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Q.
If $K_{sp}$ of $Mg(OH)_2$ is $1.2 × 10^{-11}$. Then the highest pH of the $0.1\, M$ solution of $Mg^{2+}$ ion from which $Mg(OH)_2$ is not precipitated is
Equilibrium
Solution:
As $K_{sp}$ of $Mg(OH)_2 = [Mg^{2+}] [OH^-]^2$
$\therefore \left[OH^{-}\right]^{2}=\frac{1.2\times10^{-11}}{0.1}=1.2\times10^{-10}$
$\therefore \left[OH^{-}\right]^{2}=1.1\times10^{-5}$
$\therefore \left[H^{+}\right]=10^{-14}/1.1\times10^{-5}=0.91\times10^{-9}$
$pH=-log\left[H^{+}\right]=log\,0.91\times10^{-9}=9.04$