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Common Ion Effect

What is the m...

Question

What is the maximum pH of a solution $0.10 M$ in ${M g}^{2 +}$ from which $M g {(O H)}_{2}$ will nor precipitate? Given, $K_{s p} [M g {(O H)}_{2}] = 1.2 \times 10^{- 11}$ .

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Solution

It will not precipitate when $Q_{s p} < K_{s p}$ .
$[M g^{2 +}] = 0.1 M$
$M g (O H)_{2} ⇌ M g^{2 +} + 2 O H^{-}$
$K_{s p} = [M g^{2 +}] [O H^{-}]^{2}$
$1.2 \times 10^{- 11} = (0.1) [O H^{-}]^{2}$
$[O H^{-}] = 1.09 \times 10^{- 5}$
$p O H = - l o g ([O H^{-}]) = 4.96$
As we know $p H + p O H = 14$
$p H = 14 - 4.96 = 9.04$

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