Question

The Ka value for HF is 3.5×10?4.

Calculate the change in pH when 2.0×10?2mol of NaOH is added to 0.50 L of a buffer solution that is 0.15 M in HF and 0.20 M in NaF.

Answer 1

The Ka value for HF is 3.5×10?4.

Calculate the change in pH when 2.0×10?2mol of NaOH is added to 0.50 L of a buffer solution that is 0.15 M in HF and 0.20 M in NaF.

Solution:

Ka of HF = 3.5 X 10-4

pKa = -logKa = 3.46

The pH of buffer is

pH = pKa + log[salt]/[acid]

a) initial pH

pH = 3.46 + log [0.20 / 0.15] = 3.58

b) after addition of NaOH

Moles of base added = 0.02

They will react with acid to form same moles of salt

Initial moles of acid = Molartiy X volume = 0.15 *0.5 = 0.075 moles

moles of acid after addition of base = 0.075 - 0.02 =0.055 moles

Moles of salt initially = 0.2 X 0.5 = 0.1 moles

Moles of salt after addition of NaOH = 0.1 + 0.02 = 0.12 moles

pH = 3.46 + log[salt] / [acid] = 3.46 + log [0.12/0.055] = 3.8

Change is = 3. 8 -3.58 = 0.22