A 100.0 mL sample of 0.20M HF is titrated with 0.10M KOH. Determine the pH of the solution after the addition of 100.0 mL of KOH. The Ka of HF is 3.5 x 10^-4

Answer 1

millimoles HF = mL x M = 100 mL x 0.2M = ??
mmoles KOH = 100 mL x 0.1M = ??

................HF + KOH ==> KF + H2O
initial........20....0.......0......0
add..................10..............
change........-10.....-10.....+10...-10
equil.........10......0........10....10

Equilibrium concns are as follows:
M = mmoles/mL
(HF) = 10/200 = ??
(KF) = 10/200
Plug into the Henderson-Hasselbalch equation and solve for pH.

OR, you can do it another way after working a number of these problems by recognizing that we are exactly half-way to the equivalence point. At that point, pH = pKa.