The maximum pH of a solution containing $0.10\text{M}$ $M{g}^{2+}$ from which, $Mg{\left(OH\right)}_2$ is not precipitated is:
Given: $K_{sp}$ for $Mg(OH{)}_2=1.2\times {10}^{-11}{M}^3$

The correct option is D $9.04$
The expression for the solubility product is given below:

$K_{sp}=\left[M{g}^{2+}\right][O{H}^{-}{]}^2$

Substituting values in the above expression, we get

$1.2\times {10}^{-11}=0.10\times [O{H}^{-}{]}^2$

$[O{H}^{-}{]}^2=1.2\times {10}^{-10}$

$\left[O{H}^{-}\right]=\sqrt{1.2\times {10}^{-10}}=1.09\times {10}^{-5}$

$pOH=-log\left[O{H}^{-}\right]=-log(1.09\times {10}^{-5})=4.96$

$pH=14-pOH=14-4.96=9.04$

Hence, the maximum pH of the solution should be $9.04$.