2025 OCR A-Level Chemistry A - H432 Practice Paper with Answers

\(n(\mathrm{Mg}) = 0.120 / 24.3 = 4.94 \times 10^{-3}\) mol. Mg + 2HCl → MgCl\(_2\) + H\(_2\), so \(n(\mathrm{H_2}) = 4.94 \times 10^{-3}\) mol. Volume = \(4.94 \times 10^{-3} \times 24.0 = 0.119\) dm\(^3\) = 119 cm\(^3\). A halves the amount (wrong 2:1 ratio); C doubles it; D multiplies the mass by 24.0 without dividing by M.

B (1 mark). Method: n(Mg) from m/M, 1:1 ratio to H2, multiply by 24.0 and convert dm3 to cm3. DO NOT ALLOW any other letter.

Doubling [A] with order 2 gives a factor of \(2^2 = 4\). Halving [B] with order 1 gives a factor of 0.5. Overall factor = \(4 \times 0.5 = 2\). The rate constant is unchanged as temperature is constant.

C (1 mark). Method: factor 4 for A multiplied by factor 0.5 for B. ALLOW ECF is not applicable to a single-mark item.

A localised Kekulé structure would release three times the hydrogenation enthalpy of cyclohexene (about \(-360\) kJ mol\(^{-1}\)). The measured value (about \(-208\) kJ mol\(^{-1}\)) is less exothermic, showing benzene is more stable than the Kekulé model, consistent with delocalisation. B is the wrong direction; C is false because all C–C bonds are equal in length; D is false because benzene resists bromination under these conditions.

A (1 mark). DO NOT ALLOW B, C or D.

\([\mathrm{H^+}] = \sqrt{K_a [\mathrm{HA}]} = \sqrt{1.6 \times 10^{-5} \times 0.100} = 1.26 \times 10^{-3}\) mol dm\(^{-3}\). \(\mathrm{pH} = -\log(1.26 \times 10^{-3}) = 2.90\). A treats HA as a strong acid; D is the pKa.

C (1 mark). Method: [H+] = √(Ka[HA]); accuracy: pH = 2.90.

\(E^\circ_{cell} = E^\circ(\text{positive}) - E^\circ(\text{negative}) = +0.77 - (-0.76) = +1.53\) V. Electrode potentials are not multiplied by stoichiometric coefficients, so D is wrong; B adds the signs incorrectly.

A (1 mark). Accept only +1.53 V.

Nitration needs the nitronium ion, generated from concentrated HNO\(_3\) and concentrated H\(_2\)SO\(_4\), giving nitrobenzene. Reduction with Sn/HCl gives phenylammonium chloride, and NaOH(aq) liberates phenylamine. Dilute HNO\(_3\) does not nitrate benzene; NaBH\(_4\) does not reduce nitroarenes; aryl halides do not readily undergo substitution with ammonia.

D (1 mark). DO NOT ALLOW A, B or C.

The molecule has a plane of symmetry. Distinct carbon environments: the two CH\(_3\) carbons (1), the two ring carbons bearing CH\(_3\) (1), the two ring CH carbons next to them (1) and the two remaining ring CH carbons (1). Total = 4.

B (1 mark). 6 counts only ring carbons; 8 counts every carbon.

Alkaline hydrolysis cleaves the ester bond: CH\(_3\)CH\(_2\)COOCH\(_2\)CH\(_3\) + NaOH → CH\(_3\)CH\(_2\)COO\(^-\)Na\(^+\) + CH\(_3\)CH\(_2\)OH. The acid part is propanoate (3 carbons) and the alcohol part is ethanol. A would form in acid hydrolysis; B swaps the carbon chains.

D (1 mark). DO NOT ALLOW A (carboxylic acid is deprotonated under alkaline conditions).

n(Na\(_2\)CO\(_3\)) = 0.0250 × 0.0800 = 2.00 × 10\(^{-3}\) mol. n(HCl) = 2 × 2.00 × 10\(^{-3}\) = 4.00 × 10\(^{-3}\) mol. V = n/c = 4.00 × 10\(^{-3}\) / 0.100 = 0.0400 dm\(^3\) = 40.0 cm\(^3\). Answer C.

Answer: C (1 mark). Method: moles of carbonate = 2.00 × 10^-3; use of 1:2 ratio gives 4.00 × 10^-3 mol HCl; divide by 0.100. Distractor B ignores the 1:2 ratio (20.0 cm3 would need 2.00 × 10^-3 mol); D applies the ratio twice; A has no valid derivation. ALLOW 0.0400 dm3.

The rate depends on [P]\(^2\), so halving P gives a factor of (1/2)\(^2\) = 0.25. The rate is first order in Q, so tripling Q gives a factor of 3. Overall factor = 0.25 × 3 = 0.75. Answer A.

Answer: A (1 mark). Method: (1/2)^2 × 3 = 0.75. B arises from treating P as first order; C ignores the change in Q; D ignores the change in P. k is constant because temperature is unchanged.

The Kekulé structure would release 3 × (−120) = −360 kJ mol\(^{-1}\). The measured value is −208 kJ mol\(^{-1}\). Difference = 360 − 208 = 152 kJ mol\(^{-1}\), so benzene is 152 kJ mol\(^{-1}\) more stable. Answer B.

Answer: B (1 mark). Method: 3 × 120 = 360; 360 − 208 = 152. A uses 208 − 120; C uses 2 × 120; D adds 208 and 120.

After mixing, [HX] = 0.10 mol dm\(^{-3}\) and [X\(^-\)] = 0.050 mol dm\(^{-3}\); the ratio is unaffected by dilution. [H\(^+\)] = \(K_a\)[HX]/[X\(^-\)] = 2.0 × 10\(^{-5}\) × 2 = 4.0 × 10\(^{-5}\) mol dm\(^{-3}\). pH = −log(4.0 × 10\(^{-5}\)) = 4.40. Answer B.

Answer: B (1 mark). Method: [H+] = Ka[HX]/[X-] = 4.0 × 10^-5; pH = 4.40. C is pKa (ratio ignored); D results from inverting the ratio ([H+] = 1.0 × 10^-5); A results from an incorrect factor of 4 in the ratio.

\(E^\ominus_{cell} = E^\ominus_{positive} - E^\ominus_{negative} = +0.80 - (-0.25) = 1.05\) V. The potential is not multiplied by the number of electrons. Answer B.

Answer: B (1 mark). Method: 0.80 − (−0.25) = 1.05 V. A subtracts the magnitudes incorrectly (0.80 − 0.25); D doubles the value by scaling for electrons; C has no valid derivation.

The two CH\(_3\) groups on C2 are equivalent. The distinct carbon environments are: the equivalent pair of CH\(_3\) groups, the CH, the first CH\(_2\), the second CH\(_2\), and the terminal CH\(_3\). This gives 5 environments, so 5 peaks. Answer C.

Answer: C (1 mark). Identification of the two equivalent methyl carbons leaves 5 environments from 6 carbons. D counts every carbon; A and B arise from over-merging environments.

Formation: 2C(s) + 3H\(_2\)(g) + ½O\(_2\)(g) → C\(_2\)H\(_5\)OH(l). \(\Delta H_f\) = Σ\(\Delta H_c\)(reactants) − Σ\(\Delta H_c\)(products) = [2(−394) + 3(−286)] − (−1367) = −1646 + 1367 = −279 kJ mol\(^{-1}\). Answer A.

Answer: A (1 mark). Method: use of 2 × C and 3 × H2 combustion values (−1646) minus ethanol combustion (−1367) = −279. B is a sign error; C uses 1 × H2 instead of 3; D adds all three values.

(a)(i) Experiments 1 and 2: [S2O8^2-] doubles with [I-] constant and rate doubles, so first order in S2O8^2-. Experiments 2 and 3: [I-] doubles with [S2O8^2-] constant and rate doubles, so first order in I-. (Experiment 4 confirms: 0.060/0.020 gives 3 x 0.5 = 1.5 times experiment 1 rate, 1.8 x 10^-6.)
(ii) rate = k[S2O8^2-][I-]. Using experiment 1: k = 1.2 x 10^-6 / (0.020 x 0.040) = 1.5 x 10^-3. Units: mol dm-3 s-1 / mol2 dm-6 = dm3 mol-1 s-1.
(b) ln(k2/k1) = (Ea/R)(1/T1 - 1/T2). k2/k1 = 8.0, ln 8.0 = 2.079. 1/298 - 1/318 = 3.3557 x 10^-3 - 3.1447 x 10^-3 = 2.1105 x 10^-4 K-1. Ea = 2.079 x 8.314 / 2.1105 x 10^-4 = 81 900 J mol-1 = 81.9 kJ mol-1.
(c) The rate equation contains one S2O8^2- and one I-, so the rate-determining step involves one of each (S2O8^2- + I- -> intermediate); species appearing in the rate equation are in or before the rate-determining step, and the orders show the stoichiometry of that step.
(d) Vary [I-] (e.g. by diluting with water to keep total volume constant) while keeping [S2O8^2-], thiosulfate volume, starch and temperature constant. Record time t for colour. Because a fixed amount of iodine forms in time t (it reacts with the thiosulfate until the thiosulfate is used up), rate is proportional to 1/t. Plot log(1/t) against log[I-] (or compare 1/t when [I-] is doubled); gradient 1 or doubling of 1/t shows first order. It is an approximation as it gives an average rate over the time t, not the true rate at t = 0, but the concentrations change little (a small amount of reaction only) so it approximates the initial rate. Thiosulfate must be a small, fixed amount so that [I-] and [S2O8^2-] are nearly constant and the same amount of iodine is needed each time; thiosulfate reduces the iodine back to iodide, so colour is delayed until it has been used up; starch gives a sharp end point with the iodine.

(a)(i) 3 marks
- Doubling [S2O8^2-] (expts 1 and 2) doubles the rate so first order in S2O8^2- (1)
- Doubling [I-] (expts 2 and 3) doubles the rate so first order in I- (1)
- Reasoning refers to the other concentration being constant in each comparison (1)
ALLOW expt 4 as a consistency check. IGNORE 'order = 1' without any explanation for the reasoning mark.

(a)(ii) 2 marks
- rate = k[S2O8^2-][I-] (1) ALLOW ECF from (a)(i).
- k = 1.5 x 10^-3 AND units dm3 mol-1 s-1 (1) ALLOW any experiment. ECF for units from the rate equation given.

(b) 4 marks
- Correct use of ln k or equivalent: ln(1.2 x 10^-2) = -4.423; ln(1.5 x 10^-3) = -6.502 (or ln 8 = 2.079) (1)
- 1/298 and 1/318 used with temperatures in K (1)
- Rearrangement to Ea = R x ln(k2/k1) / (1/T1 - 1/T2) or gradient method (1)
- Ea = 81.9 kJ mol-1 (1) ALLOW 81.8 to 82.0. ALLOW 82 for 3 of 4 marks (significant figures). If the final answer is 81 900 J mol-1 with the wrong or missing unit, DO NOT ALLOW the final mark. Correct answer scores 4.

(c) 2 marks
- One S2O8^2- ion and one I- ion (1)
- Because the rate equation includes both, each first order, OR the rate-determining step involves the species in the rate equation (1)
IGNORE a statement of the overall equation. DO NOT ALLOW 2I- (second order in I- would be needed).

(d)* 6 marks
Level 3 (5-6 marks): A coherent, logical account covering all three areas: (1) design/analysis for the order, (2) why the result is only an approximation, (3) the role of thiosulfate and starch. Scientific points are linked correctly, with a clear sequence.
Level 2 (3-4 marks): Covers two areas with mostly correct science, or all three with gaps or some lack of clarity.
Level 1 (1-2 marks): Some relevant points from one area, with limited detail or logic.
0 marks: no response or no creditable content.

Indicative scientific points:
- Vary [I-] only, keeping [S2O8^2-], temperature, thiosulfate and starch volumes constant; total volume kept constant using water
- Record time t for the blue-black colour
- Same amount of iodine formed at the end point, so rate is proportional to 1/t
- Plot 1/t (or log 1/t) against [I-] (or log [I-]); a straight line through the origin or gradient 1 gives first order; doubling [I-] doubles 1/t
- Approximation: gives average rate over time t, not the instantaneous rate at t = 0
- Valid because only a small proportion of reactants is used, so concentrations are almost constant
- Thiosulfate reduces the iodine to iodide, so no colour until it is used up; small fixed amount is required
- Starch forms a blue-black complex with iodine, giving a sharp, visible end point

Communication examples: logical sequence from method to graph to conclusion; correct use of terms such as 'initial rate', 'average rate' and 'proportional'.

(a) Expt 1 to 2: [BrO3-] doubles, rate doubles (x2), so first order. Expt 1 to 3: [Br-] doubles, rate doubles, so first order. Expt 1 to 4: [H+] doubles, rate x4, so second order. (Expt 5: [H+] halves, rate falls to 1/4, consistent.)

(b)(i) rate = \(k[\text{BrO}_3^-][\text{Br}^-][\text{H}^+]^2\)

(b)(ii) \(k = \dfrac{\text{rate}}{[\text{BrO}_3^-][\text{Br}^-][\text{H}^+]^2} = \dfrac{2.0\times10^{-4}}{0.010\times0.10\times(0.10)^2} = \dfrac{2.0\times10^{-4}}{1.0\times10^{-5}} = 20\). Units: mol dm\(^{-3}\) s\(^{-1}\) divided by mol\(^4\) dm\(^{-12}\) gives dm\(^9\) mol\(^{-3}\) s\(^{-1}\).

(c) rate \(= 20\times0.015\times0.040\times(0.30)^2 = 20\times5.4\times10^{-5} = 1.08\times10^{-3}\) mol dm\(^{-3}\) s\(^{-1}\).

(d) \(\ln(k_2/k_1) = \dfrac{E_a}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)\). \(\ln(71/20) = 1.267\). \(\dfrac{1}{298}-\dfrac{1}{318} = 2.1105\times10^{-4}\) K\(^{-1}\). \(E_a = 1.267\times8.314/2.1105\times10^{-4} = 4.99\times10^{4}\) J mol\(^{-1}\) = 49.9 kJ mol\(^{-1}\).

(e)(i) \(\text{BrO}_3^- + \text{Br}^- + 2\text{H}^+ \rightarrow \text{HBrO}_2 + \text{HOBr}\) (balanced for atoms and charge; one BrO3-, one Br-, two H+ as in the rate equation).

(e)(ii) Increasing temperature shifts the distribution so the area under the curve beyond \(E_a\) increases greatly (the high-energy tail is steep). A much larger proportion of molecules have energy \(\geq E_a\), so the frequency of successful collisions rises markedly.

(a) [3]
- BrO3-: first order, as doubling concentration doubles rate (Expt 1 and 2) ✓
- Br-: first order, as doubling concentration doubles rate (Expt 1 and 3) ✓
- H+: second order, as doubling concentration quadruples rate (Expt 1 and 4) ✓
ALLOW Expt 1 and 5 for H+ (halving gives one quarter). Each order must be supported by a comparison; DO NOT ALLOW order without explanation for the mark.

(b)(i) [1]
rate = k[BrO3-][Br-][H+]^2 ✓
ALLOW ECF from (a). Square brackets required.

(b)(ii) [3]
- Substitution of Expt 1 data into rearranged rate equation ✓
- k = 20 ✓ (check final answer first)
- units dm^9 mol^-3 s^-1 ✓
ALLOW ECF on k and units from the rate equation in (b)(i). ALLOW 2.0 x 10^1.

(c) [2]
- Correct substitution of concentrations (squared [H+]) and k ✓
- 1.08 x 10^-3 (mol dm^-3 s^-1) ✓
ALLOW 1.1 x 10^-3. ALLOW ECF from (b)(ii) and (b)(i), including the unit being mol dm^-3 s^-1.

(d) [3]
- Correct rearrangement/use of ln(k2/k1) or two simultaneous Arrhenius equations with T converted to 298 and 318 K ✓
- ln(71/20) = 1.267 and (1/298 - 1/318) = 2.11 x 10^-4 ✓
- Ea = 49.9 kJ mol^-1 ✓ (3 s.f.)
ALLOW 49.8 to 50.0 for final mark. Final answer 49 900 J mol^-1 scores 2 max unless converted to kJ. DO NOT ALLOW use of temperatures in degrees Celsius. ECF for unit error only once.

(e)(i) [2]
- Reactants BrO3- + Br- + 2H+ (one, one and two of each, matching rate equation) ✓
- Products HBrO2 + HOBr and equation balanced for atoms and charge ✓
ALLOW ECF for reactants from a different rate equation in (b)(i), provided the equation balances with the given products. IGNORE state symbols.

(e)(ii) [2]
- Boltzmann distribution: at higher temperature the curve shifts to higher energy / peak lowers and flattens, and the area beyond Ea is larger (a diagram with two curves at T and T+ΔT, labelled axes and Ea marked, may show this) ✓
- A much greater proportion of molecules have energy >= Ea, so more frequent successful/effective collisions ✓
ALLOW more molecules exceed Ea. IGNORE 'more collisions' alone. DO NOT ALLOW Ea decreases.

(a) A Brønsted–Lowry acid is a proton (H+) donor.

(b) \(K_a = [H^+]^2/[HA]\), assuming [H+] = [A-] and that dissociation of HA is negligible.
\([H^+]^2 = 1.35\times10^{-5} \times 0.150 = 2.025\times10^{-6}\)
\([H^+] = 1.423\times10^{-3}\) mol dm\(^{-3}\)
pH = \(-\log(1.423\times10^{-3})\) = 2.85

(c)(i) n(HA initially) = 0.0250 x 0.150 = 3.75 x 10^-3 mol
n(NaOH) = 0.0200 x 0.100 = 2.00 x 10^-3 mol
HA + OH- -> A- + H2O, so n(A-) = 2.00 x 10^-3 mol and n(HA remaining) = 3.75 x 10^-3 - 2.00 x 10^-3 = 1.75 x 10^-3 mol.

(c)(ii) \([H^+] = K_a \times [HA]/[A^-]\); the volumes cancel, so \([H^+] = 1.35\times10^{-5} \times 1.75/2.00 = 1.18\times10^{-5}\) mol dm\(^{-3}\)
pH = 4.93

(d) Buffer action: CH3CH2COOH <=> H+ + CH3CH2COO-. Added H+ reacts with the large reservoir of propanoate ions, CH3CH2COO- + H+ -> CH3CH2COOH, so equilibrium shifts left. Added OH- reacts with H+ (H+ + OH- -> H2O), removing H+; the equilibrium shifts right as propanoic acid dissociates to replace H+. In both cases [HA]/[A-] changes only slightly, so [H+] and pH are almost constant.

Determination of Ka: calibrate the pH meter with buffers of known pH. Pipette a known volume of propanoic acid into a beaker, add NaOH from a burette in small portions, stirring and recording pH after each addition (smaller portions near the rapid change). Plot pH against volume of NaOH and find the equivalence point (the mid-point of the vertical section). At half the equivalence volume, [HA] = [A-], so [H+] = Ka and pKa = pH; then Ka = 10^-pH. (Repeat/ roughly-then-accurately to improve reliability.)

(a) Proton donor ✓ [1]. ALLOW H+ donor. DO NOT ALLOW 'releases H+ ions in solution' alone.

(b) [3]
- Ka = [H+]^2/[HA] OR [H+] = sqrt(Ka x [HA]) ✓ (method)
- [H+] = 1.42 x 10^-3 (mol dm-3) ✓ (accuracy; ALLOW 1.4 x 10^-3 or more figures)
- pH = 2.85 ✓ (ALLOW ECF from the [H+] value. Correct answer with no working scores 3.)
IGNORE statement of the approximations.

(c)(i) [2]
- n(HA) initially = 3.75 x 10^-3 AND n(NaOH) = 2.00 x 10^-3 so n(A-) = 2.00 x 10^-3 mol ✓
- n(HA remaining) = 1.75 x 10^-3 mol ✓

(c)(ii) [2]
- [H+] = Ka x n(HA)/n(A-) = 1.35 x 10^-5 x 1.75/2.00 = 1.18 x 10^-5 ✓ (ECF from (i))
- pH = 4.93 ✓ (ALLOW 4.9; ECF from the [H+] value). DO NOT ALLOW use of the 0.150 mol dm-3 concentration directly with no allowance for dilution if this changes the ratio.

(d)* [6] Level marking. Indicative scientific points:
Buffer action:
1. equilibrium HA <=> H+ + A- (OR equivalent) shown
2. large reservoir of both HA and A-
3. added acid: H+ reacts with A- / equilibrium shifts left
4. added alkali: OH- reacts with H+ (to form water) and equilibrium shifts right / HA dissociates
5. ratio [HA]/[A-] almost unchanged so [H+] / pH almost constant
Determination of Ka:
6. calibrate pH meter; add NaOH from a burette in measured portions, recording pH and volume
7. plot pH against volume; identify the equivalence volume from the vertical section
8. at half the equivalence volume [HA] = [A-], so pH = pKa; Ka = 10^-pH

Level 3 (5–6 marks): Explains the buffer action for BOTH added acid and alkali with equations/equilibrium shift AND gives a complete, logically ordered method including the half-equivalence point link to Ka. Few or no errors; clear and structured.
Level 2 (3–4 marks): Explains buffer action for acid and alkali with some gaps, OR gives buffer action in full with only partial method for Ka (e.g. plots pH curve but does not link half-equivalence to Ka). Mostly organised.
Level 1 (1–2 marks): Isolated correct points, e.g. buffer resists pH change, or that pH is measured during titration; limited explanation, little structure.
Level 0 (0 marks): No creditworthy response.
Within a level, the higher mark is awarded where the response is clear, correctly sequenced and uses correct terminology.

(a) Ka = [H+][A-]/[HA]. Assuming [H+] = [A-] and [HA] = 0.200: [H+]^2 = 1.35 x 10^-5 x 0.200 = 2.70 x 10^-6, so [H+] = 1.64 x 10^-3 mol dm-3 and pH = 2.78. Assumptions: [H+] from ionisation of water is negligible, and dissociation of the acid is so small that [HA] is approximately its initial value.

(b) n(HA) initially = 0.0250 x 0.200 = 5.00 x 10^-3 mol. n(NaOH) = 0.0100 x 0.150 = 1.50 x 10^-3 mol. After reaction n(A-) = 1.50 x 10^-3 mol and n(HA) = 3.50 x 10^-3 mol. Volume cancels in the ratio: [H+] = Ka x [HA]/[A-] = 1.35 x 10^-5 x 3.50/1.50 = 3.15 x 10^-5 mol dm-3, so pH = 4.50.

(c) Initial pH = 2.78 (from (a)). The curve rises steeply at first and then flattens in the buffer region, because added OH- converts HA to A-, giving a mixture of weak acid and its conjugate base that resists pH change. At half-neutralisation (12.5 cm3), [HA] = [A-], so pH = pKa = -log(1.35 x 10^-5) = 4.87. The end-point occurs at 25.0 cm3 because n(NaOH) = n(HA) = 5.00 x 10^-3 mol. At the end-point the solution contains only propanoate (0.100 mol dm-3), a weak base: CH3CH2COO- + H2O <=> CH3CH2COOH + OH-, so pH > 7; calculation gives [OH-] = sqrt((Kw/Ka) x 0.100) = 8.6 x 10^-6, pOH = 5.07, pH = 8.9. The vertical section of the curve spans roughly pH 7 to 11, so an indicator whose range lies within it changes colour sharply at the end-point. Phenolphthalein (8.3-10.0) lies within the vertical section. Methyl orange and bromocresol green change colour in the buffer region or before it (pH below 5.4), where pH changes gradually, so the colour change would occur gradually and early, over a range of volumes well before 25.0 cm3.

(a) [3]
- Correct expression for Ka and [H+]^2 = Ka x [HA] (or equivalent) (1)
- [H+] = 1.64 x 10^-3 mol dm-3 (1)
- pH = 2.78 (1) ALLOW 2.8 or 2.784. ECF from [H+].
- Assumption stated (1) in place of the first marking point if that point is not otherwise awarded: [H+] = [A-] OR [H+] from water negligible OR [HA] at equilibrium = initial [HA] (dissociation negligible). Maximum 3 marks.

(b) [3]
- n(HA) initial = 5.00 x 10^-3 and n(NaOH) = 1.50 x 10^-3 mol (1)
- n(HA) remaining = 3.50 x 10^-3 mol and n(A-) = 1.50 x 10^-3 mol; [H+] = Ka x 3.50/1.50 = 3.15 x 10^-5 (1)
- pH = 4.50 (1) ALLOW 4.5. ECF from the previous mark.
DO NOT ALLOW use of initial moles of acid for [HA] (gives pH 4.87 or similar).

(c)* Level descriptors [6]
Level 3 (5-6 marks): Gives a coherent and complete account covering all of: initial pH value; buffer region/half-neutralisation pH = pKa = 4.87; end-point at 25.0 cm3 with pH > 7 explained by hydrolysis of the propanoate (weak base); correct indicator (phenolphthalein) chosen with reasons for rejecting both others linked to the vertical section of the curve. Logical structure; reasoning linked clearly to calculated or quoted values.
Level 2 (3-4 marks): Covers most of the points, with calculated values for at least two features (e.g. initial pH, pKa, end-point volume) and a correct indicator with some justification; may lack explanation of why the end-point is above pH 7 or of why the other indicators are unsuitable. Account is mostly logical.
Level 1 (1-2 marks): Gives some correct features (e.g. end-point volume 25.0 cm3, or phenolphthalein) with limited or no explanation or calculation. Structure may be weak.
Level 0 (0 marks): No response or no relevant chemistry.

Indicative scientific points:
- initial pH = 2.78 (ALLOW 2.8) because the acid is weak/partially dissociated
- buffer region: HA and A- both present, resisting pH change on adding OH-
- at half-neutralisation [HA] = [A-] so pH = pKa = 4.87 at 12.5 cm3
- end-point at 25.0 cm3 since n(NaOH) = n(HA)
- at end-point pH about 8.9 (ALLOW 8.5-9.5) because A- is a weak base/accepts H+ from water producing OH-
- vertical section approximately pH 7 to 11
- phenolphthalein is suitable as its range lies within the vertical section
- methyl orange and bromocresol green change colour at pH below 5.4, in the buffer region, so the colour change is gradual and not at 25.0 cm3

Communication guidance: Level 3 requires explanations that are linked to numerical values and set out in a logical order. ECF allowed from (a) for the initial pH. IGNORE sketch quality if described in words; ALLOW a labelled sketch as part of the response.

(a) \(K_a = \dfrac{[\mathrm{CH_3CH_2COO^-}][\mathrm{H^+}]}{[\mathrm{CH_3CH_2COOH}]}\)

(b) Assuming [H+] = [CH3CH2COO-] and that dissociation of the acid is negligible, so [HA] = 0.150 mol dm-3:
\([\mathrm{H^+}]^2 = 1.35\times10^{-5}\times0.150 = 2.025\times10^{-6}\)
\([\mathrm{H^+}] = 1.423\times10^{-3}\) mol dm-3
pH = \(-\log(1.423\times10^{-3}) = 2.85\)

(c) n(acid initially) = 0.0250 x 0.150 = 3.75 x 10^-3 mol
n(NaOH) = 0.0100 x 0.100 = 1.00 x 10^-3 mol
n(acid remaining) = 2.75 x 10^-3 mol; n(propanoate formed) = 1.00 x 10^-3 mol
The volume is common to both and cancels, so \([\mathrm{H^+}] = K_a \times \dfrac{[\mathrm{HA}]}{[\mathrm{A^-}]} = 1.35\times10^{-5}\times\dfrac{2.75}{1.00} = 3.71\times10^{-5}\) mol dm-3
pH = 4.43

(d) The buffer contains a large reservoir of both undissociated propanoic acid and propanoate ions. Added H+ reacts with propanoate ions: \(\mathrm{CH_3CH_2COO^-} + \mathrm{H^+} \rightarrow \mathrm{CH_3CH_2COOH}\). The equilibrium \(\mathrm{HA} \rightleftharpoons \mathrm{H^+} + \mathrm{A^-}\) shifts left, removing most of the added H+. The ratio [HA]/[A-] changes only slightly, so the pH changes only slightly.

(a) 1 mark
[CH3CH2COO-][H+] / [CH3CH2COOH] (1)
ALLOW H3O+ for H+. ALLOW RCOO-/RCOOH if defined. DO NOT ALLOW expression with water or with square brackets omitted.

(b) 3 marks
[H+]^2 = Ka x 0.150 = 2.025 x 10^-6 OR [H+] = 1.42 x 10^-3 (1)
pH = 2.85 (1) (final answer 2.85 scores 2 marks for the calculation; ALLOW 2.84 to 2.85)
Assumption: [H+] = [A-] (from the acid only; water's ionisation is negligible) AND [HA]equilibrium = [HA]initial (dissociation is negligible) (1) ALLOW either of these two assumptions if clearly stated.
ECF from an incorrect [H+] for the pH mark.

(c) 3 marks
n(HA initially) = 3.75 x 10^-3 AND n(NaOH) = 1.00 x 10^-3 (1)
n(HA remaining) = 2.75 x 10^-3 AND n(A-) = 1.00 x 10^-3 (1) (OR concentrations using total volume 35.0 cm3: 0.0786 and 0.0286 mol dm-3)
[H+] = Ka x [HA]/[A-] = 3.71 x 10^-5, so pH = 4.43 (1) (ALLOW 4.4 to 4.43)
If the ratio is inverted giving pH 5.3: maximum 2 marks, ECF applies. If NaOH not subtracted from HA: maximum 1 mark.

(d) 2 marks
H+ reacts with / is removed by propanoate ions (A-) (1) ALLOW equation CH3CH2COO- + H+ -> CH3CH2COOH.
Equilibrium HA <=> H+ + A- shifts left OR ratio [HA]/[A-] changes only slightly so [H+] / pH almost constant (1)
IGNORE 'buffer neutralises the acid'. DO NOT ALLOW reaction of H+ with the undissociated acid.

(a)(i) A Brønsted–Lowry acid is a proton donor.
(a)(ii) \(\mathrm{CH_3CH_2COOH}\) (acid) and \(\mathrm{CH_3CH_2COO^-}\) (conjugate base).

(b)(i) \(K_a = \dfrac{[\mathrm{CH_3CH_2COO^-}][\mathrm{H^+}]}{[\mathrm{CH_3CH_2COOH}]}\)

(b)(ii) Assume \([\mathrm{H^+}] = [\mathrm{CH_3CH_2COO^-}]\) (ionisation of water negligible) and that [acid] at equilibrium is approximately its initial value (dissociation negligible).
\([\mathrm{H^+}]^2 = 1.35\times10^{-5}\times0.150 = 2.025\times10^{-6}\); \([\mathrm{H^+}] = 1.42\times10^{-3}\) mol dm\(^{-3}\); pH = \(-\log(1.423\times10^{-3}) = 2.85\).

(c)(i) The mixture contains excess weak acid together with its conjugate base (propanoate ions) formed by neutralisation, so both are present in significant amounts.

(c)(ii) n(acid) initially = 0.0250 × 0.150 = 3.75 × 10\(^{-3}\) mol; n(NaOH) = 0.0200 × 0.100 = 2.00 × 10\(^{-3}\) mol.
n(propanoate) formed = 2.00 × 10\(^{-3}\) mol; n(acid) remaining = 1.75 × 10\(^{-3}\) mol.
\([\mathrm{H^+}] = K_a \times \dfrac{[\mathrm{acid}]}{[\mathrm{salt}]} = 1.35\times10^{-5}\times\dfrac{1.75}{2.00} = 1.181\times10^{-5}\) mol dm\(^{-3}\) (volumes cancel).
pH = 4.93.

(d) Added \(\mathrm{H^+}\) reacts with the propanoate ions (\(\mathrm{CH_3CH_2COO^-} + \mathrm{H^+} \rightarrow \mathrm{CH_3CH_2COOH}\)); the equilibrium shifts to the left and \([\mathrm{H^+}]\) stays almost constant because the ratio [acid]/[salt] changes little.

(e) Volume of NaOH at equivalence = (0.0250 × 0.150)/0.100 = 0.0375 dm\(^3\) = 37.5 cm\(^3\).
Initial pH about 2.85 (weak acid, partial dissociation). A gradual rise through the buffer region, with pH = pKa (4.87) at the half-equivalence point (18.75 cm\(^3\)). A steep rise (vertical section) around 37.5 cm\(^3\).
At equivalence, the solution contains propanoate ions at 0.00375/0.0625 = 0.0600 mol dm\(^{-3}\). Propanoate is a weak base: \(K_b = K_w/K_a = 7.41\times10^{-10}\); \([\mathrm{OH^-}] = \sqrt{7.41\times10^{-10}\times0.0600} = 6.67\times10^{-6}\); pOH = 5.18; pH = 8.82. So the equivalence pH is above 7.
After equivalence, pH levels off near 12.
The indicator's colour change must lie within the vertical section, so phenolphthalein (8.3–10.0) is suitable. Methyl orange (3.1–4.4) changes in the buffer region and bromothymol blue (6.0–7.6) changes before the equivalence point, so both would give a gradual colour change and an inaccurate end point.

(a)(i) proton donor ✓ [1]. ALLOW H+ donor. DO NOT ALLOW releases H+ ions in solution alone.
(a)(ii) CH3CH2COOH AND CH3CH2COO- ✓ [1]. Both needed.
(b)(i) Ka = [CH3CH2COO-][H+] / [CH3CH2COOH] ✓ [1]. ALLOW [H3O+] for [H+]. ALLOW [RCOO-]. DO NOT ALLOW round brackets or missing brackets.
(b)(ii) [H+]^2 = Ka × [acid] = 1.35 × 10^-5 × 0.150 (= 2.025 × 10^-6) ✓ (method); [H+] = 1.42 × 10^-3 and pH = 2.85 ✓ (accuracy) [2]. ALLOW 2.8 or 2.847 for the pH mark. ECF from an incorrect [H+] to the pH mark.
(c)(i) contains (excess) weak acid AND its conjugate base / propanoate ions ✓ [1]. ALLOW mixture of weak acid and its salt.
(c)(ii) n(acid) = 3.75 × 10^-3 AND n(NaOH) = 2.00 × 10^-3 ✓; n(salt) = 2.00 × 10^-3 AND n(acid) remaining = 1.75 × 10^-3 ✓; [H+] = Ka × [acid]/[salt] = 1.18 × 10^-5 ✓; pH = 4.93 ✓ [4]. ECF through the stages. Correct answer with no working scores 4. ALLOW 4.9 only if working is seen. DO NOT ALLOW use of total acid (3.75 × 10^-3) as the acid concentration.
(d) propanoate ions / conjugate base react with added H+ ✓ [1]. ALLOW equation CH3CH2COO- + H+ → CH3CH2COOH; AND equilibrium shifts to the left / acid forms, so [H+] (and pH) barely change (ORA). IGNORE neutralises.
(e)* Level 3 (5–6 marks): Describes at least four correct features of the curve with clear explanations: initial pH about 3 (weak acid); buffer region / gradual rise; equivalence volume 37.5 cm3 calculated correctly; steep rise at equivalence; equivalence pH above 7 explained by the propanoate ion (conjugate base) hydrolysing / weak base producing OH-, with a supporting calculation or figure of about 8.8; selects phenolphthalein and rejects the other two with reference to the vertical section. The answer is well structured, logically sequenced and free of contradictions.
Level 2 (3–4 marks): Gives the equivalence volume and several features of the curve, and chooses phenolphthalein with some justification. Explanations of the equivalence pH or the rejection of the other indicators may be incomplete. Reasonably structured.
Level 1 (1–2 marks): Gives isolated points such as the equivalence volume OR the correct indicator OR a general curve shape, with little or no explanation. Structure may be limited.
Level 0 (0 marks): No relevant response.
Indicative scientific points: V(equivalence) = 37.5 cm3; initial pH about 2.85; buffer region, pH = pKa (4.87) at 18.75 cm3; vertical section roughly pH 7 to 10; equivalence pH about 8.8 because propanoate is a weak base; final pH about 12–13; phenolphthalein range 8.3–10.0 lies within the vertical section; methyl orange and bromothymol blue change colour before the vertical section. ALLOW equivalence pH 8–9. ECF from an incorrect equivalence volume only where consistent with the working shown.
[Total: 17]

\(\mathrm{Mg + 2HCl \rightarrow MgCl_2 + H_2}\). \(n(\mathrm{Mg}) = 0.480/24.3 = 0.01975\) mol. Mole ratio 1:1 so \(n(\mathrm{H_2}) = 0.01975\) mol. \(V = 0.01975 \times 24.0 = 0.474\) dm\(^3\) = 474 cm\(^3\).

Answer B (1 mark). Method: n(Mg) = m/M, 1:1 ratio, multiply by 24.0 dm3 mol-1 and convert to cm3. A halves the volume (wrong ratio); C doubles it (uses 2HCl ratio wrongly); D multiplies by 24.0 only after dividing incorrectly/unit error.

Doubling [X] with second order gives \(2^2 = 4\). Halving [Y] with first order gives 0.5. Overall factor \(4 \times 0.5 = 2\).

Answer C (1 mark). Combine factor 4 for X and 0.5 for Y. B arises from treating X as first order; D from ignoring the change in Y; A from squaring the effect of Y incorrectly or ignoring X.

Sulfuric acid protonates nitric acid to give \(\mathrm{H_2NO_3^+}\), which loses water to form the nitronium ion \(\mathrm{NO_2^+}\). This is the electrophile that accepts a pair of electrons from the delocalised ring.

Answer D (1 mark). C is an intermediate in the generation of the electrophile, not the electrophile; A is a nucleophile/anion and cannot attack the electron-rich ring; B is not sufficiently electrophilic.

Equal volumes are mixed so both concentrations halve but the ratio is unchanged: [HA] : [A\(^-\)] = 2 : 1. \([\mathrm{H^+}] = K_a \times [\mathrm{HA}]/[\mathrm{A^-}]\), so \(\mathrm{pH} = \mathrm{p}K_a + \log([\mathrm{A^-}]/[\mathrm{HA}]) = 4.80 + \log 0.5 = 4.80 - 0.30 = 4.50\).

Answer B (1 mark). C ignores the ratio; D inverts the ratio (4.80 + 0.30); A subtracts 0.60 (uses ratio of 4 or doubles the log).

The more positive half-cell (Ag) is the positive electrode. \(E^\circ_{cell} = +0.80 - (-0.44) = +1.24\) V. The potential is an intensive property so the 2:1 electron balance does not alter it.

Answer A (1 mark). B subtracts magnitudes the wrong way (0.80 - 0.44); C doubles the silver value and adds (2 x 0.80 + 0.44); D is 0.80 + 0.44 - 0.08 not supported, arising from arithmetic error.

The two methyl groups on carbon 2 are equivalent. The distinct carbon environments are: the two equivalent CH\(_3\) groups, the CH, the CH\(_2\) and the terminal CH\(_3\). This gives 4 environments, so 4 peaks.

Answer C (1 mark). D counts all five carbons separately; B treats both methyl types as equivalent; A is too few.

Alkaline hydrolysis: \(\mathrm{CH_3COOCH_2CH_3 + NaOH \rightarrow CH_3COONa + CH_3CH_2OH}\). The carboxylic acid is neutralised by the alkali to form its sodium salt, and the alcohol is released.

Answer D (1 mark). A is the product of acid hydrolysis; B and C are not consistent with the stoichiometry of the reaction (an acid and its salt, or an alkoxide, would not form in aqueous alkali).

Feasible when \(\Delta G = \Delta H - T\Delta S \le 0\), so \(T \ge \Delta H/\Delta S\). Convert \(\Delta H\) to J: 40 000 J mol\(^{-1}\). \(T = 40\,000/120 = 333\) K.

Answer D (1 mark). A results from not converting kJ to J (40/120); B from 120/40; C is a distractor from dividing by 2 times or inverting units wrongly.

\(M_r(\mathrm{MgCO_3}) = 24.3 + 12.0 + 48.0 = 84.3\). \(n = 2.50 / 84.3 = 0.0297\) mol. The equation is 1:1, so \(n(\mathrm{CO_2}) = 0.0297\) mol. \(V = 0.0297 \times 24.0 = 0.712\) dm\(^3\) = 712 cm\(^3\). Answer B.

B (1 mark). Method: n = m/M then multiply by 24.0 (1:1 ratio). A halves the amount (ratio error); C doubles it; D results from dividing by 24.0 inappropriately (or mass/M error).

New rate \(= k(2[\mathrm{P}])(0.5[\mathrm{Q}])^2 = k \times 2 \times 0.25 [\mathrm{P}][\mathrm{Q}]^2 = 0.5\) times the original rate. Answer C.

C (1 mark). Factor for P is 2 and for Q is (1/2)^2 = 1/4, product 1/2. B arises from treating Q as first order; A from ignoring the change in P; D from ignoring the change in Q.

\([\mathrm{OH^-}] = 2 \times 0.0200 = 0.0400\) mol dm\(^{-3}\). \([\mathrm{H^+}] = K_w/[\mathrm{OH^-}] = 2.50 \times 10^{-13}\). \(\mathrm{pH} = -\log(2.50 \times 10^{-13}) = 12.6\). Answer D.

D (1 mark). Method: doubling for two OH- per formula unit, use of Kw, then pH = -log[H+]. B ignores the factor of 2; A is pOH of the correct concentration; C is -log 0.0200.

The oxygen lone pair delocalises into the \(\pi\) system, raising electron density of the ring, so it polarises bromine more strongly and electrophilic substitution occurs without a carrier. Answer A.

A (1 mark). B states the wrong direction of electron movement; C is incorrect because phenol has a delocalised ring and reacts by substitution; D is not the cause (acidity is weak and not catalytic here).

The molecule has a plane of symmetry through C2 and C5. Distinct environments: the two equivalent CH\(_3\) carbons; C1 and C3 (equivalent, bearing CH\(_3\)); C2; C4 and C6 (equivalent); C5. This gives 5 environments. Answer C.

C (1 mark). Award for correct counting of symmetry-equivalent carbons. D counts every carbon; A and B ignore the distinct C2 and/or C5 positions (typical of 1,4-isomer reasoning).

Alkaline hydrolysis splits the ester at the C-O bond: \(\mathrm{CH_3CH_2COOCH_2CH_3 + NaOH \rightarrow CH_3CH_2COO^-Na^+ + CH_3CH_2OH}\). The carboxylic acid is present as its sodium salt. Answer A.

A (1 mark). B is the product of acid hydrolysis; C swaps the acid and alcohol portions; D is incorrect because the alkoxide would not form in aqueous alkali and the acid would be neutralised.

\(E^\ominus_{cell} = E^\ominus_{\mathrm{positive}} - E^\ominus_{\mathrm{negative}} = +0.80 - (-0.25) = +1.05\) V. Electrode potentials are not multiplied by the number of electrons. Answer B.

B (1 mark). A subtracts magnitudes incorrectly (0.80 - 0.25); C doubles the silver value; D doubles the silver value and adds 0.25.

(a)(i) \(K_a = \frac{[H^+][X^-]}{[HX]}\).

(a)(ii) Assume [H+] = [X-] and [HX] at equilibrium = initial concentration. \([H^+]^2 = 1.35\times10^{-4}\times0.0500 = 6.75\times10^{-6}\), so \([H^+] = 2.598\times10^{-3}\) mol dm\(^{-3}\); pH = 2.59.

(b) n(HX) initially = 0.0250 × 0.0500 = 1.25 × 10\(^{-3}\) mol. n(NaOH) = 0.0100 × 0.0400 = 4.00 × 10\(^{-4}\) mol. n(X-) formed = 4.00 × 10\(^{-4}\) mol; n(HX) remaining = 8.50 × 10\(^{-4}\) mol. \([H^+] = K_a \times \frac{[HX]}{[X^-]} = 1.35\times10^{-4}\times\frac{8.50\times10^{-4}}{4.00\times10^{-4}} = 2.869\times10^{-4}\) mol dm\(^{-3}\) (volumes cancel). pH = 3.54.

(c) HX is a weak acid and is only slightly dissociated, so the buffer contains a large reservoir of HX and of X-. Added H+ is removed by X- + H+ → HX. Equilibrium HX ⇌ H+ + X- shifts left. [X-] decreases and [HX] increases only slightly, so the ratio [HX]/[X-], and hence [H+], is almost unchanged.

(d)(i) n(NaOH) = n(HX) = 1.25 × 10\(^{-3}\) mol; V = 1.25 × 10\(^{-3}\) / 0.0400 = 0.03125 dm\(^3\) = 31.25 cm\(^3\).

(d)(ii) At equivalence the solution contains only the salt NaX. X- is the conjugate base of a weak acid and accepts H+ from water: X- + H2O ⇌ HX + OH-, so [OH-] > [H+] and pH > 7 (about 8.1). The steep section of the curve spans roughly pH 7 to 10, which includes the phenolphthalein range, so phenolphthalein is suitable. Methyl orange changes colour at pH 3.1–4.4, in the buffer region before the steep section, so would give a gradual, premature end-point.

(e) At half-neutralisation [HX] = [X-], so [H+] = Ka and pH = pKa = −log(1.35 × 10\(^{-4}\)) = 3.87.

(a)(i) Ka = [H+][X-]/[HX] ✓ [1]
ALLOW [H3O+]. DO NOT ALLOW expression with square brackets missing OR with [H+]2 unless [X-] used.

(a)(ii) [H+] = √(Ka × [HX]) = 2.6 × 10^-3 (mol dm^-3) ✓ ; pH = 2.59 ✓ [2]
Check the final answer first: 2.59 scores 2. ALLOW 2.58 / 2.6 for 1 mark only if working shown (2 d.p. required). ALLOW pH from unrounded values.

(b) n(HX) initial = 1.25 × 10^-3 AND n(NaOH) = 4.00 × 10^-4 (mol) ✓ ; n(HX) remaining = 8.50 × 10^-4 AND n(X-) = 4.00 × 10^-4 (mol) ✓ ; [H+] = 2.87 × 10^-4 so pH = 3.54 ✓ [3]
Final answer 3.54 scores 3. ALLOW 3.5 for 2 marks. ECF from wrong moles of HX or X- for the final mark. Using concentrations (divide by 0.0350 dm3) is acceptable.

(c) H+ reacts with X- (conjugate base / A-) ✓ (X- + H+ → HX) ; equilibrium HX ⇌ H+ + X- shifts left / [X-] decreases and [HX] increases only slightly, so ratio [HX]/[X-] (and so [H+]) almost constant ✓ [2]
ALLOW 'large reservoirs of HX and X-'. IGNORE 'pH stays the same' without reason. DO NOT ALLOW HX reacting with the added acid.

(d)(i) 31.25 (cm3) ✓ [1]
ALLOW 31.3 or 31.

(d)(ii) Phenolphthalein AND steep section of titration curve (vertical region, approx pH 7–10) includes / matches its range ✓ ; at equivalence X- (conjugate base) reacts with / accepts H+ from water producing OH- (X- + H2O ⇌ HX + OH-) so solution is alkaline ✓ [2]
ALLOW equation as the explanation. IGNORE calculation of equivalence pH. DO NOT ALLOW 'NaOH is in excess' or 'strong base'.

(e) pH = pKa = 3.87 AND [HX] = [X-] so [H+] = Ka ✓ [1]
ALLOW 3.9. Both value and reason are required.

Total 12 marks.

Hydrogenation data: a Kekulé structure has three localised C=C bonds, so the predicted enthalpy change is \(3 \times (-120) = -360\ \text{kJ mol}^{-1}\). The measured value is \(-208\ \text{kJ mol}^{-1}\), i.e. \(152\ \text{kJ mol}^{-1}\) less exothermic. Benzene is therefore lower in energy (more stable) than the Kekulé structure predicts, which supports the delocalised model, with the six π electrons spread over the whole ring.

Reactivity with bromine: in cyclohexene the π electrons are localised between two carbons, giving high electron density that polarises Br–Br (induced dipole) and attacks the Br\(^{\delta+}\) in electrophilic addition, decolourising bromine water. In benzene the π electrons are delocalised, so the electron density between any two carbons is lower, Br₂ is not polarised enough, and the stable ring resists attack. Reaction would need a halogen carrier (e.g. FeBr₃) and would be substitution, retaining the delocalised ring.

Phenol: a lone pair on oxygen overlaps with and donates into the π system, increasing electron density in the ring (especially at the 2, 4 and 6 positions). Br₂ is polarised, so electrophilic substitution occurs readily at room temperature without a catalyst. Three hydrogens are replaced, giving 2,4,6-tribromophenol (white precipitate) plus HBr.

Level-based marking (indicative content; 6 marks).

Indicative scientific points:
1. Predicted Kekulé value \(3 \times (-120) = -360\ \text{kJ mol}^{-1}\) AND actual \(-208\ \text{kJ mol}^{-1}\) is less exothermic by 152 kJ mol⁻¹ (ALLOW ECF for arithmetic slip).
2. Benzene more stable than Kekulé structure / delocalisation energy; supports delocalised (π) model with six electrons spread over ring.
3. Cyclohexene: localised π bond, high electron density, polarises Br₂ (induced dipole), electrophilic addition.
4. Benzene: delocalised, lower electron density between carbon atoms, cannot polarise Br₂ sufficiently / addition would destroy stable delocalised ring; needs halogen carrier (ALLOW FeBr₃ / AlBr₃) and gives substitution.
5. Phenol: oxygen lone pair donated into/overlaps with ring π system, increasing electron density, so Br₂ polarised and reacts without catalyst (electrophilic substitution).
6. Product 2,4,6-tribromophenol (ALLOW white precipitate; DO NOT ALLOW addition product).

Level 3 (5–6 marks): Covers all three required bullet areas with correct calculation and clear, logical explanations linking electron density to reactivity; the answer is well organised with no significant errors. 6 marks if all points 1–6 are substantially present; 5 marks if one point is missing or one minor error.

Level 2 (3–4 marks): Addresses at least two bullet areas with mostly correct science, but with some gaps in explanation (e.g. electron density stated without link to polarisation of Br₂) or in calculation; organisation reasonably clear.

Level 1 (1–2 marks): Makes some relevant statements (e.g. calculates −360 kJ mol⁻¹, or states benzene is less reactive because of delocalisation, or names the phenol product) with limited linking; answer may lack structure.

Level 0 (0 marks): No response or no response worthy of credit.

IGNORE references to 'resonance' without delocalisation; DO NOT ALLOW 'benzene has alternating single and double bonds' as the final model; ALLOW 'electron-rich' for high electron density.

(a) \(K_a = \frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]}\)

(b) \([\mathrm{H^+}] = 10^{-2.78} = 1.66 \times 10^{-3}\) mol dm\(^{-3}\). Assuming [H+] = [A-] and that [HA] at equilibrium is approximately 0.150 mol dm\(^{-3}\) (dissociation negligible; water's ionisation ignored):
\(K_a = \frac{(1.66\times10^{-3})^2}{0.150} = 1.84 \times 10^{-5}\) mol dm\(^{-3}\).

(c)(i) n(HA) initially = 0.0250 x 0.150 = 3.75 x 10\(^{-3}\) mol. n(NaOH) = 0.0150 x 0.100 = 1.50 x 10\(^{-3}\) mol. NaOH reacts 1:1 with HA, so n(A\(^-\)) = 1.50 x 10\(^{-3}\) mol and n(HA) remaining = 3.75 x 10\(^{-3}\) - 1.50 x 10\(^{-3}\) = 2.25 x 10\(^{-3}\) mol.

(c)(ii) Volumes cancel in the ratio: \([\mathrm{H^+}] = K_a \times \frac{[\mathrm{HA}]}{[\mathrm{A^-}]} = 1.84\times10^{-5} \times \frac{2.25}{1.50} = 2.76 \times 10^{-5}\) mol dm\(^{-3}\). pH = -log(2.76 x 10\(^{-5}\)) = 4.56.

(a) Ka = [H+][A-]/[HA] (1)
ALLOW [H3O+] for [H+]. ALLOW equilibrium expression written with HA <=> H+ + A-. DO NOT ALLOW round brackets or missing square brackets if concentrations are clearly not indicated; IGNORE state symbols.

(b) [H+] = 10^-2.78 = 1.66 x 10^-3 (mol dm^-3) (1)
Ka = (1.66 x 10^-3)^2 / 0.150 = 1.84 x 10^-5 (1) (final answer; 1.83 x 10^-5 to 1.84 x 10^-5 ALLOW from rounding)
units mol dm^-3 (1) AND assumption: [H+] = [A-] (from dissociation of HA only) OR [HA]equilibrium approximately equal to [HA]initial OR dissociation of water negligible (1)
Note: the three marks are: [H+] (1), Ka value with correct units (1), assumption stated (1). Correct final answer with no working scores full marks for the first two points only. ECF for wrong [H+] only if the method (square / 0.150) is correct.

(c)(i) n(NaOH) = 0.0150 x 0.100 = 1.50 x 10^-3 mol AND n(A-) = 1.50 x 10^-3 mol (1)
n(HA) = 3.75 x 10^-3 - 1.50 x 10^-3 = 2.25 x 10^-3 mol (1)
ALLOW ECF from an incorrect n(NaOH) for the HA remaining.

(c)(ii) [H+] = Ka x n(HA)/n(A-) = 1.84 x 10^-5 x 2.25/1.50 = 2.76 x 10^-5 (1)
pH = 4.56 (1) ALLOW 4.56 only if given to 2 d.p.; ALLOW ECF from (b) and (c)(i). ALLOW 4.55 to 4.57. DO NOT ALLOW use of pH = pKa (4.74) with no ratio, which scores 0.

Total 8 marks: 1 + 3 + 2 + 2.

(a) \(K_a = \dfrac{[\mathrm{H^+}][\mathrm{C_6H_5COO^-}]}{[\mathrm{C_6H_5COOH}]}\)

(b)(i) \([\mathrm{H^+}]^2 = K_a \times [\mathrm{HA}] = 6.30\times10^{-5} \times 0.0500 = 3.15\times10^{-6}\); \([\mathrm{H^+}] = 1.775\times10^{-3}\ \mathrm{mol\ dm^{-3}}\); \(\mathrm{pH} = -\log(1.775\times10^{-3}) = 2.75\).

(b)(ii) Assumptions: [H+] = [C6H5COO-] (H+ from ionisation of water is negligible) and/or the equilibrium concentration of HA equals its initial concentration because only a very small proportion dissociates.

(c)(i) Initial n(HA) = 0.0400 dm3 x 0.100 = \(4.00\times10^{-3}\) mol; n(NaOH) = 0.0200 x 0.100 = \(2.00\times10^{-3}\) mol. NaOH reacts 1:1 with HA, so n(HA) remaining = \(2.00\times10^{-3}\) mol and n(C6H5COO-) formed = \(2.00\times10^{-3}\) mol.

(c)(ii) \([\mathrm{H^+}] = K_a \times \dfrac{[\mathrm{HA}]}{[\mathrm{A^-}]} = 6.30\times10^{-5} \times 1 = 6.30\times10^{-5}\) (the volume cancels as the amounts are equal); \(\mathrm{pH} = 4.20\).

(d) Added H+ reacts with the benzoate ions: \(\mathrm{C_6H_5COO^- + H^+ \rightarrow C_6H_5COOH}\). The equilibrium \(\mathrm{C_6H_5COOH \rightleftharpoons H^+ + C_6H_5COO^-}\) shifts to the left, removing most of the added H+. Because the buffer contains a large reservoir of both HA and A-, the ratio [HA]/[A-] changes only slightly, so [H+] and the pH remain almost constant.

(a) [H+][C6H5COO-] / [C6H5COOH] (1)
ALLOW [H3O+] for [H+]; ALLOW [A-]/[HA] forms. DO NOT ALLOW round brackets or missing brackets if the species are ambiguous. IGNORE state symbols.

(b)(i) [H+] = sqrt(6.30 x 10^-5 x 0.0500) = 1.775 x 10^-3 (1) AND pH = 2.75 (1)
Final answer 2.75 scores 2. ALLOW 2.7 or 2.8 only for 1 mark if the method is shown. ECF for an incorrect [H+] to a correct pH. Method mark: use of [H+]^2 = Ka x [HA].

(b)(ii) [H+] = [C6H5COO-] OR [H+] from water is negligible OR [HA]equilibrium = [HA]initial because dissociation is small (1)
ALLOW any one of these. DO NOT ALLOW 'temperature is constant' or 'acid is weak' without reference to dissociation.

(c)(i) n(benzoic acid) = 2.00 x 10^-3 mol (1) AND n(benzoate) = 2.00 x 10^-3 mol (1)
ECF from an incorrect initial amount. Working: n(HA) initially = 4.00 x 10^-3; n(NaOH) = 2.00 x 10^-3. One mark for both initial amounts correct with a wrong subtraction.

(c)(ii) pH = 4.20 (1)
ALLOW 4.2. ALLOW ECF from (c)(i) via [H+] = Ka x n(HA)/n(A-). Allow use of concentrations (each 0.0333 mol dm-3) giving the same answer.

(d) H+ reacts with / is removed by benzoate ions (A-) OR C6H5COO- + H+ -> C6H5COOH (1)
equilibrium shifts to the left AND ratio [HA]/[A-] changes little / [A-] is in large excess compared with the H+ added so pH changes little (1)
ALLOW 'conjugate base'. DO NOT ALLOW 'the pH stays exactly the same' OR 'the buffer neutralises all the acid'. IGNORE reference to added alkali.

(a) n(methyl benzoate) = 6.80 / 136.0 = 0.0500 mol. Reaction is 1:1, so theoretical n(product) = 0.0500 mol. Theoretical mass = 0.0500 x 181.0 = 9.05 g. Percentage yield = 6.15 / 9.05 x 100 = 67.96% = 68.0%.

(b)(i) HNO3 + 2H2SO4 -> NO2+ + H3O+ + 2HSO4- (equivalent: HNO3 + H2SO4 -> NO2+ + H2O + HSO4-).
(b)(ii) The pi electrons of the ring attack NO2+ (curly arrow from the ring to N). The intermediate is a partially delocalised ring cation with the horseshoe covering the ring carbons except the sp3 carbon bearing H and NO2. A curly arrow from the C-H bond back into the ring restores aromaticity, and H+ is released; H+ + HSO4- -> H2SO4 regenerates the catalyst.

(c)(i) Heat under reflux with tin and concentrated hydrochloric acid; then add excess aqueous sodium hydroxide to liberate the amine from its salt. (c)(ii) The lone pair on nitrogen is delocalised into the benzene ring, so it is less available to accept a proton than the lone pair on ethylamine, where the alkyl group is electron-donating.

(d) Methyl 3-nitrobenzoate has 8 carbons, all in different environments, so 8. Methyl 4-nitrobenzoate has a plane of symmetry, so the ring carbons pair up: 6 environments.

(e) Dissolve the crude solid in the minimum volume of hot solvent so that a saturated solution forms and the yield is high. Filter hot to remove insoluble impurities, using a pre-heated funnel to prevent early crystallisation. Cool slowly, then in ice, so crystals form while soluble impurities stay in solution because their concentration is low. Filter under reduced pressure, wash with a little ice-cold solvent, then dry. Place a small sample in a capillary tube and heat slowly in a melting point apparatus, recording the range from first liquid to complete melting. Compare with the data-book value: a pure sample melts sharply (range about 1-2 degrees C) at the literature value; impurities depress and broaden the range. Repeat the recrystallisation until the melting point is constant.

(a) 3 marks
Correct final answer 68.0 / 67.96 / 68 (%) = 3 marks. ALLOW 67.9 to 68.1.
Otherwise: n(methyl benzoate) = 6.80/136.0 = 0.0500 mol (1 mark, method) AND theoretical mass = 0.0500 x 181.0 = 9.05 g (1 mark, method) AND 6.15/9.05 x 100 (1 mark, accuracy). ECF from an incorrect intermediate. DO NOT ALLOW a final answer above 100%.

(b)(i) 1 mark
HNO3 + 2H2SO4 -> NO2+ + H3O+ + 2HSO4- OR HNO3 + H2SO4 -> NO2+ + H2O + HSO4-. ALLOW H2NO3+ route if it is balanced and leads to NO2+. IGNORE state symbols.

(b)(ii) 3 marks
1. Curly arrow from the delocalised ring (circle or double bond) to N of NO2+ (arrow must start at the ring, not at an atom). [1]
2. Correct intermediate: ring with horseshoe covering the five sp2 carbons, positive charge inside the horseshoe, with H and NO2 on the same carbon (meta to CO2CH3). [1]
3. Curly arrow from the C-H bond into the ring, giving the product AND H+ released (H+ + HSO4- -> H2SO4 shown or stated). [1]
DO NOT ALLOW the horseshoe to extend over the sp3 carbon. ALLOW the NO2 on any ring position shown consistently in the intermediate only if the final product is methyl 3-nitrobenzoate.

(c)(i) 2 marks
Tin AND concentrated hydrochloric acid (1 mark) AND heat under reflux (1 mark). The final addition of excess NaOH(aq) is required for the free amine: if it is absent, a maximum of 1 mark. ALLOW Fe/HCl. ALLOW H2 with Ni catalyst as a different route for a maximum of 1 mark. DO NOT ALLOW LiAlH4 or NaBH4.

(c)(ii) 1 mark
Lone pair on N is delocalised into the (benzene) ring so it is less available to accept H+ / less able to donate OR ORA for ethylamine, where the alkyl group is electron-donating. DO NOT ALLOW 'electron withdrawing nitro group' as nitro has been reduced.

(d) 2 marks
methyl 3-nitrobenzoate = 8 (1 mark) AND methyl 4-nitrobenzoate = 6 (1 mark).

(e)* 6 marks, levels of response (communication and scientific content)
Level 3 (5-6 marks): A complete, logically ordered account covering recrystallisation AND melting point, with the reasoning given for most key steps. The steps follow a coherent sequence with no significant errors.
Level 2 (3-4 marks): Describes most stages of recrystallisation and the melting point test, with some reasoning. There may be gaps in the sequence or some missing explanations. The account is mostly clear.
Level 1 (1-2 marks): Describes some steps with little or no reasoning, or covers only one of recrystallisation and melting point. The sequence may be unclear.
0 marks: no response or no response worthy of credit.

Indicative scientific points:
- Dissolve in the minimum volume of hot solvent (to give a saturated solution / maximise the yield).
- Hot filtration to remove insoluble impurities (pre-heated funnel to avoid early crystallisation).
- Cool slowly then in an ice bath so that crystals form; soluble impurities remain in the solution as their concentration is low.
- Filter under reduced pressure (Buchner funnel); wash with a small volume of ice-cold solvent; dry.
- Melting point: pack a small amount in a capillary tube, heat slowly near the melting point, record the temperature range from first liquid to all liquid.
- Compare with the data-book value: a pure compound melts sharply (narrow range) at the literature value; impurities lower and broaden the melting range.
- Repeat the recrystallisation / melting point until the value is constant.
IGNORE references to the solvent being chosen as a specific solvent unless it is incompatible. DO NOT ALLOW 'dissolve in cold solvent' or 'heat until dry'.

(a)(i) Concentrated sulfuric acid acts as a catalyst (it increases the rate of reaching equilibrium). Reflux allows the mixture to be heated at the boiling point for a long time because vapours of volatile ethanol (and ester) are condensed and returned to the flask, so none is lost and no flammable vapour escapes.

(ii) n(acid) = 5.01 / 167.0 = 0.0300 mol. Reaction is 1:1 so the theoretical n(ester) = 0.0300 mol. Actual n(ester) = 4.68 / 195.0 = 0.0240 mol. Percentage yield = 0.0240 / 0.0300 x 100 = 80.0%.

(b)(i) \(\mathrm{C_9H_9NO_4 + 6[H] \rightarrow C_9H_{11}NO_2 + 2H_2O}\)

(ii) Carbons: CH3, CH2, C=O, ring C bearing the ester, ring C bearing NH2, and two pairs of equivalent ring CH carbons (positions 2/6 and 3/5) gives 7 environments.

(iii) CH3: 3, triplet (next to 2 H). CH2: 2, quartet (next to 3 H). Ring H next to COOEt: 2, doublet. Ring H next to NH2: 2, doublet. NH2: 2, singlet (no splitting; broad, exchanges).

(c) Indicative content: dissolve the crude solid in the minimum volume of hot solvent; filter hot to remove insoluble impurities; allow filtrate to cool slowly so crystals form, soluble impurities stay in the cold solvent; vacuum (Buchner) filtration to collect crystals; wash with a small volume of ice-cold solvent; dry. Measure the melting point of the dry solid (capillary tube in a heated apparatus) and compare with the data-book value: a pure solid melts sharply at the literature value; impurities lower the melting point and broaden the range.

(a)(i) Catalyst (1) AND reflux: vapours condensed and returned to flask / prevents loss of volatile ethanol or ester / allows prolonged heating at boiling point (1). ALLOW 'speeds up reaction'. IGNORE 'to increase yield' unqualified. DO NOT ALLOW 'dehydrating agent' alone.

(a)(ii) n(acid) = 5.01/167.0 = 0.0300 mol (1); n(ester) = 4.68/195.0 = 0.0240 mol (1); percentage yield = 80.0 (%) (1). ALLOW 80 for the final mark. Correct answer scores 3. Alternative: theoretical mass = 0.0300 x 195.0 = 5.85 g, then 4.68/5.85 x 100 (ALLOW equivalent method).

(b)(i) C9H9NO4 on the left and C9H11NO2 on the right (1); 6[H] on the left AND 2H2O on the right (1). ALLOW ECF for balancing if one formula is incorrect only if it is used consistently. DO NOT ALLOW 4[H] or other imbalance.

(b)(ii) 7 (1).

(b)(iii) Any three of the following rows correct scores marks as: integration column all correct 3:2:2:2:2 (1); splitting for CH3 triplet AND CH2 quartet (1); both ring H doublets AND NH2 singlet (1). ALLOW integration in the ratio 3:2:2:2:2 or the whole-number multiple. ALLOW ring H doublets in either order. ALLOW NH2 as broad singlet.

(c)* Level 3 (5–6 marks): A coherent, logically ordered account covering all stages of recrystallisation (minimum hot solvent; hot filtration for insoluble impurities; slow cooling to crystallise leaving soluble impurities in solution; vacuum filtration, washing with cold solvent and drying) AND a melting point method with correct interpretation (sharp range at literature value = pure; lower and broader range = impure). Few or no errors; clear sequence.
Level 2 (3–4 marks): Describes most stages of recrystallisation in order, with some reasoning, AND makes a basic statement about melting point and purity; OR covers recrystallisation fully but melting point only partly. Some omissions in detail or sequence.
Level 1 (1–2 marks): Gives isolated points about recrystallisation or melting point with little detail, reasoning or sequence.
Level 0 (0 marks): No response or no response worthy of credit.
Indicative points: dissolve in minimum hot solvent; hot filtration; cool slowly / ice bath; vacuum filtration; wash with small volume of cold solvent; dry; melting point in capillary tube; compare with data-book value; pure = sharp, impure = lower and broader range. DO NOT ALLOW a response that uses the boiling point instead of the melting point.

(a) \(K_a = [H^+]^2/[HA]\), so \([H^+]^2 = 1.35\times10^{-5}\times0.200 = 2.70\times10^{-6}\); \([H^+] = 1.643\times10^{-3}\) mol dm\(^{-3}\); pH = \(-\log(1.643\times10^{-3}) = 2.78\). Assumptions: [H\(^+\)] = [A\(^-\)] (water's ionisation negligible) and dissociation of the acid is negligible so [HA] at equilibrium = 0.200 mol dm\(^{-3}\).

(b) n(HA) initial = 0.0250 × 0.200 = 5.00×10\(^{-3}\) mol; n(NaOH) = 0.0100 × 0.100 = 1.00×10\(^{-3}\) mol. After neutralisation n(HA) = 4.00×10\(^{-3}\) mol and n(A\(^-\)) = 1.00×10\(^{-3}\) mol. \([H^+] = K_a\,[HA]/[A^-] = 1.35\times10^{-5}\times4.00 = 5.40\times10^{-5}\) mol dm\(^{-3}\) (the volume cancels). pH = 4.27.

(c) The buffer contains a large reservoir of both CH\(_3\)CH\(_2\)COOH and CH\(_3\)CH\(_2\)COO\(^-\). Added H\(^+\) is removed by the propanoate ions, shifting the equilibrium to the left, so [H\(^+\)] and pH change very little.

(a) [2 marks]
• \([H^+] = \sqrt{1.35\times10^{-5}\times0.200} = 1.64\times10^{-3}\) mol dm\(^{-3}\) ✓ — ALLOW 1.6×10\(^{-3}\) or more figures.
• pH = 2.78 ✓ — ECF from the candidate's [H\(^+\)] if the pH is correctly calculated; answer must be to 2 d.p. Correct answer scores both marks.
• Guidance: the assumption [H\(^+\)] = [A\(^-\)] and [HA]\(_{eq}\) ≈ initial [HA] is not separately credited.

(b) [3 marks]
• Moles: n(HA) = 5.00×10\(^{-3}\) and n(NaOH) = 1.00×10\(^{-3}\); so n(HA) remaining = 4.00×10\(^{-3}\) mol AND n(A\(^-\)) = 1.00×10\(^{-3}\) mol ✓
• \([H^+] = K_a \times [HA]/[A^-] = 5.40\times10^{-5}\) mol dm\(^{-3}\) ✓ — ALLOW use of moles in place of concentrations OR use of concentrations (0.1143 and 0.02857 mol dm\(^{-3}\)). ECF from first mark.
• pH = 4.27 ✓ — ALLOW 4.3 only if working shown; ECF from second mark.

(c) [1 mark]
• Added H\(^+\) reacts with / is removed by propanoate ions (A\(^-\)) AND equilibrium shifts to the left (forming CH\(_3\)CH\(_2\)COOH) so [H\(^+\)] / pH stays (almost) constant ✓
• IGNORE 'neutralises the acid'. DO NOT ALLOW reaction of H\(^+\) with propanoic acid.

(a) \(K_a = \dfrac{[\mathrm{CH_3CH_2COO^-}][\mathrm{H^+}]}{[\mathrm{CH_3CH_2COOH}]}\)

(b) Assume \([\mathrm{H^+}] = [\mathrm{CH_3CH_2COO^-}]\) and that dissociation of the acid is negligible, so \([\mathrm{HA}]_{eq} \approx 0.150\).
\([\mathrm{H^+}]^2 = K_a \times [\mathrm{HA}] = 1.35\times10^{-5} \times 0.150 = 2.025\times10^{-6}\)
\([\mathrm{H^+}] = 1.423\times10^{-3}\) mol dm\(^{-3}\)
pH = 2.85

(c) n(acid) = 0.0250 × 0.150 = \(3.75\times10^{-3}\) mol
n(NaOH) = 0.0100 × 0.100 = \(1.00\times10^{-3}\) mol
After reaction: n(HA) = \(2.75\times10^{-3}\) mol; n(A\(^-\)) = \(1.00\times10^{-3}\) mol.
The volume cancels: \([\mathrm{H^+}] = K_a \times \dfrac{2.75}{1.00} = 3.71\times10^{-5}\) mol dm\(^{-3}\)
pH = 4.43

(d) The buffer contains a large reservoir of propanoate ions (A\(^-\)) and undissociated propanoic acid. Added H\(^+\) reacts with propanoate ions: \(\mathrm{CH_3CH_2COO^-} + \mathrm{H^+} \rightarrow \mathrm{CH_3CH_2COOH}\). The ratio [HA]/[A\(^-\)] changes only slightly, so the pH is almost constant.

(e) The salt formed is basic by hydrolysis, so the equivalence point is above pH 7 and the steep vertical section of the curve spans the colour-change range 8.3–10.0 of phenolphthalein, giving a sharp end-point.

(a) [1]
\(K_a = \dfrac{[\mathrm{CH_3CH_2COO^-}][\mathrm{H^+}]}{[\mathrm{CH_3CH_2COOH}]}\) ✓
ALLOW C2H5COO- / H3O+ used consistently. DO NOT ALLOW round brackets in place of square brackets.

(b) [3]
[H+] = square root of (1.35×10^-5 × 0.150) ✓ (method)
= 1.4(2)×10^-3 mol dm-3 ✓
pH = 2.85 ✓ (2 d.p. required)
Assumption: [H+] = [A-] AND dissociation of the acid is negligible / [HA] at equilibrium = initial [HA] ✓
Maximum 3. ALLOW ECF from an incorrect [H+] for the pH mark.

(c) [3]
n(HA) initially = 3.75×10^-3 mol AND n(NaOH) = 1.00×10^-3 mol ✓
n(HA) remaining = 2.75×10^-3 mol AND n(A-) = 1.00×10^-3 mol ✓
[H+] = Ka × [HA]/[A-] = 3.71×10^-5 so pH = 4.43 ✓
ALLOW ECF from the second mark. Using total volume 35.0 cm3 is allowed. DO NOT ALLOW use of 25.0 cm3 only for both.

(d) [2]
H+ reacts with / is removed by propanoate ions (A-) AND CH3CH2COO- + H+ -> CH3CH2COOH ✓
Equilibrium shifts left / ratio [HA]/[A-] changes very little (large reservoirs) so pH almost unchanged ✓
ALLOW reversible arrow in the equation. DO NOT ALLOW 'pH stays the same'.

(e) [1]
Equivalence point is above pH 7 / in the alkaline region (salt of weak acid and strong base) AND lies within the colour-change range of phenolphthalein (vertical section covers pH 8.3–10) ✓
IGNORE 'because it is a weak acid' alone.

(a)(i) \(K_a = [\mathrm{H^+}][\mathrm{A^-}]/[\mathrm{HA}]\).

(a)(ii) Assume [H\(^+\)] = [A\(^-\)] and that [HA] at equilibrium \(\approx\) 0.200 mol dm\(^{-3}\) (dissociation negligible; water's contribution to [H\(^+\)] ignored).
\([\mathrm{H^+}]^2 = K_a \times [\mathrm{HA}] = 1.51\times10^{-5} \times 0.200 = 3.02\times10^{-6}\)
\([\mathrm{H^+}] = 1.738\times10^{-3}\) mol dm\(^{-3}\)
pH = \(-\log(1.738\times10^{-3}) = 2.76\).

(b) n(HA) initially = 0.0250 \(\times\) 0.200 = 5.00 \(\times\) 10\(^{-3}\) mol.
n(NaOH) = 0.0100 \(\times\) 0.150 = 1.50 \(\times\) 10\(^{-3}\) mol.
NaOH is limiting, so n(A\(^-\)) formed = 1.50 \(\times\) 10\(^{-3}\) mol and n(HA) remaining = 5.00 \(\times\) 10\(^{-3}\) \(-\) 1.50 \(\times\) 10\(^{-3}\) = 3.50 \(\times\) 10\(^{-3}\) mol.
Both species are in the same total volume (35.0 cm\(^3\)), so the volume cancels in the ratio.
\([\mathrm{H^+}] = K_a \times [\mathrm{HA}]/[\mathrm{A^-}] = 1.51\times10^{-5} \times 3.50/1.50 = 3.52\times10^{-5}\) mol dm\(^{-3}\)
pH = 4.45.

(c) The buffer contains a large reservoir of butanoate ions (conjugate base) and undissociated butanoic acid. Added H\(^+\) is removed by reaction with butanoate ions: A\(^-\) + H\(^+\) \(\rightarrow\) HA. The equilibrium HA \(\rightleftharpoons\) H\(^+\) + A\(^-\) shifts to the left. Because the quantities of HA and A\(^-\) are large compared with the amount of H\(^+\) added, the ratio [HA]/[A\(^-\)] changes only slightly, so [H\(^+\)] and pH remain almost constant.

(a)(i) \(K_a = [\mathrm{H^+}][\mathrm{CH_3CH_2CH_2COO^-}]/[\mathrm{CH_3CH_2CH_2COOH}]\) ✓ [1]
ALLOW [H\(_3\)O\(^+\)] for [H\(^+\)]. ALLOW HA / A\(^-\) notation. DO NOT ALLOW round brackets or missing square brackets.

(a)(ii) [2]
\([\mathrm{H^+}] = \sqrt{1.51\times10^{-5}\times0.200} = 1.74\times10^{-3}\) (mol dm\(^{-3}\)) ✓
pH = 2.76 ✓
Assumption required: [H\(^+\)] = [A\(^-\)] AND dissociation of the acid is negligible, so [HA] at equilibrium = 0.200 mol dm\(^{-3}\).
ALLOW ECF from an incorrect [H\(^+\)]. Correct pH with working but no assumption stated = 1 mark only.

(b) [3]
n(HA) = 5.00 \(\times\) 10\(^{-3}\) mol AND n(NaOH) = n(A\(^-\)) = 1.50 \(\times\) 10\(^{-3}\) mol ✓
n(HA) remaining = 3.50 \(\times\) 10\(^{-3}\) mol ✓
\([\mathrm{H^+}] = K_a \times 3.50/1.50 = 3.52\times10^{-5}\) AND pH = 4.45 ✓
ALLOW the third mark for correct use of the ratio with concentrations (each divided by 0.0350 dm\(^3\)). ALLOW pH = pKa + log([A\(^-\)]/[HA]) = 4.82 + log(1.50/3.50) = 4.45. ECF from n(HA) remaining. If acid and alkali amounts are inverted, max 1 (first mark). Correct answer without working = 3.

(c) [2]
Added H\(^+\) reacts with / is removed by butanoate ions (conjugate base / A\(^-\)) AND equilibrium shifts to the left / HA forms ✓
Ratio [HA]/[A\(^-\)] (or concentrations of HA and A\(^-\)) change little (because large reservoirs) so [H\(^+\)] / pH almost constant ✓
ALLOW equation A\(^-\) + H\(^+\) \(\rightarrow\) HA. IGNORE 'pH stays the same' without reference to the ratio. DO NOT ALLOW reaction with butanoic acid, or 'the buffer neutralises the acid'.

(a) \(K_a = \dfrac{[\mathrm{H^+}][\mathrm{CH_3CH_2COO^-}]}{[\mathrm{CH_3CH_2COOH}]}\)

(b) Assuming [H+] = [CH3CH2COO-] and [HA]equilibrium = [HA]initial: \([\mathrm{H^+}]^2 = 1.35\times10^{-5}\times0.150 = 2.025\times10^{-6}\); \([\mathrm{H^+}] = 1.42\times10^{-3}\) mol dm\(^{-3}\); pH = 2.85.

(c) n(HA) initially = 0.0250 x 0.150 = 3.75 x 10^-3 mol; n(NaOH) = 0.0100 x 0.100 = 1.00 x 10^-3 mol. After reaction n(HA) = 2.75 x 10^-3 mol; n(A-) = 1.00 x 10^-3 mol. \([\mathrm{H^+}] = K_a \times [\mathrm{HA}]/[\mathrm{A^-}] = 1.35\times10^{-5}\times 2.75 = 3.71\times10^{-5}\) (the volume cancels). pH = 4.43.

(d) Added H+ reacts with the propanoate ions (the conjugate base): \(\mathrm{CH_3CH_2COO^- + H^+ \rightarrow CH_3CH_2COOH}\). The buffer contains a large reservoir of A-, so [A-] falls and [HA] rises only slightly; the ratio [HA]/[A-] and so [H+] and pH change very little.

(e) At the equivalence point the solution contains sodium propanoate; the propanoate ion is a weak base, so the pH at equivalence is greater than 7 (about 8.8). The vertical section of the curve lies roughly at pH 7 to 11. Phenolphthalein (pKIn 9.3, colour change about pH 8.3-10.3) changes colour within this vertical section, so it is suitable. Methyl orange (pKIn 3.7) would change colour gradually in the buffer region well before the equivalence point, so it is unsuitable.

(a) Ka = [H+][CH3CH2COO-] / [CH3CH2COOH] (1)
ALLOW [A-] and [HA]. DO NOT ALLOW missing square brackets or an expression with the terms inverted.

(b) [H+]^2 = Ka x [HA] = 1.35e-5 x 0.150 = 2.025e-6, so [H+] = 1.42e-3 (1) AND pH = 2.85 (1) (final answer 2.85 scores 2 marks). Assumption (1): [H+] = [A-] (negligible ionisation of water) OR [HA] at equilibrium = initial [HA] / ionisation of HA is negligible. ALLOW 2.8 or 2.847 if working shown (ECF from slip in [H+]).

(c) n(HA) initially 3.75e-3 AND n(NaOH) 1.00e-3 mol (1); n(HA) remaining 2.75e-3 AND n(A-) 1.00e-3 mol (1); [H+] = Ka x [HA]/[A-] = 3.71e-5 (1); pH = 4.43 (1). ALLOW 4.4 or 4.430. ECF between marks. Correct answer with no working scores 4. IGNORE use of total volume of 35.0 cm3 if it cancels.

(d) Equation: CH3CH2COO- + H+ -> CH3CH2COOH (1) ALLOW A- + H+ -> HA. Added H+ is removed by / reacts with the conjugate base (propanoate ions) (1). [HA]/[A-] (ratio) changes little so [H+] / pH remains (almost) constant (1). DO NOT ALLOW 'pH stays the same' without 'almost'/'little' qualification; DO NOT ALLOW equilibrium shift argument given as only reason without ratio or large reservoir of A-.

(e) Phenolphthalein chosen (1) [DO NOT ALLOW if methyl orange chosen]. Equivalence point pH > 7 because the salt/propanoate ion is a weak base OR the end point lies in the vertical section above pH 7 (1). pKIn / colour-change range of phenolphthalein lies within the vertical section; methyl orange changes colour in the buffer region before equivalence (1). ORA for methyl orange. ALLOW ECF for the third mark only if a reasoned choice is consistent.

(a) \(K_a = \dfrac{[\mathrm{CH_3CH_2COO^-}][\mathrm{H^+}]}{[\mathrm{CH_3CH_2COOH}]}\)

(b) \([\mathrm{H^+}]^2 = K_a \times [\mathrm{HA}] = 1.35\times10^{-5} \times 0.150 = 2.025\times10^{-6}\), so \([\mathrm{H^+}] = 1.423\times10^{-3}\) mol dm\(^{-3}\); pH = \(-\log(1.423\times10^{-3}) = 2.85\).

(c) n(acid) initially = 0.0250 × 0.150 = 3.75 × 10\(^{-3}\) mol; n(NaOH) = 0.0150 × 0.100 = 1.50 × 10\(^{-3}\) mol. After reaction n(HA) = 2.25 × 10\(^{-3}\) mol and n(A\(^-\)) = 1.50 × 10\(^{-3}\) mol. The volume is common, so the ratio of moles can be used: \([\mathrm{H^+}] = K_a \times \dfrac{[\mathrm{HA}]}{[\mathrm{A^-}]} = 1.35\times10^{-5}\times\dfrac{2.25}{1.50} = 2.025\times10^{-5}\) mol dm\(^{-3}\); pH = 4.69.

(d) The buffer contains a large reservoir of both propanoic acid (HA) and propanoate ions (A\(^-\)). Added H\(^+\) combines with A\(^-\): \(\mathrm{CH_3CH_2COO^-} + \mathrm{H^+} \rightarrow \mathrm{CH_3CH_2COOH}\). The equilibrium \(\mathrm{HA} \rightleftharpoons \mathrm{H^+} + \mathrm{A^-}\) shifts to the left, removing most of the added H\(^+\). The ratio [HA]/[A\(^-\)] changes only slightly, so [H\(^+\)] and pH change very little.

(a) [1]
Ka = [CH3CH2COO-][H+] / [CH3CH2COOH] (1)
ALLOW [RCOO-] / [HA] / [A-] if defined. ALLOW [H3O+] for [H+]. DO NOT ALLOW round brackets or missing square brackets. IGNORE state symbols.

(b) [3]
[H+]^2 = Ka x [HA] = 1.35e-5 x 0.150 (1) [method: assumes [H+] = [A-] and [HA] at equilibrium = 0.150]
[H+] = 1.42e-3 mol dm-3 (1)
pH = 2.85 (1)
Final answer 2.85 scores 3. ALLOW ECF from an incorrect [H+] for the pH mark. DO NOT ALLOW pH to 1 or 3 d.p. for the final mark (2.8 or 2.847).

(c) [4]
n(HA) initially = 3.75e-3 mol AND n(NaOH) = 1.50e-3 mol (1)
n(HA) remaining = 2.25e-3 mol AND n(A-) = 1.50e-3 mol (1)
[H+] = Ka x [HA]/[A-] = 1.35e-5 x 2.25e-3/1.50e-3 = 2.03e-5 (mol dm-3) (1)
pH = 4.69 (1)
Final answer 4.69 scores 4. ALLOW ECF through the calculation from an incorrect mole value; ECF for the last mark from an incorrect [H+]. ALLOW working with concentrations (total volume 40.0 cm3) giving the same ratio. DO NOT ALLOW use of initial 3.75e-3 mol as [HA] for the second mark.

(d) [3]
Contains (large reservoir of) CH3CH2COOH/HA AND CH3CH2COO-/A- (1)
Added H+ reacts with CH3CH2COO-/A- (to form CH3CH2COOH) / equilibrium HA <=> H+ + A- shifts left (1)
Ratio [HA]/[A-] (and so [H+]) changes little / pH remains almost constant (1)
ALLOW equation CH3CH2COO- + H+ -> CH3CH2COOH. ALLOW 'ORA' for equilibrium-shift arguments written in terms of le Chatelier. IGNORE 'pH stays constant' without reference to the ratio. DO NOT ALLOW H+ reacting with the acid, or 'the buffer neutralises all the H+'.

(a) \(K_a = \dfrac{[\text{H}^+][\text{C}_6\text{H}_5\text{COO}^-]}{[\text{C}_6\text{H}_5\text{COOH}]}\)

(b) \([\text{H}^+]^2 = K_a[\text{HA}] = 6.30\times10^{-5}\times0.0500 = 3.15\times10^{-6}\), so \([\text{H}^+] = 1.775\times10^{-3}\ \text{mol dm}^{-3}\); pH = 2.75. Approximations: [H+] = [A-] (water's ionisation negligible) and equilibrium [HA] = initial [HA] (dissociation negligible).

(c) n(HA) initially = 0.0250 x 0.0500 = 1.25 x 10^-3 mol; n(NaOH) = 0.0100 x 0.0500 = 5.00 x 10^-4 mol. After reaction n(HA) = 7.50 x 10^-4 mol and n(A-) = 5.00 x 10^-4 mol. [H+] = Ka x [HA]/[A-] = 6.30 x 10^-5 x 1.50 = 9.45 x 10^-5 mol dm^-3 (volumes cancel). pH = 4.02.

(d) Added H+ reacts with the conjugate base: \(\text{C}_6\text{H}_5\text{COO}^- + \text{H}^+ \rightarrow \text{C}_6\text{H}_5\text{COOH}\). The large reservoir of A- means [A-] falls and [HA] rises only slightly, so the ratio [HA]/[A-], and hence [H+], changes little.

(e)(i) At equivalence n(A-) = 1.25 x 10^-3 mol in 50.0 cm^3, so [A-] = 0.0250 mol dm^-3. Kb = Kw/Ka = 1.00 x 10^-14 / 6.30 x 10^-5 = 1.587 x 10^-10. [OH-] = sqrt(1.587 x 10^-10 x 0.0250) = 1.99 x 10^-6 mol dm^-3; pOH = 5.70; pH = 14 - 5.70 = 8.30 (alternatively [H+] = Kw/[OH-] = 5.02 x 10^-9, pH 8.30).

(ii) Phenolphthalein. The equivalence point pH (8.30) is basic because the salt of a weak acid hydrolyses; the indicator's colour change range (about 8.3 to 10.0, pKIn 9.3) lies within the steep vertical section of the curve at and just after equivalence. Methyl orange changes at pH well below this, and bromothymol blue changes mostly before the equivalence point is reached.

(a) Ka = [H+][C6H5COO-] / [C6H5COOH] (1)
ALLOW [A-] and [HA] if defined. ALLOW H3O+. DO NOT ALLOW missing square brackets.

(b) [H+] = sqrt(6.30x10^-5 x 0.0500) = 1.78x10^-3 (1); pH = 2.75 (1); both approximations: [H+] = [A-] (ignoring H+ from water) AND [HA]eq = [HA]initial (negligible dissociation) (1)
Correct final answer scores 2 for the calculation. ALLOW 1.775x10^-3. Approximations mark requires both.

(c) n(HA) initial = 1.25x10^-3 mol AND n(NaOH) = 5.00x10^-4 mol (1); n(HA) remaining = 7.50x10^-4 AND n(A-) = 5.00x10^-4 mol (1); [H+] = Ka x [HA]/[A-] = 9.45x10^-5 (1); pH = 4.02 (1)
ALLOW ECF from an incorrect mole ratio for the last two marks. ALLOW 4.0 only if working is shown and the penultimate mark is awarded; final mark requires 2 dp. Final answer 4.02 scores 4.

(d) A- + H+ -> HA (1) [ALLOW C6H5COO- + H+ -> C6H5COOH]; A- is in large excess/the ratio [HA]/[A-] changes little so [H+]/pH is almost constant (1)
IGNORE 'neutralises the acid'. DO NOT ALLOW HA reacting with the added acid.

(e)(i) [A-] = 1.25x10^-3 / 0.0500 = 0.0250 mol dm^-3 (1); Kb = Kw/Ka = 1.59x10^-10 AND [OH-] = sqrt(Kb x [A-]) = 1.99x10^-6 (1); pH = 8.30 (1)
ALLOW a method via Ka: [H+] = sqrt(Ka x Kw / [A-]) = 5.0x10^-9. ALLOW 8.3. ECF from an incorrect [A-] (e.g. 0.05 mol dm^-3 giving pH 8.45) for the last two marks.

(ii) Phenolphthalein (1); equivalence point pH (about 8.3) lies in the range of the indicator colour change / pKIn close to the pH at the equivalence point, within the vertical section of the curve (1)
OR ORA for the other indicators: methyl orange and bromothymol blue change colour at too low a pH/before the equivalence point. The second mark is dependent on correct identification. IGNORE statements that no indicator is suitable.

(a)(i) Kc = [ester][H2O] / ([acid][alcohol]).
(ii) At equilibrium: ester = 0.667 mol, water = 0.667 mol, acid = 1.00 - 0.667 = 0.333 mol, methanol = 0.333 mol. Volume V cancels because moles of reactants equal moles of products. Kc = (0.667 x 0.667) / (0.333 x 0.333) = 0.445/0.111 = 4.00 (to 3 s.f.). No units.

(b)(i) Ka = [H+]^2/[HA]. [H+]^2 = 1.35 x 10^-5 x 0.200 = 2.70 x 10^-6, so [H+] = 1.64 x 10^-3 mol dm-3. pH = -log(1.64 x 10^-3) = 2.78. Assumptions: [H+] = [A-] (water ionisation negligible) AND the equilibrium [HA] is approximately equal to the initial [HA] (dissociation is negligible).
(ii) n(acid) initially = 0.0250 x 0.200 = 5.00 x 10^-3 mol; n(NaOH) = 0.0100 x 0.100 = 1.00 x 10^-3 mol. After reaction n(HA) = 4.00 x 10^-3 mol, n(A-) = 1.00 x 10^-3 mol. [H+] = Ka x [HA]/[A-] = 1.35 x 10^-5 x 4.00 = 5.40 x 10^-5 mol dm-3. pH = 4.27.

(c)(i) Experiments 1 and 2: [ester] doubles, [OH-] constant, rate doubles, so first order in ester. Experiments 1 and 3: [OH-] doubles, [ester] constant, rate doubles, so first order in OH-.
(ii) Rate = k[ester][OH-]. k = 2.4 x 10^-4 / (0.10 x 0.10) = 2.4 x 10^-2 dm3 mol-1 s-1.

(d)(i) Three environments: CH3 of ethyl group, CH2, and OCH3.
(ii) The CH2 signal is a quartet. The adjacent CH3 has 3 equivalent protons, so by the n+1 rule the CH2 signal splits into 3 + 1 = 4 peaks.

(a)(i) [CH3CH2COOCH3][H2O] / ([CH3CH2COOH][CH3OH]) (1)
ALLOW formulae in any unambiguous form. DO NOT ALLOW round brackets used in place of square brackets for concentration. IGNORE state symbols.

(a)(ii) Equilibrium moles of acid and methanol = 0.333 each (1) AND water = 0.667 (1) AND Kc = 4.00 / 4 (1) with no units stated, or 'no units' (see guidance).
ALLOW 4.0 up to 3 s.f. Correct final answer scores 3. ALLOW volume to be shown and cancelled. If units are given, DO NOT ALLOW the final mark. ECF from incorrect equilibrium moles for the final mark only.

(b)(i) [H+]^2 = Ka x [HA] OR [H+] = sqrt(1.35 x 10^-5 x 0.200) (1) AND [H+] = 1.64 x 10^-3 (1) AND pH = 2.78 (1)
Correct final answer scores 3 marks. ALLOW 2.8 and 2.79 if working shows 1.6 x 10^-3 (rounding). Assumption (any one; counts within the second or third mark only if stated, otherwise withhold the third mark if no assumption is given): [H+] = [A-] OR dissociation of HA negligible / [HA] at equilibrium = initial [HA] OR ionisation of water negligible. NOTE: Total 3 marks = method (1), [H+] (1), pH and assumption together (1). Award the third mark for pH 2.78 with one valid assumption; pH correct without an assumption = 2 marks.

(b)(ii) n(HA) = 5.00 x 10^-3 AND n(NaOH) = 1.00 x 10^-3 (1) AND n(HA) remaining = 4.00 x 10^-3, n(A-) = 1.00 x 10^-3 (1) AND [H+] = 1.35 x 10^-5 x 4 = 5.40 x 10^-5 so pH = 4.27 (1)
ALLOW 4.3. Correct final answer scores 3. ECF for the pH mark from incorrect moles of HA and A-. DO NOT ALLOW use of initial moles of HA (pH 4.87 scores max 1).

(c)(i) First order with respect to methyl propanoate AND first order with respect to OH- (1) AND reasoning: doubling the concentration of one reactant, with the other constant, doubles the rate (1)
Both orders correct needed for first mark. ALLOW comparison of experiments 1 and 2, 1 and 3, stated or implied.

(c)(ii) k = 2.4 x 10^-4 / (0.10 x 0.10) = 2.4 x 10^-2 (1) AND units dm3 mol-1 s-1 (1)
ALLOW ECF for k from incorrect orders in (c)(i) with units consistent with that rate equation. ALLOW any correct experiment used.

(d)(i) 3 (1)

(d)(ii) Quartet (1) AND n+1 rule: adjacent CH3 carbon has 3 equivalent protons so 3 + 1 = 4 peaks (1)
ALLOW 'splits into 4 peaks'. DO NOT ALLOW reference to the OCH3 group as the neighbour. IGNORE relative peak heights 1:3:3:1.

Hydrogenation data: Kekulé benzene has three localised C=C bonds, so predicted enthalpy change = 3 × (−120) = −360 kJ mol\(^{-1}\). Measured value = −208 kJ mol\(^{-1}\), i.e. 152 kJ mol\(^{-1}\) less exothermic. Benzene is therefore more stable than the Kekulé structure predicts, which supports a delocalised model in which the six p-electrons form a ring of pi electron density above and below the plane of the ring.

Bromine water: cyclohexene has a localised C=C with high electron density that polarises Br\(_2\) and undergoes electrophilic addition, so it decolourises bromine water. In benzene the pi electron density is delocalised over six carbon atoms, so the electron density between any two carbon atoms is lower; Br\(_2\) is not polarised enough to be attacked, so no reaction and the colour persists.

Experiments 2 and 3: with FeBr\(_3\) as halogen carrier, Br\(_2\) is converted into a much stronger electrophile, Br\(^+\) (\(\mathrm{Br_2 + FeBr_3 \rightarrow Br^+ + FeBr_4^-}\)), which can attack the stable benzene ring in electrophilic substitution. In phenol a lone pair on oxygen is donated into the pi system, so the ring has higher electron density, polarises Br\(_2\) and reacts readily without a carrier. Substitution occurs at the 2, 4 and 6 positions, giving 2,4,6-tribromophenol (white precipitate).

Level 3 (5–6 marks): All three bullet points addressed with correct, relevant science: calculates −360 and compares with −208 (stability/delocalisation), explains why benzene's delocalised ring does not react with bromine water while cyclohexene does, and links the difference between Experiments 2 and 3 to oxygen lone pair donation and the need for a stronger electrophile, naming 2,4,6-tribromophenol. The response is well organised, logical and uses correct terminology throughout. Few or no errors.

Level 2 (3–4 marks): Two of the three bullet points addressed well, or all three addressed with some omissions or lack of detail (e.g. no calculation shown, or no mention of electrophile generation). Structure is mostly logical with some use of correct terminology.

Level 1 (1–2 marks): Limited valid points, e.g. only states that benzene is stable/delocalised or that phenol is more reactive, with little explanation. Organisation may be weak.

0 marks: No response or no response worthy of credit.

Indicative scientific points:
- Kekulé prediction = 3 × (−120) = −360 kJ mol^-1 (ALLOW ECF in later comments)
- actual −208 kJ mol^-1 is 152 kJ mol^-1 less exothermic OR benzene is more stable than predicted
- stability explained by delocalised pi electrons / ring of electron density (IGNORE 'resonance' alone)
- cyclohexene has localised C=C with high electron density, polarises Br2, electrophilic addition, decolourises
- benzene's electron density lower between carbon atoms because spread over ring, so Br2 not polarised enough / no reaction
- FeBr3 (halogen carrier) generates stronger electrophile Br+ (ALLOW equation Br2 + FeBr3 → Br+ + FeBr4^-)
- phenol: lone pair on O donated/delocalised into ring, increasing electron density of ring
- greater electron density polarises/attracts Br2 so no catalyst needed
- product 2,4,6-tribromophenol (white precipitate); DO NOT ALLOW bromophenol without positions or addition products

Communication: logical sequence linking evidence to explanation to be placed in the upper part of a level.