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

The correct option is C $5\times {10}^{-7}$

$pH\nless pOH=14$

$pOH=14-12=2$ (givem $pH=12)$

$pOH=-log\left(OH\right)$

$\left(O{H}^{-}\right)=10-pOH={10}^{-2}$ .... (1)

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

At equilibrium $Ba+2=x$

and $OH=2x$

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

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

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

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

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

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