Question

If the maximum concentration of $PbC{l}_2$ in water is $0.01$ M at $298$ K, its maximum concentration in $0.1$ M NaCl will be:

• A. $4\times {10}^{-3}$ M
• B. $0.4\times {10}^{-4}$ M
• C. $4\times {10}^{-2}$ M
• D. $4\times {10}^{-4}$ M

Answer 1

Answer: (D)

$4\times {10}^{-4}$ M

$PbC{l}_2\to P{b}^{2+}+2C{l}^{-}$

the concentration of $P{b}^{2+}$ ion is 0.01M

the concentration of $C{l}^{-}$ ion is $2\times 0.01=0.02M$

the solubility of $PbC{l}_2$ is

$K_{sp}=\left[P{b}^{2+}\right][C{l}^{-}{]}^2$ = $\left(0.01\right){\left(0.02\right)}^2=4\times {10}^{-6}$

when $PbC{l}_2$ is dissolved in 1M of $NaCl$ since there is no common ion of $C{l}^{-}$ the solubility of $PbC{l}_2$ will effect the concentration of $C{l}^{-}$ in 0.1M of $NaCl$

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

$⟹4\times {10}^{-6}=[P{b}^{2+}{\left]\right[0.1]}^2$

$⟹\left[P{b}^{2+}\right]=\frac{4\times {10}^{-6}}{0.01}=0.0004M$

The maximum concentraction of $PbC{l}_2$ in 0.1M of $NaCl$ is $4\times {10}^{-4}M$

Hence, the correct option is $\text{C}$