S5 Chemistry Paper: PH Of Acids, Titration And Heat Of Neutralisation, Electrolysis, Rate Of Reaction, Redox And Bonding (100 Marks)

S5 Chemistry Paper: pH of Acids, Titration and Heat of Neutralisation, Electrolysis, Rate of Reaction, Redox and Bonding (100 marks)

2026-11-13

S5 Chemistry – acids, electrolysis and rates

S5 Chemistry paper: 6 questions and 24 parts on pH and dilution, titration and heat of neutralisation, electrolysis, reaction rate, oxidation numbers and bonding.

Download　Question paper PDF　|　Answer key PDF

Paper details　S5 Chemistry practice paper with 6 questions and 24 parts on acid pH and dilution, titration with heat of neutralisation, electrolysis calculations, reaction rate, oxidation numbers and chemical bonding.

Section A: Acids, Bases and pH (18 marks)

A hydrochloric acid solution has a concentration of 0.050 mol/dm3 at 25 C.

#	Question	Answer and marking notes
1	Calculate the pH of the solution.	Answer: pH = -log[H+] = -log(0.050) = 1.30. (Marking: formula 2, answer 2)
2	If the solution is diluted ten times, find the new pH.	Answer: [H+] becomes 0.005 mol/dm3, pH = -log(0.005) = 2.30. (Marking: concentration 2, pH 3)
3	What volume of 0.10 mol/dm3 sodium hydroxide is needed to neutralise 25.0 cm3 of the acid?	Answer: moles of H+ = 0.050 x 0.0250 = 0.00125 mol, volume = 0.00125 / 0.10 = 12.5 cm3. (Marking: moles 2, volume 3)
4	Explain why the pH rises by only 1 while the hydrogen ion concentration falls tenfold.	Answer: pH is the negative logarithm of [H+], so a tenfold drop in concentration raises pH by exactly 1; the link is logarithmic, not linear. (Marking: log definition 2, explanation 2)

Section B: Titration and Heat of Neutralisation (16 marks)

25.0 cm3 of sodium hydroxide of unknown concentration is titrated with 0.100 mol/dm3 hydrochloric acid; 20.0 cm3 of acid is used to reach the end point, and the reaction warms the mixture.

#	Question	Answer and marking notes
5	Find the number of moles of hydrochloric acid used.	Answer: n = 0.100 x 20.0 / 1000 = 0.00200 mol. (Marking: formula 2, answer 2)
6	Find the concentration of the sodium hydroxide solution.	Answer: 1 : 1 mole ratio, [NaOH] = 0.00200 / 0.0250 = 0.0800 mol/dm3. (Marking: ratio 2, concentration 2)
7	If the same solution is made up to 250.0 cm3, find the mass of sodium hydroxide in it.	Answer: n = 0.0800 x 0.2500 = 0.0200 mol, mass = 0.0200 x 40.0 = 0.80 g. (Marking: moles 2, mass 2)
8	Given the heat of neutralisation is 57.1 kJ/mol, total volume 45.0 cm3, density 1.0 g/cm3 and specific heat capacity 4.18 J/gC, find the temperature rise.	Answer: heat = 0.00200 x 57.1 x 1000 = 114.2 J; dT = 114.2 / (4.18 x 45.0) = 0.61 C. (Marking: heat 2, rise 2)

Section C: Electrolysis and Faraday Calculations (18 marks)

Copper(II) sulphate solution is electrolysed with a current of 2.0 A for 1800 s using inert carbon electrodes.

#	Question	Answer and marking notes
9	Find the charge passed and the number of moles of electrons.	Answer: Q = It = 2.0 x 1800 = 3600 C; n(e-) = 3600 / 96500 = 0.03731 mol. (Marking: charge 2, moles 3)
10	Find the mass of copper deposited at the cathode.	Answer: Cu2+ + 2e- -> Cu, n(Cu) = 0.03731 / 2 = 0.01865 mol, mass = 0.01865 x 63.5 = 1.18 g. (Marking: half equation 2, mass 3)
11	Find the volume of oxygen at the anode at room conditions (molar volume 24000 cm3/mol).	Answer: n(O2) = 0.03731 / 4 = 0.009326 mol, volume = 0.009326 x 24000 = 224 cm3. (Marking: ratio 2, volume 2)
12	If copper plates are used for both electrodes, explain why the blue colour of the solution hardly changes.	Answer: copper dissolves at the anode to give Cu2+, while Cu2+ is reduced at the cathode at the same rate, so the copper ion concentration stays constant and the colour does not fade. (Marking: anode reaction 2, constant concentration 2)

Section D: Rate of Reaction (16 marks)

Equal masses of magnesium ribbon are added to excess hydrochloric acid of 0.5 mol/dm3 and 1.0 mol/dm3; 60 cm3 of hydrogen is collected in 120 s and 60 s respectively.

#	Question	Answer and marking notes
13	Find the average rate in the 0.5 mol/dm3 experiment in cm3/s.	Answer: rate = 60 / 120 = 0.50 cm3/s. (Marking: formula 2, answer 2)
14	Find the average rate in the 1.0 mol/dm3 experiment and compare the two.	Answer: rate = 60 / 60 = 1.00 cm3/s; doubling the concentration doubles the rate. (Marking: rate 2, comparison 2)
15	Explain why the slope of the volume-time curve decreases with time.	Answer: the concentration of reactants falls as the reaction proceeds, so collisions become less frequent and the rate decreases until the curve flattens. (Marking: falling concentration 2, collision frequency 2)
16	Give two ways to speed up the reaction and explain each.	Answer: (1) raise the temperature so more particles have energy above the activation energy; (2) cut the ribbon into small pieces to increase the surface area available for collisions. (Marking: 1 each for method and reason)

Section E: Oxidation Numbers and Redox (16 marks)

The following parts involve assigning oxidation numbers and identifying redox changes.

#	Question	Answer and marking notes
17	Find the oxidation number of manganese in potassium manganate(VII), KMnO4.	Answer: K is +1 and four O atoms give -8, so Mn is +7. (Marking: oxygen value 2, result 2)
18	Find the oxidation number of chromium in potassium dichromate, K2Cr2O7.	Answer: 2(+1) + 2x + 7(-2) = 0 gives x = +6. (Marking: equation 2, result 2)
19	In Zn + Cu2+ -> Zn2+ + Cu, identify the oxidising agent, the reducing agent and the number of electrons transferred.	Answer: Zn is oxidised so it is the reducing agent; Cu2+ is reduced so it is the oxidising agent; two electrons are transferred per zinc atom. (Marking: roles 2, electrons 2)
20	Explain why acidified potassium manganate(VII) acts as an oxidising agent and state its colour change.	Answer: manganese is at its highest oxidation number +7, so it can only be reduced and it takes electrons from other species; the purple colour fades to almost colourless Mn2+. (Marking: highest oxidation number 2, colour change 2)

Section F: Chemical Bonding and Structure (16 marks)

The following parts compare the bonding, structure and properties of different substances.

#	Question	Answer and marking notes
21	State the type of bonding in sodium chloride and in carbon dioxide.	Answer: sodium chloride is ionic (electron transfer between metal and non-metal); carbon dioxide is covalent (shared electron pairs between non-metal atoms). (Marking: 2 each)
22	Compare the structures of diamond and graphite, and explain why only graphite conducts electricity.	Answer: each carbon in diamond is bonded tetrahedrally to four others in a giant covalent structure with no free electrons; graphite is layered, each carbon bonds to three others and the fourth electron is delocalised and can move along the layer. (Marking: structures 2, conduction reason 2)
23	State the bond between the nitrogen atoms in N2 and the number of lone pairs on oxygen in water.	Answer: the two nitrogen atoms in N2 share three pairs of electrons, a triple bond; oxygen in water has two lone pairs plus two bonding pairs, with a bond angle of about 104.5 degrees. (Marking: triple bond 2, lone pairs 2)
24	Explain why solid sodium chloride does not conduct electricity but the molten salt does.	Answer: in the solid the ions are held in a fixed lattice and cannot move; on melting the lattice breaks down, so Na+ and Cl- ions are free to move and carry charge. (Marking: solid 2, molten 2)

Find a tutor — free Tutor Registration →

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