Question
What is the solubility of solid zinc hydroxide at a pH of 13? Given that
Zn(OH)2(s) $\rightleftharpoons$ Zn(OH)2(aq): K1 = 10–6 M
Zn(OH)2(aq) $\rightleftharpoons$ Zn(OH)+ (aq) + OH– (aq): K2 = 10–7 M
Zn(OH)+ (aq) $\rightleftharpoons$ Zn2+ (aq) + OH– (aq): K3 =10–4 M
Zn(OH)2(aq) + OH– (aq) $\rightleftharpoons$ Zn(OH)3− (aq): K4 = 103 M–1
Zn(OH)3− (aq) + OH– (aq) $\rightleftharpoons$Zn(OH)42− (aq): K5 = 10 M–1
Options
Solution
$S = \left\lbrack Zn(OH)_{2}(aq) \right\rbrack + \left\lbrack Zn(OH)^{+} \right\rbrack + \left\lbrack Zn^{2 +} \right\rbrack + \left\lbrack Zn(OH)_{3}^{-} \right\rbrack + \left\lbrack Zn(OH)_{4}^{2 -} \right\rbrack$
$= K_{1} + \frac{K_{2}.K_{1}}{\left\lbrack OH^{-} \right\rbrack} + \frac{K_{3}.K_{2}.K_{1}}{\left\lbrack OH^{-} \right\rbrack^{2}} + K_{4}K_{1}\left\lbrack OH^{-} \right\rbrack + K_{5}K_{4}K_{1}\left\lbrack OH^{-} \right\rbrack^{2}$
$= 10^{- 6} + \frac{10^{- 7} \times 10^{- 6}}{0.1} + \frac{10^{- 4} \times 10^{- 7} \times 10^{- 6}}{(0.1)^{2}} + 10^{3} \times 10^{- 6} \times 0.1 + 10 \times 10^{3} \times 10^{- 6} \times (0.1)^{2} $$$= 10^{- 6} + 10^{- 12} + 10^{- 15} + 10^{- 4} \approx 2 \times 10^{- 4}\text{ M}$$
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