Calculate the molar solubility of Ni(OH)_{2} | vedclass

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Question

(A) Let the molar solubility of $Ni(OH)_{2}$ be $S \, mol/L$.The dissolution reaction is: $Ni(OH)_{2}(s) \rightleftharpoons Ni^{2+}(aq) + 2OH^{-}(aq)$

Vedclass · Accepted Answer

$2.0 	imes 10^{-13} \, M$

Vedclass · Answer

(A) Let the molar solubility of $Ni(OH)_{2}$ be $S \, mol/L$.The dissolution reaction is: $Ni(OH)_{2}(s) ightleftharpoons Ni^{2+}(aq) + 2OH^{-}(aq)$.From the reaction,$Ni^{2+}$ concentration is $S$ and $OH^{-}$ concentration from $Ni(OH)_{2}$ is $2S$.Since $NaOH$ is a strong electrolyte,it provides $0.10 \, M$ $OH^{-}$ ions.Total $[OH^{-}] = (0.10 + 2S) \, M \approx 0.10 \, M$ (since $S$ is very small).The solubility product expression is $K_{sp} = [Ni^{2+}][OH^{-}]^{2}$.Substituting the values: $2.0 	imes 10^{-15} = (S)(0.10)^{2}$.$S = \frac{2.0 	imes 10^{-15}}{0.01} = 2.0 	imes 10^{-13} \, M$.

Calculate the molar solubility of $Ni(OH)_{2}$ in $0.10 \, M$ $NaOH$. The solubility product constant $(K_{sp})$ of $Ni(OH)_{2}$ is $2.0 \times 10^{-15}$.

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