What is the maximum pH of a 0.1M $M{g}^{2+}$, the solution from which Mg $(OH{)}_2$ will not be precipitated?

$[K_{sp}=1.2\times {10}^{-11}]$

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

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

Ksp=([Mg2+][OH−]2K_{sp} = (\left [ Mg^{2+} \right ]\left [OH^- \right ]^2

Therefore

$\left[M{g}^{2+}\right]{\left[O{H}^{-}\right]}^2=1.2\times {10}^{-11}$

Substituting values in the above expression, we get

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

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

${\left[O{H}^{-}\right]}^2=\sqrt{1.2\times {10}^{-10}}=1.09\times {10}^{-5}$\

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

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

1.2×1011=0.10×[OH−]21.2\times10^{11}=0.10\times\left[OH^-\right]^2