20 mL of 0.2 M NaOH is added to 50 mL of  0.2 M CH₃COOH. Hence, (pH- pK_a) is

20 mL of 0.2 M NaOH = 4 millimoles

50 mL of 0.2 M CH₃COOH = 10 millimoles

          CH₃COOH + NaOH $\rightleftharpoons$ CH₃COONa + H₂O

Initial          10             4                 0                   0

After           6              0                 4                   0

Resultant mixture is a buffer containing 6 millimoles of CH₃COOH and 4 millimoles of CH₃COONa.

$\therefore \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left[{\mathrm{CH}}_3\mathrm{COONa}\right]}{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]}\Rightarrow \mathrm{pH}-{\mathrm{pK}}_{\mathrm{a}}=\log \frac{2}{3}$