Question

The solubility of $PbC{l}_2$ in water is $0.01Mat{25}^{\circ }\text{C}$. Its maximum concentration in $0.1MNaCl$ will be:

• A. $2\times {10}^{-3}M$
• B. $1\times {10}^{-4}M$
• C. $1.6\times {10}^{-2}M$
• D. $4\times {10}^{-4}M$

Answer 1

The correct option is D $4\times {10}^{-4}M$

$K_{sp}$ of $PbC{l}_2=4s^3=4\times (0.01{)}^3=4\times {10}^{-6}$
In $NaCl$ solution, the solubility product of $PbC{l}_2$ is $K_{sp}=\left[P{b}^{2+}\right][C{l}^{-}{]}^2$
or $4\times {10}^{-6}=\left[P{b}^{2+}\right][0.1{]}^2\therefore [P{b}^{2+}]=4\times {10}^{-4}M$
Common ion $C{l}^{-}$, decrease the solubilty of $PbC{l}_2$ in NaCl.