HKDSE Chemistry Paper 1: Acid-Base Titration and Chemical Cells

Level: S6 (HKDSE) | Subject: Chemistry (Compulsory Part) | Time: 150 minutes | Total: 100 marks | Parts: 30

Paper at a glance

Paper	Time	Marks	Calculator
Paper 1 (Compulsory part)	2 h 30 min	100	Allowed

Questions

Question 1 (15 marks) – Acid-base concepts, pH calculations and indicator choice

(a) Define an acid and a base in terms of proton transfer and give one example of each. (3 marks)

(b) How many times greater is the hydrogen ion concentration at pH 2 than at pH 5? Explain your reasoning. (3 marks)

(c) What is the pH of 0.010 mol/dm3 hydrochloric acid, and what is the pH after tenfold dilution? (3 marks)

(d) A solution has pH 3.3. Calculate its hydrogen ion concentration, giving the unit. (3 marks)

(e) Why is phenolphthalein preferred to methyl orange for a titration of hydrochloric acid with 0.100 mol/dm3 sodium hydroxide? (3 marks)

Question 2 (18 marks) – Titration calculations and experimental error

(a) 25.0 cm3 of 0.100 mol/dm3 sodium hydroxide solution is exactly neutralised by dilute hydrochloric acid. Calculate the amount, in moles, of sodium hydroxide used. (4 marks)

(b) The titration used 20.0 cm3 of the dilute hydrochloric acid. Calculate the concentration of the acid. (4 marks)

(c) A further 250 cm3 of 0.100 mol/dm3 sodium hydroxide solution is taken. Calculate the mass of sodium hydroxide it contains (Mr = 40). (4 marks)

(d) During the titration the inside of the conical flask is washed down with distilled water. What effect does this have on the volume of acid needed? Explain. (3 marks)

(e) If the burette is rinsed with the hydrochloric acid being tested instead of distilled water, how does the calculated concentration change? Explain. (3 marks)

Question 3 (17 marks) – Preparing a soluble salt, yield and purity

(a) Describe the steps used to prepare copper(II) sulfate solution from copper(II) oxide and dilute sulfuric acid. (3 marks)

(b) Why is an excess of copper(II) oxide used? (3 marks)

(c) 50.0 cm3 of 1.00 mol/dm3 sulfuric acid reacts with excess copper(II) oxide. Calculate the theoretical mass of anhydrous copper(II) sulfate formed (Mr = 160). (4 marks)

(d) The experiment gave only 6.40 g of anhydrous copper(II) sulfate. Calculate the percentage yield. A separate 5.00 g sample contains only 4.60 g of ammonium sulfate; calculate its percentage purity. (4 marks)

(e) How can anhydrous copper(II) sulfate be used to test a liquid for water? State the observation. (3 marks)

Question 4 (17 marks) – Chemical cells: the zinc-copper cell and electromotive force

(a) Write the half-equations, with state symbols, for the two electrodes of a zinc-copper cell. (3 marks)

(b) State the direction of electron flow when the cell operates and identify the electrode that is negative. (3 marks)

(c) The standard electrode potential of Zn2+/Zn is −0.76 V and that of Cu2+/Cu is +0.34 V. Calculate the electromotive force of the cell. (4 marks)

(d) What is the purpose of the salt bridge, and in which direction do ions move inside it while the cell operates? (4 marks)

(e) After the cell has discharged for some time, how do the masses of the two electrodes change? Write the overall equation. (3 marks)

Question 5 (17 marks) – Redox reactions and electrolysis calculations

(a) Explain oxidation and reduction in terms of electron transfer and state why they always occur together. (3 marks)

(b) Copper(II) sulfate solution is electrolysed with a current of 2.00 A for 1800 s. Calculate the quantity of charge passed. (4 marks)

(c) Using the same electrolysis, calculate the amount of electrons transferred and the mass of copper deposited at the cathode (one mole of electrons is 96500 C, Cu = 63.5). (4 marks)

(d) What are the products at the anode and at the cathode during the electrolysis of copper(II) sulfate solution with inert electrodes? (3 marks)

(e) Compare a chemical cell with an electrolytic cell in terms of energy conversion. (3 marks)

Question 6 (16 marks) – Applying titration: finding the ethanoic acid content of vinegar

(a) 25.0 cm3 of vinegar is diluted to 250 cm3 before titration. Explain the purpose of this step. (3 marks)

(b) Each 25.0 cm3 portion of the diluted solution needs an average of 21.6 cm3 of 0.100 mol/dm3 sodium hydroxide. Calculate the amount, in moles, of sodium hydroxide used. (4 marks)

(c) Ethanoic acid reacts with sodium hydroxide in a one to one ratio. Calculate the amount, in moles, and the mass of ethanoic acid in the 250 cm3 of diluted solution (Mr = 60.0). (4 marks)

(d) The 25.0 cm3 of vinegar has a mass of 25.0 g. Calculate the percentage by mass of ethanoic acid in the sample. (3 marks)

(e) Suggest one way of improving the accuracy of this experiment. (2 marks)

Answers and marking notes

Question 1 – Acid-base concepts, pH calculations and indicator choice

(a) An acid is a proton donor (one mark); a base is a proton acceptor (one mark); hydrochloric acid donates protons while sodium hydroxide accepts them (one mark). [3 marks]

(b) The two solutions differ by three pH units (one mark); each unit corresponds to a tenfold change in hydrogen ion concentration (one mark); the ratio is therefore ten to the power three, that is 1000 times (one mark). [3 marks]

(c) Hydrochloric acid is a strong acid and ionises completely, so the hydrogen ion concentration is 0.010 mol/dm3 (one mark); the pH is 2 (one mark); after tenfold dilution the concentration is 0.0010 mol/dm3 and the pH is 3 (one mark). [3 marks]

(d) The hydrogen ion concentration is 10 raised to the power minus 3.3 (one mark); the value is about 5.0 × 10⁻⁴ (one mark); the unit is mol/dm3 (one mark). [3 marks]

(e) The equivalence point of a strong acid and strong alkali titration is at pH 7 (one mark); phenolphthalein changes colour between pH 8.3 and 10, which lies within the sharp rise of the titration curve (one mark); methyl orange has already changed by pH 3.1 to 4.4, so the end point would be read too early and too small (one mark). [3 marks]

Question 2 – Titration calculations and experimental error

(a) Amount in moles equals concentration multiplied by volume (one mark); 25.0 cm3 is 0.0250 dm3 (one mark); 0.100 multiplied by 0.0250 (one mark); the answer is 0.00250 mol (one mark). [4 marks]

(b) The equation HCl + NaOH → NaCl + H2O shows a ratio of one to one (one mark); the acid therefore also contains 0.00250 mol (one mark); the volume 20.0 cm3 is 0.0200 dm3 (one mark); the concentration is 0.125 mol/dm3 (one mark). [4 marks]

(c) The amount is 0.100 multiplied by 0.250, which is 0.0250 mol (one mark); mass equals moles multiplied by molar mass (one mark); substituting 0.0250 and 40 (one mark); the answer is 1.00 g (one mark). [4 marks]

(d) The added water does not change the amount of sodium hydroxide in the flask (one mark); it only dilutes the solution and does not change how much acid is needed (one mark); the titration result is therefore unchanged (one mark). [3 marks]

(e) Residual acid left in the burette makes the solution delivered slightly more concentrated (one mark); the same burette reading then delivers more moles of acid (one mark); the calculated concentration comes out too high (one mark). [3 marks]

Question 3 – Preparing a soluble salt, yield and purity

(a) Excess copper(II) oxide is added to warm dilute sulfuric acid and stirred (one mark); the mixture is filtered to remove leftover oxide (one mark); the filtrate is blue copper(II) sulfate solution (one mark). [3 marks]

(b) To make sure that all the sulfuric acid reacts (one mark); the excess solid guarantees that no acid remains in the solution (one mark); the leftover oxide is removed by filtration and does not contaminate the product (one mark). [3 marks]

(c) The amount of acid is 1.00 multiplied by 0.0500, which is 0.0500 mol (one mark); the equation shows a ratio of one to one, so 0.0500 mol of copper(II) sulfate forms (one mark); the mass is 0.0500 multiplied by 160 (one mark); the answer is 8.00 g (one mark). [4 marks]

(d) Percentage yield is 6.40 divided by 8.00, multiplied by 100 (one mark); the answer is 80.0 per cent (one mark); percentage purity is 4.60 divided by 5.00, multiplied by 100 (one mark); the answer is 92.0 per cent (one mark). [4 marks]

(e) A little anhydrous copper(II) sulfate is added to the liquid (one mark); the anhydrous salt is white and turns blue when water is present (one mark); heat is also released, showing that hydration has occurred (one mark). [3 marks]

Question 4 – Chemical cells: the zinc-copper cell and electromotive force

(a) Zinc electrode: Zn(s) → Zn²⁺(aq) + 2e⁻ (one mark); copper electrode: Cu²⁺(aq) + 2e⁻ → Cu(s) (one mark); the electrons balance, so the two half-equations can be added directly (one mark). [3 marks]

(b) Electrons flow from the zinc electrode through the external circuit to the copper electrode (one mark); the zinc electrode is the negative electrode and is oxidised (one mark); the copper electrode is the positive electrode and is reduced (one mark). [3 marks]

(c) The electromotive force equals the potential of the positive electrode minus that of the negative electrode (one mark); substituting +0.34 and −0.76 (one mark); subtracting a negative value adds, giving 0.34 plus 0.76 (one mark); the answer is 1.10 V (one mark). [4 marks]

(d) The salt bridge completes the circuit and keeps both half-cells electrically neutral (one mark); the zinc half-cell develops a positive charge as zinc ions form, so anions move into it (one mark); the copper half-cell becomes negative as copper ions are removed, so cations move into it (one mark); without a salt bridge the circuit is broken and the current stops quickly (one mark). [4 marks]

(e) The zinc electrode dissolves and loses mass (one mark); copper is deposited on the copper electrode, which gains mass (one mark); the overall reaction is Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) (one mark). [3 marks]

Question 5 – Redox reactions and electrolysis calculations

(a) Oxidation is loss of electrons and an increase in oxidation number (one mark); reduction is gain of electrons and a decrease in oxidation number (one mark); the electrons lost must be accepted by another species, so the two changes occur together (one mark). [3 marks]

(b) Quantity of charge equals current multiplied by time (one mark); substituting 2.00 and 1800 (one mark); the unit is the coulomb (one mark); the answer is 3600 C (one mark). [4 marks]

(c) The amount of electrons is 3600 divided by 96500 (one mark); that is about 0.0373 mol (one mark); each copper ion needs two electrons, so the copper is 0.0373 divided by two (one mark); the mass is about 1.18 g (one mark). [4 marks]

(d) Copper is deposited at the cathode (one mark); oxygen is released at the anode (one mark); hydroxide ions lose electrons at the anode to form oxygen and water (one mark). [3 marks]

(e) A chemical cell converts chemical energy into electrical energy and the reaction is spontaneous (one mark); an electrolytic cell converts electrical energy into chemical energy and needs an external supply (one mark); in both, oxidation and reduction still take place at the electrodes (one mark). [3 marks]

Question 6 – Applying titration: finding the ethanoic acid content of vinegar

(a) The acid in vinegar is fairly concentrated, so direct titration would need an inconveniently large volume of alkali (one mark); after dilution each titre falls within the range of the burette (one mark); the colour change at the end point is also easier to judge in a more dilute solution (one mark). [3 marks]

(b) Amount in moles equals concentration multiplied by volume (one mark); 21.6 cm3 is 0.0216 dm3 (one mark); 0.100 multiplied by 0.0216 (one mark); the answer is 0.00216 mol (one mark). [4 marks]

(c) Each 25.0 cm3 portion contains 0.00216 mol of acid (one mark); the 250 cm3 solution is ten such portions, so it holds 0.0216 mol (one mark); the mass is 0.0216 multiplied by 60.0 (one mark); the answer is 1.296 g (one mark). [4 marks]

(d) The percentage is 1.296 divided by 25.0, multiplied by 100 (one mark); the answer is 5.18 per cent (one mark); three significant figures are kept, matching the precision of the burette readings (one mark). [3 marks]

(e) Repeat the titration at least three times with the same solution and average the titres (one mark); or place the flask on a white tile and accept the end point only when the pink colour persists for 30 seconds (one mark). [2 marks]

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