Question

The solubility product of ${Mg\left(OH\right)}_2$ is $5\times {10}^{-19}$ at ${25}^{o}C$, pH of saturated solution of ${Mg\left(OH\right)}_2$ will be:

Answer 1

$Mg(OH{)}_2\rightleftharpoons M{g}^{2+}+2O{H}^{-}$

$S$ $2S$

$K_{sp}=\left[M{g}^{2+}\right][O{H}^{-}{]}^2=(S\left)\right(2S{)}^2=4S^3=5\times {10}^{-19}$

So $S=5\times {10}^{-7}$ means $\left[O{H}^{-}\right]=2S={10}^{-6}$

So $pOH=-\log \left(\right[O{H}^{-}\left]\right)=6$

As we know $pH+pOH=14$

So $pH=8$