1. The pH at the midpoint of the titration (the half-equivalence point) will be equal to the pKa of hydrofluoric acid - the Ka of HF is 6.6 x 10-4, and pKa = -log (Ka) so the pKa of HF is 3.18.
2. The pH at the equivalence point will be determined by the fluoride ion concentration.
The reaction during the titration is HF + KOH --> H2O + KF
The K+ in solution won't have an effect on the pH of the solution, but the F- will as it is the conjugate base of a weak acid.
The reaction at the equivalence point is F- + H2O ↔ HF + OH-
We need the molarity of the fluoride ion in order to complete an ICE table.
0.0284(0.416) = 0.01181... moles of HF = moles of NaOH at the equivalence point
0.01181... moles NaOH/0.418 M = 0.0283 L NaOH used
The molarity of the F- at the equivalence point will be the moles of F- divided by the TOTAL volume at this point.
0.01181.../(0.0283 + 0.0284) = 0.208 M
We also need the Kb for F-, which is the Kw/Ka = 1.0 x 10-14 / (6.6 x 10-4) = 1.5 x 10-11
From the ICE table we will get
x2 / (0.208 - x) = 1.5 x 10-11 the "-x" in the denominator can be ignored as the equilibrium constant is very small
solving this we get x = 1.78 x 10-6 M = [OH-]
pOH = -log (1.8 x 10-6) = 5.75
pH = 14 - pOH = 8.25
3. After 42.4 mL of NaOH have been added we have gone way past the equivalence point, so there is a lot of excess NaOH
42.4 - 28.3 = 14.1 mL excess
.0141 L (0.418) = 0.00589.. moles NaOH excess
0.00589.. moles / (0.0424 + 0.0284) = 0.0832 M NaOH at this point
[OH-] = [NaOH]
pOH = -log (0.0832) = 1.08
pH = 14 - pOH = 12.92
