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If $\mathrm{pH}$ of a saturated solution of $\mathrm{Ba}(\mathrm{OH})_2$ is 12 , the value of its $K_{s p}$ is
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Verified Answer
The correct answer is:
$5.00 \times 10^{-7} \mathrm{M}^3$
Given, $\mathrm{pH}$ of $\mathrm{Ba}(\mathrm{OH})_2=12$
$$
\begin{aligned}
\therefore \quad\left[\mathrm{H}^{+}\right] & =\left[1 \times 10^{-12}\right] \\
\text { and }\left[\mathrm{OH}^{-}\right] & =\frac{1 \times 10^{-14}}{1 \times 10^{-12}}\left[\because\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=1 \times 10^{-14}\right]
\end{aligned}
$$
$$
\begin{aligned}
&=1 \times 10^{-2} \mathrm{~mol} / \mathrm{L} \\
& \mathrm{Ba}(\mathrm{OH})_2 \longrightarrow \mathrm{Ba}^{2+}+2 \mathrm{OH}^{-} \\
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2 \\
&=[\mathrm{s}][2 \mathrm{~s}]^2 \\
&=\left[\frac{1 \times 10^{-2}}{2}\right]\left(1 \times 10^{-2}\right)^2 \\
&=0.5 \times 10^{-6}=5.0 \times 10^{-7} \mathrm{M}^3
\end{aligned}
$$
$$
\begin{aligned}
\therefore \quad\left[\mathrm{H}^{+}\right] & =\left[1 \times 10^{-12}\right] \\
\text { and }\left[\mathrm{OH}^{-}\right] & =\frac{1 \times 10^{-14}}{1 \times 10^{-12}}\left[\because\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=1 \times 10^{-14}\right]
\end{aligned}
$$
$$
\begin{aligned}
&=1 \times 10^{-2} \mathrm{~mol} / \mathrm{L} \\
& \mathrm{Ba}(\mathrm{OH})_2 \longrightarrow \mathrm{Ba}^{2+}+2 \mathrm{OH}^{-} \\
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2 \\
&=[\mathrm{s}][2 \mathrm{~s}]^2 \\
&=\left[\frac{1 \times 10^{-2}}{2}\right]\left(1 \times 10^{-2}\right)^2 \\
&=0.5 \times 10^{-6}=5.0 \times 10^{-7} \mathrm{M}^3
\end{aligned}
$$
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