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Q.

The solubility of $M g {(O H)}_{2}$ in a buffer of $p H = 10$ is found to be 0.0232 gm/lit. Molar solubility of $M g {(O H)}_{2}$ in pure water would be (in mol/1.it)

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a

$10^{- 5}$

b

$3.2 \times 10^{- 8}$

c

$3.2 \times 10^{- 5}$

d

$10^{- 4}$

answer is D.

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Detailed Solution

$[{OH}^{-}] = 10^{- 4}$

$K_{s p} = (\frac{0.0232}{58}) \times 10^{- 8}$

$= 4 \times 10^{- 12}$

Now, $4 s^{3} = 4 \times 10^{- 12}$

$s = 10^{- 4}$

Hence, option D is correct.

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