Question

A saturated solution of BaF₂ at 25°C is prepared by equilibrating solidBaF₂ with water. Powdered NaF is then dissolved in the solution until the solubility ofBaF₂ is 1.0% of that in H₂O alone. The solubility product K_sp ofBaF₂ is$\mathbf{1}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}$ at 25°C. Calculate the concentration of fluoride ion in the solution after addition of the powdered NaF.

Short answer

$\left[F^{-}\right]=1.50\times {10}^{-4}mol/L$

Step-by-step solution

Molar solubility of BaF2 in pure water

$$\begin{gathered} Ba{F}_2\left(s\right)\to B{a}^{2+}\left(aq\right)+2F^{-}\left(aq\right) \\ -s2s \end{gathered}$$

$\begin{array}{rcl}{K}_{sp}& =& s\times {\left(2s\right)}^2\\ 1.7\times {10}^{-6}& =& 4s^3\\ s& =& 7.51\times {10}^{-3}mol/L\end{array}$

Hence, in pure water the molar solubility is $\begin{array}{rcl}& & 7.51\times {10}^{-3}mol/L\end{array}$.

Molar solubility of BaF2 in NaF

The molar solubility of BaF₂ in NaF is 1.0% of its solubility in pure water.

Molar solubility in NaF:

$\begin{array}{rcl}s& =& 7.51\times {10}^{-3}\times \frac{1.0}{100}\\ & =& 7.51\times {10}^{-5}mol/L\end{array}$

Concentration of fluoride ion after adding NaF

$\begin{array}{rcl}\left[F^{-}\right]& =& 2\times 7.51\times {10}^{-5}mol/L\\ & =& 1.50\times {10}^{-4}mol/L\end{array}$