Question

The solubility product of a sparingly soluble salt $BaS{O}_4$​ is $1.1\times {10}^{-10}$ at ${25}^{o}C$ . What will be the solubility of $BaS{O}_4$​ in presence of $0.02M$ $H_{2}S{O}_4$ ? (assume complete dissociation of $H_{2}S{O}_4)$

• A. $2.25\times {10}^{-9}{\text{mol L}}^{-1}$
• B. $5.5\times {10}^{-9}{\text{mol L}}^{-1}$
• C. $7.5\times {10}^{-9}{\text{mol L}}^{-1}$
• D. $1.5\times {10}^{-9}{\text{mol L}}^{-1}$

Answer 1

The correct option is B $5.5\times {10}^{-9}{\text{mol L}}^{-1}$

$BaS{O}_4$ is a sparingly soluble salt , whereas $H_{2}S{O}_4$ is a strong acid.
Let the solubility of $BaS{O}_4$ be $s$. Then

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

Sulphuric acid 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 $H_{2}S{O}_4$ is highly soluble .
So $s+0.02\approx 0.02$, because of the sapringly 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
$s=\frac{1.1\times {10}^{-10}}{0.02}=5.5\times {10}^{-9}mol{L}^{-1}$