Question

$PbC{l}_2$ has maximum concentration of $1.0\times {10}^{-3}$ M in its saturated aq. solution at ${25}^0$C. Its solubility in $0.1$M NaCl solution will be :

• A. $4\times {10}^{-7}$M
• B. $4\times {10}^{-9}$M
• C. $2\times {10}^{-7}$M
• D. $2\times {10}^{-9}$M

Answer 1

The correct option is A $4\times {10}^{-7}$M

In saturated aqueous solution :

Maximum concentration in saturated aqueous solution is the solubility (S) of $PbC{l}_2$

$S=1.0\times {10}^{-3}M$

$\left[PbC{l}_2\right]=\left[P{b}^{2+}\right]=S=1.0\times {10}^{-3}M$

$\left[C{l}^{-}\right]=2\times \left[PbC{l}_2\right]=2\times S=2\times 1.0\times {10}^{-3}M=2.0\times {10}^{-3}M$

$K_{sp}=\left[P{b}^{2+}\right]\times [C{l}^{-}{]}^2$

$K_{sp}=1.0\times {10}^{-3}\times [2.0\times {10}^{-3}{]}^2$

$K_{sp}=4.0\times {10}^{-9}$

In 0.1 M NaCl solution:

$\left[C{l}^{-}\right]=\left[NaCl\right]=0.1M$

Assuming $C{l}^{-}$ ion from $PbC{l}_2$ is negligible

$K_{sp}=\left[P{b}^{2+}\right]\times [C{l}^{-}{]}^2$

$4.0\times {10}^{-9}=\left[P{b}^{2+}\right]\times [0.1{]}^2$

$\left[P{b}^{2+}\right]=4.0\times {10}^{-7}M$