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Solubility

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Question

How do you calculate the solubility (moles/L) of $M g (O H)_{2}$ in water. The $K_{s p} = 5.0 \times 10^{- 11}$ ?

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Solution

$M g (O H)_{2} (s) ⇌ M g^{2 +} + 2 O H^{-}$
$s$ $2 s$
And $K_{s p} = [M g^{2 +}] [^{-} O H]^{2}$

Solubility of $M g (O H)_{2}$ , then by definition, $S = [M g^{2 +}]$ , and $2 S = [O H^{-}]$

And thus $K_{s p} = S \times (2 S)^{2} = 4 S^{3}$
$S = \sqrt[3]{\frac{K_{s p}}{4}}$
$S = \sqrt[3]{\frac{5.0 \times 10^{- 11}}{4}} = 2.32 \times 10^{- 4} ˙ m o l ˙ L^{- 1}$

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