Question 22.10: Use the equilibrium-constant data for zinc hydroxide, Zn(OH)......
Use the equilibrium-constant data for zinc hydroxide, Zn(OH)_{2}(s), in Table 22.3 to calculate its solubility in a solution buffered at pH = 7.0, 10.0, and 14.0 at 25°C.
| TABLE 22.3 Equilibrium constants for the formation of metal hydroxy complex ions in water at 25°C | |
| Reaction equation | K_{f} |
| Al(OH)_{3}(s) + OH^−(aq) ⇋ [Al(OH)_{4}]^−(aq) | 40 |
| Pb(OH)_{2}(s) + OH^−(aq) ⇋ [Pb(OH)_{3}]^−(aq) | 0.08 |
| Zn(OH)_{2}(s) + 2\ OH^−(aq) ⇋ [Zn(OH)_{4}]^{2−}(aq) | 0.05\ M^{–1} |
| Cr(OH)_{3}(s) + OH^−(aq) ⇋ [Cr(OH)_{4}]^−(aq) | 0.04 |
| Sn(OH)_{2}(s) + OH^−(aq) ⇋ [Sn(OH)_{3}]^−(aq) | 0.01 |
The two relevant equilibrium expressions are
Zn(OH)_{2}(s) ⇋ Zn^{2+}(aq) + 2\ OH^−(aq)and
Zn(OH)_{2}(s) + 2\ OH^−(aq) ⇋ [Zn(OH)_{4}]^{2−}(aq)and the corresponding equilibrium-constant expressions are
K_{sp} = [Zn^{2+}][OH^−]^2 = 1.0 × 10^{−15}\ M^3and
K_{f} =\frac{[Zn(OH)_{4}^{2–}]}{[OH^–]^{2}} = 0.05\ M^{–1}The solubility of Zn(OH)_{2}(s) at any pH is given by
s = [Zn^{2+}] + [Zn(OH)_{4}^{2–}]At pH = 7.0, [OH^-] = 1.0 × 10^{-7}\ M, so
[Zn^{2+}] =\frac{1.0 × 10^{–15}\ M^3}{(1.0 × 10^{–7} M)^2} =0.10\ Mand
[Zn(OH)_{4}^{2–}] = (0.05\ M^{−1})(1.0 × 10^{−7}\ M)^2 = 5 × 10^{−16}\ MThe total solubility of Zn(OH)_{2}(s) is given by
s = [Zn^{2+}] + [Zn(OH)_{4}^{2–}] = 0.10\ M + 5 × 10^{−16}\ M = 0.10\ MAt pH = 10.0, [OH^-] = 1.0 × 10^{-4}\ M, so
[Zn^{2+}] =\frac{1.0 × 10^{–15}\ M^3}{(1.0 × 10^{–4}\ M)^2} =1.0 × 10^{−7}\ Mand
[Zn(OH)_{4}^{2–}] = (0.05\ M^{−1})(1.0 × 10^{−4}\ M)^2 = 5 × 10^{−10}\ MThus,
s = [Zn^{2+}] + [Zn(OH)_{4}^{2–}] =1.0 × 10^{−7}\ M + 5 × 10^{−10}\ M = 1.0 × 10^{−7}\ MAt pH = 14.0, [OH^-] = 1.0\ M, so
[Zn^{2+}] =\frac{1.0 × 10^{–15}\ M^3}{(1.0\ M)^2} =1.0 × 10^{−15}\ Mand
[Zn(OH)_{4}^{2–}] = (0.05\ M^{−1})(1.0\ M)^2 = 0.05\ MThus,
s = [Zn^{2+}] + [Zn(OH)_{4}^{2–}] =1.0 × 10^{−15}\ M + 0.05\ M = 0.05\ MNotice that the solubility of Zn(OH)_{2}(s) increases with increasing pH, but not as dramatically as in the case of Al(OH)_{3}(s), because the value of K_{f} for the complexation reaction (Table 22.3) is not as large as that of Al(OH)_{3}(s).