\(K_{sp} = [Mg^{+2}][OH^-]^2\)
\(1\times 10^{-12} = [0.01][OH^-]^2\)
\([OH^-]^2=1\times 10^{-10} \,\,\, [OH^-] = 10^{-5}\)
\(\therefore \,\,\,\,[{H^ + }]\,\, = \,\,\frac{{{{10}^{_{ - 14}}}}}{{{{10}^{ - 5}}}} = \,\,{10^{ - 9}}\)
\(pH= - log[H^+] = -log [10^{-9}] = 9\)