Saturated aqueous Mg(OH)2 has K_sp = 4.0 x 10^-12 at 298 K. Taking log 2 = 0.301, its pH is

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Saturated aqueous $Mg (OH) X_{2}$ has $K_{s p} = 4.0 \times 1 0^{- 12}$ at 298 K. Taking $lo g 2 = 0.301$ , its pH is

K_{s p} = 4 s^{3}

gives

s = 1.0 \times 1 0^{- 4} M

, so

[OH X^{-}] = 2 s = 2.0 \times 1 0^{- 4} M

. pOH = 4 - 0.301 = 3.699, pH = 14 - 3.699 = 10.30.