Chemistry - Ionic Equilibrium Question with Solution | TestHub

Given, $\mathrm{pH}$ of $\mathrm{Ba}(\mathrm{OH}{)}_2=12$ $$\begin{array}{rl}\therefore \left[{\mathrm{H}}^{+}\right]& =\left[1\times 10^{-12}\right]\\ \text{ and }\left[{\mathrm{OH}}^{-}\right]& =\frac{1\times 10^{-14}}{1\times 10^{-12}}\left[\because \left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]=1\times 10^{-14}\right]\end{array}$$ $$\begin{array}{rl}& =1\times 10^{-2}\mathrm{mol}/\mathrm{L}\\ & \mathrm{Ba}(\mathrm{OH}{)}_2⟶{\mathrm{Ba}}^{2+}+2{\mathrm{OH}}^{-}\\ & {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ba}}^{2+}\right]{\left[{\mathrm{OH}}^{-}\right]}^2\\ & =[\mathrm{s}][2\mathrm{s}{]}^2\\ & =\left[\frac{1\times 10^{-2}}{2}\right]{\left(1\times 10^{-2}\right)}^2\\ & =0.5\times 10^{-6}=5.0\times 10^{-7}{\mathrm{M}}^3\end{array}$$