Question

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

Answer 1

Given,
Molar mass of $PbC{l}_2=278\text{g/mol}$
Mass of the residue = $2.78\text{g}$
Upon evaporation of saturated solution, all water will be escaped and only dissolved $PbC{l}_2$ will be left i.e., solubility $=\frac{\frac{2.78}{278}}{1}={10}^{-2}\text{mol/L}$
For,
$PbC{l}_2\rightleftharpoons P{b}^{2+}+2C{l}^{-}$
$\text{at equilibrium}1-ss2s$

for $PbC{l}_2{K}_{sp}=\left[s\right][2s{]}^2=4s^3$
$\therefore y\times {10}^{-6}=4({10}^{-2}{)}^3=4\times {10}^{-6}$
$\Rightarrow y=4$