OCR A Chemistry H432/01 · 2026 Predicted Paper 1 — MARK SCHEME
A Level Chemistry A (OCR H432/01)
Breadth in Chemistry · Modules 1, 2, 3 & 4
2026 PREDICTED PAPER 1
MARK SCHEME
Total marks: 100 Marking symbols used:
Section A: 15 marks (1 each) ✓ Mark point awarded
Section B: 85 marks (structured) [ALLOW] Alternative acceptable answer
[DO NOT ACCEPT] Common error — no marks
NOTE Examiner guidance
General marking guidance
Answers must demonstrate understanding, not just recall. In calculations, award marks for correct
method even if arithmetic error leads to wrong answer (error carried forward). For equations, accept
correct multiples of equations shown. For extended responses (Q25), mark holistically against the
levels descriptors shown.
© Predicted Paper Mark Scheme — For Revision Use Only Page 1
, OCR A Chemistry H432/01 · 2026 Predicted Paper 1 — MARK SCHEME
SECTION A — Multiple Choice Answers
Q Ans Working / Rationale
1 A 0.25 mol N₂O₄ = 0.25 mol molecules; B: 1.60/32 = 0.05 mol; C: 1 mol at STP; D: 0.50 mol.
Answer = C (1 mol).
2 C n = 5.30/106 = 0.0500 mol; c = 0.0500/0.250 = 0.200 mol dm⁻³
3 C Ionic lattices are NOT soluble in non-polar solvents (like-dissolves-like; polar solvent
needed).
4 C NH₃: 3 bonding pairs + 1 lone pair → pyramidal, 107°. H ₂O = 104.5°; BF ₃ = 120°; CH ₄ =
109.5°
5 B X = 578 kJ mol⁻¹. Si has a lower 1st IE than Al (above it) due to 3p vs 3s subshell; B = 578
is the expected dip.
6 D MgO + H₂O → Mg(OH)₂; slightly alkaline (pH ~10–12) → purple/violet UI.
7 D Cl₂ oxidises I⁻ to I₂; iodine dissolves in hexane as purple/violet layer.
8 B Definition of standard enthalpy of formation: elements in standard states → 1 mol
compound.
9 A Bonds broken: H–H (436) + Cl–Cl (243) = 679; Bonds made: 2×H–Cl = 864. ΔH = 679 −
864 = −185 kJ mol⁻¹
10 C Higher T → more molecules have E ≥ Ea. Area is constant (same number of molecules).
Peak shifts right.
11 B Exothermic → lower T favours products. Fewer moles on RHS (2 vs 4) → higher P favours
NH₃. Catalyst has NO effect on yield.
12 B Butane (n-butane) and 2-methylpropane (isobutane) = 2 structural isomers.
13 B Propagation step 1: Cl• + CH₄ → CH₃• + HCl
14 B Markovnikov: H adds to less substituted C (CH₂ end); Br adds to more substituted C → 2-
bromopropane.
15 B Elimination requires KOH in ethanol (alcoholic), heat; produces alkene by removal of HBr.
© Predicted Paper Mark Scheme — For Revision Use Only Page 2
, OCR A Chemistry H432/01 · 2026 Predicted Paper 1 — MARK SCHEME
SECTION B — Structured Question Mark Scheme
Question 16 Amount of Substance & Redox Titration [11 marks]
[SPEC] 2.1.2 – Moles | 2.1.3 – Concentration; titration calculations
Part Mark scheme — acceptable answers Marks
16(a)(i) ✓ n = 7.84 ÷ 392 = 0.0200 mol (1) 1
Part Mark scheme — acceptable answers Marks
16(a)(ii) ✓ [Fe²⁺] = 0.0200 ÷ 0.250 = 0.0800 mol dm⁻³ (1) 1
NOTE: 1 mol of salt gives 1 mol Fe²⁺; ecf from (a)(i) accepted.
Part Mark scheme — acceptable answers Marks
16(b)(i) ✓ n(Fe²⁺) = 0.0800 × (25.0 ÷ 1000) = 0.00200 mol (2.00 × 10⁻³ 1
mol) (1)
[ALLOW] ecf from (a)(ii) accepted
Part Mark scheme — acceptable answers Marks
16(b)(ii) ✓ From equation, ratio Fe²⁺ : MnO₄⁻ = 5 : 1 (1) 1
✓ n(MnO₄⁻) = 0.00200 ÷ 5 = 4.00 × 10⁻⁴ mol
NOTE: Award 1 mark for correct ratio AND correct calculation. ecf
accepted.
Part Mark scheme — acceptable answers Marks
16(b)(iii) ✓ c = n ÷ V = (4.00 × 10⁻⁴) ÷ (22.40 ÷ 1000) (1) 2
✓ = 0.0179 mol dm⁻³ (to 3 s.f.) (1)
[ALLOW] 0.01786 accepted if rounded correctly
[DO NOT ACCEPT] Any answer not expressed to 3 s.f. loses the final
mark
Part Mark scheme — acceptable answers Marks
16(c) ✓ KMnO₄ reacts with / oxidises organic impurities / reducing 2
agents in air or water over time (1)
✓ so its concentration is not accurately known / changes on
standing (1)
© Predicted Paper Mark Scheme — For Revision Use Only Page 3
A Level Chemistry A (OCR H432/01)
Breadth in Chemistry · Modules 1, 2, 3 & 4
2026 PREDICTED PAPER 1
MARK SCHEME
Total marks: 100 Marking symbols used:
Section A: 15 marks (1 each) ✓ Mark point awarded
Section B: 85 marks (structured) [ALLOW] Alternative acceptable answer
[DO NOT ACCEPT] Common error — no marks
NOTE Examiner guidance
General marking guidance
Answers must demonstrate understanding, not just recall. In calculations, award marks for correct
method even if arithmetic error leads to wrong answer (error carried forward). For equations, accept
correct multiples of equations shown. For extended responses (Q25), mark holistically against the
levels descriptors shown.
© Predicted Paper Mark Scheme — For Revision Use Only Page 1
, OCR A Chemistry H432/01 · 2026 Predicted Paper 1 — MARK SCHEME
SECTION A — Multiple Choice Answers
Q Ans Working / Rationale
1 A 0.25 mol N₂O₄ = 0.25 mol molecules; B: 1.60/32 = 0.05 mol; C: 1 mol at STP; D: 0.50 mol.
Answer = C (1 mol).
2 C n = 5.30/106 = 0.0500 mol; c = 0.0500/0.250 = 0.200 mol dm⁻³
3 C Ionic lattices are NOT soluble in non-polar solvents (like-dissolves-like; polar solvent
needed).
4 C NH₃: 3 bonding pairs + 1 lone pair → pyramidal, 107°. H ₂O = 104.5°; BF ₃ = 120°; CH ₄ =
109.5°
5 B X = 578 kJ mol⁻¹. Si has a lower 1st IE than Al (above it) due to 3p vs 3s subshell; B = 578
is the expected dip.
6 D MgO + H₂O → Mg(OH)₂; slightly alkaline (pH ~10–12) → purple/violet UI.
7 D Cl₂ oxidises I⁻ to I₂; iodine dissolves in hexane as purple/violet layer.
8 B Definition of standard enthalpy of formation: elements in standard states → 1 mol
compound.
9 A Bonds broken: H–H (436) + Cl–Cl (243) = 679; Bonds made: 2×H–Cl = 864. ΔH = 679 −
864 = −185 kJ mol⁻¹
10 C Higher T → more molecules have E ≥ Ea. Area is constant (same number of molecules).
Peak shifts right.
11 B Exothermic → lower T favours products. Fewer moles on RHS (2 vs 4) → higher P favours
NH₃. Catalyst has NO effect on yield.
12 B Butane (n-butane) and 2-methylpropane (isobutane) = 2 structural isomers.
13 B Propagation step 1: Cl• + CH₄ → CH₃• + HCl
14 B Markovnikov: H adds to less substituted C (CH₂ end); Br adds to more substituted C → 2-
bromopropane.
15 B Elimination requires KOH in ethanol (alcoholic), heat; produces alkene by removal of HBr.
© Predicted Paper Mark Scheme — For Revision Use Only Page 2
, OCR A Chemistry H432/01 · 2026 Predicted Paper 1 — MARK SCHEME
SECTION B — Structured Question Mark Scheme
Question 16 Amount of Substance & Redox Titration [11 marks]
[SPEC] 2.1.2 – Moles | 2.1.3 – Concentration; titration calculations
Part Mark scheme — acceptable answers Marks
16(a)(i) ✓ n = 7.84 ÷ 392 = 0.0200 mol (1) 1
Part Mark scheme — acceptable answers Marks
16(a)(ii) ✓ [Fe²⁺] = 0.0200 ÷ 0.250 = 0.0800 mol dm⁻³ (1) 1
NOTE: 1 mol of salt gives 1 mol Fe²⁺; ecf from (a)(i) accepted.
Part Mark scheme — acceptable answers Marks
16(b)(i) ✓ n(Fe²⁺) = 0.0800 × (25.0 ÷ 1000) = 0.00200 mol (2.00 × 10⁻³ 1
mol) (1)
[ALLOW] ecf from (a)(ii) accepted
Part Mark scheme — acceptable answers Marks
16(b)(ii) ✓ From equation, ratio Fe²⁺ : MnO₄⁻ = 5 : 1 (1) 1
✓ n(MnO₄⁻) = 0.00200 ÷ 5 = 4.00 × 10⁻⁴ mol
NOTE: Award 1 mark for correct ratio AND correct calculation. ecf
accepted.
Part Mark scheme — acceptable answers Marks
16(b)(iii) ✓ c = n ÷ V = (4.00 × 10⁻⁴) ÷ (22.40 ÷ 1000) (1) 2
✓ = 0.0179 mol dm⁻³ (to 3 s.f.) (1)
[ALLOW] 0.01786 accepted if rounded correctly
[DO NOT ACCEPT] Any answer not expressed to 3 s.f. loses the final
mark
Part Mark scheme — acceptable answers Marks
16(c) ✓ KMnO₄ reacts with / oxidises organic impurities / reducing 2
agents in air or water over time (1)
✓ so its concentration is not accurately known / changes on
standing (1)
© Predicted Paper Mark Scheme — For Revision Use Only Page 3
