2025 HSC Chemistry Exam Paper Solutions

Check your answers to the 2025 Chemistry HSC. Our Chemistry team has written complete, step-by-step solutions to the 2025 HSC Chemistry exam. Use this page to see where you picked up marks and how to improve for trials and beyond.

2025 HSC Chemistry exam paper solutions

These are the responses to the 2025 Chemistry paper, which will be linked here on the NESA website once available.

You can also find a list of all the latest HSC Chemistry Exam Paper Solutions here.

Section 1: Multiple choice questions

Question	Answer	Explanation
1	B	A strong acid is completely ionised in water, so the solution will not contain intact acid HA. The diagram only shows H+ and A− ions, indicating a strong acid is illustrated. A concentrated solution contains a large number of moles of acid per volume. Only a few ions are illustrated, suggesting solution is not concentrated.
2	B	The reaction scheme shows conversion of a secondary alcohol into a ketone. This type of reaction is oxidation.
3	A
4	A	I− ions react with Pb2+ ions to form yellow lead(II) iodide precipitate but not with Na+ ions. Ammonium, nitrate and acetate ions do not form any precipitate with either Pb2+ or Na+.
5	C	Since only PCl3 and Cl2 are introduced into the vessel, PCl3 and Cl2 will combine to form PCl5. This means the concentration of Cl2 is initially non-zero and will decrease as the reaction progresses. This eliminates options B and D. As the vessel is closed and the reaction is reversible, the system will reach dynamic equilibrium. This means the concentration of Cl2 will eventually be a constant non-zero value.
6	D	[H3O+] = [HCl] = 0.25 mol L−1 pH = −log10[H3O+] = −log10(0.25) = 0.60 (2 s.f.)
7	C	The two complexes are different colours, hence would absorb different wavelengths of visible light. The unique wavelengths absorbed can be measured using UV-visible spectrophotometry.
8	D	When an alkene undergoes an addition reaction, the double bond breaks open to allow atoms to be added, without the elimination of any atoms. Hence the difference in molar mass between Y and X corresponds to the molar mass of the unknown substance added. MM(X) + MM(Unknown substance) = MM(Y) MM(Unknown substance) = MM(Y) − MM(X) The molecular ion peaks of X and Y correspond to the molar mass of X and Y. MM(Unknown substance) = 92 − 72 = 20 g mol−1 This is closest to the molar mass of HF.
9	A	The water-soluble carbohydrate chain and fat-soluble side chain of saponin allows it to function as a surfactant. The water-soluble region allows it to interact with water while the fat-soluble region allows it to interact with non-polar grease and oil.
10	D	Butanoic acid is a covalent molecule, and all covalent molecules exhibit dispersion forces. The carboxyl functional group makes butanoic acid polar, hence butanoic acid will also be able to form dipole-dipole forces. The hydrogen atom bonded to oxygen in the carboxyl functional group allows hydrogen bonding to form with the lone pair of electrons on the oxygen. Covalent bonds are intramolecular bonds, which means they exist between atoms within a molecule, not between molecules.
11	A	A Brønsted-Lowry acid is a proton donor and a Brønsted-Lowry base is a proton acceptor. X is the conjugate base of propanoic acid (a weak acid). Hence it can accept a proton to produce propanoic acid. Y is ethanamine, a weak organic base. It can accept a proton to form ammonium ion. Hence both X and Y are Brønsted-Lowry bases.
12	B	The equilibrium expression is given by: The concentration of a species is raised to the power of its respective stoichiometric coefficient in the equation. Pure solids are excluded from the equilibrium expression because their concentrations remain constant at a given temperature.
13	C	Combustion of octane releases heat energy, hence the reaction is exothermic and enthalpy decreases (i.e., ΔH < 0). The combustion of octane at 100 °C is: C8H18(l) + 25/2 O2(g) → 8CO2(g) + 9H2O(g) There are more gaseous particles as products form, hence entropy increases (i.e., ΔS > 0).
14	B	Hence, Keq is lower in experiment 1. A larger Keq in experiment 2 means more products are present and equilibrium has shifted in the forward endothermic direction compared to reaction 1. According to Le Chatelier’s principle, equilibrium shifts in the endothermic direction when temperature is increased as the additional heat is absorbed by the endothermic reaction to minimise the disturbance. Thus, the temperature in reaction 2 is higher than reaction 1.
15	D	The first reaction in the sequence is a hydrogenation, which involves breaking of the carbon-carbon double bond to allow the addition of H2. The −OH group does not participate in the reaction so is unchanged. This gives product X as propan-1-ol. The second reaction in the sequence is oxidation. Since propan-1-ol is a primary alcohol, the oxidation product (Y) can be propanal or propanoic acid. However, since the question states Y forms bubbles of gas with aqueous sodium carbonate, it must be propanoic acid, as carboxylic acids react with metal carbonates to form carbon dioxide gas. The last reaction in the sequence is an esterification. Reaction of propanoic acid and methanol gives product Z as methyl propanoate.
16	B	The polymerisation reaction results in elimination of n − 1 H2O molecules, where n = number of monomer units. Since 1000 monomer units joined together, 999 water molecules are eliminated. Hence MM(polymer) = MM(monomer) × 1000 − MM(H2O) × 999 = (12.01 × 3 + 1.008 × 6 + 16.00 × 3) × 1000 − (1.008 × 2 + 16.00) × 999 = 72080.016 g mol−1 Alternative: Polymer structure is H—(—OCH2CH2CO—)1000—OH Hence MM(polymer) = (12.01 × 3 + 1.008 × 4 + 16.00 × 2) × 2 + (1.008 × 2 + 16.00) = 72080.016 g mol−1
17	A
18	B	The question states potassium chromate is yellow. Hence the solution will initially be yellow and change to red at the endpoint. Silver ions can precipitate with both chloride and chromate ions. Ag+(aq) + Cl−(aq) ⇌ AgCl(s) 2Ag+(aq) + CrO42−(aq) ⇌ Ag2CrO4(s) The colour change occurs when chromate ions in solution form silver chromate precipitate. To be a suitable indicator, chromate ions should only precipitate out after all the chloride ions have been precipitated. This occurs if silver chromate is more soluble than silver chloride.
19		Sodium acetate dissociates to give acetate ions, which undergo neutralisation with HCl. CH3COO−(aq) + HCl(aq) → CH3COOH(aq) + Cl−(aq) n(HCl)initial = c × V = 0.1 × 0.500 = 0.05 mol n(CH3COO−)initial = 0.1 mol HCl is the limiting reactant. After neutralisation, 0.05 mol of CH3COO− remain and 0.05 mol of CH3COOH forms. The following equilibrium between acetic acid and acetate ions exists in solution: CH3COOH(aq) + H2O(l) ⇌ CH3COO−(aq) + H3O+(aq) Since equal moles CH3COOH and CH3COO− are present, a buffer solution will be present. The Henderson–Hasselbalch equation can be used to approximate the pH of a buffer solution: pH = pKa + log10([A−]/[HA]) The concentrations immediately after neutralisation are: [CH3COOH] and [CH3COO−] = 0.05 / 0.5 = 0.1 mol L−1 The concentrations after dilution are: [CH3COOH] and [CH3COO−] = 0.05 / 0.1 = 0.5 mol L−1 (note: adjust if your intended final volume is 0.1 L; if not, keep your original figure) When [HA] and [A−] are equal, log10([A−]/[HA]) = 0, and pH = pKa. Hence, the pH remains at 4.8.Calculation proof: [H3O+]equilibrium = 10−4.8 mol L−1 Ka = [CH3COO−][H3O+] / [CH3COOH] = (0.1 + 10−4.8)(10−4.8)/(0.1 − 10−4.8) = 1.5853956 × 10−5 Then considering dilution:Since Ka is small, assume change in concentration of CH3COOH and CH3COO− is small. Ka = (0.05)(x)/(0.05) = 1.5853956 × 10−5 x = 1.5853956 × 10−5 = [H3O+] pH = −log10[H3O+] = −log10(1.5853956 × 10−5) = 4.8
20	B	Ag2C2O4(s) ⇌ 2Ag+(aq) + C2O42−(aq) Ksp = [Ag+]2[C2O42−] [Ag+]dilute = 0.11 × 10−6 mol L−1 [Ag+]saturated = 2000 × 0.11 × 10−6 = 2.2 × 10−4 mol L−1 [C2O42−] = ½ × 2.2 × 10−4 Ksp = (2.2 × 10−4)2 × (½ × 2.2 × 10−4) = 5.324 × 10−12

Section 2: Long response questions

Question 21

Reaction condition X: UV light

IUPAC name of organic product: 2-bromobutane

Question 22

Both qualitative and quantitative analyses are essential for assessing water quality, as they offer complementary insights.

Qualitative analysis identifies which chemical species are present in a water sample, such as toxic heavy metals. This helps to quickly determine if there is potential contamination of the water and whether further quantitative testing or remediation is necessary.

Quantitative analysis measures how much of a chemical species is present. This allows comparison with environmental or health standards (e.g., drinking water guidelines) to evaluate whether contaminant levels are within safe limits or pose potential risks.

Question 23

Filter mixture after step 3, before adding BaCl₂. There may be insoluble components in fertiliser which inflate the measured mass of the precipitate.

Dry the precipitate until constant mass is achieved, instead of just 20 minutes in the sun. Twenty minutes may not be sufficient for all the water to evaporate. The water will inflate the measured mass of the precipitate.

Other answers could include:

• Add BaCl₂ until no more precipitate forms, rather than adding a fixed amount. This ensures all the sulfate ions will be precipitated and therefore mass measured.
• Add acid, such as HNO₃(aq), after step 3, before the addition of BaCl₂. This removes basic ions such as carbonates, phosphates and hydroxides, which would otherwise precipitate with the Ba²⁺ ions, thus increasing the mass of the precipitate.
• Pre-weigh the filter paper before filtration. The mass of the filter paper will inflate the mass of the precipitate.

Question 24 (a)

Structural formula:

Shape: Linear

Question 24 (b)

n(C₂H₂) = m / MM = 65.0 / 26.04 = 2.496159754 mol

n(C₂H₃Cl) = n(C₂H₂) = 2.496159754 mol (1:1)

m(C₂H₃Cl) = n × MM = 2.496159754 × 62.50 = 156.00998 g = 156 g (3 s.f.)

Question 25

Since propan-1-ol and butanoic acid react in a 1:1 ratio, the smaller moles of propan-1-ol makes it the limiting reactant.

n(ester)_theoretical = n(propan-1-ol) = 0.267 mol (1:1)

m(ester)_theoretical = n × MM = 0.267 × 130.2 = 34.7634 g

m(ester)_actual = d × V = 12.2 × 0.873 = 10.6506 g

%yield = (m(ester)_actual / m(ester)_theoretical) × 100 = (10.6506 / 34.7634) × 100 = 30.6% (3 s.f.)

Question 26 (a)

Question 26 (b)

As the halogen in the haloethanoic acid varies down Group 17, the pK_a increases.

Since pK_a = −log₁₀K_a, a larger pK_a corresponds to a smaller K_a.

K_a = [X−CH₂COO⁻][H₃O⁺] / [X−CH₂COOH]. Hence a smaller K_a indicates a less ionised acid.

Thus, as the halogen in the haloethanoic acid varies down Group 17, the strength of the acid decreases.

Question 27 (a)

The combustion of petrol obtained from crude oil can result in the formation of SO₂(g). This is a result of the oxidation of sulfur impurities found in crude oil. Sulfur dioxide can be further oxidised to sulfur trioxide: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). Sulfur trioxide can react with water in the atmosphere to form sulfuric acid, which falls to the earth as acid rain. Acid rain can leach metals such as aluminium from soil, causing harm to aquatic life.

Other answers can include:

Combustion of petrol can release toxic products such as CO(g) or C(s) when insufficient oxygen is supplied.

Combustion of petrol releases CO₂(g) into the atmosphere, which contributes to global warming.

Question 27 (b)

Question 27 (c)

Question 28 (a)

Question 28 (b)

The polar amide linkages in Kevlar chains enable the formation of strong hydrogen bonds between adjacent chains. In contrast, polystyrene chains are non-polar and interact only through dispersion forces.

As a result, the intermolecular forces between Kevlar chains are much stronger and require more energy to overcome than those in polystyrene. This makes Kevlar more difficult to pull apart, giving it high strength, hardness, and a high melting point.

Polystyrene, on the other hand, is relatively softer, has lower tensile strength, and melts at a lower temperature. Furthermore, the bulky benzene side groups in polystyrene restrict chain flexibility, contributing to its brittleness.

Question 29

The addition of an inert gas such as argon will not disturb the equilibrium of the reaction. This is because the addition of an inert gas will not change the partial pressure of NO₂(g) and N₂O₄(g) as the volume is fixed; as such there is no change in the reaction quotient, hence Q = K as the reaction is initially at equilibrium.

Since the system is still at equilibrium, the rate of the forward exothermic reaction is equal to the rate of the reverse endothermic reaction. Therefore, there will be no change in temperature in the system as the heat produced in the forward reaction will be absorbed by the endothermic reverse reaction.

Question 30 (a)

Since phosgene is a highly toxic gas, it should be handled in a certified fume hood that can capture and remove phosgene vapour at the source, preventing it from accumulating in the breathing zone and throughout the lab.

Respiratory protection such as a properly fitted full-face respirator can be worn to provide additional protection against inhalation of vapours.

Question 30 (b)

Excess carbon monoxide gas is used to increase the yield of phosgene. A higher concentration of carbon monoxide gas in the system disturbs the equilibrium. According to Le Chatelier’s principle, the equilibrium shifts in the forward direction to remove excess carbon monoxide and to minimise the disturbance. As a result, more phosgene is produced.

A catalyst is used to increase the rate of phosgene production. A catalyst provides an alternate reaction pathway of lower activation energy; hence a greater proportion of collisions can exceed the activation energy and be successful.

Question 31 (a)

Suppose the molecular formula of hydrazine is N₂H₄. The complete combustion equation would be: N₂H₄(g) + 3O₂(g) → 2NO₂(g) + 2H₂O(g).

Using Avogadro’s Law, which states that the volume of gas is proportional to the number of moles of gas present at a constant temperature and pressure, we get the stoichiometric ratio of n(hydrazine):n(O₂):n(NO₂):n(H₂O) = 1:3:2:2 which matches the volumes of gas provided 1 L: 3 L: 2 L: 2 L.

Question 31 (b)

Question 32

Ions present are Mg²⁺, Na⁺, NO₃⁻, OH⁻, CO₃²⁻. Na⁺ and NO₃⁻ are always soluble so won’t form precipitates. Thus, possible precipitates are Mg(OH)₂(s) and MgCO₃(s).

Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH⁻(aq)    K_sp = [Mg²⁺][OH⁻]² = 5.61 × 10⁻¹²

MgCO₃(s) ⇌ Mg²⁺(aq) + CO₃²⁻(aq)    K_sp = [Mg²⁺][CO₃²⁻] = 6.82 × 10⁻⁶

Since K_sp of Mg(OH)₂ is significantly smaller than K_sp of MgCO₃, Mg(OH)₂ is less soluble and would precipitate out before MgCO₃.

[Mg²⁺] = 0.006 mol L⁻¹
[OH⁻] = 0.010 mol L⁻¹

Q(Mg(OH)₂) = [Mg²⁺][OH⁻]² = 0.006 × (0.010)² = 6 × 10⁻⁷ > K_sp(Mg(OH)₂)

Hence Mg(OH)₂ will precipitate.

Assuming precipitation of Mg(OH)₂ goes to completion: Mg²⁺(aq) + 2OH⁻(aq) → Mg(OH)₂(s)

n(Mg²⁺)_initial = 0.006 mol
n(OH⁻)_initial = 0.010 mol

OH⁻ is the limiting reactant and Mg²⁺ is in excess

n(Mg²⁺) remaining = 0.006 − 0.005 = 0.001 mol
[Mg²⁺] remaining = 0.001 mol L⁻¹

Since K_sp of Mg(OH)₂ is small and there is excess common ion Mg²⁺ in solution, we can assume concentration of Mg²⁺ does not change.

[CO₃²⁻] = 0.002 mol L⁻¹

Q(MgCO₃) = [Mg²⁺][CO₃²⁻] = 0.001 × 0.002 = 2 × 10⁻⁶ < K_sp(MgCO₃)

Hence MgCO₃ will not precipitate.

Question 33

HCl(aq) + KOH(aq) → KCl(aq) + H₂O(l)

V(KOH) = 7.15 × 10⁻³ = 0.00715 L

n(KOH) = c × V = 0.10 M × 0.00715 L = 0.000715 mol

n(HCl)_excess in 20 mL aliquot = 0.000715 mol (1:1 ratio)

n(HCl)_excess in 100 mL = 0.000715 mol × 5 = 0.003575 mol

n(HCl)_initial = c × V = 0.550 M × 0.1 L = 0.0550 mol

n(HCl)_reacted = n(HCl)_initial − n(HCl)_excess = 0.0550 − 0.003575 = 0.051425 mol

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)

n(CaCO₃) = ½ × n(HCl)_reacted = ½ × 0.051425 = 0.0257125 mol

m(CaCO₃) = n × MM = 0.0257125 mol × (100.09 g mol⁻¹) = 2.573564125 g

%m/m(CaCO₃) = (2.573564125 / 3.0) × 100 = 85.8%  (round to your desired s.f.)
Therefore, Brand X was used as 85.5% is the closest value.

Question 34 (a)

Graph indicates acid (HA) is monoprotic, hence reacts with NaOH in a 1:1.

n(NaOH) = c × V = 0.10 × 0.024 = 0.0024 mol = n(HA)

[HA]_initial = n/V = 0.0024 / 0.010 = 0.24 mol L⁻¹

[H₃O⁺]_equilibrium = 10^−2.5 mol L⁻¹

Alternative method using half-equivalence point:

At the half-equivalence point, the concentration of HA and A⁻ will be equal, hence K_a = [H₃O⁺] and pK_a = pH.

Half-equivalence point occurs at 0.012 L. At this point, the pH = 4.4, so pK_a = 4.4.

K_a = 10^−pK_a = 10^−4.4 = 4 × 10⁻⁵ (1 s.f.)

Question 34 (b)

As excess NaOH is added to the flask, the 0.10 mol L⁻¹ solution is diluted by the salt solution produced from the neutralisation. This means the concentration of NaOH will continually be less than 0.10 mol L⁻¹.

Since pH = 14 − pOH, if [OH⁻] = [NaOH] < 0.10 mol L⁻¹, then pH < 14 − (−log₁₀0.010) < 13.

Question 35

The energy profile diagram provided indicates the forward reaction is exothermic as the enthalpy of products is less than the enthalpy of reactants.

When a system is cooled from 80 °C to 0 °C, it disturbs the equilibrium. According to Le Chatelier’s principle, when a system is cooled it will minimise the disturbance by favouring the exothermic reaction to replenish the heat lost. Hence, the exothermic forward reaction is favoured, producing more pink [Co(H₂O)₆]²⁺ and thus a pink solution.

The colour change can also be explained using collision theory. When a system is cooled, both the rate of the forward and reverse reactions decrease, as there is less kinetic energy in the system, reducing the proportion of reactants that can collide with enough energy to exceed the activation energy. However, the decrease in the rate of the endothermic reverse reaction will be greater as the endothermic reaction has the greater activation energy (E_a2) in comparison to the forward exothermic reaction (E_a1). As a result, it experiences a greater proportional decrease in the number of collisions that exceed the activation energy, hence rate(reverse) < rate(forward). This means the rate at which pink [Co(H₂O)₆]²⁺ is produced is greater than the rate of production of blue [CoCl₄]²⁻, producing a pink solution at equilibrium.

Question 36

IR spectrum:

The IR spectrum would have a broad signal between 2500–3000 cm⁻¹ due to the O–H bond of the carboxyl group and a sharp signal between 1680–1750 cm⁻¹ due to the C=O bond of the carboxyl group.

¹³C NMR spectrum:

The ¹³C NMR spectrum would show 3 signals as there are three unique carbon environments in the molecule, as labelled A, B and C in the diagram below.

“A” carbon is directly bonded to electronegative oxygen, hence will be most deshielded and produce the most downfield signal between 160–185 ppm.

“B” carbon will produce a signal between 20–50 ppm. It is more upfield than “A” carbon as it is adjacent to a carbonyl group and not directly bonded to an electronegative atom.

“C” carbon will produce a signal between 5–40 ppm. It is the most upfield as it is the furthest from the electronegative oxygen.

¹H NMR spectrum:

The ¹H NMR spectrum would show 3 signals as there are three unique hydrogen environments in the molecule, as labelled a, b and c in the diagram below.

“a” proton is directly bonded to electronegative oxygen, so would produce the most downfield signal. The proton is acidic so is readily exchangeable and the adjacent carbon is not bonded to any protons, hence the signal will be a broad singlet.

“b” protons are bonded to a carbon that is adjacent to a C=O and has a neighbouring carbon with 3 protons that are in a different environment to itself. Thus “b” protons will be a quartet between 2.1–4.5 ppm.

“c” protons are bonded to a carbon that is adjacent to a CH₂ group. Thus “c” protons would be a triplet between 0.7–2.1 ppm.

The signal integrals for H_a:H_b:H_c would be 1:2:3, which correspond to the number of protons in each environment.

Mass spectrum:

The mass spectrum would have a parent peak at m/z 74, which corresponds to the intact molecular ion that has a molar mass of 74 g mol⁻¹.

There will be fragment peaks with m/z values less than 74. Possible fragment peaks could include:

Question 37

The formula C₅H₁₀ for Compound A matches the general formula of an alkene C_nH_2n. It is also mentioned that compound A undergoes a reaction with H⁺/H₂O which are the reaction conditions for hydration (it requires a dilute acid catalyst like sulfuric acid and H₂O), which is an addition reaction that alkenes can undergo.

When an alkene undergoes a hydration, it produces an alcohol. It is noted that two products B and C are produced, indicating the alkene is asymmetrical. Since Compound B does not oxidise in the presence of acidified dichromate, it is a tertiary alcohol. This indicates one of the carbon atoms in the C=C in Compound A must be bonded to an alkyl group.

Compound C does react with acidified dichromate, indicating it is a primary or secondary alcohol which could produce either a carboxylic acid product or ketone when oxidised respectively. Since Compound D, the product of oxidation from Compound C, does not react with Na₂CO₃, Compound D is not a carboxylic acid and is a ketone indicating Compound C must be a secondary alcohol. This indicates the carbons in the C=C in Compound A are not terminal carbons.

Hence compound A is 2-methylbut-2-ene.

(Compound B is 2-methylbutan-2-ol, compound C is 3-methylbutan-2-ol and compound D is 3-methylbutan-2-one)