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]$

$=1 \times 10^{-2}\, \mathrm{mol} / L$

$B a(O H)_{2} \rightarrow \frac{B a^{2+}}{S}+_{2 S}^{2 O H}$

$K_{n} p=\left[B a^{2+}\right]\left[O H^{-}\right]^{2}=[S][2 S]^{2}$

$=\left[\frac{1 \times 10^{-2}}{2}\right]\left(1 \times 10^{-2}\right)^{2}$

$=0.5 \times 10^{-6}=5.0 \times 10^{-6}\, M^{3}$