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The ubility product of $\mathrm{Mg}(\mathrm{OH})_{2}$ is $1.0 \times 10^{-12}$. Concentrated aqueous $\mathrm{NaOH}$ ution is added to a $0.01 \mathrm{M}$ aqueous ution of $\mathrm{MgCl}_{2}$. The $\mathrm{pH}$ at which precipitation occur is-
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The correct answer is:
$9.0$
$\quad \mathrm{MgCl}_{2} \longrightarrow \mathrm{Mg}^{+2}+2 \mathrm{Cl}^{-}$
$$
\begin{array}{c}
\quad ? \\
0.01 \mathrm{M} \\
\mathrm{K}_{\mathrm{p}}=\mathrm{Q}=\left[\mathrm{Mg}^{+2}\right]\left[\mathrm{OH}^{-}\right]^{2} \\
10^{-12}=[0.01]\left[\mathrm{OH}^{-}\right]^{2} \\
{\left[\mathrm{OH}^{-}\right]^{2}=10^{-10}} \\
{\left[\mathrm{OH}^{-}\right]=10^{-5}} \\
\mathrm{pOH}=5 \quad \therefore \mathrm{pH}=9
\end{array}
$$
$$
\begin{array}{c}
\quad ? \\
0.01 \mathrm{M} \\
\mathrm{K}_{\mathrm{p}}=\mathrm{Q}=\left[\mathrm{Mg}^{+2}\right]\left[\mathrm{OH}^{-}\right]^{2} \\
10^{-12}=[0.01]\left[\mathrm{OH}^{-}\right]^{2} \\
{\left[\mathrm{OH}^{-}\right]^{2}=10^{-10}} \\
{\left[\mathrm{OH}^{-}\right]=10^{-5}} \\
\mathrm{pOH}=5 \quad \therefore \mathrm{pH}=9
\end{array}
$$
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