Chemistry 9701/43 May/June 2025: question paper and mark scheme
Cambridge AS & A Level Chemistry (9701), Paper 4 A Level Structured Questions, May/June 2025. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Both calcium carbonate, CaCO3, and barium carbonate, BaCO3, decompose when heated to form the metal oxide and a gas.
(a) Write an equation for the thermal decomposition of CaCO3.
…… [1]
(b) State which of CaCO3 and BaCO3 decomposes at a lower temperature.
Explain your answer.
The compound that decomposes at a lower temperature is …… .
explanation ……
[2]
(c) Calcium oxide, CaO, reacts with water to form compound A.
Barium oxide, BaO, reacts with water to form compound B.
(i) Identify A.
…… [1]
(ii) Explain why A is less soluble than B.
…… [3]
[Total: 7]
Mark scheme
1(a) CaCO3 → CaO + CO2 [1] 1
1(b) M1: calcium carbonate AND calcium ion / Ca2+ smaller / ionic radius bigger down group 2
M2: carbonate ion / anion / CO32– is more distorted/polarised by Ca2+
1(c)(i) calcium hydroxide / Ca(OH)2 1
1(c)(ii) M1: A / calcium hydroxide has more exothermic ΔHlatt AND ΔHhyd than B / barium hydroxide 3
M2: difference in ΔHlatt is greater for A / calcium hydroxide than for B / barium hydroxide
M3: ΔHsol is less exothermic for A / calcium hydroxide than for B / barium hydroxide
Three experiments are carried out to investigate the reaction of nitrogen oxide, NO, with chlorine.
2NO + Cl 2 2NOCl
The rate equation for this reaction is shown.
rate = k [NO]2[Cl 2]
(a) Under the conditions used in experiments 1 and 2, the value of k is 26.4.
(i) The rate of the reaction is measured in mol dm–3 s–1. State the units of k.
units of k = …… [1]
(ii) In experiment 1, the initial concentrations of NO and Cl 2 are equal.
The initial rate of the reaction in experiment 1 is 2.57 × 10–6 mol dm–3 s–1.
Calculate the initial concentration of NO.
Show your working.
initial concentration of NO = …… mol dm–3 [2]
(iii) In experiment 2, the initial concentrations of NO and Cl 2 are both ten times greater than the initial concentrations used in experiment 1.
Calculate the initial rate of the reaction in experiment 2.
(b) Experiment 3 uses a large excess of NO.
The initial concentration of Cl 2 is 2.00 × 10– 4 mol dm–3.
(i) The graph of [Cl 2] against time shows that the reaction has a constant half-life, t ½.
Explain this observation.
…… [1]
(ii) Under the conditions used in experiment 3, the value of the rate constant is 105.6.
Show that t ½ of [Cl 2] is 6.56 × 10–3 s under these conditions.
[1]
(iii) Calculate the time taken, in s, for [Cl 2] to fall to 1.25 × 10–5 mol dm–3 in experiment 3.
time = …… s [1]
(c) Sulfur dioxide, SO2, reacts very slowly with oxygen in the atmosphere, forming sulfur trioxide,
SO3. This reaction is much faster in the presence of NO.
Explain the role of NO in this process.
Include chemical equations in your answer.
…… [3]
[Total: 10]
Mark scheme
2(a)(i) mol–2 dm6 s–1 1
2(a)(ii) M1: 2.57 × 10–6 = 26.4 × [concentration]3 2
M2: [concentration] = 4.60 × 10–3
2(a)(iii) 2.57 × 10–3 1
2(b)(i) overall 1st order / pseudo 1st order 1
OR 1st order with respect to Cl2 AND concentration of NO doesn’t change
2(b)(ii) ln2 / 105.6 or 0.693 / 105.6 1
2(b)(iii) 0.0262 / 2.62 × 10–2 1
2(c) M1: homogeneous catalyst OR catalyst because it is regenerated 3
M2: 2NO + O2 → 2NO2
M3: NO2 + SO2 → NO + SO3
(a) Chromium(III) hydroxide, Cr(OH)3, is only slightly soluble in water. The value of the solubility product, Ksp, of Cr(OH)3 is 1.0 × 10–33 at 298 K.
(i) Complete the expression for Ksp of Cr(OH)3. Include the units.
Ksp =
units = ……
[2]
(ii) Calculate the solubility, in g dm–3, of Cr(OH)3 in pure water at 298 K.
Show your working.
solubility = …… g dm–3 [3]
(iii) Cr(OH)3 is less soluble in 0.100 mol dm–3 NaOH than it is in pure water.
Explain this observation.
…… [1]
(b) The value of the acid dissociation constant, Ka, of butanoic acid, CH3CH2CH2COOH, is
1.51 × 10–5 at 298 K.
(i) Calculate the pH of 0.100 mol dm–3 CH3CH2CH2COOH at 298 K.
Show your working.
pH = …… [2]
(ii) Calculate the pH of 0.100 mol dm–3 NaOH at 298 K.
pH = …… [1]
(iii) 5.00 cm3 of 0.100 mol dm–3 NaOH is added to 10.00 cm3 of 0.100 mol dm–3
CH3CH2CH2COOH.
Calculate the pH of the resulting solution.
Show your working.
pH = …… [2]
(c) 80.0 cm3 of an aqueous solution containing 0.704 g of CH3CH2CH2COOH is shaken with
100 cm3 of benzene, C6H6.
There is 0.556 g of CH3CH2CH2COOH in the 100 cm3 of C6H6 at equilibrium.
Calculate the partition coefficient, Kpc, of CH3CH2CH2COOH between C6H6 and water.
Show your working.
Kpc = …… [2]
[Total: 13]
Mark scheme
3(a)(i) M1: Ksp = [Cr3+] [OH–]3 2
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: [CH3CH2CH2COOH] = [CH3CH2CH2COO–] 2
M2: pH = 4.82
3(c) M1: 0.148 g is left in water 2
M2: Kpc = 3.01 OR 0.333
Table 4.1 gives the enthalpy changes of hydration, ΔHhyd, of three ions, F–, K+ and Ca2+.
Table 4.1
ion ΔHhyd / kJ mol–1
F– –506
K+ –322
Ca2+ –1650
(a) (i) Define enthalpy change of hydration.
…… [1]
(ii) Explain the relative magnitudes of the enthalpy changes of hydration of K+ and Ca2+.
…… [2]
(iii) Define lattice energy.
…… [1]
(iv) The lattice energy, ΔHlatt, of calcium fluoride, CaF2, is –2602 kJ mol–1.
Calculate the enthalpy change of solution, ΔHsol, in kJ mol–1, of CaF2.
ΔHsol of CaF2 = …… kJ mol–1 [2]
(b) The formation of CaF2 at 298 K is shown.
Ca(s) + F2(g) CaF2(s) ΔH o = –1214 kJ mol–1
ΔG o = –1162 kJ mol–1
Calculate the entropy change, ΔS o , in J K–1 mol–1, for this reaction.
ΔS o = …… J K–1 mol–1 [2]
[Total: 8]
Mark scheme
4(a)(i) energy change when one mole of gaseous ions dissolves in water 1
4(a)(ii) M1: K+ has larger radius AND K+ has smaller charge 2
M2: Ca2+ has larger attraction for H2O
4(a)(iii) energy change when one mole of ionic solid forms from gaseous ions 1
4(a)(iv) M1: –1650–(2 × 506) + 2602 2
M2: –60
4(b) M1: ΔG = ΔH –TΔS AND T = 298 2
M2: –174.5 [1]
Copper is a transition element.
(a) (i) Complete the electronic configurations of a Cu+ ion and a Cu2+ ion.
Cu+ ion: [Ar] ……
Cu2+ ion: [Ar] ……
[1]
(ii) Explain why transition elements have variable oxidation states.
…… [1]
(b) Aqueous copper(II) sulfate, CuSO4, contains the [Cu(H2O)6]2+ complex ion.
(i) A few drops of NH3(aq) are added to CuSO4(aq).
Describe any observations made.
…… [1]
(ii) Write an equation for the reaction taking place.
…… [1]
(c) (i) An excess of NH3(aq) is added to CuSO4(aq).
Describe any further observations made.
…… [1]
(ii) Write an equation for the reaction taking place.
…… [1]
(iii) State the name for the type of reaction taking place.
…… [1]
(d) Copper metal can be oxidised by acidified KMnO4. The relevant half-equations and their standard electrode potentials, E o , are shown.
Cu2+ + 2e– Cu E o = +0.34 V
MnO4– + 8H+ + 5e– Mn2+ + 4H2O E o = +1.52 V
(i) A MnO4–/Mn2+ electrode is constructed using 0.0020 mol dm–3 MnO4–, 1.0 mol dm–3 Mn2+ and 1.0 mol dm–3 H+. The temperature used is 298 K.
Use the Nernst equation to show that the E value for this MnO4–/Mn2+ electrode is +1.49 V.
[2]
(ii) An electrochemical cell is constructed using a standard Cu2+/Cu electrode and the
MnO4–/Mn2+ electrode described in (d)(i).
Calculate the value of Ecell.
Ecell = …… V [1]
(iii) Write an equation for the reaction taking place in the electrochemical cell described in
(d)(ii).
…… [1]
(iv) Complete the sentences for the electrochemical cell described in (d)(ii).
The …… electrode is the negative electrode. Electrons flow from the
…… electrode to the …… electrode when the cell is in use.
[1]
(e) A solution containing [Cu(H2O)6]2+ is electrolysed for 5.00 hours using a constant electric current. 0.764 g of copper metal is formed at the cathode. No other reduction reaction takes place.
Calculate the electric current, in A, used. Give your answer to three significant figures.
current = …… A [3]
[Total: 15]
Mark scheme
5(a)(i) (Ar) 3d10 AND (Ar) 3d9 1
5(a)(ii) similar energy of 3d and 4s subshells 1
5(b)(i) pale blue precipitate 1
5(b)(ii) [Cu(H2O)6]2+ + 2OH– → Cu(OH)2(H2O)4 + 2H2O 1
5(c)(i) deep blue solution 1
5(c)(ii) [Cu(H2O)6]2+ + 4NH3 → [Cu(NH3)4(H2O)2]2+ + 4H2O 1
5(c)(iii) ligand exchange 1
5(d)(i) M1: E = Eo +(0.059 / z) log(ox/red) 2
M2: E = 1.52 + (0.059 / 5) × log (0.002)
5(d)(ii) 1.15 1
5(d)(iii) 5Cu + 2MnO4– + 16H+ → 5Cu2+ + 2Mn2+ + 8H2O 1
5(d)(iv) Cu2+ / Cu, Cu2+ / Cu, MnO4– / Mn2+ / platinum 1
5(e) M1: moles of copper = 0.0120 mol 3
M2: coulombs required = 2322 C
M3: current required = 0.129 A
(a) Iron forms complex ions with the monodentate ligand, CN–.
(i) Complex ion A contains one Fe3+ ion and six CN– ligands.
Complex ion B contains one Fe2+ ion and six CN– ligands.
State the formulae of these two complex ions. Include the overall charge of each complex ion.
complex ion A ……
complex ion B ……
[1]
(ii) Explain why a solution containing complex ion A and a solution containing complex ion B are different colours.
…… [2]
(iii) In complex ion A, the carbon atom of each CN– ligand bonds to the Fe3+ ion.
State the type of bonding involved.
…… [1]
(b) Complex ions have different geometries.
Complex ion A is octahedral.
Ag+ ions form a linear complex with ammonia.
Ni atoms form a tetrahedral complex with carbon monoxide molecules. The carbon atom in the monodentate carbon monoxide ligand bonds to the nickel atom.
Pd2+ ions form a square planar complex with chloride ions.
Complete Fig. 6.1 to show the geometry of each of these four ions, using three-dimensional bonds where necessary. Label one bond angle on each complex ion.
complex ion A Ag+ with ammonia
Fe Ag
Ni with carbon monoxide Pd2+ with chloride
Ni Pd
Fig. 6.1
[4]
[Total: 8]
Mark scheme
6(a)(i) [Fe(CN)6]3– AND [Fe(CN)6]4– 1
6(a)(ii) M1: different ΔE 2
M2: absorption of different frequency of visible light
OR wavelength of visible light
6(a)(iii) dative covalent / coordinate 1
6(b) 4
Each correct structure = [1]
A hydrocarbon is known to be either compound D or compound E.
D E
CH2CH3 CH2CH3
CH2CH3
CH2CH3
Fig. 7.1
(a) Give the systematic name of E.
…… [1]
(b) The proton (1H) NMR spectra of D and E are compared. They are very similar.
The proton (1H) NMR spectrum of D is shown in Fig. 7.2.
10 9 8 7 6 5 4 3 2 1 0
chemical shift δ / ppm
Fig. 7.2
(i) Suggest a suitable solvent for obtaining the spectrum in Fig. 7.2.
…… [1]
(ii) The proton (1H) NMR spectrum of E is obtained twice, once before and once after shaking with D2O.
Describe any differences between these two spectra. Explain your answer.
…… [1]
(iii) Complete Table 7.1 for the proton (1H) NMR spectrum of D.
Table 7.1
chemical shift number of 1H atoms group responsible splitting pattern δ / ppm responsible for the peak for the peak
1.3
2.7
7.1
[3]
Table 7.2
chemical shift range environment of proton example δ / ppm alkane –CH3, –CH2–, >CH– 0.9–1.7 alkyl next to C=O CH3–C=O, –CH2–C=O, >CH–C=O 2.2–3.0
alkyl next to aromatic ring CH3–Ar, –CH2–Ar, >CH–Ar 2.3–3.0 alkyl next to electronegative atom CH3–O, –CH2–O, –CH2–Cl 3.2–4.0 attached to alkene =CHR 4.5–6.0 attached to aromatic ring H–Ar 6.0–9.0 aldehyde HCOR 9.3–10.5 alcohol ROH 0.5–6.0 phenol Ar–OH 4.5–7.0 carboxylic acid RCOOH 9.0–13.0
alkyl amine R–NH– 1.0–5.0 aryl amine Ar–NH2 3.0–6.0 amide RCONHR 5.0–12.0
(iv) Compounds D and E can be distinguished by carbon-13 NMR spectroscopy.
State the number of peaks in each spectrum.
The carbon-13 NMR spectrum of D has …… peaks.
The carbon-13 NMR spectrum of E has …… peaks.
[1]
(c) Compound D can be oxidised to compound F by alkaline KMnO4 followed by dilute acid.
D F
CH2CH3 COOH
CH2CH3 COOH
Fig. 7.3
(i) Write an equation for this reaction using molecular formulae for D and F. The products of this reaction are F, water and carbon dioxide.
Use [O] to represent one atom of oxygen from the oxidising agent.
…… [1]
(ii) F reacts with an excess of SOCl 2 to form compound G. The molecular formula of G is
C8H4Cl 2O2.
G reacts with ethane-1,2-diol, HOCH2CH2OH, to form a mixture of products that includes compounds J, molecular formula C10H8O4, and K, molecular formula C18H12Cl 2O6.
Draw the structures of compounds G, J and K in Fig. 7.4.
G, C8H4Cl 2O2
J, C10H8O4
K, C18H12Cl 2O6
Fig. 7.4
[3]
[Total: 11]
Mark scheme
7(a) 1,3-diethylbenzene 1
7(b)(i) CDCl3 1
7(b)(ii) no difference AND no H that can be exchanged with D 1
7(b)(iii) 3
δ / ppm number of group name of
protons responsible splitting pattern
1.3 6 CH3 triplet
2.7 4 CH2 quartet / quadruplet
7.1 4
Each correct column = [1]
7(b)(iv) 5 AND 6 1
7(c)(i) C10H14 + 12[O] → C8H6O4 + 2CO2 + 4H2O 1
7(c)(ii) 3
compound
G
compound
J
compound
K
Each correct structure = [1]
Bromine reacts with methylbenzene in the dark in the presence of a suitable catalyst to form HBr and compound L, C7H7Br. L is one of three isomers that can form in this reaction.
(a) (i) The mechanism for the reaction involves methylbenzene reacting with a Br+ ion. This ion is produced when bromine reacts with the catalyst.
Complete the equation for the reaction of Br2 with the catalyst.
Br2 + …… Br+ + …… [1]
(ii) One of the isomers of L forms in much smaller amounts than the other two isomers.
Draw the structure of this isomer and explain why it forms in the smallest amount.
[2]
(iii) Complete the mechanism in Fig. 8.1 for the reaction between methylbenzene and the
Br+ ion.
Include all relevant curly arrows and charges.
intermediate
CH3
Br+ + ……
Fig. 8.1
[3]
(b) Chlorobutane and chlorobenzene are added separately to samples of warm aqueous AgNO3.
One of the chloro-compounds reacts slowly and the other does not react.
(i) Identify the chloro-compound that reacts and describe any observations.
…… [1]
(ii) Write two equations to explain any observations in (b)(i).
…… [2]
(iii) Explain the difference in reactivity of chlorobutane and chlorobenzene with warm aqueous AgNO3.
…… [2]
[Total: 11]
Mark scheme
8(a)(i) Br2 + Al Br3 → Br+ + Al Br4– 1
8(a)(ii) 2
• reference to methyl group having an effect
• methyl group is 2,4,6 directing
Correct structure and one correct point [1] correct structure and two correct points [2]
8(a)(iii) 3
→ →
M1: first curly arrow from inside hexagon towards Br+
M2: correct intermediate
M3: second curly arrow from bond into hexagon AND organic product, H+
8(b)(i) chlorobutane AND white precipitate 1
8(b)(ii) M1: C4H9Cl + H2O → C4H9OH + H+ + Cl – 2
M2: Cl – + Ag+ → AgCl [1]
8(b)(iii) M1: delocalisation of lone pair of Cl into benzene ring / π system 2
M2: strengthens C–Cl bond / causes C–Cl bond to have partial double bond nature
(a) Phenylamine, C6H5NH2, and propylamine, CH3CH2CH2NH2, can be produced by different reduction reactions.
(i) Identify an organic compound that can be converted into C6H5NH2 by a reduction reaction. State the reagents and conditions for this reaction.
organic compound ……
reagents ……
conditions ……
[2]
(ii) Identify an organic compound that can be converted into CH3CH2CH2NH2 by a reduction reaction. State the reagent for this reaction.
organic compound ……
reagent ……
[2]
(iii) Identify a single test that will distinguish between C6H5NH2 and CH3CH2CH2NH2 by producing a white precipitate with only one of these amines.
Draw the structure of the compound that is precipitated.
testing reagent ……
amine that gives a precipitate ……
structure of the compound that is precipitated:
[2]
(b) Describe the relative basicities of C6H5NH2, CH3CH2CH2NH2 and NH3.
Explain your answer.
…… < …… < ……
least basic most basic
[4]
[Total: 10]
Mark scheme
9(a)(i) • nitrobenzene / C6H5NO2 2
• Sn and HCl
• concentrated HCl and heat / boil / reflux
Two correct point [1] all three correct points [2]
9(a)(ii) M1: propylamide / propenamide / C2H5CONH2 OR propanenitrile / C2H5CN 2
M2: LiAl H4
9(a)(iii) M1: bromine (aq) AND phenylamine / C6H5NH2 2
M2: correct structure
9(b) M1: phenylamine ammonia propylamine 4
M2: a base has a lone pair that can accept a proton / H+
M3: the lone pair on the nitrogen atom of phenylamine is delocalised into the benzene ring / π-system
M4: propylamine has an electron donating alkyl group
Propanoic acid, methanoic acid and ethanedioic acid are all weak acids.
(a) Draw the displayed formula of ethanedioic acid.
[1]
(b) The three acids, propanoic acid, methanoic acid and ethanedioic acid, can be distinguished using a combination of two chemical tests. Neither testing reagent is an acid–base indicator.
Identify two suitable testing reagents and complete Table 10.1 to show the observations from each test.
reagent 1 …… reagent 2 ……
Table 10.1
observation when treated observation when treated with reagent 1 with reagent 2
propanoic acid
methanoic acid
ethanedioic acid
[4]
(c) When propanoic acid is treated with chlorine gas in the presence of ultraviolet light, a mixture of products is formed. One of these products is 2,2-dichloropropanoic acid.
Explain why 2,2-dichloropropanoic acid is stronger than propanoic acid. Refer to the structure of each compound in your answer.
…… [2]
[Total: 7]
Important values, constants and standards
molar gas constant R = 8.31 J K–1 mol–1
Faraday constant F = 9.65 × 104 C mol–1
Avogadro constant L = 6.02 × 1023 mol–1 electronic charge e = –1.60 × 10–19 C
molar volume of gas Vm = 22.4 dm3 mol–1 at s.t.p. (101 kPa and 273 K)
Vm = 24.0 dm3 mol–1 at room conditions
ionic product of water K w = 1.00 × 10–14 mol2 dm–6 (at 298 K (25 °C)) specific heat capacity of water c = 4.18 kJ kg–1 K–1 (4.18 J g–1 K–1)
The Periodic Table of Elements
Group
1 2 13 14 15 16 17 18
1 2
H He hydrogen helium
Key 1.0 4.0
3 4 atomic number 5 6 7 8 9 10
Li Be atomic symbol B C N O F Ne lithium beryllium name boron carbon nitrogen oxygen fluorine neon
6.9 9.0 relative atomic mass 10.8 12.0 14.0 16.0 19.0 20.2
11 12 13 14 15 16 17 18
Na Mg Al Si P S Cl Ar sodium magnesium aluminium silicon phosphorus sulfur chlorine argon
23.0 24.3 3 4 5 6 7 8 9 10 11 12 27.0 28.1 31.0 32.1 35.5 39.9
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
at www.cambridgeinternational.org after the live examination series.
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr potassium calcium scandium titanium vanadium chromium manganese iron cobalt nickel copper zinc gallium germanium arsenic selenium bromine krypton
39.1 40.1 45.0 47.9 50.9 52.0 54.9 55.8 58.9 58.7 63.5 65.4 69.7 72.6 74.9 79.0 79.9 83.8
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
24
Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe rubidium strontium yttrium zirconium niobium molybdenum technetium ruthenium rhodium palladium silver cadmium indium tin antimony tellurium iodine xenon
85.5 87.6 88.9 91.2 92.9 95.9 – 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3
55 56 57–71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 lanthanoids
Cs Ba Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn caesium barium hafnium tantalum tungsten rhenium osmium iridium platinum gold mercury thallium lead bismuth polonium astatine radon
132.9 137.3 178.5 180.9 183.8 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 – – –
87 88 89–103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 actinoids
Fr Ra Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og francium radium rutherfordium dubnium seaborgium bohrium hassium meitnerium darmstadtium roentgenium copernicium nihonium flerovium moscovium livermorium tennessine oganesson
– – – – – – – – – – – – – – – – –
57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 lanthanoids La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu lanthanum cerium praseodymium neodymium promethium samarium europium gadolinium terbium dysprosium holmium erbium thulium ytterbium lutetium
138.9 140.1 140.9 144.2 – 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.1 175.0
89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 actinoids Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr actinium thorium protactinium uranium neptunium plutonium americium curium berkelium californium einsteinium fermium mendelevium nobelium lawrencium
– 232.0 231.0 238.0 – – – – – – – – – – –
Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download
To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge
Mark scheme
10(a) 1
10(b) M1: reagent 1:Fehling’s OR Tollens’ reagent 4
M2: acidified MnO4–
observation when treated observation when treated
with reagent 1 with reagent 2
propanoic acid no change no change
methanoic acid red ppt OR silver mirror colourless
ethanedioic acid no change colourless
M3 / M4: One completely correct column [1] both completely correct columns [2]
10(c) M1: chlorine atoms are electron withdrawing / electronegative 2
M2: weakening O–H bond OR stabilising conjugate base
Checked your answers against these? Let AI mark the whole paper, on your own working.
Examiners follow the generic marking principles at the front of the scheme, then each question's points: method (M), accuracy (A) and independent (B) marks, error carried forward where the scheme allows it, and positive marking, so you're credited for what you got right.
Check your answers against it
Done this paper? Upload your script and CIE Buddy's AI marker marks it question by question against this exact scheme, puts the ticks on your own working and says what each lost mark needed.
Where can I find the 9701/43 May/June 2025 mark scheme?+
On this page, free: the mark scheme for AS & A Level Chemistry 9701/43, May/June 2025 (9701_s25_ms_43), and the question paper (9701_s25_qp_43), to read or download, with each question's marking points written out underneath.
How many marks is 9701/43 out of?+
Chemistry 9701/43 for May/June 2025 is out of 100 marks, with 2 hours to answer it, across 10 questions.
Can I get my 9701/43 answers marked?+
Yes. Upload your answers to 9701/43/M/J/25 as a PDF or photos, and CIE Buddy marks every question against this paper's official mark scheme, using Cambridge's marking rules. You get your score, each lost mark explained on your own script, and the topics to revise. It's a guide, not an official Cambridge grade.