S5 Chemistry Practice Paper: Moles, Titration, PH, Redox And Electrolysis (100 Marks)

S5 Chemistry Practice Paper: Moles, Titration, pH, Redox and Electrolysis (100 marks)

2026-11-12

Chemistry – S5 DSE Practice

S5 chemistry paper: 6 questions, 26 parts, 100 marks, 60 minutes on moles, titration, pH, redox, electrolysis, rates and equilibrium, with full answers and marking notes.

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Part 1: Moles and Solution Concentration (16 marks)

Answer the questions below, showing your working and units.

#	Question	Answer and marking notes
1	(a) Calculate the molar mass of sodium hydroxide and the mass required to prepare 250.0 cm3 of 0.100 M solution.	Answer: Molar mass 40.0 g per mol. Moles = 0.100 x 0.2500 = 0.025 mol; mass = 0.025 x 40.0 = 1.00 g. Marking: molar mass 1 mark, moles 2 marks, mass 1 mark with unit.
2	(b) Find the number of moles of hydrogen chloride in 20.0 cm3 of 0.125 M hydrochloric acid.	Answer: 20.0 cm3 = 0.0200 dm3; moles = 0.125 x 0.0200 = 2.50 x 10^-3 mol. Marking: unit conversion 2 marks, calculation 2 marks.
3	(c) Find the mass of hydrogen chloride in 250.0 cm3 of 0.50 M hydrochloric acid.	Answer: Moles = 0.50 x 0.2500 = 0.125 mol; mass = 0.125 x 36.46 = 4.56 g. Marking: moles 2 marks, mass 2 marks.
4	(d) State three correct techniques when making a solution up to the mark in a volumetric flask.	Answer: add distilled water to about two thirds and mix; add the last drops with a dropper up to the mark; read with the eye level with the bottom of the meniscus, then invert to mix. Marking: 1 mark per point, up to 4.

Part 2: Acid-Base Titration Calculations (17 marks)

All titrations below go to complete neutralisation; state the mole ratio used.

#	Question	Answer and marking notes
5	(a) What volume of 0.125 M hydrochloric acid neutralises 25.0 cm3 of 0.100 M sodium hydroxide?	Answer: Ratio 1:1. Moles of sodium hydroxide = 0.100 x 0.0250 = 2.50 x 10^-3 mol, so the same moles of acid; volume = 2.50 x 10^-3 / 0.125 x 1000 = 20.0 cm3. Marking: ratio 1, moles 2, volume 2.
6	(b) What volume of 0.100 M hydrochloric acid reacts with 25.0 cm3 of 0.0500 M sodium carbonate?	Answer: Na2CO3 + 2HCl gives 2NaCl + H2O + CO2, ratio 1:2. Moles of carbonate = 1.25 x 10^-3 mol, so acid = 2.50 x 10^-3 mol; volume = 25.0 cm3. Marking: ratio 2, moles 1, volume 1.
7	(c) What volume of 0.0500 M sulfuric acid neutralises 20.0 cm3 of 0.100 M sodium hydroxide?	Answer: H2SO4 + 2NaOH gives Na2SO4 + 2H2O, ratio 1:2. Sodium hydroxide = 2.00 x 10^-3 mol, acid = 1.00 x 10^-3 mol; volume = 20.0 cm3. Marking: ratio 2, calculation 2.
8	(d) Which type of indicator suits a strong acid against a strong base, and how long must the end point colour persist before a reading is recorded?	Answer: Phenolphthalein (colourless to pale pink) or methyl orange (red to orange) works because the sharp pH change is wide; the colour must persist for at least half a minute so that a local excess does not give a false end point. Marking: indicator 2, half-minute rule with reason 2.

Part 3: pH and Acid Strength (17 marks)

Assume 25 degrees Celsius with Kw = 1.0 x 10^-14; show your working.

#	Question	Answer and marking notes
9	(a) Calculate the pH of 0.010 M hydrochloric acid.	Answer: Hydrochloric acid is strong and fully ionised, so the hydrogen ion concentration is 0.010 M; pH = -log(0.010) = 2.00. Marking: concentration 2, pH 2.
10	(b) Calculate the pOH and pH of 0.0010 M sodium hydroxide.	Answer: pOH = -log(0.0010) = 3.00; pH = 14.00 – 3.00 = 11.00. Marking: pOH 2, pH 1.
11	(c) 10.0 cm3 of 0.50 M hydrochloric acid is diluted to 250.0 cm3. Find the new concentration and pH.	Answer: Concentration = 0.50 x 10.0 / 250.0 = 0.020 M; pH = 1.70. Marking: concentration 2, pH 1.
12	(d) How do the pH values of 0.10 M hydrochloric acid and 0.10 M ethanoic acid differ, and why?	Answer: Hydrochloric acid has pH 1.00 while ethanoic acid is about 2.87, so the weak acid has the higher pH. Ethanoic acid is only partly ionised, so its hydrogen ion concentration is far below 0.10 M, whereas the strong acid is fully ionised. Marking: comparison 2, partial ionisation 2.
13	(e) Equal moles of the two acids react with excess magnesium ribbon. Which gives more hydrogen?	Answer: The total amount is the same. A weak acid keeps ionising as it is used up, so equal moles of monobasic acid supply equal moles of hydrogen ions, and the same volume of hydrogen is collected. Marking: equal total 2, continued ionisation 1.

Part 4: Redox and Half-Equations (17 marks)

Answer the questions, showing how each oxidation number is worked out.

#	Question	Answer and marking notes
14	(a) Find the oxidation number of manganese in potassium manganate(VII) and of chromium in potassium dichromate(VI).	Answer: In KMnO4, potassium is +1 and each oxygen -2, so manganese = -[1 + 4(-2)] = +7; in K2Cr2O7, 2(+1) + 2x + 7(-2) = 0 gives chromium = +6. Marking: 2 marks each.
15	(b) Write the balanced half-equation for the reduction of manganate(VII) to manganese(II) in acidic solution.	Answer: MnO4- + 8H+ + 5e- gives Mn2+ + 4H2O. Atoms and charge balance: the left side carries +7 minus 5, which is +2, matching the right side. Marking: species 2, balancing 2.
16	(c) What is the mole ratio of manganate(VII) to iron(II) in acidic solution?	Answer: The ratio is 1:5, because each manganate(VII) ion accepts 5 electrons while each iron(II) ion releases only 1. Marking: ratio 2, electron transfer 1.
17	(d) 25.0 cm3 of 0.0200 M manganate(VII) exactly oxidises 25.0 cm3 of iron(II) solution. Find the concentration of iron(II).	Answer: Manganate(VII) = 0.0200 x 0.0250 = 5.00 x 10^-4 mol; iron(II) = 5 x 5.00 x 10^-4 = 2.50 x 10^-3 mol; concentration = 0.100 M. Marking: moles 1, ratio 1, concentration 1.
18	(e) Why does a manganate(VII) titration need no added indicator?	Answer: The manganate(VII) ion is deep purple while manganese(II) is almost colourless, so the first slight excess after the end point leaves a persistent pale pink colour; the reagent acts as its own indicator. Marking: colour change 2, self-indicator 1.

Part 5: Electrolysis and Quantitative Calculation (17 marks)

A current of 2.00 A passes through copper(II) sulfate solution for 30.0 minutes; take the Faraday constant as 96500 C per mol.

#	Question	Answer and marking notes
19	(a) Find the charge passed and the mass of copper deposited at the cathode.	Answer: Charge = 2.00 x 1800 = 3600 C; moles of electrons = 3.73 x 10^-2; Cu2+ + 2e- gives Cu, so copper = 1.87 x 10^-2 mol and mass = 1.19 g. Marking: charge 1, electrons 2, mass 2.
20	(b) If oxygen forms at the anode, find its volume at room temperature (molar volume 24.0 dm3 per mol) in cm3.	Answer: 2H2O gives O2 + 4H+ + 4e-, so oxygen = 9.33 x 10^-3 mol; volume = 9.33 x 10^-3 x 24.0 x 1000 = 224 cm3. Marking: ratio 2, volume 2.
21	(c) How does the direction of electron flow in the wires compare with conventional current?	Answer: They are opposite. Electrons leave the negative terminal and travel through the external circuit to the cathode, while conventional current is defined as positive charge flow, from the positive terminal through the circuit to the negative one. Marking: opposite 2, description 2.
22	(d) In the purification of copper by electrolysis, why is the anode impure copper and the cathode pure copper?	Answer: The impure anode dissolves to give copper ions, while unreactive impurities such as gold and silver fall away as anode sludge; copper ions are preferentially discharged onto the pure cathode, giving high purity copper. Marking: anodic dissolution 2, cathodic deposition and impurity separation 2.

Part 6: Reaction Rate and Chemical Equilibrium (16 marks)

Answer the questions below, giving units where a calculation is involved.

#	Question	Answer and marking notes
23	(a) The concentration of a reactant falls from 0.80 M to 0.20 M in 120 s. Find the average rate.	Answer: Change in concentration = 0.60 M; average rate = 0.60 / 120 = 5.0 x 10^-3 M per second. Marking: change 2, rate with unit 2.
24	(b) Use collision theory to explain how a catalyst and a higher temperature increase the rate.	Answer: A catalyst provides an alternative route of lower activation energy, so more molecules have enough energy and effective collisions increase; raising the temperature increases average kinetic energy so a much larger fraction of molecules exceed the activation energy, and collisions are also more frequent. Marking: catalyst 2, temperature 2, each with activation energy and effective collisions.
25	(c) At 460 degrees, H2 + I2 gives 2HI reaches equilibrium with 0.20 M H2, 0.20 M I2 and 1.60 M HI. Find Kc.	Answer: Kc = [HI]^2 / ([H2][I2]) = 1.60^2 / (0.20 x 0.20) = 2.56 / 0.040 = 64.0, and Kc has no unit. Marking: expression 2, value 2.
26	(d) For N2 + 3H2 giving 2NH3 at equilibrium, which way does the equilibrium shift when pressure is increased, and what happens to the ammonia yield?	Answer: The equilibrium shifts towards the side with fewer moles of gas, which is to the right, because four moles of gas become two; the ammonia yield therefore rises. Marking: direction 2, mole comparison and yield 2.

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