IB Chemistry SL Paper 2 Mock Paper (10 questions, 50 marks, 75 minutes, calculator allowed)

2026-10-31

IB Chemistry

IB Chemistry SL Paper 2 mock paper: 10 questions, 50 marks, 75 minutes, calculator allowed, covering acid-base titration, enthalpy of neutralisation, electrolysis and Faraday calculations, the equilibrium constant Kc, orders of reaction and rate equations, pH and dilution, an iodometric vitamin C titration, esterification and percentage yield, molar gas volume and bond enthalpies, each with worked answers and marking notes.

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Section A (Questions 1 to 5, 25 marks)

Answer ALL questions. A calculator may be used. Show your working in calculations, give numerical answers to three significant figures unless told otherwise, and state units in every final answer. All data are examples and this is not a past paper of any examination board.

#	Question	Answer and marking notes
1	25.0 cm3 of sulfuric acid of unknown concentration was titrated with 0.100 mol dm-3 sodium hydroxide; 24.8 cm3 of the alkali was needed to reach the end point. (a) Write the balanced equation for the neutralisation. (b) Calculate the amount, in moles, of sodium hydroxide used. (c) Calculate the concentration of the sulfuric acid in mol dm-3 (three significant figures). (d) State the colour change at the end point when methyl orange is used.	(a) H2SO4 + 2NaOH -> Na2SO4 + 2H2O (or 2H+ + 2OH- -> 2H2O). (b) n(NaOH) = 0.100 x 0.0250 = 2.5000e-03 mol. (c) From the equation n(H2SO4) = n(NaOH) / 2 = 1.2500e-03 mol, so c = 1.2500e-03 / 0.0248 = 0.0504 mol dm-3, about 0.050 mol dm-3. (d) Methyl orange is red in acid and yellow in alkali; near the end point (pH about 4 to 6) it turns from red to orange-yellow, and one extra drop turns it clearly yellow. Marking notes: 1 mark (a), 1 mark (b), 2 marks (c), 1 mark (d).
2	50.0 cm3 of 1.00 mol dm-3 hydrochloric acid was mixed with 50.0 cm3 of 1.00 mol dm-3 sodium hydroxide in a polystyrene cup. The temperature rose from 24.3 to 31.0 degrees C. Assume the solution has a density of 1.00 g cm-3 and a specific heat capacity of 4.18 J g-1 K-1. (a) Calculate the temperature rise of the mixture. (b) Calculate the heat released, in J. (c) Calculate the enthalpy change of neutralisation in kJ mol-1, per mole of water formed. (d) Explain why the experimental value is usually smaller in magnitude than the literature value (about -57.3 kJ mol-1).	(a) The rise is 31.0 – 24.3 = 6.7 K. (b) Mass = 50.0 + 50.0 = 100.0 g, so q = mcDT = 100.0 x 4.18 x 6.7 = 2800.6 J. (c) n(H2O) = 0.0500 x 1.00 = 0.0500 mol, so the enthalpy change is 2800.6 / 0.0500 = -56012 J mol-1, about -56.0 kJ mol-1. (d) Heat is lost to the surroundings and absorbed by the cup and thermometer, so the measured temperature rise is too small and the calculated heat release is underestimated. Marking notes: 1 mark (a), 2 marks (b), 1 mark (c), 1 mark (d).
3	A current of 2.50 A was passed through copper(II) sulfate solution for 30.0 minutes using inert electrodes. Take the Faraday constant as 96500 C mol-1 and the molar volume as 24.0 dm3 mol-1 at room conditions. (a) Calculate the charge passed, in C. (b) Calculate the mass of copper deposited at the cathode, in g. (c) Write the anode reaction and calculate the volume of oxygen released there, in cm3. (d) Explain why the blue colour of the solution fades during electrolysis.	(a) Q = It = 2.50 x 1800 = 4500 C. (b) n(e-) = 4500 / 96500 = 0.04663 mol. Since Cu2+ + 2e- -> Cu, n(Cu) = 0.02332 mol and m = 63.5 x 0.02332 = 1.48 g. (c) Anode: 2H2O -> O2 + 4H+ + 4e-. n(O2) = 0.04663 / 4 = 0.01166 mol, so V = 0.01166 x 24.0 x 1000 = 280 cm3. (d) Copper(II) ions are continuously reduced to copper at the cathode, so their concentration falls and the blue colour of the hydrated ions fades. Marking notes: 1 mark (a), 2 marks (b), 1 mark (c), 1 mark (d).
4	1.00 mol of hydrogen and 1.00 mol of iodine vapour were mixed in a 1.00 dm3 vessel and heated to a constant temperature. At equilibrium the amount of hydrogen iodide was 1.56 mol. The reaction is H2(g) + I2(g) ⇌ 2HI(g). (a) State the amount of each substance at equilibrium. (b) Write the expression for Kc in terms of concentrations. (c) Calculate Kc at that temperature. (d) State how the amount of HI changes if the volume of the vessel is halved at constant temperature.	(a) Forming 1.56 mol of HI uses 0.78 mol of each reactant, so at equilibrium there is 0.22 mol of H2, 0.22 mol of I2 and 1.56 mol of HI. (b) Kc = [HI]2 / ([H2][I2]). (c) With a volume of 1.00 dm3 the concentrations are 1.56, 0.22 and 0.22 mol dm-3, so Kc = 1.562 / (0.22 x 0.22) = 50.3. (d) There are equal numbers of gas moles on both sides, so halving the volume does not shift the position of equilibrium and the amount of HI stays the same. Marking notes: 2 marks (a), 1 mark (b), 1 mark (c), 1 mark (d).
5	Initial rate data for the reaction A + 2B -> products are: Experiment 1: [A] = 0.10 mol dm-3, [B] = 0.10 mol dm-3, rate = 2.0e-04 mol dm-3 s-1; Experiment 2: [A] = 0.20 mol dm-3, [B] = 0.10 mol dm-3, rate = 4.0e-04 mol dm-3 s-1; Experiment 3: [A] = 0.10 mol dm-3, [B] = 0.20 mol dm-3, rate = 8.0e-04 mol dm-3 s-1. (a) Determine the order with respect to A. (b) Determine the order with respect to B. (c) Write the rate equation and calculate the rate constant k, with units. (d) Explain why the orders cannot be deduced from the coefficients of the equation.	(a) Comparing 1 and 2, doubling [A] at constant [B] doubles the rate from 2.0e-04 to 4.0e-04, so the reaction is first order in A. (b) Comparing 1 and 3, doubling [B] at constant [A] increases the rate fourfold from 2.0e-04 to 8.0e-04, and 2 to the power n equals 4, so it is second order in B. (c) rate = k[A][B]2. Using experiment 1, k = 2.0e-04 / (0.10 x 0.010) = 0.20 mol-2 dm6 s-1. (d) Orders come from experimental rate data and depend on the mechanism, in particular the number of species in the rate-determining step; the overall equation only shows the stoichiometry. Marking notes: 1 mark (a), 1 mark (b), 2 marks (c), 1 mark (d).

Section B (Questions 6 to 10, 25 marks)

#	Question	Answer and marking notes
6	Answer at 25 degrees C, where Kw = 1.0 x 10-14. (a) Calculate the pH of 0.010 mol dm-3 hydrochloric acid. (b) 10.0 cm3 of this acid is diluted with water to a total volume of 1.00 dm3. Calculate the pH. (c) Calculate the pH of 0.010 mol dm-3 sodium hydroxide solution. (d) 25.0 cm3 of 0.10 mol dm-3 hydrochloric acid is mixed with 20.0 cm3 of 0.10 mol dm-3 sodium hydroxide. Calculate the pH of the mixture to three significant figures.	(a) [H+] = 0.010 mol dm-3 because the acid is strong, so pH = 2.00. (b) Diluting 100 times gives [H+] = 1.0 x 10-4 mol dm-3 and pH = 4.00. (c) [OH-] = 0.010 mol dm-3, so pOH = 2.00 and pH = 12.00. (d) n(H+) = 0.0025 mol and n(OH-) = 0.0020 mol, leaving 5.0000e-04 mol of H+ in 45.0 cm3, so [H+] = 0.01111 mol dm-3 and pH = 1.95. Marking notes: 1 mark (a), 1 mark (b), 1 mark (c), 2 marks (d).
7	A 25.0 cm3 sample of a vitamin C drink was titrated with 0.0200 mol dm-3 iodine solution; the mean titre was 18.6 cm3. The reaction is C6H8O6 + I2 -> C6H6O6 + 2HI, one mole of the vitamin reacting with one mole of iodine. The molar mass of vitamin C is 176.1 g mol-1. (a) Calculate the amount of iodine used, in moles. (b) Calculate the mass of vitamin C in the 25.0 cm3 sample, in mg. (c) Calculate the concentration of vitamin C in the drink, in g dm-3. (d) Explain why the titration must be done quickly and when the starch indicator should be added.	(a) n(I2) = 0.0200 x 0.0186 = 3.720e-04 mol. (b) The ratio is 1:1, so n(vitamin C) = 3.720e-04 mol and m = 176.1 x 3.720e-04 = 0.06551 g, that is 65.5 mg. (c) Each 25.0 cm3 contains 0.06551 g, so the concentration is 2.62 g dm-3. (d) Vitamin C is easily oxidised by dissolved oxygen, so the titration must be quick to avoid a low result. Starch should be added only near the end point when the solution is pale yellow, because excess iodine forms a very dark complex with starch and can decompose it, making the end point hard to judge. Marking notes: 1 mark (a), 2 marks (b), 1 mark (c), 1 mark (d).
8	10.0 g of ethanoic acid (molar mass 60.0 g mol-1) was heated with excess ethanol and a few drops of concentrated sulfuric acid. After distillation 12.80 g of ethyl ethanoate (molar mass 88.0 g mol-1) was collected. (a) Write the balanced equation and name the organic product. (b) Calculate the amount of ethanoic acid, in moles. (c) Calculate the theoretical mass of the ester, in g. (d) Calculate the percentage yield. (e) State two roles of concentrated sulfuric acid in this reaction.	(a) CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O; the product is ethyl ethanoate. (b) n = 10.0 / 60.0 = 0.1667 mol. (c) The ratio is 1:1, so the theoretical mass is 0.1667 x 88.0 = 14.67 g. (d) Percentage yield = 12.80 / 14.67 x 100% = 87.3%. (e) It acts as a catalyst, speeding up the approach to equilibrium, and as a dehydrating agent that removes water so the equilibrium shifts to the right. Marking notes: 1 mark each for (a) to (e).
9	2.40 g of magnesium ribbon was added to excess dilute hydrochloric acid and the hydrogen evolved was collected (Ar: Mg = 24.3; molar volume 24.0 dm3 mol-1 at room conditions). (a) Write the balanced equation. (b) Calculate the amount of magnesium, in moles. (c) Calculate the volume of hydrogen collected at room conditions, in dm3 and cm3. (d) Calculate the mass of magnesium chloride formed, in g (Mr = 95.3). (e) Suggest why the volume actually collected is slightly less than the calculated value.	(a) Mg + 2HCl -> MgCl2 + H2. (b) n(Mg) = 2.40 / 24.3 = 0.0988 mol. (c) The ratio is 1:1, so n(H2) = 0.0988 mol and V = 0.0988 x 24.0 = 2.37 dm3, which is 2370 cm3. (d) n(MgCl2) = 0.0988 mol, so m = 0.0988 x 95.3 = 9.41 g. (e) Some hydrogen dissolves in the water or escapes before collection, air remains in the syringe and delivery tube, and the reaction warms the gas above room temperature, so the measured volume differs slightly from the calculated one. Marking notes: 1 mark each for (a) to (e).
10	The combustion of methane is CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g). Average bond enthalpies in kJ mol-1: C-H 414, O=O 498, C=O 799, O-H 464. (a) List the bonds broken and their numbers. (b) List the bonds formed and their numbers. (c) Calculate the total energy needed to break the bonds and the total energy released in forming them. (d) Calculate the enthalpy change of the reaction in kJ mol-1 and state whether it is exothermic. (e) Explain why this value differs from the standard enthalpy of combustion, which refers to liquid water.	(a) Broken: four C-H bonds and two O=O bonds. (b) Formed: two C=O bonds in carbon dioxide and four O-H bonds in two water molecules. (c) Energy needed = 4 x 414 + 2 x 498 = 1656 + 996 = 2652 kJ mol-1; energy released = 2 x 799 + 4 x 464 = 1598 + 1856 = 3454 kJ mol-1. (d) The enthalpy change is 2652 – 3454 = -802 kJ mol-1; the negative sign shows the reaction is exothermic. (e) Bond enthalpies are average values for gaseous species and the water here is gaseous, whereas the standard enthalpy of combustion refers to liquid water, so condensation releases about 44 kJ mol-1 for each mole of water, a difference of roughly 88 kJ mol-1. Marking notes: 1 mark each for (a) to (e).

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