The $\mathrm{pH}$ of saturated aqueous solution of $\mathrm{Ba}(\mathrm{OH}{)}_2$ is 10 . If the ${\mathrm{K}}_{\text{sp }}$ of $\mathrm{Ba}(\mathrm{OH}{)}_2$ is $5\times {10}^{-13}$, then the concentration of ${\mathrm{Ba}}^{2+}$ ions in the solution is

 $\mathrm{pH}+\mathrm{pOH}=14$

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

$\mathrm{pOH}=-\log \left(\mathrm{OH}\right)$ $\left({\mathrm{OH}}^{-}\right)=10-\mathrm{pOH}={10}^{-2}\dots .$. (1)

$\mathrm{Ba}(\mathrm{OH}{)}_2\to {\mathrm{Ba}}^{+2}+2{\mathrm{OH}}^{-}$

$\mathrm{At}$ equilibrium ${\mathrm{Ba}}^{+2}=\mathrm{x}$

and $\mathrm{OH}=2\mathrm{x}$

Since $2\mathrm{x}={10}^{-2}$ as eq (1)

$\therefore \mathrm{x}=\frac{{10}^{-2}}{2}=0.5\times {10}^{-2}$

$\mathrm{Ksp}=\left[{\mathrm{Ba}}^{+2}\right]\times {\left[{\mathrm{OH}}^{-}\right]}^2$

$=\left[0.5\times {10}^{-2}\right]{\left[{10}^{-2}\right]}^2$

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

$=5\times {10}^{-5}$

Hence the correct option is (B) $5\times {10}^{-5}$.