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Common Ion Effect

Calculate the...

Question

Calculate the pH at which $M g {(O H)}_{2}$ begins to precipitate from a solution containing $0.10 M$ ${M g}^{2 +}$ ions.
$K_{s p} M g {(O H)}_{2} = 1.0 \times 10^{- 11}$

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Solution

Equilibrium precipitate for magnesium:

$[M g^{2 +}] = 0.1 M$ $, K_{s p} = 1 \times 10^{- 11}$
$M g^{2 +} + 2 O H^{-} ⇌ M g {(O H)}_{2} K_{s p} = \frac{[M g^{2 +}] {[O H^{-}]}^{2}}{M g {(O H)}_{2}}$
$K_{s p} \times M g {(O H)}_{2} = 1.0 \times 10^{- 11} = [{M g}^{2 +}] {[{O H}^{-}]}^{2} = (0.10) {[{O H}^{-}]}^{2}$
$\Rightarrow {[{O H}^{-}]}^{2} = \frac{1.0 \times 10^{- 11}}{0.10} = 1 \times 10^{- 10}$
$\Rightarrow [{O H}^{-}] = 1 \times 10^{- 5}$
$[H^{+}] = \frac{1 \times 10^{- 14}}{1 \times 10^{- 5}} = 1 \times 10^{- 9}$
$p H = - log [H^{+}] = - log (1 \times 10^{- 9}) = 9.0$
$p H = 9$

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