Question

Solid $N{a}_{2}S{O}_4$ is slowly added to a solution which is $0.020M$ in $Ba{\left(N{O}_3\right)}_2$ and $0.020M$ in $Pb{\left(N{O}_3\right)}_2$. Assume that there is no increase in volume on adding $N{a}_{2}S{O}_4$. There preferential precipitation takes place. What is the concentration of $B{a}^2+$ when $PbS{O}_4$ starts to precipitate? $[K_{sp}\left(BaS{O}_4\right)=1.0\times 10^{-10}$ and $K_{sp}\left(PbS{O}_4\right)=1.6\times 10^{-8}]$

• A. $5.0\times 10^{-9}M$
• B. $8.0\times 10^{-7}M$
• C. $1.25\times 10^{-4}M$
• D. $1.95\times 10^{-8}M$

Answer 1

Answer: (C)

When $BaS{O}_4$ begins to precipitate
$\left[S{O}_4^{2-}\right]=\frac{K_{sp}\left[BaS{O}_4\right]}{\left[B{a}^{2+}\right]}=\frac{1.0\times 10^{-10}}{0.020}$
$=5.0\times 10^{-9}M$
When $PbS{O}_4$ begins to precipitate
$\left[S{O}_4^{2-}\right]=\frac{K_{sp}\left(PbS{O}_4\right)}{\left[P{b}^{2+}\right]}=\frac{1.6\times 10^{-8}}{0.020}$
$=8.0\times 10^{-7}M$
[ $S{O}_4^{2-}$ ]is less for $BaS{O}_4$ precipitation so $BaS{O}_4$ precipitates first when $PbS{O}_4$ begins to precipitate $\left[S{O}_4^{2-}\right]=8.0\times 10^{-7}M$ at that point concentration of $B{a}^{2+}=\frac{K_{sp}\left(BaS{O}_4\right)}{\left[S{O}_4^{2-}\right]}$
$=\frac{1.0\times 10^{-10}}{8.0\times 10^{-7}}=1.25\times 10^{-4}M$