A buffer solution was formed by mixing 20.0 cm^3 of sodium hydroxide solution of concentration 0.100 mol dm^–3 with 25.0 cm^3 of ethanoic acid of concentration 0.150 mol dm^–3. CH3COOH + NaOH---CH3COONa + H2O Calculate the pH of this buffer solution.

Ka for ethanoic acid = 1.74 × 10–5 mol dm^–3 also given in question. moles = concentration x volume / 1000Ka = [H+] [A-]/[HA]pH = -log [H+]
initial moles:NaOH: 20x0.1/1000 = 0.002molCH₃COOH: 25 x 0.15 / 1000 = 0.00375
NaOH< CH₃COOH and react in 1:1 ratio, so all NaOH reacts.
final moles:CH₃COO- Na+ = initial moles NaOH = 0.002molCH₃COOH = 0.00375 - 0.002 = 0.00175
final concentrations (remember to use total volume of 45 cm³)[CH₃COO- Na+] = 1000 x 0.002 / 45 = 0.04444 moldm^-3 = [A-][CH₃COOH] = 1000 x 0.00175 /45 = 0.03889 moldm^-3 = [HA]from Ka:[H+] = Ka [HA] / [A-] = 1.74 x 10 ^-5 x 0.03889 / 0.04444 = 1.523 x 10 ^-5pH = -log [H+] = 4.82