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Coagulation

At 25oC, th...

Question

At $25^{o} C$ , the solubility product of $M g (O H)_{2}$ is $1.0 \times 10^{- 11}$ . At which $p H$ , will $M g^{2 +}$ ions start precipitating in the form of $M g (O H)_{2}$ from a solution of 0.001 M $M g^{2 +}$ ions?

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Solution

The correct option is D 10

$K_{s p} = [M g^{2 +}] [O H^{-}]^{2}$

$[O H^{-}] = \sqrt{\frac{K_{s p}}{[M g^{2 +}]}} = 10^{- 4}$

$p O H = 4$ and $p H = 10$

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25^{\circ} C

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