Chemistry - Ionic Equilibrium Question with Solution | TestHub

Given, $\mathrm{pH}$ of $\mathrm{Ba}(\mathrm{OH}{)}_2=12$ $\begin{array}{rl}\therefore \mathrm{pOH}& =14-\mathrm{pH}\\ & =14-12=2\end{array}$ We know that, $\begin{array}{rl}\mathrm{pOH}& =-\log \left[{\mathrm{OH}}^{-}\right]\\ 2& =-\log \left[{\mathrm{OH}}^{-}\right]\\ \left[{\mathrm{OH}}^{-}\right]& =\mathrm{antilog}(-2)\\ \left[{\mathrm{OH}}^{-}\right]& =1\times 10^{-2}\end{array}$ $\mathrm{Ba}(\mathrm{OH}{)}_2$ dissolves in water as $\begin{array}{rl}& \mathrm{Ba}(\mathrm{OH}{)}_2(s)\rightleftharpoons {\mathrm{Ba}}_s^{2+}+2{\mathrm{OH}}_2^{-}\\ & s\mathrm{mol}{\mathrm{L}}^{-1}\\ & \therefore \left[{\mathrm{OH}}^{-}\right]=2s=1\times 10^{-2}\\ & \left[{\mathrm{Ba}}^{2+}\right]=\frac{\left[{\mathrm{OH}}^{-}\right]}{2}=\frac{1\times 10^{-2}}{2}\\ & K_{\text{sp }}=\left[{\mathrm{Ba}}^{2+}\right]{\left[{\mathrm{OH}}^{-}\right]}^2\end{array}$ $\begin{array}{rl}& =\left(\frac{1\times 10^{-2}}{2}\right){\left(1\times 10^{-2}\right)}^2\\ & =0.5\times 10^{-6}=5\times 10^{-7}\end{array}$