Question

The solubility product of lead bromide is $8\times {10}^{-5}$. If the salt is $80$% dissociated in saturated solution, find the solubility of the salt.

Answer 1

Let $S$ be the solubility of the salt
Degree of dissociation of the salt$=0.8$
$Pb{Br}_2\rightleftharpoons {Pb}^{2+}+2{Br}^{-}$
$0.8S$ $2\times 0.8S$
$K_{sp}=\left[{Pb}^{2+}\right]{\left[{Br}^{-}\right]}^2$
$=\left(0.8S\right)\times {\left(1.6S\right)}^2=2.048S^3$
or $S^3=\frac{8\times {10}^{-5}}{2.048}=3.906\times {10}^{-5}$
$S=\sqrt[3]{33.906\times {10}^{-5}}=3.39\times {10}^{-2}mol{L}^{-1}$
Mol. mass of $Pb{Br}_2=367$
Solubility of $Pb{Br}_2=3.39\times {10}^{-2}\times 367=12.44g{L}^{-1}$