pH of a saturated solution of Ba(OH)_2 is 12. The value of...

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