Question

The molar solubility of PbCl${}_2$ in $0.20$ M Pb(NO${}_3$)${}_2$ solution is:
$(K_{sp}$ for $PbC{l}_2=1.6\times {10}^{-5})$

• A. $1.7\times {10}^{-4}M$
• B. $9.2\times {10}^{-3}M$
• C. $1.7\times {10}^{-5}M$
• D. $4.6\times {10}^{-3}M$
• E. $8.5\times {10}^{-5}M$

Answer 1

The correct option is D $4.6\times {10}^{-3}M$

$\left[P{b}^{2+}\right]=\left[Pb\right(N{O}_3{)}_2]=0.20M$
The expression for the solubility product is $K_{SP}=\left[P{b}^{2+}\right][C{l}^{-}{]}^2$

Substitute values in the expression for the solubility product.

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

$1.6\times {10}^{-5}=0.20\times [C{l}^{-}{]}^2$

$[C{l}^{-}{]}^2=8\times {10}^{-5}$

$\left[C{l}^{-}\right]=8.9\times {10}^{-3}M$

The molar solubility of $PbC{l}_2$ in 0.20 M $Pb(N{O}_3{)}_2$

$=\frac{\left[C{l}^{-}\right]}{2}=\frac{8.9\times {10}^{-3}M}{2}=4.6\times {10}^{-3}M$