S5 Chemistry Paper: Acids And PH, Titration, Redox And Electrolysis, Rates Of Reaction And Chemical Equilibrium

S5 Chemistry Paper: Acids and pH, Titration, Redox and Electrolysis, Rates of Reaction and Chemical Equilibrium

2026-11-13

S5 Chemistry – acids, titration and equilibrium

S5 Chemistry practice paper: 6 questions, 24 parts, 100 marks, 60 minutes on acids and pH, titration and back titration, redox and electrolysis, rates and equilibrium.

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Paper details　S5 Chemistry practice paper with 6 questions and 24 parts on acids and pH, titration, redox, electrolysis, reaction rates and chemical equilibrium, with marking points.

Section A: Acids, Bases and pH (16 marks)

Show your working and give pH and pOH to two decimal places.

#	Question	Answer and marking notes
1	What are the pH and pOH at 298 K of 0.010 mol dm-3 hydrochloric acid, a strong acid that ionises completely?	Answer: pH = -log(0.010) = 2.00 and pOH = 14.00 – 2.00 = 12.00. (Marking: pH 2, pOH 2)
2	The solution in the previous question is diluted 100 times with water. Find the pH of the diluted solution.	Answer: new concentration = 0.010 / 100 = 1.0 x 10^-4 mol dm-3, so pH = 4.00. (Marking: concentration 2, pH 2)
3	What are the pOH and pH at 298 K of 0.0050 mol dm-3 potassium hydroxide solution, a strong base?	Answer: pOH = -log(0.0050) = 2.30 and pH = 14.00 – 2.30 = 11.70. (Marking: pOH 2, pH 2)
4	What volume of 0.050 mol dm-3 sodium hydroxide solution is needed to neutralise 25.0 cm3 of 0.100 mol dm-3 hydrochloric acid completely?	Answer: moles of acid = 0.100 x 0.0250 = 0.0025 mol; the base reacts 1 to 1, so volume = 0.0025 / 0.050 = 0.0500 dm3, that is 50.0 cm3. (Marking: moles 2, volume 2)

Section B: Titration and Concentration (18 marks)

Give all amounts in mol and concentrations in mol dm-3 to three significant figures.

#	Question	Answer and marking notes
5	25.0 cm3 of sodium carbonate solution is titrated with 0.100 mol dm-3 hydrochloric acid and 24.0 cm3 is used to reach the end point. The two substances react in a mole ratio of 1 to 2. Find the concentration of the sodium carbonate solution.	Answer: n(HCl) = 0.100 x 0.0240 = 2.40 x 10^-3 mol, so n(Na2CO3) = 1.20 x 10^-3 mol and the concentration is 1.20 x 10^-3 / 0.0250 = 0.0480 mol dm-3. (Marking: n(HCl) 1, mole ratio 2, concentration 1)
6	Using the previous question, find the mass of sodium carbonate in that 25.0 cm3 of solution. The molar mass of sodium carbonate is 106 g mol-1.	Answer: mass = 1.20 x 10^-3 x 106 = 0.127 g. (Marking: moles 2, mass 2)
7	A 1.200 g sample containing calcium carbonate is treated with 50.0 cm3 of 0.500 mol dm-3 hydrochloric acid in excess. The remaining acid needs 20.0 cm3 of 0.250 mol dm-3 sodium hydroxide solution for back titration. Find the mass fraction of calcium carbonate in the sample, molar mass 100.1 g mol-1.	Answer: total n(HCl) = 0.0250 mol and n(NaOH) = 5.00 x 10^-3 mol, so acid reacting with the carbonate = 0.0200 mol, giving n(CaCO3) = 0.0100 mol and mass 1.001 g; the mass fraction is 1.001 / 1.200 x 100 = 83.4 percent. (Marking: excess acid 1, reacting acid 1, moles 1, mass and percentage 2)
8	10.0 cm3 of 2.00 mol dm-3 strong acid solution is diluted with water to a total volume of 250.0 cm3. Find the concentration and pH of the diluted solution.	Answer: concentration = 2.00 x 10.0 / 250.0 = 0.0800 mol dm-3; the acid ionises completely, so pH = -log(0.0800) = 1.10. (Marking: concentration 3, pH 2)

Section C: Redox and Electrolysis (16 marks)

Use the Faraday constant 96,500 C mol-1 and molar gas volume 24.0 dm3 mol-1.

#	Question	Answer and marking notes
9	Find the oxidation number of manganese in the permanganate ion (MnO4-) and of chromium in potassium dichromate (K2Cr2O7).	Answer: in MnO4- oxygen is -2, so manganese is -1 – 4(-2) = +7; in K2Cr2O7 potassium is +1 and oxygen is -2, so chromium is (0 – 2(+1) – 7(-2)) / 2 = +6. (Marking: manganese 2, chromium 2)
10	A current of 2.00 A passes through copper sulphate solution for 30.0 minutes. Find the mass of copper deposited at the cathode, molar mass 63.5 g mol-1.	Answer: charge = 2.00 x 1800 = 3600 C, so moles of electrons = 3600 / 96,500 = 0.0373 mol; each copper ion takes two electrons, so n(Cu) = 0.0187 mol and the mass is 1.18 g. (Marking: charge 1, electrons 1, moles 1, mass 1)
11	If the same quantity of charge is used to electrolyse water, find the moles and the room-temperature volume of oxygen produced at the anode.	Answer: 1 mol of oxygen needs 4 mol of electrons, so n(O2) = 0.0373 / 4 = 9.33 x 10^-3 mol and the volume is 9.33 x 10^-3 x 24.0 = 0.224 dm3, that is 224 cm3. (Marking: electron ratio 2, moles 1, volume 1)
12	In a reaction iron(II) ions are converted to iron(III) ions. State whether this is oxidation or reduction, with the number of electrons transferred per ion and the change in oxidation number.	Answer: the oxidation number rises from +2 to +3, which is oxidation, and each ion loses 1 electron. (Marking: judgement 2, electrons 1, oxidation number change 1)

Section D: Rates of Reaction (16 marks)

Give rates in mol dm-3 s-1 and show the formula used.

#	Question	Answer and marking notes
13	In a 2.0 dm3 vessel the amount of a reactant falls from 0.50 mol to 0.30 mol in 25 s. Find the average rate of consumption of the reactant.	Answer: the concentration change is (0.50 – 0.30) / 2.0 = 0.10 mol dm-3, so the rate is 0.10 / 25 = 4.0 x 10^-3 mol dm-3 s-1. (Marking: concentration change 2, rate 2)
14	The rate of a reaction roughly doubles for every 10 K rise in temperature. If the temperature rises from 300 K to 330 K, by what factor does the rate increase?	Answer: the temperature rises by 30 K, which is three intervals of 10 K, so the rate becomes about 2 to the power 3 = 8 times the original. (Marking: number of intervals 2, factor 2)
15	The activation energy of a reaction is 40 kJ mol-1 and falls to 25 kJ mol-1 when a catalyst is added. Find the percentage fall in activation energy and state whether the rate increases or decreases.	Answer: the fall is 40 – 25 = 15 kJ mol-1, that is 15 / 40 x 100 = 37.5 percent; a lower activation energy lets more particles react, so the rate increases. (Marking: fall 1, percentage 2, direction 1)
16	A reaction is first order with respect to A and the rate constant is 0.020 s-1. Find the initial rate when the concentration of A is 0.50 mol dm-3, and the rate when the concentration of A is doubled.	Answer: rate = k[A] = 0.020 x 0.50 = 0.010 mol dm-3 s-1; for a first order reaction the rate is proportional to concentration, so doubling it gives 0.020 mol dm-3 s-1. (Marking: formula 1, initial rate 2, rate after doubling 1)

Section E: Chemical Equilibrium (18 marks)

Equilibrium constants have no units; write the expression before substituting.

#	Question	Answer and marking notes
17	In a 1.0 dm3 vessel, 1.0 mol of hydrogen reacts with 1.0 mol of iodine vapour to form hydrogen iodide. At equilibrium 1.6 mol of hydrogen iodide is present. Find Kc.	Answer: forming 1.6 mol of hydrogen iodide uses 0.80 mol of each reactant, leaving 0.20 mol of each, that is 0.20 mol dm-3; Kc = 1.6 squared / (0.20 x 0.20) = 64. (Marking: equilibrium concentrations 2, expression 1, value 1)
18	In the formation of ammonia from nitrogen and hydrogen, the reactants total 4 mol of gas and the product 2 mol, and the reaction is exothermic. If pressure is increased at constant temperature, state the direction of the shift, the effect on the ammonia yield and whether Kc changes.	Answer: a higher pressure shifts the equilibrium towards the side with fewer moles of gas, that is towards the product, so the ammonia yield rises; Kc is unchanged at constant temperature. (Marking: direction 2, yield 1, Kc 1)
19	In a 2.0 dm3 vessel, 2.0 mol of sulphur dioxide reacts with 1.0 mol of oxygen to form sulphur trioxide, and 1.2 mol of sulphur trioxide is present at equilibrium. Find Kc.	Answer: forming 1.2 mol of sulphur trioxide uses 1.2 mol of sulphur dioxide and 0.60 mol of oxygen, leaving 0.80 mol and 0.40 mol, that is 0.40, 0.20 and 0.60 mol dm-3; Kc = 0.60 squared / (0.40 squared x 0.20) = 11.25. (Marking: moles 2, concentrations 1, value 2)
20	Using the previous question, a catalyst is added at constant volume and then the temperature is raised. Explain separately whether Kc changes and why.	Answer: a catalyst speeds up both directions equally and only shortens the time to reach equilibrium, so Kc is unchanged; raising the temperature shifts the equilibrium in the endothermic direction, and since this reaction is exothermic it shifts left and Kc decreases. (Marking: catalyst 2, temperature direction 2, conclusion 1)

Section F: Integrated Applications (16 marks)

Show formulas and units, and give values to three significant figures.

#	Question	Answer and marking notes
21	A solution has pH 3.00 at 298 K. Find its hydrogen ion concentration and hydroxide ion concentration, taking Kw as 1.0 x 10^-14.	Answer: [H+] = 10^-3 = 1.0 x 10^-3 mol dm-3 and [OH-] = 1.0 x 10^-14 / 1.0 x 10^-3 = 1.0 x 10^-11 mol dm-3. (Marking: 2 each)
22	Name two main gaseous pollutants that cause acid rain and give one measure to reduce acid rain for each.	Answer: the main pollutants are sulphur dioxide from burning sulphur-containing fuel and nitrogen oxides from high-temperature combustion and vehicle exhaust; measures include low-sulphur fuel or flue gas desulphurisation, and catalytic converters on vehicles or greater use of public transport. (Marking: two pollutants 1 each, two measures 1 each)
23	A current of 3.00 A passes through nickel sulphate solution for 20.0 minutes for electroplating. Find the mass of nickel plated on the cathode, molar mass 58.7 g mol-1.	Answer: charge = 3.00 x 1200 = 3600 C, so moles of electrons = 0.0373 mol, giving n(Ni) = 0.0187 mol and a mass of 1.09 g. (Marking: charge 1, electrons 1, moles 1, mass 1)
24	25.0 cm3 of industrial waste water needs 18.0 cm3 of 0.0200 mol dm-3 sodium hydroxide solution for neutralisation. Find the moles of hydrogen ions in the sample and the acid concentration expressed as hydrochloric acid.	Answer: n(NaOH) = 0.0200 x 0.0180 = 3.60 x 10^-4 mol, so the moles of hydrogen ions are the same; the concentration is 3.60 x 10^-4 / 0.0250 = 0.0144 mol dm-3. (Marking: moles 2, 1 to 1 ratio 1, concentration 1)

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