IGCSE Chemistry 0620 Paper 4 Practice Paper: Acids, Bases, Salts and Titration Calculations (Extended Theory, 100 marks)

Level: Year 10-11 (IGCSE) | Subject: Chemistry 0620 | Time: 60 minutes | Total: 100 marks | Parts: 34

Paper at a glance

Paper	Time	Marks	Calculator
Paper 4 (Extended, structured)	1 h 15 min	75	Not allowed

Questions

Question 1 (16 marks) – Acids, alkalis, pH and neutralisation calculations

(a) Define an acid in terms of ions and name one ion produced when hydrogen chloride dissolves in water. (2 marks)

(b) State the colour of universal indicator and of blue litmus paper in an acidic solution. (2 marks)

(c) 25.0 cm3 of 0.100 mol/dm3 sodium hydroxide solution is exactly neutralised by 0.125 mol/dm3 hydrochloric acid. Calculate the volume of acid required. (3 marks)

(d) 10.0 cm3 of 0.10 mol/dm3 hydrochloric acid is diluted with water to a total volume of 250 cm3. Calculate the hydrogen ion concentration and the pH of the diluted solution (give pH to two decimal places). (3 marks)

(e) 0.48 g of magnesium is added to 500 cm3 of 0.100 mol/dm3 hydrochloric acid. Deduce which reactant is in excess and calculate the volume of hydrogen formed at room temperature and pressure (molar volume 24 000 cm3/mol). (3 marks)

(f) Classify sodium oxide, sulfur dioxide and aluminium oxide as acidic, basic or amphoteric oxides, naming each one. (3 marks)

Question 2 (18 marks) – Preparing a soluble salt: copper(II) sulfate crystals

(a) When preparing copper(II) sulfate, why is an excess of copper(II) oxide added instead of the calculated amount? (2 marks)

(b) Write the equation for the reaction between copper(II) oxide and dilute sulfuric acid, and describe how the unreacted solid is separated from the solution. (3 marks)

(c) 25.0 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 = 159.6). (3 marks)

(d) The experiment gives 3.20 g of CuSO4.5H2O crystals (Mr = 249.6). Calculate the percentage yield based on the hydrated salt. (4 marks)

(e) Explain why the solution must not be evaporated to dryness to obtain the crystals, and state one step used to obtain larger crystals. (3 marks)

(f) Describe the test for sulfate ions in solution and the expected observation. (3 marks)

Question 3 (18 marks) – Titration, strong and weak acids

(a) State the difference between a weak acid and a strong acid. (2 marks)

(b) 0.010 mol/dm3 hydrochloric acid has pH 2.00, while ethanoic acid of the same concentration has pH about 3.4. Explain the difference. (3 marks)

(c) 25.0 cm3 of 0.100 mol/dm3 sodium hydroxide solution is titrated with sulfuric acid until the end point. Given that the acid concentration is 0.0625 mol/dm3, calculate the volume required and state the mole ratio used. (4 marks)

(d) Write the ionic equation for the reaction between sodium hydroxide and sulfuric acid, and explain why it is a neutralisation reaction. (3 marks)

(e) Phenolphthalein is suitable for this titration. State two conditions that govern the choice of indicator. (3 marks)

(f) Assume the titration at the end point produces 0.00125 mol of sodium sulfate. Calculate the mass of sodium sulfate formed (Mr = 142.1). (3 marks)

Question 4 (16 marks) – Solubility of salts and precipitation reactions

(a) State three solubility rules, giving one example of each. (3 marks)

(b) Silver nitrate solution is mixed with sodium chloride solution. State the observation, name the precipitate and give the ionic equation. (3 marks)

(c) 50.0 cm3 of 0.200 mol/dm3 silver nitrate solution reacts with excess sodium chloride solution. Calculate the mass of precipitate formed (Mr of AgCl = 143.5). (4 marks)

(d) If 50.0 cm3 of 0.200 mol/dm3 silver nitrate is mixed with 25.0 cm3 of 0.200 mol/dm3 sodium chloride instead, calculate the mass of precipitate and state which ion is left in excess. (3 marks)

(e) State the flame colours seen when sodium, potassium and copper(II) ions are tested in a flame test. (3 marks)

Question 5 (16 marks) – Applications of neutralisation: soil and fertilisers

(a) A farmer adds slaked lime to acidic soil. Explain the chemistry behind this treatment. (2 marks)

(b) A field needs 0.050 mol of calcium oxide per square metre. Calculate the mass of calcium oxide required per square metre (Mr = 56). (3 marks)

(c) Ammonia reacts with nitric acid to form ammonium nitrate. If 0.500 mol of ammonia reacts completely, calculate the mass of ammonium nitrate formed (Mr = 80). (3 marks)

(d) Compare the percentage by mass of nitrogen in ammonium nitrate and in ammonium sulfate, and use the result to say which fertiliser is more concentrated. (4 marks)

(e) State two problems that may result from applying too much lime. (2 marks)

(f) State two safety precautions when handling concentrated sulfuric acid and give the reason for each. (2 marks)

Question 6 (16 marks) – Practical investigation: acid with calcium carbonate

(a) 0.50 g of calcium carbonate reacts with excess dilute hydrochloric acid. Calculate the theoretical volume of carbon dioxide formed (molar volume 24 000 cm3/mol). (3 marks)

(b) If 2.0 mol/dm3 hydrochloric acid is used, calculate the volume needed to react completely with 0.0050 mol of calcium carbonate. (3 marks)

(c) The experiment collects 108 cm3 of carbon dioxide. Calculate the percentage yield and suggest one reason why it is below the theoretical value. (4 marks)

(d) Describe the test for the gas collected and write the ionic equation for the reaction of carbonate with acid. (3 marks)

(e) If the same mass of calcium carbonate powder is used instead, state how the rate of reaction and the total volume of gas collected change. (3 marks)

Answers and marking notes

Question 1 – Acids, alkalis, pH and neutralisation calculations

(a) An acid produces hydrogen ions, H+, when dissolved in water (one mark); the solution contains hydrogen ions and chloride ions (one mark). [2 marks]

(b) Universal indicator turns red, orange or yellow (one mark); blue litmus turns red (one mark). [2 marks]

(c) n(NaOH) = 0.0250 x 0.100 = 0.00250 mol (one mark); the ratio is 1 to 1 so n(HCl) = 0.00250 mol (one mark); V = 0.00250 / 0.125 = 0.0200 dm3 = 20.0 cm3 (one mark). [3 marks]

(d) n(H+) = 0.010 x 0.10 = 0.0010 mol (one mark); concentration = 0.0010 / 0.250 = 0.0040 mol/dm3 (one mark); pH = -log(0.0040) = 2.40 (one mark). [3 marks]

(e) n(Mg) = 0.48 / 24 = 0.020 mol and n(HCl) = 0.500 x 0.100 = 0.050 mol; 0.020 mol of magnesium needs only 0.040 mol of acid so the acid is in excess (one mark); n(H2) = 0.020 mol (one mark); V = 0.020 x 24 000 = 480 cm3 (one mark). [3 marks]

(f) Sodium oxide is a basic oxide and reacts with acids (one mark); sulfur dioxide is an acidic oxide and reacts with alkalis (one mark); aluminium oxide is amphoteric and reacts with both acids and alkalis (one mark). [3 marks]

Question 2 – Preparing a soluble salt: copper(II) sulfate crystals

(a) To make sure all the acid is neutralised (one mark); the excess copper(II) oxide is insoluble and can be removed by filtration (one mark). [2 marks]

(b) CuO + H2SO4 gives CuSO4 + H2O (one mark for the equation, one mark for balancing); filter the mixture so the solid stays on the filter paper (one mark). [3 marks]

(c) n(H2SO4) = 0.0250 x 1.00 = 0.0250 mol (one mark); the ratio is 1 to 1 so n(CuSO4) = 0.0250 mol (one mark); mass = 0.0250 x 159.6 = 3.99 g (one mark). [3 marks]

(d) Theoretical n(CuSO4.5H2O) = 0.0250 mol (one mark); theoretical mass = 0.0250 x 249.6 = 6.24 g (one mark); percentage yield = 3.20 / 6.24 x 100 per cent (one mark) = 51.3 per cent (one mark). [4 marks]

(e) Evaporating to dryness drives off the water of crystallisation and decomposes the salt or leaves impurities (one mark); heat until the solution is nearly saturated then cool it slowly so crystals grow (two marks). [3 marks]

(f) Acidify with dilute hydrochloric acid (one mark) then add barium chloride solution (one mark); a white precipitate shows that sulfate ions are present (one mark). [3 marks]

Question 3 – Titration, strong and weak acids

(a) A strong acid ionises completely in water (one mark); a weak acid ionises only partially so many undissociated molecules remain (one mark). [2 marks]

(b) Hydrochloric acid ionises completely so the hydrogen ion concentration is 0.010 mol/dm3 and pH = -log(0.010) = 2.00 (one mark); ethanoic acid ionises only partly (one mark) so the hydrogen ion concentration is much lower and the pH is higher (one mark). [3 marks]

(c) n(NaOH) = 0.0250 x 0.100 = 0.00250 mol (one mark); the ratio H2SO4 to NaOH is 1 to 2 (one mark) so n(H2SO4) = 0.00125 mol (one mark); V = 0.00125 / 0.0625 = 0.0200 dm3 = 20.0 cm3 (one mark). [4 marks]

(d) H+ + OH- gives H2O (two marks); hydrogen ions and hydroxide ions combine to form water, so it is neutralisation (one mark). [3 marks]

(e) The colour change range of the indicator must match the pH at the end point (two marks); the colour change must be sharp enough to see clearly (one mark). [3 marks]

(f) Mass = 0.00125 x 142.1 (one mark) = 0.178 g (two marks, including three significant figures and the unit). [3 marks]

Question 4 – Solubility of salts and precipitation reactions

(a) All nitrates are soluble, for example potassium nitrate (one mark); most chlorides are soluble but silver chloride is insoluble (one mark); most carbonates are insoluble but sodium and potassium carbonates are soluble (one mark). [3 marks]

(b) A white precipitate forms (one mark); it is silver chloride (one mark); Ag+ + Cl- gives AgCl (one mark). [3 marks]

(c) n(AgNO3) = 0.0500 x 0.200 = 0.0100 mol (one mark); the ratio is 1 to 1 so n(AgCl) = 0.0100 mol (one mark); mass = 0.0100 x 143.5 (one mark) = 1.44 g (one mark). [4 marks]

(d) n(Ag+) = 0.0100 mol and n(Cl-) = 0.0050 mol so chloride is limiting (one mark); n(AgCl) = 0.0050 mol and mass = 0.0050 x 143.5 = 0.72 g (one mark); silver ions remain in excess, 0.0050 mol (one mark). [3 marks]

(e) Sodium ions give a yellow flame (one mark); potassium ions give a lilac flame (one mark); copper(II) ions give a blue-green flame (one mark). [3 marks]

Question 5 – Applications of neutralisation: soil and fertilisers

(a) Slaked lime is an alkali that neutralises the acid in the soil (one mark) so the pH rises to the range suitable for crops (one mark). [2 marks]

(b) Mass = moles x molar mass (one mark) = 0.050 x 56 (one mark) = 2.8 g (one mark). [3 marks]

(c) The ratio is 1 to 1 so n(NH4NO3) = 0.500 mol (one mark); mass = 0.500 x 80 (one mark) = 40.0 g (one mark). [3 marks]

(d) Ammonium nitrate contains 28 / 80 x 100 per cent = 35.0 per cent nitrogen (two marks); ammonium sulfate contains 28 / 132 x 100 per cent = 21.2 per cent (one mark); ammonium nitrate is the more concentrated fertiliser (one mark). [4 marks]

(e) The soil becomes alkaline so some nutrients become insoluble and cannot be absorbed (one mark); soil structure or microbial activity may be harmed (one mark). [2 marks]

(f) Wear goggles and gloves so splashes cannot cause serious burns to the eyes or skin (one mark); work in a fume cupboard to avoid breathing corrosive vapour (one mark). [2 marks]

Question 6 – Practical investigation: acid with calcium carbonate

(a) n(CaCO3) = 0.50 / 100 = 0.0050 mol (one mark); n(CO2) = 0.0050 mol (one mark); V = 0.0050 x 24 000 = 120 cm3 (one mark). [3 marks]

(b) The equation shows 2 mol of acid per 1 mol of carbonate (one mark); n(HCl) = 0.010 mol (one mark); V = 0.010 / 2.0 = 0.0050 dm3 = 5.0 cm3 (one mark). [3 marks]

(c) Percentage yield = 108 / 120 x 100 per cent (two marks) = 90.0 per cent (one mark); some gas dissolves in water or escapes before collection (one mark). [4 marks]

(d) Bubble the gas through limewater, which turns milky (one mark); CO3 2- + 2H+ gives H2O + CO2 (two marks). [3 marks]

(e) Powder has a larger surface area so collisions are more frequent and the rate increases (two marks); the total volume of gas is unchanged because the moles of carbonate are the same (one mark). [3 marks]

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