A buffer contains 0.10 mol of propionic acid (C2H5COOH) and 0.27 mol of sodium propionate (C2H5COONa) in 1.20 L.?

A buffer contains 0.10 mol of propionic acid (C₂H₅COOH) and 0.27 mol of sodium propionate (C₂H₅COONa) in 1.20 L. What is the pH of this buffer?

Refer to: https://www.chm.uri.edu/weuler/chm112/refmater/KaTable.html
Ka for C₂H₅COOH = 1.3 × 10⁻⁵

[C₂H₅COO⁻]/[C₂H₅COOH] = (Moles of C₂H₅COO⁻)/(Moles of C₂H₅COOH) = 0.27/0.10

The dissociation of propionic acid:
C₂H₅COOH(aq) + H₂O(ℓ) ⇌ C₂H₅COO⁻(aq) + H₃O⁺(aq)   Ka = 1.3 × 10⁻⁵

Henderson-Hasselbach equation:
pH = pKa + log([C₂H₅COO⁻]/[C₂H₅COOH])
pH = -log(1.3 × 10⁻⁵) + log(0.27/0.10) = 5.32

====
What is the pH of the buffer after the addition of 0.02 mol of NaOH?

Addition of each mole of NaOH converts 1 mole of C₂H₅COOH to 1 mole of C₂H₅COO⁻.
Moles of C₂H₅COOH after addition = 0.10 - 0.02 = 0.08
Moles of C₂H₅COO⁻ after addition = 0.27 + 0.02 = 0.29

[C₂H₅COO⁻]/[C₂H₅COOH] = (Moles of C₂H₅COO⁻)/(Moles of C₂H₅COOH) = 0.29/0.08

Apply Henderson-Hasselbach equation:
pH = pKa + log([C₂H₅COO⁻]/[C₂H₅COOH])
pH = -log(1.3 × 10⁻⁵) + log(0.29/0.08) = 5.45

====
What is the pH of the buffer after the addition of 0.02 mol of HI?

Addition of each mole of HI converts 1 mole of C₂H₅COO⁻ to 1 mole of C₂H₅COOH.
Moles of C₂H₅COOH after addition = 0.10 + 0.02 = 0.12
Moles of C₂H₅COO⁻ after addition = 0.27 - 0.02 = 0.25

[C₂H₅COO⁻]/[C₂H₅COOH] = (Moles of C₂H₅COO⁻)/(Moles of C₂H₅COOH) = 0.25/0.12

Apply Henderson-Hasselbach equation:
pH = pKa + log([C₂H₅COO⁻]/[C₂H₅COOH])
pH = -log(1.3 × 10⁻⁵) + log(0.25/0.12) = 5.20