Question

If equal volumes of $Ba{Cl}_2$ and $NaF$ solutions are mixed, which of these combinations will not give a precipitate?
Substance | $K_{sp}$
$Ba{F}_2$ | $1.7\times {10}^{-7}$

• A. $0.004M$ $Ba{Cl}_2$ and $0.02M$ $NaF$
• B. $0.010M$ $Ba{Cl}_2$ and $0.015M$ $NaF$
• C. $0.015M$ $Ba{Cl}_2$ and $0.010M$ $NaF$
• D. $0.020M$ $Ba{Cl}_2$ and $0.002M$ $NaF$

Answer 1

The correct option is D $0.020M$ $Ba{Cl}_2$ and $0.002M$ $NaF$

A) $\left[{Ba}^{2+}\right]{\left[F^{-}\right]}^2=0.04\times {\left(0.02\right)}^2=1.6\times {10}^{-5}>K_{sp}$ of ${BaF}_2$

${BaF}_2$ will precipitate.

B) $\left[{Ba}^{2+}\right]{\left[F^{-}\right]}^2=0.01\times {\left(0.015\right)}^2=2.25\times {10}^{-6}>K_{sp}$ of ${BaF}_2$

So ${BaF}_2$ will precipitate.

C) $\left[{Ba}^{2+}\right]{\left[F^{-}\right]}^2=\left[0.015\right]\times {\left[0.01\right]}^2=1.5\times {10}^{-6}>K_{sp}$ of ${BaF}_2$

${BaF}_2$ will precipitate.

D) $\left[{Ba}^{2+}\right]{\left[F^{-}\right]}^2=0.02\times {\left(0.002\right)}^2=8\times {10}^{-8}<K_{sp}$ of ${BaF}_2$

So ${BaF}_2$ will not precipitate.

Since for precipitate to start the product of $\left[{Ba}^{2+}\right]$ and square of $\left[F^{-}\right]$ should be greater than $K_{sp}$ of ${BaF}_2$.