Question

What is the pH of a solution prepared by mixing 100. mL of 0.0500 M HCl with 300. mL of 0.500 M HF? [Ka(HF) = 7.1 × 10–4]

I believe that I'd use M1V1 + M2V2 = M3V3 but I don't know where to go after that. Thanks!

Answer 1

I don't think what you propose will work. HF is a weak acid but most of the H^+ comes from the HCl. Some smaller amount comes from the HF. I would do it this way.
(HCl) = 0.05 x (100/400) = ?
HCl ==> H^+ + Cl^- so H^+ from this source is approx 0.012M.
(HF) initial ix 0.5 x (300/400) = approx 0.38. You need to recalculate everything because my numbers are just close estimates.
......HF ==> H^+ + F^-
I..0.38M...0.12....0
C.....-x.....+x....+x
E...0.38-x..0.12+x..x

Plug the E line into the Ka expression for HF and solve for x. Find total H, which is 0.125+x) and convert to pH.
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