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 95 mL of KOH. The Ka of HF is 3.5 × 10-4.

Answer 1

Given:

M(HF) = 0.2 M

V(HF) = 100 mL

M(KOH) = 0.1 M

V(KOH) = 95 mL

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

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

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.1 M * 95 mL = 9.5 mmol

We have:

mol(HF) = 20 mmol

mol(KOH) = 9.5 mmol

9.5 mmol of both will react

excess HF remaining = 10.5 mmol

Volume of Solution = 100 + 95 = 195 mL

[HF] = 10.5 mmol/195 mL = 0.0538M

[F-] = 9.5/195 = 0.0487M

They form acidic buffer

acid is HF

conjugate base is F-

Ka = 3.5*10^-4

pKa = - log (Ka)

= - log(3.5*10^-4)

= 3.456

use:

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

= 3.456+ log {4.872*10^-2/5.385*10^-2}

= 3.412

Answer: 3.41