IGCSE Chemistry Paper 4 (Theory / Extended)

Topic 7: Acids, Bases and Salts -- Mock Exam 1

Duration1 hour 15 minutes

Total Marks80

Questions7

Time Remaining75:00

Syllabus0620

Instructions

Answer all questions in the spaces provided.
Show all working for calculations.
Use appropriate scientific terminology.
Your answers will be automatically graded when you submit.

Question Navigation

Question 1 -- Characteristic Properties of Acids

Total: 12 marks

A student at a school in Kolkata is given four samples: magnesium ribbon, black copper(II) oxide powder, solid sodium carbonate and aqueous sodium hydroxide. She adds dilute hydrochloric acid to each in turn and records what she sees.

(a) [4]

Write a word equation for the reaction of dilute hydrochloric acid with each of the four substances, and state one observation you would make in each case. [4]

Model Answer -- 1(a)

magnesium + hydrochloric acid → magnesium chloride + hydrogen; rapid effervescence, the ribbon dissolves and the mixture warms up [1]

copper(II) oxide + hydrochloric acid → copper(II) chloride + water; the black solid dissolves on warming to give a blue–green solution [1]

sodium carbonate + hydrochloric acid → sodium chloride + water + carbon dioxide; vigorous fizzing and the solid disappears [1]

sodium hydroxide + hydrochloric acid → sodium chloride + water; no visible change but the temperature of the mixture rises [1]

⚠ If you missed marks here: The commonest slip is adding hydrogen to the oxide or hydroxide reactions. Only a metal above hydrogen in the reactivity series releases hydrogen; a base gives a salt and water only. A second frequent loss is writing that carbon dioxide is given off with the hydroxide instead of the carbonate.

1 mark -- magnesium chloride hydrogen effervescence ribbon dissolves 1 mark -- copper oxide chloride water black solid dissolves blue green solution 1 mark -- sodium carbonate chloride water carbon dioxide fizzing 1 mark -- sodium hydroxide chloride water no visible change temperature rises

(b) [4]

(i) State what all acids produce when they dissolve in water, and define a base in terms of protons. [2]

(ii) Write the ionic equation, with state symbols, for the neutralisation of any acid by any alkali. [1]

(iii) Hydrogen chloride gas dissolved in the solvent methylbenzene has no effect on dry litmus paper. Explain why. [1]

Model Answer -- 1(b)

(i) All acids release hydrogen ions, H⁺, into aqueous solution, and it is these ions that give every acid its characteristic reactions [1]

(i) A base is a proton acceptor: it takes an H⁺ ion from an acid; a base that dissolves in water to give OH⁻ ions is called an alkali [1]

(ii) H⁺(aq) + OH⁻(aq) → H₂O(l) — all other ions are spectators [1]

(iii) Methylbenzene is non-polar, so the hydrogen chloride stays as covalent molecules and does not split up; with no free hydrogen ions present the solution shows no acidic properties [1]

⚠ If you missed marks here: Writing that acids produce hydrogen gas confuses the ion with the element — hydrogen gas only appears later when the acid meets a reactive metal. In the ionic equation, marks are lost for omitting state symbols or for including the spectator ions. For part (iii), saying that hydrogen chloride is a base in methylbenzene is wrong; it is simply not ionised.

1 mark -- hydrogen ions aqueous solution characteristic reactions 1 mark -- base proton acceptor alkali soluble hydroxide ions 1 mark -- H OH water state symbols spectator ions 1 mark -- non polar covalent molecules not split free hydrogen ions

Copy and complete a table giving the colour of litmus, methyl orange and thymolphthalein in a dilute acid and in a dilute alkali. Then state which single indicator would be the least useful for judging the exact end point of a titration between hydrochloric acid and sodium hydroxide, and why. [4]

Model Answer -- 1(c)

Litmus: red in acid, blue in alkali (purple when neutral) [1]

Methyl orange: red in acid, yellow in alkali [1]

Thymolphthalein: colourless in acid, blue in alkali [1]

Litmus is the least useful because its colour change is gradual and spreads over a wide pH range, so no sharp end point can be seen; methyl orange and thymolphthalein each change over a narrow range and give a distinct single-drop change [1]

⚠ If you missed marks here: Many candidates reverse the acid and alkali colours of methyl orange — remember that its alkaline colour is yellow, matching the yellow end of the universal indicator acid range only by coincidence. Writing that thymolphthalein is pink confuses it with phenolphthalein.

1 mark -- litmus red acid blue alkali 1 mark -- methyl orange red acid yellow alkali 1 mark -- thymolphthalein colourless acid blue alkali 1 mark -- litmus gradual wide range no sharp end point narrow

Question 2 -- The pH Scale and Acid Strength

Total: 12 marks

A laboratory technician in Manchester prepares four solutions, each of concentration 0.10 mol/dm³: hydrochloric acid, ethanoic acid, sodium hydroxide and aqueous ammonia. She measures the pH of each with a calibrated pH meter and obtains the values 1.0, 2.9, 11.1 and 13.0, but the labels have fallen off.

(a) [4]

Match each of the four solutions to its pH value, and explain your reasoning for the two acids. [4]

Model Answer -- 2(a)

Hydrochloric acid is pH 1.0 and ethanoic acid is pH 2.9 [1]

Sodium hydroxide is pH 13.0 and aqueous ammonia is pH 11.1 [1]

Hydrochloric acid is a strong acid, fully dissociated, so at 0.10 mol/dm³ the hydrogen ion concentration is also 0.10 mol/dm³ and the pH is 1 [1]

Ethanoic acid is a weak acid and only partially dissociates, so its hydrogen ion concentration is far lower at the same concentration and its pH is correspondingly higher [1]

⚠ If you missed marks here: If you matched the acids by concentration alone you missed the whole point — the concentrations are identical, so only the degree of dissociation can explain the difference. Do not say that ethanoic acid is more dilute; it is equally concentrated but weaker.

1 mark -- hydrochloric 1.0 ethanoic 2.9 1 mark -- sodium hydroxide 13.0 ammonia 11.1 1 mark -- strong fully dissociated hydrogen ion concentration equals pH 1 1 mark -- weak partially dissociates lower hydrogen ion concentration higher pH

(b) [5]

(i) Write equations, using the correct type of arrow in each case, for the dissociation of hydrochloric acid and of ethanoic acid in water. Explain your choice of arrows. [3]

(ii) Excess magnesium ribbon is added to 50 cm³ of each acid. Compare the rate of fizzing and the total volume of hydrogen produced, and justify each comparison. [2]

Model Answer -- 2(b)

(i) HCl(aq) → H⁺(aq) + Cl⁻(aq) using a single arrow because dissociation is complete [1]

(i) CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) using a reversible arrow [1]

(i) The reversible arrow shows an equilibrium lying well to the left, so at any moment most of the ethanoic acid is present as undissociated molecules [1]

(ii) The hydrochloric acid fizzes faster because its higher hydrogen ion concentration gives more frequent successful collisions with the magnesium surface [1]

(ii) The total volume of hydrogen is the same in both, because the two acids contain the same number of moles of acid and the weak acid keeps dissociating as its hydrogen ions are used up [1]

⚠ If you missed marks here: A very common error is to claim that the weak acid produces less hydrogen in total. It produces the same amount, only more slowly, because dissociation continues until all the acid has reacted. Using a single arrow for ethanoic acid throws away an easy mark.

1 mark -- HCl single arrow complete dissociation chloride 1 mark -- ethanoic reversible arrow ethanoate equilibrium 1 mark -- equilibrium left undissociated molecules most 1 mark -- hydrochloric faster higher hydrogen ion concentration frequent collisions 1 mark -- same total volume moles acid weak keeps dissociating

Describe an experiment, other than measuring pH, that would show clearly that hydrochloric acid is a stronger acid than ethanoic acid. State the variables you would control and the result you would expect. [3]

Model Answer -- 2(c)

Suitable method: connect each solution in turn into a simple circuit with a lamp and an ammeter and compare the current, or measure the time for a fixed mass of magnesium to disappear in each acid [1]

Control variables: the same concentration of each acid, the same volume, the same temperature, and the same mass, surface area and form of magnesium (or the same electrodes at the same separation) [1]

Expected result: the hydrochloric acid gives the larger current (brighter lamp) or the shorter reaction time, because it contains a far greater concentration of mobile ions and of hydrogen ions [1]

⚠ If you missed marks here: Answers that say only compare them and see which is stronger score nothing — the examiner needs a measurable quantity (current, time, temperature rise) and a stated expected outcome. Forgetting to state that the concentrations must be equal invalidates the whole comparison, because otherwise concentration and strength are confounded.

1 mark -- conductivity circuit ammeter lamp current time magnesium disappear 1 mark -- control same concentration volume temperature mass surface area 1 mark -- hydrochloric larger current shorter time greater mobile ions

Question 3 -- Classifying Oxides

Total: 12 marks

A group of students in Johannesburg is given five oxides labelled A to E: sodium oxide, zinc oxide, sulfur dioxide, carbon monoxide and aluminium oxide. They test each one with dilute hydrochloric acid and with hot concentrated sodium hydroxide solution, and where possible they shake it with water and measure the pH.

(a) [4]

Classify each of the five oxides as acidic, basic, amphoteric or neutral, and in each case state the experimental evidence from the tests described that supports your classification. [4]

Model Answer -- 3(a)

Sodium oxide is basic: it reacts with the acid to give a salt and water, has no reaction with the alkali, and dissolves in water to give a high pH of about 13 [1]

Sulfur dioxide is acidic: it has no reaction with the acid, reacts with the sodium hydroxide to give a salt, and dissolves in water to give a low pH [1]

Zinc oxide and aluminium oxide are both amphoteric: each dissolves in the acid to give a salt and water and also dissolves in hot concentrated alkali, giving a zincate or an aluminate [1]

Carbon monoxide is neutral: it reacts with neither the acid nor the alkali and does not change the pH of water [1]

⚠ If you missed marks here: Students often call an amphoteric oxide neutral because it seems to sit in the middle. The two are opposites in behaviour: amphoteric means it reacts with both an acid and an alkali, neutral means it reacts with neither. Evidence must be quoted from the tests, not just asserted.

1 mark -- sodium oxide basic reacts acid salt water pH 13 1 mark -- sulfur dioxide acidic reacts sodium hydroxide salt low pH 1 mark -- zinc aluminium amphoteric both acid alkali zincate aluminate 1 mark -- carbon monoxide neutral neither reaction unchanged pH

(b) [4]

(i) Write a balanced symbol equation for the reaction of zinc oxide with dilute hydrochloric acid. [1]

(ii) Zinc oxide also dissolves in hot concentrated sodium hydroxide to form sodium zincate, Na₂ZnO₂, and water. Write the balanced equation. [1]

(iii) Explain what the two reactions together tell you about zinc oxide, and name one other oxide that behaves in the same way. [2]

Model Answer -- 3(b)

(i) ZnO + 2HCl → ZnCl₂ + H₂O [1]

(ii) ZnO + 2NaOH → Na₂ZnO₂ + H₂O [1]

(iii) Because it forms a salt with an acid (acting as a base) and also forms a salt with an alkali (acting as an acidic oxide), zinc oxide is amphoteric [1]

(iii) Aluminium oxide, Al₂O₃, behaves in the same way, dissolving in acids to give aluminium salts and in hot alkali to give aluminates [1]

⚠ If you missed marks here: Unbalanced equations are the biggest loss here — check that you have two HCl and two NaOH. Naming a hydroxide such as sodium hydroxide as the second amphoteric example is wrong; you need an amphoteric oxide such as aluminium oxide (or the amphoteric hydroxides of zinc and aluminium if hydroxides are asked for).

1 mark -- ZnO 2HCl ZnCl2 H2O balanced 1 mark -- ZnO 2NaOH Na2ZnO2 H2O balanced 1 mark -- forms salt acid base salt alkali acidic amphoteric 1 mark -- aluminium oxide Al2O3 aluminates same behaviour

Sulfur dioxide released by burning high-sulfur coal has damaged the marble surfaces of historic buildings in Agra. Explain, with the help of equations, how an acidic oxide causes this damage, and suggest one method of reducing the sulfur dioxide released from a power station. [4]

Model Answer -- 3(c)

Sulfur dioxide is a non-metal oxide and therefore acidic; it dissolves in rainwater to form sulfurous acid, SO₂ + H₂O → H₂SO₃, and further oxidation gives sulfuric acid [1]

The resulting rain has a pH well below the natural value of about 5.6, so it is described as acid rain [1]

Marble is calcium carbonate, and acid + carbonate gives salt + water + carbon dioxide: CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂, so the stone is eaten away and the surface detail is lost [1]

Reduction method: pass the flue gases through a spray of calcium oxide or calcium carbonate slurry (flue gas desulfurisation), which neutralises the acidic oxide, or remove sulfur from the fuel before burning [1]

⚠ If you missed marks here: Writing that the acid rain simply dissolves the marble misses the chemistry mark; marble is insoluble in pure water and it is the neutralisation reaction with the acid that destroys it. Suggesting that a tall chimney solves the problem is not accepted — that only moves the pollution elsewhere.

1 mark -- sulfur dioxide acidic dissolves rainwater sulfurous sulfuric acid 1 mark -- acid rain pH below natural 5.6 1 mark -- marble calcium carbonate sulfate water carbon dioxide eaten away 1 mark -- flue gas desulfurisation calcium oxide carbonate remove sulfur fuel

Question 4 -- Preparing a Soluble Salt from an Insoluble Base

Total: 12 marks

A student in Colombo is asked to prepare a pure dry sample of hydrated copper(II) sulfate crystals, CuSO₄·5H₂O, starting from dilute sulfuric acid.

(a) [3]

(i) The student first suggests adding copper metal to the dilute sulfuric acid. Explain why this will not work. [1]

(ii) Name a suitable copper compound to use instead, and state why the excess-solid method is appropriate for it. [2]

Model Answer -- 4(a)

(i) Copper lies below hydrogen in the reactivity series, so it cannot displace hydrogen from the acid and no reaction occurs [1]

(ii) Copper(II) oxide (or copper(II) carbonate, or copper(II) hydroxide) should be used, because it is a base that reacts readily with the warm acid to give copper(II) sulfate [1]

(ii) The compound is insoluble in water, so an excess can be added to make certain that all the acid is used up and the unreacted surplus can then simply be filtered off [1]

⚠ If you missed marks here: Saying copper is unreactive is too vague — you must refer to its position below hydrogen in the reactivity series. If you chose copper(II) sulfate itself as the starting solid you have assumed the product, which gains no credit.

1 mark -- copper below hydrogen reactivity series cannot displace no reaction 1 mark -- copper oxide carbonate hydroxide base reacts warm acid sulfate 1 mark -- insoluble excess added all acid used filtered off

(b) [5]

Describe, in a numbered sequence of steps, how the student should obtain pure dry crystals of hydrated copper(II) sulfate. Include how she knows when to stop adding the solid, and state the purpose of each stage. [5]

Model Answer -- 4(b)

Warm the dilute sulfuric acid gently in a beaker (this speeds up the reaction, but do not boil it) [1]

Add copper(II) oxide a little at a time, stirring, until some solid remains undissolved even after stirring — this shows that all the acid has been neutralised and the acid is the limiting reactant [1]

Filter the mixture; the unreacted excess copper(II) oxide is left as residue on the filter paper and the blue copper(II) sulfate solution passes through as the filtrate [1]

Heat the filtrate in an evaporating basin until it is saturated, testing by dipping a cold glass rod in and looking for crystals forming on it; do not evaporate to dryness [1]

Leave the saturated solution to cool slowly so that large well-formed crystals grow, then pour off the mother liquor and dry the crystals between filter papers or in a warm oven [1]

⚠ If you missed marks here: Two marks are commonly thrown away here. Evaporating to dryness destroys the water of crystallisation and leaves a white powder, not blue crystals. Drying in a hot oven or over a Bunsen does the same, so the drying must be gentle. You must also make clear how you know when to stop adding solid.

1 mark -- warm dilute sulfuric acid beaker gently not boil 1 mark -- add copper oxide excess stirring until solid remains undissolved acid used 1 mark -- filter residue unreacted excess filtrate blue solution 1 mark -- evaporate saturated glass rod crystals not dryness 1 mark -- cool slowly large crystals dry filter papers warm

(i) Write a balanced symbol equation, with state symbols, for the reaction between copper(II) oxide and dilute sulfuric acid. [2]

(ii) The student calculates a theoretical yield of 6.24 g of hydrated crystals but obtains only 5.30 g. Calculate the percentage yield and suggest one reason, other than a chemical error, for the shortfall. [2]

Model Answer -- 4(c)

(i) CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l) [1]

(i) State symbols correct throughout: solid oxide, aqueous acid, aqueous salt, liquid water [1]

(ii) Percentage yield = 5.30/6.24 × 100 = 84.9 % (accept 85 %) [1]

(ii) Reason: some product remains dissolved in the mother liquor poured away, or some solution is left behind on the filter paper and glassware, or crystals are lost when scraped from the basin [1]

⚠ If you missed marks here: Dividing the wrong way round gives a yield above 100 %, which is impossible — always sanity-check the answer. Suggesting that some mass was destroyed in the reaction is never accepted, because mass is conserved; every loss must be a physical transfer loss.

1 mark -- CuO H2SO4 CuSO4 H2O balanced 1 mark -- state symbols solid aqueous liquid 1 mark -- 5.30 6.24 100 84.9 85 percentage yield 1 mark -- product left mother liquor filter paper glassware crystals lost

Question 5 -- Titration and Concentration Calculations

Total: 10 marks

A student in Dubai prepares a soluble salt by the titration route. She pipettes 25.0 cm³ of aqueous sodium hydroxide into a conical flask, adds two drops of methyl orange, and runs dilute hydrochloric acid of concentration 0.150 mol/dm³ from a burette until the colour changes. Her concordant titres are 22.40 cm³, 22.45 cm³ and 22.35 cm³.

(a) [4]

(i) Name the piece of apparatus used to measure the 25.0 cm³ of alkali accurately, and state why a measuring cylinder would not be suitable. [2]

(ii) State the colour change seen at the end point. [1]

(iii) Explain why the titration is repeated until concordant titres are obtained. [1]

Model Answer -- 5(a)

(i) A (volumetric) pipette, used with a pipette filler [1]

(i) A measuring cylinder is graduated too coarsely and has a much larger uncertainty, so it cannot deliver a fixed volume to the nearest 0.05 cm³ as the calculation requires [1]

(ii) Methyl orange changes from yellow in the alkali to red as soon as the acid is in the smallest excess (an orange shade is seen exactly at the end point) [1]

(iii) Repeating and using only titres that agree within 0.10 cm³ shows the results are reliable and allows anomalous readings, including the rough titration, to be discarded before the mean is calculated [1]

⚠ If you missed marks here: Writing burette for the 25 cm³ alkali confuses the two roles: the burette measures the variable volume of the solution being added, the pipette the fixed volume in the flask. Saying the colour goes red to yellow reverses the direction, because the flask starts alkaline.

1 mark -- pipette volumetric filler 1 mark -- measuring cylinder coarse large uncertainty cannot fixed volume 1 mark -- yellow alkali red acid excess orange end point 1 mark -- concordant agree 0.10 reliable discard anomalous rough mean

(b) [4]

(i) Calculate the mean titre from the three concordant results. [1]

(ii) Calculate the number of moles of hydrochloric acid used. [1]

(iii) Using the equation NaOH + HCl → NaCl + H₂O, calculate the concentration of the sodium hydroxide in mol/dm³ and then in g/dm³. (Aₕ: Na 23, O 16, H 1) [2]

Model Answer -- 5(b)

(i) Mean titre = (22.40 + 22.45 + 22.35)/3 = 22.40 cm³ [1]

(ii) Moles HCl = concentration × volume in dm³ = 0.150 × 22.40/1000 = 3.36 × 10⁻³ mol [1]

(iii) The ratio is 1 : 1, so moles NaOH = 3.36 × 10⁻³ mol; concentration = 3.36 × 10⁻³ ÷ (25.0/1000) = 0.1344 mol/dm³, which rounds to 0.134 mol/dm³ [1]

(iii) Mₕ(NaOH) = 23 + 16 + 1 = 40, so concentration = 0.1344 × 40 = 5.38 g/dm³ [1]

⚠ If you missed marks here: The single biggest source of lost marks is failing to convert cm³ to dm³ by dividing by 1000; a factor-of-1000 error usually shows up as a concentration in the hundreds. When converting to g/dm³ you multiply by the relative formula mass — dividing gives an absurdly small figure.

1 mark -- mean titre 22.40 average three 1 mark -- moles HCl 0.150 22.40 1000 3.36 1 mark -- ratio 1 1 divide 0.0250 0.134 mol 1 mark -- Mr 40 multiply 5.38 g dm3

Explain how the student would now obtain a pure dry sample of sodium chloride from this reaction, and state why the sample would be unusable if she simply evaporated the contents of the conical flask. [2]

Model Answer -- 5(c)

Repeat the experiment using exactly 25.0 cm³ of the alkali and 22.40 cm³ of the acid but with no indicator added, then evaporate the solution to saturation and leave it to crystallise [1]

The mixture in the conical flask contains methyl orange, which would remain in the solid and colour it, so the salt would not be pure [1]

⚠ If you missed marks here: Answers that say just filter the mixture score nothing — sodium chloride is soluble, so there is nothing to filter. The key idea is that the exact volumes are now known, so the titration can be repeated without the contaminating indicator.

1 mark -- repeat same volumes without indicator evaporate crystallise 1 mark -- methyl orange remains colours solid not pure

Question 6 -- Solubility Rules and Insoluble Salts

Total: 12 marks

A technician in Auckland has bottles of aqueous barium chloride, sodium sulfate, lead(II) nitrate, potassium iodide, silver nitrate and sodium chloride. She is asked to prepare a dry sample of barium sulfate for use as a standard, and to predict what happens when various pairs of the other solutions are mixed.

(a) [4]

State whether each of the following is soluble or insoluble in water, and in each case quote the solubility rule that you used: (i) potassium carbonate; (ii) calcium carbonate; (iii) silver chloride; (iv) ammonium sulfate. [4]

Model Answer -- 6(a)

(i) Potassium carbonate is soluble — all sodium, potassium and ammonium salts are soluble without exception [1]

(ii) Calcium carbonate is insoluble — all carbonates are insoluble except those of sodium, potassium and ammonium [1]

(iii) Silver chloride is insoluble — all chlorides are soluble except silver chloride and lead(II) chloride [1]

(iv) Ammonium sulfate is soluble — all ammonium salts are soluble, and in any case most sulfates are soluble [1]

⚠ If you missed marks here: The rule must be quoted, not just the answer. A very common error is to forget that the sodium, potassium and ammonium rule overrides every other rule, so ammonium carbonate and potassium sulfide are both soluble even though most carbonates and sulfides are not.

1 mark -- potassium carbonate soluble sodium ammonium all salts 1 mark -- calcium carbonate insoluble carbonates except sodium potassium ammonium 1 mark -- silver chloride insoluble chlorides except silver lead 1 mark -- ammonium sulfate soluble ammonium salts sulfates most

(b) [4]

(i) Name the two solutions from the technician's bottles that should be mixed to prepare barium sulfate. [1]

(ii) Describe how the dry solid would be obtained from the mixture, and give a reason for each step. [2]

(iii) Write the ionic equation, with state symbols, for the precipitation. [1]

Model Answer -- 6(b)

(i) Aqueous barium chloride mixed with aqueous sodium sulfate [1]

(ii) Filter the mixture to separate the insoluble barium sulfate from the solution of the soluble by-product sodium chloride [1]

(ii) Wash the residue on the filter paper with distilled water to rinse away the soluble spectator ions clinging to it, then dry it in a warm oven or between filter papers [1]

(iii) Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s) [1]

⚠ If you missed marks here: Washing with tap water instead of distilled water contaminates the sample with dissolved calcium and chloride ions. In the ionic equation, marks are lost for including the sodium and chloride spectator ions or for omitting the state symbols that show two solutions giving a solid.

1 mark -- barium chloride sodium sulfate mixed 1 mark -- filter separate insoluble residue soluble sodium chloride solution 1 mark -- wash distilled water spectator ions dry warm oven filter papers 1 mark -- Ba2 SO4 BaSO4 aq s ionic

For each of the following mixtures, state whether a precipitate forms and name it if it does: (i) lead(II) nitrate + potassium iodide; (ii) silver nitrate + sodium chloride; (iii) sodium chloride + potassium iodide; (iv) barium chloride + sodium chloride. [4]

Model Answer -- 6(c)

(i) A bright yellow precipitate of lead(II) iodide forms, because all iodides are soluble except those of lead and silver [1]

(ii) A white precipitate of silver chloride forms, which darkens in sunlight; it is one of the two insoluble chlorides [1]

(iii) No precipitate: the possible products, sodium iodide and potassium chloride, are both soluble [1]

(iv) No precipitate: barium chloride and sodium chloride are both soluble, so swapping the partners produces nothing new that is insoluble [1]

⚠ If you missed marks here: Candidates often write barium sulfate for part (iv) out of habit — check that a sulfate ion is actually present before predicting it. When no precipitate forms you must say so explicitly and explain that both possible products are soluble; leaving the space blank scores nothing.

1 mark -- lead iodide yellow precipitate iodides except lead silver 1 mark -- silver chloride white precipitate darkens sunlight 1 mark -- no precipitate sodium iodide potassium chloride both soluble 1 mark -- no precipitate barium sodium chlorides soluble nothing insoluble

Question 7 -- Water of Crystallisation

Total: 10 marks

A student in Lagos heats 5.00 g of blue hydrated copper(II) sulfate crystals, CuSO₄·xH₂O, in a crucible until the colour has completely changed and the mass no longer falls. The mass of the residue is 3.20 g. (Aₕ: Cu 64, S 32, O 16, H 1)

(a) [4]

(i) Explain what is meant by water of crystallisation. [1]

(ii) State the colour change seen during the heating and write the equation for the change, using a reversible arrow. [2]

(iii) Explain why the student heats the crucible until the mass no longer falls. [1]

Model Answer -- 7(a)

(i) Water of crystallisation is water that is chemically combined into the crystal lattice of a salt in a fixed molar ratio, shown after a dot in the formula [1]

(ii) The blue crystals turn white as the water is driven off [1]

(ii) CuSO₄·5H₂O(s) ⇌ CuSO₄(s) + 5H₂O(l) — adding water to the white solid turns it blue again and releases heat [1]

(iii) Heating to constant mass shows that all the water of crystallisation has been driven off; if any remained the mass would still be falling and the calculated value of x would be too small [1]

⚠ If you missed marks here: Describing water of crystallisation as water the crystals are wet with loses the mark — surface moisture is not chemically combined and is not in a fixed ratio. The reverse reaction turning white copper(II) sulfate blue is the standard test for the presence of water, so learn it in both directions.

1 mark -- water chemically combined crystal lattice fixed ratio dot formula 1 mark -- blue crystals turn white water driven off 1 mark -- CuSO4 5H2O reversible arrow anhydrous returns blue 1 mark -- constant mass all water removed otherwise x too small

(b) [4]

Use the results to calculate the value of x in CuSO₄·xH₂O. Show all your working clearly. [4]

Model Answer -- 7(b)

Mass of water lost = 5.00 − 3.20 = 1.80 g [1]

Mₕ(CuSO₄) = 64 + 32 + (4 × 16) = 160, so moles of anhydrous salt = 3.20/160 = 0.0200 mol [1]

Mₕ(H₂O) = 18, so moles of water = 1.80/18 = 0.100 mol [1]

Ratio CuSO₄ : H₂O = 0.0200 : 0.100 = 1 : 5, therefore x = 5 and the formula is CuSO₄·5H₂O [1]

⚠ If you missed marks here: The classic error is to use the original 5.00 g as the mass of anhydrous salt; the residue left in the crucible is the anhydrous salt and the difference is the water. Dividing the smaller number of moles by the larger inverts the ratio and gives a fractional answer such as 0.2, which should immediately look wrong.

1 mark -- mass water lost 5.00 3.20 1.80 1 mark -- Mr CuSO4 160 moles 3.20 0.0200 1 mark -- Mr water 18 moles 1.80 0.100 1 mark -- ratio 1 5 x equals five formula

Explain why crystals of a hydrated salt prepared in the laboratory must be dried gently between filter papers rather than in a hot oven, and state what would be observed if this precaution were ignored. [2]

Model Answer -- 7(c)

Strong heating would drive off the water of crystallisation, changing the chemical composition of the salt and lowering its mass, so the product would no longer be the hydrated compound required [1]

The blue crystals would turn to a white or very pale powder, and the sample would lose its regular crystalline shape [1]

⚠ If you missed marks here: Saying the crystals would melt or burn is not accepted. The change is dehydration, not decomposition of the sulfate, and the visible evidence is the loss of the blue colour and of the crystal shape.

1 mark -- strong heating drives water crystallisation composition mass changes 1 mark -- blue crystals white pale powder lose crystalline shape

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