The solubility product of $A l (O H)_{3}$ is $2.7 \times 10^{- 11}$ . Calculate its solubility in $g L^{- 1}$ and also find out pH of this solution. (Atomic mass of $A l = 27 u$ ).

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Solution

$S$ be the solubility of $A l (O H)_{3}$

$K_{s p} = [A l^{3 +}] [O H^{-}]^{3} = (S) (3 S)^{3} = 27 S^{4}$
$S^{4} = \frac{K_{s p}}{27} = \frac{27 * 10^{- 11}}{27 \times 10} = 1 \times 10^{- 12}$
$S = 1 \times 10^{- 3} m o l$ $L^{- 1}$ .
(i) Solubility of $A l (O H)_{3}$ : Molar mass of $A l (O H)_{3}$ is $78 g$ . Therefore, solubility of $A l (O H)_{3}$ in
$g L^{- 1} = 1 \times 10^{3} \times 78 g L^{- 1} = 78 \times 10^{- 3} g L^{- 1} = 7.8 \times 10^{- 2} g L^{- 1}$
(ii) pH of the solution: $S = 1 \times 10^{- 3} m o l$ $L^{- 1}$
$[O H] = 3 S = 3 \times 1 \times 10^{- 3} = 3 \times 10^{- 3}$
$p [O H] = 3 - log 3$
$p H = 14 - p O H = 11 + log 3 = 11.4771$ .

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