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$\mathrm{Mg}(\mathrm{OH})_{2}$ is precipitated when $\mathrm{NaOH}$ is added to a solution of $\mathrm{Mg}^{2+}$. If the final concentration of $\mathrm{Mg}^{2+}$ is $10^{-10} \mathrm{M}$, the concentration of $\mathrm{OH}^{-}(\mathrm{M})$ in the solution is
[Solubility product for $\left.\mathrm{Mg}(\mathrm{OH})_{2}=5.6 \times 10^{-12}\right]$
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[Solubility product for $\left.\mathrm{Mg}(\mathrm{OH})_{2}=5.6 \times 10^{-12}\right]$
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Verified Answer
The correct answer is:
$0.24$
$\mathrm{K}_{\text {sp }} \mathrm{Mg}(\mathrm{OH})_{2}=\left[\mathrm{Mg}^{+2}\right]\left[\mathrm{OH}^{-}\right]^{2}$
$5.6 \times 10^{-12}=\left[10^{-10}\right]\left[\mathrm{OH}^{-}\right]^{2}$
$\left[\mathrm{OH}^{-}\right]=\sqrt{5.6 \times 10^{-2}}=0.24 \mathrm{M}$
$5.6 \times 10^{-12}=\left[10^{-10}\right]\left[\mathrm{OH}^{-}\right]^{2}$
$\left[\mathrm{OH}^{-}\right]=\sqrt{5.6 \times 10^{-2}}=0.24 \mathrm{M}$
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