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December 21, 2014

December 21, 2014

Posted by **Kelley** on Sunday, April 4, 2010 at 9:31pm.

- chemistry -
**DrBob222**, Sunday, April 4, 2010 at 10:12pmLet solubility = S

CaF2 ==> Ca^+2 + 2F^-

Ksp = (Ca^+2)(F^-)^2

HF ==> H^+ + F^-

Ka = (H^+)(F^-)/(HF)

pH = pKa(formic acid) + log[(HCOO^-)/(HCOOH)

mole balance: 2S =(Ca^+2) = (F^-) +(HF)

1. Use the Henderson-Hasselbalch equation (equation 3 above) to solve for (H^+) in the buffered solution.

2. Plug (H^+) into Ka for HF (equation 2 above) and solve for (HF) in terms of (F^-).

3. Substitute this for HF into the mole balance equation (equation 4) and solve for (F^-) in terms of S, then substitute into Ksp equation (1 above) and solve for S.

It isn't as complicated as it sounds. I'll be glad to check your work if you care to post it. If you do post the work, be sure to post what you are using for pKa formic acid, Ksp CaF2, and Ka for HF.

- chemistry -
**Julia**, Monday, April 5, 2010 at 7:02pmhi kelley =]

- chemistry -
**Kelley**, Monday, April 5, 2010 at 9:25pmi found the pH of the buffer solution to be 3.5, the Ka for the buffer solution being 1.9x10^-4

the Ka for HF is 6.9x10^-4, so i found the concentration of HF to be .413(F^-1)

for the Ksp i used 1.5x10^-10, so i got S to be 3.5x10^-4.

the answer in my textbook says its supposed to be 1.8x10^-4, so i dont know where i went wrong.

- chemistry -
**DrBob222**, Tuesday, April 6, 2010 at 1:09amUsing your values I obtained 4.2 x 10^-4 M for the solubility and I worked it as above plus another way of doing the problem and obtained identical values. I think the value you have been given is incorrect---this is why I think so.

What is the solubility of CaF2 in water (no acid to increase the solubility)?

CaF2 ==> Ca^+2 + 2F^-

Solubility CaF2 = S

(Ca^+2) = S

(F^-) = 2S

Plug into Ksp

1.5 x 10^-10 = (S)(2S)^2 = 4S^3

Solve for S and I get 3.35 x 10^-4 M. We KNOW that adding an acid increases the solubility because the H^+ ties up F^- as HF and that pulls more CaF2 into solution (Le Chatelier's principle). So your answer and my answer show a higher solubility in acid than with no acid, as it should. BUT an answer of 1.8 x 10^-4 is LESS solubility and that can't be. It MUST be more soluble in acid solution than in a neutral solution. I looked this up in a quant book and they give an example of CaF2 in an acid solution (but they used a higher concn of acid and different values for Ksp etc). One thing I've noticed is that Ksp is quoted as several values depending upon where I look and Ka values varied. I assume the answer of 1.8 x 10^-4 is in M and not some other unit?

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