Question

The solubility product $\left(K_{sp}\right)$ of $Ba{SO}_4$ is $1.5\times {10}^{-9}$. Calculate the solubility of barium sulphate in pure water and in $0.1M$ $Ba{Cl}_2$.

Answer 1

$Ba{SO}_4\left(s\right)⟶{Ba}^{2+}\left(aq\right)+{SO}_4^{2-}\left(aq\right)$
$\therefore K_{sp}=\left[{Ba}^{2+}\right]\left[{SO}_4^{2-}\right]=x$
Then, $1.5\times {10}^{-9}=x\times x;x^2=15\times {10}^{-10}or3.87\times {10}^{-5}$
Then, solubility of $Ba{SO}_4$ in pure water is $3.87\times {10}^{-5}$.
Let the solubility of $Ba{SO}_4$ in $0.1MBa{Cl}_2$ be ${}^{\prime }{s}^{\prime }$
$\underset{Allsolid}{Ba{SO}_4\left(s\right)}\Rightarrow \underset{0.1M}{{Ba}^{2+}\left(aq\right)}+{SO}_4^{-}\left(aq\right)$
Initial (from $BaC{l}_2$) $0$
At equilibrium $(0.1M+s)$ $s$
Hence, $1.5\times {10}^{-9}=(s+0.1)\times s=s\times 0.1$(Since $s<<1$)
$s=1.5\times {10}^{-8}$
Thus the solubility of $Ba{SO}_4$ in presence of $0.1M$ $Ba{Cl}_2$ is $1.5\times {10}^{-8}$