Question

Find the solubility of a sparingly soluble salt, $BaS{O}_4$ in $0.02M$ $N{a}_{2}S{O}_4$ solution ?
$(K_{sp}$ for $BaS{O}_4=1.1\times {10}^{-10})$

• A. $5.5\times {10}^{-9}$
• B. $2\times {10}^{-9}$
• C. $1.1\times {10}^{-9}$
• D. $3.2\times {10}^{-6}$

Answer 1

The correct option is A $5.5\times {10}^{-9}$

$BaS{O}_4$ is a sparingly soluble salt , whereas $N{a}_{2}S{O}_4$ is a highly soluble salt.
Let the solubility of $BaS{O}_4$ be $s$. Then

$BaS{O}_4\rightleftharpoons B{a}^{2+}+S{O}_4^{2-}$
$t=t_{eq}c-sss$

Sodium sulphate ($N{a}_{2}S{O}_4$) is a strong electrolyte and is completely ionised. It shall provide $S{O}_4^{2-}$ ion concentration= $0.02M$.

$\left[B{a}^{2+}\right]=s$
$\left[S{O}_4^{2-}\right]=(s+0.02)M$

$K_{sp}=\left[B{a}^{2+}\right]\left[S{O}_4^{2-}\right]=s\times (s+0.02)$

Since $BaS{O}_4$ is a highly soluble .So $s+0.02\approx 0.02$, because of the sparingly soluble salt $\left(BaS{O}_4\right)$ where $K_{sp}<<{10}^{-3}$
Given $K_{sp}=1.1\times {10}^{-10}$
$\Rightarrow 1.1\times {10}^{-10}=0.02\times s$
or
$s=\frac{1.1\times {10}^{-10}}{0.02}=5.5\times {10}^{-9}$