Question

Calculate the solubility of a sparingly soluble salt, $BaS{O}_4$ in $0.01M$ $N{a}_{2}S{O}_4$ solution ?
$(K_{sp}$ for $BaS{O}_4=1.1\times {10}^{-10})$

• A. $3.5\times {10}^{-7}mol{L}^{-1}$
• B. $2.9\times {10}^{-5}mol{L}^{-1}$
• C. $1.6\times {10}^{-7}mol{L}^{-1}$
• D. $1.1\times {10}^{-8}mol{L}^{-1}$

Answer 1

The correct option is D $1.1\times {10}^{-8}mol{L}^{-1}$

$BaS{O}_4$ 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= $0.01M$.

$\left[B{a}^{2+}\right]=s$
$\left[S{O}_4^{2-}\right]=(s+0.01)M$

$K_{sp}=\left[B{a}^{2+}\right]\left[S{O}_4^{2-}\right]=s\times (s+0.01)$

Since $BaS{O}_4$ is a highly soluble .So $s+0.01\approx 0.01$, because of the sparingly soluble salt $\left(BaS{O}_4\right)$ where $K_{sp}<<{10}^{-3}$
Given $K_{sp}=1.1\times {10}^{-10}$
$\Rightarrow 1.1\times {10}^{-10}=0.01\times s$
or
$s=\frac{1.1\times {10}^{-10}}{0.01}=1.1\times {10}^{-8}mol{L}^{-1}$