Question

If equal volumes of $BaC{l}_2$ and $NaF$ solutions are mixed, which of the following combinations will give a precipitate?
$(K_{sp}$ of $Ba{F}_2=1.7\times {10}^{-7})$

• A. 0.0040 M $BaC{l}_2$ and 0.02 M $NaF$
• B. 0.01 M $BaC{l}_2$ and 0.015 M $NaF$
• C. 0.015 M $BaC{l}_2$ and 0.01 M $NaF$
• D. 0.02 M $BaC{l}_2$ and 0.002 M $NaF$

Answer 1

Answer: (A), (B), (C)

The correct options are
A 0.0040 M $BaC{l}_2$ and 0.02 M $NaF$
B 0.01 M $BaC{l}_2$ and 0.015 M $NaF$
C 0.015 M $BaC{l}_2$ and 0.01 M $NaF$

Ionic product, $Q_{sp}$ for $Ba{F}_2=\left[B{a}^{2+}\right][F^{-}{]}^2$
When ionic product exceeds the solubility product, precipitation occurs. The important thing to note here is that the actual concentrations would be half of the initial ones since equal volumes of the solutions are mixed.
$a.Q_{sp}=2\times {10}^{-3}\times ({10}^{-2}{)}^2=2\times {10}^{-7}>K_{sp};$ precipitation occurs
$b.Q_{sp}=5\times {10}^{-3}\times (7.5\times {10}^{-2}{)}^2=2.8\times {10}^{-7}>K_{sp};$ precipitation occurs
$c.Q_{sp}=7.5\times {10}^{-3}\times (5\times {10}^{-2}{)}^2=1.875\times {10}^{-7}>K_{sp};$ precipitation occurs
$d.Q_{sp}={10}^{-2}\times ({10}^{-3}{)}^2={10}^{-8}<K_{sp};$
precipitation does not occur.