d
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 \(\left[O H^{-}\right]=\frac{1 \times 10^{-14}}{1 \times 10^{-19}}\)\(\left[\therefore\left[H^{+}\right]\left[O H^{-}\right]=1 \times 10^{-14}\right]\)

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

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

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

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

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