IGCSE Chemistry 0620 Paper 4: Acids, Bases, Salts and Titration Calculations

Level: Year 10-11 (IGCSE) | Subject: Chemistry 0620 | Time: 75 minutes | Total: 100 marks | Parts: 30

Paper at a glance

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

Questions

Question 1 (16 marks) – Acid-base definitions, pH and oxides

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

(b) Explain what pH measures and state how many times greater the hydrogen ion concentration is at pH 2 than at pH 5. (3 marks)

(c) Write the equation, with state symbols, for the neutralisation of hydrochloric acid by sodium hydroxide and name both products. (3 marks)

(d) Classify each of these oxides and state its acid-base character: Na2O, SO2, Al2O3, CO. (4 marks)

(e) At the same concentration hydrochloric acid conducts electricity better than ethanoic acid. Explain why. (3 marks)

Question 2 (17 marks) – Titration procedure and mole calculations

(a) Why is an indicator used in a titration? Name the indicator commonly used for a strong acid and strong alkali titration and state its colour change. (3 marks)

(b) 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)

(c) If 20.0 cm3 of the dilute hydrochloric acid was used, calculate the concentration of the acid. (4 marks)

(d) A further 250 cm3 of 0.100 mol/dm3 sodium hydroxide solution is taken. Calculate the mass of sodium hydroxide it contains (Mr = 40). (3 marks)

(e) During a 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)

Question 3 (17 marks) – Preparing a soluble salt: copper(II) sulfate

(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) How are copper(II) sulfate crystals obtained from the filtrate? (3 marks)

(d) 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)

(e) The experiment gave only 6.40 g of anhydrous copper(II) sulfate. Calculate the percentage yield and give one reason for the loss. (4 marks)

Question 4 (17 marks) – Insoluble salts, precipitation and ionic equations

(a) State three solubility rules, one each for chlorides, sulfates and carbonates. (3 marks)

(b) Sodium sulfate solution is mixed with barium chloride solution. Write the full equation and the net ionic equation, and state what is seen. (4 marks)

(c) Why is the precipitate washed with distilled water? (3 marks)

(d) 50.0 cm3 of 0.200 mol/dm3 barium chloride solution reacts with excess sodium sulfate. Calculate the mass of barium sulfate formed (Mr = 233). (4 marks)

(e) How would you test a solution for sulfate ions? Give the reagents, the observation and how interference is avoided. (3 marks)

Question 5 (16 marks) – Strong and weak acids, pH and dilution

(a) Write the expression that defines pH and state what pH 7 means at 25 degrees Celsius. (3 marks)

(b) What is the pH of 0.010 mol/dm3 hydrochloric acid, and what is the pH after it is diluted ten times? (4 marks)

(c) A solution has pH 3.3. Calculate its hydrogen ion concentration. (4 marks)

(d) Explain how the partial ionisation of a weak acid affects its pH and its rate of reaction with a metal. (3 marks)

(e) Universal indicator is added to solutions of pH 1, pH 7 and pH 13. State the colour seen in each. (2 marks)

Question 6 (17 marks) – Ammonia, ammonium salts and fertiliser purity

(a) Explain why aqueous ammonia is alkaline and write the equation for the reaction of ammonia with water. (3 marks)

(b) 25.0 cm3 of 0.0500 mol/dm3 sulfuric acid is completely neutralised by ammonia solution. Calculate the amount, in moles, of ammonia needed. (4 marks)

(c) 20.0 cm3 of the ammonia solution was needed. Calculate the concentration of the ammonia solution. (4 marks)

(d) A 5.00 g sample of ammonium sulfate fertiliser contains only 4.60 g of ammonium sulfate. Calculate the percentage purity of the sample. (3 marks)

(e) How would you test a solid for an ammonium salt? Give the reagent and the observation. (3 marks)

Answers and marking notes

Question 1 – Acid-base definitions, pH and oxides

(a) An acid is a substance that donates protons, that is hydrogen ions (one mark); a base is a substance that accepts protons (one mark); for example hydrochloric acid donates protons and sodium hydroxide accepts them (one mark). [3 marks]

(b) pH is the negative logarithm of the hydrogen ion concentration (one mark); a difference of one pH unit means a tenfold difference in hydrogen ion concentration (one mark); the concentration at pH 2 is 1000 times that at pH 5 (one mark). [3 marks]

(c) HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) (one mark); the products are sodium chloride (one mark); and water (one mark). [3 marks]

(d) Na2O is a basic oxide (one mark); SO2 is an acidic oxide (one mark); Al2O3 is amphoteric (one mark); CO is a neutral oxide that does not form a salt (one mark). [4 marks]

(e) A strong acid ionises completely in water and gives more ions (one mark); a weak acid ionises only partly so there are fewer ions (one mark); the more ions present, the better the solution conducts (one mark). [3 marks]

Question 2 – Titration procedure and mole calculations

(a) The indicator changes colour at the end point, showing that neutralisation is complete (one mark); phenolphthalein is the usual choice for a strong acid and strong alkali titration (one mark); it is colourless in acid and pink in alkali (one mark). [3 marks]

(b) 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]

(c) The equation is HCl + NaOH → NaCl + H2O, so the ratio is 1 to 1 (one mark); the acid therefore contains 0.00250 mol (one mark); the volume is 20.0 cm3, that is 0.0200 dm3 (one mark); the concentration is 0.125 mol/dm3 (one mark). [4 marks]

(d) 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); the answer is 1.00 g (one mark). [3 marks]

(e) 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]

Question 3 – Preparing a soluble salt: copper(II) sulfate

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

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

(c) The filtrate is heated to evaporate it to about half its volume (one mark); cooling makes the crystals form (one mark); the crystals are filtered off and dried between filter papers (one mark). [3 marks]

(d) The amount of acid is 1.00 multiplied by 0.0500, which is 0.0500 mol (one mark); the equation shows a 1 to 1 ratio, 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]

(e) Percentage yield is the actual mass divided by the theoretical mass, multiplied by 100 (one mark); substituting 6.40 and 8.00 (one mark); the answer is 80.0 per cent (one mark); some crystals dissolve during filtering or are left on the filter paper (one mark). [4 marks]

Question 4 – Insoluble salts, precipitation and ionic equations

(a) Most chlorides are soluble, but silver chloride and lead(II) chloride are not (one mark); most sulfates are soluble, but barium sulfate and lead(II) sulfate are not (one mark); most carbonates are insoluble, but the sodium, potassium and ammonium salts are soluble (one mark). [3 marks]

(b) Na2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2NaCl(aq) (one mark); the net ionic equation is Ba²⁺(aq) + SO4²⁻(aq) → BaSO4(s) (one mark); a white precipitate forms (one mark); the sodium and chloride ions are spectator ions (one mark). [4 marks]

(c) To remove soluble impurities sticking to the surface of the precipitate (one mark); otherwise those impurities would contaminate the product when it dries (one mark); after washing it must be filtered again and dried before weighing (one mark). [3 marks]

(d) The amount of barium chloride is 0.200 multiplied by 0.0500, which is 0.0100 mol (one mark); the ratio is 1 to 1, so 0.0100 mol of barium sulfate forms (one mark); the mass is 0.0100 multiplied by 233 (one mark); the answer is 2.33 g (one mark). [4 marks]

(e) First add dilute hydrochloric acid to acidify the solution (one mark); then add barium chloride solution (one mark); a white precipitate shows sulfate ions are present, and the acid removes interference from carbonate ions (one mark). [3 marks]

Question 5 – Strong and weak acids, pH and dilution

(a) pH is the negative logarithm of the hydrogen ion concentration (one mark); pH 7 means a hydrogen ion concentration of 1 × 10⁻⁷ mol/dm3 (one mark); at 25 degrees Celsius the solution is neutral (one mark). [3 marks]

(b) Hydrochloric acid is a strong acid that 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 (one mark); the pH is 3 (one mark). [4 marks]

(c) The hydrogen ion concentration is 10 raised to the power minus 3.3 (one mark); this works out to about 5.0 × 10⁻⁴ mol/dm3 (one mark); the unit must be given (one mark); a lower pH means a higher hydrogen ion concentration, which agrees with the value (one mark). [4 marks]

(d) A weak acid ionises only partly, so at the same concentration it gives a lower hydrogen ion concentration (one mark); its pH is therefore higher (one mark); with fewer hydrogen ions the reaction with a metal is slower (one mark). [3 marks]

(e) pH 1 gives red (one mark); pH 7 gives green and pH 13 gives purple (one mark). [2 marks]

Question 6 – Ammonia, ammonium salts and fertiliser purity

(a) Ammonia accepts a proton from a water molecule (one mark); ammonium ions and hydroxide ions are formed (one mark); the equation is NH3 + H2O ⇌ NH4⁺ + OH⁻ (one mark). [3 marks]

(b) The amount of acid is 0.0500 multiplied by 0.0250, which is 0.00125 mol (one mark); the equation 2NH3 + H2SO4 → (NH4)2SO4 shows a ratio of 2 to 1 (one mark); the ammonia needed is 2 multiplied by 0.00125 (one mark); the answer is 0.00250 mol (one mark). [4 marks]

(c) Concentration equals moles divided by volume (one mark); 20.0 cm3 is 0.0200 dm3 (one mark); substituting 0.00250 and 0.0200 (one mark); the answer is 0.125 mol/dm3 (one mark). [4 marks]

(d) Percentage purity is 4.60 divided by 5.00, multiplied by 100 (one mark); the answer is 92.0 per cent (one mark); the remaining 8.0 per cent is impurity, so the dose must be corrected for purity (one mark). [3 marks]

(e) Add sodium hydroxide solution and warm the mixture (one mark); a gas with a pungent smell is released (one mark); the gas turns damp red litmus paper blue, which shows it is ammonia (one mark). [3 marks]

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