Question

50 mL of 0.100 M HF is titrated with 55 mL of 0.100 M KOH. What is the pH of the solution. Ka of HF = 6.8 x 10-4

Answer 1

Given:

M(HF) = 0.1 M

V(HF) = 50 mL

M(KOH) = 0.1 M

V(KOH) = 55 mL

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

mol(HF) = 0.1 M * 50 mL = 5 mmol

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

mol(KOH) = 0.1 M * 55 mL = 5.5 mmol

We have:

mol(HF) = 5 mmol

mol(KOH) = 5.5 mmol

5 mmol of both will react

excess KOH remaining = 0.5 mmol

Volume of Solution = 50 + 55 = 105 mL

[OH-] = 0.5 mmol/105 mL = 0.0048 M

use:

pOH = -log [OH-]

= -log (4.762*10^-3)

= 2.3222

use:

PH = 14 - pOH

= 14 - 2.3222

= 11.6778

Answer: 11.68