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If $pH$ of a saturated solution of $Ba(OH)_2$ is $12,$ the value of its $K_{sp}$ is
$4.00 \times 10^{-6} \, M^3$
$4.00 \times 10^{-7} \, M^3$
$5.00 \times 10^{-7} \,M^3$
$5.00 \times 10^{-6} \, M^3$
Solution
Given, $pH$ of $B a(O H)_{2}=12$
so, $\quad p O H=2$
$\therefore \quad\left[H^{+}\right]=\left[1 \times 10^{-12}\right]$
${K_{w}=\left(H^{+}\right)\left(O H^{-}\right)}$
${K_{w}=1 \times 10^{-14}}$
${O H^{-}=\frac{K_{w}}{H^{+}}}$
and $\left[O H^{-}\right]=\frac{1 \times 10^{-14}}{1 \times 10^{-19}}$$\left[\therefore\left[H^{+}\right]\left[O H^{-}\right]=1 \times 10^{-14}\right]$
$=1 \times 10^{-2}\, \mathrm{mol} / L$
$B a(O H)_{2} \rightarrow \frac{B a^{2+}}{S}+_{2 S}^{2 O H}$
$K_{n} p=\left[B a^{2+}\right]\left[O H^{-}\right]^{2}=[S][2 S]^{2}$
$=\left[\frac{1 \times 10^{-2}}{2}\right]\left(1 \times 10^{-2}\right)^{2}$
$=0.5 \times 10^{-6}=5.0 \times 10^{-6}\, M^{3}$
