Question

Solid $Na _{2} SO _{4}$ is slowly added to a solution which is $0.020\, M$ in $Ba \left( NO _{3}\right)_{2}$ and $0.020\, M$ in $Pb \left( NO _{3}\right)_{2}$. Assume that there is no increase in volume on adding $Na _{2} SO _{4}$. There preferential precipitation takes place. What is the concentration of $Ba ^{2}+$ when $PbSO _{4}$ starts to precipitate? $\left[ K _{ sp }\left( BaSO _{4}\right)=1.0 \times 10^{-10}\right.$ and $\left.K _{ sp }\left( PbSO _{4}\right)=1.6 \times 10^{-8}\right]$

• 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 $BaSO _{4}$ begins to precipitate
${\left[ SO _{4}^{2-}\right]=\frac{ K _{ sp }\left[ BaSO _{4}\right]}{\left[ Ba ^{2+}\right]}=\frac{1.0 \times 10^{-10}}{0.020}}$
$=5.0 \times 10^{-9} M$
When $PbSO _{4}$ begins to precipitate
${\left[ SO _{4}^{2-}\right]=\frac{ K _{ sp }\left( PbSO _{4}\right)}{\left[ Pb ^{2+}\right]}=\frac{1.6 \times 10^{-8}}{0.020}} $
$=8.0 \times 10^{-7} M$
[ $SO _{4}^{2-}$ ]is less for $BaSO _{4}$ precipitation so $BaSO _{4}$ precipitates first when $PbSO _{4}$ begins to precipitate $\left[ SO _{4}^{2-}\right]=8.0 \times 10^{-7} M$ at that point concentration of $Ba ^{2+}=\frac{ K _{ sp }\left( BaSO _{4}\right)}{\left[ SO _{4}^{2-}\right]}$
$=\frac{1.0 \times 10^{-10}}{8.0 \times 10^{-7}}=1.25 \times 10^{-4} M$