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If the solubility product of $\mathrm{Ni}(\mathrm{OH})_2$ is $4.0 \times 10^{-15}$, the solubility (in $\mathrm{mol} \mathrm{L}^{-1}$ ) is
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Verified Answer
The correct answer is:
$1.0 \times 10^{-5}$
Given, solubility product $\left(K_{\mathrm{sp}}\right)$ of $\mathrm{Ni}(\mathrm{OH})_2$ is
$$
=4.0 \times 10^{-15}
$$
Solubility $(s)=$ ?
$$
\begin{aligned}
& \quad \mathrm{Ni}(\mathrm{OH})_2 \rightleftharpoons \mathrm{Ni}^{2+}+2 \mathrm{OH}^{-} \\
& {[\mathrm{s}] \mathrm{mol} / \mathrm{L} 25 \mathrm{~mol} / \mathrm{L} } \\
& K_{\text {sp }}=[s][2 s]^2 \\
& K_{\text {sp }}=4 s^3 \\
& s=3 \sqrt{\frac{K_{\text {sp }}}{4}}=3 \sqrt{\frac{4 \times 10^{-15}}{4}}=1 \times 10^{-5} \mathrm{~mol} / \mathrm{L}
\end{aligned}
$$
$$
=4.0 \times 10^{-15}
$$
Solubility $(s)=$ ?
$$
\begin{aligned}
& \quad \mathrm{Ni}(\mathrm{OH})_2 \rightleftharpoons \mathrm{Ni}^{2+}+2 \mathrm{OH}^{-} \\
& {[\mathrm{s}] \mathrm{mol} / \mathrm{L} 25 \mathrm{~mol} / \mathrm{L} } \\
& K_{\text {sp }}=[s][2 s]^2 \\
& K_{\text {sp }}=4 s^3 \\
& s=3 \sqrt{\frac{K_{\text {sp }}}{4}}=3 \sqrt{\frac{4 \times 10^{-15}}{4}}=1 \times 10^{-5} \mathrm{~mol} / \mathrm{L}
\end{aligned}
$$
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