Question

What is the solubility of a sparignly soluble salt, $PbS{O}_4$ in 0.01 $M$ $N{a}_{2}S{O}_4$ solution ?
$(K_{sp}$ for $PbS{O}_4=1.6\times {10}^{-8})$

• A. $1.6\times {10}^{-6}mol{L}^{-1}$
• B. $2.6\times {10}^{-9}mol{L}^{-1}$
• C. $3.5\times {10}^{-7}mol{L}^{-1}$
• D. $0.4\times {10}^{-9}mol{L}^{-1}$

Answer 1

The correct option is A $1.6\times {10}^{-6}mol{L}^{-1}$

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

$PbS{O}_4\rightleftharpoons P{b}^{2+}+S{O}_4^{2-}$
$t=t_{eq}c-sss$

Sodium sulphate is a strong electrolyte and is completely ionised. It shall provide $S{O}_4^{2-}$ ion concentration= $0.01M$.

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

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

Since $N{a}_{2}S{O}_4$ is a highly soluble .So $s+0.01\approx 0.01$, because of the sparingly soluble salt $\left(PbS{O}_4\right)$ where $K_{sp}<<{10}^{-3}$
Given $K_{sp}=1.6\times {10}^{-8}$
$\Rightarrow 1.6\times {10}^{-8}=0.01\times s$
or
$s=\frac{1.6\times {10}^{-8}}{0.01}=1.6\times {10}^{-6}mol{L}^{-1}$