Equilibria

8(a)(i) M1: CH CO H and CH CO – 3 3 2 3 2 M2: due to buffering action / acting as a buffer solution M3: CH CO H reacts with NaOH / OH– (forming CH CO – and water) 3 2 3 2 OR OH- reacts with H+ and equilibrium CH CO H ⇌ CH CO – + H+ shifts to the right 3 2 3 2 8(a)(ii) identifying CH CO – is present (with water) at the equivalence point 1 3 2 OR CH CO – react with water forming OH– 3 2 OR titrating a weak acid with a strong base © UCLES 2021 Page 15 of 17 8(b) M1: moles MnO – = 0.025 × 0.0201 = 5.025 × 10–4 3 4 moles V2+ = 5.025 × 10–4 × 5 / 3 = 8.375 × 10–4 M2: moles VO – = 8.375 × 10–4 3 mass of NH VO = 116.9 × 8.375 × 10–4 = 0.0979 g 4 3 M3: % Purity of NH VO = 100 × 0.0979 / 0.15 = 65.3 must be 3 sf 4 3 © UCLES 2021 Page 16 of 17

8(a)(i) M1: CH CO H and CH CO – 3 3 2 3 2 M2: due to buffering action / acting as a buffer solution M3: CH CO H reacts with NaOH / OH– (forming CH CO – and water) 3 2 3 2 OR OH- reacts with H+ and equilibrium CH CO H ⇌ CH CO – + H+ shifts to the right 3 2 3 2 8(a)(ii) identifying CH CO – is present (with water) at the equivalence point 1 3 2 OR CH CO – react with water forming OH– 3 2 OR titrating a weak acid with a strong base © UCLES 2021 Page 15 of 17 8(b) M1: moles MnO – = 0.025 × 0.0201 = 5.025 × 10–4 3 4 moles V2+ = 5.025 × 10–4 × 5 / 3 = 8.375 × 10–4 M2: moles VO – = 8.375 × 10–4 3 mass of NH VO = 116.9 × 8.375 × 10–4 = 0.0979 g 4 3 M3: % Purity of NH VO = 100 × 0.0979 / 0.15 = 65.3 must be 3 sf 4 3 © UCLES 2021 Page 16 of 17

2(a)(i H +Cl ( CH ) CO − 1 K =   2 3 2  [1] a ( ) Cl CH CO H  2 3 2  2(a)(ii) pK = – logK OR K = 10–pKa [1] 1 a a a 2(a)(iii) [H+] = 10–4.0 = 1 × 10–4 [1] 1 © UCLES 2021 Page 5 of 16 2(a)(iv) • [HCl] = 1 × 10–4 ecf 2(a)(iii) 2 • K = 10–4.52 = 3.02 × 10–5 a • [Cl(CH ) CO H] = (1 × 10–4)2 / 3.02 × 10–5 2 3 2 = 3.3 × 10–4 ecf [ HCl ] 1×10 −4 • = =0.302 min 2sf ecf Cl ( CH ) CO H 3.3×10 −4  2 3 2  2(b) M1 [H+] = 10–5 OR 1 × 10–5 3 K = [H+][A–] / [HA] OR pH = pKa + log [A–] / [HA] a   [1] M2 moles of A– = (1.35 × 10–5)(5 / 74) / (1 × 10–5) moles of A– = 0.0912 [1] ecf M3 mass of sodium propanoate = 0.0912 × 96 = 8.76 [1] min 2sf ecf 2(c) all of the (sodium) propanoate (ion) has been 2 protonated / converted to (propanoic) acid / neutralised [1] H+ is in excess / H+ is 0.1 mol dm–3 (from the H SO ) [1] 2 4 Question Answer Marks

2(a) M1 reference to equilibrium not being established / not at equilibrium [1] 2 M2 concentration in water too high or conc in octan-1-ol too low owtte [1] 2(b)(i) 1.1 / 1.13 / 1.127 / 1.1272 [1] 1 2(b)(ii) M1 6.760 = (0.762 – x) / x [1] 3 M2 7.76x = 0.762 x = 0.0982 g (correct value scores 2) [1] M3 0.0982 / 74 = 0.0013 / 0.00133 mol [1] 2(c)(i) 1.0x10–9 [1] 2 mol3 dm–9 [1] 2(c)(ii) yellow precipitate / yellow solid / yellow crystals [1] 2 common ion effect / K has been exceeded [1] sp Question Answer Marks

2(a)(i H +Cl ( CH ) CO − 1 K =   2 3 2  [1] a ( ) Cl CH CO H  2 3 2  2(a)(ii) pK = – logK OR K = 10–pKa [1] 1 a a a 2(a)(iii) [H+] = 10–4.0 = 1 × 10–4 [1] 1 © UCLES 2021 Page 5 of 16 2(a)(iv) • [HCl] = 1 × 10–4 ecf 2(a)(iii) 2 • K = 10–4.52 = 3.02 × 10–5 a • [Cl(CH ) CO H] = (1 × 10–4)2 / 3.02 × 10–5 2 3 2 = 3.3 × 10–4 ecf [ HCl ] 1×10 −4 • = =0.302 min 2sf ecf Cl ( CH ) CO H 3.3×10 −4  2 3 2  2(b) M1 [H+] = 10–5 OR 1 × 10–5 3 K = [H+][A–] / [HA] OR pH = pKa + log [A–] / [HA] a   [1] M2 moles of A– = (1.35 × 10–5)(5 / 74) / (1 × 10–5) moles of A– = 0.0912 [1] ecf M3 mass of sodium propanoate = 0.0912 × 96 = 8.76 [1] min 2sf ecf 2(c) all of the (sodium) propanoate (ion) has been 2 protonated / converted to (propanoic) acid / neutralised [1] H+ is in excess / H+ is 0.1 mol dm–3 (from the H SO ) [1] 2 4 Question Answer Marks

1(a)(i) oxidising agent [1] 2 H O + 2H+ + 2I– → 2H O + I [1] 2 2 2 2 1(a)(ii) M1: K (93.8) = [I (cyclohexane)] ÷ [I (aq)] 2 pc 2 2 93.8 = (0.390 / 15) ÷ (x / 20) M2: mass of I (aq), x = 5.54 × 10–3 (g) ecf 2 1(a)(iii) • K would be lower 2 pc • hexan-2-one is more polar (than cyclohexane) OR hexan-2-one is polar AND cyclohexane is non-polar • I is (therefore) less soluble in hexan-2-one 2 All three correct for two marks 1(b)(i) enthalpy change when one mole of a solute AND dissolves in water to form a solution of infinite dilution 1 1(b)(ii) –(–629) + (–322) + (–293) = (+)14 (kJ mol–1) 1 1(b)(iii) (cationic) charge density decreases Li+ to K+ [1] 2 so lattice energies become less negative / less exothermic AND because less attraction between ions [1] 1(c)(i) M1: moles of thiosulfate = 0.02230 × 0.150 = 3.345 × 10–3 2 M2: [Cu2+] = 2 × ½ × 3.345 × 10–3 ÷ 0.0250 = 0.134 (mol dm–3) ecf 1(c)(ii) starch 1 1(c)(iii) 1 3d xy 1(d) Reduction of Fe3+: 2Fe3+ + 2I– → 2Fe2+ + I [1] 2 2 Regeneration of Fe3+: 2Fe2+ + S O 2– → 2Fe3+ + 2SO 2– [1] 2 8 4 © UCLES 2022 Page 5 of 12 1(e) M1: ([Hg2+]) = 1.00 × 10–7 ÷ 454.4 = 2.20 × 10–10 (mol dm–3) 3 M2: K = [Hg2+][I–]2 = 4[Hg2+]3 = 4.26 × 10–29 ecf sp M3: units = mol3 dm–9 ecf Question Answer Marks

4(a)(i) M1 all five points plotted correctly 2 M2 best-fit straight line (ruler) with negative gradient drawn 4(a)(ii) M1 gradient correctly calculated OR gradient working seen 2 M2 gradient = –So So evaluated correctly ecf So = (+)160 ± 10 (J K–1 mol–1) 4(b)(i) 2HCO –  CO 2– + CO + H O 1 3 3 2 2 © UCLES 2022 Page 8 of 16 4(b)(ii) M1 ionic radius of M+ / cationic radius increases 2 OR charge density of ion / M+ decreases down Group 1 M2 less distortion / polarisation of the anion / HCO – 3 OR CO bond / C-O / C=O less weakened 4(c)(i) M1 solution which resists changes in pH when 2 opposes / resists change in pH M2 when small amount of acid / H+ or alkali / base / OH– is added 4(c)(ii) M1 (with acid) HCO – + H+ → H CO OR HCO – + H O+ → H CO + H O 2 3 2 3 3 3 2 3 2 M2 (with alkali) H CO + OH– → HCO – + H O 2 3 3 2 4(c)(iii) M1 K = 10–6.35 = 4.47  10–7 3 a M2 [H+] = 4.47  10–7 / 14.1 = 3.17  10–8 ecf M3 pH = –log [H+] = 7.5 ecf from a calculated [H+] min 2sf Question Answer Marks

4(a)(i) ratio of concentration of the solute in two solvents at equilibrium [1] 1 4(a)(ii) 3.50 = (1.62 / 100) / (0.38 / x) [1] 2 x= 82 (cm3) (82.0987654) [1] ecf M1 min 2sf 4(b)(i) pH resists change when small amount of acid or base added 1 4(b)(ii) CH CH CH COO– / salt of butanoic acid / sodium butanoate 1 3 2 2 AND NaOH 4(c)(i) K = [Al 3+][OH–]3 1 sp 4(c)(ii) [OH-] = 3  2.47  10–9 OR 7.41  10–9 [1] 3 K = [2.47  10–9][7.41  10–9]3 sp = 1.01  10–33 min 2sf ecf from 1st mark [1] mol4 dm–12 [1] © UCLES 2022 Page 8 of 16

4(a)(i) 27 x4 = 2  10–39 1 x = 9.28  10–11 mol dm–3 4(a)(ii) [Fe3+][0.02]3 = 2  10–39 [1] 2 [Fe3+] = 2.5  10–34 [1] 4(a)(iii) common ion (effect) 1 4(b)(i) two species that differ by one H+ ion 1 4(b)(ii) HBrO BrO– 1 H O+ H O 3 2 4(b)(iii) [acid] = 4  10–2 2 [H+]2 = 8.0  10–11 [1] pH = 5.05 [1] 4(b)(iv) 8.70 1 Question Answer Marks

4(a)(i) ratio of concentration of the solute in two solvents at equilibrium [1] 1 4(a)(ii) 3.50 = (1.62 / 100) / (0.38 / x) [1] 2 x= 82 (cm3) (82.0987654) [1] ecf M1 min 2sf 4(b)(i) pH resists change when small amount of acid or base added 1 4(b)(ii) CH CH CH COO– / salt of butanoic acid / sodium butanoate 1 3 2 2 AND NaOH 4(c)(i) K = [Al 3+][OH–]3 1 sp 4(c)(ii) [OH-] = 3  2.47  10–9 OR 7.41  10–9 [1] 3 K = [2.47  10–9][7.41  10–9]3 sp = 1.01  10–33 min 2sf ecf from 1st mark [1] mol4 dm–12 [1] © UCLES 2022 Page 8 of 16

6(a)(i) SiO OR Al O OR silica OR alumina OWTTE 1 2 2 3 6(a)(ii) Cd2+ AND R of M (= 2/5) = 0.40 / same R as in the Table 6.1 1 f f 6(b) metal cations are less soluble in butan-1-ol (than in water) 1 OR metal cations weaker ion-dipole forces with butan-1-ol 6(c) H O AND SCN– 1 2 6(d)(i) C H O K + KOH → C H O K + H O OR C H O K + OH– → C H O K– + H O 1 8 5 4 8 4 4 2 2 8 5 4 8 4 4 2 © UCLES 2023 Page 16 of 17 6(d)(ii) M1 initially: moles of KOH = 0.150  20.0 ÷ 1000 = 3.0  10–3 1 AND moles of acid HA = 0.100  50.0÷1000 = 5.0  10–3 M2 at equilibrium: moles of salt KA = 3.0  10–3 ecf 1 AND moles of acid HA = 5.0  10–3 – moles of NaOH = 2.0  10–3 ecf M3 K = 10–5.40 = 3.98  10–6 1 a 3.98  10–6 = [H+] [3.0  10–3] ÷ [2.0  10–3] [H+] = 2.65  10–6 ecf M4 pH = –log(2.65  10–6) = 5.58 / 5.6 ecf on a calculated and identified [H+] 1 ALLOW alternative approach using Henderson-Hasselbalch equation for M3 and M4 © UCLES 2023 Page 17 of 17

2(a) (K =)[H+][OH–] OR (K =)[H3O+][OH–] [1] 1 w w 2(b)(i) M1: pH values 1.5 AND 12.5 [1] 2 M2: conc of HCl = 10–1.5 = 0.0316 ecf [1] min 2sf 2(b)(ii) sodium chloride / NaCl AND 7 [1] 1 2(b)(iii) (mixture Y) 1 to 3 AND (mixture Z) 11 to 13 [1] 1 2(c)(i) H SO > CH CCl COOH > CH CH COOH [1] u / c 3 2 4 3 2 3 2 explanation • H SO is fully dissociated / strong acid 2 4 • CH CH COOH / CH CCl COOH are partly dissociated / weak acids 3 2 3 2 • Cl / chlorine is electron-withdrawing / electronegative • alkyl group (in CH3CH2COOH) is electron donating • stabilises / destabilises anion OR weakens / strengthens O-H bond (linked correctly) • correct reference to release of / donation of / form H+ / proton any two [1] any four [2] 2(c)(ii) • H O+ and H O 2 3 2 • H SO and HSO – 2 4 4 • HSO – and SO 2– 4 4 any one correct pair [1] all three correct pairs [2] 2(d)(i) (the) ratio of the concentrations (of a solute between) 1 two solvents / two liquids (at) equilibrium [1] 2(d)(ii) mass = 5-x / x = 7.84 OR mass = x / 5-x = 7.84 1 mass = 0.5656 g mass = 4.43 g [1] min 2sf © UCLES 2023 Page 6 of 18 2(d)(iii) Any numbers in which volume of water is 7.84 times volume of hexane e.g. 78.4 cm3 water 10 cm3 hexane [1] u / c 1 e.g mass = 0.566 g 177.4 cm3 water 22.6 cm3 hexane ALLOW reverse ratio volume of hexane is 7.84 times volume of water when consistent with d(ii) 2(d)(iv) Q is CH (CH ) OH 1 3 2 4 AND it is least polar / contains a large non-polar hydrocarbon chain / stronger id-id forces with hexane in Q OWTTE [1] Question Answer Marks

2(a) resists pH change when small amount of acid or alkali is added [1] 1 2(b)(i) C H COOH + NaOH → C H COO–Na+ + H O [1] 1 6 5 6 5 2 2(b)(ii) C H COO–Na+ + HNO → C H COOH + NaNO [1] 1 6 5 3 6 5 3 2(c) [H+] = 7.08  10–5 [1] 3 [C H COOH] = 8.20  10–3 [1] 6 5 [C H COO–Na+] = 7.31  10–3 [1] 6 5 © UCLES 2023 Page 5 of 14 2(d) neutralisation is exothermic [1] 2 All of the C H COOH has reacted AND excess KOH 1] 6 5 2(e)(i) If [X] = [Mg2+] = [Mg(C H COO–) ] 2 6 5 2 4X3 = 1.76  10–7 so X = 3.53  10–3 [1] 0.940 g dm–3 [1] 2(e)(ii) lower than AND common ion effect [1] 1 Question Answer Marks

2(a) (K =)[H+][OH–] OR (K =)[H3O+][OH–] [1] 1 w w 2(b)(i) M1: pH values 1.5 AND 12.5 [1] 2 M2: conc of HCl = 10–1.5 = 0.0316 ecf [1] min 2sf 2(b)(ii) sodium chloride / NaCl AND 7 [1] 1 2(b)(iii) (mixture Y) 1 to 3 AND (mixture Z) 11 to 13 [1] 1 2(c)(i) H SO > CH CCl COOH > CH CH COOH [1] u / c 3 2 4 3 2 3 2 explanation • H SO is fully dissociated / strong acid 2 4 • CH CH COOH / CH CCl COOH are partly dissociated / weak acids 3 2 3 2 • Cl / chlorine is electron-withdrawing / electronegative • alkyl group (in CH3CH2COOH) is electron donating • stabilises / destabilises anion OR weakens / strengthens O-H bond (linked correctly) • correct reference to release of / donation of / form H+ / proton any two [1] any four [2] 2(c)(ii) • H O+ and H O 2 3 2 • H SO and HSO – 2 4 4 • HSO – and SO 2– 4 4 any one correct pair [1] all three correct pairs [2] 2(d)(i) (the) ratio of the concentrations (of a solute between) 1 two solvents / two liquids (at) equilibrium [1] 2(d)(ii) mass = 5-x / x = 7.84 OR mass = x / 5-x = 7.84 1 mass = 0.5656 g mass = 4.43 g [1] min 2sf © UCLES 2023 Page 6 of 18 2(d)(iii) Any numbers in which volume of water is 7.84 times volume of hexane e.g. 78.4 cm3 water 10 cm3 hexane [1] u / c 1 e.g mass = 0.566 g 177.4 cm3 water 22.6 cm3 hexane ALLOW reverse ratio volume of hexane is 7.84 times volume of water when consistent with d(ii) 2(d)(iv) Q is CH (CH ) OH 1 3 2 4 AND it is least polar / contains a large non-polar hydrocarbon chain / stronger id-id forces with hexane in Q OWTTE [1] Question Answer Marks

2(a)(i) ● conjugate base of acid I = OH– 1 ● acid II = HNO ● conjugate base of acid II = NO – 2 2 2(a)(ii) lone pair on the N can be donated to a proton /H+ 1 OR lone pair on the N can accept / gain a proton /H+ OR lone pair on the N can form a dative bond to a proton/H+ 2(a)(iii) H O + CH COOH → H O+ + CH COO– 1 2 3 3 3 2(b)(i) (K ) = [H+][OH–] ALLOW (K ) = K  K 1 w w a b 2(b)(ii) endothermic 1 AND equilibrium position moves right OR as water dissociates more OR K increases with temperature w 2(b)(iii) M1 (K increases with temperature so) pH of neutral solution decreases 1 w OR (from graph) K = 1.50  10–14 = [H+]2 w ∴ neutral pH = –½ log (1.50  10–14) = 6.91 OR [H+] = 10–7.00 ∴ [OH–] = 1.50  10–14 / 10–7.00 = 1.50  10–7 M2 pH 7 is therefore above neutral pH / is alkaline 1 OR [OH–] > [H+] (so alkaline) 2(c)(i) H 2 O(l) particles / molecules has more randomness / disorder 1 OR H 2 O(l) has more ways to arrange particles / energy (than in solid) 2(c)(ii) H O(g) particles / molecules has much more randomness / disorder 1 2 OR H O(g) has many more ways to arrange particles / energy (than in liquid) 2 2(c)(iii) +6030 / (70.1 – 48.0) = 272.85 / 272.9 / 273 K (which is 0 °C) 1 2(d)(i) (+)1.62 V 1 © Cambridge University Press & Assessment 2024 Page 7 of 13 2(d)(ii) [OH–] = 10–14/10–11 = 1  10–3 1 E = +0.40 – ½  0.059  log (10–3)2 = +0.40 – 0.059(–3) = +0.577 (V) min 2sf 1 Question Answer Marks

6(a)(i) species / molecules / pair that differ (by the presence or absence) of a H+ ion / proton 1 6(a)(ii) HCOOH + H 2 O ⇌ H 3 O+ + HCOO– 1 6(b)(i) [H+] [CH CH CH COO] 1 K  3 2 2 a [CH CH CH COOH] 3 2 2 © Cambridge University Press & Assessment 2024 Page 9 of 15 6(b)(ii) M1 K = 10–4.82 = 1.51  10–5 OR [H+] = 10–3.25 = 5.62  10–4 2 a M2 [HA] = (10–3.25)2 ÷ 1.51  10–5 = 0.021 / 0.0209 / 0.02089 (mol dm–3) ecf min 2sf 6(c)(i) M1 a solution that resists / opposes / minimises changes in pH 2 M2 when small amounts of acid / H+ and base / alkali / OH– are added to it 6(c)(ii) Method 1 3 M1 [H+] = 10–5.70 OR 2.00  10–6 OR 1.99526  10–6 M2 moles CH COONa = 0.400  0.125 = 5.00  10–2 3 moles CH COOH = (5.00  10–2  2.00  10–6) ÷ 1.78  10–5 OR 5.60  10–3 3 eqm M3 [CH COOH] = 5.60  10–3  1000 / 600 3 initial [CH COOH] = 0.00933–0.00936 (mol dm–3) ecf min 2sf 3 initial Method 2 M1 pK = –log(1.78  10–5) OR 4.75 a M2 pH = pK + log[B–] / [HA] a 5.7 = 4.75 + log (0.05 / (0.6x) 0.95 = log (0.05 / (0.6x) M3 100.95 = (0.05 / (0.6x) 5.348x = 0.05 x = 0.00935-0.00937 (mol dm–3) ecf min 2sf 6(d)(i) HCOOH → CO + 2H+ + 2e– 1 2 6(d)(ii) M1 Q = 3.75  40  60 OR 9000 C AND use of 96500 OR 1.6  10–19  6.02  1023 2 M2 moles of oxygen = 9000 ÷ 386000 = 0.0233 volume of O = 0.0233  24000 = 559.6 / 559.8 / 560 cm3 ecf min 2sf 2 ALLOW [use of Q ÷ (1.6  10–19  6.02  1023)] = 560.6 cm3 © Cambridge University Press & Assessment 2024 Page 10 of 15

1(a)(i) conjugate acid conjugate base 1 H PO – HPO 2– [1] BOTH 2 4 4 1(a)(ii) allows only small changes in pH / resists changes in pH [1] 2 when small amounts of acid or alkali/base / H+ or OH– are added [1] 1(a)(iii) Na+ HPO 2– + H+ → Na+H PO – + Na+ 2 2 4 2 4 or HPO 2– + H+ → H PO – [1] 4 2 4 Na+H PO – + NaOH → Na+ HPO 2– + H O 2 4 2 4 2 or H PO – + OH– → HPO 2– + H O [1] 2 4 4 2 1(a)(iv) HCO – / hydrogen carbonate [1] 1 3 1(b)(i) either 10–13.09 or [H+]= 8.1  10–14 seen [1] 2 8.1x10–14  [OH–] = 1  10–14 seen [1] 110−14 OR [OH–] = seen 8.110−14 ALLOW alternative method pOH = 0.91 [1] [OH–] = 10–0.91 seen [1] 1(b)(ii) 2 moles of OH– in one mole of E / X(OH) 1 2 OR there are two hydroxide ions in each formula unit of E / X(OH) [1] 2 1(b)(iii) 0.0615  0.25 = 0.0154 moles of E 1 RFM = 2.63 / 0.0154 = 170.8 / 171 OR RAM = 34 – 2.63 / 0.0154 = 137.1 /137 AND barium hydroxide / Ba(OH) [1] 2 © Cambridge University Press & Assessment 2024 Page 5 of 14 1(c) K = [Mg2+][OH–]2 OR K = 1.4  10-4  (2  1.4  10-4)2 [1] 3 sp sp K = 1.10  10–11 [1] min 2sf sp correct answer [2] mol3 dm–9 [1] ecf from M1 1(d) M1: H and H less exothermic (down the group) 3 latt hyd OR H and H more exothermic/more negative for Mg [1] latt/LE hyd M2: H changes more than H (down the group) latt hyd OR H changes more / faster for Mg [1] latt M3: H becomes more exothermic (down the group) sol OR H becomes less exothermic / less negative for Mg [1] sol mark independently Question Answer Marks

3(a)(i) [H+] = 10–12.35 OR [H+] = 4.47  10–13 [1] 2 K / 4.47  10–13 [1] w OR pOH = 1.65 [1] [OH–] = 10–1.65 [1] 3(a)(ii) K = (0.0112)(0.0224)2 OR K = [Ca2+][OH–]2 [1] 3 sp sp answer 5.62  10–6 [1] mol3 dm–9 [1] © Cambridge University Press & Assessment 2024 Page 6 of 12 3(a)(iii) white solid / white ppt AND pH between 7.01 and 12.34 [1] 3 common ion effect [1] hydroxide removed by precipitation OR ppt is Ca(OH) [1] 2 3(a)(iv) lattice energy and hydration energy greater for CaSO OR 3 4 lattice energy and hydration energy decrease down group [1] hydration energy decreases more / is dominant factor [1] enthalpy of solution is more endothermic for BaSO OR is more endothermic down grp [1] 4 3(b)(i) Ca + 2CH COOH → Ca(CH COO) + H [1] 1 3 3 2 2 3(b)(ii) Pair 1: CH COOH and CH COO– [1] 2 3 3 Pair 2: H O+ and H O OR H O and OH– [1] 3 2 2 3(b)(iii) [H+][CH COO−] 1 3 [1] [CH COOH] 3 3(b)(iv) [CH COO–] = 0.788, [CH COOH] = 0.270 2 3 3 [H+] = 5.96  10–6 [1] pH = 5.22 [1] 3(b)(v) CH COO– + H+ → CH COOH 2 3 3 OR Ca(CH COO) + 2H+ → 2CH COOH + Ca2+ [1] 3 2 3 CH COOH + OH– → CH COO– + H O [1] 3 3 2 Question Answer Marks

1(a)(i) conjugate acid conjugate base 1 H PO – HPO 2– [1] BOTH 2 4 4 1(a)(ii) allows only small changes in pH / resists changes in pH [1] 2 when small amounts of acid or alkali/base / H+ or OH– are added [1] 1(a)(iii) Na+ HPO 2– + H+ → Na+H PO – + Na+ 2 2 4 2 4 or HPO 2– + H+ → H PO – [1] 4 2 4 Na+H PO – + NaOH → Na+ HPO 2– + H O 2 4 2 4 2 or H PO – + OH– → HPO 2– + H O [1] 2 4 4 2 1(a)(iv) HCO – / hydrogen carbonate [1] 1 3 1(b)(i) either 10–13.09 or [H+]= 8.1  10–14 seen [1] 2 8.1x10–14  [OH–] = 1  10–14 seen [1] 110−14 OR [OH–] = seen 8.110−14 ALLOW alternative method pOH = 0.91 [1] [OH–] = 10–0.91 seen [1] 1(b)(ii) 2 moles of OH– in one mole of E / X(OH) 1 2 OR there are two hydroxide ions in each formula unit of E / X(OH) [1] 2 1(b)(iii) 0.0615  0.25 = 0.0154 moles of E 1 RFM = 2.63 / 0.0154 = 170.8 / 171 OR RAM = 34 – 2.63 / 0.0154 = 137.1 /137 AND barium hydroxide / Ba(OH) [1] 2 © Cambridge University Press & Assessment 2024 Page 5 of 14 1(c) K = [Mg2+][OH–]2 OR K = 1.4  10-4  (2  1.4  10-4)2 [1] 3 sp sp K = 1.10  10–11 [1] min 2sf sp correct answer [2] mol3 dm–9 [1] ecf from M1 1(d) M1: H and H less exothermic (down the group) 3 latt hyd OR H and H more exothermic/more negative for Mg [1] latt/LE hyd M2: H changes more than H (down the group) latt hyd OR H changes more / faster for Mg [1] latt M3: H becomes more exothermic (down the group) sol OR H becomes less exothermic / less negative for Mg [1] sol mark independently Question Answer Marks

2(a)(i) [CH COCH ] concentration stays (almost) constant / does not change 1 3 3 OR [CH COCH ] concentration does not affect rate 3 3 2(a)(ii) M1: first order w.r.t. H+ AND rate  4 / 3 and [H+]  4 / 3 (in expts 2 and 3) 2 M2: first order w.r.t. CH OH 3 AND [H+]  1.5, [CH OH]  1.5 and rate  2.25 / 1.52 (expts 1 and 2) 3 / [H+]  2, [CH OH]  1.5 and rate  3 / (2  1.5) (expts 1 and 3) 3 2(a)(iii) two quoted half-lives within range / 155–170 s AND (roughly) constant 1 2(b) M1: proposed mechanism = 4 3 M2: slow step contains only species from the rate equation/law OR slow step contains CH COCH and H+ from the rate equation/law 3 3 M3: stoichiometric / (mole) ratio / amounts AND corresponds to order of reaction / to that in the rate equation 2(c)(i) 1 © Cambridge University Press & Assessment 2025 Page 6 of 14 2(c)(ii) K = 10–3.95 = 1.122  10–4 2 a [H+]2 = (1.122  10–4  0.5) [H+] = √(1.122  10–4  0.5) = 7.49  10–3 pH = –log (7.49  10–3) pH = 2.13 min 2sf 2(d)(i) a solution that resists / minimises / opposes changes in pH 2 when small amounts of (strong) acid or base are added 2(d)(ii) M1: moles of conjugate base = 20 / 1000  1.00 = 0.0200 mol 3 AND moles of acid C left = 50 / 1000  0.500 – 0.0200 = 0.0300 mol M2: K = [H+]  0.0200 / 0.0300 a [H+] = 3 / 2  10–3.95 = 1.68  10–4 ecf from M1 M3: pH = –log (1.68  10–4) pH = 3.77 min 2sf ecf from a calculated [H+] M2 Question Answer Marks

3(a)(i) M1: K = [Cr3+] [OH–]3 2 sp M2: mol4 dm–12 3(a)(ii) M1: [OH–] = 3[Cr3+] 3 M2: 2.47  10–9 mol dm–3 M3: 2.54  10–7 g dm–3 3(a)(iii) common ion effect 1 3(b)(i) M1: [H+] = 1.23  10–3 2 M2: pH = 2.91 3(b)(ii) 13.0 1 3(b)(iii) M1: [CH CH CH COOH] = [CH CH CH COO–] 2 3 2 2 3 2 2 M2: pH = 4.82 3(c) M1: 0.148 g is left in water 2 M2: K = 3.01 OR 0.333 pc © Cambridge University Press & Assessment 2025 Page 8 of 16

5(a)(i) acid I HClO conjugate base of acid I ClO– 1 acid II H O conjugate base of acid II OH– 2 (a)(ii) M1: [H+] = 10–4.51 = 3.09  10–5 2 M2: [HClO] = (3.09  10–5)2  (3.70  10–8) = 0.0258 min 2sf (a)(iii) 3 HClO → HClO + 2 HCl 1 3 © Cambridge University Press & Assessment 2025 Page 10 of 17 5(b)(ii) hydroxide ions / NaOH / OH– AND it reacts with the acid to form the conjugate base / ethanoate ions 1 OR ethanoate ions / sodium ethanoate AND as it is the conjugate base (of ethanoic acid) 5(d)(i) M1: Use of [Ca2+] = [PO 3–] = 1.14  10–7 in the expression 3 4 OR K = (3.42  10–7)3  (2.28  10–7)2 sp M2: K = (3  1.14  10–7)3  (2  1.14  10–7)2 sp K = 2.079  10–33 sp OR 2.08  10–33 OR 2.1  10–33 min 2sf M3: units = mol5 dm–15 5(d)(ii) solubility decreases AND due to the common ion effect 1 Question Answer Marks

3(a)(i) M1: K = [Cr3+] [OH–]3 2 sp M2: mol4 dm–12 3(a)(ii) M1: [OH–] = 3[Cr3+] 3 M2: 2.47  10–9 mol dm–3 M3: 2.54  10–7 g dm–3 3(a)(iii) common ion effect 1 3(b)(i) M1: [H+] = 1.23  10–3 2 M2: pH = 2.91 3(b)(ii) 13.0 1 3(b)(iii) M1: [CH CH CH COOH] = [CH CH CH COO–] 2 3 2 2 3 2 2 M2: pH = 4.82 3(c) M1: 0.148 g is left in water 2 M2: K = 3.01 OR 0.333 pc © Cambridge University Press & Assessment 2025 Page 8 of 16

3(a)(i) M1: K = [Ni2+][IO –]2 2 sp 3 M2: units: mol3 dm–9 3(a)(ii) [Ni2+] = 2.3  10–2; [IO –] = 4.6  10–2 1 3 K = (2.3  10–2)  (4.6  10-2)2 = 4.87  10–5 sp 3(b)(i) Eo = 1.19 – (0.34) = (+)0.85 (V) AND (positive electrode is) right hand side / nickel electrode / platinum / iodate 1 cell 3(b)(ii) (E would be) less positive / more negative AND as [IO –] is less than 1.0 mol dm–3 / has a lower / smaller concentration 1 cell 3 3(b)(iii) M1: E is less positive ticked 2 cell M2: [IO –(aq)] is lowered due to the common ion effect 3 OR Ni(IO ) precipitating or shown by equation 3 2 OR solubility of Ni(IO ) decreasing 3 2 3(c)(i) [H+] = 10–pH = 10–0.47 = 0.34 / 0.339 (mol dm-3) min 2sf 1 3(c)(ii) [HIO ] = 1.0 – 0.34 = 0.66 (mol dm–3) min 2sf 1 3eqm AND [H+] = [IO –] = 0.34 (mol dm–3) min 2sf eqm 3 eqm 3(c)(iii) K = (0.34)2 / 0.66 = 0.17– 0.18 (mol dm–3) min 2sf 1 a 3(d)(i) M1: opposes / resists change in pH / controls pH / pH kept within a small range 2 M2: when small amount of acid / H+ or alkali / base / OH– is added 3(d)(ii) M1: k = {2.1  10–2}/{0.5  (1  10–3)2  0.012} = 4.20  108 min 2sf 2 M2: mol–4 dm12 s–1 3(d)(iii) rate will be (0.03 / 0.01)2 = 9 times as fast, so rate = 9  2.1  10–2 1 = 0.19 (0.189) (mol dm–3 s–1) min 2sf © Cambridge University Press & Assessment 2025 Page 10 of 17

3(a)(i) two species that differ by one proton / H+ 1 3(a)(ii) H PO – AND PO 3– 1 2 4 4 3(b)(i) M1: [H+] = 2.51  10–4 2 M2: [CH CH COOH] = 4.67  10–3 3 2 3(b)(ii) 3.98  10–11 1 3(b)(iii) [HCl] = 2.51  10–4 1 3(b)(iv) M1: top box ticked AND weaker acid 3 M2: alkyl group is electron donating M3: O–H bond strengthened / anion is destabilised 3(c)(i) NaOH AND CH CH COOH + OH– → CH CH COO– + H O 1 3 2 3 2 2 3(c)(ii) H SO AND conjugate base of CH CH COOH is not present 1 2 4 3 2 3(d)(i) 3.69  10–5 1 © Cambridge University Press & Assessment 2025 Page 8 of 15 3(d)(ii) M1: K = [Mn2+][OH–]2 2 sp M2: units = mol3 dm–9 3(d)(iii) 2.0  10–13 1 Question Answer Marks

3(a) 2 3(b) HO(CH ) COOH + 2PCl → Cl (CH ) COCl + 2POCl + 2HCl 2 2 3 5 2 3 3 organic product Cl (CH ) COCl [1] 2 3 rest of equation [1] © Cambridge University Press & Assessment 2026 Page 12 of 18 3(c)(i) [H+] = (K  [HA])½ = (1.91  10–5  0.150)½ OR 1.69  10–3 (mol dm–3) 2 a pH = –log [H+] = 2.77 min 2sf 3(c)(ii) (a solution that) resists / opposes / minimises changes in pH 2 when small (amounts / volumes) of acid / H+ and base / alkali / OH– are added to it 3(c)(iii) HO(CH ) COO– 1 2 3 3(c)(iv) M1: moles of HA initially = (0.150  25.0 / 1000) OR 3.75  10–3 mol 3 AND moles of NaOH / moles of A– = (0.200  10.0 / 1000) OR 2.00  10–3 mol M2: remaining moles of HA = 3.75  10–3 – 2.00  10–3 OR 1.75  10–3 mol M3: [H+] = K [HA]/[A–] = 1.91  10–5  a © Cambridge University Press & Assessment 2026 Page 13 of 18 ( ( 1 2 . 7 . 0 5 0   1 1 0 0 − − 3 3   0 0 . 0 . 0 3 3 5 5 ) ) = 1.67  10–5 AND pH = –log 1.67  10–5 = 4.78 min 2sf 10 3(d) (an excess of) acidified K Cr O / KMnO AND heat / boil / hot / reflux 1 2 2 7 4 3(e)(i) (the) ratio of the concentrations of a solute between two solvents at equilibrium 1 3(e)(ii) M1: [acid(ether)] = 0.0375 mol dm–3 AND [acid(aq)] = 0.2775 mol dm–3 2 M2: K = [acid(ether)] / [acid(aq)] = 0.135 pc

2(a)(i) –log[H+] 1 [ ] essential. CON –log(H+), –log [H+] , –ln[H+]. e 2(a)(ii) [H+][OH–] / K xK 1 [ ] essential but do not penalise in (i) and a b (ii). I 1  10–14. 2(b)(i) (10–2.9) 1 0.0013/0.00126/1.3x10–3/1.26x10–3 2(b)(ii) 2bi2 / 0.105 0.001262 / 0.105 ecf bi 1 1.5  10–5 / 1.51  10–5 / 1.53  10–5 ecf 2(c)(i) M1 [CH CH CH COOH] = 0.15 3 If [CH CH CH COOH] = 0.45 [0] but then 3 2 2 3 2 2 M2 [CH CH CH COO–Na+] in solution D = 0.036 / 0.0359 [salt] in mixture = 0.108 [1] ecf 3 2 2 [salt] in solution C = 0.162 [1] ecf VERIFY M3 [CH CH CH COO–Na+] in solution C = 0.054 / 0.0538 / 0.0539 / 0.0540 with working. 3 2 2 Another possible approach goes via moles of acid to get moles of salt, M1 moles acid = 0.045 mol M2 moles salt = 0.0108 / 0.0107 mol M3 original [salt] = 0.0538 / 0.0539 / 0.0540 2(c)(ii) CH CH CH COO–Na+ + H+ → CH CH CH COOH + Na+ 1 3 2 2 3 2 2 or CH CH CH COO– + H+ → CH CH CH COOH 3 2 2 3 2 2 or CH CH CH COO– + H SO → CH CH CH COOH + HSO – 3 2 2 2 4 3 2 2 4 or 2CH CH CH COO–Na+ + H SO → 2CH CH CH COOH + Na SO 3 2 2 2 4 3 2 2 2 4 2(c)(iii) CH CH CH COOH + OH– → CH CH CH COO– + H O 1 3 2 2 3 2 2 2 or CH CH CH COOH + KOH → CH CH CH COOK + H O 3 2 2 3 2 2 2 or CH CH CH COOH + K+OH– → CH CH CH COO–K+ + H O 3 2 2 3 2 2 2 © Cambridge University Press & Assessment 2026 Page 9 of 21 Trace ID: PC-F9EA0E463D Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P97a01p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/41 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 2(d) M1 2,2-dichlorobutanoic and 2 chlorine is electron withdrawing / negative inductive effect / electronegative I references to C=O, O, alkyl groups. M2 weakens O–H bond / stabilises anion / stabilises conjugate base / stabilises A –O–H bond and OH bond but CON – carboxylate ion OH bond, C–O–H bond. Question Answer Marks Guidance

4(a) (the) ratio of the concentrations of a solute / substance / compound 1 CON if in terms of concentration of ‘solutes’. between two (immiscible) solvents / two liquids (at) equilibrium 4(b) EITHER (3 Bullet 3 subsumes bullet 2 → [1]. Solution 1 Bullet 4 subsumes bullets 1,2,3 → [2]. M1M2 • Initial moles of (COOH) = 2.7 / 90 or 0.03 A ecf for bullet 4 from calculated values of 2 • moles of NaOH = 0.0165  0.200 or 0.0033 bullets 1 and 3, for example using the incorrect • moles of (COOH) 2 in ether = 0.0033 / 2 or 0.00165 M r / the wrong ratio. • moles of (COOH) in water = 0.0300 – 0.00165 or 0.02835 ecf 2 NO ecf for M3 from bullets 1 and 2 ● ✓● ✓ M3 K pc = 0.00165 / 0.02835 = 0.058(2) min 2sf ecf from bullets 3 & 4 only for K pc = [0.033 / 0.03]. or 3) Solution 2 M1M2 • moles of NaOH = 0.0165  0.200 or 0.0033 • moles of (COOH) in ether = 0.0033 / 2 or 0.00165 2 • mass of (COOH) in ether = 0.00165  90 or 0.1485 g 2 • mass of (COOH) in water = 2.70 – 0.1485 or 2.5515 g ecf 2 ● ✓● ✓ A reverse ratio = 17(.2) [3]. M3 K = (0.1485 ÷ 100) ÷ (2.5515 ÷ 100) = 0.058(2) min 2sf ecf from pc bullets 3 & 4 only © Cambridge University Press & Assessment 2026 Page 14 of 25 Trace ID: PC-9D76E91B48 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P97a01p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/42 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 4(c) 2 I in terms of hydrophilic / phobic. • K would be higher / larger (if K < 1 in 4(b)) pc pc (for butanedioic acid) (ecf from 4(b)) ora A bullet 1 K would be lower (if K > 1 pc pc in 4(b)). • butanedioic acid is less polar / more non-polar ora A ‘organic solvent’ / ‘ether’ for ethoxyethane. • so less soluble in water / more soluble in ethoxyethane ora or dissolves less in water / dissolves more in ethoxyethane I bullet 2 ‘butanedioic acid is non-polar’ alone. ● ✓✓ I bullet 3 concentration of butanedioic acid is lower in water. Bullet 2 / 3 – comparison needed. © Cambridge University Press & Assessment 2026 Page 15 of 25 Trace ID: PC-9D76E91B48 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P97a01p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/42 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

2(a)(i) –log[H+] 1 [ ] essential. CON –log(H+), –log [H+] , –ln[H+]. e 2(a)(ii) [H+][OH–] / K xK 1 [ ] essential but do not penalise in (i) and a b (ii). I 1  10–14. 2(b)(i) (10–2.9) 1 0.0013/0.00126/1.3x10–3/1.26x10–3 2(b)(ii) 2bi2 / 0.105 0.001262 / 0.105 ecf bi 1 1.5  10–5 / 1.51  10–5 / 1.53  10–5 ecf 2(c)(i) M1 [CH CH CH COOH] = 0.15 3 If [CH CH CH COOH] = 0.45 [0] but then 3 2 2 3 2 2 M2 [CH CH CH COO–Na+] in solution D = 0.036 / 0.0359 [salt] in mixture = 0.108 [1] ecf 3 2 2 [salt] in solution C = 0.162 [1] ecf VERIFY M3 [CH CH CH COO–Na+] in solution C = 0.054 / 0.0538 / 0.0539 / 0.0540 with working. 3 2 2 Another possible approach goes via moles of acid to get moles of salt, M1 moles acid = 0.045 mol M2 moles salt = 0.0108 / 0.0107 mol M3 original [salt] = 0.0538 / 0.0539 / 0.0540 2(c)(ii) CH CH CH COO–Na+ + H+ → CH CH CH COOH + Na+ 1 3 2 2 3 2 2 or CH CH CH COO– + H+ → CH CH CH COOH 3 2 2 3 2 2 or CH CH CH COO– + H SO → CH CH CH COOH + HSO – 3 2 2 2 4 3 2 2 4 or 2CH CH CH COO–Na+ + H SO → 2CH CH CH COOH + Na SO 3 2 2 2 4 3 2 2 2 4 2(c)(iii) CH CH CH COOH + OH– → CH CH CH COO– + H O 1 3 2 2 3 2 2 2 or CH CH CH COOH + KOH → CH CH CH COOK + H O 3 2 2 3 2 2 2 or CH CH CH COOH + K+OH– → CH CH CH COO–K+ + H O 3 2 2 3 2 2 2 © Cambridge University Press & Assessment 2026 Page 9 of 21 Trace ID: PC-F2CDA313A1 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P97a01p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/43 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 2(d) M1 2,2-dichlorobutanoic and 2 chlorine is electron withdrawing / negative inductive effect / electronegative I references to C=O, O, alkyl groups. M2 weakens O–H bond / stabilises anion / stabilises conjugate base / stabilises A –O–H bond and OH bond but CON – carboxylate ion OH bond, C–O–H bond. Question Answer Marks Guidance

2(a) M1 (strong acid) fully ionised / dissociated and 2 I fully / partially dissolves. (weak acid) partially ionised / dissociated I ‘not fully’ for ‘partially’ dissociated / ionised M2 to give / donate H+ / proton or is a proton donor (mentioned once) I ‘strongly dissociates’, ‘dissociates easily’. 2(b) pair 1: HOOCCOOH and HOOCCOO– and 1 CON Na+ salts pair 2: HOOCCOO– and –OOCCOO– (question asks for organic pairs) . pairs can be in either order A other representations if correct, e.g. (COOH) , (COO–) , (OOCCOO)2–, 2 2 C O 2– but not OOCCOO2–. 2 4 Trace ID: PC-0F078E40FE Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P97a01p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/44 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 2(c)(i) M1 Two reactants that will work (ignore stoichiometry) 2 I Ca2+ as a reactant. M2 Correct balanced equation M2 dependent on M1. e.g: I state symbols. Ca + 2CH COOH → Ca(CH COO) + H or 3 3 2 2 CaO + 2CH COOH → Ca(CH COO) + H O or pass other correct examples to TL. 3 3 2 2 Ca(OH) + 2CH COOH → Ca(CH COO) + 2H O or 2 3 3 2 2 I where a chemical reaction is not CaCO + 2CH COOH → Ca(CH COO) + CO + H O or 3 3 3 2 2 2 happening, e.g. Ca(HCO ) + 2CH COOH → Ca(CH COO) + 2CO + 2H O 3 2 3 3 2 2 2 CaCl + 2CH COOH → Ca(CH COO) + 2 3 3 2 2HCl. 2(c)(ii) (a solution that) resists / opposes / minimises changes in pH 1 A keeps pH almost constant. when small amounts / volumes A pH does not change much. of acid / H+ or base / alkali / OH– are added to it A ‘does not change significantly’ for resists. A ‘controls pH’ as it is in 2(c)(iii) I ‘minimises pH’ alone. CON ‘kept constant’ / ‘maintains pH’ but A maintains pH in a small range. I moderate for ‘small’. 2(c)(iii) M1 CH COOH + OH– → CH COO– + H O 2 If full salt formula used, must be correct. 3 3 2 A use of ‘⇌’ but read in forward direction. M2 CH 3 COO– + H+ → CH 3 COOH or A use of HCl, NaOH, etc. Ca(CH 3 COO) 2 + 2H+ → 2CH 3 COOH + Ca2+ I CH 3 COOH → CH 3 COO– + H+ and/or H+ + OH– → H O. 2 If more than two eqns given, mark answers on answer lines. © Cambridge University Press & Assessment 2026 Page 9 of 21 Trace ID: PC-0F078E40FE Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P97a01p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/44 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 2(d) M1 In NaOH : [H+] = 10–11.34 or 4.57  10–12 3 11.6 scores [3]. 11.04 [2] with working and one mistake M2 In NaOH: [OH–] = 10–14 ÷ 4.57  10–12 or 2.1910–3 seen In Ba(OH) : [OH–] = 2.19  10–3 × 2 or 4.376  10–3 ecf throughout 2 A use of pOH: M3 In Ba(OH) : [H+] =10–14 ÷ 4.376  10–3 or 2.29  10–12 pOH = 14 – 11.34 or 2.66. [1] 2 pH = 11.6(41) min 1dp [OH-] = 10-2.66 or 2.188  10-3 Alternative method: [OH-] in Ba(OH) 2 = 2  2.188  10-3 M1 [H+] = 10–11.34 or 4.57  10–12 or 4.376  10-3 [1] M2 4.57  10–12 ÷ 2 = 2.29  10–12 M3 pH = 11.6(41) (No mention of [OH–]) new pOH = -log(4.376  10-3) or 2.36 new pH = 14 – 2.36 = 11.64 [1] 2(e)(i) (the) ratio of the concentrations of a solute / substance / compound 1 A eqm for equilibrium. between two (immiscible) solvents / liquids (at) equilibrium I equations. A ‘Ratio’ may be given using mathematical expression. the equilibrium constant that relates ‘Solvents / liquids’ may be clearly implied. the concentration of a solute / substance / compound (partitioned) I ‘solubility’, ‘species’. between two (immiscible) solvents / liquids I ‘solutions’ for solvent. 2(e)(ii) (0.68 / 5.32 =) 0.13 / 0.128 / 0.1278 min 2sf 1 Only answer 0.1278195489. 2(e)(iii) • less than / lower (ecf if K from 2(e)(ii) is the wrong way up) 3 (inverse K = 7.82) pc pc • solute / A must be polar / form H-bonds / ionic (as more soluble in water) Bullets 2, 3 and 4 are unconditional. • cyclohexane is non-polar / less polar than butan-1-ol or No ecf, marks are unconditional cyclohexane does not form H-bonds throughout. • A is less soluble in cyclohexane (than in butan-1-ol) or less (A) will dissolve in cyclohexane or more (A) will dissolve in water (than cyclohexane) ✓✓✓ ⚫ © Cambridge University Press & Assessment 2026 Page 10 of 21 Trace ID: PC-0F078E40FE Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P97a01p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/44 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/