What is the molar solubility of CaF₂ in a 0.100 M solution of HF? K_afor HF = 7.2 × 10^–4.

We have a 0.100 M solution of HF\(\left( {{K_a} = 7.2 \cdot {{10}^{ - 4}}} \right)\).

Now, let us calculate the concentration of F^-

\(\begin{array}{*{20}{c}}{{K_a} = \frac{{\left[ {{H^ + }} \right] \cdot \left[ {{F^ - }} \right]}}{{[HF]}}}\\{7.2 \cdot {{10}^{ - 4}} = \frac{{x \cdot x}}{{0.100 - x}}}\\{{x^2} = 7.2 \cdot {{10}^{ - 5}} - 7.2 \cdot {{10}^{ - 4}}x}\\{0 = {x^2} + 7.2 \cdot {{10}^{ - 4}}x - 7.2 \cdot {{10}^{ - 5}}}\\{x = 8.13 \cdot {{10}^{ - 3}}}\\{\left[ {{F^ - }} \right] = x = 8.13 \cdot {{10}^{ - 3}}}\end{array}\)

Finally, we will find the molar solubility of CaF₂

\({\rm{Ca}}{{\rm{F}}_2}({\rm{s}}) \to {\rm{C}}{{\rm{a}}^{2 + }}({\rm{aq}}) + 2{{\rm{F}}^ - }({\rm{aq}})\)

\(\begin{array}{*{20}{c}}{{K_{sp}} = \left[ {C{a^{2 + }}} \right] \cdot {{\left[ {{F^ - }} \right]}^2}}\\{\left[ {C{a^{2 + }}} \right] = \frac{{{K_{sp}}}}{{{{[F]}^2}}}}\\{ = \frac{{4.0 \cdot {{10}^{ - 11}}}}{{{{\left( {8.13 \cdot {{10}^{ - 3}}} \right)}^2}}}}\\{ = 6.05 \cdot {{10}^{ - 7}}{\rm{M}}}\end{array}\)