IB Chemistry SL Paper 2 Practice Paper | 10 Calculation Questions, 100 Marks

2026-11-01

IB Chemistry SL

IB Chemistry SL Paper 2 practice paper: amount of substance and gas volumes, titration, enthalpy of neutralisation, empirical and molecular formulae, rate data, the equilibrium constant Kc, electrolysis calculations, pH and dilution, limiting reagents and the percentage of carbon in ethanol; 10 questions for 100 marks with full answers and marking notes.

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Paper details　Level: IB SL (S5-S6) | Subject: Chemistry | Time: 1 hour 30 minutes | Marks: 100 | Questions: 10

Section A: Amount of substance, titration and enthalpy of neutralisation (Questions 1-3, 32 marks)

1. Show all working. Relative atomic masses: H 1.0, C 12.0, O 16.0, Na 23.0, Mg 24.3, Cl 35.5, Cu 63.5, Zn 65.4; molar volume at room conditions 24.0 dm3/mol; Faraday constant 96500 C/mol.

#	Question	Answer and marking notes
1	A 4.86 g strip of magnesium reacts with dilute hydrochloric acid: Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g). (a) Calculate the amount, in mol, of magnesium. (2) (b) Calculate the amount of HCl required. (2) (c) Calculate the volume of hydrogen formed at room conditions. (2) (d) Calculate the mass of magnesium chloride that could be formed. (3) (e) Only 17.8 g of magnesium chloride is obtained. Calculate the percentage yield. (3)	(a) n = 4.86 / 24.3 = 0.2 mol. (b) n(HCl) = 2 x 0.2 = 0.4 mol. (c) V = 0.2 x 24.0 = 4.8 dm3. (d) Mr(MgCl2) = 24.3 + 2(35.5) = 95.3; m = 0.2 x 95.3 = 19.06 g. (e) yield = 17.8 / 19.06 x 100% = 93.4%.
2	25.0 cm3 of sodium hydroxide solution is titrated with 0.100 mol dm-3 hydrochloric acid; 23.40 cm3 is required: NaOH + HCl -> NaCl + H2O. (a) Calculate the amount of HCl used. (2) (b) Calculate the amount of NaOH. (2) (c) Calculate the concentration of the NaOH solution. (3) (d) The original solution had a volume of 250 cm3. Calculate the mass of NaOH it contained. (3)	(a) n(HCl) = 0.100 x 23.4/1000 = 0.002 mol. (b) 1:1 ratio, so n(NaOH) = 0.002 mol. (c) [NaOH] = 0.002 / 0.0250 = 0.094 mol dm-3. (d) n = 0.094 x 0.250 = 0.023 mol; m = 0.023 x 40.0 = 0.94 g.
3	50.0 cm3 of 0.500 mol dm-3 hydrochloric acid is mixed with 50.0 cm3 of 0.500 mol dm-3 sodium hydroxide. The temperature rises from 22.4 C to 35.9 C. Assume the mixture has density 1.00 g cm-3 and specific heat capacity 4.18 J g-1 K-1. (a) Calculate the amount of water formed. (2) (b) Calculate the heat released. (3) (c) Calculate the enthalpy change of neutralisation in kJ mol-1. (3) (d) Give two reasons why the experimental value is less negative than the data book value. (2)	(a) Acid and alkali are each 0.500 x 0.0500 = 0.025 mol, so 0.025 mol of water is formed. (b) q = 100.0 x 4.18 x 13.5 = 5,643 J = 5.643 kJ. (c) dH = -5.643 / 0.025 = -225.7 kJ mol-1. (d) Heat is lost to the surroundings, the container and thermometer absorb heat, mixing is incomplete, and the real density and specific heat capacity differ. Answers to (a), (b), (c): 0.025 mol, 5,643 J, -225.7 kJ mol-1.

Section B: Formulae, rates of reaction and equilibrium (Questions 4-6, 28 marks)

1. Show all working. Relative atomic masses: H 1.0, C 12.0, O 16.0, Na 23.0, Mg 24.3, Cl 35.5, Cu 63.5, Zn 65.4; molar volume at room conditions 24.0 dm3/mol; Faraday constant 96500 C/mol.

#	Question	Answer and marking notes
4	An organic compound contains 40.0%% carbon, 6.7%% hydrogen and 53.3%% oxygen by mass; its relative molecular mass is 180. (a) Using a 100 g sample, calculate the amount of each element. (3) (b) Deduce the empirical formula. (2) (c) Deduce the molecular formula. (3)	(a) n(C) = 40.0 / 12.0 = 3.33; n(H) = 6.7 / 1.0 = 6.7; n(O) = 53.3 / 16.0 = 3.33 mol. (b) The ratio simplifies to 1 : 2 : 1, so the empirical formula is CH2O, formula mass 30. (c) 180 / 30 = 6, so the molecular formula is C6H12O6.
5	A metal reacts with dilute acid. The volume of gas collected (cm3) is measured every 20 s: 0 s 0, 20 s 32, 40 s 52, 60 s 66, 80 s 74, 100 s 78, 120 s 80. (a) Calculate the average rate over the first 40 s. (2) (b) Calculate the average rate between 60 s and 100 s. (2) (c) Explain why the rate falls as the reaction proceeds. (3) (d) State how doubling the acid concentration and raising the temperature each affect the initial rate. (3)	(a) 52 / 40 = 1.30 cm3 s-1. (b) (78 – 66) / 40 = 0.30 cm3 s-1. (c) The concentration of reactants falls as the reaction proceeds, so collisions become less frequent; the surface area of the metal also decreases. (d) Doubling the acid concentration doubles the number of particles per unit volume and increases the collision frequency, so the initial rate rises; raising the temperature increases the average kinetic energy and the fraction of effective collisions. Answers to (a) and (b): 1.30 and 0.30 cm3 s-1.
6	In a 2.0 dm3 closed vessel, N2(g) + 3H2(g) <=> 2NH3(g) reaches equilibrium with 0.60 mol N2, 0.90 mol H2 and 1.20 mol NH3 present. (a) Calculate the equilibrium concentration of each species. (3) (b) Calculate Kc. (3) (c) State the units of Kc. (1) (d) State the effect of increasing pressure on the yield of ammonia and on Kc. (3)	(a) [N2] = 0.60 / 2.0 = 0.30; [H2] = 0.90 / 2.0 = 0.45; [NH3] = 1.20 / 2.0 = 0.60 mol dm-3. (b) Kc = 0.60^2 / (0.30 x 0.45^3) = 13.17, i.e. about 13.2. (c) Units mol-2 dm6. (d) Increasing pressure shifts the equilibrium to the side with fewer moles of gas (the right), so the yield of ammonia rises; Kc depends only on temperature, so it is unchanged.

Section C: Electrolysis and pH (Questions 7-8, 20 marks)

1. Show all working. Relative atomic masses: H 1.0, C 12.0, O 16.0, Na 23.0, Mg 24.3, Cl 35.5, Cu 63.5, Zn 65.4; molar volume at room conditions 24.0 dm3/mol; Faraday constant 96500 C/mol.

#	Question	Answer and marking notes
7	A current of 2.00 A is passed through copper(II) sulfate solution for 30.0 minutes; oxygen is produced at the anode. (a) Calculate the charge passed. (2) (b) Calculate the amount of electrons. (2) (c) Calculate the mass of copper deposited at the cathode. (3) (d) Calculate the volume of oxygen produced at the anode at room conditions, in cm3. (3)	(a) Q = 2.00 x 1800 = 3,600 C. (b) n(e-) = 3600 / 96500 = 0.037 mol. (c) Cu2+ + 2e- -> Cu, so n(Cu) = 0.037 / 2 = 0.019 mol and m = 0.019 x 63.5 = 1.18 g. (d) 2H2O -> O2 + 4H+ + 4e-, so n(O2) = 0.037 / 4 = 0.009 mol and V = 0.009 x 24.0 x 1000 = 224 cm3.
8	(a) Calculate the pH of 0.0100 mol dm-3 hydrochloric acid. (2) (b) Calculate the pH of 0.0010 mol dm-3 sodium hydroxide solution. (2) (c) 25.0 cm3 of 0.100 mol dm-3 hydrochloric acid is diluted with water to 250 cm3. Calculate the pH after dilution. (3) (d) Explain why 0.100 mol dm-3 ethanoic acid does not have pH 1.00. (3)	(a) [H+] = 0.0100, pH = -log(0.0100) = 2.00. (b) [OH-] = 0.0010, pOH = 3.00, pH = 14.00 – 3.00 = 11.00. (c) The amount is unchanged, so c = 0.100 x 25.0 / 250 = 0.0100 mol dm-3 and pH = 2.00. (d) Ethanoic acid is a weak acid and is only partly ionised, so [H+] is much lower than 0.100 mol dm-3 and the pH is higher (about 2.9).

Section D: Limiting reagents and organic chemistry (Questions 9-10, 20 marks)

1. Show all working. Relative atomic masses: H 1.0, C 12.0, O 16.0, Na 23.0, Mg 24.3, Cl 35.5, Cu 63.5, Zn 65.4; molar volume at room conditions 24.0 dm3/mol; Faraday constant 96500 C/mol.

#	Question	Answer and marking notes
9	6.50 g of zinc is added to 100 cm3 of 2.00 mol dm-3 hydrochloric acid: Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g). (a) Calculate the amount of zinc and of HCl. (2) (b) Identify the limiting reagent and justify your answer. (2) (c) Calculate the volume of hydrogen formed at room conditions. (3) (d) Calculate the mass of zinc chloride formed. (3)	(a) n(Zn) = 6.50 / 65.4 = 0.099 mol; n(HCl) = 2.00 x 0.100 = 0.2 mol. (b) All the zinc needs 2 x 0.099 = 0.199 mol HCl, which is less than 0.2 mol, so zinc is limiting. (c) V = 0.099 x 24.0 = 2.39 dm3. (d) Mr(ZnCl2) = 136.4, so m = 0.099 x 136.4 = 13.56 g.
10	(a) Name the two structural isomers with molecular formula C4H10. (2) (b) State the conditions and the type of reaction for the conversion of ethene and steam into ethanol. (3) (c) Calculate the relative molecular mass of ethanol, C2H5OH, and the percentage by mass of carbon in it. (3) (d) Explain why addition polymerisation of ethene produces no other small molecules. (2)	(a) Butane (straight chain) and 2-methylpropane (branched). (b) Phosphoric acid catalyst, about 300 C, high pressure; hydration (addition): CH2=CH2 + H2O -> C2H5OH. (c) Mr = 2(12.0) + 6(1.0) + 16.0 = 46; %C = 24.0 / 46 x 100% = 52.2%. (d) In addition polymerisation the double bonds open and the units link together, so every atom enters the polymer chain and no small molecule is eliminated.

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