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Solubility Product

The solubilit...

Question

The solubility of $A_{2} B_{3}$ is $s$ $m o l L^{- 1}$ . Its solubility product is 108 $s^{n}$ . The value of $^{'} n^{'}$ will be:

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Solution

$A_{2} B_{3} ⇌ 2 A^{3 +} + 3 B^{2 -}$
$K s p = 108 S^{n}$
$108 S^{n} = {(2 S)}^{2} {(3 S)}^{3}$
$108 S^{n} = 4 S^{2} \times {27 S}^{3}$
$108 S^{n} = 108 S^{5}$
$∴ n = 5$

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