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  4. Ksp for Al(OH)3 is 8.5 × 10-32. What is its solubility in moles/litre?

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Q. $K_{sp}$ for $Al(OH)_3$ is $8.5 \times 10^{-32}$. What is its solubility in moles/litre?

UP CPMTUP CPMT 2011Equilibrium

Solution:

$Al (OH)_3 \to \underset{\text{S}}{Al^3+ } + \underset{\text{3S}}{3OH^- }$
$K_{sp} = (S)(3S)^2 = 27 S^4 $
$27S^4 = 8.5 \times 10^{-32}$
$S=\sqrt[4]{8.5\times 10^{-32} /27}$
$=(0.31 \times 10^{-32})^{1/4}$
$=0.749 \times 10^{-8}$
$ = 7.49 \times 10^{-7}$ moles /litre.