IB Chemistry SL Paper 2 Structured Questions Practice Paper | 6 questions, 50 marks, 1 hour 15 minutes

2026-10-31

IB Chemistry SL Paper 2

IB Chemistry SL Paper 2 practice paper: 6 questions, 50 marks, 75 minutes; Section A data-based questions on rates, titration, calorimetry and weak acid pH, Section B extended questions on electrolysis and esterification, with solutions.

Download　Question paper PDF　|　Answer key PDF

Paper details　Level: IB SL | Subject: Chemistry | Time: 1 hour 15 minutes | Marks: 50 | Questions: 6

Section A: data-based questions (30 marks)

Structured questions: show your working and give correct units; answer every part of each question. Total 50 marks.

#	Question	Answer and marking notes
1	At 25 °C, marble chips (calcium carbonate) are added to 50.0 cm³ of 1.00 mol dm⁻³ hydrochloric acid and the volume of carbon dioxide is measured. Time / s: 0, 20, 40, 60, 80, 100; volume / cm³: 0, 24, 40, 50, 54, 55. (a) Calculate the average rate over the first 20 s (cm³ s⁻¹). (2 marks) (b) Calculate the average rate between 40 s and 60 s. (1 mark) (c) Explain why the rate decreases with time. (3 marks) (d) State how the initial rate is obtained from the data, and give one change that raises the initial rate, with a reason. (2 marks)	Answer: (a) Mean rate = 24 cm³ / 20 s = 1.2 cm³ s⁻¹. (b) Rate between 40 s and 60 s = (50 − 40) cm³ / 20 s = 0.5 cm³ s⁻¹. (c) The concentration of hydrogen ions falls as the acid is used up, so collisions with the carbonate surface become less frequent; the surface is also consumed, so the contact area decreases, which lowers the collision frequency further. (d) Take the gradient of the first part of the curve (0 to 20 s). Increasing the acid concentration, powdering the marble or raising the temperature raises the number of effective collisions per unit time, so the initial rate increases.
2	25.0 cm³ of ethanoic acid of unknown concentration is titrated with 0.100 mol dm⁻³ sodium hydroxide using phenolphthalein; the mean titre is 23.60 cm³. (a) Calculate the concentration of the acid (mol dm⁻³). (3 marks) (b) Calculate the mass of ethanoic acid in 250.0 cm³ of this solution (g; Mr = 60.05). (3 marks) (c) Explain why phenolphthalein is used instead of methyl orange. (2 marks)	Answer: (a) n(NaOH) = 0.100 × 23.60 / 1000 = 0.00236 mol; the acid reacts 1:1 with the alkali, so n(CH₃COOH) = 0.00236 mol and c = 0.00236 / 25.0 × 1000 = 0.0944 mol dm⁻³. (b) In 250.0 cm³, n = 0.0944 × 0.2500 = 0.0236 mol, mass = 0.0236 × 60.05 = 1.42 g. (c) At the end point the solution contains sodium ethanoate and is weakly alkaline (pH about 8 to 9), which matches the range of phenolphthalein (pH 8.2 to 10.0). Methyl orange changes at pH 3.1 to 4.4, so it would change before the equivalence point and give a large negative error.
3	1.15 g of ethanol is burned in a simple calorimeter and heats 150 g of water from 21.3 °C to 43.7 °C. (Specific heat capacity of water = 4.18 J g⁻¹ K⁻¹; Mr of ethanol = 46.07; literature ΔHc = −1367 kJ mol⁻¹) (a) Calculate the heat absorbed by the water (kJ). (2 marks) (b) Calculate the experimental molar enthalpy of combustion of ethanol (kJ mol⁻¹). (3 marks) (c) Calculate the experimental value as a percentage of the literature value, and give two reasons why it is lower. (2 marks)	Answer: (a) q = 150 × 4.18 × 22.4 = 14045 J = 14.04 kJ. (b) n(C₂H₅OH) = 1.15 / 46.07 = 0.0250 mol; ΔHc = − 14.04 / 0.0250 = −562.6 kJ mol⁻¹. (c) Percentage = 562.6 / 1367 × 100% = 41.2%; values are low because heat is lost to the surroundings and the calorimeter, combustion is incomplete (carbon monoxide or soot forms), and some water escapes as vapour, carrying away latent heat.
4	At 25 °C, Ka of 0.100 mol dm⁻³ ethanoic acid is 1.74 × 10⁻⁵. (a) Write the expression for Ka. (1 mark) (b) Calculate the hydrogen ion concentration at equilibrium (mol dm⁻³). (2 marks) (c) Calculate the pH of the solution. (1 mark) (d) Calculate the degree of ionisation of the acid (%). (1 mark) (e) A little solid sodium ethanoate is added. State and explain how the pH and the degree of ionisation change. (2 marks)	Answer: (a) Ka = [H⁺][CH₃COO⁻] / [CH₃COOH]. (b) Let [H⁺] = [CH₃COO⁻] = x; x² / 0.100 = 1.74 × 10⁻⁵, so x = 1.32 × 10⁻³ mol dm⁻³. (c) pH = −log(1.32 × 10⁻³) = 2.88. (d) Degree of ionisation = 1.32 × 10⁻³ / 0.100 × 100% = 1.3%. (e) The ethanoate ion concentration rises, so the equilibrium shifts to the left (common ion effect) and some hydrogen ions are removed, so the pH rises; the degree of ionisation falls because ionisation is suppressed.

Section B: extended questions (20 marks)

Structured questions: show your working and give correct units; answer every part of each question. Total 50 marks.

#	Question	Answer and marking notes
5	A steady current of 2.50 A is passed through copper(II) sulfate solution for 30.0 minutes using copper electrodes. (Faraday constant = 96 485 C mol⁻¹; Ar of Cu = 63.55) (a) Calculate the charge passed (C). (2 marks) (b) Calculate the number of moles of electrons. (1 mark) (c) Calculate the mass of copper deposited at the cathode (g). (3 marks) (d) Write the anode reaction and explain why the blue colour of the solution fades. (2 marks) (e) If the current efficiency is only 92%, what mass of copper is actually deposited? (2 marks)	Answer: (a) Q = I × t = 2.50 × 1800 = 4500 C. (b) n(e⁻) = 4500 / 96 485 = 0.04664 mol. (c) Cu²⁺ + 2e⁻ → Cu, so n(Cu) = 0.04664 / 2 = 0.02332 mol and mass = 0.02332 × 63.55 = 1.48 g. (d) Anode: Cu(s) → Cu²⁺(aq) + 2e⁻. Copper dissolving at the anode replaces the ions deposited at the cathode, so the total amount of copper ions should stay the same; in practice uneven migration and convection plus the current efficiency below 100% make the blue colour fade slightly. (e) Actual mass = 1.48 × 0.92 = 1.36 g.
6	12.0 g of ethanoic acid is heated with excess ethanol and a concentrated sulfuric acid catalyst to give 14.5 g of ethyl ethanoate. (Mr: ethanoic acid 60.05; ethanol 46.07; ethyl ethanoate 88.11) (a) Calculate the number of moles of ethanoic acid. (1 mark) (b) Calculate the theoretical yield of ethyl ethanoate (g). (2 marks) (c) Calculate the percentage yield. (2 marks) (d) Calculate the atom economy of this reaction. (2 marks) (e) Give three ways to raise the yield of ethyl ethanoate, with a brief reason for each. (3 marks)	Answer: (a) n = 12.0 / 60.05 = 0.2 mol. (b) The ratio is 1:1, so the theoretical yield = 0.2 × 88.11 = 17.6 g. (c) Percentage yield = 14.5 / 17.6 × 100% = 82.4%. (d) Atom economy = 88.11 / (60.05 + 46.07) × 100% = 83%. (e) (i) Remove the water formed with a Dean-Stark trap or by distillation so the equilibrium shifts right; (ii) use excess ethanol to raise the concentration of a reactant; (iii) distil the ester out of the mixture so hydrolysis cannot reverse the reaction.

Find a tutor — free

A TutorZone practice paper; original questions and passages, not publisher material.