What is the difference in pH for 1/3 and 2/3 stages of neutralisation of 0.1 M CH3COOH with 0.1 M NaOH.?

CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) ….. Ka = 1.8 × 10⁻⁵

Henderson Hasselbalch equation :
pH = pKa + log([CH₃COO⁻]/[CH₃COOH])

In the 1/3 stage of the neutralization :
1/3 of CH₃COOH is changed to CH₃COO⁻.
Hence, [CH₃COO⁻]/[CH₃COOH] = (1/3) / [1 - (1/3)] = 1/2
pH = pKa + log([CH₃COO⁻]/[CH₃COOH])
pH = -log(1.8 × 10⁻⁵) + log(1/2)
pH = 4.44

In the 2/3 stage of the neutralization :
2/3 of CH₃COOH is changed to CH₃COO⁻.
Hence, [CH₃COO⁻]/[CH₃COOH] = (2/3) / [1 - (2/3)] = 2
pH = pKa + log([CH₃COO⁻]/[CH₃COOH])
pH = -log(1.8 × 10⁻⁵) + log(2)
pH = 5.05