Question

When one litre of a saturated solution of $PbC{l}_2$ $\left({\text{molar mass=278 g mol}}^{-1}\right)$ is evaporated, the residue is found to weight $2.78\text{g}$. If $K_{sp}$ of $PbC{l}_2$ is $y\times {10}^{-6}$, then $y$ is:

Answer 1

It is given that one litre of a saturated solution of $PbC{l}_2$ is evaporated, the residue weighs $2.78\text{g}$
First , we need to calculate the solubility (S) of $PbC{l}_2$.
$S=\frac{\text{mole}}{\text{volume of solution(L)}}=\frac{2.78}{278\times 1}=0.01M$
The expression for the solubility product of $PbC{l}_2$ is,
$PbC{l}_2\rightleftharpoons P{b}^{2+}+2C{l}^{-}$
$K_{sp}\left[P{b}^{2+}\right][C{l}^{-}{]}^2=S\times (2S{)}^2=4S^3$
Substiuting values in the above expression, we get
$y\times {10}^{-6}=4\times {10}^{-6}$
Hence, $y=4$.