Question

When one litre of a saturated solution of $PbC{l}_2$ (molecular weight $=278$ g/mol) is evaporated, the residue is found to weigh $2.78$ g. If $K_{sp}$ of $PbC{l}_2$ is represented as $y\times {10}^{-6}$, then find the value of $y$.

Answer 1

It is given that one litre of a saturated solution of $PbC{l}_2$ is evaporated, the residue weighs $2.78$ g.

First we need to calculate the solubility $\left(S\right)$ of $PbC{l}_2$.
$S=\frac{2.78}{278\times 1}=0.01$ M
The expression for the solubility product of $PbC{l}_2$is
$K_{sp}=\left[P{b}^{2+}\right][C{l}^{-}{]}^2=S\times (2S{)}^2=4S^3$
Substituting values in the above expression, we get
$y\times {10}^{-6}=4({10}^{-2}{)}^3=4\times {10}^{-6}$
Hence, $y=4$
Thus, the value of $y$ is $4$.