Question

A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH. Determine the pH of the solution after the addition of 200.0mL of KOH. Ka of HFis 3.5 x 10^-4 Answer needs to be solved with milli moles

Answer 1

mmoles of HF present = M*V = 100 mL * 0.20 M =20 mmol
mmoles of KOH added = M*V = 200 mL * 0.10 M = 20 mmol

20 mmol of each will react to form 20 mmoles of F-

HF + KOH ---> F- + K+ + H2O

[F-] = number of moles / volume
= 20mmoles/(100+200) mL
= 0.067 M

this F- with water will form:
F- + H2O ---> HF + OH-
x x

Kb of F- = 10^-14/(3.5*10^-4)
= 2.86*10^-11

Kb = x*x/[F-]
2.86*10^-11 = x^2/(0.067)
x = 1.384*10^-6 M
so,
[OH-] = 1.384*10^-6 M
pOH = -log [OH-]
= -log (1.384*10^-6)
= 5.86

pH = 14-pOH = 14-5.86 = 8.14

Answer: 8.14