May I be able to get help on this one as well please?

Question

A solution was prepared with 0.380 mol of pyridinium fluoride (C5H5NHF) and enough water to make a 1.00 L. Pyridine (C5H5N) has a 𝐾b=1.70×10^−9 and HF has a 𝐾a=6.30×10^−4.

Calculate the pH of the solution.

pH=

Answer 1

Pyridinium fluoride is the salt of a weak base (pyridinium) and a weak acid (HF).

C5H5NH⁺ + H2O ==> H3O⁺ + C5H5N Ka = 1x10^-14 / 1.70x10^-9 = 5.88x10^-6

F- + H2O ==> HF + OH- Kb = 1x10^-14 / 6.30x10^-4 = 1.59x10^-11

Comparing the Ka and Kb values, we conclude that C5H5NH⁺ is a stronger acid than F^- is a base, so solution should have a pH < 7 (acidic)

Without going into the derivation (which you can find in a chemistry text, or online), we can use

pH = 7 + 1/2 (pKa - pKb)

pH = ?

pKa for HF = 3.20

pKb for C5H5N = 8.77

pH = 7 + 1/2 (3.20 - 8.77)

pH = 7 + 1/2 (-5.57)

pH = 7 + -2.79

pH = 4.215