.

Ksp = (S)(0.40)2 = 3.2x10–11 S = 2.0x10–10 M Molar solubility of CaF2 in 0.20 M NiF2 is 2.0x10–10 M.

The solubility of CaF2 decreases when either calcium or fluoride ions are already present in the solution.

19.9A Plan: First, write the solubility reaction for the salt. Then, check the ions produced when the salt dissolves to see if they will react with acid. Three cases are possible: 1) If OH− is produced, then addition of acid will neutralize the hydroxide ions and shift the solubility equilibrium toward the products. This causes more salt to dissolve. Write the solubility and neutralization reactions. 2) If the anion from the salt is the conjugate base of a weak acid, it will react with the added acid in a neutralization reaction. Solubility of the salt increases as the anion is neutralized. Write the solubility and neutralization reactions.

19-13

3) If the anion from the salt is the conjugate base of a strong acid, it does not react with a strong acid. The solubility of the salt is unchanged by the addition of acid. Write the solubility reaction.

Solution: a) Calcium fluoride, CaF2 Solubility reaction: CaF2(s) Ca2+(aq) + 2F−(aq)

Fluoride ion is the conjugate base of HF, a weak acid. Thus, it will react with H3O+ from the strong acid, HNO3. Neutralization reaction: F−(aq) + H3O+(aq) → HF(aq) + H2O(l)

The neutralization reaction decreases the concentration of fluoride ions, which causes the solubility equilibrium to shift to the right and more CaF2 dissolves. The solubility of CaF2 increases with the addition of HNO3.

b) Zinc sulfide, ZnS Solubility reaction: ZnS(s) + H2O(l) Zn2+(aq) + HS−(aq) + OH−(aq)

Two anions are formed because the sulfide ion from ZnS reacts almost completely with water to form HS− and OH−.

The hydroxide ion reacts with the added acid: Neutralization reaction: OH−(aq) + H3O+(aq) → 2H2O(l)

In addition, the hydrogen sulfide ion, the conjugate base of the weak acid H2S, reacts with the added acid: Neutralization reaction: HS−(aq) + H3O+(aq) → H2S(aq) + H2O(l)

Both neutralization reactions decrease the concentration of products in the solubility equilibrium, which causes a shift to the right, and more ZnS dissolves. The addition of HNO3 will increase the solubility of ZnS.

c) Silver iodide, AgI Solubility reaction: AgI(s) Ag+(aq) + I−(aq)

The iodide ion is the conjugate base of a strong acid, HI. So, I− will not react with added acid. The solubility of AgI will not change with added HNO3.

19.9B Plan: First, write the solubility reaction for the salt. Then, check the ions produced when the salt dissolves to see if

they will react with acid. Three cases are possible: 1) If OH− is produced, then addition of acid will neutralize the hydroxide ions and shift the solubility equilibrium toward the products. This causes more salt to dissolve. Write the solubility and neutralization reactions. 2) If the anion from the salt is the conjugate base of a weak acid, it will react with the added acid in a neutralization reaction. Solubility of the salt increases as the anion is neutralized. Write the solubility and neutralization reactions. 3) If the anion from the salt is the conjugate base of a strong acid, it does not react with a strong acid. The solubility of the salt is unchanged by the addition of acid. Write the solubility reaction.

Solution: a) Silver cyanide, AgCN Solubility reaction: AgCN(s) Ag+(aq) + CN−(aq)

Cyanide ion is the conjugate base of HCN, a weak acid. Thus, it will react with H3O+ from the strong acid, HBr. Neutralization reaction: CN−(aq) + H3O+(aq) → HCN(aq) + H2O(l)

The neutralization reaction decreases the concentration of cyanide ions, which causes the solubility equilibrium to shift to the right and more AgCN dissolves. The solubility of AgCN increases with the addition of HBr.

b) copper(I) chloride, CuCl Solubility reaction: CuCl (s) Cu+(aq) + Cl−(aq)

The chloride ion is the conjugate base of a strong acid, HCl. So, Cl− will not react with added acid. The solubility of CuCl will not change with added HBr.

c) Magnesium phosphate, Mg3(PO4)2 Solubility reaction: Mg3(PO4)2(s) Mg2+(aq) + 2PO4

3−(aq) Phosphate ion is the conjugate base of HPO4

2−, a weak acid. Thus, it will react with H3O+ from the strong acid, HBr.

Neutralization reaction: PO43−(aq) + H3O+(aq) → HPO4

2−(aq) + H2O(l) HPO4

2−, in turn, is the conjugate base of H2PO4−, a weak acid. Thus, it will react with H3O+ from the strong acid,

HBr. Neutralization reaction: HPO4

2−(aq) + H3O+(aq) → H2PO4−(aq) + H2O(l)

H2PO4−, in turn, is the conjugate base of H3PO4, a weak acid. Thus, it will react with H3O+ from the strong acid,

HBr.

19-14

Neutralization reaction: H2PO4−(aq) + H3O+(aq) → H3PO4(aq) + H2O(l)

Each of these neutralization reactions ultimately decreases the concentration of phosphate ions, which causes the solubility equilibrium to shift to the right and more Mg3(PO4)2 dissolves. The solubility of Mg3(PO4)2 increases with the addition of HBr.

19.10A Plan: First, write the solubility equilibrium equation and ion-product expression. Use the given concentrations of

calcium and phosphate ions to calculate Qsp. Compare Qsp to Ksp. If Ksp < Qsp, precipitation occurs. If Ksp ≥ Qsp then, precipitation will not occur. Solution: Write the solubility equation: Ca3(PO4)2(s) 3Ca2+(aq) + 2PO4

3−(aq) and ion-product expression: Qsp = [Ca2+]3[PO4

3−]2 [Ca2+] = [PO4

3–] = 1.0x10–9 M Qsp = [Ca2+]3[PO4

3−]2 = (1.0x10–9)3(1.0x10–9)2 = 1.0x10–45 Compare Ksp and Qsp. Ksp = 1.2x10–29 > 1.0x10–45 = Qsp Precipitation will not occur because concentrations are below the level of a saturated solution as shown by the value of Qsp.

19.10B Plan: First, write the solubility equilibrium equation and ion-product expression. Find the concentrations of the

lead and sulfide ions in the final mixture. Use these concentrations to calculate Qsp. Compare Qsp to Ksp. If Ksp < Qsp, precipitation occurs. If Ksp ≥ Qsp then, precipitation will not occur. Solution: Write the solubility equation: PbS(s) Pb2+(aq) + S2−(aq) and ion-product expression: Qsp = [Pb2+][S2−] 25 L of a solution containing Pb2+ are mixed with 0.500 L of a solution containing S2−. The final volume is 25.5 L.

Amount (mol) of Pb2+ = (25 L) �0.015 g Pb2+

1 L� � 1 mol Pb2+

207.2 g Pb2+� = 0.0018 mol Pb2+

Molarity of Pb2+ = 0.0018 mol Pb2+

25.5 L = 7.1x10–5 M

Amount (mol) of S2−: (0.500 L) �0.10 mol S2–

1 L� = 0.050 mol S2−

Molarity of S2− = 0.050 mol S2–

25.5 L = 2.0x10–3 M

Qsp = [Pb2+][S2−] = (7.1x10–5)(2.0x10–3) = 1.4x10–7 Compare Ksp and Qsp. Ksp = 3x10–25 < 1.4x10–7 = Qsp Precipitation will occur because concentrations are above the level of a saturated solution as shown by the value of Qsp.

19.11A Plan: First, write the solubility equilibrium equation and ion-product expression. For b) use the given amounts of

nickel (II) and hydroxide ions to calculate Qsp. Compare Qsp to Ksp. For c) check the ions produced when the salt dissolves to see if they will react with acid or if the hydroxide ion of a strong base is part of the ion-product expression, in which case, it will influence the solubility of the solid through the common ion effect. Solution: a) Write the solubility equation:

Ni(OH)2(s) Ni2+(aq) + 2OH−(aq) Scene 3 has the same relative number of ions as in the formula of Ni(OH)2. The Ni2+ and OH− ions are in a 1:2 ratio in Scene 3.

b) Write the ion-product expression: Qsp = [Ni2+][OH−]2 Calculate Ksp using Scene 3: Ksp = [Ni2+][OH−]2 = [2][4]2 = 32 Calculate Qsp using Scene 1: Qsp = [Ni2+][OH−]2 = [3][4]2 = 48 Calculate Qsp using Scene 2: Qsp = [Ni2+][OH−]2 = [4][2]2 = 16 Qsp exceeds Ksp in Scene 1 (48 > 32) so additional solid will form in Scene 1.

19-15

c) Hydroxide ion is one of the products of the solubility equilibrium reaction. The hydroxide ion reacts with added acid: Neutralization reaction: OH−(aq) + H3O+(aq) → 2H2O(l) The neutralization reaction decreases the concentration of OH−(aq) in the solubility equilibrium, which causes a shift to the right, and more Ni(OH)2(s) dissolves. Addition of base (OH–) shifts the equilibrium to the left due to the common-ion effect and the mass of Ni(OH)2 increases.

19.11B Plan: First, write the solubility equilibrium equation and ion-product expression. For b) use the given amounts of

lead(II) and chloride ions to calculate Qsp. Compare Qsp to Ksp. For c) check the ions produced when the salt dissolves to see if they will react with acid or if the anion of the acid is part of the ion-product expression, in which case, it will influence the solubility of the solid through the common ion effect. Solution: a) Write the solubility equation:

PbCl2(s) Pb2+(aq) + 2Cl−(aq) Scene 1 has the same relative number of ions as in the formula of PbCl2. The Pb2+ and Cl− ions are in a 1:2 ratio in Scene 1.

b) Write the ion-product expression: Qsp = [Pb2+][Cl−]2 Calculate Qsp using Scene 1: Qsp = [Pb2+][Cl−]2 = [3][6]2 = 108 Calculate Qsp using Scene 2: Qsp = [Pb2+][Cl−]2 = [4][5]2 = 100 Calculate Ksp using Scene 3: Ksp = [Pb2+][Cl−]2 = [5][5]2 = 125 Qsp exceeds Ksp in Scene 3 (125 > 108) so additional solid will form in Scene 3. c) Chloride ion is one of the products of the solubility equilibrium reaction. Added chloride ion (from HCl) will shift the reaction to the left, due to the common ion effect. As a result of the reaction shifting to the left, the mass of solid PbCl2 will increase.

19.12A Plan: Compare the Ksp values for the two salts. Since CaSO4 is more soluble, calculate the concentration of

sulfate ions in equilibrium with the Ca2+ concentration. Solution: The solubility equilibrium for CaSO4 is CaSO4(s) Ca2+(aq) + SO4

2−(aq) and Ksp = [Ca2+][SO4

2−] = 2.4x10−5

sp24 2+

KSO =

[Ca ]−

= 52.4x10

0.025 M

−

= 9.6x10−4 M

19.12B Plan: Compare the Ksp values for the two salts. Since BaF2 is more soluble, calculate the concentration of fluoride

ions in equilibrium with the Ba2+ concentration. Solution: The solubility equilibrium for BaF2 is BaF2(s) Ba2+(aq) + 2F−(aq) and Ksp = [Ba2+][F−]2 = 1.5x10−6

[F−] =�Ksp

[Ba2+] = �1.5x10–6

(0.020) = 8.7x10−3 M

19.13A Plan: Write the complex-ion formation equilibrium reaction. Calculate the initial concentrations of Fe(H2O)6

3+ and CN−. The approach to complex ion equilibria problems is slightly different than for solubility equilibria because formation constants are generally large while solubility product constants are generally very small. The best mathematical approach is to assume that the equilibrium reaction goes to completion and then calculate back to find the equilibrium concentrations of reactants. So, assume that all of the Fe(H2O)6

3+ reacts to form Fe(CN)63−

and calculate the concentration of Fe(CN)63− formed from the given concentrations of Fe(H2O)6

3+ and CN− and the concentration of the excess reactant. Then use the complex ion formation equilibrium to find the equilibrium concentration of Fe(H2O)6

3+.

19-16

Solution: Equilibrium reaction: Fe(H2O)6

3+(aq) + 6CN−(aq) Fe(CN)63−(aq) + 6H2O(l)

Initial concentrations from a simple dilution calculation: Mf = M iV i/Vf

[Fe(H2O)63+] =

( )( )23.1x10 25.5 mL

25.5 + 35.0 mL

M−

= 0.0130661157 M

[CN−] = ( )( )1.5 35.0 mL

25.5 + 35.0 mLM

= 0.867768595 M

Set up a reaction table: Concentration (M) Fe(H2O)6

3+(aq) + 6CN−(aq) Fe(CN)63−(aq) + 6H2O(l)

Initial 0.0130661157 0.867768595 0 — Change −0.0130661157 + x − 6(0.013066115) +0.013066115 — Equilibrium x 0.789371905 0.013066115 —

Kf = 4.0x1043 = 36

632 6

Fe(CN)

Fe(H O) CN

−

+ −

= [ ]

[ ][ ]60.0130661

x 0.789371905

x = 1.35019x10−45 = 1.4x10−45 M The concentration of Fe(H2O)6

3+ at equilibrium is 1.4x10−45 M. This concentration is so low that it is impossible to calculate it using the initial concentrations minus a variable x. The variable would have to be so close to the initial concentration that the initial concentration of x cannot be calculated to enough significant figures (43 in this case) to get a difference in concentrations of 1x10−45 M. Thus, the approach above is the best to calculate the very low equilibrium concentration of Fe(H2O)6

3+. 19.13B Plan: Write the complex-ion formation equilibrium reaction. Calculate the initial concentration of Al(H2O)6

3+. The approach to complex ion equilibria problems is slightly different than for solubility equilibria because formation constants are generally large while solubility product constants are generally very small. The best mathematical approach is to assume that the equilibrium reaction goes to completion and then calculate back to find the equilibrium concentrations of reactants. So, assume that all of the Al(H2O)6

3+ reacts to form AlF63− and

calculate the concentration of AlF63− formed from the given concentrations of Al(H2O)6

3+ and F−. Then use the complex ion formation equilibrium to find the equilibrium concentration of Al(H2O)6

3+. Solution: Equilibrium reaction: Al(H2O)6

3+(aq) + 6F−(aq) AlF63−(aq) + 6H2O(l)

Initial concentrations:

[Al(H2O)63+] =

2.4 g AlCl3 x 1 mol AlCl3 133.33 g AlCl3

0.250 L = 0.072 M

[F−] = 0.560 M Set up a reaction table:

Concentration (M) Al(H2O)63+(aq) + 6F−(aq) AlF6

3−(aq) + 6H2O(l) Initial 0.072 0.560 0 — Change −0.072 + x − 6(0.072) +0.072 — Equilibrium x 0.128 0.072 —

Kf = 4x1019 =3

663+

2 6

AlF

Al(H O) F

−

−

= (0.072)

(x)(0.128)6 x = 4.0927x10−16 = 4x10−16 M

The concentration of Al(H2O)63+ at equilibrium is 4x10−16 M.

19.14A Plan: Write equations for the solubility equilibrium and formation of the silver-ammonia complex ion. Add the

two equations to get the overall reaction. Set up a reaction table for the overall reaction with the given value for initial [NH3]. Write equilibrium expressions from the overall balanced reaction and calculate Koverall from Kf and

19-17

Ksp values. Insert the equilibrium concentration values from the reaction table into the equilibrium expression and calculate solubility.

Solution: Equilibria:

AgBr(s) Ag+(aq) + Br−(aq) Ksp = 5.0x10–13 Ag+(aq) + 2NH3(aq) Ag(NH3)2

+(aq) Kf = 1.7x107 AgBr(s) + 2NH3(aq) Ag(NH3)2

+(aq) + Br−(aq) Koverall = KspKf Set up reaction table: Concentration (M) AgBr(s) + 2NH3(aq) Ag(NH3)2

+(aq) + Br−(aq) Initial — 1.0 0 0 Change — – 2S +S +S Equilibrium — 1.0 – 2S S S

Koverall = 3 2

23

Ag(NH ) Br

NH

+ −

= KspKf = (5.0x10–13)(1.7x107) = 8.5x10–6

Calculate the solubility of AgBr:

Koverall = [ ][ ]

[ ]21.0 2−

S S

S = 8.5x10–6

Assume that 1.0 – 2S is approximately equal to 1.0, which appears to be a good assumption based on the fact that 1.0 >> 8.5x10–6, Koverall. [ ][ ][ ]21.0

S S = 8.5x10–6

S = 2.9154759x10–3 = 2.9x10–3 M The solubility of AgBr in ammonia is less than its solubility in hypo (sodium thiosulfate).

Since the formation constant for Ag(NH3)2+ is less than the formation constant of Ag(S2O3)2

3−, the addition of ammonia will increase the solubility of AgBr less than the addition of thiosulfate ion increases its solubility.

19.14B Plan: Write equations for the solubility equilibrium and formation of the lead(II)-hydroxide complex ion. Add the

two equations to get the overall reaction. Set up a reaction table for the overall reaction with the given value for initial [OH−]. Write equilibrium expressions from the overall balanced reaction and calculate Koverall from Kf and Ksp values. Insert the equilibrium concentration values from the reaction table into the equilibrium expression and calculate solubility.

Solution: Equilibria:

PbCl2(s) Pb2+(aq) + 2Cl−(aq) Ksp = 1.7x10–5 Pb2+(aq) + 3OH−(aq) Pb(OH)3

−(aq) Kf = 8x1013 PbCl2(s)+ 3OH−(aq) Pb(OH)3

−(aq) + 2Cl−(aq) Koverall = KspKf Set up reaction table: Concentration (M) PbCl2 (s) + 3OH− (aq) Pb(OH)3

− (aq) + 2Cl− Initial — 0.75 0 0 Change — – 3S +S +2S Equilibrium — 0.75 – 3S S 2S

Koverall =

23

3

Pb(OH) Cl

OH

− −

−

= KspKf = (1.7x10–5)(8x1013) = 1x109

Calculate the solubility of PbCl2:

Koverall = [S][2S]2

�0.75 – 3S�3 = 1x109

4S3

�0.75 – 3S�3 = 1x109 Take the cube root of both sides of the equation

19-18

√43 𝑆0.75 – 3S

= 1x103

S = 472.4704 – 1889.8816S S = 0.2499 = 0.25 M

CHEMICAL CONNECTIONS BOXED READING PROBLEMS B19.1 Plan: Consult Figure 19.5 for the colors and pH ranges of the indicators. Solution: Litmus paper indicates the pH is below 7. The result from thymol blue, which turns yellow at a pH above 2.5,

indicates that the pH is above 2.5. Bromphenol blue is the best indicator as it is green in a fairly narrow range of 3.5 < pH < 4. Methyl red turns red below a pH of 4.3. Therefore, a reasonable estimate for the rainwater pH is 3.5 to 4.

B19.2 Plan: Find the volume of the rain received by multiplying the surface area of the lake by the depth of rain. Find

the volume of the lake before the rain. Express both volumes in liters. The pH of the rain is used to find the molarity of H3O+; this molarity multiplied by the volume of rain gives the moles of H3O+. The moles of H3O+ divided by the volume of the lake plus rain gives the molarity of H3O+ and the pH of the lake.

Solution: a) To find the volume of rain, multiply the surface area in square inches by the depth of rain. Convert the volume in in3 to cm3 and then to L.

Volume (L) of rain = ( ) ( )2 33 2 3

34.840x10 yd 36 in 2.54 cm 1 mL 10 L10.0 acres 1.00 in

1 acre 1 yd 1 in 1 mL1 cm

−

= 1.027902x106 L At pH = 4.20, [H3O+] = 10–4.20 = 6.3095734x10–5 M

Moles of H3O+ = ( )5

6 6.3095734x10 mol1.027902x10 LL

−

= 64.8562 = 65 mol

b) Volume (L) of the lake =

( ) ( )2 33 2 3

34.840x10 yd 36 in 12 in 2.54 cm 1 mL 10 L10.0 acres 10.0 ft

1 acre 1 yd 1 ft 1 in 1 mL1 cm

−

= 1.23348x108 L Total volume of lake after rain = 1.23348x108 L + 1.027902x106 L = 1.243759x108 L

[H3O+] = 38

mol H O 64.8562 mol = L 1.243759x10 L

+

= 5.214531x10–7 M

pH = –log (5.214531x10–7) = 6.2827847 = 6.28 c) Each mol of H3O+ requires one mole of HCO3

− for neutralization.

Mass (g) = (64.8562 mol H3O+) 3 3

3 3

1 mol HCO 61.02 g HCO1 mol H O 1 mol HCO

− −

+ −

= 3.97575x103 = 4.0x103 g HCO3–

END–OF–CHAPTER PROBLEMS 19.1 The purpose of an acid-base buffer is to maintain a relatively constant pH in a solution. 19.2 The weak-acid component neutralizes added base and the weak-base component neutralizes added acid so that the pH of the buffer solution remains relatively constant. The components of a buffer do not neutralize one another when they are a conjugate acid-base pair.

19-19

19.3 The presence of an ion in common between two solutes will cause any equilibrium involving either of them to shift in accordance with Le Châtelier’s principle. For example, addition of NaF to a solution of HF will cause the equilibrium HF(aq) + H2O(l) H3O+(aq) + F− (aq) to shift to the left, away from the excess of F− , the co mmon ion. 19.4 a) Buffer 3 has equal, high concentrations of both HA and A− . It has the highest buffering capacity. b) All of the buffers have the same pH range. The practical buffer range is pH = pKa ± 1, and is independent of concentration. c) Buffer 2 has the greatest amount of weak base and can therefore neutralize the greatest amount of added acid. 19.5 A buffer with a high capacity has a great resistance to pH change. A high buffer capacity results when the weak

acid and weak base are both present at high concentration. Addition of 0.01 mol of HCl to a high capacity buffer will cause a smaller change in pH than with a low capacity buffer, since the ratio [HA]/[A−] will change less.

19.6 Only the concentration of the buffer components (c) has an affect on the buffer capacity. In theory, any

conjugate pair (of any pKa) can be used to make a high capacity buffer. With proper choice of components, it can be at any pH. The buffer range changes along with the buffer capacity, but does not determine it. A high capacity buffer will result when comparable quantities (i.e., buffer-component ratio < 10:1) of weak acid and weak base are dissolved so that their concentrations are relatively high.

19.7 The buffer-component ratio refers to the ratio of concentrations of the acid and base that make up the buffer.

When this ratio is equal to 1, the buffer resists changes in pH with added acid to the same extent that it resists changes in pH with added base. The buffer range extends equally in both the acidic and basic direction. When the ratio shifts with higher [base] than [acid], the buffer is more effective at neutralizing added acid than base so the range extends further in the acidic than basic direction. The opposite is true for a buffer where [acid] > [base]. Buffers with a ratio equal to 1 have the greatest buffer range. The more the buffer-component ratio deviates from 1, the smaller the buffer range.

19.8 pKa (formic acid) = 3.74; pKa (acetic acid) = 4.74. Formic acid would be the better choice, since its pKa is closer

to the desired pH of 3.5. If acetic acid were used, the buffer-component ratio would be far from 1:1 and the buffer’s effectiveness would be lower. The NaOH serves to partially neutralize the acid and produce its conjugate base.

19.9 Plan: Remember that the weak-acid buffer component neutralizes added base and the weak-base buffer

component neutralizes added acid. Solution: a) The buffer-component ratio and pH increase with added base. The OH− reacts with HA to decrease its

concentration and increase [NaA]. The ratio [NaA]/[HA] thus increases. The pH of the buffer will be more basic because the concentration of base, A−, has increased and the concentration of acid, HA, decreased. b) Buffer-component ratio and pH decrease with added acid. The H3O+ reacts with A− to decrease its concentration and increase [HA]. The ratio [NaA]/[HA] thus decreases. The pH of the buffer will be more acidic because the concentration of base, A−, has decreased and the concentration of acid, HA, increased.

c) Buffer-component ratio and pH increase with the added sodium salt. The additional NaA increases the concentration of both NaA and HA, but the relative increase in [NaA] is greater. Thus, the ratio increases and the solution becomes more basic. Whenever base is added to a buffer, the pH always increases, but only slightly if the amount of base is not too large.

d) Buffer-component ratio and pH decrease. The concentration of HA increases more than the concentration of NaA, so the ratio is less and the solution is more acidic. 19.10 a) pH would increase by a small amount. b) pH would decrease by a small amount. c) pH would increase by a very small amount. d) pH would increase by a large amount.

19-20

19.11 Plan: The buffer components are propanoic acid and propanoate ion, the concentrations of which are known. The sodium ion is a spectator ion and is ignored because it is not involved in the buffer. Write the propanoic acid- dissociation reaction and its Ka expression. Set up a reaction table in which x equals the amount of acid that dissociates; solving for x will result in [H3O+], from which the pH can be calculated. Alternatively, the pH can be calculated from the Henderson-Hasselbalch equation. Solution:

Concentration (M) CH3CH2COOH(aq) + H2O(l) CH3CH2COO−(aq) + H3O+(aq) Initial 0.15 — 0.35 0 Change – x — +x +x Equilibrium 0.15 – x — 0.35 + x x Assume that x is negligible with respect to both 0.15 and 0.35 since both concentrations are much larger than Ka.

Ka = 1.3x10–5 = 3 3 2

3 2

H O CH CH COO

CH CH COOH

+ −

=

[ ][ ][ ]

x 0.35 x0.15 x

+

− =

[ ][ ][ ]

x 0.350.15

x = [H3O+] = Ka =3 2

3 2

CH CH COOH

CH CH COO−

= ( )5 0.151.3x100.35

−

= 5.57143x10–6 = 5.6x10–6 M

Check assumption: Percent error = (5.6x10–6/0.15)100% = 0.0037%. The assumption is valid. pH = –log [H3O+] = –log (5.57143x10–6) = 5.2540 = 5.25 Another solution path to find pH is using the Henderson-Hasselbalch equation:


K pKa = –log (1.3x10–5) = 4.886

3 2

3 2

[CH CH COO ]pH = 4.886 + log

[CH CH COOH]

−

= [0.35] 4.886 + log[0.15]

pH = 5.25398 = 5.25 19.12 C6H5COOH(aq) + H2O(l) C6H5COO–(aq) + H3O+(aq)

Ka = 6.3x10–5 = 3 6 5

6 5

H O C H COO

C H COOH

+ −

= [ ][ ][ ]x33.0

x28.0x−+

= [ ][ ][ ]33.0

28.0x

x = [H3O+] = (6.3x10–5)(0.33/0.28) = 7.425x10–5 = 7.4x10–5 M Check assumption: Percent error = (7.425x10–5/0.28)100% = 0.026%. The assumption is valid. pH = –log [H3O+] = –log (7.425x10–5) = 4.1293 = 4.13 19.13 Plan: The buffer components are HNO2 and NO2

−, the concentrations of which are known. The potassium ion is a spectator ion and is ignored because it is not involved in the buffer. Write the HNO2 acid-dissociation reaction and its Ka expression. Set up a reaction table in which x equals the amount of acid that dissociates; solving for x will result in [H3O+], from which the pH can be calculated. Alternatively, the pH can be calculated from the Henderson-Hasselbalch equation. Solution:

Concentration (M) HNO2(aq) + H2O(l) NO2−(aq) + H3O+(aq)

Initial 0.55 — 0.75 0 Change −x — +x +x Equilibrium 0.55 − x — 0.75 + x x Assume that x is negligible with respect to both 0.55 and 0.75 since both concentrations are much larger than Ka.

Ka = 7.1x10–4 = [ ]

3 2

2

H O NO

HNO

+ − =

[ ][ ][ ]

x 0.75 x0.55 x

+

− = [ ][ ][ ]

x 0.750.55

x = [H3O+] = Ka [ ]2

2

HNO

NO −

= ( ) [ ][ ]

4 0.557.1x10

0.75− = 5.2066667x10–4 = 5.2x10–4 M

Check assumption: Percent error = (5.2066667x10–4/0.55)100% = 0.095%. The assumption is valid. pH = −log [H3O+] = −log (5.2066667x10–4) = 3.28344 = 3.28

19-21

Verify the pH using the Henderson-Hasselbalch equation.


K pKa = –log(7.1x10–4) = 3.149

2

2

[NO ]pH = 3.149 + log

[HNO ]

−

= [0.75] 3.149 + log[0.55]

pH = 3.2837 = 3.28 19.14 HF(aq) + H2O(l) F–(aq) + H3O+(aq)

Ka = 6.8x10−4 =[ ]

3H O F

HF

+ − =

[ ][ ][ ]

x 0.25 x0.20 x

+

− = [ ][ ][ ]

x 0.250.20

[H3O+] = Ka [ ]HF

F−

= (6.8x10−4)(0.20/0.25) = 5.44x10−4 = 5.4x10−4 M

Check assumption: Percent error = (5.44x10−4/0.20)100% = 0.27%. The assumption is valid. pH = −log [H3O+] = −log (5.44x10−4) = 3.2644 = 3.26 Verify the pH using the Henderson-Hasselbalch equation. 19.15 Plan: The buffer components are formic acid, HCOOH, and formate ion, HCOO−, the concentrations of which are

known. The sodium ion is a spectator ion and is ignored because it is not involved in the buffer. Write the HCOOH acid-dissociation reaction and its Ka expression. Set up a reaction table in which x equals the amount of acid that dissociates; solving for x will result in [H3O+], from which the pH can be calculated. Alternatively, the pH can be calculated from the Henderson-Hasselbalch equation.

Solution: Ka = ap10− K = 10−3.74 = 1.8197x10−4 Concentration (M) HCOOH(aq) + H2O(l) HCOO−(aq) + H3O+(aq) Initial 0.45 — 0.63 0 Change −x — +x +x Equilibrium 0.45 − x — 0.63 + x x Assume that x is negligible because both concentrations are much larger than Ka.

Ka = 1.8197x10−4 = [ ]

3H O HCOO

HCOOH

+ − =

[ ][ ][ ]

x 0.63 x0.45 x

+

− = [ ][ ][ ]

x 0.630.45

x = [H3O+] = Ka [ ]HCOOH

HCOO−

= ( ) [ ][ ]

4 0.451.8197x10

0.63− = 1.29979x10−4 = 1.3x10−4 M

Check assumption: Percent error = (1.29979x10−4/0.45)100% = 0.029%. The assumption is valid. pH = −log [H3O+] = −log (1.29979x10−4) = 3.886127 = 3.89

Verify the pH using the Henderson-Hasselbalch equation.


K

[HCOO ]pH = 3.74 + log[HCOOH]

−

= [0.63] 3.74 + log[0.45]

pH = 3.8861 = 3.89 19.16 HBrO(aq) + H2O(l) BrO−(aq) + H3O+(aq)

Ka = ap10− K = 10−8.64 = 2.2908677x10−9

Ka = 2.2908677x10−9 = [ ]3H O BrO

HBrO

+ − = [ ][ ][ ]x95.0

x68.0x−+

= [ ][ ][ ]95.0

68.0x

19-22

x = [H3O+] = Ka [ ]HBrO

BrO− = (2.2908677x10−9)(0.95/0.68) = 3.2004769x10−9 = 3.2x10−9 M

Check assumption: Percent error = (3.2004769x10−9/0.68)100% = 0.00000047%. The assumption is valid. pH = −log [H3O+] = −log (3.2004769x10−9) = 8.4947853 = 8.49 Verify the pH using the Henderson-Hasselbalch equation. 19.17 Plan: The buffer components phenol, C6H5OH, and phenolate ion, C6H5O−, the concentrations of which are

known. The sodium ion is a spectator ion and is ignored because it is not involved in the buffer. Write the C6H5OH acid-dissociation reaction and its Ka expression. Set up a reaction table in which x equals the amount of acid that dissociates; solving for x will result in [H3O+], from which the pH can be calculated. Alternatively, the pH can be calculated from the Henderson-Hasselbalch equation.

Solution: Ka = ap10− K = 10–10.00 = 1.0x10–10 Concentration (M) C6H5OH(aq) + H2O(l) C6H5O−(aq) + H3O+(aq) Initial 1.2 — 1.3 0 Change –x — +x +x Equilibrium 1.2 – x — 1.3 + x x

Assume that x is negligible with respect to both 1.0 and 1.2 because both concentrations are much larger than Ka.

Ka = 1.0x10–10 = 3 6 5

6 5

H O C H O

C H OH

+ −

= [ ][ ][ ]

x 1.3 x1.2 x

+

− = [ ][ ][ ]

x 1.31.2

x = [H3O+] = Ka 6 5

6 5

C H OH

C H O−

= ( )10 1.21.0x101.3

−

= 9.23077x10–11 M

Check assumption: Percent error = (9.23077x10–11/1.2)100% = 7.7x10–9%. The assumption is valid. pH = –log (9.23077x10–11) = 10.03476 = 10.03 Verify the pH using the Henderson-Hasselbalch equation:


K

6 5

6 5

[C H O ]pH = 10.00 + log

[C H OH]

−

= [1.3] 10.00 + log[1.2]

pH = 10.03 19.18 H3BO3(aq) + H2O(l) H2BO3

–(aq) + H3O+(aq) Ka = ap10− K = 10–9.24 = 5.7543994x10–10

Ka = 5.7543994x10–10 = 3 2 3

3 3

H O H BO

H BO

+ −

= [ ][ ][ ]

x 0.82 x0.12 x

+

− = [ ][ ][ ]

x 0.820.12

x = [H3O+] = Ka 3 3

2 3

H BO

H BO −

= (5.7543994x10–10)(0.12/0.82) = 8.4210723x10–11 M

Check assumption: Percent error = (8.4210723x10–11/0.12)100% = 7.0x10–8%. The assumption is valid. pH = –log [H3O+] = –log (8.4210723x10–11) = 10.0746326 = 10.07 Verify the pH using the Henderson-Hasselbalch equation. 19.19 Plan: The buffer components ammonia, NH3, and ammonium ion, NH4

+, the concentrations of which are known. The chloride ion is a spectator ion and is ignored because it is not involved in the buffer. Write the NH4

+ acid- dissociation reaction and its Ka expression. Set up a reaction table in which x equals the amount of acid that dissociates; solving for x will result in [H3O+], from which the pH can be calculated. Alternatively, the pH can be calculated from the Henderson-Hasselbalch equation. The Ka of NH4

+ will have to be calculated from the pKb.

19-23

Solution: 14 = pKa + pKb

pKa = 14 – pKb = 14 – 4.75 = 9.25 Ka = ap10− K = 10–9.25 = 5.62341325x10–10

Concentration (M) NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq)

Initial 0.15 — 0.25 0 Change –x — +x +x Equilibrium 0.15 – x — 0.25 + x x


Ka = .62341325x10–10 = 3 3

4

NH H O

NH

+

+

= [ ]

[ ]30.25 x H O

0.15 x

+ + −

= [ ]

[ ]30.25 H O

0.15

+

X = [H3O+] = Ka 4

3

NH

NH

+

= ( )10 0.155.62341325x100.25

−

= 3.374048x10–10 M

Check assumption: Percent error = (3.374048x10–10/0.15)100% = 2x10–7%. The assumption is valid. pH = –log [H3O+] = –log [3.374048x10–10] = 9.4718 = 9.47 Verify the pH using the Henderson-Hasselbalch equation.


K

3+

4

[NH ]pH = 9.25 + log

[NH ]

= [0.25] 9.25 + log[0.15]

pH = 9.47 19.20 Kb = bp10 K− = 10–3.35 = 4.4668359x10–4

The base component is CH3NH2 and the acid component is CH3NH3+. Neglect Cl–. Assume + x and – x are

negligible. CH3NH2(aq) + H2O(l) CH3NH3

+(aq) + OH–(aq)

Kb = 4.4668359x10–4 = 3 3

3 2

CH NH OH

CH NH

+ −

= [ ][ ]

0.60 x OH

0.50 x

− + −

= [ ][ ]

0.60 OH

0.50

−

[OH–] = Kb 3 2

3 3

CH NH

CH NH +

= (4.4668359x10–4)(0.50/0.60) = 3.7223632x10–4 M

Check assumption: Percent error = (3.7223632x10–4/0.50)100% = 0.074%. The assumption is valid. pOH = –log [OH–] = –log (3.7223632x10–4) = 3.429181254 pH = 14.00 – pOH = 14.00 – 3.429181254 = 10.57081875 = 10.57 Verify the pH using the Henderson-Hasselbalch equation. 19.21 Plan: The buffer components are HCO3

− from the salt KHCO3 and CO32− from the salt K2CO3. Choose the Ka

value that corresponds to the equilibrium with these two components. The potassium ion is a spectator ion and is ignored because it is not involved in the buffer. Write the acid-dissociation reaction and its Ka expression. Set up a reaction table in which x equals the amount of acid that dissociates; solving for x will result in [H3O+], from which the pH can be calculated. Alternatively, the pH can be calculated from the Henderson-Hasselbalch equation.

Solution: a) Ka1 refers to carbonic acid, H2CO3, losing one proton to produce HCO3

−. This is not the correct Ka because H2CO3 is not involved in the buffer. Ka2 is the correct Ka to choose because it is the equilibrium constant for the loss of the second proton to produce CO3

2− from HCO3−.

19-24

b) Set up the reaction table and use Ka2 to calculate pH. Concentration (M) HCO3

−(aq) + H2O(l) CO32−(aq) + H3O+(aq)

Initial 0.22 — 0.37 0 Change –x — +x +x Equilibrium 0.22 – x — 0.37 + x x


Ka = 4.7x10−11 = 2

3 3

3

H O CO

HCO

+ −

−

= [ ][ ][ ]

x 0.37 x0.22 x

+

− = [ ][ ][ ]

x 0.370.22

[H3O+] = Ka 3

23

HCO

CO

−

−

= ( )11 0.224.7x100.37

−

= 2.79459x10−11 M

Check assumption: Percent error = (2.79459x10−11/0.22)100% = 1.3x10-8%. The assumption is valid. pH = –log [H3O+] = –log (2.79459x10−11) = 10.5537 = 10.55 Verify the pH using the Henderson-Hasselbalch equation.


K pKa = –log (4.7x10–11) = 10.328

2

3

3

[CO ]pH = 10.328 + log

[HCO ]

−

−

= [0.37] 10.328 + log[0.22]

pH = 10.55 19.22 a) The conjugate acid-base pair is related by Ka2 (6.3x10–8).

b) Assume that x is negligible with respect to both 0.50 and 0.40 because both concentrations are much larger than Ka. The acid component is H2PO4

– and the base component is HPO42–. Neglect Na+. Assume + x and – x are

negligible. H2PO4

–(aq) + H2O(l) HPO42–(aq) + H3O+(aq)

Ka = 6.3x10–8 = 2

3 4

2 4

H O HPO

H PO

+ −

−

= [ ][ ][ ]

x 0.40 x0.50 x

+

− =

[ ][ ][ ]

x 0.400.50

[H3O+] = Ka 2 4

24

H PO

HPO

−

−

= (6.3x10–8)(0.50/0.40) = 7.875x10–8 M

Check assumption: Percent error = (7.875x10–8/0.50)100% = 1.6x10–5%. The assumption is valid. pH = –log [H3O+] = –log (7.875x10–8) = 7.103749438 = 7.10 Verify the pH using the Henderson-Hasselbalch equation. 19.23 Plan: Given the pH and Ka of an acid, the buffer-component ratio can be calculated from the Henderson- Hasselbalch equation. Convert Ka to pKa. Solution: pKa = –log Ka = –log (1.3x10–5) = 4.8860566


K

[Pr ]5.44 = 4.8860566 + log[HPr]

−

0.5539467 = [Pr ]log[HPr]

−

Raise each side to 10x.

19-25

[Pr ][HPr]

−

= 3.5805 = 3.6

19.24 pKa = –log Ka = –log (7.1x10–4) = 3.14874165


K

2

2

[NO ]2.95 = 3.14874165 + log

[HNO ]

−

–0.19874165 = 2

2

[NO ]log

[HNO ]

−


2

2

[NO ][HNO ]

−

= 0.632788 = 0.63

19.25 Plan: Given the pH and Ka of an acid, the buffer-component ratio can be calculated from the Henderson- Hasselbalch equation. Convert Ka to pKa. Solution: pKa = –log Ka = –log (2.3x10–9) = 8.63827


K

[BrO ]7.95 = 8.63827 + log[HBrO]

−

–0.68827 = [BrO ]log[HBrO]

−


[BrO ][HBrO]

−

= 0.204989 = 0.20

19.26 Given the pH and Ka of an acid, the buffer-component ratio can be calculated from the Henderson-Hasselbalch equation. pKa = –log Ka = –log (1.8x10–5) = 4.744727495


K

3

3

[CH COO ]4.39 = 4.744727495 + log

[CH COOH]

−

–0.35473 = 3

3

[CH COO ]log

[CH COOH]

−


3

3

[CH COO ][CH COOH]

−

= 0.441845 = 0.44

19.27 Plan: Determine the pKa of the acid from the concentrations of the conjugate acid and base, and the pH of the

solution. This requires the Henderson-Hasselbalch equation. Set up a reaction table that shows the stoichiometry of adding the strong base NaOH to the weak acid in the buffer. Calculate the new concentrations of the buffer components and use the Henderson-Hasselbalch equation to find the new pH.

19-26

Solution:


K

a[A ]3.35 = p + log[HA]

K−

= a[0.1500] p + log[0.2000]

K

3.35 = pKa – 0.1249387 pKa = 3.474939 = 3.47

Determine the moles of conjugate acid (HA) and conjugate base (A–) using (M)(V) = moles.

Moles of HA = ( ) 0.2000 mol HA0.5000 L1 L

= 0.1000 mol HA

Moles of A– = ( ) 0.1500 mol A0.5000 L1 L

−

= 0.07500 mol A–

The reaction is: HA(aq) + NaOH(aq) → Na+(aq) + A–(aq) + H2O(l) Initial 0.1000 mol 0.0015 mol 0.07500 mol Change –0.0015 mol –0.0015 mol + 0.0015 mol Final 0.0985 mol 0 mol 0.0765 mol

NaOH is the limiting reagent. The addition of 0.0015 mol NaOH produces an additional 0.0015 mol A– and consumes 0.0015 mol of HA.

Then:

[A–] = 0.0765 mol A0.5000 L

−

= 0.153 M A–

[HA] = 0.0985 mol HA0.5000 L

= 0.197 M HA


K

[0.153]pH = 3.474939 + log[0.197]

= 3.365164 = 3.37

19.28 Determine the pKa of the acid from the concentrations of the conjugate acid and base and the pH of the solution. This requires the Henderson-Hasselbalch equation.


K

a +[B]8.88 = p + log

[BH ]

K = a[0.40] p + log[0.25]

K

8.88 = pKa + 0.20411998 pKa = 8.67588 = 8.68

Determine the moles of conjugate acid (BH+) and conjugate base (B) using (M)(V) = moles. Moles BH+ = (0.25 L)(0.25 mol BH+/L) = 0.0625 mol BH+ Moles B = (0.25 L)(0.40 mol B/L) = 0.10 mol B The reaction is: B(aq) + HCl(aq) → BH+(aq) + Cl–(aq) + H2O(l) Initial 0.10 mol 0.0020 mol 0.0625 mol Change –0.0020 mol –0.0020 mol +0.0020 mol Final 0.098 mol 0 mol 0.0645 mol

HCl is the limiting reagent. The addition of 0.0020 mol HCl produces an additional 0.0020 mol BH+ and consumes 0.0020 mol of B.

19-27

Then:

[B] = 0.098 mol B0.25 L

= 0.392 M B

[BH+] = +0.0645 mol BH

0.25 L = 0.258 M BH+


K

[0.392]pH = 8.67588 + log[0.258]

= 8.857546361 = 8.86

19.29 Plan: Determine the pKa of the acid from the concentrations of the conjugate acid and base, and the pH of the

solution. This requires the Henderson-Hasselbalch equation. Set up a reaction table that shows the stoichiometry of adding the strong base Ba(OH)2 to the weak acid in the buffer. Calculate the new concentrations of the buffer components and use the Henderson-Hasselbalch equation to find the new pH.

Solution:


K

a[Y ]8.77 = p + log[HY]

−

K = a[0.220] p + log[0.110]

K

8.77 = pKa + 0.3010299957 pKa = 8.46897 = 8.47

Determine the moles of conjugate acid (HY) and conjugate base (Y–) using (M)(V) = moles.

Moles of HY = ( ) 0.110 mol HY0.350 L1 L

= 0.0385 mol HY

Moles of Y– = ( ) 0.220 mol Y0.350 L1 L

−

= 0.077 mol Y–

The reaction is: 2HY(aq) + Ba(OH)2(aq) → Ba2+(aq) + 2Y–(aq) + 2H2O(l) Initial 0.0385 mol 0.0015 mol 0.077 mol Change –0.0030 mol –0.0015 mol +0.0030 mol Final 0.0355 mol 0 mol 0.0800 mol

Ba(OH)2 is the limiting reagent. The addition of 0.0015 mol Ba(OH)2 will produce 2 x 0.0015 mol Y– and consume 2 x 0.0015 mol of HY.

Then:

[Y–] = 0.0800 mol Y0.350 L

−

= 0.228571 M Y–

[HY] = 0.0355 mol HY0.350 L

= 0.101429 M HY

a[Y ]pH = p + log[HY]

−

K

[0.228571]pH = 8.46897 + log[0.101429]

= 8.82183 = 8.82

19.30 Determine the pKa of the acid from the concentrations of the conjugate acid and base and the pH of the solution.

This requires the Henderson-Hasselbalch equation.


K

19-28

a +[B]9.50 = p + log

[BH ]

K = a[1.05] p + log

[0.750]

K

9.50 = pKa + 0.1461280357 pKa = 9.353872 = 9.35

Determine the moles of conjugate acid (BH+) and conjugate base (B). Moles of BH+ = (0.500 L)(0.750 mol BH+/L) = 0.375 mol BH+ Moles of B = (0.500 L)(1.05 mol B/L) = 0.525 mol B The reaction is: B(aq) + HCl(aq) → BH+(aq) + Cl– (aq) + H2O(l) Initial 0.525 mol 0.0050 mol 0.375 mol Change –0.0050 mol –0.0050 mol +0.0050 mol Final 0.520 mol 0 mol 0.380 mol

HCl is the limiting reagent. The addition of 0.0050 mol HCl will produce 0.0050 mol BH+ and consume 0.0050 mol of B.

Then

[B] = 0.520 mol B0.500 L

= 1.04 M B

[BH+] = +0.380 mol BH

0.500 L = 0.760 M BH+

a +

[B]pH = p + log[BH ]

K

[ ][1.04]pH = 9.353872 + log0.760

= 9.490092 = 9.49

19.31 Plan: The hydrochloric acid will react with the sodium acetate, NaC2H3O2, to form acetic acid, HC2H3O2. Calculate the number of moles of HCl and NaC2H3O2. Set up a reaction table that shows the stoichiometry of the reaction of HCl and NaC2H3O2. All of the HCl will be consumed to form HC2H3O2, and the number of moles of C2H3O2

− will decrease. Find the new concentrations of NaC2H3O2 and HC2H3O2 and use the Henderson- Hasselbalch equation to find the pH of the buffer. Add 0.15 to find the pH of the buffer after the addition of the KOH. Use the Henderson-Hasselbalch equation to find the [base]/[acid] ratio needed to achieve that pH. Solution:

a) Initial moles of HCl = ( )30.452mol HCl 10 L 204mL

L 1mL

−

= 0.092208 mol HCl

Initial moles of NaC2H3O2 = ( )2 3 20.400 mol NaC H O0.500L

L

= 0.200 mol NaC2H3O2

HCl + NaC2H3O2 → HC2H3O2 + NaCl Initial 0.092208 mol 0.200 mol 0 mol Change –0.092208 mol –0.092208 mol +0.092208 mol Final 0 mol 0.107792 mol 0.092208 mol Total volume = 0.500 L + (204 mL)(10–3 L/1 mL) = 0.704 L

[HC2H3O2] = 0.092208 mol0.704 L

= 0.1309773 M

[C2H3O2−] = 0.107792 mol

0.704 L = 0.1531136 M

pKa = –log Ka = –log (1.8x10–5) = 4.744727495

2 3 2a

2 3 2

[C H O ]pH = p + log

[HC H O ]

−

K

19-29

[0.1531136]pH = 4.744727495 + log[0.1309773]

= 4.812545 = 4.81

b) The addition of base would increase the pH, so the new pH is (4.81 + 0.15) = 4.96. The new [C2H3O2

−]/[ HC2H3O2] ratio is calculated using the Henderson-Hasselbalch equation.

2 3 2a

2 3 2


[HC H O ]

−

K

2 3 2

2 3 2

[C H O ]4.96 = 4.744727495 + log

[HC H O ]

−

0.215272505 = 2 3 2

2 3 2

[C H O ]log

[HC H O ]

−

2 3 2

2 3 2

[C H O ][HC H O ]

−

= 1.64162

From part a), we know that [HC2H3O2] + [C2H3O2−] = (0.1309773 M + 0.1531136 M) = 0.2840909 M. Although

the ratio of [C2H3O2−] to [HC2H3O2] can change when acid or base is added, the absolute amount does not

change unless acetic acid or an acetate salt is added. Given that [C2H3O2

−]/[ HC2H3O2] = 1.64162 and [HC2H3O2] + [C2H3O2−] = 0.2840909 M, solve for

[C2H3O2−] and substitute into the second equation.

[C2H3O2−] = 1.64162[HC2H3O2] and [HC2H3O2] + 1.64162[HC2H3O2] = 0.2840909 M

[HC2H3O2] = 0.1075441 M and [C2H3O2−] = 0.176547 M

Moles of C2H3O2− needed = (0.176547 mol C2H3O2

–/L)(0.500 L) = 0.0882735 mol Moles of C2H3O2

− initially = (0.1531136 mol C2H3O2–/L)(0.500 L) = 0.0765568 mol

This would require the addition of (0.0882735 mol – 0.0765568 mol) = 0.0117167 mol C2H3O2−

The KOH added reacts with HC2H3O2 to produce additional C2H3O2−:

HC2H3O2 + KOH → C2H3O2–

+ K+ + H2O(l) To produce 0.0117167 mol C2H3O2

– would require the addition of 0.0117167 mol KOH.

Mass (g) of KOH = ( ) 56.11 g KOH0.0117167 mol KOH1 mol KOH

= 0.657424 = 0.66 g KOH

19.32 a) The sodium hydroxide will react with the sodium bicarbonate, NaHCO3, to form carbonate ion, CO3

2–: NaOH + NaHCO3 → 2Na+ + CO3

2– + H2O Calculate the number of moles of NaOH and NaHCO3. All of the NaOH will be consumed to form CO3

2–, and the number of moles of NaHCO3 will decrease. The HCO3

– is the important part of NaHCO3.

Initial moles NaOH = ( )30.10 mol NaOH 10 L 10.7 mL

L 1 mL

−

= 0.00107 mol NaOH

Initial moles HCO3– = ( )

330.050 mol HCO 10 L 50.0 mL

L 1 mL

− −

= 0.0025 mol HCO3–

NaOH + NaHCO3 → 2Na+ + CO3

2– + H2O Initial 0.00107 mol 0.0025 mol 0 mol Change –0.00107 mol –0.00107 mol +0.00107 mol Final 0 mol 0.00143 mol 0.00107 mol Total volume = (50.0 mL + 10.7 mL)(10–3 L/1 mL) = 0.0607 L

[HCO3–] = 0.00143 mol

0.0607 L = 0.023558484 M

19-30

[CO32–] = 0.00107 mol

0.0607 L = 0.017627677 M

pKa = –log Ka = –log (4.7x10–11) = 10.32790214

2

3a

3

[CO ]pH = p + log

[HCO ]

−

−

K

[0.017627677]pH = 10.32790214 + log[0.023558484]

= 10.2019 = 10.20

b) The addition of acid would decrease the pH, so the new pH is (10.20 – 0.07) = 10.13. The new [CO3

2–]/[HCO3–] ratio is calculated using the Henderson-Hasselbalch equation.

2

3a

3

[CO ]pH = p + log

[HCO ]

−

−

K

2

3

3

[CO ]10.13 = 10.32790214 + log

[HCO ]

−

−

–0.19790214 = 2

3

3

[CO ] log

[HCO ]

−

−

2

3

3

[CO ] [HCO ]

−

− = 0.63401

From part a), we know that [HCO3–] + [CO3

2–] = (0.023558484 M + 0.017627677 M) = 0.041185254 M. Although the ratio of [CO3

2–] to [HCO3–] can change when acid or base is added, the absolute amount does not

change unless acetic acid or an acetate salt is added. Given that [CO3

2–]/[ HCO3–] = 0.63401 and [HCO3

–] + [CO32–] = 0.041185254 M, solve for [CO3

2–] and substitute into the second equation.

[CO32–] = 0.63401[HCO3

–] and [HCO3–] + 0.63401[HCO3

–] = 0.041185254 M [HCO3

–] = 0.025205019 M and [CO32–] = 0.015980234 M

Moles of HCO3– needed = (0.025205019 mol HCO3

–/L)(10–3 L/1 mL)(25.0 mL) = 0.0006301255 mol Moles of HCO3

– initially = (0.023558484 mol HCO3–/L)(10–3 L/1 mL)(25.0 mL)

= 0.000588962 mol This would require the addition of (0.0006301255 mol – 0.000588962 mol) = 0.0000411635 mol HCO3

– The HCl added reacts with CO3

2– to produce additional HCO3−:

CO32– + HCl → HCO3

– + Cl–

To produce 0.0000411635 mol HCO3– would require the addition of 0.0000411635 mol HCl.

Mass (g) of HCl = ( ) 36.46 g HCl0.0000411635 mol HCl1 mol HCl

= 0.0015008 = 0.0015 g HCl

19.33 Plan: Select conjugate pairs with Ka values close to the desired [H3O+]. Solution:

a) For pH ≈ 4.5, [H3O+] = 10–4.5 = 3.2x10–5 M. Some good selections are the HOOC(CH2)4COOH/ HOOC(CH2)4COOH− conjugate pair with Ka equal to 3.8x10–5 or C6H5CH2COOH/C6H5CH2COO− conjugate pair with Ka equal to 4.9x10–5. From the base list, the C6H5NH2/C6H5NH3

+ conjugate pair comes close with Ka = Kw/Kb = 1.0x10–14/4.0x10–10 = 2.5x10–5. b) For pH ≈ 7.0, [H3O+] = 10–7.0 = 1.0x10–7 M. Two choices are the H2PO4

−/HPO42− conjugate pair with Ka of

6.3x10–8 and the H2AsO4−/HAsO4

2− conjugate pair with Ka of 1.1x10–7. 19.34 Select conjugate pairs that have Ka or Kb values close to the desired [H3O+] or [OH–].

a) For [H3O+] ≈ 1x10–9 M, the HOBr/OBr– conjugate pair comes close with Ka equal to 2.3 x10–9. From the base list, the NH3/NH4

+ conjugate pair comes close with Ka = Kw/Kb = 1.0x10–14/1.76x10–5 = 5.7x10–10.

19-31

b) For [OH–] ≈ 3x10–5 M, the NH3/NH4 + conjugate pair comes close; also, it is possible to choose

[H3O+] = 1.0x10–14/3x10–5 = 3.3x10–10; the C6H5OH/C6H5O– comes close with Ka = 1.0x10–10. 19.35 Plan: Select conjugate pairs with pKa values close to the desired pH. Convert pH to [H3O+] for easy comparison to Ka values in the Appendix. Determine an appropriate base by [OH–] = Kw/[H3O+]. Solution:

a) For pH ≈ 3.5 ([H3O+] = 10–pH = 10–3.5 = 3.2x10–4), the best selection is the HOCH2CH(OH)COOH/ HOCH2CH(OH)COOH – conjugate pair with a Ka = 2.9x10–4. The CH3COOC6H4COOH/CH3COOC6H4COO – pair, with Ka = 3.6x10–4, is also a good choice. The [OH–] = Kw/[H3O+] = 1.0x10–14/3.2x10–4 = 3.1x10–11, results in no reasonable Kb values from the Appendix. b) For pH ≈ 5.5 ([H3O+] = 10–pH = 3x10–6), no Ka1 gives an acceptable pair; the Ka2 values for adipic acid, malonic acid, and succinic acid are reasonable. The [OH–] = Kw/[H3O+] = 1.0x10–14/3x10–6 = 3x10–9; the Kb selection is C5H5N/C5H5NH+.

19.36 Select conjugate pairs that have Ka or Kb values close to the desired [H3O+] or [OH–]. a) For [OH–] ≈ 1x10–6 M, no Kb values work. The Ka values are [H3O+] = Kw/[OH–] =

1.0x10–14/1x10–6 = 1x10–8, giving the following acceptable pairs H2PO4–/HPO4

2– or HC6H5O72–/C6H5O7

3–

or HOCl/OCl–. b) For [H3O+] ≈ 4x10–4 M, the HF/F– conjugate pair comes close with Ka equal to 6.8x10–4. From the base list, [OH–] = 1.0x10–14/4x10–4 = 2.5x10–11, there are no reasonable choices. 19.37 The value of the Ka from the Appendix: Ka = 2.9x10–8 pKa = –log (2.9x10–8) = 7.5376

a[ClO ]pH = p + log[HClO]

K−

a) pH = [0.100]7.5376 + log[0.100]

= 7.5376 = 7.54

b) pH = [0.150]7.5376 + log[0.100]

= 7.71369 = 7.71

c) pH = [0.100]7.5376 + log[0.150]

= 7.3615 = 7.36

d) The reaction is NaOH + HClO → Na+ + ClO– + H2O. The original moles of HClO and OCl– are both = (0.100 mol/L)(1.0 L) = 0.100 mol NaOH + HClO → Na+ + ClO– + H2O Initial 0.0050 mol 0.100 mol 0.100 mol Change –0.0050 mol –0.0050 mol + 0.0050 mol Final 0 mol 0.095 mol 0.105 mol

pH = [0.105]7.5376 + log[0.095]

= 7.5811 = 7.58

19.38 Plan: Given the pH and Ka of an acid, the buffer-component ratio can be calculated from the Henderson- Hasselbalch equation. Convert Ka to pKa. Solution: The value of the Ka from the Appendix: Ka = 6.3x10–8 (We are using Ka2 since we are dealing with the

equilibrium in which the second hydrogen ion is being lost.) pKa = –log Ka = –log (6.3x10–8) = 7.200659451 Use the Henderson-Hasselbalch equation:

2

4a

2 4

[HPO ]pH = p + log

[H PO ]

−

−

K

19-32

2

4

2 4

[HPO ]7.40 = 7.200659451 + log

[H PO ]

−

−

0.19934055 = 2

4

2 4

[HPO ]log

[H PO ]

−

−

2

4

2 4

[HPO ][H PO ]

−

− = 1.582486 = 1.6

19.39 You need to know the pKa value for the indicator. Its transition range is approximately pKa ± 1. If the indicator is a diprotic acid, it will have two transition ranges, one for each of the two H3O+ ions lost. 19.40 To see a distinct color in a mixture of two colors, you need one color to be about 10 times the intensity of

the other. For this to take place, the concentration ratio [HIn]/[In–] needs to be greater than 10:1 or less than 1:10. This will occur when pH = pKa – 1 or pH = pKa + 1, respectively, giving a transition range of about two units.

19.41 The addition of an acid-base indicator does not affect the pH of the test solution because the concentration of indicator is very small. 19.42 The equivalence point in a titration is the point at which the number of moles of OH– equals the number of moles

of H3O+ (be sure to account for stoichiometric ratios, e.g., one mol of Ca(OH)2 produces two moles of OH–). The end point is the point at which the added indicator changes color. If an appropriate indicator is selected, the end point is close to the equivalence point, but not normally the same. Using an indicator that changes color at a pH after the equivalence point means the equivalence point is reached first. However, if an indicator is selected that changes color at a pH before the equivalence point, then the end point is reached first.

19.43 a) The reactions are: OH–(aq) + H3PO4(aq) → H2PO4

–(aq) + H2O(l) Ka1 = 7.2x10–3 OH–(aq) + H2PO4

–(aq) → HPO42–(aq) + H2O(l) Ka2 = 6.3x10–8

The correct order is C, B, D, A. Scene C shows the solution before the addition of any NaOH. Scene B is halfway to the first equivalence point; Scene D is halfway to the second equivalence point and Scene A is at end of the titration.

b) Scene B is the second scene in the correct order. This is halfway towards the first equivalence point when there are equal amounts of the acid and conjugate base, which constitutes a buffer.

2 4a

3 4

[H PO ]pH = p + log

[H PO ]K

−

Determine the pKa using pKa = –log (7.2x10–3) = 2.142668 [3]pH = 2.1426675 + log[3]

= 2.1426675 = 2.14

c) 10.00 mL of NaOH is required to reach the first half equivalence point. Therefore, an additional 10.00 mL of NaOH is required to reach the first equivalence point, for a total of 20 mL for the first equivalence point. An additional 20.00 mL of NaOH will be required to reach the second equivalence point where only HPO4

2– remains. A total of 40.00 mL of NaOH is required to reach Scene A. 19.44 a) The initial pH is lowest for the flask solution of the strong acid, followed by the weak acid and then the weak base. In other words, strong acid–strong base < weak acid–strong base < strong acid–weak base in terms of initial pH. b) At the equivalence point, the moles of H3O+ equal the moles of OH–, regardless of the type of titration. However, the strong acid–strong base equivalence point occurs at pH = 7.00 because the resulting cation-anion combination does not react with water. An example is the reaction NaOH + HCl → H2O + NaCl. Neither Na+ nor Cl– ions dissociate in water. The weak acid–strong base equivalence point occurs at pH > 7, because the anion of the weak acid is weakly basic, whereas the cation of the strong base does not react with water. An example is the reaction

19-33

HCOOH + NaOH → HCOO− + H2O + Na+. The conjugate base, HCOO−, reacts with water according to this reaction: HCOO− + H2O → HCOOH + OH−.

The strong acid–weak base equivalence point occurs at pH < 7, because the anion of the strong acid does not react with water, whereas the cation of the weak base is weakly acidic. An example is the reaction HCl + NH3 → NH4

+ + Cl−. The conjugate acid, NH4+, dissociates slightly in water: NH4

+ + H2O → NH3 + H3O+. In rank order of pH at the equivalence point, strong acid–weak base < strong acid–strong base < weak acid–strong base. 19.45 In the buffer region, comparable amounts of weak acid and its conjugate base are present. At the equivalence point, the predominant species is the conjugate base. In a strong acid–weak base titration, the weak base and its conjugate acid are the predominant species present. 19.46 At the very center of the buffer region of a weak acid–strong base titration, the concentration of the weak acid and its conjugate base are equal. If equal values for concentration are put into the Henderson-Hasselbalch equation, the [base]/[acid] ratio is 1, the log of 1 is 0, and the pH of the solution equals the pKa of the weak acid.


K

apH = p + log 1K 19.47 The titration curve for a diprotic acid has two “breaks,” i.e., two regions where the pH increases sharply. For a monoprotic acid, only one break occurs. 19.48 Plan: Indicators have a pH range that is approximated by pKa ± 1. Find the pKa of the indicator by using the relationship pKa = –log Ka. Solution: The pKa of cresol red is –log (3.5x10–9) = 8.5, so the indicator changes color over an approximate range of 8.5 ± 1 or 7.5 to 9.5. 19.49 Indicators have a pH range that is approximated by pKa ± 1. The pKa of ethyl red is –log (3.8x10–6) = 5.42, so the indicator changes color over an approximate range of 4.4 to 6.4. 19.50 Plan: Choose an indicator that changes color at a pH close to the pH of the equivalence point. Solution: a) The equivalence point for a strong acid–strong base titration occurs at pH = 7.0. Bromthymol blue is an indicator that changes color around pH 7.

b) The equivalence point for a weak acid–strong base is above pH 7. Estimate the pH at equivalence point from equilibrium calculations. At the equivalence point, all of the HCOOH and NaOH have been consumed; the solution is 0.050 M HCOO−. (The volume doubles because equal volumes of base and acid are required to reach the equivalence point. When the volume doubles, the concentration is halved.) The weak base HCOO– undergoes a base reaction: Concentration, M COOH–(aq) + H2O(l) HCOOH(aq) + OH–(aq) Initial 0.050 M __ 0 0 Change –x +x +x Equilibrium 0.050 – x x x

The Ka for HCOOH is 1.8x10–4, so Kb = 1.0x10–14/1.8x10–4 = 5.5556x10–11

Kb = 5.5556x10–11 = [ ][ ]

[ ]−−

HCOOOHHCOOH

= [ ][ ]

[ ]x050.0xx−

= [ ][ ][ ]050.0

xx

[OH–] = x = 1.666673x10–6 M pOH = –log (1.666673x10–6) = 5.7781496 pH = 14.00 – pOH = 14.00 – 5.7781496 = 8.2218504 = 8.22 Choose thymol blue or phenolphthalein.

19-34

19.51 a) Determine the Ka (of the conjugate acid) from the Kb for CH3NH2. Ka = Kw/Kb = (1.0x10–14)/(4.4x10–4) = 2.2727x10–11 An acid-base titration of two components of equal concentration and at a 1:1 ratio gives a solution of the conjugates with half the concentration. In this case, the concentration of CH3NH3

+ = 0.050 M.

Ka = 2.2727x10–11 = 3 3 2

3 3

H O CH NH

CH NH

+

+

= 2x

0.050 x− =

2x0.050

x = [H3O+] = 1.0659972x10–6 M pH = –log [H3O+] = –log (1.0659972x10–6) = 5.97224 = 5.97 Either methyl red or alizarin is acceptable. b) This is a strong acid–strong base titration; thus, the equivalence point is at pH = 7.00. The best choice would be bromthymol blue; alizarin might be acceptable. 19.52 Plan: Choose an indicator that changes color at a pH close to the pH of the equivalence point. Solution: a) The equivalence point for a weak base–strong acid is below pH 7. Estimate the pH at equivalence point from equilibrium calculations. At the equivalence point, the solution is 0.25 M (CH3)2NH2

+. (The volume doubles because equal volumes of base and acid are required to reach the equivalence point. When the volume doubles, the concentration is halved.) Ka = Kw/Kb = (1.0x10–14)/(5.9x10–4) = 1.69491525x10–11

Concentration, M (CH3)2NH2+(aq) + H2O(l) (CH3)2NH(aq) + H3O+(aq)

Initial 0.25 M __ 0 0 Change –x +x +x Equilibrium 0.25 – x x x

Ka = 1.69491525x10–11 = 3 3 2

3 2 2

H O (CH ) NH

(CH ) NH

+

+

= 2x

0.25 x− =

2x0.25

x = [H3O+] = 2.0584674x10–6 M pH = –log [H3O+] = –log (2.0584674x10–6) = 5.686456 = 5.69 Methyl red is an indicator that changes color around pH 5.7. b) This is a strong acid–strong base titration; thus, the equivalence point is at pH = 7.00. Bromthymol blue is an indicator that changes color around pH 7. 19.53 a) Determine the Kb (of the conjugate base) from the Ka for C6H5COOH. Kb = Kw/Ka = (1.0x10–14)/(6.3x10–5) = 1.5873x10–10 An acid-base titration of two components of equal concentration and at a 1:1 ratio gives a solution of the conjugates with half the concentration. In this case, the concentration of C6H5COO– = 0.125 M.

Kb = 1.5873x10–10 = 6 5

6 5

C H COOH OH

C H COO

−

−

= [ ][ ]

[ ]x x

0.125 x− =

[ ][ ][ ]

x x0.125

[OH–] = x = 4.4543518x10–6 M pOH = –log (4.4543518x10–6) = 5.351215485 pH = 14.00 – pOH = 14.00 – 5.351215485 = 8.64878 = 8.65 The choices are phenolphthalein or thymol blue.

b) The titration will produce a 0.25 M NH3 solution at the equivalence point. Use the Kb for NH3 from the Appendix.

Kb = 1.76x10–5 = 4

3

NH OH

NH

+ −

=

[ ][ ][ ]

x x0.25 x−

= [ ][ ][ ]x x0.25

[OH–] = x = 2.0976177x10–3 M pOH = –log (2.0976177x10–3) = 2.67827 pH = 14.00 – pOH = 14.00 – 2.67827 = 11.32173 = 11.32 The best choice would be alizarin yellow R; alizarin might be acceptable.

19-35

19.54 Plan: The reaction occurring in the titration is the neutralization of H3O+ (from HCl) by OH− (from NaOH): HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) or, omitting spectator ions:

H3O+(aq) + OH−(aq) → 2H2O(l) For the titration of a strong acid with a strong base, the pH before the equivalence point depends on the excess concentration of acid and the pH after the equivalence point depends on the excess concentration of base. At the equivalence point, there is not an excess of either acid or base so the pH is 7.0. The equivalence point occurs when 40.00 mL of base has been added. Use (M)(V) to determine the number of moles of acid and base. Note that the NaCl product is a neutral salt that does not affect the pH.

Solution: The initial number of moles of HCl = (0.1000 mol HCl/L)(10–3 L/1 mL)(40.00 mL) = 4.000x10–3 mol HCl a) At 0 mL of base added, the concentration of hydronium ion equals the original concentration of HCl. pH = –log (0.1000 M) = 1.0000 b) Determine the moles of NaOH added: Moles of added NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(25.00 mL) = 2.500x10–3 mol NaOH HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) Initial 4.000x10–3 mol 2.500x10–3 mol – 0 Change –2.500x10–3 mol –2.500x10–3 mol – +2.500x10–3 mol Final 1.500x10–3 mol 0 2.500x10–3 mol The volume of the solution at this point is [(40.00 + 25.00) mL](10–3 L/1 mL) = 0.06500 L The molarity of the excess HCl is (1.500x10–3 mol HCl)/(0.06500 L) = 0.023077 M pH = –log (0.023077) = 1.6368 c) Determine the moles of NaOH added: Moles of added NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(39.00 mL) = 3.900x10–3 mol NaOH

HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) Initial 4.000x10–3 mol 4.900x10–3 mol – 0 Change –3.900x10–3 mol –3.900x10–3 mol – +3.900x10–3 mol Final 1.000x10–4 mol 0 3.900x10–3 mol The volume of the solution at this point is [(40.00 + 39.00) mL](10–3 L/1 mL) = 0.07900 L The molarity of the excess HCl is (1.00x 10–4mol HCl)/(0.07900 L) = 0.0012658 M pH = –log (0.0012658) = 2.898 d) Determine the moles of NaOH added: Moles of added NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(39.90 mL) = 3.990x10–3 mol NaOH HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) Initial 4.000x10–3 mol 3.990x10–3 mol – 0 Change –3.990x10–3 mol –3.990x10–3 mol – +3.990x10–3 mol Final 1.000x10–5 mol 0 3.900x10–3 mol The volume of the solution at this point is [(40.00 + 39.90) mL](10–3 L/1 mL) = 0.07990 L The molarity of the excess HCl is (1.0x10–5 mol HCl)/(0.07990 L) = 0.000125156 M pH = –log (0.000125156) = 3.903

e) Determine the moles of NaOH added: Moles of added NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(40.00 mL) = 4.000x10–3 mol NaOH HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) Initial 4.000x10–3 mol 4.000x10–3 mol – 0 Change –4.000x10–3 mol –4.000x10–3 mol – +4.000x10–3 mol Final 0 0 4.000x10–3 mol

The NaOH will react with an equal amount of the acid and 0.0 mol HCl will remain. This is the equivalence point of a strong acid–strong base titration, thus, the pH is 7.00. Only the neutral salt NaCl is in solution at the equivalence point.

f) The NaOH is now in excess. It will be necessary to calculate the excess base after reacting with the HCl. The excess strong base will give the pOH, which can be converted to the pH. Determine the moles of NaOH added: Moles of added NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(40.10 mL) = 4.010x10–3 mol NaOH The HCl will react with an equal amount of the base, and 1.0x10–5 mol NaOH will remain. HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) Initial 4.000x10–3 mol 4.010x10–3 mol – 0

19-36

Change –4.000x10–3 mol –4.000x10–3 mol – +4.000x10–3 mol Final 0 1.000x10–5 mol 4.000x10–3 mol The volume of the solution at this point is [(40.00 + 40.10) mL](10–3 L/1 mL) = 0.08010 L The molarity of the excess NaOH is (1.0x10–5 mol NaOH)/(0.08010 L) = 0.00012484 M pOH = –log (0.00012484) = 3.9036 pH = 14.00 – pOH = 14.00 – 3.9036 = 10.09637 = 10.10 g) Determine the moles of NaOH added: Moles of NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(50.00 mL) = 5.000x10–3 mol NaOH The HCl will react with an equal amount of the base, and 1.000x10–3 mol NaOH will remain. HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) Initial 4.000x10–3 mol 5.000x10–3 mol – 0 Change –4.000x10–3 mol –4.000x10–3 mol – +4.000x10–3 mol Final 0 1.000x10–3 mol 4.000x10–3 mol The volume of the solution at this point is [(40.00 + 50.00) mL](10–3 L/ 1 mL) = 0.09000 L The molarity of the excess NaOH is (1.000x10–3 mol NaOH)/(0.09000 L) = 0.011111 M pOH = –log (0.011111) = 1.95424 pH = 14.00 – pOH = 14.00 – 1.95424 = 12.04576 = 12.05 19.55 The reaction occurring in the titration is the neutralization of OH− (from KOH) by H3O+ (from HBr):

HBr(aq) + KOH(aq) → H2O(l) + KBr(aq) H3O+(aq) + OH−(aq) → 2 H2O(l) For the titration of a strong base with a strong acid, the pH before the equivalence point depends on the excess concentration of base and the pH after the equivalence point depends on the excess concentration of acid. At the equivalence point, there is not an excess of either acid or base so pH is 7.0. The equivalence point occurs when 30.00 mL of acid has been added. The initial number of moles of KOH = (0.1000 mol KOH/L)(10–3 L/1 mL)(30.00 mL) = 3.000x10–3 mol KOH a) At 0 mL of acid added, the concentration of hydroxide ion equals the original concentration of KOH. pOH = –log (0.1000 M) = 1.0000 pH = 14.00 – pOH = 14.00 – 1.0000 = 13.00 b) Determine the moles of HBr added: Moles of added HBr = (0.1000 mol HBr/L)(10–3 L/1 mL)(15.00 mL) = 1.500x10–3 mol HBr The HBr will react with an equal amount of the base, and 1.500x10–3 mol KOH will remain. The volume of the solution at this point is [(30.00 + 15.00) mL](10–3 L/1 mL) = 0.04500 L The molarity of the excess KOH is (1.500x10–3 mol KOH)/(0.04500 L) = 0.03333 M pOH = –log (0.03333) = 1.4772 pH = 14.00 – pOH = 14.00 – 1.4772 = 12.5228 = 12.52 c) Determine the moles of HBr added: Moles of added HBr = (0.1000 mol HBr/L)(10–3 L/1 mL)(29.00 mL) = 2.900x10–3 mol HBr The HBr will react with an equal amount of the base, and 1.00x10–4 mol KOH will remain. The volume of the solution at this point is [(30.00 + 29.00) mL](10–3 L/1 mL) = 0.05900 L The molarity of the excess KOH is (1.00x10–4 mol KOH)/(0.05900 L) = 0.0016949 M pOH = –log (0.0016949) = 2.7708559 pH = 14.00 – pOH = 14.00 – 2.7708559 = 11.2291441 = 11.23 d) Determine the moles of HBr added: Moles of added HBr = (0.1000 mol HBr/L)(10–3 L/1 mL)(29.90 mL) = 2.990x10–3 mol HBr The HBr will react with an equal amount of the base, and 1.0x10–5 mol KOH will remain.

The volume of the solution at this point is [(30.00 + 29.90) mL](10–3 L/1 mL) = 0.05990 L The molarity of the excess KOH is (1.0x10–5 mol KOH)/(0.05990 L) = 0.000166945 M pOH = –log (0.000166945) = 3.7774266 pH = 14.00 – pOH = 14.00 – 3.7774266 = 10.2225734 = 10.2

e) Determine the moles of HBr added: Moles of added HBr = (0.1000 mol HBr/L)(10–3 L/1 mL)(30.00 mL) = 3.000x10–3 mol HBr The HBr will react with an equal amount of the base and 0.0 mol KOH will remain. This is the equivalence point of a strong acid–strong base titration; thus, the pH is 7.00. f) The HBr is now in excess. It will be necessary to calculate the excess base after reacting with the HCl. The excess strong acid will give the pH.

19-37

Determine the moles of HBr added: Moles of added HBr = (0.1000 mol HBr/L)(10–3 L/1 mL)(30.10 mL) = 3.010x10–3 mol HBr The HBr will react with an equal amount of the base, and 1.0x10–5 mol HBr will remain. The volume of the solution at this point is [(30.00 + 30.10) mL](10–3 L/1 mL) = 0.06010 L The molarity of the excess HBr is (1.0x10–5 mol HBr)/(0.06010 L) = 0.000166389 M pH = –log (0.000166389) = 3.778875 = 3.8 g) Determine the moles of HBr added: Moles of added HBr = (0.1000 mol HBr/L)(10–3 L/1 mL)(40.00 mL) = 4.000x10–3 mol HBr The HBr will react with an equal amount of the base, and 1.000x10–3 mol HBr will remain. The volume of the solution at this point is [(30.00 + 40.00) mL](10–3 L/1 mL) = 0.07000 L The molarity of the excess HBr is (1.000x10–3 mol HBr)/(0.07000 L) = 0.0142857 M pH = –log (0.0142857) = 1.845098 = 1.85 19.56 Plan: This is a titration between a weak acid and a strong base. The pH before addition of the base is dependent on the Ka of the acid (labeled HBut). Prior to reaching the equivalence point, the added base reacts with the acid to form butanoate ion (labeled But–). The equivalence point occurs when 20.00 mL of base is added to the acid because at this point, moles acid = moles base. Addition of base beyond the equivalence point is simply the addition of excess OH−. Solution: a) At 0 mL of base added, the concentration of [H3O+] is dependent on the dissociation of butanoic acid: HBut + H2O H3O+ + But– Initial 0.100 M 0 0 Change –x +x +x Equilibrium 0.100 – x x x

Ka = 1.54x10–5 = [ ]

3H O But

HBut

+ − =

2x0.1000 x−

= 2x

0.1000

x = [H3O+] = 1.2409674x10–3 M pH = –log [H3O+] = –log (1.2409674x10–3) = 2.9062 = 2.91 b) The initial number of moles of HBut = (M)(V) = (0.1000 mol HBut/L)(10–3 L/1 mL)(20.00 mL) = 2.000x10–3 mol HBut Determine the moles of NaOH added: Moles of added NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(10.00 mL) = 1.000x10–3 mol NaOH The NaOH will react with an equal amount of the acid, and 1.000x10–3 mol HBut will remain. An equal number of moles of But– will form. HBut(aq) + NaOH(aq) → H2O(l) + But–(aq) + Na+(aq) Initial 2.000x10–3 mol 1.000x10–3 mol – 0 – Change –1.000x10–3 mol –1.000x10–3 mol – +1.000x10–3 mol – Final 1.000x10–3 mol 0 1.000x10–3 mol The volume of the solution at this point is [(20.00 + 10.00) mL](10–3 L/1 mL) = 0.03000 L The molarity of the excess HBut is (1.000x10–3 mol HBut)/(0.03000 L) = 0.03333 M The molarity of the But– formed is (1.000x10–3 mol But– /(0.03000 L) = 0.03333 M Using a reaction table for the equilibrium reaction of HBut: HBut + H2O H3O+ + But– Initial 0.03333 M – 0 0.03333 M Change –x +x +x Equilibrium 0.03333 – x x 0.03333 + x

Ka = 1.54x10–5 = [ ]

3H O But

HBut

+ − =

( )x 0.0333 x0.03333 x

+

− =

( )x 0.033330.03333

x = [H3O+] = 1.54x10–5 M pH = –log [H3O+] = –log (1.54x10–5) = 4.812479 = 4.81

19-38

c) Determine the moles of NaOH added: Moles of added NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(15.00 mL) = 1.500x10–3 mol NaOH The NaOH will react with an equal amount of the acid, and 5.00x10–4 mol HBut will remain, and 1.500x10–3 moles of But– will form. HBut(aq) + NaOH(aq) → H2O(l) + But–(aq) + Na+(aq) Initial 2.000x10–3 mol 1.500x10–3 mol – 0 – Change –1.500x10–3 mol –1.500x10–3 mol – +1.500x10–3 mol – Final 5.000x10–4 mol 0 1.500x10–3 mol The volume of the solution at this point is [(20.00 + 15.00) mL](10–3 L/1 mL) = 0.03500 L The molarity of the excess HBut is (5.00x10–4 mol HBut)/(0.03500 L) = 0.0142857 M The molarity of the But– formed is (1.500x10–3 mol But–)/(0.03500 L) = 0.042857 M Using a reaction table for the equilibrium reaction of HBut: HBut + H2O H3O+ + But– Initial 0.0142857 M – 0 0.042857 M Change –x +x +x Equilibrium 0.0142857 – x x 0.042857 + x

Ka = 1.54x10–5 = [ ]

3H O But

HBut

+ − =

( )x 0.042857 x0.0142857 x

+

− =

( )x 0.0428570.0142857

x = [H3O+] = 5.1333x10–6 M pH = –log [H3O+] = –log (5.1333x10–6) = 5.2896 = 5.29 d) Determine the moles of NaOH added: Moles of added NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(19.00 mL) = 1.900x10–3 mol NaOH The NaOH will react with an equal amount of the acid, and 1.00x10–4 mol HBut will remain, and 1.900x10–3 moles of But– will form. HBut(aq) + NaOH(aq) → H2O(l) + But–(aq) + Na+(aq) Initial 2.000x10–3 mol 1.900x10–3 mol – 0 – Change –1.900x10–3 mol –1.900x10–3 mol – +1.900x10–3 mol – Final 1.000x10–4 mol 0 1.900x10–3 mol The volume of the solution at this point is [(20.00 + 19.00) mL](10–3 L/1 mL) = 0.03900 L The molarity of the excess HBut is (1.00x10–4 mol HBut)/(0.03900 L) = 0.0025641 M The molarity of the But– formed is (1.900x10–3 mol But–)/(0.03900 L) = 0.0487179 M Using a reaction table for the equilibrium reaction of HBut: HBut + H2O H3O+ + But– Initial 0.0025641 M – 0 0.0487179 M Change –x – +x +x Equilibrium 0.0025641 – x +x 0.0487179 + x

Ka = 1.54x10–5 = [ ]

3H O But

HBut

+ − =

( )x 0.0487179 x0.0025641 x

+

− =

( )x 0.04871790.0025641

x = [H3O+] = 8.1052632x10–7 M pH = –log [H3O+] = –log (8.1052632x10–7) = 6.09123 = 6.09 e) Determine the moles of NaOH added: Moles of addedNaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(19.95 mL) = 1.995x10–3 mol NaOH The NaOH will react with an equal amount of the acid, and 5x10–6 mol HBut will remain, and 1.995x10–3 moles of But– will form. HBut(aq) + NaOH(aq) → H2O(l) + But–(aq) + Na+(aq) Initial 2.000x10–3 mol 1.995x10–3 mol – 0 – Change –1.995x10–3 mol –1.995x10–3 mol – +1.995x10–3 mol – Final 5.000x10–6 mol 0 1.995x10–3 mol The volume of the solution at this point is [(20.00 + 19.95) mL](10–3 L/1 mL) = 0.03995 L The molarity of the excess HBut is (5x10–6 mol HBut)/(0.03995 L) = 0.000125156 M The molarity of the But– formed is (1.995x10–3 mol But–)/(0.03995 L) = 0.0499374 M Using a reaction table for the equilibrium reaction of HBut:

19-39

HBut + H2O H3O+ + But– Initial 0.000125156 M – 0 0.0499374 M Change –x – +x +x Equilibrium 0.000125156 – x x 0.0499374 + x

Ka = 1.54x10–5 = [ ]

3H O But

HBut

+ − =

( )x 0.0499374 x0.000125156 x

+

− =

( )x 0.04993740.000125156

x = [H3O+] = 3.859637x10–8 M pH = –log [H3O+] = –log (3.859637x10–8) = 7.41345 = 7.41 f) Determine the moles of NaOH added: Moles of added NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(20.00 mL) = 2.000x10–3 mol NaOH The NaOH will react with an equal amount of the acid, and 0 mol HBut will remain, and 2.000x10–3 moles of But– will form. This is the equivalence point.

HBut(aq) + NaOH(aq) → H2O(l) + But–(aq) + Na+(aq) Initial 2.000x10–3 mol 2.000x10–3 mol – 0 – Change –2.000x10–3 mol –2.000x10–3 mol – +2.000x10–3 mol – Final 0 0 2.000x10–3 mol

The Kb of But– is now important. The volume of the solution at this point is [(20.00 + 20.00) mL](10–3 L/1 mL) = 0.04000 L

The molarity of the But– formed is (2.000x10–3 mol But–)/(0.04000 L) = 0.05000 M Kb = Kw/Ka = (1.0x10–14)/(1.54x10–5) = 6.49351x10–10 Using a reaction table for the equilibrium reaction of But–: But– + H2O HBut + OH– Initial 0.05000 M – 0 0 Change –x – +x +x Equilibrium 0.05000 – x x x

Kb = 6.49351x10–10 = [ ]HBut OH

But

−

−

= [ ][ ]

[ ]x x

0.05000 x− =

[ ][ ][ ]

x x0.05000

[OH–] = x = 5.6980304x10–6 M pOH = –log (5.6980304x10–6) = 5.244275238 pH = 14.00 – pOH = 14.00 – 5.244275238 = 8.7557248 = 8.76 g) After the equivalence point, the excess strong base is the primary factor influencing the pH. Determine the moles of NaOH added: Moles of added NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(20.05 mL) = 2.005x10–3 mol NaOH

The NaOH will react with an equal amount of the acid, 0 mol HBut will remain, and 5x10–6 moles of NaOH will be in excess. There will be 2.000x10–3 mol of But– produced, but this weak base will not affect the pH compared to the excess strong base, NaOH.

HBut(aq) + NaOH(aq) → H2O(l) + But–(aq) + Na+(aq) Initial 2.000x10–3 mol 2.005x10–3 mol – 0 – Change –2.000x10–3 mol –2.000x10–3 mol – +2.000x10–3 mol – Final 0 5.000x10–6 mol 2.000x10–3 mol

The volume of the solution at this point is [(20.00 + 20.05) mL](10–3 L/1 mL) = 0.04005 L The molarity of the excess OH– is (5x10–6 mol OH–)/(0.04005 L) = 1.2484x10–4 M pOH = –log (1.2484x10–4) = 3.9036 pH = 14.00 – pOH = 14.00 – 3.9036 = 10.0964 = 10.10

h) Determine the moles of NaOH added: Moles of added NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(25.00 mL) = 2.500x10–3 mol NaOH The NaOH will react with an equal amount of the acid, 0 mol HBut will remain, and 5.00x10–4 moles of NaOH will be in excess. HBut(aq) + NaOH(aq) → H2O(l) + But–(aq) + Na+(aq) Initial 2.000x10–3 mol 2.500x10–3 mol – 0 –

19-40

Change –2.000x10–3 mol –2.000x10–3 mol – +2.000x10–3 mol – Final 0 5.000x10–4 mol 2.000x10–3 mol The volume of the solution at this point is [(20.00 + 25.00) mL](10–3 L/1 mL) = 0.04500 L The molarity of the excess OH– is (5.00x10–4 mol OH–/(0.04500 L) = 1.1111x10–2 M pOH = –log (1.1111x10–2) = 1.9542 pH = 14.00 – pOH = 14.00 – 1.9542 = 12.0458 = 12.05 19.57 This is a titration between a weak base and a strong acid. The pH before addition of the acid is dependent on the Kb of the base ((CH3CH2)3N)). Prior to reaching the equivalence point, the added acid reacts with base to form (CH3CH2)3NH+ ion. The equivalence point occurs when 20.00 mL of acid is added to the base because at this point, moles acid = moles base. Addition of acid beyond the equivalence point is simply the addition of excess H3O+. The initial number of moles of (CH3CH2)3N = (0.1000 mol (CH3CH2)3N)/L)(10–3 L/1 mL)(20.00 mL) = 2.000x10–3 mol (CH3CH2)3N

a) Since no acid has been added, only the weak base (Kb) is important.

Kb = 5.2x10–4 = ( )

( )

+3 2 3

3 2 3

CH CH NH OH

CH CH N

−

=

[ ][ ][ ]x1000.0

xx−

= [ ][ ][ ]1000.0

xx

[OH–] = x = 7.2111x10–3 M pOH = –log (7.2111x10–3) = 2.141998 pH = 14.00 – pOH = 14.00 – 2.141998 = 11.8580 = 11.86 b) Determine the moles of HCl added: Moles of added HCl = (0.1000 mol HCl/L)(10–3 L/1 mL)(10.00 mL) = 1.000x10–3 mol HCl The HCl will react with an equal amount of the base, and 1.000x10–3 mol (CH3CH2)3N will remain; an equal number of moles of (CH3CH2)3NH+ will form. The volume of the solution at this point is [(20.00 + 10.00) mL](10–3 L/1 mL) = 0.03000 L The molarity of the excess (CH3CH2)3N is (1.000x10–3 mol (CH3CH2)3N)/(0.03000 L) = 0.03333 M The molarity of the (CH3CH2)3NH+ formed is (1.000x10–3 mol (CH3CH2)3NH+)/(0.03000 L) = 0.03333 M

Kb = 5.2x10–4 = ( )

( )3 2 3

3 2 3

CH CH NH OH

CH CH N

+ −

=

[ ][ ][ ]x03333.0

x0333.0x−+

= [ ][ ][ ]03333.0

0333.0x

[OH–] = x = 5.2x10–4 M pOH = –log (5.2x10–4) = 3.283997 pH = 14.00 – pOH = 14.00 – 3.283997 = 10.7160 = 10.72 c) Determine the moles of HCl added: Moles of added HCl = (0.1000 mol HCl/L)(10–3 L/1 mL)(15.00 mL) = 1.500x10–3 mol HCl The HCl will react with an equal amount of the base, and 5.00x10–4 mol (CH3CH2)3N will remain; and 1.500x10–3 moles of (CH3CH2)3NH+ will form. The volume of the solution at this point is [(20.00 + 15.00) mL](10–3 L/1 mL) = 0.03500 L The molarity of the excess (CH3CH2)3N is (5.00x10–4 mol (CH3CH2)3N)/(0.03500 L) = 0.0142857 M The molarity of the (CH3CH2)3NH+ formed is (1.500x10–3 mol (CH3CH2)3NH+)/(0.03500 L) = 0.0428571 M

Kb = 5.2x10–4 = ( )

( )3 2 3

3 2 3

CH CH NH OH

CH CH N

+ −

=

[ ][ ][ ]x0142857.0

x0428571.0x−+

= [ ][ ][ ]0142857.0

0428571.0x

[OH–] = x = 1.7333x10–4 M pOH = –log (1.7333x10–4) = 3.761126 pH = 14.00 – pOH = 14.00 – 3.761126 = 10.23887 = 10.24

d) Determine the moles of HCl added: Moles of added HCl = (0.1000 mol HCl/L)(10–3 L/1 mL)(19.00 mL) = 1.900x10–3 mol HCl The HCl will react with an equal amount of the base, and 1.00x10–4 mol (CH3CH2)3N will remain; and 1.900x10–3 moles of (CH3CH2)3NH+ will form. The volume of the solution at this point is [(20.00 + 19.00) mL](10–3 L/1 mL) = 0.03900 L

19-41

The molarity of the excess (CH3CH2)3N is (1.00x10–4 mol (CH3CH2)3N)/(0.03900 L) = 0.002564103 M The molarity of the (CH3CH2)3NH+ formed is (1.900x10–3 mol (CH3CH2)3NH+)/(0.03900 L) = 0.0487179 M

Kb = 5.2x10–4 = ( )

( )3 2 3

3 2 3

CH CH NH OH

CH CH N

+ −

=

[ ][ ][ ]x 0.0487179 x0.002564103 x

+−

= [ ][ ][ ]x 0.04871790.002564103

[OH–] = x = 2.73684x10–5 M pOH = –log (2.73684x10–5) = 4.56275 pH = 14.00 – pOH = 14.00 – 4.56275 = 9.43725 = 9.44 e) Determine the moles of HCl added: Moles of added HCl = (0.1000 mol HCl/L)(10–3 L/1 mL)(19.95 mL) = 1.995x10–3 mol HCl The HCl will react with an equal amount of the base, and 5x10–6 mol (CH3CH2)3N will remain; and 1.995x10–3 moles of (CH3CH2)3NH+ will form. The volume of the solution at this point is [(20.00 + 19.95) mL](10–3 L/1 mL) = 0.03995 L

The molarity of the excess (CH3CH2)3N is (5x10–6 mol (CH3CH2)3N)/(0.03995 L) = 0.000125156 M The molarity of the (CH3CH2)3NH+ formed is (1.995x10–3 mol (CH3CH2)3NH+)/(0.03995 L) = 0.0499374 M

Kb = 5.2x10–4 = ( )

( )3 2 3

3 2 3

CH CH NH OH

CH CH N

+ −

=

[ ][ ][ ]x 0.0499374 x0.000125156 x

+

− =

[ ][ ][ ]x 0.04993740.000125156

[OH–] = x = 1.303254x10–6 M pOH = –log (1.303254x10–6) = 5.88497 pH = 14.00 – pOH = 14.00 – 5.88497 = 8.11503 = 8.1 f) Determine the moles of HCl added: Moles of added HCl = (0.1000 mol HCl/L)(10–3 L/1 mL)(20.00 mL) = 2.000x10–3 mol HCl The HCl will react with an equal amount of the base, and 0 mol (CH3CH2)3N will remain; and 2.000x10–3 moles of (CH3CH2)3NH+ will form. This is the equivalence point. The volume of the solution at this point is [(20.00 + 20.00) mL](10–3 L/1 mL) = 0.04000 L The molarity of the (CH3CH2)3NH+ formed is (2.000x10–3 mol (CH3CH2)3NH+)/(0.04000 L) = 0.05000 M Ka = Kw/Kb = (1.0x10–14)/(5.2x10–4) = 1.9231x10–11

Ka = 1.9231x10–11 = ( )

( )3 3 2 3

3 2 3

H O CH CH N

CH CH NH

+

+

= [ ][ ]

[ ]x x

0.05000 x− =

[ ][ ][ ]

x x0.05000

x = [H3O+] = 9.80587x10–7 M pH = –log [H3O+] = –log (9.80587x10–7) = 6.0085 = 6.01 g) After the equivalence point, the excess strong acid is the primary factor influencing the pH. Determine the moles of HCl added: Moles of added HCl = (0.1000 mol HCl/L)(10–3 L/1 mL)(20.05 mL) = 2.005x10–3 mol HCl The HCl will react with an equal amount of the base, and 0 mol (CH3CH2)3N will remain, and 5x10–6 moles of HCl will be in excess. The volume of the solution at this point is [(20.00 + 20.05) mL](10–3 L/1 mL) = 0.04005 L The molarity of the excess H3O+ is (5x10–6 mol H3O+)/(0.04005 L) = 1.2484x10–4 M pH = –log (1.2484x10–4) = 3.9036 = 3.90

h) Determine the moles of HCl added: Moles of added HCl = (0.1000 mol HCl/L)(10–3 L/1 mL)(25.00 mL) = 2.500x10–3 mol HCl The HCl will react with an equal amount of the base, and 0 mol (CH3CH2)3N will remain, and 5.00x10–4 mol of HCl will be in excess. The volume of the solution at this point is [(20.00 + 25.00) mL](10–3 L/1 mL) = 0.04500 L The molarity of the excess H3O+ is (5.00x10–4 mol H3O+)/(0.04500 L) = 1.1111x10–2 M pH = –log (1.1111x10–2) = 1.9542 = 1.95 19.58 Plan: Use (M)(V) to find the initial moles of acid and then use the mole ratio in the balanced equation to find

19-42

moles of base; dividing moles of base by the molarity of the base gives the volume. At the equivalence point, the conjugate base of the weak acid is present; set up a reaction table for the base dissociation in which x = the amount of dissociated base. Use the Kb expression to solve for x from which pOH and then pH is obtained. Solution: a) The balanced chemical equation is: NaOH(aq) + CH3COOH(aq) → Na+(aq) + CH3COO–(aq) + H2O(l) The sodium ions on the product side are written as separate species because they have no effect on the pH of the solution. Calculate the volume of NaOH needed: Volume (mL) of NaOH =

( )3

33

3

0.0520 mol CH COOH 10 L 1 mol NaOH L 1 mL42.2 mLL 1 mL 1 mol CH COOH 0.0372 mol NaOH 10 L

−

−

= 58.989247 = 59.0 mL NaOH Determine the moles of initially CH3COOH present:

Moles of CH3COOH = ( )3

30.0520 mol CH COOH 10 L 42.2 mLL 1 mL

−

= 0.0021944 mol CH3COOH

At the equivalence point, 0.0021944 mol NaOH will be added so the moles acid = moles base. The NaOH will react with an equal amount of the acid, 0 mol CH3COOH will remain, and 0.0021944 moles of CH3COO– will be formed.

CH3COOH(aq) + NaOH(aq) → H2O(l) + CH3COO–(aq) + Na+(aq) Initial 0.0021944 mol 0.0021944 mol – 0 – Change –0.0021944 mol –0.0021944 mol – +0.0021944 mol – Final 0 0 0.0021944 mol Determine the liters of solution present at the equivalence point: Volume = [(42.2 + 58.989247) mL](10–3 L/1 mL) = 0.101189247 L Concentration of CH3COO– at equivalence point: Molarity = (0.0021944 mol CH3COO–)/(0.101189247 L) = 0.0216861 M Calculate Kb for CH3COO–: Ka CH3COOH = 1.8x10–5 Kb = Kw/Ka = (1.0x10–14)/(1.8x10–5) = 5.556x10–10 Using a reaction table for the equilibrium reaction of CH3COO–: CH3COO– + H2O CH3COOH + OH– Initial 0.0216861 M – 0 0 Change –x +x +x Equilibrium 0.0216861 – x x x Determine the hydroxide ion concentration from the Kb, and then determine the pH from the pOH.

Kb = 5.556x10–10 = 3

3

CH COOH OH

CH COO

−

−

= [ ][ ]

[ ]x x

0.0216861 x− =

[ ][ ][ ]

x x0.0216861

[OH–] = x = 3.471138x10–6 M pOH = –log (3.471138x10–6) = 5.459528 pH = 14.00 – pOH = 14.00 – 5.459528 = 8.54047 = 8.540

b) In the titration of a diprotic acid such as H2SO3, two OH– ions are required to react with the two H+ ions of each acid molecule. Because of the large difference in Ka values, each mole of H+ is titrated separately, so H2SO3 molecules lose one H+ before any HSO3

– ions do. The balanced chemical equation for the neutralization of H2SO3 at the first equivalence point is:

NaOH(aq) + H2SO3(aq) → Na+(aq) + HSO3–(aq) + H2O(l)

The sodium ions on the product side are written as separate species because they have no effect on the pH of the solution.

Calculate the volume of NaOH needed to reach the first equivalence point: Volume (mL) of NaOH =

19-43

( )3

2 33

2 3

0.0850 mol H SO 10 L 1 mol NaOH L 1 mL28.9 mLL 1 mL 1 mol H SO 0.0372 mol NaOH 10 L

−

−

= 66.034946 = 66.0 mL NaOH Determine the moles of HSO3

– produced at the first equivalence point:

Moles of HSO3– = ( )

232 3 3

2 3

0.0850 mol H SO 1 mol HSO10 L 28.9 mLL 1 mL 1 mol H SO

−−

= 0.0024565 mol HSO3–

Determine the liters of solution present at the first equivalence point: Volume = [(28.9 + 66.034946) mL](10–3 L/1 mL) = 0.094934946 L Determine the concentration of HSO3

– at the first equivalence point: Molarity = (0.0024565 moles HSO3

–)/(0.094934946 L) = 0.0258756 M HSO3

– is an amphoteric substance. To calculate the pH at the first equivalence point, determine whether HSO3– is

a stronger acid or a stronger base. The Ka for HSO3

– is 6.5 x 10-8. Calculate Kb for HSO3

–: Ka1 for H2SO3 = 1.4x10–2 Kb = Kw/Ka1 = (1.0x10–14)/(1.4x10–2) = 7.142857x10–13

Because HSO3– is a stronger acid than it is a base (Ka is larger than Kb for HSO3

–), at the first equilvance point, it will behave as an acid and donate a hydrogen ion.

For the first equivalence point: Using a reaction table for the equilibrium reaction of HSO3

–: HSO3

– + H2O H3O+ + SO32–

Initial 0.0258756 M 0 0 Change –x +x +x Equilibrium 0.0258756 – x x x

Determine the hydronium ion concentration from the Ka, and then determine the pH from the hydronium ion concentration.

Ka = 6.5x10–8 = �H3O+��SO3

2-�

�HSO3- �

= [ ][ ]

[ ]x x

0.0258756 x− =

[ ][ ][ ]

x x0.0258756

(assuming that x is small compared with 0.0258756 M) [H3O+] = x = 4.101114x10–5 M pH at first equivalence point = –log (4.101114x10–5) = 4.3870982 = 4.387

The balanced chemical equation for the neutralization of HSO3– at the second equivalence point is:

NaOH(aq) + HSO3–(aq) → Na+(aq) + SO3

2–(aq) + H2O(l) A total of 66.034946 mL were required to reach the first equivalence point. It will require an equal volume of NaOH to reach the second equivalence point from the first equivalence point. (2 x 66.0 mL = 132 mL) Determine the total volume of solution present at the second equivalence point.

Volume = [(28.9 + 66.034946 + 66.034946) mL](10–3 L/1 mL) = 0.160969892 L 0.0024565 mol HSO3

– were present at the first equivalence point. An equal number of moles of SO32– will be

present at the second equivalence point. Determine the concentration of SO3

2– at the second equivalence point. Molarity = (0.0024565 moles SO3

2–)/(0.160969892 L) = 0.0152606 M

SO32– does not have a hydrogen ion to donate, so it acts as a base and accepts a hydrogen. Because it acts as a

base, we must calculate its Kb. Calculate Kb for SO3

2–: Ka HSO3– (Ka2)= 6.5x10–8

Kb = Kw/Ka = (1.0x10–14)/(6.5x10–8) = 1.53846x10–7

For the second equivalence point:

Using a reaction table for the equilibrium reaction of SO32–:

SO32– + H2O HSO3

– + OH– Initial 0.0152606 M 0 0 Change –x +x +x

19-44

Equilibrium 0.0152606 – x x x Determine the hydroxide ion concentration from the Kb, and then determine the pH from the pOH.

Kb = 1.53846x10–7 = 3

23

HSO OH

SO

− −

−

= [ ][ ]

[ ]x x

0.0152606 x− =

[ ][ ][ ]

x x0.0152606

(assuming x is small compared with 0.0152606 M) [OH–] = x = 4.84539x10–5 M pOH = –log (4.84539x10–5) = 4.31467 pH = 14.00 – pOH = 14.00 – 4.31467 = 9.68533 = 9.685 19.59 a) The balanced chemical equation is: K+(aq) + OH–(aq) + HNO2(aq) → K+(aq) + NO2

–(aq) + H2O(l) The potassium ions on the product side are written as separate species because they have no effect on the pH of the solution. Calculate the volume of NaOH needed: Volume (mL) of KOH =

( )3

23

2

0.0390 mol HNO 10 L 1 mol KOH L 1 mL23.4 mLL 1 mL 1 mol HNO 0.0588 mol KOH 10 L

−

−

= 15.5204 = 15.5 mL KOH Determine the moles of HNO2 present:

Moles of HNO2 = ( )3

20.0390 mol HNO 10 L 23.4 mLL 1 mL

−

= 0.0009126 mol HNO2

At the equivalence point, 0.0009126 mol KOH will be added so the moles acid = moles base. The KOH will react with an equal amount of the acid, 0 mol HNO2 will remain, and 0.0009126 moles of NO2

– will be formed.

HNO2(aq) + KOH(aq) → H2O(l) + NO2–(aq) + K+(aq)

Initial 0.0009126 mol 0.0009126 mol – 0 – Change –0.0009126 mol –0.0009126 mol – +0.0009126 mol – Final 0 0 0.0009126mol Determine the liters of solution present at the equivalence point: Volume = [(23.4 + 15.5204) mL] (10–3 L/1 mL) = 0.0389204 L Concentration of NO2

– at equivalence point: Molarity = (0.0009126 mol NO2

–)/(0.0389204 L) = 0.023447858 M Calculate Kb for NO2

–: Ka HNO2 = 7.1x10–4 Kb = Kw/Ka = (1.0x10–14)/(7.1x10–4) = 1.40845x10–11 Using a reaction table for the equilibrium reaction of NO2

–: NO2

– + H2O HNO2 + OH– Initial 0.023447858 M – 0 0 Change –x +x +x Equilibrium 0.023447858 – x x x Determine the hydroxide ion concentration from the Kb, and then determine the pH from the pOH.

Kb = 1.40845x10–11 = [ ]2

2

HNO OH

NO

−

−

= [ ][ ]

[ ]x x

0.023447858 x− =

[ ][ ][ ]

x x0.023447858

[OH–] = x = 5.7468x10–7 M pOH = –log (5.7468x10–7) = 6.240574 pH = 14.00 – pOH = 14.00 – 6.240574 = 7.759426 = 7.76 b) The balanced chemical equations are: KOH(aq) + H2CO3(aq) → K+(aq) + HCO3

–(aq) + H2O(l) KOH(aq) + HCO3

-(aq) → K+(aq) + CO32–(aq) + H2O(l)

The potassium ions on the product side are written as separate species because they have no effect on the pH of the solution. Calculate the volume of KOH needed:

19-45

Volume (mL) of KOH =

( )3

2 33

2 3

0.130 mol H CO 10 L 1 mol KOH L 1 mL17.3 mLL 1 mL 1 mol H CO 0.0588 mol KOH 10 L

−

−

= 38.248299 = 38.2 mL KOH It will require an equal volume to reach the second equivalence point (76.4 mL). Determine the moles of HCO3

– produced:

Moles = ( )3

2 3 3

2 3

0.130 mol H CO 1 mol HCO10 L 17.3 mLL 1 mL 1 mol H CO

−−

= 0.002249 mol HCO3–

An equal number of moles of CO32– will be present at the second equivalence point.

Determine the liters of solution present at the first equivalence point: Volume = [(17.3 + 38.248299) mL](10–3 L/1 mL) = 0.055548 L Determine the liters of solution present at the second equivalence point: Volume = [(17.3 + 38.248299 + 38.248299) mL](10–3 L/1 mL) = 0.0937966 L Concentration of HCO3

– at equivalence point: Molarity = (0.002249 mol HCO3

–)/(0.055548 L) = 0.0404875 M Concentration of CO3

2– at equivalence point: Molarity = (0.002249 mol CO3

2–)/(0.0937966 L) = 0.023977 M Calculate Kb for HCO3

–: Ka H2CO3 = 4.5x10–7 Kb = Kw/Ka = (1.0x10–14)/(4.5x10–7) = 2.222x10–8 Calculate Kb for CO3

2–: Ka HCO3– = 4.7x10–11

Kb = Kw/Ka = (1.0x10–14)/(4.7x10–11) = 2.1276596x10–4 Determine the hydroxide ion concentration from the Kb, and then determine the pH from the pOH. For the first equivalence point:

Kb = 2.222x10–8= 2 3

3

H CO OH

HCO

−

−

= [ ][ ]

[ ]x x

0.0404875 x− =

[ ][ ][ ]

x x0.0404875

[OH–] = x = 2.999387x10–5 M pOH = – log (2.999387x10–5) = 4.522967495 pH = 14.00 – pOH = 14.00 – 4.522967495 = 9.4770 = 9.48 For the second equivalence point:

Kb = 2.1276596x10–4 = 3

23

HCO OH

CO

− −

−

= [ ][ ]

[ ]x x

0.023977 x− =

[ ][ ][ ]

x x0.023977

[OH–] = x = 2.2586477x10–3 M pOH = – log (2.2586477x10–3) = 2.6461515 pH = 14.00 – pOH = 14.00 – 2.6461515 = 11.3538 = 11.35 19.60 Plan: Use (M)(V) to find the initial moles of base and then use the mole ratio in the balanced equation to find moles of acid; dividing moles of acid by the molarity of the acid gives the volume. At the equivalence point, the conjugate acid of the weak base is present; set up a reaction table for the acid dissociation in which x = the amount of dissociated acid. Use the Ka expression to solve for x from which pH is obtained. Solution: a) The balanced chemical equation is: HCl(aq) + NH3(aq) → NH4

+(aq) + Cl–(aq) The chloride ions on the product side are written as separate species because they have no effect on the pH of the solution. Calculate the volume of HCl needed: Volume (mL) of HCl =

( )3

33

3

0.234 mol NH 10 L 1 mol HCl L 1 mL65.5 mLL 1 mL 1 mol NH 0.125 mol HCl 10 L

−

−

= 122.616 = 123 mL HCl

19-46

Determine the moles of NH3 present:

Moles = ( )3

30.234 mol NH 10 L 65.5 mLL 1 mL

−

= 0.015327 mol NH3

At the equivalence point, 0.015327 mol HCl will be added so the moles acid = moles base. The HCl will react with an equal amount of the base, 0 mol NH3 will remain, and 0.015327 moles of NH4

+ will be formed.

HCl(aq) + NH3(aq) → NH4+(aq) + Cl–(aq)

Initial 0.015327 mol 0.015327 mol 0 – Change –0.015327 mol –0.015327 mol +0.015327 mol – Final 0 0 0.015327 mol Determine the liters of solution present at the equivalence point: Volume = [(65.5 + 122.616) mL](10–3 L/1 mL) = 0.188116 L Concentration of NH4

+ at equivalence point: Molarity = (0.015327 mol NH4

+)/(0.188116 L) = 0.081476 M Calculate Ka for NH4

+: Kb NH3 = 1.76x10–5 Ka = Kw/Kb = (1.0x10–14)/(1.76x10–5) = 5.6818x10–10 Using a reaction table for the equilibrium reaction of NH4

+: NH4

+ + H2O NH3 + H3O+ Initial 0.081476 M – 0 0 Change –x +x +x Equilibrium 0.081476 – x x x

Determine the hydrogen ion concentration from the Ka, and then determine the pH.

Ka= 5.6818x10–10 = 3 3

4

H O NH

NH

+

+

= [ ][ ]

[ ]x x

0.081476 x− =

[ ][ ][ ]

x x0.081476

x = [H3O+] = 6.803898x10–6 M pH = –log [H3O+] = –log (6.803898x10–6) = 5.1672 = 5.17 b) The balanced chemical equation is: HCl(aq) + CH3NH2(aq) → CH3NH3

+(aq) + Cl–(aq) The chloride ions on the product side are written as separate species because they have no effect on the pH of the solution. Calculate the volume of HCl needed:

Volume (mL) of HCl =

( )3

3 23

3 2

1.11 mol CH NH 10 L 1 mol HCl L 1 mL21.8 mLL 1 mL 1 mol CH NH 0.125 mol HCl 10 L

−

−

= 193.584 = 194 mL HCl Determine the moles of CH3NH2 present:

Moles = ( )3

3 21.11 mol CH NH 10 L 21.8 mLL 1 mL

−

= 0.024198 mol CH3NH2

At the equivalence point, 0.024198 mol HCl will be added so the moles acid = moles base. The HCl will react with an equal amount of the base, 0 mol CH3NH2 will remain, and 0.024198 moles of CH3NH3

+ will be formed. HCl(aq) + CH3NH2(aq) → CH3NH3

+(aq) + Cl–(aq) Initial 0.024198 mol 0.024198 mol 0 – Change –0.024198 mol –0.024198 mol +0.024198 mol – Final 0 0 0.024198 mol Determine the liters of solution present at the equivalence point: Volume = [(21.8 + 193.584) mL](10–3 L/1 mL) = 0.215384 L Concentration of CH3NH3

+ at equivalence point: Molarity = (0.024198 mol CH3NH3

+)/(0.215384 L) = 0.1123482 M Calculate Ka for CH3NH3

+: Kb CH3NH2 = 4.4x10–4

19-47

Ka = Kw/Kb = (1.0x10–14)/(4.4x10–4) = 2.2727x10–11 Using a reaction table for the equilibrium reaction of CH3NH3

+: CH3NH3

+ + H2O CH3NH2 + H3O+ Initial 0.1123482 M – 0 0 Change –x +x +x Equilibrium 0.1123482 – x x x Determine the hydrogen ion concentration from the Ka, and then determine the pH.

Ka = 2.2727x10–11 = 3 3 2

3 3

H O CH NH

CH NH

+

+

= [ ][ ]

[ ]x x

0.1123482 x− =

[ ][ ][ ]

x x0.1123482

x = [H3O+] = 1.5979x10–6 M pH = –log [H3O+] = –log (1.5979x10–6) = 5.7964 = 5.80 19.61 a) The balanced chemical equation is: HNO3(aq) + C5H5N(aq) → C5H5NH+(aq) + NO3

–(aq) The nitrate ions on the product side are written as separate species because they have no effect on the pH of the solution. Calculate the volume of HNO3 needed: Volume (mL) of HNO3 =

( )5 5 33

5 5 3

0.0750 mol C H N 1 mol HNO L 1 mL2.65 LL 1 mol C H N 0.447 mol HNO 10 L−

= 444.63087 = 445 mL HNO3 Determine the moles of C5H5NH+ produced:

Moles of C5H5NH+ = ( )5 5 5 5

5 5

0.0750 mol C H N 1 mol C H NH2.65 L

L 1 mol C H N

+

= 0.19875 mol C5H5NH+

Determine the liters of solution present at the equivalence point: Volume = 2.65 L + (444.63087mL)(10–3 L/1 mL) = 3.09463 L

Concentration of C5H5NH+ at equivalence point: Molarity = (0.19875 mol C5H5NH+)/(3.09463 L) = 0.064224 M Calculate Ka for C5H5NH+: Kb C5H5N = 1.7x10–9 Ka = Kw/Kb = (1.0x10–14)/(1.7x10–9) = 5.88235x10–6 Determine the hydrogen ion concentration from the Ka, and then determine the pH.

Ka = 5.88235x10–6 = 3 5 5

5 5

H O C H N

C H NH

+

+

= [ ][ ]

[ ]x x

0.064224 x− =

[ ][ ][ ]

x x0.064224

x = [H3O+] = 6.1464x10–4 M pH = –log [H3O+] = –log (6.1464x10–4) = 3.211379 = 3.21 b) The balanced chemical equations are: HNO3(aq) + H2NCH2CH2NH2(aq) → H2NCH2CH2NH3

+(aq) + NO3–(aq)

HNO3(aq) + H2NCH2CH2NH3+(aq) → H3NCH2CH2NH3

2+(aq) + NO3–(aq)

The nitrate ions on the product side are written as separate species because they have no effect on the pH of the solution. Calculate the volume of HNO3 needed: Volume (mL) of HNO3 =

( )2 2 3 2 33

2 2 3 2 3

0.250 mol H NCH CH NH 1 mol HNO L 1 mL0.188 LL 1 mol H NCH CH NH 0.447 mol HNO 10 L−

= 105.1454 = 105 mL HCl It will require an equal volume to reach the second equivalence point. (210. mL) Determine the moles of H2NCH2CH2NH3

+ produced:

Moles of H2NCH2CH2NH3+ = ( )2 2 3 2 2 2 3 3

2 2 3 2

0.250 mol H NCH CH NH 1 mol H NCH CH NH0.188 L

L 1 mol H NCH CH NH

+

= 0.0470 mol H2NCH2CH2NH3+

An equal number of moles of H3NCH2CH2NH32+ will be present at the second equivalence point.

19-48

Determine the liters of solution present at the first equivalence point: Volume = 0.188 L + (105.1454 mL)(10–3 L/1 mL) = 0.293145 L Determine the liters of solution present at the second equivalence point: Volume = 0.188 L + 2(105.1454 mL)(10–3 L/1 mL) = 0.39829 L Concentration of H2NCH2CH2NH3

+ at equivalence point: Molarity = (0.0470 mol H2NCH2CH2NH3

+)/(0.293145 L) = 0.16033 M Concentration of H3NCH2CH2NH3

2+ at equivalence point: Molarity = (0.0470 mol H3NCH2CH2NH3

2+)/(0.39829 L) = 0.11800 M Calculate Ka for H2NCH2CH2NH3

+: Kb H2NCH2CH2NH2 = 8.5x10–5 Ka = Kw/Kb = (1.0x10–14)/(8.5x10–5) = 1.17647x10–10 Calculate Ka for H3NCH2CH2NH3

2+: Kb H2NCH2CH2NH3+ = 7.1x10–8

Ka = Kw/Kb = (1.0x10–14)/(7.1x10–8) = 1.40845x10–7

Determine the hydrogen ion concentration from the Ka, and then determine the pH for the first equivalence point.

Ka = 1.17647x10–10 = 3 2 2 3 2

2 2 3 3

H O H NCH CH NH

H NCH CH NH

+

+

= [ ][ ]

[ ]x x

0.16033 x− =

[ ][ ][ ]

x x0.16033

x = [H3O+] = 4.3430780x10–6 M pH = –log [H3O+] = –log (4.3430780x10–6) = 5.36220 = 5.36 Determine the hydrogen ion concentration from the Ka, and then determine the pH for the second equivalence point.

Ka = 1.40845x10–7 = 3 2 2 3 3

23 2 3 3

H O H NCH CH NH

H NCH CH NH

+ +

+

= [ ][ ]

[ ]x x

0.11800 x− =

[ ][ ][ ]

x x0.11800

x = [H3O+] = 1.2891745x10–4 M pH = –log [H3O+] = –log (1.2891745x10–4) = 3.889688 = 3.89 19.62 M2X(s) 2M+(aq) + X2–(aq)

Ksp = [M+]2[X2–], assuming M2X is a strong electrolyte. S = molar solubility = 5x10–5 M [M+] = 2S = 1x10–4 M [X2–] = S = 5x10–5 M The actual Ksp is lower than the calculated value because the assumption that M2X is a strong electrolyte (i.e., exists as M+ + X2–) is in error to some degree. There would be some (probably significant) amount of ion pairing to form MX–(aq), M2X(aq), etc., which reduces the effective concentrations of the ions. 19.63 Fluoride ion in BaF2 is the conjugate base of the weak acid HF. The base hydrolysis reaction of fluoride ion F−(aq) + H2O(l) HF(aq) + OH−(aq)

therefore is influenced by the pH of the solution. As the pH increases, [OH–] increases and the equilibrium shifts to the left to decrease [OH–] and increase [F−]. As the pH decreases, [OH–] decreases and the equilibrium shifts to the right to increase [OH–] and decrease [F−]. The changes in [F−] influence the solubility of BaF2.

Chloride ion is the conjugate base of a strong acid so it does not react with water. Thus, its concentration is not influenced by pH, and solubility of BaCl2 does not change with pH. 19.64 To use Ksp for comparing solubilities, the Ksp expressions must be of the same mathematical form. Stated differently, AgCl and AgBr are both 1:1 electrolytes, while Ag2CrO4 is a 2:1 electrolyte. 19.65 Consider the reaction AB(s) A+(aq) + B−(aq), where Qsp = [A+][B−]. If Qsp > Ksp, then there are more ions dissolved than expected at equilibrium, and the equilibrium shifts to the left and the compound AB precipitates. The excess ions precipitate as solid from the solution. 19.66 Plan: Write an equation that describes the solid compound dissolving to produce its ions. The ion-product expression follows the equation Ksp = [Mn+]p[Xz−]q where p and q are the subscripts of the ions in the compound’s formula.

19-49

Solution: a) Ag2CO3(s) 2Ag+(aq) + CO3

2−(aq) Ion-product expression: Ksp = [Ag+]2[CO3

2−] b) BaF2(s) Ba2+(aq) + 2F−(aq) Ion-product expression: Ksp = [Ba2+][F−]2 c) CuS(s) + H2O(l) Cu2+(aq) + HS−(aq) + OH−(aq) Ion-product expression: Ksp = [Cu2+][HS−][OH−] 19.67 a) Fe(OH)3(s) Fe3+(aq) + 3OH–(aq) Ion-product expression: Ksp = [Fe3+][OH–]3 b) Ba3(PO4)2(s) 3Ba2+(aq) + 2PO4

3–(aq) Ion-product expression: Ksp = [Ba2+]3[PO4

3–]2 c) SnS(s) + H2O(l) Sn2+(aq) + HS−(aq) + OH−(aq) Ion-product expression: Ksp = [Sn2+][HS–][OH–] 19.68 Plan: Write an equation that describes the solid compound dissolving to produce its ions. The ion-product expression follows the equation Ksp = [Mn+]p[Xz−]q where p and q are the subscripts of the ions in the compound’s formula. Solution: a) CaCrO4(s) Ca2+(aq) + CrO4

2−(aq) Ion-product expression: Ksp = [Ca2+][CrO4

2−] b) AgCN(s) Ag+(aq) + CN−(aq) Ion-product expression: Ksp = [Ag+][CN−] c) NiS(s) + H2O(l) Ni2+(aq) + HS−(aq) + OH−(aq) Ion-product expression: Ksp = [Ni2+][HS−][OH−] 19.69 a) PbI2(s) Pb2+(aq) + 2I−(aq) Ion-product expression: Ksp = [Pb2+][I−]2 b) SrSO4(s) Sr2+(aq) + SO4

2−(aq) Ion-product expression: Ksp = [Sr2+][SO4

2−] c) CdS(s) + H2O(l) Cd2+(aq) + HS−(aq) + OH−(aq) Ion-product expression: Ksp = [Cd2+][HS−][OH−] 19.70 Plan: Write an equation that describes the solid compound dissolving in water and then write the ion-product expression. Write a reaction table, where S is the molar solubility of Ag2CO3. Substitute the given solubility, S, into the ion-expression and solve for Ksp. Solution: Concentration (M) Ag2CO3(s) 2Ag+(aq) + CO3

2−(aq) Initial — 0 0 Change — +2S +S Equilibrium — 2S S S = [Ag2CO3] = 0.032 M so [Ag+] = 2S = 0.064 M and [CO3

2−] = S = 0.032 M Ksp = [Ag+]2[CO3

2−] = (0.064)2(0.032) = 1.31072x10–4 = 1.3x10–4 19.71 Write a reaction table, where S is the molar solubility of ZnC2O4: Concentration (M) ZnC2O4(s) Zn2+(aq) + C2O4

2−(aq) Initial — 0 0 Change — +S +S Equilibrium — S S S = [ZnC2O4] = 7.9x10–3 M so [Zn2+] = [C2O4

2−] = S = 7.9x10–3 M Ksp = [Zn2+][C2O4

2−] = (7.9x10–3)( 7.9x10–3) = 6.241x10–5 = 6.2x10–5

19-50

19.72 Plan: Write an equation that describes the solid compound dissolving in water and then write the ion-product expression. Write a reaction table, where S is the molar solubility of Ag2Cr2O7. Substitute the given solubility, S, converted from mass/volume to molarity, into the ion-expression and solve for Ksp. Solution: The solubility of Ag2Cr2O7, converted from g/100 mL to M is:

Molar solubility = S = 3

2 2 7 2 2 73

2 2 7

8.3x10 g Ag Cr O 1 mol Ag Cr O1 mL100 mL 431.8 g Ag Cr O10 L

−

−

= 0.00019221862 M

The equation for silver dichromate, Ag2Cr2O7, is: Concentration (M) Ag2Cr2O7(s) 2Ag+(aq) + Cr2O7

2−(aq) Initial — 0 0 Change — +2S +S Equilibrium — 2S S

2S = [Ag+] = 2(0.00019221862 M) = 0.00038443724 M S = [Cr2O7

2–] = 0.00019221862 M Ksp = [Ag+]2[Cr2O7

2–] = (2S)2(S) = (0.00038443724)2(0.00019221862) = 2.8408x10–11 = 2.8x10–11 19.73 The equation and ion-product expression for calcium sulfate, CaSO4, is: CaSO4(s) Ca2+(aq) + SO4

2−(aq) Ksp = [Ca2+][SO42−]

The solubility of CaSO4, converted from g/100 mL to M is:

Molar solubility = S = 4 43

4

0.209 g CaSO 1 mol CaSO1 mL100 mL 136.14 g CaSO10 L−

= 0.015351844 M

Since one mole of CaSO4 dissociates to form one mole of Ca2+, the concentration of Ca2+ is S = 0.015350716 M. The concentration of SO4

2− is S = 0.015350716 M because one mole of CaSO4 dissociates to form one mole of SO4

2−. Ksp = [Ca2+][SO4

2−] = (S)(S) = (0.015351844)( 0.015351844) = 2.35679x10–4 = 2.36x10–4 19.74 Plan: Write the equation that describes the solid compound dissolving in water and then write the ion-product expression. Set up a reaction table that expresses [Sr2+] and [CO3

2−] in terms of S, substitute into the ion-product expression, and solve for S. In part b), the [Sr2+] that comes from the dissolved Sr(NO3)2 must be included in the reaction table. Solution: a) The equation and ion-product expression for SrCO3 is: SrCO3(s) Sr2+(aq) + CO3

2−(aq) Ksp = [Sr2+][CO32−]

The solubility, S, in pure water equals [Sr2+] and [CO32–]

Write a reaction table, where S is the molar solubility of SrCO3: Concentration (M) SrCO3(s) Sr2+(aq) + CO3

2−(aq) Initial — 0 0 Change — +S +S Equilibrium — S S Ksp = 5.4x10–10 = [Sr2+][CO3

2−] = [S][S] = S2 S = 2.32379x10–5 = 2.3x10–5 M b) In 0.13 M Sr(NO3)2, the initial concentration of Sr2+ is 0.13 M. Equilibrium [Sr2+] = 0.13 + S and equilibrium [CO3

2−] = S where S is the solubility of SrCO3. Concentration (M) SrCO3(s) Sr2+(aq) + CO3

2−(aq) Initial — 0.13 0 Change — +S +S Equilibrium — 0.13 + S S Ksp = 5.4x10–10 = [Sr2+][CO3

2−] = (0.13 + S)S This calculation may be simplified by assuming S is small and setting 0.13 + S = 0.13.

Ksp = 5.4x10–10 = (0.13)S S = 4.1538x10–9 = 4.2x10–9 M

19-51

19.75 The equation and ion-product expression for BaCrO4 is: BaCrO4(s) Ba2+(aq) + CrO4

2–(aq) Ksp =[Ba2+][CrO42–]

a) The solubility, S, in pure water equals [Ba2+] and [CrO42–]

Ksp = 2.1x10–10 = [Ba2+][CrO42–] = S2

S = 1.4491x10–5 = 1.4x10–5 M b) In 1.5x10–3 M Na2CrO4, the initial concentration of CrO4

2– is 1.5x10–3 M. Equilibrium [Ba2+] = S and equilibrium [CrO4

2–] = 1.5x10–3 + S where S is the solubility of BaCrO4. Ksp = 2.1x10–10 = [Ba2+][CrO4

2–] = S(1.5x10–3 + S) Assume S is small so 1.5x10–3 + S = 1.5x10–3 Ksp = 2.1x10–10 = S(1.5x10–3) S = 1.4x10–7 M 19.76 Plan: Write the equation that describes the solid compound dissolving in water and then write the ion-product expression. Set up a reaction table that expresses [Ca2+] and [IO3

−] in terms of S, substitute into the ion-product expression, and solve for S. The [Ca2+] that comes from the dissolved Ca(NO3)2 and the [IO3

−] that comes from NaIO3 must be included in the reaction table. Solution: a) The equilibrium is: Ca(IO3)2(s) Ca2+(aq) + 2IO3

–(aq). From the Appendix, Ksp(Ca(IO3)2) = 7.1x10–7. Write a reaction table that reflects an initial concentration of Ca2+ = 0.060 M. In this case, Ca2+ is the common ion. Concentration (M) Ca(IO3)2(s) Ca2+(aq) + 2IO3

−(aq) Initial — 0.060 0 Change — +S +2S Equilibrium — 0.060 + S 2S

Assume that 0.060 + S ≈ 0.060 because the amount of compound that dissolves will be negligible in comparison to 0.060 M.

Ksp = [Ca2+][IO3−]2 = (0.060)(2S)2 = 7.1x10–7

S = 1.71998x10–3 = 1.7x10–3 M Check assumption: (1.71998x10–3 M)/(0.060 M) x 100% = 2.9% < 5%, so the assumption is good. S represents both the molar solubility of Ca2+ and Ca(IO3)2, so the molar solubility of Ca(IO3)2 is 1.7x10–3 M. b) Write a reaction table that reflects an initial concentration of IO3

− = 0.060 M. IO3− is the common ion.

Concentration (M) Ca(IO3)2(s) Ca2+(aq) + 2IO3−(aq)

Initial — 0 0.060 Change — +S +2S Equilibrium — S 0.060 + 2S The equilibrium concentration of Ca2+ is S, and the IO3

− concentration is 0.060 + 2S. Assume that 0.060 + 2S ≈ 0.060

Ksp = [Ca2+][IO3−]2 = (S)(0.060)2 = 7.1x10–7

S = 1.97222x10–4 = 2.0x10–4 M Check assumption: (1.97222x10–4 M)/(0.060 M) x 100% = 0.3% < 5%, so the assumption is good. S represents both the molar solubility of Ca2+ and Ca(IO3)2, so the molar solubility of Ca(IO3)2 is 2.0x10–4 M. 19.77 The equilibrium is: Ag2SO4(s) 2Ag+(aq) + SO4

2–(aq). From the Appendix, Ksp(Ag2SO4) = 1.5x10–5. a) Write a reaction table that reflects an initial concentration of Ag+ = 0.22 M. In this case, Ag+ is the common ion. Concentration (M) Ag2SO4(s) 2Ag+(aq) + SO4

2–(aq) Initial — 0.22 0 Change — +2S +S Equilibrium — 0.22 + 2S S Assume that 0.22 + 2S ≈ 0.22 because the amount of compound that dissolves will be negligible in comparison to 0.22 M. Ksp = [Ag+]2[SO4

2–] = (0.22)2(S) = 1.5x10–5 S = 3.099174x10–4 = 3.1x10–4 Check assumption: (3.099174x10–4 M)/(0.22 M) x 100% = 1.4% < 5%, so the assumption is good. S represents the molar solubility of Ag2SO4(s): 3.1x10–4 M.

19-52

b) Write a reaction table that reflects an initial concentration of SO42– = 0.22 M. In this case, SO4

2– is the common ion. Concentration (M) Ag2SO4(s) 2Ag+(aq) + SO4

2–(aq) Initial — 0 0.22 Change — +2S +S Equilibrium — 2S 0.22 + S The equilibrium concentration of Ag+ is 2S, and the SO4

2– concentration is 0.22 + S. Assume that 0.22 + S ≈ 0.22.

Ksp = [Ag+]2[SO42–] = (2S)2(0.22) = 1.5x10–5

S = 4.1286x10–3 = 4.1x10–3 Check assumption: (4.1286x10–3 M)/(0.22 M) x 100% = 1.9% < 5%, so the assumption is good. S represents the molar solubility of Ag2SO4, so the molar solubility of Ag2SO4 is 4.1x10–3 M. 19.78 Plan: The larger the Ksp, the larger the molar solubility if the number of ions are equal. Solution: a) Mg(OH)2 with Ksp = 6.3x10–10 has higher molar solubility than Ni(OH)2 with Ksp = 6x10–16. b) PbS with Ksp = 3x10–25 has higher molar solubility than CuS with Ksp = 8x10–34. c) Ag2SO4

with Ksp = 1.5x10–5 has higher molar solubility than MgF2 with Ksp = 7.4x10–9. 19.79 The larger the Ksp, the larger the molar solubility if the number of ions are equal. a) SrSO4 with Ksp = 3.2x10–7 has higher molar solubility than BaCrO4 with Ksp = 2.1x10–10. b) CaCO3 with Ksp = 3.3x10–9 has higher molar solubility than CuCO3 with Ksp = 3x10–12. c) Ba(IO3)2

with Ksp = 1.5x10–9 has higher molar solubility than Ag2CrO4 with Ksp = 2.6x10–12. 19.80 Plan: The larger the Ksp, the more water soluble the compound if the number of ions are equal. Solution: a) CaSO4 with Ksp = 2.4x10–5 is more water soluble than BaSO4 with Ksp = 1.1x10–10. b) Mg3(PO4)2 with Ksp = 5.2x10–24 is more water soluble than Ca3(PO4)2 with Ksp = 1.2x10–29. c) PbSO4 with Ksp = 1.6x10–8 is more water soluble than AgCl with Ksp = 1.8x10–10. 19.81 The larger the Ksp, the more water soluble the compound if the number of ions are equal. a) Ca(IO3)2 with Ksp = 7.1x10–7 is more water soluble than Mn(OH)2 with Ksp = 1.6x10–13. b) SrCO3 with Ksp = 5.4x10–10 is more water soluble than CdS with Ksp = 1.0x10–24. c) CuI with Ksp = 1x10–12 is more water soluble than AgCN with Ksp = 2.2x10–16. 19.82 Plan: If a compound contains an anion that is the weak conjugate base of a weak acid, the concentration of that anion, and thus the solubility of the compound, is influenced by pH. Solution: a) AgCl(s) Ag+(aq) + Cl−(aq) The chloride ion is the anion of a strong acid, so it does not react with H3O+. The solubility is not affected by pH. b) SrCO3(s) Sr2+(aq) + CO3

2−(aq) The strontium ion is the cation of a strong base, so pH will not affect its solubility. The carbonate ion is the conjugate base of a weak acid and will act as a base: CO3

2−(aq) + H2O(l) HCO3−(aq) + OH−(aq)

and HCO3−(aq) + H2O(l) H2CO3(aq) + OH−(aq)

The H2CO3 will decompose to CO2(g) and H2O(l). The gas will escape and further shift the equilibrium. Changes in pH will change the [CO3

2−], so the solubility of SrCO3 is affected. Solubility increases with addition of H3O+ (decreasing pH). A decrease in pH will decrease [OH−], causing the base equilibrium to shift to the right which decreases [CO3

2−], causing the solubility equilibrium to shift to the right, dissolving more solid. 19.83 a) CuBr(s) Cu+(aq) + Br−(aq) The bromide ion is the anion of a strong acid, so it does not react with H3O+. At high pH the copper ion may precipitate. Cu+(aq) + OH–(aq) CuOH(s)

19-53

b) Ca3(PO4)2(s) 3Ca2+(aq) + 2PO43–(aq)

The calcium ion is the cation of a strong base so pH will not affect its solubility. PO4

3– is the anion of a weak acid, so the following equilibria would be present. PO4

3–(aq) + nH2O(l) HnPO4(3 – n)–(aq) + nOH–(aq) (n = 1,2,3)

Since these involve OH–, the solubility will change with changing pH. Solubility increases with addition of H3O+ (decreasing pH). A decrease in pH will decrease [OH−], causing the base equilibrium to shift to the right which decreases [PO4

3−], causing the solubility equilibrium to shift to the right, dissolving more solid. 19.84 Plan: If a compound contains an anion that is the weak conjugate base of a weak acid, the concentration of that anion, and thus the solubility of the compound, is influenced by pH. Solution: a) Fe(OH)2(s) Fe2+(aq) + 2OH−(aq) The hydroxide ion reacts with added H3O+: OH−(aq) + H3O+(aq) → 2H2O(l)

The added H3O+ consumes the OH−, driving the equilibrium toward the right to dissolve more Fe(OH)2. Solubility increases with addition of H3O+ (decreasing pH).

b) CuS(s) + H2O(l) Cu2+(aq) + HS−(aq) + OH−(aq) Both HS− and OH− are anions of weak acids, so both ions react with added H3O+. Solubility increases with addition of H3O+ (decreasing pH). 19.85 a) PbI2(s) Pb2+(aq) + 2I−(aq) The iodide ion is the anion of a strong acid, so it does not react with H3O+. Thus, the solubility does not increase in acid solution. At high pH the lead ion may precipitate. b) Hg2(CN)2(s) Hg2

2+(aq) + 2CN−(aq) At high pH the mercury(I) ion may precipitate. CN− is the anion of a weak acid, so the equilibrium would be CN−(aq) + H2O(l) HCN(aq) + OH−(aq) Since this involves OH−, it would shift with changing pH. Solubility increases with addition of H3O+

(decreasing pH). 19.86 Plan: Find the initial molar concentrations of Cu2+ and OH–. The molarity of the KOH is calculated by converting

mass to moles and dividing by the volume. Put these concentrations in the ion-product expression, solve for Qsp, and compare Qsp with Ksp. If Qsp > Ksp, precipitate forms.

Solution: The equilibrium is: Cu(OH)2(s) Cu2+(aq) + 2OH–(aq). The ion-product expression is Ksp = [Cu2+][OH−]2 and,

from the Appendix, Ksp equals 2.2x10–20.

[Cu2+] = 3 2

3 2

3 2

1.0x10 mol Cu(NO ) 1 mol CuL 1 mol Cu(NO )

− +

= 1.0x10–3 M Cu2+

[OH–] = 0.075 g KOH 1 mol KOH 1 mol OH1.0 L 56.11 g KOH 1 mol KOH

−

= 1.33666x10–3 M OH–

Qsp = [Cu2+][OH−]2 = (1.0x10–3)(1.33666x10–3)2 = 1.786660x10–9 Qsp is greater than Ksp (1.8x10–9 > 2.2x10–20), so Cu(OH)2 will precipitate. 19.87 The ion-product expression for PbCl2 is Ksp = [Pb2+][Cl−]2 and, from the Appendix, Ksp equals 1.7x10–5. To decide if a precipitate will form, calculate Qsp with the given quantities and compare it to Ksp.

[Pb2+] = 2

3 2

3 2

0.12 mol Pb(NO ) 1 mol PbL 1 mol Pb(NO )

+

= 0.12 M Pb2+

[Cl–] = 33.5 mg NaCl 10 g 1 mol NaCl 1 mol Cl

0.250 L 1 mg 58.45 g NaCl 1 mol NaCl

− −

= 2.3952x10–4 M Cl–

Qsp = [Pb2+][Cl−]2 = (0.12)(2.3952x10–4)2 = 6.8843796x10–9 Qsp is smaller than Ksp (6.9x10–9 < 1.7x10–5), so PbCl2 will not precipitate.

19-54

19.88 Plan: Find the initial molar concentrations of Ba2+ and IO3

–. The molarity of the BaCl2 is calculated by converting mass to moles and dividing by the volume. Put these concentrations in the ion-product expression, solve for Qsp, and compare Qsp with Ksp. If Qsp > Ksp, precipitate forms.

Solution: The equilibrium is: Ba(IO3)2(s) Ba2+(aq) + 2IO3

–(aq). The ion-product expression is Ksp = [Ba2+][IO3−]2 and,

from the Appendix, Ksp equals 1.5x10–9.

[Ba2+] = 3 2

2 23

2 2

7.5 mg BaCl 1 mol BaCl10 g 1 mL 1 mol Ba500. mL 1 mg 208.2 g BaCl 1 mol BaCl10 L

− +

−

= 7.204611x10–5 M Ba2+

[IO3–] = 3 3

3

0.023 mol NaIO 1 mol IOL 1 mol NaIO

−

= 0.023 M IO3–

Qsp = [Ba2+][IO3−]2 = (7.204611x10–5)(0.023)2 = 3.81124x10–8

Since Qsp > Ksp (3.8x10–8 > 1.5x10–9), Ba(IO3)2 will precipitate. 19.89 The ion-product expression for Ag2CrO4 is Ksp = [Ag+]2[CrO4

2–] and, from the Appendix, Ksp equals 2.6x10–12. To decide if a precipitate will form, calculate Qsp with the given quantities and compare it to Ksp.

[Ag+] =5

3 33

3 3

2.7x10 g AgNO 1 mol AgNO1 mL 1 mol Ag15.0 mL 169.9 g AgNO 1 mol AgNO10 L

− +

−

= 1.0594467x10–5 M Ag+

[CrO42–] =

4 22 4 4

2 4

4.0x10 mol K CrO 1 mol CrOL 1 mol K CrO

− −

= 4.0x10–4 M IO3–

Qsp = [Ag+]2[ CrO42–]= (1.0594467x10–5)2(4.0x10–4) = 4.4897x10–14

Since Qsp < Ksp (4.5x10–14 < 2.6x10–12), Ag2CrO4 will not precipitate.

19.90 Original moles of Ca2+ = ( )5 2+ 2+

2+9.7x10 g Ca 1 mol Ca104 mL

mL 40.08 g Ca

−

= 2.5170x10–4 mol Ca2+

Moles of C2O42– added = ( )

232 2 4 2 4

2 2 4

0.1550 mol Na C O 1molC O10 L 100.0 mLL 1 mL 1mol Na C O

−−

= 0.01550 mol

C2O42– The Ca2+ is limiting leaving 0 M, and after the reaction there will be (0.01550 – 0.00025170) = 0.0152483 mol of

C2O42– left in a total volume of 104 + 100.0 mL = 204 mL.

[C2O42–] = 2

30.0152483 mol C O 1 mL

204 mL 10 L−

= 0.0747466 M C2O42–

Concentration (M) CaC2O4•H2O(s) Ca2+(aq) + C2O42–(aq) + H2O(l)

Initial — 0 0.0747466 — Change — +S +S — Equilibrium — S 0.0747466 + S — Assume that 0.0747466 + S ≈ 0.0747466 because the amount of compound that dissolves will be negligible in comparison to 0.0747466 M. The Ksp from the Appendix is: 2.3x10–9 Ksp = [Ca2+][C2O4

2–] = (S)(0.0747466) = 2.3x10–9 S = 3.07706x10–8 = 3.1x10–8 Check assumption: (3.07706x10–8 M)/(0.0747466 M) x 100% = 0.00004% < 5%, so the assumption is good. S represents both the molar solubility of Ca2+ and CaC2O4•H2O(s), so the concentration of Ca2+ is 3.1x10–8 M. 19.91 Plan: When Fe(NO3)3 and Cd(NO3)2 mix with NaOH, the insoluble compounds Fe(OH)3 and Cd(OH)2 form. The compound with the smaller value of Ksp precipitates first. Calculate the initial concentrations of Fe3+ and Cd2+ from the dilution formula MconcVconc = MdilVdil. Use the ion-product expressions to find the minimum OH– concentration required to cause precipitation of each compound. Solution: a) Fe(OH)3 will precipitate first because its Ksp (1.6x10–39) is smaller than the Ksp for Cd(OH)2 at 7.2x10–15.

19-55

The precipitation reactions are: Fe3+(aq) + 3OH–(aq) → Fe(OH)3(s) Ksp = [Fe3+][ OH–]3 Cd2+(aq) + 2OH–(aq) → Cd(OH)2(s) Ksp = [Cd2+][ OH–]2 The concentrations of Fe3+ and Cd2+ in the mixed solution are found from MconcVconc = MdilVdil [Fe3+] = [(0.50 M)(50.0 mL)]/[(50.0 + 125) mL] = 0.142857 M Fe3+ [Cd2+] = [(0.25 M)(125 mL)]/[(50.0 + 125) mL] = 0.178571 M Cd2+ The hydroxide ion concentration required to precipitate the metal ions comes from the metal ion concentrations and the Ksp.

[OH–]Fe = sp3 3Fe

K+

= [ ]

393

1.6x100.142857

−

= 2.237x10–13 = 2.2x10–13 M

[OH–]Cd = sp2Cd

K+

= [ ]

157.2x100.178571

−

= 2.0079864x10–7 = 2.0x10–7 M

A lower hydroxide ion concentration is required to precipitate the Fe3+. b) The two ions are separated by adding just enough NaOH to precipitate the iron(III) hydroxide, but precipitating no more than 0.01% of the cadmium. The Fe3+ is found in the solid precipitate while the Cd2+ remains in the solution.

c) A hydroxide concentration between the values calculated in part a) will work. The best separation would be when Qsp = Ksp for Cd(OH)2. This occurs when [OH−] = 2.0x10–7 M. 19.92 The metal ion can act as a Lewis acid and bond to one or more negatively charged ligands. If the total negative charge of the ligands exceeds the positive charge on the metal ion, the complex will be negative.

19.93 Cd(H2O)42+(aq) + I–(aq) CdI(H2O)3

+(aq) + H2O(l) Kf1 = ( )

( )2 3

22 4

CdI H O

Cd H O I

+

+ −

CdI(H2O)3+(aq) + I–(aq) CdI2(H2O)2(aq) + H2O(l) Kf2 =

( )( )

2 2 2

2 3

CdI H O

CdI H O I+ −

CdI2(H2O)2(aq) + I–(aq) CdI3(H2O)–(aq) + H2O(l) Kf3 = ( )

( )3 2

2 2 2

CdI H O

CdI H O I

−

−

CdI3(H2O)–(aq) + I–(aq) CdI42–(aq) + H2O(l) Kf4 =

( )

24

3 2

CdI

CdI H O I

−

− −

Overall: Cd(H2O)2+4(aq) + 4I–(aq) CdI4

2–(aq) + 4H2O(l) Kf = ( )

24

422 4

CdI

Cd H O I

−

+ −

Kf = ( )

( )2 3

22 4

CdI H O

Cd H O I

+

+ −

x ( )

( )2 2 2

2 3

CdI H O

CdI H O I+ −

x ( )

( )3 2

2 2 2

CdI H O

CdI H O I

−

−

x

( )

24

3 2

CdI

CdI H O I

−

− −

= Kf1 x Kf2 x Kf3 x Kf4 19.94 In the context of this equilibrium only, the increased solubility with added OH− appears to be a violation of Le

Châtelier’s principle. Adding OH− should cause the equilibrium to shift towards the left, decreasing the solubility of PbS. Before accepting this conclusion, other possible equilibria must be considered. Lead is a metal ion and hydroxide ion is a ligand, so it is possible that a complex ion forms between the lead ion and hydroxide ion:

19-56

Pb2+(aq) + nOH−(aq) Pb(OH)n2 − n(aq)

This decreases the concentration of Pb2+, shifting the solubility equilibrium to the right to dissolve more PbS. 19.95 Plan: In many cases, a hydrated metal complex (e.g., Hg(H2O)4

2+) will exchange ligands when placed in a solution of another ligand (e.g., CN−). Solution: Hg(H2O)4

2+(aq) + 4CN−(aq) Hg(CN)42−(aq) + 4H2O(l)

Note that both sides of the equation have the same “overall” charge of −2. The mercury complex changes from +2 to −2 because water is a neutral molecular ligand, whereas cyanide is an ionic ligand. 19.96 Zn(H2O)4

2+(aq) + 4CN− (aq) Zn(CN)42–(aq) + 4H2O(l)

19.97 Plan: In many cases, a hydrated metal complex (e.g., Ag(H2O)4

+) will exchange ligands when placed in a solution of another ligand (e.g., S2O3

2−). Solution: The two water ligands are replaced by two thiosulfate ion ligands. The +1 charge from the silver ion plus the –4 charge from the two thiosulfate ions gives a net charge on the complex ion of −3. Ag(H2O)2

+(aq) + 2S2O32−(aq) Ag(S2O3)2

3−(aq) + 2H2O(l) 19.98 Al(H2O)6

3+(aq) + 6F–(aq) AlF63–(aq) + 6H2O(l)

19.99 Plan: Write the formation reaction and the Kf expression. The initial concentrations of Ag+ and S2O3

2− may be determined from MconcVconc = MdilVdil. Set up a reaction table and use the limiting reactant to find the amounts of species in the mixture, assuming a complete reaction. A second reaction table is then written, with x representing the amount of complex ion that dissociates. Use the Kf expression to solve for x. Solution: Ag+(aq) + 2S2O3

2−(aq) Ag(S2O3)23−(aq)

[Ag+] = (0.044 M)(25.0 mL)/((25.0 + 25.0) mL) = 0.022 M Ag+ [S2O3

2−] = (0.57 M)(25.0 mL)/((25.0 + 25.0) mL) = 0.285 M S2O32−

The reaction gives: Concentration (M) Ag+(aq) + 2S2O3

2−(aq) → Ag(S2O3)23−(aq)

Initial 0.022 0.285 0 Change −0.022 −2(0.022) +0.022 1:2:1 mole ratio Equilibrium 0 0.241 0.022 To reach equilibrium: Concentration (M) Ag+(aq) + 2S2O3

2−(aq) Ag(S2O3)23−(aq)

Initial 0 0.241 0.022 Change +x +2x −x Equilibrium + x 0.241 + 2x 0.022 − x Kf is large, so [Ag(S2O3)2

3−] ≈ 0.022 M and [S2O32−]equil ≈ 0.241 M

Kf = 4.7x1013 = ( ) 3

2 3 222

2 3

Ag S O

Ag S O

−

+ −

= [ ]

[ ][ ]20.022

x 0.241

x = [Ag+] = 8.0591778x10−15 = 8.1x10−15 M 19.100 The reaction between SCN− and Fe3+ produces the red complex FeSCN2+. One can assume from the much larger concentration of SCN− and large Kf that all of the Fe3+ ions react to form the complex. Calculate the initial concentrations of SCN− and Fe3+ and write a reaction table in which x is the concentration of FeSCN2+ formed.

[Fe3+]initial = ( )( )( )

( )( ) ( )33 3

3 3

0.0015M Fe NO 0.50 L 1 mol Fe1 mol Fe NO0.50 0.50 L

+ +

= 0.00075 M Fe3+

19-57

[SCN−]initial = ( )( )

( )( )0.20 M KSCN 0.50 L 1 mol SCN

1 mol KSCN0.50 0.50 L

− +

= 0.10 M SCN−

Set up a reaction table: Concentration (M) Fe3+(aq) + SCN−(aq) FeSCN2+ Initial 7.5x10−4 0.10 0 Change −x −x +x Equilibrium 7.5x10−4 − x 0.10 − x x It is reasonable to assume that x is much less than 0.10, so 0.10 − x ≈ 0.10. However, it is not reasonable to assume that 0.00075 − x ≈ 0.00075, because x may be significant in relation to such a small number. The equilibrium expression and the constant, from the problem, are:

Kf = 8.9x102 = 2

3

FeSCN

Fe SCN

+

+ −

= [ ]

[ ]4

x

7.5x10 x 0.10 x− − − =

[ ][ ]4

x

7.5x10 x 0.10− −

x = (7.5x10−4 − x)(0.10)(8.9x102) = (7.5x10-4 − x)(89) x = 6.675x10−2 − 89x x = 7.416667x10−4 From the reaction table, [Fe3+]eq = 7.5x10−4 – x. Therefore, [Fe3+]eq = 7.5x10−4 − 7.416667x10−4 = 8.33333x10−6 = 1x10−5 M. 19.101 Plan: Write the ion-product equilibrium reaction and the complex-ion equilibrium reaction. Add the two reactions to yield an overall reaction; multiply the two constants to obtain Koverall. Write a reaction table where S = [Cr(OH)3]dissolved = [Cr(OH)4

−]. Solution:

Solubility-product: Cr(OH)3(s) Cr3+(aq) + 3OH−(aq) Ksp = 6.3x10–31 Complex-ion Cr3+(aq) + 4OH−(aq) Cr(OH)4

−(aq) Kf = 8.0x1029

Overall: Cr(OH)3(s) + OH−(aq) Cr(OH)4

−(aq) K = KspKf = 0.504 At pH 13.0, the pOH is 1.0 and [OH−] = 10–1.0 = 0.1 M.

Reaction table: Concentration (M) Cr(OH)3(s) + OH−(aq) Cr(OH)4

−(aq) Initial —— 0.1 0 Change —— – S +S Equilibrium —— 0.1 – S S

Assume that 0.1 – S ≈ 0.1.

Koverall = 0.504 = ( )4Cr OH

OH

−

−

= [ ][ ]0.1

S

S = [Cr(OH)4−] = 0.0504 = 0.05 M

19.102 Write the ion-product equilibrium reaction and the complex-ion equilibrium reaction. Add the two reactions to yield an overall reaction; multiply the two constants to obtain Koverall. Write a reaction table where S = [AgI]dissolved = [Ag(NH3)2

+]. Solubility-product: AgI(s) Ag+(aq) + I−(aq) Ksp = 8.3x10–17 Complex-ion: Ag+(aq) + 2NH3(aq) Ag(NH3)2

+(aq) Kf = 1.7x107 Overall: AgI(s) + 2NH3(aq) Ag(NH3)2

+(aq) + I−(aq) Koverall = Ksp x Kf = (8.3x10–17)(1.7x107) = 1.411x10–9 Reaction table: Concentration (M) AgI(s) + 2NH3(aq) Ag(NH3)2

+(aq) + I−(aq) Initial —— 2.5 0 0 Change —— –2S +S +S Equilibrium —— 2.5 – 2S S S

19-58

Assume that 2.5 – 2S ≈ 2.5 because Koverall is so small.

Koverall = 1.411x10–9 = ( )3 2

23

Ag NH I

NH

+ −

=

[ ][ ][ ]22.5 2−

S S

S =

[ ][ ][ ]22.5

S S S = 9.3908x10–5 = 9.4x10–5 M

19.103 Plan: First, calculate the initial moles of Zn2+ and CN–, then set up reaction table assuming that the reaction first goes to completion, and then calculate back to find the reactant concentrations. Solution: The complex formation equilibrium is: Zn2+(aq) + 4CN−(aq) Zn(CN)4

2−(aq) Kf = 4.2x1019

Moles of Zn2+ = ( )2

22

2 2

1 mol ZnCl 1 mol Zn0.84 g ZnCl136.31 g ZnCl 1 mol ZnCl

+

= 0.0061624 mol Zn2+

Moles of CN– = ( )30.150 mol NaCN 10 L 1 mol CN245 mL

L 1 mL 1 mol NaCN

− −

= 0.03675 mol CN–

The Zn2+ is limiting because there are significantly fewer moles of this ion, thus, [Zn2+] = 0.

Moles of CN– reacting = ( )22

4 mol CN0.0061624 mol Zn1 mol Zn

−+

+

= 0.0246496 mol CN–

Moles of CN– remaining are: 0.03675 – 0.0246496 = 0.0121004 mol CN–

[CN–] = ( )

( ) 3

0.0121004 mol CN 1 mL245 mL 10 L

−

−

= 0.0493894 M CN–

The Zn2+ will produce an equal number of moles of the complex with the concentration:

[Zn(CN)42−] =

( ) 224

3 2

1 mol Zn CN0.0061624 mol Zn 1 mL245 mL 10 L 1 mol Zn

−+

− +

= 0.025153 M Zn(CN)4

2−

Concentration (M) Zn2+(aq) + 4CN−(aq) Zn(CN)42−(aq)

Initial 0 0.0493894 0.025153 Change +x +4x –x Equilibrium x 0.0493894 + 4x 0.025153 – x

Assume the –x and the +4x do not significantly change the associated concentrations.

Kf = 4.2x1019 = ( ) 2

442

Zn CN

Zn CN

−

+ −

= [ ]

[ ][ ]40.025153 x

x 0.0493894 4x

−

+ =

[ ][ ][ ]4

0.025153

x 0.0493894

x = 1.006481x10–16 = 1.0x10–16 [Zn2+] = 1.0x10–16 M Zn2+ [Zn(CN)4

2−] = 0.025153 – x = 0.025153 – 1.0x10–16 = 0.025153 = 0.025 M Zn(CN)42–

[CN−] = 0.0493894 + 4x =0.0493894 + 4(1.0x10–16) = 0.0493894 = 0.049 M CN– 19.104 The complex formation equilibrium is: Co2+(aq) + 4OH−(aq) Co(OH)4

2−(aq) Kf = 5x109 First, calculate the initial moles of Co2+ and OH-, then set up reaction table assuming that the reaction first goes to completion and then calculate back to find reactant concentrations.

Moles of Co2+ = ( )( ) ( )( ) ( )

23 2

3 23 32 2

1 mol Co NO 1 mol Co2.4 g Co NO182.95 g Co NO 1 mol Co NO

+

= 0.013118 mol Co2+

Moles of OH− = ( )0.22 mol KOH 1 mol OH0.350 LL 1 mol KOH

−

= 0.077 mol OH−

The Co2+ is limiting, thus, [Co2+] = 0, and the moles of OH− remaining are: [0.077 − 4(0.013118)]

19-59

[OH−] = ( )( )

0.077 4 0.013118 molOH

0.350 L

− − = 0.07008 M OH−

The Co2+ will produce an equal number of moles of the complex with the concentration:

[Co(OH)42−] =

( ) 224

2

1 mol Co OH0.013118 mol Co0.350 L 1 mol Co

−+

+

= 0.03748 M Co(OH)4

2−

Concentration (M) Co2+(aq) + 4OH−(aq) Co(OH)42−(aq)

Initial 0 0.07008 0.03748 Change +x +4x −x Equilibrium x 0.07008 + 4x 0.03748 − x Assume the −x and the + 4x do not significantly change the associated concentrations.

Kf = 5x109 = ( ) 2

442

Co OH

Co OH

−

+ −

= [ ]

[ ][ ]40.03748 x

x 0.07008 4x

−

+ =

[ ][ ][ ]4

0.03748

x 0.07008

x = 3.1078x10−7 = 3.1x10−7 [Co2+] = 3.1x10−7 M Co2+ [Co(OH)4

2−] = 0.03748 − x = 0.037479689 = 0.037 M Co(OH)42−

[OH−] = 0.07008 + 4 x = 0.070078756 = 0.070 M OH− 19.105 Plan: The NaOH will react with the benzoic acid, C6H5COOH, to form the conjugate base benzoate ion, C6H5COO−. Calculate the number of moles of NaOH and C6H5COOH. Set up a reaction table that shows the stoichiometry of the reaction of NaOH and C6H5COOH. All of the NaOH will be consumed to form C6H5COO−, and the number of moles of C6H5COOH will decrease. Find the new concentrations of C6H5COOH and C6H5COO− and use the Henderson-Hasselbalch equation to find the pH of this buffer. Once the pH of the benzoic acid/benzoate buffer is known, the Henderson-Hasselbalch equation can be used to find the ratio of formate ion and formic acid that will produce a buffer of that same pH. From the ratio, the volumes of HCOOH and NaOH are calculated. Solution: The Ka for benzoic acid is 6.3x10−5 (from the Appendix). The pKa is −log (6.3x10−5) = 4.201. The reaction of benzoic acid with sodium hydroxide is: C6H5COOH(aq) + NaOH(aq) → Na+(aq) + C6H5COO−(aq) + H2O(l)

Moles of C6H5COOH = ( )3

6 50.200 mol C H COOH 10 L 475 mLL 1 mL

−

= 0.0950 mol C6H5COOH

Moles of NaOH = ( )32.00 mol NaOH 10 L 25 mL

L 1 mL

−

= 0.050 mol NaOH

NaOH is the limiting reagent: The reaction table gives:

C6H5COOH(aq) + NaOH(aq) → Na+(aq) + C6H5COO−(aq) + H2O(l) Initial 0.0950 mol 0.050 mol — 0 — Reacting −0.050 mol −0.050 mol + 0.050 mol Final 0.045 mol 0 mol 0.050 mol The concentrations after the reactions are:

[C6H5COOH] = ( )

6 53

0.045 mol C H COOH 1 mL475 25 mL 10 L−

+

= 0.090 M C6H5COOH

[C6H5COO−] = ( )

6 53

0.050 mol C H COO 1 mL475 25 mL 10 L

−

−

+

= 0.10 M C6H5COO−

19-60

Calculating the pH from the Henderson-Hasselbalch equation:

6 5a

6 5

[C H CO ]pH = p + log

[C H COOH]K

−

[ ]0.10= 4.201 + log

[0.090]

= 4.24676 = 4.2

Calculations on formic acid (HCOOH) also use the Henderson-Hasselbalch equation. The Ka for formic acid is 1.8x10−4 and the pKa = −log (1.8x10−4) = 3.7447.

The formate to formic acid ratio may now be determined:

a[HCOO ]pH = p + log[HCOOH]

K−

[HCOO ]4.24676 = 3.7447 + log[HCOOH]

−

0.50206 = [ ][HCOO ]logHCOOH

−

[ ][HCOO ]HCOOH

−

= 3.177313

[HCOO−] = 3.177313 [HCOOH] Since the conjugate acid and the conjugate base are in the same volume, the mole ratio and the molarity ratios are identical. Moles HCOO− = 3.177313 mol HCOOH The total volume of the solution is (500. mL)(10−3 L/1 mL) = 0.500 L Let Va = volume of acid solution added, and Vb = volume of base added. Thus: Va + Vb = 0.500 L

The reaction between the formic acid and the sodium hydroxide is: HCOOH(aq) + NaOH(aq) → HCOONa(aq) + H2O(l) The moles of NaOH added equal the moles of HCOOH reacted and the moles of HCOONa formed. Moles NaOH = (2.00 mol NaOH/L)(Vb) = 2.00Vb mol Total moles HCOOH = (0.200 mol HCOOH/L)(Va) = 0.200Va mol The stoichiometric ratios in this reaction are all 1:1. Moles HCOOH remaining after the reaction = (0.200Va − 2.00Vb) mol Moles HCOO− = moles HCOONa = moles NaOH = 2.00 Vb Using these moles and the mole ratio determined for the buffer gives: Moles HCOO− = 3.177313 mol HCOOH 2.00Vb mol = 3.177313(0.200Va − 2.00Vb) mol 2.00Vb = 0.6354626Va − 6.354626Vb 8.354626 Vb = 0.6354626 Va The volume relationship given above gives Va = (0.500 − Vb) L. 8.354626 Vb = 0.6354626 (0.500 − Vb) 8.354626 Vb = 0.3177313 − 0.6354626 Vb 8.9900886 Vb = 0.3177313 Vb = 0.0353424 = 0.035 L NaOH Va = 0.500 − 0.0353424 = 0.4646576 = 0.465 L HCOOH Limitations due to the significant figures lead to a solution with only an approximately correct pH. 19.106 pKa = – log Ka = – log 6.3x10–8 = 7.200659. The Ka comes from Appendix C; it is Ka2

for phosphoric acid.

2

4a

2 4

[HPO ]pH = p + log

[H PO ]K

−

−

2

4

2 4

[HPO ]7.00 = 7.200659 + log

[H PO ]

−

−

19-61

–0.200659 = 2

4

2 4

[HPO ] log

[H PO ]

−

−

24

2 4

[HPO ] [H PO ]

−

− = 0.63000

Since they are equimolar, 24

2 4

HPO

H PO

−

−

V

V = 0.63000

and VHPO42– + VH2PO4

– = 100. mL

so (0.63000)VH2PO4– + VH2PO4

– = 100. mL

VH2PO4– = 61 mL and VHPO4

2– = 39 mL 19.107 Plan: A formate buffer contains formate (HCOO−) as the base and formic acid (HCOOH) as the acid. The Henderson-Hasselbalch equation gives the component ratio, [HCOO−]/[HCOOH]. The ratio is used to find the volumes of acid and base required to prepare the buffer. Solution: From the Appendix, the Ka for formic acid is 1.8x10–4 and the pKa = −log (1.8x10–4) = 3.7447.

a) a[HCOO ]pH = p + log[HCOOH]

K−

[HCOO ]3.74 = 3.7447 + log[HCOOH]

−

−0.0047 = [ ][HCOO ]logHCOOH

−

[ ][HCOO ]HCOOH

−

= 0.989236 = 0.99

b) To prepare solutions, set up equations for concentrations of formate and formic acid with x equal to the volume, in L, of 1.0 M HCOOH added. The equations are based on the neutralization reaction between HCOOH and NaOH that produces HCOO−.

HCOOH(aq) + NaOH(aq) → HCOO−(aq) + Na+(aq) + H2O(l)

[HCOO–] = ( ) ( )0.700 x L NaOH 1 mol HCOO1.0M NaOH0.700 L solution 1 mol NaOH

− −

[HCOOH] = ( ) x L HCOOH1.0 M HCOOH0.700 L solution

– ( ) ( )0.700 x L NaOH 1 mol HCOO1.0 M NaOH0.700 L solution 1 mol NaOH

− −

The component ratio equals 0.99 (from part a)). Simplify the above equations and plug into ratio:

[ ]HCOO

HCOOH

− =

( )( )

0.700 x M HCOO0.700

x 0.700 x M HCOOH0.700

− −

− −

= 0.700 x

2 x 0.700−

− = 0.989236

Solving for x: x = 0.46751 = 0.468 L Mixing 0.468 L of 1.0 M HCOOH and 0.700 − 0.468 = 0.232 L of 1.0 M NaOH gives a buffer of pH 3.74. c) The final concentration of HCOOH from the equation in part b):

[HCOOH] = ( ) 0.468 L HCOOH1.0 M HCOOH0.700 L solution

– ( ) 0.232 L NaOH 1 mole HCOO1.0 M NaOH0.700 L solution 1 mole NaOH

−

= 0.33714 = 0.34 M HCOOH

19-62

19.108 This is because Ka depends on temperature (like all other equilibrium constants). In this case, since the pH drops as the temperature increases, Ka must increase with temperature, indicating that the dissociation reaction is endothermic. 19.109 H2SO4 is a strong acid and will be completely ionized: H2SO4(aq) + 2H2O(l) → SO4

2−(aq) + 2H3O+(aq). Calculate the moles of H3O+(aq) from the H2SO4 in the 8.0x103 lb of water and then the amount of sodium acetate trihydrate (NaC2H3O2•3H2O) that will be required to neutralize that amount of H3O+(aq).

Mass (g) of water = ( )3 1 kg 1000 g8.0x10 lb water2.205 lb 1 kg

= 3.628118x106 g H2O

6mass of soluteppm = x10mass of solution

62 46

mass of H SO10 ppm = x103.628118x10 g

Mass (g) of H2SO4 = 36.28118 g

Moles of H3O+ = ( ) 32 42 4

2 4 2 4

2 mol H O1 mol H SO36.28118 g H SO

98.08 g H SO 1 mol H SO

+

= 0.7398 mol H3O+

The reaction between H3O+ and the base sodium acetate is: H3O+(aq) + NaC2H3O2(aq) → H2O(l) + HC2H3O2(aq) + Na+(aq)

Mass (lb) of NaC2H3O2•3H2O required to neutralize the H2SO4 =

( ) 2 3 2 2 2 3 2 23

2 3 2 2 2 3 2 23

1 mol NaC H O •3H O 136.08 g NaC H O •3H O 1 kg 2.205 lb0.7398 mol H O1 mol NaC H O •3H O 1000 g NaC H O •3H O 1 kg1 mol H O

++

= 0.22198 lb NaC2H3O2•3H2O Now consider the acetic acid. Calculate the amount of acetic acid in 8.0x103 lb or 3.628118x106 g H2O.

Moles of acetic acid = ( )6 2 3 22

2 3 2

1 mol HC H O0.015%3.628118x10 g H O100% 60.05 g HC H O

= 9.0627 mol

Find the amount of C2H3O2– necessary to maintain a pH of 5.

2 3 2a

2 3 2


[HC H O ]K

−

2 3 2[C H O ]5.0 = 4.7447 + log

[9.0627 mol]

−

2 3 2[C H O ]0.2553 = log

[9.0627 mol]

−

2 3 2[C H O ]1.800 =

[9.0627 mol]

−

16.31286 mol of C2H3O2− (NaC2H3O2•3H2O) will be required to maintain the pH.

Mass (lb) of NaC2H3O2•3H2O required =

( ) 2 3 2 23 2

2 3 2 2

136.08 g NaC H O •3H O 1 kg 2.205 lb16.31286 mol CH COONa•3H O1 mol NaC H O •3H O 1000 g 1 kg

= 4.89478 lb

Total amount of NaC2H3O2•3H2O required = 0.22198 lb + 4.89478 lb = 5.11676 = 5.1 lb 19.110 Plan: The minimum urate ion concentration necessary to cause a deposit of sodium urate is determined by the Ksp

for the salt. Convert solubility in g/100. mL to molar solubility and calculate Ksp. Substituting [Na+] and Ksp into the ion-product expression allows one to find [Ur−].

Solution: Molar solubility of NaUr:

19-63

[NaUr] = 30.085 g NaUr 1 mL 1 mol NaUr

100. mL 190.10 mol NaUr10 L−

= 4.4713309x10–3 M NaUr

4.4713309x10–3 M NaUr = [Na+] = [Ur−] Ksp = [Na+][Ur−] = (4.4713309x10–3) (4.4713309x10–3) = 1.99927998x10–5 M When [Na+] = 0.15 M: Ksp = 1.99927998x10–5 M = [0.15][Ur−]

[Ur−] = 1.33285x10–4 The minimum urate ion concentration that will cause precipitation of sodium urate is 1.3x10–4 M. 19.111 a) K = [CO2(aq)]/[CO2(g)] = 3.1x10–2 [CO2(aq)] = K[CO2(g)] = (3.1x10–2)(4x10–4) = 1.24x10–5 M = 1x10–5 M CO2 b) K = [Ca2+][HCO3

–]2/[CO2(aq)] = (x)(2x)2/(1.24x10–5 – x) = 1x10–12 Neglect –x x = 1.458x10–6 = 1x10–6 M Ca2+ c) K = [CO2(aq)]/[CO2(g)] = 3.1x10–2 [CO2(aq)] = K[CO2(aq)] = (3.1x10–2)(2x4x10–4) = 2.48x10–5 M K = [Ca2+][HCO3

–]2/[CO2(aq)] = (x)(2x)2/(2.48x10–5 – x) = 1x10–12 Neglect –x x = 1.837x10–6 = 2x10–6 M Ca2+ 19.112 The buffer is made by starting with phosphoric acid and neutralizing some of the acid by adding sodium hydroxide: H3PO4(aq) + OH−(aq) → H2PO4

−(aq) + H2O(l) Present initially is (0.50 L)(1.0 M H3PO4) = 0.50 mol H3PO4. Adding 0.80 mol NaOH converts all the phosphoric acid to dihydrogen phosphate ions (0.50 mol) and 0.30 mol NaOH are left. The remaining OH− will react with the dihydrogen phosphate: H2PO4

−(aq) + OH−(aq) → HPO42−(aq) + H2O(l)

The 0.50 mol H2PO4− reacts with the 0.30 mol OH− to produce 0.30 mol HPO4

2−. 0.20 mol H2PO4− will remain.

The pH is determined from the equilibrium involving the conjugate pair HPO42−/H2PO4

−. H2PO4

−(aq) + H2O(l) HPO42−(aq) + H3O+(aq) Ka = 6.3x10−8

24

a2 4

[HPO ]pH = p + log

[H PO ]K

−

−

= −log (6.3x10−8) + log

24

2 4

0.30 mol HPO0.50 L

0.20 mol H PO0.50 L

−

−

= 7.37675 = 7.38

19.113 Plan: Substitute the given molar solubility of KCl into the ion-product expression to find the Ksp of KCl.

Determine the total concentration of chloride ion in each beaker after the HCl has been added. This requires the moles originally present and the moles added. Determine a Qsp value to see if Ksp is exceeded. If Qsp < Ksp, nothing will precipitate.

Solution: a) The solubility equilibrium for KCl is: KCl(s) K+(aq) + Cl−(aq)

The solubility of KCl is 3.7 M. Ksp = [K+][Cl−] = (3.7)(3.7) = 13.69 = 14 b) Find the moles of Cl−: Original moles from the KCl:

Moles of K+ = moles of Cl− = ( )33.7 mol KCl 10 L 1 mol Cl ion100. mL

1 L 1 mL 1 mol KCl

− −

= 0.37 mol Cl−

Original moles from the 6.0 M HCl in the first beaker:

19-64

Moles of Cl− = ( )36.0 mol HCl 10 L 1 mol Cl100. mL

1 L 1 mL 1 mol HCl

− −

= 0.60 mol Cl−

This results in (0.37 + 0.60) mol = 0.97 mol Cl−. Original moles from the 12 M HCl in the second beaker:

Moles of Cl− = ( )312 mol HCl 10 L 1 mol Cl100. mL

1 L 1 mL 1 mol HCl

− −

= 1.2 mol Cl−

This results in (0.37 + 1.2) mol = 1.57 mol Cl− Volume of mixed solutions = (100. mL + 100. mL)(10–3 L/1 mL) = 0.200 L After the mixing: [K+] = (0.37 mol K+)/(0.200 L) = 1.85 M K+ From 6.0 M HCl in the first beaker: [Cl−] = (0.97 mol Cl−)/(0.200 L) = 4.85 M Cl− From 12 M HCl in the second beaker: [Cl−] = (1.57 mol Cl−)/(0.200 L) = 7.85 M Cl− Determine a Qsp value to see if Ksp is exceeded. If Qsp < Ksp, nothing will precipitate. From 6.0 M HCl in the first beaker: Qsp = [K+][Cl−] = (1.85)(4.85) = 8.9725 = 9.0 < 14, so no KCl will precipitate. From 12 M HCl in the second beaker: Qsp = [K+][Cl−] = (1.85)(7.85) = 14.5225 = 15 > 14, so KCl will precipitate. The mass of KCl that will precipitate when 12 M HCl is added: Equal amounts of K and Cl will precipitate. Let x be the molarity change. Ksp = [K+][Cl−] = (1.85 – x)(7.85 – x) = 13.69 x = 0.08659785 = 0.09 This is the change in the molarity of each of the ions.

Mass (g) of KCl = ( )0.08659785 mol K 1 mol KCl 74.55 g KCl0.200 LL 1 mol KCl1 mol K

+

+

= 1.291174 = 1 g KCl

19.114 [NH3] + [NH4

+] = 0.15. If [NH3] = 0.01 M, then [NH4+] = 0.14 M.

Kb = 1.76x10−5 (from the Appendix) Ka = Kw/Kb = 1.0x10−14/1.76x10−5 = 5.6818x10−10 pH = pKa + log [NH3]/[NH4

+] pH = −log Ka + log [NH3]/[NH4

+] pH = −log (5.6818x10−10) + log [0.01]/[0.14] pH = 8.099386 = 8.10 19.115 Determine the solubility of MnS:

S = 4

34.7x10 g MnS 1 mL 1 mol MnS

100 mL 87.00 g MnS10 L

−

−

= 5.402298850x10–5 M

MnS(s) + H2O(l) Mn2+(aq) + HS−(aq) + OH−(aq) Ksp = [Mn2+][HS–][OH–] = S3 = (5.402298850x10–5)3 = 1.5767x10–13 = 1.6x10–13 19.116 Plan: Use the Henderson-Hasselbalch equation to find the ratio of [HCO3

–]/[H2CO3] that will produce a buffer with a pH of 7.40 and a buffer of 7.20. Solution: a) Ka1 = 4.5x10–7

pKa = – log Ka = – log (4.5x10–7) = 6.34679

3a

2 3

[HCO ]pH = p + log

[H CO ]

−

K

3

2 3

[HCO ]7.40 = 6.34679 + log

[H CO ]

−

19-65

1.05321 = 3

2 3

[HCO ]log

[H CO ]

−

3

2 3

[HCO ][H CO ]

−

= 11.30342352

2 3

3

[H CO ][HCO ]−

= 0.0884688 = 0.088

b) 3a

2 3

[HCO ]pH = p + log

[H CO ]

−

K

3

2 3

[HCO ]7.20 = 6.34679 + log

[H CO ]

−

0.85321 = 3

2 3

[HCO ]log

[H CO ]

−

3

2 3

[HCO ][H CO ]

−

= 7.131978

2 3

3

[H CO ][HCO ]−

= 0.14021 = 0.14

19.117 Plan: The buffer components will be TRIS, (HOCH2)3CNH2, and its conjugate acid TRISH+, (HOCH2)3CNH3

+. The conjugate acid is formed from the reaction between TRIS and HCl. Since HCl is the limiting reactant in this problem, the concentration of conjugate acid will equal the starting concentration of HCl, 0.095 M. The concentration of TRIS is the initial concentration minus the amount reacted. Once the concentrations of the TRIS-TRISH+ acid-base pair are known, the Henderson-Hasselbalch equation can be used to find the pH. Solution:

Moles of TRIS = ( ) 1 mol TRIS43.0 g TRIS121.14 g TRIS

= 0.354961 mol

Moles of HCl added = ( )0.095 mol HCl 1.00 LL

= 0.095 mol HCl = mol TRISH+

(HOCH2)3CNH2(aq) + HCl(aq) (HOCH2)3CNH3

+(aq) + Cl−(aq) Initial 0.354961 mol 0.095 mol 0 0

Reacting −0.095 mol −0.095 mol +0.095 mol — Final 0.259961 mol 0 mol 0.095 mol Since there is 1.00 L of solution, the moles of TRIS and TRISH+ equal their molarities. pKa of TRISH+ = 14 – pKb = 14 – 5.91 = 8.09

a +[TRIS]pH = p + log

[TRISH ]

K = [0.259961] 8.09 + log

[0.095]

= 8.527185 = 8.53

19-66

19.118

19.119 Zinc sulfide, ZnS, is much less soluble than manganese sulfide, MnS. Convert ZnCl2 and MnCl2 to ZnS and MnS by saturating the solution with H2S; [H2S]sat’d = 0.10 M. Adjust the pH so that the greatest amount of ZnS will precipitate and not exceed the solubility of MnS as determined by Ksp(MnS). Ksp(MnS) = [Mn2+][HS−][OH−] = 3x10−11

[Mn2+] = [MnCl2] = 0.020 M [HS−] is calculated using the Ka1 expression: Concentration: H2S(aq) + H2O(l) H3O+(aq) + HS−(aq)

Initial 0.10 mol 0 0 Reacting −x +x +x Final 0.10 − x x x

Ka1 = 9x10−8 = [ ]

3

2

H O HS

H S

+ − =

[ ]3H O HS

0.10 x

+ −

− Assume 0.10 – x = 0.10.

[H3O+][HS−] = 9x10−9 [HS−] = 9x10−9/[H3O+] Substituting [Mn2+] and [HS−] into the Ksp(MnS) above gives: Ksp(MnS) = [Mn2+][HS−][OH−] = 3x10−11 Ksp(MnS) = [Mn2+](9x10−9/[H3O+])[OH−] = 3x10−11

Substituting Kw/[H3O+] for [OH−]:

Ksp(MnS) = 3x10−11 = 9

2 w+ +

3 3

9x10MnH O H O

K−+

3x10−11 x [H3O+]2 = ( ) ( )2 9wMn 9x10 K+ −

[H3O+] = 2 9

w11

Mn 9x10

3x10

K+ −

−

=

( ) ( ) ( )9 14

11

0.020 9x10 1.0x10

3x10

− −

− = 2.4494897x10−7

pH = −log [H3O+] = −log (2.4494897x10−7) = 6.610924 = 6.6 Maintain the pH below 6.6 to separate the ions as their sulfides. 19.120 a) Since Ka (−COOH) > Kb (−NH2), the proton will be transferred from the −COOH to the −NH2, producing

−COO− and −NH3+.

b) pKa = −log Ka = −log (4.47x10−3) = 2.34969

19-67

3 2a

3 2

[+NH CH COO ]pH = p + log

[+NH CH COOH]

−

K

3 2

3 2

[+NH CH COO ]5.5 = 2.34969 + log

[+NH CH COOH]

−

3 2

3 2

NH CH COO

NH CH COOH

+ −

+

= 1.41355x103 = 1x103

c)

CH C

CH2

OH

O

CH2

CH2

CH2

NH3

H3N CH C

CH2

O

O

CH2

CH2

CH2

NH3

H2N CH C

CH2

O

O

CH2

CH2

CH2

NH2

H3N

pH = 1 pH = 7 pH = 13 d) at pH 1: D, at pH 7: A, at pH 13: B 19.121 a) The equilibrium is: MCl2(s) M2+(aq) + 2Cl–(aq). The ion-product expression is Ksp = [M2+][Cl−]2. Based on the picture, the ion concentrations are:

[M2+] = ( )

6

3

1.0x10 mol3 spheres1 sphere 1 mL

250.0 mL 10 L

−

−

= 1.2x10−5 M

[Cl−] = ( )

6

3

1.0x10 mol10 spheres1 sphere 1 mL

250.0 mL 10 L

−

−

= 4.0x10−5 M

Ksp = [M2+][Cl−]2 = [1.2x10−5][ 4.0x10−5]2 = 1.92x10−14 = 1.9x10−14 b) M2+ is a common ion for M(NO3)2 and MCl2. If M(NO3)2 is added to the solution, [M2+] is increased

and, according to Le Chatalier’s principle, the solubility equilibrium will shift to the left, precipitating more MCl2. The number of Cl– particles decreases, the mass of MCl2 increases, and the Ksp value remains the same. 19.122 The equilibrium is: Ca5(PO4)3OH(s) 5Ca2+(aq) + 3PO4

3−(aq) + OH−(aq) Ksp = 6.8x10–37 = [Ca2+]5[PO4

3−]3[OH−] = (5S)5(3S)3(S) = 84375S9 S = 2.7166444x10–5 = 2.7x10–5 M Solubility = (2.7166444x10–5 mol/L)(502.32 g/mol) = 0.013646248 = 0.014 g/L Ca5(PO4)3OH The equilibrium is: Ca5(PO4)3F(s) 5Ca2+(aq) + 3PO4

3−(aq) + F−(aq) Ksp = 1.0x10–60 = [Ca2+]5[PO4

3−]3[F−] = (5S)5(3S)3(S) = 84375S9 S = 6.1090861x10–8 = 6.1x10–8 M Solubility = (6.1090861x10–8 mol/L)(504.31 g/mol) = 3.0808732x10–5 = 3.1x10–5 g/L Ca5(PO4)3F

19-68

19.123 Plan: An indicator changes color when the buffer-component ratio of the two forms of the indicator changes from a value greater than 1 to a value less than 1. The pH at which the ratio equals 1 is equal to pKa. The midpoint in the pH range of the indicator is a good estimate of the pKa of the indicator. Solution: pKa = (3.4 + 4.8)/2 = 4.1 Ka = 10–4.1 = 7.943x10–5 = 8x10–5 19.124

Due to the large range of [H3O+], this plot is difficult to prepare and does not easily show the end point. A logarithmic scale (pH vs. mL OH– added) shows this more clearly. 19.125 Plan: A spreadsheet will help you to quickly calculate ∆pH/∆V and average volume for each data point. At the

equivalence point, the pH changes drastically when only a small amount of base is added, therefore, ∆pH/∆V is at a maximum at the equivalence point.

Solution: a) Example calculation: For the first two lines of data: ∆pH = 1.22 – 1.00 = 0.22; ∆V = 10.00 – 0.00 = 10.00

pH∆∆V

= 0.2210.00

= 0.022 Vaverage(mL) = (0.00+ 10.00)/2 = 5.00

V(mL) pH pH∆∆V

Vaverage(mL)

0.00 1.00 10.00 1.22 0.022 5.00 20.00 1.48 0.026 15.00 30.00 1.85 0.037 25.00 35.00 2.18 0.066 32.50 39.00 2.89 0.18 37.00 39.50 3.20 0.62 39.25 39.75 3.50 1.2 39.63 39.90 3.90 2.67 39.83 39.95 4.20 6 39.93 39.99 4.90 18 39.97 40.00 7.00 200 40.00 40.01 9.40 200 40.01 40.05 9.80 10 40.03 40.10 10.40 10 40.08 40.25 10.50 0.67 40.18 40.50 10.79 1.2 40.38 41.00 11.09 0.60 40.75

0

0.05

0.1

0 10 20 30 40 50

[H3O

+ ]

mL NaOH added

19-69

45.00 11.76 0.17 43.00 50.00 12.05 0.058 47.50

60.00 12.30 0.025 55.00 70.00 12.43 0.013 65.00 80.00 12.52 0.009 75.00 b)

19.126 Check to see if the concentration of Ca(OH)2 exceeds the Ksp. M Ca(OH)2 = (6.5x10–9 mol Ca(OH)2)/(10.0 L) = 6.5x10–10 M Ca(OH)2 Determine the concentration of a saturated calcium hydroxide solution from the Ksp. Ca(OH)2(s) Ca2+(aq) + 2OH–(aq) Ksp = 6.5x10–6 = [Ca2+][OH–]2 = (S)(2S)2 = 4S3

S = 6

3 6.5x104

−

= 0.01175667 = 0.012 M

Thus, the solution is less than saturated so the Ksp does not affect the concentration of Ca(OH)2. M OH– from Ca(OH)2 = (6.5x10–10 M Ca(OH)2)(2 mol OH–/1 mol Ca(OH)2) = 1.3x10–9 M OH– Pure water has 1x10–7 M OH–, thus the contribution from the Ca(OH)2 is not significant. pH of pure water = 7.0. 19.127 Use HLac to indicate lactic acid and Lac− to indicate the lactate ion. The Henderson-Hasselbalch equation gives the pH of the buffer. Determine the final concentrations of the buffer components from MconcVconc = MdilVdil. Determine the pKa of the acid from the Ka. pKa = –log Ka = –log (1.38x10−4) = 3.86012 Determine the molarity of the diluted buffer component as Mdil = MconcVconc/Vdil. [HLac] = [(0.85 M) (225 mL)]/[(225 + 435) mL] = 0.28977 M HLac [Lac−] = [(0.68 M) (435 mL)]/[(225 + 435) mL] = 0.44818 M Lac−

a[Lac ]pH = p + log[HLac]

K−

[0.44818]pH = 3.86012 + log[0.28977]

= 4.049519 = 4.05

19.128 The ion-product equilibrium reaction is:

CaF2(s) Ca2+(aq) + 2F–(aq) F– is a weak base with the following equilibrium reaction: F−(aq) + H2O(l) HF(aq) + OH−(aq)

(I) Pure water: There is no common-ion effect and the pH is neutral. (II) 0.01M HF: Because of the common-ion effect, less CaF2 would dissolve in this solution than in pure water.

(III) 0.01M NaOH: Additional OH– ions shift the base equilibrium reaction to the left, producing more F−. The additional F− shifts the ion-product equilibrium to the left so less CaF2 would dissolve.

19-70

(IV) 0.01M HCl: H+ ions remove OH– ions from solution so the base equilibrium reaction shifts to the right, consuming F−. This shifts the ion-product equilibrium to the right so that more CaF2 dissolves in this solution than in pure water. (V) 0.01M Ca(OH)2: Because of the common-ion effect, less CaF2 would dissolve in this solution than in pure water. Additional OH– ions shift the base equilibrium reaction to the left, producing more F−. The additional F− shifts the ion-product equilibrium to the left so less CaF2 would dissolve.

a) 0.01M HCl b) 0.01M Ca(OH)2 19.129 a) The NaOH is a strong base, so it dissociates completely. NaOH(s) → Na+(aq) + OH–(aq) 0.050 mol 0.050 mol 0.050 mol The OH– ions from NaOH will react with HClO. OH–(aq) + HClO(aq) → H2O(l) + ClO–(aq) 0.050 mol 0.13 mol 0 –0.050 mol –0.050 mol +0.050 mol 0 0.080 mol 0.050 mol The initial amount of HClO is 0.13 mol – 0.050 mol = 0.08 mol The initial amount of ClO– is 0.050 mol ClO–. The volume of the solution is (500. mL)(10–3 L/1 mL) = 0.500 L [HClO]i = (0.080 mol HClO)/(0.500 L) = 0.16 M HClO [ClO–]I = (0.050 mol ClO–)/(0.500 L) = 0.10 M OCl– Concentration (M) HClO(aq) + H2O H3O+(aq) + ClO–(aq) Initial 0.16 — 0.10 Change –x +x +x Equilibrium 0.16 – x x 0.10 + x x is small, so [HClO] ≈ 0.16 M HClO and [ClO–] ≈ 0.10 M ClO–

Ka = [ ]

3H O ClO

HClO

+ −

[H3O+] = [ ]a HClO

ClO

K−

[H3O+] = ( ) ( )

( )

83.0x10 0.16

0.10

−

= 4.8x10–8 M H3O+

[OH–] = Kw/[H3O+] = (1.0x10–14)/(4.8x10–8) = 2.08333x10–7 = 2.1x10–7 M OH– [Na+] = (0.050 mol)/(0.500 L) = 0.10 M Na+ b) pH = –log [H3O+] = –log (4.8x10–8 M) = 7.31875876 = 7.32

c) If 0.0050 mol HCl is added then the ClO– will react with the H+ to form HClO. H+(aq) + ClO–(aq) → HClO(aq)

0.0050 mol 0.050 mol 0.080 mol –0.0050 mol –0.0050 mol +0.0050 mol 0 0.045 mol 0.085 mol [HClO]i = (0.085 mol HClO)/(0.500 L) = 0.17 M HClO [ClO–]I = (0.045 mol ClO–)/(0.500 L) = 0.090 M OCl–

Concentration (M) HClO(aq) + H2O H3O+(aq) + ClO–(aq) Initial 0.17 — 0.090 Change –x +x +x Equilibrium 0.17 – x x 0.090 + x x is small, so [HClO] ≈ 0.17 M and [ClO–] ≈ 0.090 M

19-71

[H3O+] = [ ]a HClO

ClO

K−

[H3O+] = ( ) ( )

( )

83.0x10 0.17

0.090

−

= 5.6667x10–8 M H3O+

pH = –log [H3O+] = –log (5.6667x10–8 M) = 7.246669779 = 7.25 19.130 In both cases the equilibrium is: CaCO3(s) Ca2+(aq) + CO3

2–(aq) Ksp = [Ca2+][CO3

2–] = S2 At 10°C Ksp = 4.4x10–9 = [Ca2+][CO3

2–] = S2 S = 6.6332496x10–5 = 6.6x10–5 M CaCO3 At 30°C Ksp = 3.1x10–9 = [Ca2+][CO3

2–] = S2 S = 5.5677644x10–5 = 5.6x10–5 M CaCO3 19.131 Hg2C2O4(s) Hg2

2+(aq) + C2O42–(aq)

Ksp = 1.75x10–13 = [Hg22+][C2O4

2–] = (0.13 + S)S ≈ (0.13) S S = 1.3461538x10–12 = 1.3x10–12 M 19.132 [H+] = 10–9.5 = 3.16227766x10–10 M H+ pOH = 14.0 – pH = 14.0 – 9.5 = 4.5 [OH–] = 10–4.5 = 3.16227766x10–5 M OH–

[HCO3–] =

33 3

3

65.0 mg HCO 1 mol HCO10 gL 1 mg 61.02 g HCO

− −−

−

= 1.0652245x10–3 M HCO3–

[CO32–] =

2 233 3

23

26.0 mg CO 1 mol CO10 gL 1 mg 60.01 g CO

− −−

−

= 4.3326112x10–4 M CO32–

Alkalinity = [HCO3–] + 2[CO3

2–] + [OH–] – [H+] Alkalinity = (1.0652245x10–3) + 2(4.3326112x10–4) + (3.16227766x10–5) – (3.16227766x10–10) Alkalinity = 1.9633692x10–3 = 1.96x10–3 M 19.133 Plan: To determine which species are present from a buffer system of a polyprotic acid, check the pKa values for

the one that is closest to the pH of the buffer. The two components involved in the equilibrium associated with this Ka are the principle species in the buffer. Use the Henderson-Hasselbalch equation to find the ratio of the phosphate species that will produce a buffer with a pH of 7.4.

Solution: For carbonic acid, pKa1 [−log (8x10−7) = 6.1] is closer to the pH of 7.4, so H2CO3 and HCO3

− are the species present. For phosphoric acid, pKa2 [−log (2.3x10−7) = 6.6] is closest to the pH, so H2PO4

− and HPO42− are the

principle species present. H2PO4

−(aq) + H2O(l) HPO42−(aq) + H3O+(aq)

2

4a

2 4

[HPO ]pH = p + log

[H PO ]K

−

−

2

4

2 4

[HPO ]7.4 = 6.6383 + log

[H PO ]

−

−

24

2 4

HPO

H PO

−

−

= 5.77697 = 5.8

19.134 Plan: Mercury sulfide, HgS, is much less soluble than nickel sulfide, NiS. Adjust the pH so that the greatest amount of HgS will precipitate and not exceed the solubility of NiS as determined by Ksp of NiS.

19-72

Determine the minimum pH needed to cause the initial precipitation of NiS. Use the Ka expression for H2S to express [HS−] in terms of [H3O+]; use the Kw expression for water to express [OH−] in terms of [H3O+]. Solution: NiS(s) + H2O(l) Ni2+(aq) + HS−(aq) + OH−(aq) Ksp (NiS) = 3x10−16 = [Ni2+][HS−][OH−] [Ni2+] = 0.15 M, so [HS−] and [OH−] must be found. From [H2S] = 0.050 M and Ka1 in the Appendix: H2S (aq) + H2O(l) HS−(aq) + H3O+(aq)

Ka1 = [ ]

3

2

H O HS

H S

+ − =

[ ]3H O HS

0.050 x

+ −

− =

[ ]3H O HS

0.050

+ − = 9x10−8

[HS−][H3O+] = 4.5x10−9 or [HS−] = 4.5x10−9/[H3O+] [OH−] = Kw/[H3O+] Ksp (NiS) = 3x10−16 = [Ni2+][HS−][OH−]

Ksp (NiS) = 3x10−16 = ( )9 14

3 3

4.5x10 1.0x100.15[H O] [H O]

− −

+ +

3x10−16 = ( ) ( ) ( )

( )9 14

23

0.15 4.5x10 1.0x10

[H O]

− −

+

[H3O+]2 = [(0.15)(4.5x10−9)(1.0x10−14)]/(3x10−16) = 2.25x10−8 [H3O+] = 1.5x10−4 M pH = −log [H3O+] = −log (1.5x10−4 M) = 3.8239 = 3.8 19.135 a) Combine the separate equilibria to produce the desired equilibrium. The K values are in the Appendix. 2AgCl(s) 2Ag+(aq) + 2Cl–(aq) K' = (Ksp)2 = (1.8x10–10)2 = 3.24x10–20 2Ag+(aq) + CrO4

2–(aq) Ag2CrO4(s) K" = 1/Ksp = 1/(2.6x10–12) = 3.8462x1011 2AgCl(s) + CrO4

2–(aq) Ag2CrO4(s) + 2Cl–(aq) K = K'K" = 1.2462x10–8 = 1.2x10–8 b) Since the above reaction has such a small K, it lies far to the left as written. c) The mixing of equal amounts of equal molar solutions would precipitate all the AgCl, thus the silver ion concentration comes entirely from the Ksp of AgCl. Ksp = 1.8x10–10 = [Ag+][Cl–] = S2 S = [Ag+] = 1.34164x10–5 M = 1.3x10–5 M Ag+ Use the Ksp for silver chromate. Ksp = 2.6x10–12 = [Ag+]2[CrO4

2–] [CrO4

2–] = (2.6x10–12)/(1.34164x10–5)2 = 0.01444 = 0.014 M If the chromate ion concentration exceeds 0.014 M, Ag2CrO4 will precipitate. 19.136 Plan: Find the moles of quinidine initially present in the sample by dividing its mass in grams by the molar mass. Use the molar ratio between quinidine and HCl to find the moles of HCl that would react with the moles of quinidine and subtract the reacted HCl from the initial moles of HCl to find the excess. Use the molar ratio between HCl and NaOH to find the volume of NaOH required to react with the excess HCl. Then use the molar ratio between NaOH and quinidine to find the volume of NaOH required to react with the quinidine. Solution: a) To find the concentration of HCl after neutralizing the quinidine, calculate the concentration of quinidine and the amount of HCl required to neutralize it, remembering that the mole ratio for the neutralization is 2 mol HCl/1 mol quinidine.

Moles of quinidine = ( )310 g 1 mol quinidine33.85 mg quinidine

1 mg 324.41 g quinidine

−

= 1.0434327x10-4 mol quinidine

Moles of HCl excess = ( )310 L 0.150 mol HCl6.55 mL

1 mL L

−

19-73

– ( )4 2 mol HCl1.0434327 x10 mol quinidine1 mol quinidine

−

= 7.7381346x10–4 mol HCl

Volume (mL) of NaOH needed = ( )43

1 mol NaOH 1 L 1 mL7.7381346x10 mol HCl1 mol HCl 0.0133 mol NaOH 10 L

−−

= 58.18146 = 58.2 mL NaOH solution b) Use the moles of quinidine and the concentration of the NaOH to determine the milliliters.

Volume = ( )43

1 mol NaOH 1 L 1 mL1.0434327x10 mol quinidine1 mol quinidine 0.0133 mol NaOH 10 L

−−

= 7.84536 = 7.85 mL NaOH solution c) When quinidine (QNN) is first acidified, it has the general form QNH+NH+. At the first equivalence point, one of the acidified nitrogen atoms has completely reacted, leaving a singly protonated form, QNNH+. This form of quinidine can react with water as either an acid or a base, so both must be considered. If the concentration of quinidine at the first equivalence point is greater than Kb1, then the [OH–] at the first equivalence point can be estimated as:

[OH–] = b1 b2K K = ( ) ( )6 104.0x10 1.0x10− − = 2.0x10–8 M

[H3O+] = Kw/[OH–] = (1.0x10–14)/(2.0x10–8) = 5.0x10–7 M pH = –log [H3O+] = –log (5.0x10–7 M) = 6.3010 = 6.30 19.137 K values from the Appendix: H2C2O4(aq) H+(aq) + HC2O4

–(aq) Ka1 = 5.6x10–2 HC2O4

–(aq) H+(aq) + C2O42–(aq) Ka2 = 5.4x10–5

H2C2O4(aq) 2H+(aq) + C2O4

2–(aq) K = Ka1Ka2 = 3.024x10–6 K = [H+]2[C2O4

2–]/[H2C2O4] [C2O4

2–] = K[H2C2O4]/[H+]2 a) At pH = 5.5: [H+] = 10–5.5 = 3.162x10–6 M [C2O4

2–] = K[H2C2O4]/[H+]2 = (3.024x10–6)(3.0x10–13)/(3.162x10–6)2 [C2O4

2–] = 9.07359x10–8 M Q = [Ca2+][C2O4

2–] Q = (2.6x10–3)(9.07359x10–8) = 2.3591x10–10 = 2.4x10–10 < Ksp = No precipitate b) At pH = 7.0: [H+] = 10–7.0 = 1x10–7 M [C2O4

2–] = K[H2C2O4]/[H+]2 = (3.024x10–6)(3.0x10–13)/(1x10–7)2 [C2O4

2–] = 9.072x10–5 M Q = (2.6x10–3) (9.072x10–5) = 2.35872x10–7 = 2.4x10–7 > Ksp = Precipitate forms c) The higher pH would favor precipitation. 19.138 Plan: The Henderson-Hasselbalch equation demonstrates that the pH changes when the ratio of acid to base in

the buffer changes (pKa is constant at a given temperature). Solution:

a[A ]pH = p + log[HA]

K−

The pH of the A−/HA buffer cannot be calculated because the identity of “A” and, thus, the value of pKa are unknown. However, the change in pH can be described:

∆final initial

[A ] [A ]pH = log log[HA] [HA]

− − −

Since both [HA] and [A−] = 0.10 M, initial

[A ]log[HA]

−

= 0 because [HA] = [A−], and log (1) = 0

So the change in pH is equal to the concentration ratio of base to acid after the addition of H3O+.

19-74

Consider the buffer prior to addition to the medium. H3O+ (aq) + A–(aq) → HA(aq)

0.0010 mol 0.10 mol 0.10 mol –0.0010 mol –0.0010 mol +0.0010 mol 0 0.099 mol 0.101 mol

When 0.0010 mol H3O+ is added to 1 L of the undiluted buffer, the [A−]/[HA] ratio changes from 0.10/0.10 to (0.099)/(0.101). The change in pH is:

∆pH = log (0.099/0.101) = –0.008686 If the undiluted buffer changes 0.009 pH units with addition of 0.0010 mol H3O+, how much can the buffer be diluted and still not change by 0.05 pH units (∆pH < 0.05)?

Let x = fraction by which the buffer can be diluted. Assume 0.0010 mol H3O+ is added to 1 L.

log [ ][ ]baseacid

= log ( )( )0.10x 0.00100.10x 0.0010

− +

= –0.05

( )( )0.10x 0.00100.10x 0.0010

− +

= 10–0.05 = 0.89125

0.10x – 0.0010 = 0.89125 (0.10x + 0.0010) x = 0.173908 = 0.17 The buffer concentration can be decreased by a factor of 0.17, or 170 mL of buffer can be diluted to 1 L of medium. At least this amount should be used to adequately buffer the pH change.

19.139 a) Ka = 6.8x10–4 = [ ]

3H O F

HF

+ − =

2x0.2500 x−

≈ 2x

0.2500

x = [H3O+] = 0.0130384 M pH = –log [H3O+] = –log (0.0130384 M) = 1.8847757 = 1.88

b) Volume (mL) of Na = ( )3

310 L 0.2500 mol HF 1 mol NaOH 1 L 1 mL35.00 mL1 mL 1 L 1 mol HF 0.1532 mol NaOH 10 L

−

−

= 57.11488 = 57.11 mL NaOH

c) Moles of HF (initial) = ( )310 L 0.2500 mol HF35.00 mL

1 mL 1 L

−

= 8.750x10–3 mol HF

Moles of NaOH added = ( )( )310 L 0.1532 mol NaOH57.11488 0.50 mL

1 mL 1 L

− −

= 8.6734x10–3 mol NaOH

Moles of F– formed = moles NaOH Moles of HF remaining = (8.750x10–3 – 8.6734x10–3) mol = 7.66x10–5 mol HF Volume of solution = (35.00 + 57.11488 – 0.50)(10–3 L/1 mL) = 0.091615 L [HF] = (7.66x10–5 mol HF)/(0.091615 L) = 0.00083611 M HF [F–] = (8.6734x10–3 mol F–)/(0.091615 L) = 0.09467 M F– pKa = –log Ka = –log (6.8x10–4) = 3.1675


K−

[0.09467]pH = 3.1675 + log

[0.0008361]

pH = 5.2215 = 5.22 d) At this point there are 8.750x10–3 mol of F– in (35.00 + 57.11488) mL of solution.

Molarity of F– = ( )

3

38.750x10 mol F 1 mL

35.00 57.11488 mL 10 L

− −

−

+

= 0.09499 M F–

Kb = Kw/Ka = (1.0x10–14)/(6.8x10–4) = 1.470588x10–11

19-75

Kb = 1.470588x10–11 = [ ]HF OH

F

−

−

= 2x

0.09499 x− ≈

2x0.09499

x = [OH–] = 1.1819x10–6 M [H3O+] = Kw/[OH–] = (1.0x10–14)/(1.1819x10–6) = 8.4609527x10–9 M pH = –log [H3O+] = –log (8.4609527x10–9 M) = 8.07258 = 8.07

e) Moles of NaOH excess = ( )310 L 0.1532 mol NaOH0.50 mL

1 mL 1 L

−

= 7.66x10–5 mol NaOH

Volume of solution = (35.00 + 57.11488 + 0.50)(10–3 L/1 mL) = 0.092615 L [OH–] = (7.66x10–5 mol F–)/(0.092615 L) = 8.271x10–4 M OH– The excess OH– will predominate and essentially control the pH. [H3O+] = Kw/[OH–] = (1.0x10–14)/(8.271x10–4) = 1.2090436x10–11 M pH = –log [H3O+] = –log (1.2090436x10–11 M) = 10.917558 = 10.92 19.140 a) The formula is Hg2Cl2 which simplifies to the empirical formula HgCl. b) The equilibrium is: Hg2Cl2(s) Hg2

2+(aq) + 2Cl−(aq) Ksp = 1.5x10–18 Ksp = [Hg2

2+][Cl−]2 = (S)(2S)2 = 4S3 = 1.5x10–18 S = 7.2112479x10–7 = 7.2x10–7 M c) [Hg2

2+] = Ksp/[Cl−]2 =

18

23

1.5x10

0.20 lb NaCl 1 kg 10 g 1 mol NaCl 1 mol Cl 1 gal 1.057 qtgal 2.205 lb 1 kg 58.44 g NaCl 1 mol NaCl 4 qt 1 L

−

−

= 8.9174129x10–18 = 8.9x10–18 M Hg22+

d) Use the value of S for a saturated solution (see part b)).

Mass (g) of Hg2Cl2 = ( )37 3

32 2 2 23 3

2 2

7.2112479x10 mol Hg Cl 1 L 10 m 472.1 g Hg Cl4900 kmL 1 km 1 mol Hg Cl10 m

−

−

= 1.6681708x1012 = 1.7x1012 g Hg2Cl2 e) Use the value determined in part c). Mass (g) of Hg2Cl2 =

( )318 2 3

32 2 2 2 22 3 3

2 22

8.9174129x10 mol Hg 1 mol Hg Cl 1 L 10 m 472.1 g Hg Cl4900 kmL 1 km 1 mol Hg Cl1 mol Hg 10 m

− +

+ −

= 20.62856 = 21 g Hg2Cl2 19.141 a) CaF2 with Ksp = 3.2x10–11 will precipitate before BaF2 with Ksp = 1.5x10–6. b) Add KF until [F–] is such that the CaF2 precipitates but just lower than the concentration required to precipitate BaF2. c) Determine the barium concentration after mixing from MconcVconc = MdilVdil. [Ba2+] = [(0.090 M)(25.0 mL)]/[(25.0 + 35.0) mL] = 0.0375 M Use the barium ion concentration and the Ksp to find the fluoride ion concentration. [F–]2 = Ksp/[Ba2+] = (1.5x10–6)/(0.0375) = 4.0x10–5 M [F–] = 6.324555x10–3 = 6.3x10–3 M or less 19.142 Plan: Use the ideal gas law to calculate the moles of CO2 dissolved in water. Use the Ka expression for H2CO3 to find the [H3O+] associated with that CO2 concentration. Solution: Carbon dioxide dissolves in water to produce H3O+ ions: CO2(g) CO2(aq) CO2(aq) + H2O(l) H2CO3(aq) H2CO3(aq) H3O+(aq) + HCO3

−(aq) Ka1 = 4.5x10–7

19-76

The molar concentration of CO2, [CO2], depends on how much CO2(g) from the atmosphere can dissolve in pure water. Since air is not pure CO2, account for the volume fraction of air (0.040 L/100 L) when determining the moles.

Volume (L) of CO2 = (88 mL) �10-3 L1 mL � �

0.040%100% � = 3.520x10–5 L CO2

Moles of dissolved CO2 = PVRT

= (1 atm)�3.520 x 10-5 L�

�0.0821 L•atmmol•K��(273+25)K�

= 1.438743x10–6 mol CO2

[CO2] = (1.438743x10–6 mol CO2)/[(100 mL)(10–3 L/1 mL)] = 1.438743x10–5 M CO2

Ka1 = 4.5x10–7 = 3 3

2 3

H O HCO

H CO

+ −

=

[ ]3 3

2

H O HCO

CO

+ − Let x = [H3O+] = [HCO3

−]

4.5x10–7 = [x][x]

�1.438743x10–5 −x� Assume that x is small compared to 1.438743x10–5

4.5x10–7 =[x][x]

�1.438743x10–5�

x = 2.544473x10–6 Check assumption that x is small compared to 1.438743x10–5:

2.544473 x 10-6

1.438743 x 10-5(100)= 18% error, so the assumption is not valid.

Since the error is greater than 5%, it is not acceptable to assume x is small compared to 1.438743x10–5, and it is necessary to use the quadratic equation. 4.5x10–7 = [x][x]

�1.438743x10–5 −x�

x2 + 4.5x10–7x – 6.474344x10–12 = 0 a = 1 b = 4.5x10–7 c = –6.474344x10–12

x = 2b b 4ac

2a− ± −

x = -4.5 x 10-7 ± ��4.5 x 10-7�

2- 4(1)(–6.474344 x 10-12)

2(1)

x = 2.329402x10–6 M = [H3O+] pH = –log (2.329402x10–6) = 5.632756 = 5.63 19.143 a) For H2CO3, pKa = –log Ka pKa1 = –log 4.5x10–7 = 6.3468 pKa2 = –log 4.7x10–11 = 10.3279 pKa1 = 6.35 and pKa2 = 10.33. Since pKa1 > pH < pKa2, the base in the first dissociation (HCO3

–) and the acid in the second dissociation (also HCO3

–) will predominate. b) pH = pKa + log [base]/[acid] 8.5 = 6.35 + log [HCO3

–]/[H2CO3] [HCO3

–]/[H2CO3] = 1.4125x102 = 1x102 M pH = pKa + log [base]/[acid] 8.5 = 10.33 + log [CO3

2–]/[HCO3–]

[CO32–]/[HCO3

–] = 1.4791x10–2 = 1x10–2 M c) In deep water, animals can exist but plants, which depend on light for photosynthesis, cannot. Photosynthesis converts carbon dioxide to oxygen; animals convert oxygen to carbon dioxide. Near the surface, plants remove carbon dioxide (which, in water, can be represented as the weak acid H2CO3) and thus the pH is higher than in deep water, where higher concentrations of carbon dioxide (H2CO3) accumulate. Also, at greater depths, the pressure is higher and so is the concentration of CO2 (Henry’s law).

19-77

19.144 Initial concentrations of Pb2+ and Ca(EDTA)2− before reaction based on mixing 100. mL of 0.10 M Na2Ca(EDTA) with 1.5 L blood:

[Pb2+] = 2 6 2

3 2120 g Pb 1 mL 1.5 L blood 10 g 1 mol Pb

100 mL 1.6 L mixture 1 g10 L 207.2 g Pb

+ − +

− +

µ µ

= 5.42953668x10–6 M Pb2+

MconcVconc = MdilVdil [Ca(EDTA)2–] = MconcVconc/Vdil = [(0.10 M (100 mL)(10–3 L/1 mL)]/(1.6 L) = 6.25x10–3 M Set up a reaction table assuming the reaction goes to completion: Concentration (M) [Ca(EDTA)]2−(aq) + Pb2+(aq) [Pb(EDTA)]2−(aq) + Ca2+(aq) Initial 6.25x10–3 5.42953668x10–6 0 0 React –5.42953668x10–6 –5.42953668x10–6 +5.42953668x10–6 +5.42953668x10–6 6.24457x10–3 0 5.42953668x10–6 5.42953668x10–6 Now set up a reaction table for the equilibrium process: Concentration (M) [Ca(EDTA)]2−(aq) + Pb2+(aq) [Pb(EDTA)]2−(aq) + Ca2+(aq) Initial 6.24457x10–3 0 5.42953668x10–6 5.42953668x10–6 Change +x +x –x –x Equilibrium 6.24457x10–3 + x x 5.4295366x10–6 – x 5.4295366x10–6 – x

Kc = 2.5x107 = ( )

( )

2 2

2 2

Pb EDTA Ca

Ca EDTA Pb

− +

− +

= [ ]

6 6

3

5.42953668x10 5.42953668x10

6.24457x10 x

− −

−

x = [Pb2+] = 1.8883522x10–16 M

Mass (µg) of Pb2+ in 100 mL = ( )16 2 3 2

2 61.8883522x10 mol Pb 10 L 207.2 g Pb 1 g100 mL

L 1 mL 1 mol Pb 10 g

− + − +

+ −

µ

= 3.9126658x10–9 µg Pb2+ The final concentration is 3.9x10–9 µg/100 mL. 19.145 Assume that pKa is in the center of the range, and calculate the Ka from the average pKa. Average pKa (center of range) = (7.9 + 6.5)/2 = 7.2 Ka = 10–7.2 = 6.3095734x10–8 = 6x10–8 There is only one digit after the decimal point in the pKa values; thus, there is only one significant figure. 19.146 Plan: Convert the solubility of NaCl from g/L to mol/l (molarity). Use the solubility to find the Ksp value for NaCl. Find the moles of Na+ and Cl− in the original solution; find the moles of added Cl− (from the added HCl). The molarity of the Na+ and Cl− ions are then found by dividing moles of each by the total volume after mixing. Using the molarities of the two ions, determine a Q value and compare this value to the Ksp to determine if precipitation will occur. Solution:

Concentration (M) of NaCl = 317 g NaCl 1 mol NaCl

L 58.44 g NaCl

= 5.42436687 M NaCl

Determine the Ksp from the molarity just calculated. NaCl(s) Na+(aq) + Cl−(aq)

Ksp = [Na+][Cl−] = S2 = (5.42436687)2 = 29.42375594 = 29.4

Moles of Cl− initially = ( )5.42436687 mol NaCl 1 mol Cl0.100 LL 1 mol NaCl

−

= 0.542436687 mol Cl−

This is the same as the moles of Na+ in the solution.

Moles of Cl− added = ( )38.65 mol HCl 10 L 1 mol Cl28.5 mL

L 1 mL 1 mol HCl

− −

= 0.246525 mol Cl−

19-78

0.100 L of saturated solution contains 0.542 mol each Na+ and Cl−, to which you are adding 0.246525 mol of additional Cl− from HCl.

Volume of mixed solutions = 0.100 L + (28.5 mL)(10–3 L/1 mL) = 0.1285 L Molarity of Cl− in mixture = [(0.542436687 + 0.246525) mol Cl−]/(0.1285 L) = 6.13978 M Cl− Molarity of Na+ in mixture = (0.542436687 mol Na+)/(0.1285 L) = 4.22130 M Na+ Determine a Q value and compare this value to the Ksp to determine if precipitation will occur. Qsp = [Na+][Cl−] = (4.22130)(6.13978) = 25.9179 = 25.9 Since Qsp < Ksp, no NaCl will precipitate. 19.147 Plan: A buffer contains a weak acid conjugate base pair. A Ka expression is used to calculate the pH of a weak acid while a Kb expression is used to calculate the pH of a weak base. The Henderson-Hasselbalch equation is used to calculate the pH when both the weak acid and conjugate base are present (a buffer). Solution: a) For the solution to be a buffer, both HA and A− must be present in the solution. This situation occurs in A and D. b) Scene A: The amounts of HA and A− are equal.


K−

[A ][HA]

−

= 1 when the amounts of HA and A− are equal

apH = p + log 1K pH = pKa = –log (4.5x10–5) = 4.346787 = 4.35 Scene B: Only A− is present at a concentration of 0.10 M. The Kb for A− is needed. Kb = Kw/Ka = 1.0x10–14/4.5x10–5 = 2.222x10–10 A−(aq) + H2O(l) OH−(aq) + HA(aq) Initial: 0.10 M 0 0 Change: –x –x –x Equilibrium: 0.10 – x x x

Kb = 2.222x10–10 = [ ]HA OH

A

−

−

Kb = 2.222x10–10 = [ ][ ]

[ ]x x

0.10 x− Assume that x is small compared to 0.10

Kb = 2.222x10–10 = ( )( )( )x x0.10

x = 4.7138095x10–6 M OH− Check assumption: (4.7138095x10–6/0.10) x 100% = 0.005% error, so the assumption is valid. [H3O]+ = Kw/[OH−] = (1.0x10–14)/(4.7138095x10–6) = 2.1214264x10–9 M H3O+ pH = –log [H3O+] = –log (2.1214264x10–9) = 8.67337 = 8.67

Scene C: This is a 0.10 M HA solution. The hydrogen ion, and hence the pH, can be determined from the Ka. Concentration HA(aq) + H2O(l) H3O+(aq) + A–(aq) Initial 0.10 M — 0 0 Change –x +x +x Equilibrium 0.10 – x x x

19-79

(The H3O+ contribution from water has been neglected.)

Ka = 4.5x10–5 = [ ]

3H O A

HA

+ −

Ka = 4.5x10–5 = ( )( )

( )x x

0.10 x− Assume that x is small compared to 0.10.

Ka = 4.5x10–5 = ( )( )( )x x0.10

[H3O+] = x = 2.12132x10–3 Check assumption: (2.12132x10–3/0.10) x 100% = 2% error, so the assumption is valid. pH = –log [H3O+] = –log (2.12132x10–3) = 2.67339 = 2.67 Scene D: This is a buffer with a ratio of [A–]/[HA] = 5/3.


K−

pH = –log (4.5x10–5) + log 53

= 4.568636 = 4.57

c) The initial stage in the titration would only have HA present. The amount of HA will decrease, and the amount of A– will increase until only A– remains. The sequence will be: C, A, D, and B. d) At the equivalence point, all the HA will have reacted with the added base. This occurs in scene B. 19.148 a) The dissolution of MZ will produce equal amounts of M2+ and Z2–. The only way unequal amounts of these ions could be present would be either if one of the ions were already present or if one of the ions were removed from the solution. Distilled water will neither add nor remove ions, thus the M2+ and Z2– must be equal; as in Scene B. b) Using box B; there are 4(2.5x10–6M) = 1.0x10–5 M for each ion. Ksp = [M2+][Z2–] = (1.0x10–5)(1.0x10–5) = 1.0x10–10 c) The addition of Na2Z would increase the Z2– and shifts the equilibrium to the left, resulting in fewer ions of M2+. There will be more Z2– than M2+. This occurs in box C. d) Lowering the pH will protonate some Z2– (the weak base CO3

2–). This will decrease the Z2– concentration and and shift the equilibrium to the right, resulting in more M2+. This occurs in box A. 19.149 a) Ag+ ions come from the dissolution of AgCl(s). Ksp = [Ag+][Cl–] = 1.8x10–10

[Ag+] = 101.8x10

Cl

−

−

AgCl(s) Ag+(aq) + Cl– (aq) Ksp = 1.8x10–10 Ag+(aq) + 2Cl– (aq) AgCl2

–(aq) Kf = 1.8x105

AgCl(s) + Cl–(aq) AgCl2

–(aq) K = KspKf = (1.8x10–10)(1.8x105) = 3.24x10–5

K = 3.24x10–5 = 2AgCl

Cl

−

−

[AgCl2–] = (3.24x10–5)[Cl–] = (3.2x10–5)[Cl–]

b) [Ag+] = [AgCl2–]

101.8x10

Cl

−

−

= (3.24x10–5)[Cl–]

19-80

[Cl–]2 = 10

51.8x103.24x10

−

−

[Cl–] = 2.3570226x10–3 = 2.4x10–3 M Cl– c) At low Cl– ion concentration, Ag+ ions are present in solution. As Cl– ion concentration increases, more AgCl(s) is formed as the solubility equilibrium is shifted to the left and the solubility of AgCl decreases. At even higher Cl– ion concentrations, AgCl2

– ions are present in solution as the formation equilibrium is shifted to the right.

d) The solubility of AgCl(s) = [Ag+] + [AgCl2

–] You can use either equation from part a) to calculate [Ag+] and [AgCl2

–]. [Ag+] = [AgCl2

–] = (3.24x10–5)(2.3570226x10–3) = 7.6367532x10–8 = 7.6x10–8 M The solubility of AgCl(s) = (7.6367532x10–8 + 7.6367532x10–8) M = 1.52735x10–7 = 1.5x10–7 M 19.150 Co2+(aq) + EDTA4−(aq) [Co(EDTA)]2−(aq) Kf = 1016.31 = 2.0417379x1016

Kf = ( )2

2 4

Co EDTA

Co EDTA

−

+ −

Moles of Co2+ (initial) = ( )2 30.048 mol Co 10 L 50.0 mL

L 1 mL

+ −

= 0.0024 mol Co2+

a) Moles of EDTA added = ( )4 30.050 mol EDTA 10 L 25.0 mL

L 1 mL

− −

= 0.00125 mol EDTA

The moles of EDTA added equals the moles of [Co(EDTA)]2− formed. The EDTA is limiting so no EDTA is left after the reaction and the remaining Co2+: Co2+ = (0.0024 – 0.00125) = 0.00115 mol Co2+ Total volume = (50.0 + 25.0) mL (10–3 L/1 mL) = 0.0750 L [Co2+] = (0.00115 mol Co2+)/(0.0750 L) = 0.015333 M Co2+ [Co(EDTA)2-] = (0.00125 mol [Co(EDTA)]2−)/(0.0750 L) = 0.016667 M To reach equilibrium the concentrations of the species involved are: [EDTA4−] = x [Co2+] = 0.015333 + x [Co(EDTA)2–] = 0.016667 – x

Kf = 2.0417379x1016 = ( )2

2 4

Co EDTA

Co EDTA

−

+ −

= [ ][ ][ ]

0.016667 x0.01533 x x

−

+ =

[ ][ ][ ]

0.0166670.01533 x

x = 5.324947x10–17 [EDTA] = 5.3x10–17 M [Co2+] = 0.015 M

[Cl─]

AgC

l solubility

Ag+ AgCl2─

19-81

b) Moles of EDTA added = ( )4 30.050 mol EDTA 10 L 75.0 mL

L 1 mL

− −

= 0.00375 mol EDTA

The Co2+ is limiting so no Co2+ is left. The original moles Co2+ equals the moles of complex formed. Moles of EDTA remaining = (0.00375 – 0.0024) mol = 0.00135 mol EDTA Total volume = (50.0 + 75.0) mL (10−3 L/1 mL) = 0.1250 L [EDTA] = (0.00135 mol EDTA)/(0.1250 L) = 0.0108 M EDTA [Co(EDTA)2–] = (0.0024 mol [Co(EDTA)]2−)/(0.1250 L) = 0.0192 M To reach equilibrium the concentrations of the species involved are: [EDTA4–] = 0.0108 + x [Co2+] = x [Co(EDTA)2–] = 0.0192 – x

Kf = 2.0417379x1016 = ( )2

2 4

Co EDTA

Co EDTA

−

+ −

= [ ]

[ ][ ]0.0192 x

x 0.0108 x−

+ =

[ ][ ][ ]

0.0192x 0.0108

x = 8.7071792x10–17 [EDTA4–] = 00108 + x = 0.0108 M [Co2+] = 8.7x10–17 M

Documents

CHAPTER 19 IONIC EQUILIBRIA IN AQUEOUS SYSTEMS...will have more conjugate base particles than weak acid particles. The buffer in scene 3 has 6 conjugate base particles to 3 weak acid