Answer: A, B, C
pH = 4.80 + log(0.10/0.20) = 4.499 (a true). Dilution keeps the ratio (b true). 0.05 mol NaOH gives HA = 0.15, A- = 0.15 so pH = pKa (c true). 0.10 mol HCl uses up all 0.10 mol A-, leaving HA = 0.30 mol and excess nothing of salt: buffer destroyed, formula not applicable (d false).