DSE Chemistry Practice Paper: Acids, Bases, Redox And Equilibrium (12 Questions, 100 Marks, 75 Minutes)

DSE Chemistry Practice Paper: Acids, Bases, Redox and Equilibrium (12 questions, 100 marks, 75 minutes)

2026-11-03

DSE Chemistry: Acids, Bases and Redox

DSE Chemistry practice paper on acids, bases, redox and equilibrium: 12 questions, 100 marks, 75 minutes, with full worked answers and marking points.

Download　Question paper PDF　|　Answer key PDF

Paper details　Level: S6 | Subject: Chemistry | Time: 75 minutes | Marks: 100 | Questions: 12

Section A: Mole calculations, pH and titration (Questions 1-4, 24 marks)

Answer all questions. Write down the formula, substitute values and give units.

#	Question	Answer and marking notes
1	24.5 g of pure sulfuric acid (H2SO4, relative molecular mass 98.0) is dissolved in water and made up to 500 cm3. (a) Calculate the number of moles of sulfuric acid. (b) Calculate the molar concentration in mol dm-3. (6 marks)	(a) Moles = 24.5 / 98.0 = 0.250 mol. (b) Concentration = 0.250 / 0.500 = 0.500 mol dm-3. Marks: 3 moles, 3 concentration (units required).
2	A hydrochloric acid solution has a concentration of 0.010 mol dm-3. (a) Calculate its pH. (b) The solution is diluted 100 times; calculate the pH of the diluted solution. (6 marks)	(a) pH = -log(0.010) = 2.00. (b) After dilution the concentration is 1.0 × 10-4 mol dm-3, so pH = 4.00; each tenfold dilution raises the pH by 1. Marks: 3 + 3.
3	25.0 cm3 of sodium hydroxide solution (0.100 mol dm-3) requires 12.5 cm3 of sulfuric acid solution to reach the end point: H2SO4 + 2NaOH → Na2SO4 + 2H2O. (a) Calculate the moles of NaOH. (b) Calculate the moles of H2SO4. (c) Calculate the molar concentration of the acid. (6 marks)	(a) Moles of NaOH = 0.100 × 0.0250 = 2.50 × 10-3 mol. (b) The equation gives a 1:2 ratio, so moles of H2SO4 = 1.25 × 10-3 mol. (c) Concentration = 1.25 × 10-3 / 0.0125 = 0.100 mol dm-3. Marks: 2 + 2 + 2.
4	Give the oxidation number of the stated element in each species: (a) Mn in KMnO4; (b) Cr in K2Cr2O7; (c) S in H2SO4. (6 marks)	(a) In KMnO4, K is +1 and O is -2, so Mn is +7. (b) In K2Cr2O7, Cr is +6. (c) In H2SO4, S is +6. Marks: 2 each.

Section B: Redox, equilibrium constants and enthalpy (Questions 5-8, 32 marks)

Show full working in calculations; write explanations in complete sentences.

#	Question	Answer and marking notes
5	A 6.5 g piece of zinc is added to excess copper(II) sulfate solution: Zn + Cu2+ → Zn2+ + Cu (relative atomic masses: Zn 65.0, Cu 63.5). (a) Calculate the moles of zinc. (b) Calculate the moles of electrons transferred. (c) Calculate the mass of copper deposited. (8 marks)	(a) Moles of zinc = 6.5 / 65.0 = 0.10 mol. (b) Each zinc atom loses two electrons, so 0.20 mol of electrons is transferred. (c) Moles of copper = 0.10 mol; mass = 0.10 × 63.5 = 6.35 g. Marks: 3 + 2 + 3.
6	In a 1.0 dm3 vessel, 1.0 mol of hydrogen reacts with 1.0 mol of iodine vapour: H2(g) + I2(g) ⇌ 2HI(g). At equilibrium the concentration of hydrogen iodide is 1.60 mol dm-3. (a) Calculate the equilibrium concentrations of H2 and I2. (b) Write the expression for Kc. (c) Calculate the value of Kc. (8 marks)	(a) 1.60 mol dm-3 of HI is formed, so 0.80 mol dm-3 each of H2 and I2 reacts; both equilibrium concentrations are 0.20 mol dm-3. (b) Kc = [HI]2 / ([H2][I2]). (c) Kc = 1.60² / (0.20 × 0.20) = 64. Marks: 3 + 2 + 3.
7	50.0 cm3 of 1.0 mol dm-3 hydrochloric acid is mixed with 50.0 cm3 of 1.0 mol dm-3 sodium hydroxide solution and the temperature rises by 6.7 °C. Assume the density of the solution is 1.0 g cm-3 and its specific heat capacity is 4.2 J g-1 K-1. (a) Calculate the heat released. (b) Calculate the moles of water formed. (c) Calculate the enthalpy change of neutralisation in kJ mol-1. (8 marks)	(a) Heat released = 100 × 4.2 × 6.7 = 2814 J. (b) Moles of water = 1.0 × 0.0500 = 0.0500 mol. (c) Enthalpy change = -2814 / 0.0500 = -56 280 J mol-1, about -56.3 kJ mol-1. Marks: 3 + 3 + 2.
8	In a reaction the concentration of a reactant falls from 0.80 mol dm-3 to 0.20 mol dm-3 in 60 seconds. (a) Calculate the average rate over the first 60 seconds in mol dm-3 s-1. (b) Explain why the rate decreases with time. (c) State the effect of adding a catalyst on the rate and on the position of equilibrium. (8 marks)	(a) Average rate = (0.80 – 0.20) / 60 = 0.010 mol dm-3 s-1. (b) The concentration of the reactant falls with time, so collisions become less frequent and the rate decreases. (c) A catalyst provides a pathway of lower activation energy, speeding up both the forward and reverse reactions, but it does not change the position of equilibrium. Marks: 3 + 2 + 3.

Section C: Weak acids, buffers and redox titration (Questions 9-12, 44 marks)

Answer the long questions in points; give numerical answers to three significant figures.

#	Question	Answer and marking notes
9	A weak acid HA has a concentration of 0.10 mol dm-3 and Ka = 1.8 × 10-5. (a) Write the expression for Ka and, assuming the degree of dissociation is small, calculate [H+]. (b) Calculate the pH of the solution. (c) Calculate the degree of dissociation as a percentage. (11 marks)	(a) Ka = [H+][A-] / [HA]; with [H+] = [A-] = x, x² / 0.10 = 1.8 × 10-5, so x = 1.34 × 10-3 mol dm-3. (b) pH = -log(1.34 × 10-3) = 2.87. (c) Degree of dissociation = 1.34 × 10-3 / 0.10 × 100% = 1.34%. Marks: 4 + 4 + 3.
10	A buffer solution is made by mixing equal volumes of 0.10 mol dm-3 ethanoic acid and 0.10 mol dm-3 sodium ethanoate (Ka of CH3COOH = 1.8 × 10-5). (a) Calculate the pH of the buffer. (b) 0.010 mol of hydrogen chloride gas is added to 1.0 dm3 of the buffer with no change in volume; calculate the new pH. (c) Explain why the pH changes only slightly when a small amount of strong acid is added. (11 marks)	(a) [acid] = [salt], so pH = pKa = -log(1.8 × 10-5) = 4.74. (b) After adding the strong acid, [acid] is about 0.11 and [salt] about 0.09, so pH = 4.74 + log(0.09 / 0.11) = 4.66. (c) The added H+ ions combine with ethanoate ions to form the weak acid, so [H+] rises only very slightly and the pH hardly changes. Marks: 3 + 4 + 4.
11	25.0 cm3 of iron(II) sulfate solution (0.100 mol dm-3) is titrated with 0.0200 mol dm-3 potassium permanganate solution in acidic medium: MnO4- + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O. (a) Calculate the moles of Fe2+. (b) Calculate the moles of MnO4- required. (c) Calculate the volume of potassium permanganate solution in cm3. (11 marks)	(a) Moles of Fe2+ = 0.100 × 0.0250 = 2.50 × 10-3 mol. (b) The ratio is 1:5, so moles of MnO4- = 5.00 × 10-4 mol. (c) Volume = 5.00 × 10-4 / 0.0200 = 0.0250 dm3 = 25.0 cm3. Marks: 4 + 4 + 3.
12	The industrial manufacture of ammonia is N2(g) + 3H2(g) ⇌ 2NH3(g), with a negative ΔH. (a) State the effect of increasing the pressure on the yield of ammonia and explain it in terms of moles of gas. (b) State the effects of raising the temperature on the yield and on the rate of reaction. (c) Can a catalyst increase the equilibrium yield? Explain. (11 marks)	(a) There are 4 moles of gas on the left and 2 on the right; increasing the pressure shifts the equilibrium to the right, so the yield of ammonia increases. (b) Raising the temperature shifts the equilibrium to the left (the endothermic direction), lowering the yield but increasing the rate. (c) A catalyst speeds up the approach to equilibrium but does not change the equilibrium constant, so it cannot increase the yield. Marks: 4 + 4 + 3.

Find a tutor — free Tutor Registration →

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