The solubility of CaF2 is increased due to the presence of H^+ because H^+ confines with the F^- to form the weakly ionized HF. The easy to do this is as follows.
First, determine the value of H^+) from the buffer part of the problem.
pH = pKa + log (base)/(acid).
You know pKa HF(or you can calculate it for formic acid), and you're given base of 0.20M and acid of 0.45M. Solve for pH and convert to H^+. Let S stand for solubility CaF2 so
Ksp = (Ca^2+)(F^-)2
Ksp = (S)(2S)^2
Since a good portion of the H^+ will take away the F^- to form HF, we want to know (F^-), and that fraction is alpha 1 = ka/(ka+H^+).
Then Ksp = (S)(2S*alpha1). Substitute and solve for S.
What is the solubility of CaF2 in a buffer solution containing 0.45M formic acid and 0.20M sodium formate?
Given: Ka for formic acid=1.7x10^-4
Ka for hydrofluoric acid=6.8x10^-4
Ksp for CaF2=3.4x10^-11
Thank you so much. I don't have much idea on this.I just calculated the concentration of H+ by using Ka of formic acid (buffer)
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