$K_{sp} = [Mg^{+2}][OH^-]^2$
$1\times 10^{-12} = [0.01][OH^-]^2$
$[OH^-]^2=1\times 10^{-10} \,\,\, [OH^-] = 10^{-5}$
$\therefore \,\,\,\,[{H^ + }]\,\, = \,\,\frac{{{{10}^{_{ - 14}}}}}{{{{10}^{ - 5}}}} = \,\,{10^{ - 9}}$
$pH= - log[H^+] = -log [10^{-9}] = 9$
${H_2}{O_2}\, + \,{H_2}O\, \rightleftharpoons \,{H_3}{O^ + }\, + \,HO_2^ - $