Ka = (1) [H+] =  = 2.03 × 10–3 (1) pH = 2.69 (1)  pH should be given to 2 decimal places penalise answer to 1 d.p. once in question 5 (d) moles OH– added = 1.875 × 10–3 = moles A– = moles HA left (1) or [A–] = [HA] Ka = [H+] or pH = pKa (1)  pH = 4.56 (1) 3  [13] 14. Penalise pH given to 1 dp first time it would have scored only (a) (i) Kw = [H+] [OH–] (1) (ii) pH = [H+] (1) –log or in words or below unless contradiction (iii) [H+] = (1) Calculation: = 2.34 × 10–7 pH = 6.63 or 6.64 (1)  pure water [H+] = [OH–] (1) 5  Explanation: (b) (i) [OH–] = 0.150 [H+] = 10–14/0.15 = 6.66 × 10–14  or pOH = 0.82  pH = 13.18 (1) or pH= 13.17 (ii) moles OH– = (35 × 10–3) × 0.150 = 5.25 × 10–3 (1)a moles H+ = (40 × 10–3) × 0.120 = 4.8(0) × 10–3 (1)b  excess moles of OH– = 4.5 × 10–4 (1)c  [OH–] = (4.5(0) × 10–4) × 1000/75d (1)e [H+] = = 1.66 × 10–12 or pOH = 2.22 pH = 11.78 (1)f  or 11.77 8 (c) (i) Ka = (1) (ii) [H+] = 1.80 × 10–2 × 0.150 = 2.70 × 10–3 (1) Ka = = 4.86 × 10–5 (1) mol dm–3 (1) 5 or = 4.95 × 10–5
Ka = (1)  [H+] = = 2.03 × 10–3 (1)  pH = 2.69 (1) pH should be given to 2 decimal places penalise answer to 1 d.p. once in question 5 (d) moles OH– added = 1.875 × 10–3 = moles A– = moles HA left (1) or [A–] = [HA]  Ka = [H+] or pH = pKa (1)  pH = 4.56 (1) 3 [13] 14. Penalise pH given to 1 dp first time it would have scored only (a) (i) Kw = [H+] [OH–] (1) (ii) pH = –log [H+] (1) or in words or below unless contradiction (iii) Calculation: [H+] = (1) = 2.34 × 10–7  pH = 6.63 or 6.64 (1) Explanation: pure water  [H+] = [OH–] (1) 5 (b) (i) [OH–] = 0.150  [H+] = 10–14/0.15 = 6.66 × 10–14 or pOH = 0.82  pH = 13.18 (1) or pH= 13.17 (ii) moles OH– = (35 × 10–3) × 0.150 = 5.25 × 10–3 (1)a moles H+ = (40 × 10–3) × 0.120 = 4.8(0) × 10–3 (1)b  excess moles of OH– = 4.5 × 10–4 (1)c  [OH–] = (4.5(0) × 10–4) × 1000/75d (1)e [H+] = = 1.66 × 10–12 or pOH = 2.22  pH = 11.78 (1)f or 11.77 8 (c) (i) Ka = (1) (ii) [H+] = 1.80 × 10–2 × 0.150 = 2.70 × 10–3 (1) Ka = = 4.86 × 10–5 (1) mol dm–3 (1) 5 or = 4.95 × 10–5