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Solubility Product

What is the p...

Question

What is the pH at which $M g (O H)_{2}$ begins to precipitate from a solution containing $0.1$ M $M g^{2 +}$ ions? $[K_{s p} f o r M g (O H)_{2} = 1.0 \times 10^{- 11}]$

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Solution

The correct option is C $9$
$K_{s p}$ for $M g (O H)_{2}$ = 1.0 $\times 10^{- 11}$
Therefore, $[M g^{2 +}] [O H^{-}]^{2} = (0.1) \times [O H^{-}]^{2} = 1.0 \times 10^{- 11}$
or, $[O H^{-}]^{2} = \frac{1 \times 10^{- 11}}{0.1} = 10^{- 10}$
or, $[O H^{-}] = 10^{- 5}$
$\to [H^{+}] = \frac{10^{- 14}}{10^{- 5}} = 10^{- 9}$
or, pH = $9$

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