Question
A 0.1 M solution of [Cu(NH3)4]+ is stirred with an excess of potassium cyanide sufficient to convert all the ammonium complex to the corresponding cuprocyanide complex [Cu(CN)4]−3 and in addition to provide the solution with an excess of CN– equal to 0.2 M. Calculate the maximum pH of the solution when the final solution is treated with hydrogen sulphide to maintain [H2S] = 0.1 M and the precipitation of cuprous sulphide is prevented. The instability constant for [Cu(CN)4]−3 is 6.4 × 10−15, Ka,overall of H2S = 1.6 × 10−21, Ksp of Cu2S = 2.56 × 10−27.
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Solution
From the question, $\left\lbrack Cu(CN) \right\rbrack_{4}^{3 -} = 0.1\text{ M and }\left\lbrack CN^{-} \right\rbrack = 0.2\text{ M}$
$\therefore\left\lbrack Cu^{+} \right\rbrack = \frac{K_{\text{Instab}}.\left\lbrack Cu(CN)_{4}^{3 -} \right\rbrack}{\left\lbrack CN^{-} \right\rbrack^{4}} = \frac{6.4 \times 10^{- 15} \times 0.1}{(0.2)^{4}} = 4 \times 10^{- 13}\text{ M}$
Now, $\left\lbrack S^{2 -} \right\rbrack = \frac{K_{sp}\left( Cu_{2}S \right)}{\left\lbrack Cu^{+} \right\rbrack^{2}} = \frac{2.56 \times 10^{- 27}}{\left( 4 \times 10^{- 13} \right)^{2}} = 1.6 \times 10^{- 2}\text{ M}$
$\therefore\left\lbrack H^{+} \right\rbrack = \sqrt{\frac{K_{a} \times \left\lbrack H_{2}S \right\rbrack}{\left\lbrack S^{2 -} \right\rbrack}} = \sqrt{\frac{1.6 \times 10^{- 21} \times 0.1}{1.6 \times 10^{- 2}}} = 10^{- 10}\text{ M and }P^{H} = 10.0$
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