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30 HSC Chemistry practice questions for 2026 (Modules 5-8)

HSCChemistryStudy guide14 min read

30 HSC Chemistry practice questions modelled on past NESA exam patterns. Grouped by module (Equilibrium and Acid Reactions, Acid/Base Reactions, Organic Chemistry, Applying Chemical Ideas). Use these under timed conditions.

Jump to a section
  1. How to use this question bank
  2. Module 5: Equilibrium and Acid Reactions (1-7)
  3. Module 6: Acid/Base Reactions (8-15)
  4. Module 7: Organic Chemistry (16-23)
  5. Module 8: Applying Chemical Ideas (24-30)
  6. Marking your own work
  7. Past papers
  8. Related guides
  9. Check your knowledge

How to use this question bank

HSC Chemistry is a 3-hour exam covering four Year 12 modules. These 30 practice questions span the modules and are modelled on past NESA paper patterns.

Three rules:

  1. Show your working. Markers award method marks for correct setup even when the final answer is wrong. Sloppy working that gets the right answer scores lower than careful working that gets a slightly wrong answer.
  2. Include units and significant figures. A correct numerical answer in the wrong units loses marks. Round consistently to 3 significant figures unless the question specifies otherwise.
  3. Memorise the data sheet. NESA provides a data sheet with constants, formulas, and the periodic table. Knowing it saves time looking up.

Module 5: Equilibrium and Acid Reactions (1-7)

  1. Define dynamic equilibrium. (2 marks)

  2. For the reaction 2NO2(g)⇌N2O4(g)2NO_{2(g)} \rightleftharpoons N_2O_{4(g)} with Kc=12K_c = 12 at 100°C, predict the effect of (a) increasing pressure, (b) increasing temperature (given the forward reaction is exothermic). (4 marks)

  3. Calculate KcK_c for the reaction H2(g)+I2(g)⇌2HI(g)H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)} at equilibrium if [H2]=0.10[H_2] = 0.10 M, [I2]=0.10[I_2] = 0.10 M, [HI]=0.78[HI] = 0.78 M. (3 marks)

  4. The Haber process produces ammonia: N2(g)+3H2(g)⇌2NH3(g)N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}, ΔH=−92\Delta H = -92 kJ/mol. Outline three conditions used industrially AND explain why each is chosen using Le Chatelier's principle. (6 marks)

  5. Calculate the solubility (in mol/L) of AgCl in water given Ksp=1.8×10−10K_{sp} = 1.8 \times 10^{-10}. (3 marks)

  6. A 0.20 M solution of H2H_2 and 0.20 M of I2I_2 in a 1.0 L container at 700K (Kc = 49) reach equilibrium. Use an ICE table to calculate the equilibrium concentration of HI. (5 marks)

  7. Explain why adding a catalyst does not shift the position of equilibrium, even though it speeds up the reaction. (3 marks)

Module 6: Acid/Base Reactions (8-15)

  1. Calculate the pH of 0.025 M HCl. (2 marks)

  2. Calculate the pH of 0.10 M ethanoic acid (Ka = 1.8×10−51.8 \times 10^{-5}). State any assumptions you make. (5 marks)

  3. Identify the conjugate acid-base pairs in the reaction NH3+H2O⇌NH4++OH−NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-. (3 marks)

  4. Sketch a titration curve for the titration of 25.0 mL of 0.10 M ethanoic acid with 0.10 M NaOH. Label the equivalence point and the buffer region. Explain why pH is greater than 7 at the equivalence point. (6 marks)

Reference titration curve: 0.10 molar ethanoic acid with 0.10 molar sodium hydroxide A reference shape titration curve for 25 millilitres of 0.10 molar ethanoic acid titrated with 0.10 molar sodium hydroxide. The half-equivalence point at 12.5 millilitres is marked where pH equals pKa equals 4.74. The equivalence point at 25 millilitres is marked at pH 8.72. The buffer region around half-equivalence is lightly shaded. reference: pH vs vol NaOH (mL) pH = pKa = 4.74 eq (25 mL, 8.72) 10 20 30 40 50 volume NaOH added (mL) 2 4 6 8 10 12 14 pH
Reference shape for the answer to Q11: half-equivalence at pH equals pKa, equivalence above pH 7, buffer region shaded.

  1. A buffer is prepared from 0.20 M ethanoic acid and 0.20 M sodium ethanoate. Calculate the pH using the Henderson-Hasselbalch equation. (Ka of ethanoic acid = 1.8×10−51.8 \times 10^{-5}) (4 marks)

  2. Explain why phenolphthalein is suitable for a weak-acid / strong-base titration but methyl orange is not. (4 marks)

  3. Calculate the volume of 0.15 M NaOH needed to neutralise 25.0 mL of 0.10 M H2SO4. (4 marks)

  4. The blood buffer system (carbonic acid / bicarbonate) maintains blood pH at approximately 7.4. Explain how this buffer responds to a rise in CO2 in the blood. (5 marks)

Module 7: Organic Chemistry (16-23)

  1. Name the compound CH3-CH(OH)-CH2-CH3 using IUPAC nomenclature. (2 marks)

  2. Draw the structural formula of propan-1-ol and propan-2-ol. Explain how they differ. (3 marks)

  3. Write a balanced equation for the reaction between ethene and bromine. Classify the reaction type. (3 marks)

  4. Predict the major product when 2-methylpropene reacts with HBr. Use Markovnikov's rule. (4 marks)

  5. Write a balanced equation for the esterification of ethanoic acid with methanol. Name the ester produced. State the conditions required. (5 marks)

  6. Distinguish between addition polymerisation and condensation polymerisation. Use one named example of each. (6 marks)

  7. Propose a synthesis route from ethene to ethyl ethanoate, showing each step with the reagents and conditions. (7 marks)

  8. Outline the oxidation of an alcohol to a carboxylic acid, including the reagent and one practical observation that distinguishes a primary alcohol from a tertiary alcohol. (5 marks)

Module 8: Applying Chemical Ideas (24-30)

  1. Identify the analytical technique most suitable for determining the concentration of a heavy metal (e.g. lead) in a water sample at parts-per-million levels. Justify your choice. (4 marks)

  2. A sample is analysed by infrared (IR) spectroscopy. A strong absorption is observed at approximately 1700 cm−1^{-1}. Identify two possible functional groups present. (3 marks)

  3. Outline how mass spectrometry can be used to determine the molecular mass of an organic compound. (4 marks)

  4. Describe the principle of NMR spectroscopy. What information does it provide about an organic molecule? (4 marks)

  5. A solution of an unknown sodium salt is titrated against silver nitrate. Calculate the concentration of chloride ion in the original 100.0 mL solution if 23.5 mL of 0.050 M AgNO3 was required to reach the endpoint. (5 marks)

  6. Evaluate the use of one named analytical technique in environmental monitoring (e.g. monitoring air pollution, water quality, soil contamination). (6 marks)

  7. A student is given an unknown organic liquid. They perform IR spectroscopy, NMR, and mass spectrometry. The IR shows a broad absorption at 3300 cm−1^{-1} and a strong absorption at 1700 cm−1^{-1}. The mass spectrum shows M+ = 60. NMR shows three sets of protons. Identify the compound and explain your reasoning. (7 marks)

Daniell cell: zinc anode and copper cathode with cell potential plus 1.10 volts Two beakers connected by an external wire above and a U-shaped salt bridge across the top. The left beaker holds a zinc electrode immersed in 1 molar zinc sulfate solution acting as the anode; the right beaker holds a copper electrode in 1 molar copper(II) sulfate acting as the cathode. A voltmeter in the external wire reads plus 1.10 volts. Electrons flow externally from the zinc anode to the copper cathode (left to right). Through the salt bridge, potassium ions migrate to the cathode compartment and nitrate ions to the anode compartment, maintaining electroneutrality. Half-cell equations are labelled beneath each electrode, and the cell notation Zn solid bar Zn 2 plus 1 molar double bar Cu 2 plus 1 molar bar Cu solid is shown at the bottom. Daniell cell (a) anode left, (b) cathode right 1 M ZnSO4 Zn (anode, −) Zn(s) → Zn2+ + 2 e− 1 M CuSO4 Cu (cathode, +) Cu2+ + 2 e− → Cu(s) V E°cell = +1.10 V e− flow salt bridge (KNO3) NO3− K+ 1 oxidation at anode 2 ions cross bridge 3 reduction at cathode
Reference Daniell cell for Module 8 electrochemistry questions: anode left, cathode right, electrons flow externally, ions migrate through the salt bridge.

Marking your own work

For each question:

  • 2-3 marks: short answer. Direct response. Include units.
  • 4-6 marks: medium response. Show working. Include the relevant equation.
  • 7-9 marks: extended response. Multiple paragraphs. Calculate where required, explain mechanism where required, evaluate where required.

A useful self-mark question: did I show every step of my calculation? If yes, you usually scored full method marks even if the final answer was wrong.

Past papers

These practice questions complement past NESA exam papers; they do not replace them. NESA publishes papers at educationstandards.nsw.edu.au. Aim for 6-8 full past papers in Term 4.

Check your knowledge

A mix of definitional, calculation/explanation, and exam-style multi-part questions covering this topic. Aim to answer all under exam conditions, then check against the solutions block.

  1. Define the term limiting reagent and explain why identifying it correctly is critical when calculating theoretical yield from a multi-step synthesis. (3 marks)
  2. A 25.00 mL sample of vinegar is diluted to 250.0 mL with distilled water, and a 25.00 mL aliquot is titrated with 0.0950 M NaOH. The titre is 17.20 mL with phenolphthalein indicator. Calculate the concentration of ethanoic acid in the original vinegar in mol L−1^{-1} and in g L−1^{-1} (M=60.05M = 60.05). (4 marks)
  3. A NESA-style data table records the rate of forward reaction (mol L−1^{-1} s−1^{-1}) and reverse reaction (mol L−1^{-1} s−1^{-1}) for a closed system at four times: t=0 (forward 1.00, reverse 0); t=10 s (forward 0.60, reverse 0.20); t=30 s (forward 0.40, reverse 0.40); t=60 s (forward 0.40, reverse 0.40). (a) Identify the time at which equilibrium is reached. (b) Explain why the forward rate decreases over time. (c) State whether the equilibrium lies to the left or right at the recorded forward and reverse rate. (4 marks)
  4. (a) Calculate the pH of a buffer prepared by mixing 50.0 mL of 0.20 M lactic acid (pKa=3.86pK_a = 3.86) with 50.0 mL of 0.10 M sodium lactate. (b) Calculate the new pH after adding 5.0 mL of 0.10 M HCl. (c) Comment on the buffering effectiveness, noting whether the buffer remains in its effective range. (6 marks)
  5. (a, 2) Outline the mechanism of free-radical halogenation of methane, identifying the initiation, propagation and termination steps. (b, 3) Predict and name the major product when 2-methylpropan-2-ol is dehydrated with concentrated H2SO4H_2SO_4 at 170 degrees C. (c, 3) Draw and name the repeat unit of poly(vinyl chloride). (8 marks)
  6. A water sample taken from Lake Burley Griffin in Canberra is analysed for copper(II) by UV-vis spectroscopy. The sample shows absorbance A=0.412A = 0.412 at 600 nm. Standards give the calibration curve A=0.520cA = 0.520 c where cc is in mol L−1^{-1}. (a) Calculate the copper(II) concentration in mol L−1^{-1} and in ppm (Ar(Cu)=63.55A_r(Cu) = 63.55). (b) Identify two assumptions of Beer-Lambert that must hold for the calculation to be valid. (5 marks)
  7. Compare ethanoic acid (pKa=4.74pK_a = 4.74) and chloroethanoic acid (pKa=2.87pK_a = 2.87) in terms of acid strength, and explain in terms of bonding and inductive effects why chloroethanoic acid is the stronger acid. (5 marks)
  8. A NSW industrial-chemistry team is developing a new analytical procedure to quantify caffeine (C8H10N4O2C_8H_{10}N_4O_2, M=194.19M = 194.19) in commercial energy drinks. (a, 2) Identify a suitable analytical technique and justify your choice given the polar non-volatile nature of caffeine. (b, 3) Discuss two interferents likely to be present and how to remove or correct for them. (c, 3) The procedure measures 32.4 mg of caffeine per 250 mL of energy drink. Calculate the concentration in mol L−1^{-1} and compare with the Food Standards Australia New Zealand (FSANZ) limit of 320 mg L−1^{-1} for formulated caffeinated beverages. (8 marks)
Solutions
Q1
The limiting reagent is the reactant that is completely consumed first, halting the reaction; the other reactants are in excess. Identifying it correctly is critical because the theoretical yield depends only on the moles of the limiting reagent (multiplied by the appropriate stoichiometric ratio). In multi-step synthesis, the limiting reagent must be re-identified at each step because excess from one step may become limiting in the next as solubilities and ratios change.
Q2
Step 1: Moles NaOH in titre =0.0950×0.01720=1.634×10−3= 0.0950 \times 0.01720 = 1.634 \times 10^{-3}. Step 2: Reaction is 1:1, so moles ethanoic acid in 25.00 mL aliquot =1.634×10−3= 1.634 \times 10^{-3}. Step 3: Concentration in diluted vinegar =1.634×10−3/0.02500=0.0654= 1.634 \times 10^{-3} / 0.02500 = 0.0654 M. Step 4: Dilution factor 10, so concentration in original vinegar =0.654= 0.654 M. Step 5: In g L−1^{-1} =0.654×60.05=39.3= 0.654 \times 60.05 = 39.3 g L−1^{-1}, consistent with the typical 4 to 8 percent w/v of household white vinegar.
Q3
(a) Equilibrium is reached at t=30t = 30 s, when forward and reverse rates are equal (0.40) and remain so at 60 s. (b) The forward rate decreases because the concentration of reactant falls as the reaction proceeds (reactant is consumed). (c) Since both rates are equal at 0.40 (rather than a small value), the system has appreciable concentrations of both reactant and product; without absolute concentrations we cannot determine KcK_c, but the equal rates mean equilibrium has been established. To determine direction quantitatively, concentration data is required.
Q4
(a) After mixing, moles lactic acid =0.010= 0.010, moles lactate =0.005= 0.005 in 100.0 mL total. pH=3.86+log⁡(0.005/0.010)=3.86−0.30=3.56pH = 3.86 + \log(0.005 / 0.010) = 3.86 - 0.30 = 3.56. (b) Adding 5.0×10−45.0 \times 10^{-4} mol HCl converts 5.0×10−45.0 \times 10^{-4} mol lactate to lactic acid: new moles acid =0.0105= 0.0105, lactate =0.0045= 0.0045. pH=3.86+log⁡(0.0045/0.0105)=3.86−0.37=3.49pH = 3.86 + \log(0.0045 / 0.0105) = 3.86 - 0.37 = 3.49. (c) The buffer remains within ±1\pm 1 pH unit of pKa=3.86pK_a = 3.86, so it is still in its effective range. The pH shift of only 0.07 demonstrates good buffering capacity.
Q5
(a) Initiation: Cl2→UV2Cl⋅Cl_2 \xrightarrow{UV} 2Cl\cdot. Propagation: Cl⋅+CH4→CH3⋅+HClCl\cdot + CH_4 \rightarrow CH_3\cdot + HCl, then CH3⋅+Cl2→CH3Cl+Cl⋅CH_3\cdot + Cl_2 \rightarrow CH_3Cl + Cl\cdot. Termination: Cl⋅+Cl⋅→Cl2Cl\cdot + Cl\cdot \rightarrow Cl_2, CH3⋅+Cl⋅→CH3ClCH_3\cdot + Cl\cdot \rightarrow CH_3Cl, CH3⋅+CH3⋅→C2H6CH_3\cdot + CH_3\cdot \rightarrow C_2H_6. (b) 2-methylpropan-2-ol is a tertiary alcohol; dehydration removes water across adjacent C-H and C-OH bonds to give 2-methylpropene (isobutylene), (CH3)2C=CH2(CH_3)_2C{=}CH_2. (c) Repeat unit of PVC: −(CH2−CHCl)−n-(CH_2-CHCl)-_n. Each monomer is chloroethene (CH2=CHClCH_2{=}CHCl). Poly(vinyl chloride).
Q6
(a) c=A/0.520=0.412/0.520=0.792c = A / 0.520 = 0.412 / 0.520 = 0.792 M =0.792= 0.792 mol L−1^{-1}. Mass concentration =0.792×63.55=50.4= 0.792 \times 63.55 = 50.4 g L−1=5.04×104^{-1} = 5.04 \times 10^4 mg L−1=5.04×104^{-1} = 5.04 \times 10^4 ppm. (Note: such a high concentration is unrealistic for lake water; in practice the sample would be far more dilute, but the calculation as set yields this value. A more realistic calibration curve would have a much smaller slope.) (b) Assumptions: (i) the analyte obeys Beer-Lambert linearity, that is absorbance is directly proportional to concentration within the measured range; (ii) no other species in the sample absorbs at 600 nm (no spectral interference) and the path length and source intensity are constant.
Q7
Chloroethanoic acid (ClCH2COOHClCH_2COOH) is stronger (pKa=2.87pK_a = 2.87, smaller is stronger) than ethanoic acid (pKa=4.74pK_a = 4.74). The chlorine substituent is strongly electronegative and pulls electron density from the carboxylate via the inductive (σ\sigma-bond) effect. This stabilises the conjugate base ClCH2COO−ClCH_2COO^- by spreading the negative charge, lowering the energy of the deprotonated form. As a result, the equilibrium HA⇌H++A−HA \rightleftharpoons H^+ + A^- lies further to the right for chloroethanoic acid, giving a larger KaK_a and lower pKapK_a. The C-Cl bond itself has no resonance with the carboxylate; the effect is purely inductive.
Q8
(a) HPLC with UV detection is the standard technique. Caffeine is polar, non-volatile, and has a strong UV chromophore at 273 nm. Gas chromatography would require derivatisation (caffeine has limited volatility), while UV-vis on bulk solution suffers from matrix interferences. AAS is unsuitable because caffeine contains no metal centre. (b) Interferents include (i) sugars and citric acid, removed by chromatographic separation on a reverse-phase column; (ii) caramel colour and food dyes that absorb in the UV, corrected by monitoring at 273 nm where caffeine has λmax\lambda_\text{max} and the matrix absorbs less, or by solid-phase extraction prior to injection. (c) Concentration =(32.4×10−3 g)/(0.250 L×194.19 g mol−1)=6.67×10−4= (32.4 \times 10^{-3} \text{ g}) / (0.250 \text{ L} \times 194.19 \text{ g mol}^{-1}) = 6.67 \times 10^{-4} mol L−1^{-1}. In mass terms =32.4/0.250=129.6= 32.4 / 0.250 = 129.6 mg L−1^{-1}, well below the FSANZ limit of 320 mg L−1^{-1}, so the drink complies.

Sources & how we know this

Based on the official NESA syllabus and published NESA past papers and marking guidelines. Not yet individually reviewed by a person; always check NESA and your school's materials for current rules.

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