${K_{sp}} = [M{g^{ + 2 }}]{[O{H^ - }]^2}$
$1.0 \times {10^{ - 11}} = {10^{ - 3}} \times {[O{H^ - }]^2}$
$[O{H^ - }] = \sqrt {\frac{{{{10}^{ - 11}}}}{{{{10}^{ - 3}}}}} = {10^{ - 4}}$
$\therefore \,pOH = 4$
$\therefore pH + pOH = 14$
$\therefore \,pH = 10$
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