[thegoodaaron]

I have the following question to answer:

Calculate the pH of a solution formed by mixing 172.0 mL of 0.100 M NaF and 208.0 mL of 0.025 M HCl.

The information I can take away from this right away is that the dominant species in the solution are H⁺ and F^-. I can then calculate their initial concentrations. However, at this point I am stuck. How am I to determine how much HF will be reformed in the solution?

[Borek]

Assume F^- is stoichiometrically protonated by added HCl. Then it is Henderson-Hasselbalch equation.

[thegoodaaron]

Does this look appropriate? I only have one possible submission left on this particular problem, and I don't want to miss points just for something minor (or for something major, I suppose).

[NaF] = 0.100M(172mL/380mL) = 0.0453M NaF
[HCl] = 0.025M(208mL/380mL) = 0.014M HCL

HCl lmiting reagent, therefore:
  NaF      +    HCl ---->    HF    +    NaCL
0.0453M     0.014M           0            ----
-0.014M     -0.014M      +0.014M

And [HF] = 0.014M
Then..
   HF   <------> H⁺ + F^-
0.014M               0      0
   -x                  +x    +x
0.014-x                x      x

K_a=[H⁺][F^-]
         [HF]
7.2e-4 = x²
      0.014 - x

x = 2.8e-2
pH = 1.55

Sound okay?

[Borek]

Nope. It is not HF that dissociaties now, you have a buffer solution (HF from protonation and excess F^-).

You start with
172 mL of 0.1M NaF - 17.2 mmol
208 mL of 0.025M HCl - 5.2 mmol

Assuming protonation is stoichiometric:
F^- + H⁺ -> HF

you have after the reaction completion:
F^- - 17.2-5.2 = 12 mmol
HF - 5.2 mmol

Put these into HH equation. You don't need anything more. You even don't need final volume, as it cancels out in the expression under logarithm:

pH = pKa + log([acid]/[base])

[acid] = n_acid/V_final
[base] = n_base/V_final

[acid]/[base] = n_acid/n_base

[thegoodaaron]

Alright, so one final check:

pH = pKa + log([base]/[acid])
pH = -log(7.2e-4) + log(12/5.2)
pH = 3.51

Sound fine?

EDIT: Wonderful, it worked! Thanks!!