[14%] When water is bound to a metal ion, its acidity increases. For example, the pKa of Zn2+(aq) is — Aqueous Equilibria Chemistry Question
Problem Context
[14%] When water is bound to a metal ion, its acidity increases. For example, the pKa of Zn2+(aq) is 8.96.
Calculate the pH of a 0.010 M solution of zinc nitrate, Zn(NO3)2.
Model Answer
The major reaction is
Zn(OH2)2+(aq) + H2O(l) <=> Zn(OH+(aq) + H3O+(aq)
Keq = 10–8.96 = 1.10 × 10-9 = [Zn(OH+][H3O+] / [Zn(OH2)2+]
Since [Zn(OH+] and [H3O+] are approximately equal, and [Zn(OH2)2+] ≈ 0.010 (since the reaction is quite unfavorable), then
[H3O+]2 = 1.10 × 10-11
[H3O+] = 3.31 × 10-6 M
pH = –log10[H3O+] = 5.48
This pH is much lower than that of pure water, so the assumption of [Zn(OH+] ≈ [H3O+] is justified.
Calculate the pH of a 0.010 M solution of zinc acetate, Zn(CH3COO)2. The pKa of CH3COOH is 4.75.
Model Answer
Now the major reaction is
Zn(OH2)2+(aq) + CH3COO–(aq) <=> Zn(OH+(aq) + CH3COOH(aq)
Keq = 10-8.96 / 10-4.75 = 6.17 × 10-5 = [Zn(OH+][CH3COOH] / ([Zn(OH2)2+][CH3COO–])
Similarly to above, [Zn(OH+] ≈ [CH3COOH]. Because the reaction is again rather unfavorable, [Zn(OH2)2+] and [CH3COO–] will be little changed from the initial solution, so [Zn(OH2)2+] = 0.5[CH3COO–]. Thus:
6.17 × 10-5 = [CH3COOH]2 / (0.5[CH3COO–]2)
3.08 × 10-5 = [CH3COOH]2 / [CH3COO–]2
5.55 × 10-3 = [CH3COOH] / [CH3COO–]
The ratio of acetate ion to acetic acid determines the pH:
pH = pKa + log([CH3COO–]/[CH3COOH])
pH = 4.75 + log(180)
pH = 7.01
Note that the initial concentration of zinc acetate was never used in this calculation! So as long as it is high enough to justify the assumptions made, the pH is the same.
Zinc hydroxide is sparingly soluble, with Ksp = 4.5 × 10-17. Calculate the pH of a solution of water saturated with solid Zn(OH)2.
Model Answer
[Zn2+] ≈ 0.5[OH–] in this solution, so:
Ksp = 4.5 × 10-17 = [Zn2+][OH–]2 = 0.5[OH–]3
[OH–] = 4.48 × 10-6 M
pOH = –log[OH–] = 5.35
pH = 14 – pOH = 8.65
Under what circumstances, if any, will a solution of zinc acetate spontaneously form a precipitate of Zn(OH)2? If precipitation is possible, specify the circumstances under which it is spontaneous. If it is not possible, justify why not.
Model Answer
From part b, the pH of the solution is 7.01 (independent of the amount of zinc acetate), so [OH–] = 1.02 × 10-7. Precipitation of Zn(OH)2 is spontaneous if Qsp > Ksp, so:
[Zn2+][OH–]2 > 4.5 × 10-17
[Zn2+][1.02 × 10-7]2 > 4.5 × 10-17
[Zn2+] > 4.3 × 10-3 M
Thus precipitation is spontaneous if the concentration of zinc acetate is greater than 0.0043 M.
Zinc also forms a complex ion, Zn(OH)4 2-, with Kf = 5.0 × 1014. Calculate the solubility of Zn(OH)2 in a solution with pH = 12.00.
Model Answer
At pH = 12.00, [OH–] = 0.010 M. From the solubility equilibrium:
[Zn2+][OH–]2 = [Zn2+][0.010]2 = 4.5 × 10-17
[Zn2+] = 4.5 × 10-13
From the complex ion equilibrium:
Kf = 5.0 × 1014 = [Zn(OH)4 2-] / ([Zn2+][OH–]4)
[Zn(OH)4 2-] = (5.0 × 1014)[4.5 × 10-13][0.010]4
[Zn(OH)4 2-] = 2.25 × 10-6
Obviously, almost all the zinc in solution is in the form of the complex ion, so 2.3 × 10-6 moles of Zn(OH)2 will dissolve per liter of pH 12 solution.