What is the solubility of CaF2 in a buffer solution containing .30 M HCHO2 and .20 M NaCHO2?
Let solubility = S
CaF2 ==> Ca^+2 + 2F^-
Ksp = (Ca^+2)(F^-)^2
HF ==> H^+ + a = (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.
hi kelley =]
i 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 don't know where i went wrong.
Using 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?
To determine the solubility of CaF2 in the given buffer solution, we need to consider the common ion effect. The common ion effect states that the solubility of a salt is decreased in a solution that already contains one of its constituent ions.
Here, the buffer solution contains HCHO2 (formic acid) and NaCHO2 (sodium formate). We can assume that HCHO2 ionizes completely in water according to the following equation:
HCHO2 (aq) → H+ (aq) + CHO2- (aq)
Since CaF2 is a slightly soluble salt, it dissociates according to the reaction:
CaF2 (s) ↔ Ca2+ (aq) + 2F- (aq)
The common ion in this case is the F- ion, which is already present in the buffer solution due to the dissociation of HCHO2. Therefore, the solubility of CaF2 will be decreased in the buffer solution compared to pure water.
To calculate the solubility of CaF2 in the buffer solution, we need the Ksp (solubility product constant) value for CaF2. The Ksp expression for CaF2 is:
Ksp = [Ca2+][F-]²
The value of Ksp for CaF2 is 3.9 x 10^-11 at 25°C.
Since the solubility of CaF2 is x, the concentrations of Ca2+ and F- ions in the buffer solution are also x.
Using the Ksp expression, we can set up an equation:
3.9 x 10^-11 = (x)(x²)
Simplifying the equation:
3.9 x 10^-11 = x³
Now, we can solve for x:
x = (3.9 x 10^-11)^(1/3)
Using a calculator, we find:
x ≈ 1.67 x 10^-4 M
Therefore, the solubility of CaF2 in the buffer solution containing 0.30 M HCHO2 and 0.20 M NaCHO2 is approximately 1.67 x 10^-4 M.