pH = 4.15 1 [16] 21. (a) Concentration of acid: m1v1 = m2v2 hence 25 × m1 = 18.2 × 0.150 OR moles NaOH = 2.73 ×10–3; 1 m1 = 18.2 × 0.150/25= 0.109; 1 (b) (i) = [H+]2 / [HA]; 1 Ka=[H+][A–]/[HA] not Ka (ii) pKa = 1 –logKa; (iii) [A–] = [HA]; 1 hence = [H+] [A–] / [HA] = [H+] Ka and = –logKa –log[H+]; 1 (c) ratio [A–] : [HA] remains constant; 1 hence as [H+] [HA] / [A–]; [H+] remains constant; 1 = Ka (d) (i) pH of 0.250 mol dm-3 HCl = 0.60 and pH of 0.150 mol dm-3 HCl = 0.82; 1 pH change = 0.22; 1 (ii) moles HCl = 30 × 0.250 × 10–3 = v × 0.150 × 10–3 = 7.50 × 10–3 OR v =30 × 0.250 × 10–3/ 0.150 × 10–3 =50; 1 water added = 50 30 = 20 cm3; 1 – [12] 22. (a) proton or H+ donor (1) 1 (b) (i) partially ionised or dissociated (1) not fully (ii) NH3 (1) not NH4OH – but allow in equation not H2O but allow in equation if both weak acid and base stated –
pH = 4.15 1 [16] 21. (a) Concentration of acid: m1v1 = m2v2 hence 25 × m1 = 18.2 × 0.150 OR moles NaOH = 2.73 ×10–3; 1 m1 = 18.2 × 0.150/25= 0.109; 1 (b) (i) Ka=[H+][A–]/[HA] not Ka = [H+]2 / [HA]; 1 (ii) pKa = –logKa; 1 (iii) [A–] = [HA]; 1 hence Ka = [H+] [A–] / [HA] = [H+] and –logKa = –log[H+]; 1 (c) ratio [A–] : [HA] remains constant; 1 hence as [H+] = Ka [HA] / [A–]; [H+] remains constant; 1 (d) (i) pH of 0.250 mol dm-3 HCl = 0.60 and pH of 0.150 mol dm-3 HCl = 0.82; 1 pH change = 0.22; 1 (ii) moles HCl = 30 × 0.250 × 10–3 = v × 0.150 × 10–3 = 7.50 × 10–3 OR v =30 × 0.250 × 10–3/ 0.150 × 10–3 =50; 1 water added = 50 – 30 = 20 cm3; 1 [12] 22. (a) proton or H+ donor (1) 1 (b) (i) partially ionised or dissociated (1) not fully (ii) NH3 (1) not NH4OH – but allow in equation not H2O – but allow in equation if both weak acid and base stated
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