Question

The solubility of $Pb{SO}_4$ in $0.01M$ ${Na}_2{SO}_4$ solution is:
($K_{sp}$ for $Pb{SO}_4=1.25\times {10}^{-9}$)

• A. $1.25\times {10}^{-7}mol{L}^{-1}$
• B. $1.25\times {10}^{-9}mol{L}^{-1}$
• C. $1.25\times {10}^{-10}mol{L}^{-1}$
• D. $1.25\times {10}^{-18}mol{L}^{-1}$

Answer 1

The correct option is A $1.25\times {10}^{-7}mol{L}^{-1}$

Let the solubility of $PbS{O}_4$ be $S$. Then
$Pb{SO}_4\rightleftharpoons {Pb}^{2+}+{S{O}_4}^{2-}$
$S$ $S$ $S$
Potassium iodide is a strong electrolyte and is completely ionised. It shall provide ${S{O}_4}^{2-}$ ion concentration $=0.01M$
$\left[{Pb}^{2+}\right]=S$
$\left[{S{O}_4}^{2-}\right]=(S+0.01)M$
$K_{sp}=\left[{Pb}^{2+}\right]\left[{S{O}_4}^{2-}\right]=S\times (S+0.01)=S^2+0.01S$
Neglecting $S^3$ and $S^2$
$1.25\times {10}^{-9}=0.01S$
or $S=\frac{1.25\times {10}^{-9}}{0.01}=1.25\times {10}^{-7}mol{L}^{-1}$