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HKDSE Chemistry Practice Paper: Acids, Bases, Titration and Electrochemistry (12 questions, 100 marks, 75 minutes)
HKDSE Chemistry: Acids, Bases and Electrochemistry
Twelve questions on pH, titration calculations, molarity, electrolysis, cells, redox titration, indicators, salt hydrolysis, rates and titration data, with full answers and marking points.
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Paper details Level: S6 | Subject: Chemistry | Time: 75 minutes | Marks: 100 | Questions: 12
Section A (32 marks)
Acid-base, concentration and titration basics: quote the formula, substitute values and give units.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A 0.010 M hydrochloric acid solution (strong acid, fully ionised). (a) Calculate the pH of the solution. (b) Calculate the pH of a 0.010 M sodium hydroxide solution. (c) 10.0 cm3 of the acid is diluted with water to a total volume of 1.00 dm3. Calculate the pH of the diluted solution. |
(a) [H+] = 0.010 M, so pH = -log 0.010 = 2.00 (b) [OH-] = 0.010 M, so pOH = 2.00 and pH = 14.00 – 2.00 = 12.00 (c) Amount before dilution = 0.010 x 0.0100 = 1.00 x 10^-4 mol; after dilution [H+] = 1.00 x 10^-4 / 1.00 = 1.00 x 10^-4 M, so pH = 4.00 |
| 2 | 25.0 cm3 of 0.100 M sodium hydroxide solution is titrated with 0.0500 M sulphuric acid. (a) Calculate the amount, in mol, of sodium hydroxide in 25.0 cm3. (b) Write the ionic equation for the neutralisation and find the amount of sulphuric acid required. (c) Calculate the volume of sulphuric acid required, in cm3. |
(a) n(NaOH) = 0.100 x 0.0250 = 2.50 x 10^-3 mol (b) H+ + OH- -> H2O. Each H2SO4 gives two H+, so n(H2SO4) = 2.50 x 10^-3 / 2 = 1.25 x 10^-3 mol (c) V = 1.25 x 10^-3 / 0.0500 = 0.0250 dm3 = 25.0 cm3 |
| 3 | 4.0 g of sodium hydroxide (molar mass 40.0 g/mol) is dissolved in water and made up to 500 cm3. (a) Calculate the molarity of the solution. (b) 25.0 cm3 of this solution is diluted to 250 cm3. Calculate the new concentration. (c) Calculate the amount, in mol, of sodium hydroxide in 25.0 cm3 of the diluted solution. |
(a) n = 4.0 / 40.0 = 0.10 mol, so c = 0.10 / 0.500 = 0.20 M (b) c1V1 = c2V2: 0.20 x 25.0 = c2 x 250, so c2 = 0.020 M (c) n = 0.020 x 0.0250 = 5.0 x 10^-4 mol |
| 4 | Two beakers contain 0.0010 M hydrochloric acid and 0.0010 M ethanoic acid (Ka = 1.8 x 10^-5). (a) Calculate the pH of the hydrochloric acid. (b) Using [H+] = square root of (Ka x c), calculate the pH of the ethanoic acid. (c) Explain why the ethanoic acid reacts more slowly with a magnesium ribbon at the same concentration. |
(a) pH = -log 0.0010 = 3.00 (b) [H+] = square root of (1.8 x 10^-5 x 0.0010) = 1.34 x 10^-4 M, so pH = 3.87 (c) Ethanoic acid is only partially ionised, so [H+] is much lower than in the hydrochloric acid; collisions with the magnesium are less frequent and bubbles form more slowly. |
Section B (36 marks)
Electrolysis, cells and redox titration: write the electrode reactions and explain the principle.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | A copper(II) sulphate solution is electrolysed with copper electrodes using a current of 0.50 A for 32 minutes 10 seconds (1930 s). (a) Calculate the quantity of charge passed. (b) Calculate the amount, in mol, of copper deposited at the cathode (1 mol of electrons = 96500 C; Cu2+ + 2e- -> Cu). (c) Calculate the mass of copper deposited (Cu = 63.5). |
(a) Q = It = 0.50 x 1930 = 965 C (b) n(e-) = 965 / 96500 = 0.0100 mol; n(Cu) = 0.0100 / 2 = 5.0 x 10^-3 mol (c) m = 5.0 x 10^-3 x 63.5 = 0.3175 g, about 0.32 g |
| 6 | Concentrated sodium chloride solution is electrolysed using graphite electrodes. (a) Write the anode reaction and name the product. (b) Write the cathode reaction and name the product. (c) Explain why Na+ is not discharged at the cathode, and state how the acidity of the solution changes. |
(a) Anode: 2Cl- -> Cl2 + 2e-, giving chlorine gas. (b) Cathode: 2H2O + 2e- -> H2 + 2OH-, giving hydrogen gas. (c) Sodium is more reactive than the hydrogen in water, so Na+ is harder to reduce and water is discharged first; OH- builds up at the cathode, so the solution becomes alkaline. |
| 7 | A zinc-copper cell has standard electrode potentials Zn2+/Zn = -0.76 V and Cu2+/Cu = +0.34 V. (a) Calculate the standard cell electromotive force. (b) State the direction of electron flow in the external circuit and write the reaction at each electrode. (c) Explain the function of the salt bridge. |
(a) Ecell = Ecathode – Eanode = 0.34 – (-0.76) = 1.10 V (b) Electrons flow from the zinc electrode through the external circuit to the copper electrode; anode (zinc): Zn -> Zn2+ + 2e-; cathode (copper): Cu2+ + 2e- -> Cu (c) Ions migrate through the salt bridge to neutralise the charge built up in each half cell, keeping both solutions electrically neutral and completing the circuit. |
| 8 | In acidic solution, 25.0 cm3 of 0.0200 M iron(II) sulphate solution is titrated with 0.0100 M potassium permanganate solution (MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O). (a) Describe the change in oxidation number of manganese. (b) Write the balanced overall ionic equation. (c) Calculate the volume of potassium permanganate solution required. |
(a) Manganese goes from +7 to +2, so each MnO4- gains five electrons and is reduced. (b) MnO4- + 5Fe2+ + 8H+ -> Mn2+ + 5Fe3+ + 4H2O (c) n(Fe2+) = 0.0200 x 0.0250 = 5.00 x 10^-4 mol; n(MnO4-) = 5.00 x 10^-4 / 5 = 1.00 x 10^-4 mol; V = 1.00 x 10^-4 / 0.0100 = 0.0100 dm3 = 10.0 cm3 |
Section C (32 marks)
Indicators, salt hydrolysis, rates and data analysis: explain your reasoning clearly.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | For each titration below, choose a suitable indicator and explain your choice. (a) Hydrochloric acid titrated with sodium hydroxide. (b) Hydrochloric acid titrated with aqueous ammonia. (c) Ethanoic acid titrated with sodium hydroxide. (d) Explain why the indicator colour-change range must overlap the steep part of the pH curve at the equivalence point. |
(a) Strong acid with strong base: equivalence pH = 7, so either phenolphthalein or methyl orange works. (b) Strong acid with weak base: equivalence pH is about 5 (acidic), so methyl orange (range 3.1-4.4) is suitable. (c) Weak acid with strong base: equivalence pH is about 9 (alkaline), so phenolphthalein (range 8.3-10.0) is suitable. (d) The observed end point is close to the equivalence point only if the indicator changes colour within the steep part of the curve, which keeps the indicator error small. |
| 10 | A 0.10 M sodium ethanoate solution (Ka of ethanoic acid = 1.8 x 10^-5). (a) Explain why the solution is alkaline. (b) Using Kb = Kw / Ka, calculate Kb of the ethanoate ion. (c) Using [OH-] = square root of (Kb x c), calculate the pH of the solution. |
(a) The ethanoate ion hydrolyses: CH3COO- + H2O ⇌ CH3COOH + OH-, producing OH- and making the solution alkaline. (b) Kb = 1.0 x 10^-14 / 1.8 x 10^-5 = 5.6 x 10^-10 (c) [OH-] = square root of (5.6 x 10^-10 x 0.10) = 7.5 x 10^-6 M, so pOH = 5.13 and pH = 14.00 – 5.13 = 8.87 |
| 11 | A 2.00 g piece of marble (calcium carbonate) is added to 50.0 cm3 of 1.0 M hydrochloric acid and the total mass loss is measured at intervals. Time / s: 0, 30, 60, 90, 120, 150 Mass loss / g: 0, 0.80, 1.40, 1.80, 2.00, 2.00 (a) Calculate the average rate of reaction from 0 to 60 s, in g/s. (b) Explain why the mass stops decreasing after 90 s. (c) Explain how the curve would change if the marble were ground into powder. |
(a) Average rate = 1.40 / 60 = 0.0233 g/s (b) The acid is the limiting reagent and is used up after 90 s, so the reaction stops and the mass no longer falls. (c) Powdering increases the surface area and the collision frequency, so the initial rate is higher and the curve steeper; the acid amount is unchanged, so the final mass loss is the same. |
| 12 | 25.0 cm3 of sodium hydroxide solution of unknown concentration is titrated with 0.100 M hydrochloric acid. Three titres are obtained: 24.25 cm3, 24.00 cm3 and 24.05 cm3. (a) State which reading should be rejected and explain why. (b) Calculate the mean titre in cm3 to 3 significant figures. (c) Calculate the concentration of the sodium hydroxide solution. |
(a) Reject 24.25 cm3 because it differs from the other two titres by more than 0.10 cm3 and is an outlier. (b) Mean = (24.00 + 24.05) / 2 = 24.025, about 24.03 cm3 (c) n(HCl) = 0.100 x 0.024025 = 2.4025 x 10^-3 mol; n(NaOH) = 2.4025 x 10^-3 mol; c = 2.4025 x 10^-3 / 0.0250 = 0.0961 M |
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