You have 30mL of .2M HF being titrated with .15M KOH. The pka is 3.17....half way to the equivalence point you have 20mL of KOH. Find the pH when you add 40mL of KOH . Find the pH when you have 100mL of KOH added.

Question

Sol:pH = pKa + log (base / acid)
= -log 3.17 + log ( .009/ .006)

Answer 1

Your problem is that you don't recognize what you have at this point.
If 20 mL is half-way to the equivalence point, then 40 mL must be AT the equivalence point; therefore, the pH at this point will be determined by the hydrolysis of the salt; i.e., the KF. The HH equation doesn't work at the equivalence point. (KF) = 6/(40+30) = about 0.09 but that's an estimate only.
.............F^- + HOH ==> HF + OH^-
I...........0.09...........0.....0
C...........-x.............x.....x
E.........0.09-x...........x.....x

Kb for F^- = (Kw/Ka for HF) = (x)(x)/(0.09-x)
Substitute and solve for x = (OH^-) and convert to pH.