If the solubility product of Ni(OH)_2 is 4.0 | vedclass

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(D) The dissociation of $Ni(OH)_2$ is given by: $Ni(OH)_2(s) \rightleftharpoons Ni^{2+}(aq) + 2OH^{-}(aq)$ Let the solubility be $s \ mol \ L^{-1}$. T

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$1.0 	imes 10^{-5}$

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(D) The dissociation of $Ni(OH)_2$ is given by: $Ni(OH)_2(s) ightleftharpoons Ni^{2+}(aq) + 2OH^{-}(aq)$ Let the solubility be $s \ mol \ L^{-1}$. Then, $[Ni^{2+}] = s$ and $[OH^{-}] = 2s$. The solubility product expression is: $K_{sp} = [Ni^{2+}][OH^{-}]^2$ $K_{sp} = (s)(2s)^2 = 4s^3$ Given $K_{sp} = 4.0 	imes 10^{-15}$. $4s^3 = 4.0 	imes 10^{-15}$ $s^3 = 1.0 	imes 10^{-15}$ $s = \sqrt[3]{1.0 	imes 10^{-15}} = 1.0 	imes 10^{-5} \ mol \ L^{-1}$

If the solubility product of $Ni(OH)_2$ is $4.0 \times 10^{-15}$, the solubility (in $mol \ L^{-1}$) is

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