Question

Assume the solubility of $PbCl_{2}$ in water is $S\,mol\,L^{- 1}$ and its solubility product is $K_{sp}$ . The relation between $K_{sp}$ and $S$ is represented as $S=\sqrt[3]{\frac{K_{sp}}{x}}$ . The value of $x$ is_____.

Answer 1

$\underset{(S)}{PbCl _2} \rightleftharpoons \underset{(S)}{Pb ^{2+}}+\underset{(2S)}{2 Cl ^{-}}$
$K_{s p}=\left[P b^{2 +}\right]\left[C l\right]^{2}$
$=S\times \left(2 S\right)^{2}=4S^{3}$
$\therefore S=\sqrt[3]{\frac{K_{S P}}{4}}$
$\therefore $ Value of $x=4$