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Question 1 -- Acids, Bases and Neutralisation
Total: 12 marks
A pharmacist in Hyderabad sells two indigestion remedies. Remedy P contains magnesium hydroxide, Mg(OH)₂, and Remedy Q contains sodium hydrogencarbonate, NaHCO₃. Both are taken to relieve the effects of excess hydrochloric acid in the stomach, where the pH can fall to about 1.5.
(a)[4]
(i) Define an acid and a base in terms of protons. [2]
(ii) Explain the difference between a base and an alkali, giving one example of each that is not mentioned in the stem. [2]
Model Answer -- 1(a)
(i) An acid is a proton donor: in aqueous solution it releases hydrogen ions, H⁺ [1]
(i) A base is a proton acceptor: it removes hydrogen ions from an acid, usually forming water [1]
(ii) An alkali is a base that is soluble in water, so it releases hydroxide ions into solution; all alkalis are bases but most bases are not alkalis [1]
(ii) Suitable examples: copper(II) oxide or iron(III) oxide as an insoluble base; potassium hydroxide or aqueous ammonia as an alkali [1]
⚠ If you missed marks here: Defining an acid as something with a low pH describes a property rather than giving a definition. Also, saying that a base is the opposite of an acid gains nothing — you must mention proton transfer. Choosing sodium hydroxide as the insoluble base example is self-contradictory.
(b)[4]
(i) Write a balanced symbol equation for the reaction of each remedy with hydrochloric acid. [2]
(ii) State one observation that would distinguish the two reactions if they were carried out in test-tubes. [1]
(iii) State the name of the type of reaction taking place. [1]
Model Answer -- 1(b)
(i) Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O [1]
(i) NaHCO₃ + HCl → NaCl + H₂O + CO₂ [1]
(ii) Only Remedy Q effervesces, because the hydrogencarbonate releases carbon dioxide, which would turn limewater milky; Remedy P dissolves quietly with no gas [1]
(iii) Neutralisation (an acid–base reaction) in both cases [1]
⚠ If you missed marks here: The hydroxide equation is frequently left unbalanced — two HCl are needed for the two OH groups. Do not write that the hydroxide also gives carbon dioxide; only carbonates and hydrogencarbonates can do that, because only they contain carbon.
(c)[4]
A student models the stomach by adding Remedy P a little at a time to 50 cm³ of dilute hydrochloric acid of pH 1.5, measuring the pH after each addition. Sketch in words how the pH changes, explain the shape of the change in terms of ions, and explain why universal indicator would be a poor choice for this investigation. [4]
Model Answer -- 1(c)
The pH rises slowly at first, then climbs steeply as the last of the acid is neutralised, and finally levels off at a mildly alkaline value as excess magnesium hydroxide is present [1]
The rise happens because hydroxide ions from the remedy remove hydrogen ions from the acid, H⁺(aq) + OH⁻(aq) → H₂O(l), lowering the hydrogen ion concentration [1]
The final value does not rise as far as it would with sodium hydroxide because magnesium hydroxide is only very slightly soluble, so few extra hydroxide ions enter the solution [1]
Universal indicator gives only a colour that has to be matched by eye to the nearest whole pH unit, and the suspension of undissolved solid makes the colour hard to judge; a pH meter gives a precise numerical value [1]
⚠ If you missed marks here: Answers that describe the pH as falling have confused which reactant is in the flask. Note also that the curve for a sparingly soluble base flattens off around pH 9 to 10 rather than approaching 14 — a point worth making explicitly for the third mark.
Question 2 -- Strength, Concentration and Conductivity
Total: 12 marks
A student in Leeds investigates four solutions, each of concentration 0.100 mol/dm³, and records the pH and the current that flows when the same pair of carbon electrodes is dipped into each and connected to a 6 V supply.
Solution
pH
Current / mA
hydrochloric acid
1.0
62
sulfuric acid
0.8
110
ethanoic acid
2.9
6
aqueous ammonia
11.1
5
(a)[4]
(i) Explain why hydrochloric acid and ethanoic acid have very different pH values even though their concentrations are identical. [2]
(ii) Explain why sulfuric acid gives roughly twice the current of hydrochloric acid. [1]
(iii) State what the low current for aqueous ammonia tells you about it. [1]
Model Answer -- 2(a)
(i) Hydrochloric acid is a strong acid and dissociates completely, so essentially all of the 0.100 mol/dm³ is present as hydrogen ions [1]
(i) Ethanoic acid is a weak acid and dissociates only partially, so its hydrogen ion concentration is roughly a hundred times lower, giving a pH about two units higher [1]
(ii) Sulfuric acid is dibasic, releasing two hydrogen ions per molecule as well as the sulfate ion, so at the same concentration it provides about twice as many mobile ions to carry the current [1]
(iii) The low current shows that aqueous ammonia contains few ions, so it is a weak base that is only partially converted to ammonium and hydroxide ions [1]
⚠ If you missed marks here: Do not say that ethanoic acid is more dilute; the concentrations are stated as equal. For the sulfuric acid mark you must refer to it being dibasic (two replaceable hydrogen atoms), not simply to it being stronger.
(b)[4]
(i) Write the equation for the dissociation of ethanoic acid in water, using the correct arrow, and explain what the arrow tells you. [2]
(ii) Write the equation for the reaction of ammonia with water, and use it to explain why aqueous ammonia turns red litmus blue. [2]
Model Answer -- 2(b)
(i) CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) [1]
(i) The reversible arrow shows that the dissociation is incomplete and reaches an equilibrium lying far to the left, so most of the acid remains as intact molecules [1]
(ii) NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq) [1]
(ii) Ammonia accepts a proton from water, releasing hydroxide ions; these make the solution alkaline, and litmus is blue in alkaline conditions [1]
⚠ If you missed marks here: Writing ammonia as NH₄OH is outdated and is not credited; show ammonia reacting with water instead. Using a single arrow for either equation loses the equilibrium mark, since both species are weak.
(c)[4]
Limescale (calcium carbonate) builds up inside kettles. A household descaler contains ethanoic acid; an industrial descaler contains hydrochloric acid. Using ideas about acid strength, explain the advantage and the disadvantage of each choice, and write an ionic equation for the removal of limescale. [4]
Model Answer -- 2(c)
Hydrochloric acid works much faster because its high hydrogen ion concentration gives frequent successful collisions with the carbonate surface [1]
However, that same high hydrogen ion concentration makes it corrosive and hazardous for household use, and it can attack the metal of the kettle itself [1]
Ethanoic acid is much safer to handle and store because it only partially dissociates, but it works more slowly and may need to be left in contact for longer or warmed [1]
⚠ If you missed marks here: A frequent error is to claim that the weak acid cannot dissolve limescale at all; it can, given time, because as its hydrogen ions are used up more molecules dissociate. In the ionic equation the calcium carbonate must be shown as a solid, since it is insoluble and does not exist as separate ions in the kettle.
Question 3 -- Oxides and the Periodic Table
Total: 12 marks
The oxides of the Period 3 elements sodium, magnesium, aluminium, silicon, phosphorus and sulfur show a clear pattern in their acid–base behaviour. A student in Singapore investigates several of them.
(a)[4]
(i) Describe how the acid–base character of the oxides changes across Period 3, naming an example of each type. [3]
(ii) Explain how this trend is related to the change in character of the elements themselves. [1]
Model Answer -- 3(a)
(i) On the left the oxides are basic: sodium oxide and magnesium oxide react with acids to give salts and water, and dissolve to give alkaline solutions [1]
(i) In the middle aluminium oxide is amphoteric, reacting with both dilute acids and hot concentrated alkalis [1]
(i) On the right the oxides are acidic: silicon(IV) oxide, phosphorus(V) oxide and sulfur dioxide react with alkalis to give salts, and the soluble ones give acidic solutions [1]
(ii) The elements themselves change from metallic on the left to non-metallic on the right; metal oxides are basic and non-metal oxides are acidic, with the borderline element aluminium giving the amphoteric oxide [1]
⚠ If you missed marks here: Simply writing basic to acidic without examples earns only part of the credit. Note also that silicon(IV) oxide is acidic even though it is insoluble in water — classification depends on reaction with acids and alkalis, not just on the pH of a solution.
(b)[4]
(i) Write balanced equations for the reaction of aluminium oxide with dilute sulfuric acid and with hot concentrated sodium hydroxide (which gives sodium aluminate, NaAlO₂, and water). [2]
(ii) Aqueous sodium hydroxide is added drop by drop to aluminium sulfate solution until it is in excess. Describe and explain what is seen. [2]
Model Answer -- 3(b)
(i) Al₂O₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂O [1]
(i) Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O [1]
(ii) A white precipitate of aluminium hydroxide forms first, because hydroxide ions react with aluminium ions: Al³⁺ + 3OH⁻ → Al(OH)₃ [1]
(ii) The precipitate then dissolves in the excess alkali to give a colourless solution, because aluminium hydroxide is amphoteric and reacts with the extra hydroxide ions to form the soluble aluminate ion [1]
⚠ If you missed marks here: Balancing the sulfuric acid equation trips up many candidates: aluminium sulfate is Al₂(SO₄)₃, so three molecules of acid and three of water are needed. In part (ii), stopping at the white precipitate loses the second mark — the redissolving in excess is the whole point.
(c)[4]
(i) Name two neutral oxides and explain what makes an oxide neutral. [2]
(ii) Carbon monoxide is a neutral oxide but is a serious pollutant, while carbon dioxide is an acidic oxide. Explain the difference in their behaviour with aqueous sodium hydroxide, and state why carbon monoxide is dangerous. [2]
Model Answer -- 3(c)
(i) Two neutral oxides: carbon monoxide (CO), dinitrogen oxide (N₂O), nitrogen monoxide (NO) or water itself [1]
(i) A neutral oxide reacts with neither acids nor alkalis to form salts, and if it dissolves in water it leaves the pH at 7 [1]
(ii) Carbon dioxide reacts with sodium hydroxide to form sodium carbonate and water, whereas carbon monoxide has no reaction at all with the alkali [1]
(ii) Carbon monoxide is toxic because it binds to haemoglobin in the blood in place of oxygen, so the blood can no longer carry enough oxygen around the body [1]
⚠ If you missed marks here: Naming an amphoteric oxide such as zinc oxide as a neutral oxide is a serious confusion: amphoteric means it reacts with both acid and alkali, whereas neutral means it reacts with neither. Do not say carbon monoxide is dangerous because it is an acid — it is not.
Question 4 -- Choosing a Method of Salt Preparation
Total: 12 marks
A technician in Nairobi has to prepare four salts for a practical class: zinc sulfate, sodium nitrate, silver chloride and copper(II) nitrate. She has available dilute sulfuric, nitric and hydrochloric acids, a range of metals, metal oxides and carbonates, and solutions of sodium chloride and silver nitrate.
(a)[4]
For each of the four salts, state which of the three general methods (excess solid with an acid, titration, or precipitation) should be used, and give the reactants you would choose. [4]
Model Answer -- 4(a)
Zinc sulfate: excess-solid method, using zinc metal (or zinc oxide or zinc carbonate) with dilute sulfuric acid, because zinc is above hydrogen in the reactivity series and the oxide and carbonate are insoluble [1]
Sodium nitrate: titration, using aqueous sodium hydroxide (or sodium carbonate) with dilute nitric acid, because all sodium compounds are soluble so no excess could be filtered off [1]
Silver chloride: precipitation, by mixing aqueous silver nitrate with aqueous sodium chloride, because silver chloride is insoluble [1]
Copper(II) nitrate: excess-solid method, using copper(II) oxide or copper(II) carbonate with dilute nitric acid, because copper metal itself will not react with a dilute acid [1]
⚠ If you missed marks here: Choosing copper metal with nitric acid is a classic trap: copper is below hydrogen, so the ordinary metal + acid route fails and an oxide or carbonate must be used. Choosing the excess-solid route for sodium nitrate is equally wrong, because nothing would be left to filter.
(b)[4]
(i) Write the balanced symbol equation, with state symbols, for the reaction of zinc with dilute sulfuric acid. [2]
(ii) Describe how the technician would know when to stop adding zinc, and explain why any unreacted zinc must be removed before crystallisation. [2]
Model Answer -- 4(b)
(i) Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g) [1]
(i) State symbols correct: solid metal, aqueous acid, aqueous salt and gaseous hydrogen [1]
(ii) Stop when the effervescence ceases and unreacted zinc remains in the beaker even after warming and stirring, which shows all the acid has been used up [1]
(ii) The excess zinc is insoluble and would be left mixed with the crystals, so the product would be impure; it is removed by filtration and the filtrate is then crystallised [1]
⚠ If you missed marks here: Forgetting the state symbols, especially the (g) on hydrogen, costs an easy mark. Saying you stop when the zinc has all dissolved describes the opposite situation — if the metal fully dissolves you cannot be sure the acid has been fully consumed.
(c)[4]
The technician tests her zinc sulfate crystals. Explain what she would observe, and what she should conclude, if (i) the crystals turn damp blue litmus paper red when moistened; (ii) the crystals give a small amount of insoluble grey residue when dissolved in distilled water. In each case suggest how the preparation should be improved. [4]
Model Answer -- 4(c)
(i) A red colour shows that unreacted sulfuric acid is still present, so the salt is contaminated with acid [1]
(i) The remedy is to add more zinc (or zinc oxide) and continue warming until solid clearly remains, guaranteeing that all the acid has reacted before filtering [1]
(ii) A grey insoluble residue is unreacted zinc metal that passed through or around the filter paper [1]
(ii) The remedy is to filter more carefully, using a properly fitted filter paper and refiltering the solution, or to allow the mixture to settle before filtering [1]
⚠ If you missed marks here: Suggesting that the acid contamination should be removed by washing the crystals with water is not a good answer, because zinc sulfate is soluble and would dissolve away with the acid, destroying the yield. Prevention at the excess stage is the marking point.
Question 5 -- Titration with a Dibasic Acid
Total: 10 marks
A student in Toronto titrates dilute sulfuric acid against a standard solution of sodium hydroxide. She pipettes 25.0 cm³ of 0.200 mol/dm³ sodium hydroxide into a conical flask with thymolphthalein indicator, and runs the sulfuric acid in from a burette. Her rough titre is 24.10 cm³ and her accurate titres are 23.40 cm³, 23.45 cm³ and 23.35 cm³.
(a)[3]
(i) Explain why the rough titre is not used in calculating the mean, and calculate the mean titre. [2]
(ii) State the colour change seen at the end point with thymolphthalein. [1]
Model Answer -- 5(a)
(i) The rough titration is carried out quickly to find the approximate end point, so acid is usually added past it and the reading is too high and unreliable [1]
(i) Mean titre = (23.40 + 23.45 + 23.35)/3 = 23.40 cm³ [1]
(ii) The solution changes from blue in the alkali to colourless as soon as the acid is in slight excess [1]
⚠ If you missed marks here: Including the rough value drags the mean upward and is penalised. Note that thymolphthalein goes blue to colourless in this direction; writing colourless to blue describes adding alkali to acid instead.
(b)[5]
The equation is H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.
(i) Calculate the moles of sodium hydroxide used. [1] (ii) Deduce the moles of sulfuric acid that reacted. [1] (iii) Calculate the concentration of the sulfuric acid in mol/dm³. [1] (iv) Calculate the mass of sodium sulfate that would be obtained if the neutralised solution were evaporated to dryness. (Mₕ(Na₂SO₄) = 142) [2]
(ii) The ratio is 1 mol acid to 2 mol alkali, so moles H₂SO₄ = 5.00 × 10⁻³ ÷ 2 = 2.50 × 10⁻³ mol [1]
(iii) Concentration = 2.50 × 10⁻³ ÷ (23.40/1000) = 0.1068, which rounds to 0.107 mol/dm³ [1]
(iv) Moles of Na₂SO₄ formed = moles of H₂SO₄ = 2.50 × 10⁻³ mol (one sulfate per acid molecule) [1]
(iv) Mass = 2.50 × 10⁻³ × 142 = 0.355 g [1]
⚠ If you missed marks here: Ignoring the 1 : 2 ratio is the single commonest error with sulfuric acid and doubles the calculated concentration. In part (iv) the number of moles of salt equals the number of moles of acid, not of alkali — using the alkali figure doubles the mass.
(c)[2]
The student notices that a drop of acid is hanging from the tip of the burette at the end of the titration, and that she rinsed the conical flask with distilled water before use. Explain the effect of each of these on the calculated concentration of the acid. [2]
Model Answer -- 5(c)
The hanging drop means the volume recorded is larger than the volume actually delivered, so the titre is too high and the calculated acid concentration comes out too low [1]
Rinsing the conical flask with distilled water has no effect: the water dilutes the alkali but does not change the number of moles of sodium hydroxide present, so the same volume of acid is needed [1]
⚠ If you missed marks here: The distilled-water rinse of the conical flask catches out most candidates, who assume any dilution must matter. What counts in the flask is the number of moles delivered by the pipette, which is unchanged. Contrast this with rinsing the burette with water, which really does dilute the acid.
Question 6 -- Precipitation and Identifying Solutions
Total: 12 marks
Four unlabelled bottles in a laboratory in Kuala Lumpur contain aqueous solutions of sodium sulfate, potassium chloride, lead(II) nitrate and sodium carbonate, all at the same concentration. A student mixes them in pairs and records which mixtures give a precipitate.
(a)[4]
State the solubility rules for (i) nitrates, (ii) chlorides, (iii) sulfates and (iv) carbonates, including all the common exceptions. [4]
Model Answer -- 6(a)
(i) All nitrates are soluble, with no exceptions [1]
(ii) All chlorides are soluble except silver chloride and lead(II) chloride (and the same pattern holds for bromides and iodides) [1]
(iii) All sulfates are soluble except barium sulfate, calcium sulfate and lead(II) sulfate [1]
(iv) All carbonates are insoluble except those of sodium, potassium and ammonium [1]
⚠ If you missed marks here: Calcium sulfate is only slightly soluble and must be listed among the sulfate exceptions. Remember that the sodium, potassium and ammonium rule always wins: ammonium carbonate is soluble even though most carbonates are not.
(b)[4]
(i) Name the precipitates formed when lead(II) nitrate is mixed with each of the other three solutions, and give the colour of each. [3]
(ii) Explain why no precipitate forms when sodium sulfate and potassium chloride are mixed. [1]
Model Answer -- 6(b)
(i) With sodium sulfate: a white precipitate of lead(II) sulfate [1]
(i) With potassium chloride: a white precipitate of lead(II) chloride, which dissolves on warming and reforms on cooling [1]
(i) With sodium carbonate: a white precipitate of lead(II) carbonate [1]
(ii) The possible products, sodium chloride and potassium sulfate, are both soluble because all sodium and potassium salts are soluble, so all the ions stay in solution [1]
⚠ If you missed marks here: A common error is to predict a yellow precipitate with the chloride; lead(II) iodide is the bright yellow one, lead(II) chloride is white. When no reaction occurs you must name the two possible products and say why each is soluble.
(c)[4]
The student wants a pure dry sample of lead(II) sulfate. (i) Write the ionic equation with state symbols. [1]
(ii) Describe the practical steps, explaining the reason for each. [3]
Model Answer -- 6(c)
(i) Pb²⁺(aq) + SO₄²⁻(aq) → PbSO₄(s) [1]
(ii) Mix the two solutions, adding one in slight excess so that all of the limiting ion is precipitated and the maximum yield is obtained [1]
(ii) Filter the mixture; the insoluble lead(II) sulfate stays as residue on the filter paper while the soluble sodium nitrate passes through as filtrate [1]
(ii) Wash the residue with distilled water to remove the soluble sodium and nitrate ions clinging to it, then dry it in a warm oven or between filter papers [1]
⚠ If you missed marks here: Marks are lost for suggesting the mixture be evaporated: that would leave the soluble by-product mixed in with the product. Note also that lead compounds are toxic, so this preparation would be done with care and the waste disposed of properly.
Question 7 -- Hydrated Salts and Their Formulae
Total: 10 marks
Washing soda crystals are hydrated sodium carbonate, Na₂CO₃·xH₂O. A student in Cardiff dissolves 2.86 g of the crystals in distilled water and makes the solution up to exactly 250 cm³ in a volumetric flask. She then titrates 25.0 cm³ portions against 0.100 mol/dm³ hydrochloric acid using methyl orange, and finds that a mean titre of 20.0 cm³ is needed. (Mₕ(Na₂CO₃) = 106; Mₕ(H₂O) = 18)
(a)[4]
(i) Define water of crystallisation and explain the difference between a hydrated and an anhydrous salt. [2]
(ii) Describe a chemical test that would show that the water driven off from a hydrated salt on heating really is water, and state the result. [2]
Model Answer -- 7(a)
(i) Water of crystallisation is water chemically combined in the crystal structure of a salt in a fixed molar ratio [1]
(i) A hydrated salt contains this water in its lattice; an anhydrous salt is the same salt with the water of crystallisation removed, usually by heating [1]
(ii) Add the liquid to white anhydrous copper(II) sulfate (or to blue cobalt(II) chloride paper) [1]
(ii) The anhydrous copper(II) sulfate turns from white to blue (or the cobalt(II) chloride paper turns from blue to pink); to show the water is also pure it would boil at exactly 100 °C at normal pressure [1]
⚠ If you missed marks here: Describing a hydrated salt as one that is wet loses the mark: the water is chemically combined in a fixed ratio, not simply on the surface. The copper(II) sulfate and cobalt(II) chloride tests show that water is present but not that it is pure — that needs a boiling or freezing point.
(b)[4]
Use the titration data to calculate the value of x in Na₂CO₃·xH₂O. The equation is Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. Show all your working. [4]
Model Answer -- 7(b)
Moles HCl in the titre = 0.100 × 20.0/1000 = 2.00 × 10⁻³ mol, so moles of Na₂CO₃ in 25.0 cm³ = 1.00 × 10⁻³ mol (ratio 1 : 2) [1]
Moles of Na₂CO₃ in the whole 250 cm³ = 1.00 × 10⁻³ × 10 = 1.00 × 10⁻² mol [1]
Mass of Na₂CO₃ present = 1.00 × 10⁻² × 106 = 1.06 g, so mass of water = 2.86 − 1.06 = 1.80 g, which is 1.80/18 = 0.100 mol [1]
Ratio Na₂CO₃ : H₂O = 0.0100 : 0.100 = 1 : 10, therefore x = 10 [1]
⚠ If you missed marks here: Two scaling steps catch people out: halving for the 1 : 2 mole ratio, and multiplying by ten to scale from the 25.0 cm³ portion to the full 250 cm³. Missing either gives x = 5 or x = 20 rather than 10, so always state the scaling factor explicitly in your working.
(c)[2]
Washing soda crystals left in an open dish in a warm dry room slowly turn into a white powder and lose mass. Name this process, explain what has happened, and state one consequence for the accuracy of the student's result if her sample had been stored this way. [2]
Model Answer -- 7(c)
The process is efflorescence: the crystals lose some of their water of crystallisation to the dry air, leaving a powder of a lower hydrate or of the anhydrous salt [1]
A sample that had partly effloresced would contain a greater proportion of sodium carbonate for the same mass, so the titre would be larger and the calculated value of x would come out too small [1]
⚠ If you missed marks here: Do not confuse efflorescence with deliquescence, which is the opposite — taking up water from the air until the solid dissolves. For the second mark you must say in which direction the calculated x would be wrong, not just that it would be inaccurate.
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