Calculate the molar solubility of \({\bf{Ba}}{{\bf{F}}_{\bf{2}}}\) in a buffer solution containing\({\bf{0}}.{\bf{20}}{\rm{ }}{\bf{M}}{\rm{ }}{\bf{HF}}{\rm{ }}{\bf{and}}{\rm{ }}{\bf{0}}.{\bf{20}}{\rm{ }}{\bf{M}}{\rm{ }}{\bf{NaF}}\) .

We have a buffer solution with \(0.20 M HF and 0.20 M{\rm{ }}Na\) .

Calculate the molar solubility of \({{\mathop{\rm BaF}\nolimits} _2}\)

First, let us calculate the concentration of \(\left[ {{F^ - }} \right]\)

\(\begin{array}{*{20}{c}}{{K_a} = \frac{{\left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right] \cdot \left[ {{{\rm{F}}^ - }} \right]}}{{[HF]}}}\\{3.5 \cdot {{10}^{ - 4}} = \frac{{x \cdot (0.20 + x)}}{{0.20 - x}}}\\{0.20x + {x^2} = 0.7 \times {{10}^{ - 4}} - 3.5 \times {{10}^{ - 4}}x}\\{0 = {x^2} + 0.20x - 0.7 \cdot {{10}^{ - 4}}}\\{x = 3.49 \times {{10}^{ - 4}}}\\{\left[ {{F^ - }} \right] = 0.20 + x \approx 0.20{\rm{M}}}\end{array}\)

Let us Calculate the molar solubility of \({{\mathop{\rm BaF}\nolimits} _2}\)

\(\begin{array}{*{20}{c}}{{K_{sp}} = \left[ {B{a^{2 + }}} \right] \cdot {{\left[ {{F^ - }} \right]}^2}}\\{\left[ {B{a^{2 + }}} \right] = \frac{{{K_{sp}}}}{{{{\left[ {{F^{ - 2}}} \right]}^2}}}}\\{ = \frac{{2.4 \times {{10}^{ - 5}}}}{{{{(0.20)}^2}}}}\\{ = 6 \times {{10}^{ - 4}}{\rm{M}}}\end{array}\)