Question

The solubility product of $Ba{SO}_4$ is $1.5\times {10}^{-9}$. Find out the solubility in (i) pure water and (ii) $0.1M$ $Ba{Cl}_2$ solution.

Answer 1

The equilibrium is:
(i) $Ba{SO}_4\rightleftharpoons {Ba}^{2+}+{SO}_4^{2-}$
Let $S$ be the solubility in $mol$ ${Litre}^{-1}$; then
$K_s=\left[{Ba}^{2+}\right]\left[{SO}_4^{2-}\right]=S^2$
or $1.5\times {10}^{-9}=S^2$
so, $S=3.87\times {10}^{-5}mol{L}^{-1}$
(ii) Let $S^{\prime }$ be the solubility of $Ba{SO}_4$ in $0.1M$ $Ba{Cl}_2$ solution.
Total ${Ba}^{2+}$ ions concentration $=(S^{\prime }+c)mol{L}^{-1}$
and ${SO}_4^{2-}$ ions concentration $=S^{\prime }mol{L}^{-1}$
So, $K_s=(S^{\prime }+c)S^{\prime }=(S^{\prime }+0.1)S^{\prime }$
or $1.5\times {10}^{-9}=(S^{\prime }+0.1)S^{\prime }$
or ${S^{\prime }}^2+0.1S^{\prime }=1.5\times {10}^{-9}$
Neglecting ${\left(S^{\prime }\right)}^2$
$0.1S^{\prime }=1.5\times {10}^{-9}$
or $S^{\prime }=1.5\times {10}^{-8}mol{L}^{-1}$