Question

A 100mL sample of.20 M HF is titrated with 0.10 M KOH. Determine the pH of the solution after the addition of 60.0 mL of KOH. Ka (HF) = 3.5*10^-4

Answer 1

Given:

M(HF) = 0.2 M

V(HF) = 100 mL

M(NaOH) = 0.1 M

V(NaOH) = 60 mL

mol(HF) = M(HF) * V(HF)

mol(HF) = 0.2 M * 100 mL = 20 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.1 M * 60 mL = 6 mmol

We have:

mol(HF) = 20 mmol

mol(NaOH) = 6 mmol

6 mmol of both will react

excess HF remaining = 14 mmol

Volume of Solution = 100 + 60 = 160 mL

[HF] = 14 mmol/160 mL = 0.0875M

[F-] = 6/160 = 0.0375M

They form acidic buffer

acid is HF

conjugate base is F-

Ka = 3.5*10^-4

pKa = - log (Ka)

= - log(3.5*10^-4)

= 3.4559

use:

pH = pKa + log {[conjugate base]/[acid]}

= 3.4559+ log {0.0375/0.0875}

= 3.09

Answer: 3.09