What is the pH of a saturated solution of Cu( | vedclass

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Question

(B) The solubility equilibrium is $Cu(OH)_2(s) \rightleftharpoons Cu^{2+}(aq) + 2OH^-(aq)$.Let the solubility be $S$. Then $[Cu^{2+}] = S$ and $[OH^-]

Vedclass · Accepted Answer

$12.3$

Vedclass · Answer

(B) The solubility equilibrium is $Cu(OH)_2(s) ightleftharpoons Cu^{2+}(aq) + 2OH^-(aq)$.Let the solubility be $S$. Then $[Cu^{2+}] = S$ and $[OH^-] = 2S$.The solubility product expression is $K_{sp} = [Cu^{2+}][OH^-]^2 = S(2S)^2 = 4S^3$.Given $K_{sp} = 4.0 	imes 10^{-6}$,we have $4S^3 = 4.0 	imes 10^{-6}$,so $S^3 = 10^{-6}$,which gives $S = 10^{-2} \ M$.Therefore,$[OH^-] = 2S = 2 	imes 10^{-2} \ M$.$pOH = -\log[OH^-] = -\log(2 	imes 10^{-2}) = 2 - \log(2) = 2 - 0.301 = 1.699 \approx 1.7$.Finally,$pH = 14 - pOH = 14 - 1.7 = 12.3$.

What is the $pH$ of a saturated solution of $Cu(OH)_2$? $(K_{sp} = 4.0 \times 10^{-6})$

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