DSE Chemistry Practice Paper: Acid-Base Titration And Electrochemistry (12 Questions, 50 Marks, 75 Minutes)

DSE Chemistry Practice Paper: Acid-Base Titration and Electrochemistry (12 Questions, 50 Marks, 75 Minutes)

2026-11-03

DSE Chemistry: Acids, Bases and Electrochemistry

DSE Chemistry: 12 original questions, 50 marks, 75 minutes on titration, pH, electrolysis, enthalpy, equilibrium and cells.

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Paper details　Level: S6 | Subject: Chemistry | Time: 75 minutes | Marks: 50 | Questions: 12

Section A: Short Questions (4 marks each)

Answer all questions. Show the formula, substitution and units.

#	Question	Answer and marking notes
1	2.00 g of sodium hydroxide is dissolved in water and made up to 250.0 cm³ of solution. Find (a) the molar concentration of the solution, (b) the concentration of hydroxide ions. (Na 23.0, O 16.0, H 1.0) (4 marks)	M(NaOH) = 23.0 + 16.0 + 1.0 = 40.0 g/mol; n = 2.00/40.0 = 0.0500 mol; c = 0.0500/0.250 = 0.200 M. Sodium hydroxide dissociates completely, so [OH⁻] = 0.200 M.
2	25.0 cm³ of 0.100 M sodium hydroxide is titrated with 0.125 M hydrochloric acid. Find (a) the volume of acid needed to reach the equivalence point, (b) the pH at the equivalence point. (4 marks)	n(NaOH) = 0.100 × 0.0250 = 2.50 × 10⁻³ mol. The reaction is 1:1, so n(HCl) = 2.50 × 10⁻³ mol and V = 2.50 × 10⁻³/0.125 = 0.0200 dm³ = 20.0 cm³. A strong acid and a strong base neutralise completely, leaving only sodium chloride, so the pH is 7.00.
3	At 25 °C, find the pH of (a) 0.0100 M hydrochloric acid, (b) 0.0100 M sodium hydroxide. (4 marks)	(a) [H⁺] = 0.0100 M, pH = −log 0.0100 = 2.00. (b) [OH⁻] = 0.0100 M, pOH = 2.00, pH = 14.00 − 2.00 = 12.00.
4	25.0 cm³ of 2.0 M hydrochloric acid is diluted with water to 250.0 cm³. Find (a) the new molar concentration, (b) the pH of the diluted solution. (4 marks)	The number of moles is unchanged on dilution: c₂ = c₁V₁/V₂ = 2.0 × 25.0/250.0 = 0.20 M; pH = −log 0.20 = 0.70.
5	A copper(II) sulphate solution is electrolysed with a current of 2.0 A for 30.0 minutes. Find (a) the quantity of charge passed, (b) the mass of copper deposited at the cathode. (Cu 63.5, Faraday constant 96500 C/mol) (4 marks)	Q = It = 2.0 × 1800 = 3600 C; n(e⁻) = 3600/96500 = 3.73 × 10⁻² mol. Since Cu²⁺ + 2e⁻ → Cu, n(Cu) = 1.87 × 10⁻² mol and m = 1.87 × 10⁻² × 63.5 ≈ 1.18 g.
6	50.0 cm³ of 1.0 M hydrochloric acid is mixed with 50.0 cm³ of 1.0 M sodium hydroxide and the temperature rises by 6.7 °C. Taking the density of the mixture as 1.0 g/cm³ and its specific heat capacity as 4.18 J/(g·K), find the molar enthalpy change of neutralisation. (4 marks)	Heat released = mcΔT = 100.0 × 4.18 × 6.7 ≈ 2800 J. The number of moles of water formed is 0.0500 mol, so ΔH = −2800/0.0500 ≈ −56 kJ/mol; the negative sign shows that heat is released.

Section B: Application Questions (4 marks each)

For equilibrium and electrochemical cells, state the temperature and write the half-equations.

#	Question	Answer and marking notes
7	At 25 °C, hydrogen and iodine vapour are sealed in a container and reach equilibrium: H₂(g) + I₂(g) ⇌ 2HI(g). The equilibrium concentrations are [H₂] = 0.20 M, [I₂] = 0.20 M and [HI] = 1.60 M. Find (a) the equilibrium constant Kc; (b) state, with an explanation, whether Kc changes if [HI] becomes 2.40 M at the same temperature. (4 marks)	(a) Kc = [HI]²/([H₂][I₂]) = 1.60²/(0.20 × 0.20) = 2.56/0.040 = 64. (b) Kc depends only on temperature; at the same temperature it stays at 64, and changing a concentration shifts the position of equilibrium without changing the equilibrium constant.
8	A zinc-copper cell operates at 25 °C. Given E°(Zn²⁺/Zn) = −0.76 V and E°(Cu²⁺/Cu) = +0.34 V, find (a) the standard cell e.m.f.; (b) identify the anode and the cathode and write the half-equation at the anode. (4 marks)	(a) E°cell = E°(cathode) − E°(anode) = 0.34 − (−0.76) = 1.10 V. (b) Zinc has the more negative electrode potential, so it is the anode and is oxidised: Zn(s) → Zn²⁺(aq) + 2e⁻; copper is the cathode, where Cu²⁺ is reduced.
9	5.00 g of calcium carbonate is decomposed by heating: CaCO₃(s) → CaO(s) + CO₂(g). 1.98 g of carbon dioxide is actually collected. Find (a) the theoretical yield of carbon dioxide, (b) the percentage yield. (Ca 40.0, C 12.0, O 16.0) (4 marks)	M(CaCO₃) = 100 g/mol, so n = 5.00/100 = 0.0500 mol. From the 1:1 equation, n(CO₂) = 0.0500 mol and the theoretical yield is 0.0500 × 44.0 = 2.20 g. Percentage yield = 1.98/2.20 × 100% = 90.0%.

Section C: Long Questions (14 marks in total)

Show full working; write explanations in complete sentences.

#	Question	Answer and marking notes
10	0.24 g of magnesium is added to excess dilute hydrochloric acid. Find (a) the amount of hydrogen produced in moles; (b) the volume of hydrogen at room temperature and pressure (molar volume 24.0 dm³/mol), in cm³; (c) the ionic equation for the reaction. (Mg 24.0) (4 marks)	(a) n(Mg) = 0.24/24.0 = 0.0100 mol. From Mg + 2H⁺ → Mg²⁺ + H₂, n(H₂) = 0.0100 mol. (b) V = 0.0100 × 24.0 = 0.240 dm³ = 240 cm³. (c) Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g).
11	25.0 cm³ of 0.100 M sodium carbonate reacts with hydrochloric acid: Na₂CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g). Find (a) the volume of 0.200 M acid needed for complete reaction; (b) the volume of carbon dioxide released at room temperature and pressure (molar volume 24.0 dm³/mol). (4 marks)	(a) n(Na₂CO₃) = 0.100 × 0.0250 = 2.50 × 10⁻³ mol. Twice as much acid is needed, i.e. 5.00 × 10⁻³ mol, so V = 5.00 × 10⁻³/0.200 = 0.0250 dm³ = 25.0 cm³. (b) n(CO₂) = 2.50 × 10⁻³ mol and V = 2.50 × 10⁻³ × 24.0 = 0.060 dm³ = 60.0 cm³.
12	A copper(II) sulphate solution is electrolysed with copper electrodes and a current of 0.50 A for 20.0 minutes. (a) Find the mass of copper dissolved from the anode. (b) Explain why the blue colour of the solution hardly changes during electrolysis. (c) State how the anode product differs if graphite electrodes are used, and explain in terms of oxidation and reduction. (Cu 63.5, 96500 C/mol) (6 marks)	(a) Q = 0.50 × 1200 = 600 C; n(e⁻) = 600/96500 = 6.22 × 10⁻³ mol. At the anode Cu → Cu²⁺ + 2e⁻, so n(Cu) = 3.11 × 10⁻³ mol and m = 3.11 × 10⁻³ × 63.5 ≈ 0.20 g. (b) For every mole of copper dissolved at the anode, one mole is deposited at the cathode, so the net change in copper ion concentration is zero and the blue colour hardly changes. (c) Graphite is not oxidised, so water is oxidised at the anode instead and oxygen is released: 2H₂O → O₂ + 4H⁺ + 4e⁻. Copper ions continue to be reduced at the cathode and are not replaced, so the blue colour fades; overall copper ions are reduced while water is oxidised.

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