The solubility product of Mg(OH)2 is 4.0x 10^-12 at 298 K. The pH of its saturated aqueous solution is

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The solubility product of $Mg (OH) X_{2}$ is $4.0 \times 1 0^{- 12}$ at 298 K. The pH of its saturated aqueous solution is

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

, so

s = (4.0 \times 1 0^{- 12} /4)^{1/3} = 1.00 e - 04 M

[OH X^{-}] = 2 s = 2.00 e - 04 M

pOH = 3.70

pH = 10.30