Ka for ethanoic acid = 1.74 × 10–5 mol dm–3 moles = concentration x volume / 1000 Ka = [H+] [A-]/[HA] pH = -log [H+] Initial moles of NaOH: 20x0.1/1000 = 0.002mol Initial moles of CH3COOH: 25 x 0.15 / 1000 = 0.00375 mol Moles of NaOH < Moles of CH3COOH and react in 1:1 ratio, so all NaOH reacts. final moles: 0.00375 - 0.002 = 0.00175 mol final concentrations [CH3COO- Na+] = 1000 x 0.002 / 45 = 0.04444 mo ldm-3 [A-][CH3COOH] = 1000 x 0.00175 /45 = 0.03889 mol dm-3 [HA]from Ka:[H+] = Ka [HA] / [A-] = 1.74 x 10 -5 x 0.03889 / 0.04444 = 1.523 x 10 -5 pH = -log [H+] = 4.82