Question

The solubility product of $Pb{Cl}_2$ at $298K$ is $1.7\times {10}^{-5}$. Calculate the solubility of $Pb{Cl}_2$ in $g/lit$ at $298K$.
Atomic weights : [$Pb=207$ and $Cl=35.5$]

Answer 1

$Pb{Cl}_2\rightleftharpoons {Pb}^{2+}+2{Cl}^{-}$
$K_{sp}=\stackrel{S}{\left[{Pb}^{2+}\right]}\stackrel{2S}{{\left[{Cl}^{-}\right]}^2}$
$=S\times {\left(2S\right)}^2$
$=4S^3$
$K_{sp}=1.7\times {10}^{-5}$
$1.7\times {10}^{-5}=4S^3$
$S=\sqrt[3]{3\frac{1.7\times {10}^{-5}}{4}}$
$=0.01619mole/litre$
Molecular of $Pb{Cl}_2=207+2\times 35.5$
$=207+71=278$
Solubility of $Pb{Cl}_2$ in $gm/litre$
$=0.01619\times 278$
$=4.50082gm/litre$