Question

Given, that the solubility product of radium sulphate $\left(Ra{SO}_4\right)$ is $4\times {10}^{-11}$. Calculate its solubility in:
(a) pure water
(b) $0.10M$ ${Na}_2{SO}_4$

Answer 1

(a) Let the solubility of $RaS{O}_4$ be $sM$

$RaS{O}_4\left(s\right)\rightleftharpoons \left[R{a}^{2+}\right]\left[S{O}_4^{2-}\right]$

$\Rightarrow s^2=K_{sp}=4\times {10}^{-11}$

$\therefore s=\sqrt{4\times {10}^{-11}}=6.32\times {10}^{-6}M$

$\therefore$ Solubility of $RaS{O}_4$ in pure water is $6.32\times {10}^{-6}M$.

(b) Let the solubility be $s^{\prime }$ in presence of $0.10M$ $N{a}_{2}S{O}_4$.

$\Rightarrow s^{\prime }\times [s^{\prime }+C_{S{O}_4^{2-}}$ from $N{a}_{2}S{O}_4]=K_{sp}$

$\Rightarrow s^{\prime }\times 0.10=4\times {10}^{-11}$ [Since $s^{\prime }<<S_{S{O}_4^{2-}}$ from $N{a}_{2}S{O}_4]$

$\Rightarrow s^{\prime }=4\times {10}^{-10}M$

$\therefore$ Solubility of $RaS{O}_4$ in $0.1M$ $N{a}_{2}S{O}_4$ is $4\times {10}^{-10}M$ .