Question

What is the pH of the a buffer solution containing 0.10 moles of acetic acid and 0.13 moles of sodium acetate in 1.0 L after adding 0.02 moles of KOH?

Answer 1

Answer – Given, moles of acetic acid = 0.10 moles , moles of acetate = 0.13 moles

Volume = 1.0 L , moles of KOH = 0.02 moles

When we added the base then it is reacted with acetic acid, of there is moles of acetic acid decrease and moles of acetate increase.

Moles of acetic acid = 0.10 moles – 0.02 moles = 0.080 moles

Moles of acetate = 0.13 moles + 0.02 moles = 0.15 moles

[CH₃COOH] = 0.080 moles / 1.0 L = 0.080 M

[CH₃COOH] = 0.150 moles / 1.0 L = 0.15 M

We know the pKa for the acetic acid is 4.74

We know the Henderson Hasselbalch equation –

pH = pKa + log [Conjugate base] / [Acid]

      = 4.74 + log 0.15 M / 0.080 M

       = 5.01