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Magazine Chapter 15: Q15.3-94E (page 878)
What is the molar solubility of CaF2 in a 0.100 M solution of HF? Kafor HF = 7.2 脳 10鈥�4.
Short Answer
The molar solubility of CaF2is\(6.05 \cdot {10^{ - 7}}{\rm{M}}\).
Step by step solution
Calculate the molar solubility of CaF2:
We have a 0.100 M solution of HF\(\left( {{K_a} = 7.2 \cdot {{10}^{ - 4}}} \right)\).
- The reaction of dissociation of HF
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
\({\rm{Ca}}{{\rm{F}}_2}({\rm{s}}) \to {\rm{C}}{{\rm{a}}^{2 + }}({\rm{aq}}) + 2{{\rm{F}}^ - }({\rm{aq}})\)
- \({K_{sp}}{\rm{\;of\;Ca}}{{\rm{F}}_2}{\rm{\;is\;}}4.0 \cdot {10^{ - 11}}\)
\(\begin{array}{*{20}{c}}{{K_{sp}} = \left[ {C{a^{2 + }}} \right] \cdot {{\left[ {{F^ - }} \right]}^2}}\\{\left[ {C{a^{2 + }}} \right] = \frac{{{K_{sp}}}}{{{{[F]}^2}}}}\\{ = \frac{{4.0 \cdot {{10}^{ - 11}}}}{{{{\left( {8.13 \cdot {{10}^{ - 3}}} \right)}^2}}}}\\{ = 6.05 \cdot {{10}^{ - 7}}{\rm{M}}}\end{array}\)
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