Question

0.5 L of a buffer solution contains 0.44 M acetic acid and 0.44 M NaCH3COO, and has a pH of 4.74. What will the pH be if 0.20 mol of HCL is added to the solution?

Answer 1

Answer – We are given, volume of buffer solution = 0.5 L , [CH₃COOH] = 0.44 M,

[NaCH₃COO] = [CH₃COO^-] = 0.44 M

First we need to calculate the moles of both acetic acid and its conjugate base.

Moles of CH₃COOH = 0.44 M * 0.5 L

                              = 0.22 moles

Moles of CH₃COO^- = 0.44 M * 0.5 L

                             = 0.22 moles

When we added 0.20 mol of HCL is added to the solution then moles of acid increased and moles of conjugate base decrease

Moles of CH₃COOH = 0.22 moles + 0.20 moles = 0.42 moles

Moles of CH₃COO^- = 0.22 moles - 0.20 moles = 0.02 moles

New molarity

[CH₃COOH] = 0.42 moles / 0.50 L = 0.84 M

[CH₃COO^-] = 0.02 moles / 0.50 L = 0.04 M

We know the pKa for acetic acid is 4.75 and we know the Henderson-Hasselbalch Equation

pH = pKa + log [CH₃COO^-] / [CH₃COOH]

      = 4.75 + log 0.04 M / 0.84 M

      = 3.43

So, the pH be if 0.20 mol of HCL is added to the solution is 3.43