pH of saturated solution of $\mathrm{Ba}(\mathrm{OH}{)}_2$ is 12.The value of solubility product of ${\mathrm{K}}_{\mathrm{sp}}\text{ of }\mathrm{Ba}(\mathrm{OH}{)}_2$ is

Given $\mathrm{pH}\text{ of }\mathrm{Ba}(\mathrm{OH}{)}_2=12$

$\mathrm{pOH}=14-\mathrm{pH}=14-12=2$

We know that, $\mathrm{pOH}=-\log \left[{\mathrm{OH}}^{-}\right]$

$2=-\log \left[{\mathrm{OH}}^{-}\right]$

$\left[{\mathrm{OH}}^{-}\right]={10}^{-2}$

$\mathrm{Ba}(\mathrm{OH}{)}_2\text{ dissolve in water as }$

$\mathrm{Ba}\left(\mathrm{OH}{)}_2\right(\mathrm{s})\rightleftharpoons {\mathrm{Ba}}^{2+}+2{\mathrm{OH}}^{-}$

$\mathrm{S} \mathrm{mol}{ \mathrm{L}}^{-1} \mathrm{S} 2\mathrm{S}$

$\left[{\mathrm{OH}}^{-}\right]=2\mathrm{s}={10}^{-2}$

${\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ba}}^{2+}\right]{\left[{\mathrm{OH}}^{-}\right]}^2=\frac{\left({10}^{-2}\right)}{2}\times {\left({10}^{-2}\right)}^2$

$=0.5\times {10}^{-6}=5\times {10}^{-7}$