IB Chemistry SL Paper 3 Practice Paper (Practical Skills, 10 Questions, 100 Marks)

2026-11-01

IB Chemistry SL

IB Chemistry SL Paper 3 practical skills: titration uncertainty, calorimetry, colorimetry, Kc, electrolysis and weak-acid pH. Answers with marking notes.

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Paper details　Level: IB DP　|　Subject: Chemistry　|　Time: 75 minutes　|　Marks: 100　|　Questions: 10

Section A: Titration and calorimetry (30 marks)

Answer questions 1 to 3. Each question carries 10 marks.

#	Question	Answer and marking notes
1	Q1 25.0 cm³ of sodium hydroxide solution is titrated with 0.100 mol dm⁻³ hydrochloric acid; the three titres are 24.55, 24.60 and 24.65 cm³. (a) Calculate the mean titre. (b) Calculate the amount of hydrochloric acid used. (c) Calculate the concentration of the sodium hydroxide (mol dm⁻³, 3 s.f.). (d) Explain why only readings within an acceptable range (≤0.10 cm³ apart) are averaged.	(a) (24.55 + 24.60 + 24.65) / 3 = 24.60 cm³ (2 marks). (b) 0.100 × 24.60 / 1000 = 2.46 × 10⁻³ mol (3 marks). (c) 1:1 ratio, so 2.46 × 10⁻³ / 0.0250 = 0.0984 mol dm⁻³ (3 marks; 3 s.f. gives 0.0984). (d) A wide spread shows random operational error (end-point judgement, air bubble, unrinsed wall); rejecting outliers before averaging reduces the effect of random error (2 marks).
2	Q2 A 50 cm³ burette (±0.05 cm³ per reading) and a 25.0 cm³ pipette (±0.06 cm³) are used; the mean titre is 24.60 cm³. (a) Calculate the absolute uncertainty of a burette volume. (b) Calculate the percentage uncertainty of the titre (1 d.p.). (c) Calculate the percentage uncertainty of the pipette. (d) Find the total percentage uncertainty and state which instrument contributes most.	(a) A burette volume is the difference of two readings: 2 × 0.05 = 0.10 cm³ (2 marks). (b) 0.10 / 24.60 × 100 = 0.4%% (3 marks). (c) 0.06 / 25.0 × 100 = 0.2%% (2 marks). (d) Total = 0.4%% + 0.2%% = 0.6%%; the burette contributes most because its percentage uncertainty is larger (3 marks).
3	Q3 50.0 cm³ of 1.00 mol dm⁻³ hydrochloric acid is mixed with 50.0 cm³ of 1.00 mol dm⁻³ sodium hydroxide; the temperature rises by 6.8 °C. Assume a density of 1.00 g cm⁻³ and a specific heat capacity of 4.18 J g⁻¹ K⁻¹. (a) Calculate the heat released (J). (b) Calculate the amount neutralised. (c) Calculate the enthalpy change of neutralisation (kJ mol⁻¹). (d) The literature value is −57.3 kJ mol⁻¹ — calculate the percentage error and give one reason why the measured value is smaller.	(a) q = 100.0 × 4.18 × 6.8 = 2,842.4 J (3 marks). (b) 1.00 × 50.0 / 1000 = 0.050 mol (2 marks). (c) ΔH = −2,842.4 / 0.050 / 1000 = -56.8 kJ mol⁻¹ (3 marks). (d) Error = (56.8 − 57.3) / 57.3 × 100 = 0.8%%; causes: heat loss to surroundings, heat absorbed by the calorimeter, incomplete mixing, or reading before the maximum temperature is reached (2 marks).

Section B: Colorimetry and stoichiometry (30 marks)

Answer questions 4 to 6. Each question carries 10 marks.

#	Question	Answer and marking notes
4	Q4 A colorimeter (path length 1 cm) gives an absorbance of 0.42 for a 0.0200 mol dm⁻³ standard; a sample diluted tenfold gives 0.315. (a) Calculate the molar absorption coefficient (dm³ mol⁻¹ cm⁻¹). (b) Calculate the concentration of the diluted sample. (c) Calculate the concentration of the original sample. (d) Explain why the meter must be zeroed with a blank before each measurement.	(a) A = εcl, so ε = 0.42 / (0.0200 × 1) = 21.0 dm³ mol⁻¹ cm⁻¹ (3 marks). (b) c = 0.315 / 21.0 = 0.0150 mol dm⁻³ (2 marks). (c) Tenfold dilution, so 0.0150 × 10 = 0.150 mol dm⁻³ (2 marks). (d) The blank contains the same solvent and reagents but no analyte, so zeroing removes absorbance from the solvent, the cuvette and the reagents; readings then reflect only the analyte (3 marks).
5	Q5 2.40 g of a copper oxide is reduced completely in hydrogen, leaving 1.92 g of copper (Cu = 63.55, O = 16.00). (a) Calculate the mass of oxygen in the oxide. (b) Calculate the amounts of Cu and O. (c) Deduce the empirical formula. (d) Write the balanced equation for the reduction.	(a) 2.40 − 1.92 = 0.48 g (2 marks). (b) n(Cu) = 1.92 / 63.55 = 0.0302 mol; n(O) = 0.48 / 16.00 = 0.0300 mol (3 marks). (c) The mole ratio is about 1:1, so the empirical formula is CuO (3 marks). (d) CuO + H₂ → Cu + H₂O (2 marks, balanced).
6	Q6 0.243 g of magnesium reacts with excess dilute hydrochloric acid (Mg = 24.3; molar volume at room temperature = 22.7 dm³ mol⁻¹). (a) Calculate the amount of magnesium. (b) Write the balanced equation. (c) Calculate the volume of hydrogen produced (cm³). (d) Give two experimental reasons for a smaller measured volume.	(a) 0.243 / 24.3 = 0.0100 mol (2 marks). (b) Mg + 2HCl → MgCl₂ + H₂ (2 marks). (c) 1:1 ratio, so 0.0100 × 22.7 = 0.227 dm³ = 227 cm³ (3 marks; unit conversion required). (d) Any two: gas escapes before collection, air left in the syringe, an oxide layer makes the reaction incomplete, the volume is read before the reaction finishes, or the temperature and pressure differ from the assumed values (3 marks).

Section C: Kinetics and equilibrium (20 marks)

Answer questions 7 and 8. Each question carries 10 marks.

#	Question	Answer and marking notes
7	Q7 In an iodine clock reaction the blue colour appears after 64 s at 20 °C and after 32 s at 30 °C. (a) Use 1/t to find the relative rate at each temperature. (b) Calculate Q10. (c) Explain the chemical meaning of Q10 being about 2. (d) Suggest one way to speed up the reaction without changing the temperature, with a reason.	(a) At 20 °C: 1 / 64 = 0.0156 s⁻¹; at 30 °C: 1 / 32 = 0.0312 s⁻¹ (2 marks). (b) Q10 = 0.0312 / 0.0156 = 2.0 (3 marks). (c) A 10 °C rise roughly doubles the rate because the mean kinetic energy and collision frequency increase, so a much larger fraction of particles exceed the activation energy (2 marks). (d) Raise the concentration of a reactant or add a catalyst: the first increases the collision frequency per unit volume, the second lowers the activation energy (3 marks; mechanism needed).
8	Q8 In a 1.00 dm³ vessel, H₂ and I₂ each start at 0.100 mol; at equilibrium [HI] = 0.156 mol dm⁻³. (a) Calculate the amount of H₂ reacted. (b) Calculate [H₂] and [I₂] at equilibrium. (c) Calculate Kc (3 s.f.). (d) State whether Kc has units.	(a) Two HI form for each H₂ consumed, so 0.078 mol (2 marks). (b) 0.100 − 0.078 = 0.022 mol dm⁻³ for both H₂ and I₂ (3 marks). (c) Kc = 0.156² / (0.022 × 0.022) = 50.3 (3 marks). (d) No units: Δn = 2 − (1 + 1) = 0, so the concentration units cancel (2 marks).

Section D: Electrolysis and acids (20 marks)

Answer questions 9 and 10. Each question carries 10 marks.

#	Question	Answer and marking notes
9	Q9 A current of 2.00 A passes through copper(II) sulphate solution for 30.0 minutes (F = 96500 C mol⁻¹; Cu = 63.55). (a) Calculate the charge passed. (b) Calculate the amount of electrons. (c) Calculate the mass of copper deposited at the cathode (g, 3 s.f.). (d) Write the cathode half-equation.	(a) Q = 2.00 × 30.0 × 60 = 3,600 C (2 marks). (b) 3,600 / 96500 = 0.0373 mol (2 marks). (c) Cu²⁺ + 2e⁻ → Cu, so 0.0373 / 2 × 63.55 = 1.19 g (3 marks). (d) Cu²⁺ + 2e⁻ → Cu (3 marks; electron number required).
10	Q10 A 0.100 mol dm⁻³ solution of the weak acid HA has pH 2.87. (a) Calculate [H⁺]. (b) Calculate Ka. (c) Calculate pKa (2 d.p.). (d) Calculate the percentage ionisation and explain why [H⁺] cannot be taken as 0.100 mol dm⁻³.	(a) [H⁺] = 10^(−2.87) = 1.35e-03 mol dm⁻³ (3 marks). (b) Ka = (1.35e-03)² / 0.100 = 1.82e-05 mol dm⁻³ (3 marks). (c) pKa = −log₁₀(1.82e-05) = 4.74 (2 marks). (d) Ionisation = 1.35e-03 / 0.100 × 100 = 1.35%%; a weak acid ionises only partly, so [H⁺] is far below the initial concentration and assuming complete ionisation would badly overestimate acidity and Ka (2 marks).

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