The solubility product of Ni(OH)_2 at 298 K | vedclass

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Question

(C) The dissociation of $Ni(OH)_2$ is given by: $Ni(OH)_2 \rightleftharpoons Ni^{2+} + 2OH^-$Let the solubility be $S \ mol \ L^{-1}$. Then $[Ni^{2+}]

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$9$

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(C) The dissociation of $Ni(OH)_2$ is given by: $Ni(OH)_2 ightleftharpoons Ni^{2+} + 2OH^-$Let the solubility be $S \ mol \ L^{-1}$. Then $[Ni^{2+}] = S$ and $[OH^-] = 2S$.The solubility product expression is: $K_{sp} = [Ni^{2+}][OH^-]^2 = S(2S)^2 = 4S^3$.Given $K_{sp} = 2 	imes 10^{-15}$,we have $4S^3 = 2 	imes 10^{-15}$,so $S^3 = 0.5 	imes 10^{-15} = 5 	imes 10^{-16}$.$S = (5 	imes 10^{-16})^{1/3} \approx 7.937 	imes 10^{-6} \ mol \ L^{-1}$.$[OH^-] = 2S = 2 	imes 7.937 	imes 10^{-6} = 1.587 	imes 10^{-5} \ mol \ L^{-1}$.$pOH = -\log[OH^-] = -\log(1.587 	imes 10^{-5}) \approx 5 - 0.2 = 4.8$.Since $pH + pOH = 14$,$pH = 14 - 4.8 = 9.2 \approx 9$.

The solubility product of $Ni(OH)_2$ at $298 \ K$ is $2 \times 10^{-15} \ mol^3 \ dm^{-9}$. The $pH$ value of its aqueous and saturated solution is

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