← Topic 7
⚡ Challenge Paper Preparation

Challenge Prep: Acids, Bases and Salts

IGCSE Chemistry 0620 — Topic 7

Topic 7 feels like the friendliest topic on the paper. Three reactions, a colour chart, a few recipes. Then the challenge paper arrives and asks whether concentrated ethanoic acid is a strong acid, why sodium chloride cannot be made by the excess-solid method, whether a solution at pH 7 must be water, and what happens to your marks when you divide by 1000 in the wrong direction. This guide hunts every one of those traps: strong versus concentrated, rate versus amount, soluble versus insoluble, and the salt name that comes from the acid, not the metal. Work through it and Topic 7 becomes the most reliable marks on the paper.

⚠️ Common Traps & Misconceptions

▼

Twelve traps that cost students marks on Topic 7 questions. Every one of them appears on challenge papers regularly.

⚠️ TRAP
Trap 1: Confusing STRONG with CONCENTRATED (and weak with dilute)
The Trap"Ethanoic acid is weak, so it must be dilute." "This bottle is concentrated, so it is a strong acid." "Diluting hydrochloric acid makes it a weak acid." All three sentences appear in scripts every single session.
The TruthStrong / weak describes the degree of dissociation — whether the acid splits completely into ions (HCl → H+ + Cl−) or only partially (CH3COOH ⇌ H+ + CH3COO−). It is a fixed property of the substance. Concentrated / dilute describes moles of solute per dm3 and is set by how much water you add. The two are completely independent: concentrated weak acids and dilute strong acids both exist.
Why It MattersCambridge sets this distinction almost every year, and it is worth 2–3 marks. Worse, using the wrong word poisons the rest of your answer — if you think dilution weakens an acid, you cannot explain why the same volume of alkali neutralises both acids.
Example Question"A bottle of glacial ethanoic acid is labelled 17 mol/dm3. A student says it is therefore a strong acid. Explain why the student is incorrect. [2]"
⚠️ TRAP
Trap 2: Thinking a weak acid does not react fully — the "less gas" error
The Trap"Ethanoic acid is weak, so it will produce less hydrogen than hydrochloric acid." Or: "a weak acid needs less sodium hydroxide to neutralise it, because there is less acid to cancel out."
The TruthWeak means slower, not less. At the same concentration and volume, the two acids contain the same number of moles. As the free H+ is consumed, the equilibrium CH3COOH ⇌ H+ + CH3COO− shifts to the right to replace it, so every acid molecule eventually gives up its proton. Same total volume of gas, same titre in a titration — only the rate and the pH differ.
Why It MattersThe examiner is testing whether you can separate concentration of H+ at a moment from total moles of acid available. The mark scheme typically awards one mark for "same volume of gas" and one for the equilibrium-shift reason. Getting the rate right and the amount wrong scores half.
Example Question"Excess magnesium is added to 50 cm3 of 0.1 mol/dm3 HCl and to 50 cm3 of 0.1 mol/dm3 CH3COOH. Sketch both curves of volume of hydrogen against time on the same axes. [3]"
⚠️ TRAP
Trap 3: Choosing the wrong salt preparation method
The Trap"To make sodium chloride, add excess sodium to hydrochloric acid and filter." "To make barium sulfate, add excess barium carbonate to sulfuric acid and evaporate." "To make copper(II) sulfate, titrate copper(II) oxide against sulfuric acid using methyl orange."
The TruthAsk two questions in order. (1) Is the target salt insoluble? If yes → precipitation, mixing two soluble solutions. (2) If soluble, is one reactant an insoluble solid? If yes → excess solid, filter off the surplus. If both reactants are solutions → titration. Sodium, potassium and ammonium salts are always titration, because every compound of those metals dissolves so there is nothing to filter.
Why It MattersThe method choice is usually worth 1 mark but it gates the other 4–5: choose wrong and every procedural step afterwards is wrong too. This is the single largest block of marks lost in Topic 7.
Example Question"Describe how you would prepare a pure dry sample of (a) potassium nitrate and (b) lead(II) iodide, starting from suitable reagents. [6]"
⚠️ TRAP
Trap 4: Skipping the excess/filtration steps, or evaporating to dryness
The Trap"Add copper(II) oxide to sulfuric acid, then evaporate the solution to dryness to get the crystals." Two errors in one sentence: no mention of excess or filtration, and a fatal evaporation instruction.
The TruthThree separate marks live here. (1) Add the solid in excess, "to make sure all the acid has reacted" — otherwise leftover acid contaminates the crystals. (2) Filter to remove the unreacted excess solid. (3) Evaporate only to the point of crystallisation, then leave to cool slowly — heating to dryness drives off the water of crystallisation and leaves an anhydrous powder instead of crystals.
Why It MattersSalt preparation questions are marked as a procedure: each named step with its reason earns a mark. Candidates who describe the chemistry perfectly but omit "excess", "filter" and "not to dryness" routinely score 2 out of 5.
Example Question"Describe how to prepare pure dry crystals of hydrated copper(II) sulfate from copper(II) carbonate and dilute sulfuric acid. [5]"
⚠️ TRAP
Trap 5: Amphoteric oxide confusion
The TrapThree versions: (1) "Amphoteric means it is neither acidic nor basic" — that is neutral. (2) "Zinc oxide is amphoteric because it reacts with acids" — only half the definition. (3) "Amphoteric oxides dissolve in water to give pH 7."
The TruthAmphoteric means it reacts with BOTH acids AND alkalis, forming a salt and water in each case. Neutral means it reacts with NEITHER. The two are opposites, not synonyms. ZnO + 2HCl → ZnCl2 + H2O and ZnO + 2NaOH → Na2ZnO2 + H2O. Both ZnO and Al2O3 are essentially insoluble in water, so they change the pH of water not at all — insolubility is unrelated to classification.
Why It MattersA one-sided definition scores zero on a 2-mark "define amphoteric" question. And the ZnO/Al2O3 versus CO/NO/N2O split is the standard multiple-choice discriminator in 7.2.
Example Question"Zinc oxide is described as amphoteric. Explain what this means and write TWO equations to support your answer. [3]"
⚠️ TRAP
Trap 6: Getting the wrong salt name from the acid
The Trap"Magnesium + sulfuric acid gives magnesium sulfide." "Calcium + nitric acid gives calcium nitride." "Sodium + hydrochloric acid gives sodium chlorine." Also: writing MgSO4 as MgSO3, or Ca(NO3)2 as CaNO3.
The TruthThe acid names the salt: hydrochloric → chloride (Cl−); sulfuric → sulfate (SO42−); nitric → nitrate (NO3−); ethanoic → ethanoate. Sulfides (S2−) and nitrides (N3−) come from the elements directly and never from these acids. Then balance the charges: Mg2+ with SO42− gives MgSO4; Ca2+ with NO3− needs two nitrates, Ca(NO3)2.
Why It MattersThis is a free mark that thousands of candidates give away. It also breaks every subsequent calculation — a wrong formula means a wrong Mr and a wrong answer, even if your method is flawless.
Example Question"Write the balanced equation for the reaction of calcium carbonate with dilute nitric acid, including state symbols. [2]"
⚠️ TRAP
Trap 7: Assuming pH 7 means pure water (and other neutrality myths)
The Trap"Universal indicator turns green, so the solution must be water." "A neutral solution contains no ions, so it cannot conduct electricity." "Neutralisation always produces a solution of pH 7."
The TruthNeutral means the H+ and OH− concentrations are equal — nothing more. Sodium chloride solution is pH 7, packed with ions, and conducts electricity well. And neutralisation does not always end at pH 7: a strong acid neutralised by a weak base (HCl + NH3) gives a slightly acidic salt solution, while a weak acid neutralised by a strong base (CH3COOH + NaOH) gives a slightly alkaline one — which is why phenolphthalein, not methyl orange, is chosen for a vinegar titration.
Why It MattersChallenge papers love a "which conclusion is valid?" question where the pH 7 reading is deliberately ambiguous. The examiner is testing whether you know that a measurement constrains the answer without determining it.
Example Question"A colourless solution has a pH of 7 and conducts electricity. A student concludes it is pure water. Comment on this conclusion. [2]"
⚠️ TRAP
Trap 8: Over-generalising the solubility rules
The Trap"All carbonates are insoluble." "All sulfates are soluble." "All chlorides are soluble." Each is nearly true, which is exactly what makes it dangerous.
The TruthCarbonates are insoluble except sodium, potassium and ammonium carbonate — and those exceptions are precisely the reagents you use to precipitate other carbonates. Sulfates are soluble except barium sulfate and lead(II) sulfate (calcium sulfate only slightly). Chlorides are soluble except silver chloride and lead(II) chloride. The only families with genuinely no exceptions are nitrates and sodium/potassium/ammonium salts.
Why It MattersThe exceptions are not obscure — they are the entire basis of the precipitation method, the barium meal, and the tests for sulfate and halide ions. A candidate who believes all sulfates are soluble cannot explain the barium sulfate test at all.
Example Question"Explain why sodium carbonate solution can be used to prepare calcium carbonate, but calcium carbonate cannot be used to prepare sodium carbonate. [2]"
⚠️ TRAP
Trap 9: Titration calculation slips — cm3 to dm3 and the mole ratio
The TrapFour recurring slips: (1) forgetting to divide volumes by 1000; (2) multiplying by 1000 instead; (3) assuming every acid–alkali pair is 1 : 1; (4) averaging the rough titre in with the concordant ones.
The Truthcm3 ÷ 1000 = dm3, written as its own line before any other arithmetic. Write the balanced equation first and read the ratio off it: NaOH + HCl is 1 : 1, but 2NaOH + H2SO4 is 2 : 1 because sulfuric acid is diprotic. And discard the rough titre, averaging only results within 0.10 cm3 of each other.
Why It MattersThese are pure arithmetic marks — the chemistry is already understood. A factor-of-1000 slip usually loses every mark after the first, and a 1 : 1 assumption with sulfuric acid doubles or halves the final answer.
Example Question"25.0 cm3 of 0.150 mol/dm3 NaOH required 18.75 cm3 of H2SO4. Calculate the concentration of the acid in g/dm3. [4]"
⚠️ TRAP
Trap 10: Using "base" and "alkali" as if they were the same word
The Trap"Copper(II) oxide is an alkali because it neutralises acid." "Add an alkali such as calcium carbonate to the soil." "All bases dissolve in water to give OH− ions."
The TruthA base is anything that neutralises an acid — metal oxides, metal hydroxides, metal carbonates, ammonia. An alkali is the subset of bases that dissolves in water and releases OH−(aq). CuO, Fe2O3 and CaCO3 are bases but not alkalis. Every alkali is a base; most bases are not alkalis.
Why It MattersThe words are not interchangeable on a mark scheme. It also matters practically: only an alkali can be used in a titration (it must be in solution), and only a non-alkali base can be used in the excess-solid method (it must be insoluble). Getting the word wrong signals you have not grasped why the two methods exist.
Example Question"Explain the difference between a base and an alkali, giving one example of a base that is not an alkali. [2]"
⚠️ TRAP
Trap 11: Applying the reactivity series rule to the wrong reaction
The Trap"Copper carbonate will not react with hydrochloric acid, because copper is below hydrogen in the reactivity series." Or the reverse: "copper reacts with dilute sulfuric acid to give copper(II) sulfate and hydrogen."
The TruthThe reactivity-series rule applies only to acid + metal: a metal must be above hydrogen to displace it. It does not apply to acid + metal oxide, acid + hydroxide or acid + carbonate — those are neutralisations and work for every metal, including copper, silver and lead. Copper metal + HCl gives nothing; copper(II) carbonate + HCl fizzes vigorously.
Why It MattersCopper(II) sulfate is the most-set salt preparation on the syllabus, and it is made from copper oxide or copper carbonate precisely because copper metal will not react. A candidate who writes "Cu + H2SO4 → CuSO4 + H2" loses the equation mark and the method mark together.
Example Question"Suggest why copper(II) sulfate is prepared from copper(II) oxide rather than from copper metal. [2]"
⚠️ TRAP
Trap 12: Indicator errors — wrong colours, and using one indicator to measure acidity
The Trap"Thymolphthalein turns pink in alkali" (that is phenolphthalein). "Methyl orange is blue in alkali" (it is yellow). "I would use litmus to find out which acid is more acidic." Also: stating an end colour without stating the change.
The TruthLearn the three named indicators exactly: litmus red (acid) / purple (neutral) / blue (alkali); methyl orange red / orange / yellow; thymolphthalein colourless / colourless / blue. A single indicator can only tell you acid or alkali — litmus is equally red at pH 6 and pH 1. To compare acidity you need universal indicator with a colour chart or a pH meter. And always write the change: "turns from blue to red".
Why It MattersNaming an indicator is a recall mark, but choosing the right one is an application mark. Questions that ask you to distinguish "which is the stronger acid" are testing whether you know the limits of a single indicator.
Example Question"A student wishes to show that 0.1 mol/dm3 ethanoic acid is a weaker acid than 0.1 mol/dm3 hydrochloric acid. State why litmus would be unsuitable and describe a better method. [3]"

🧩 Multi-Step Reasoning Walkthroughs

▼

Six challenging questions broken down step by step. Try each step yourself before revealing the next.

Walkthrough 1 — Choosing and Justifying Three Different MethodsYou are asked to prepare pure dry samples of (a) zinc nitrate, (b) ammonium sulfate and (c) barium sulfate. For each, choose the method, name suitable starting materials, write a balanced equation and give the key purification step. [9]
1

Solubility first, always

(a) Zinc nitrate — all nitrates are soluble ⇒ soluble. (b) Ammonium sulfate — all ammonium salts are soluble ⇒ soluble. (c) Barium sulfate — one of the two insoluble sulfates ⇒ INSOLUBLE. Write "sol" or "insol" beside each before doing anything else; the answer to (c) is already decided.

2

Yes → excess solid method

Zinc metal, zinc oxide and zinc carbonate are all insoluble, so I can add any of them in excess to dilute nitric acid and filter off the surplus. Choosing zinc oxide: ZnO(s) + 2HNO3(aq) → Zn(NO3)2(aq) + H2O(l). Key purification step: filter to remove the unreacted excess zinc oxide, then evaporate the filtrate to the point of crystallisation and cool.

3

No insoluble reactant exists → titration

Every ammonium compound is soluble, so there is no insoluble solid to add in excess and nothing to filter off. The method is forced: titration. Titrate 25.0 cm3 of aqueous ammonia against dilute sulfuric acid using methyl orange, record the exact titre, then repeat with the same volumes but no indicator. 2NH3(aq) + H2SO4(aq) → (NH4)2SO4(aq). Key purification step: repeating without indicator, so the crystals are not stained by dye.

4

Insoluble → precipitation

I need Ba2+ and SO42−, each supplied by a soluble compound. Use the reliable trick: take the positive ion as a nitrate and the negative ion as a sodium salt. So barium nitrate + sodium sulfate: Ba(NO3)2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaNO3(aq). Ionic: Ba2+(aq) + SO42−(aq) → BaSO4(s). Key purification step: filter, then wash the residue with distilled water to remove the soluble sodium nitrate.

5

Three tempting wrong turns

Wrong turn 1: making barium sulfate by adding barium carbonate to sulfuric acid. It would work chemically, but the product coats the unreacted carbonate and the reaction stops — and you cannot separate product from reactant, since both are insoluble solids. Wrong turn 2: making ammonium sulfate by the excess-solid method — impossible, nothing is insoluble. Wrong turn 3: evaporating the barium sulfate mixture — that would leave the sodium nitrate behind with your product.

Final Answer(a) Zinc nitrate — excess solid method. Add excess zinc oxide to warm dilute nitric acid; filter off the excess; evaporate the filtrate to the point of crystallisation; cool, filter and dry. ZnO + 2HNO3 → Zn(NO3)2 + H2O.
(b) Ammonium sulfate — titration. Titrate aqueous ammonia against dilute sulfuric acid with methyl orange; repeat without indicator; evaporate and crystallise. 2NH3 + H2SO4 → (NH4)2SO4.
(c) Barium sulfate — precipitation. Mix barium nitrate and sodium sulfate solutions; filter; wash the residue with distilled water; dry. Ba(NO3)2 + Na2SO4 → BaSO4 + 2NaNO3.
Examiner's NoteThis is the archetypal 9-mark Topic 7 question and it is marked in blocks of three: method, equation, purification step. The most common total is 5 or 6, lost by using the excess-solid method for the ammonium salt and by omitting the washing step for the precipitate. Note also that examiners accept any suitable insoluble zinc compound in (a) — metal, oxide or carbonate — but if you choose the carbonate you must show CO2 in your equation.
Walkthrough 2 — A Full Titration Calculation with a Diprotic AcidA student titrates 25.0 cm3 portions of 0.150 mol/dm3 sodium hydroxide against dilute sulfuric acid. Her titres are 19.60, 18.70, 18.80 and 18.75 cm3. Calculate the concentration of the sulfuric acid in mol/dm3 and in g/dm3, and calculate the mass of sodium sulfate that could be obtained from one titration. (Ar: H = 1, O = 16, Na = 23, S = 32) [6]
1

Discard the rough, average the concordant

19.60 is the rough titre — it sits 0.85 cm3 away from the others and is discarded. The concordant set is 18.70, 18.80 and 18.75, all within 0.10 cm3. Mean titre = (18.70 + 18.80 + 18.75) ÷ 3 = 56.25 ÷ 3 = 18.75 cm3. Averaging all four would have given 18.96 and cost every subsequent mark.

2

The ratio is 2 : 1, not 1 : 1

2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l). Sulfuric acid is diprotic — each molecule supplies two H+ ions — so one mole of acid neutralises two moles of alkali. Assuming 1 : 1 here would double the final answer.

3

Divide by 1000, then multiply

25.0 cm3 = 25.0 ÷ 1000 = 0.0250 dm3. 18.75 cm3 = 0.01875 dm3.
moles NaOH = concentration × volume = 0.150 × 0.0250 = 3.75 × 10−3 mol.

4

Halve, then divide by the acid volume

moles H2SO4 = 3.75 × 10−3 ÷ 2 = 1.875 × 10−3 mol.
concentration = moles ÷ volume = 1.875 × 10−3 ÷ 0.01875 = 0.100 mol/dm3.

5

Multiply by Mr

Mr of H2SO4 = (2 × 1) + 32 + (4 × 16) = 2 + 32 + 64 = 98.
concentration in g/dm3 = 0.100 × 98 = 9.80 g/dm3.

6

Read the ratio again

From the equation, 2 mol NaOH gives 1 mol Na2SO4, so moles of Na2SO4 = 1.875 × 10−3 mol.
Mr of Na2SO4 = (2 × 23) + 32 + (4 × 16) = 46 + 32 + 64 = 142.
mass = moles × Mr = 1.875 × 10−3 × 142 = 0.266 g.

Final AnswerMean titre = 18.75 cm3. Concentration of sulfuric acid = 0.100 mol/dm3 = 9.80 g/dm3. Mass of sodium sulfate obtainable = 0.266 g (3 s.f.).
Examiner's NoteMark schemes for this question award: mean titre [1], correct equation or ratio [1], moles of NaOH [1], moles of acid [1], concentration [1], g/dm3 conversion [1]. Crucially, most schemes allow error carried forward — so even if you slip at step 3, correct method afterwards still earns marks, provided your working is visible. Never write a bare answer. The two errors that are not recoverable are averaging in the rough titre and using a 1 : 1 ratio, because both look like correct arithmetic and cannot be traced.
Walkthrough 3 — Strong versus Weak, Six WaysTwo beakers each contain 50 cm3 of 0.100 mol/dm3 acid: A is hydrochloric acid, B is ethanoic acid. Predict and explain the comparison for (i) pH, (ii) electrical conductivity, (iii) initial rate with magnesium, (iv) total volume of hydrogen, (v) volume of 0.100 mol/dm3 NaOH needed for neutralisation, (vi) temperature rise on neutralisation. [8]
1

The single most useful move in this question

Three of these depend on the concentration of free H+ at a given moment: pH, conductivity, initial rate. Three depend on the total moles of acid present: total gas, titre, and (broadly) the heat released. Group them first, and the answers fall out automatically. Both beakers contain 0.050 × 0.100 = 5.00 × 10−3 mol of acid — identical.

2

A wins all three

(i) pH: A is lower (about 1) and B higher (about 3), because HCl is fully dissociated so its H+ concentration is far greater.
(ii) Conductivity: A conducts better, because it contains more mobile ions in solution — B keeps most of its acid as intact, uncharged molecules.
(iii) Initial rate: A fizzes faster, because a higher H+ concentration gives a greater frequency of successful collisions with the magnesium surface.

3

Identical — and this is where the marks are

(iv) Total volume of hydrogen: the same. Both contain 5.00 × 10−3 mol of monoprotic acid, so both can release 2.50 × 10−3 mol of H2. As H+ is consumed in B, the equilibrium CH3COOH ⇌ H+ + CH3COO− shifts right to replace it, so eventually every molecule reacts.
(v) Volume of NaOH: the same, 50.0 cm3 in each case, because the titration measures total moles of acid, not free H+.
(vi) Temperature rise: very similar (slightly smaller for B, since a little energy is absorbed dissociating the remaining molecules) — the same number of moles of water are formed.

4

Different gradient, same plateau

Curve A rises steeply and levels off early. Curve B rises more gently and reaches its plateau later. Both plateaus sit at exactly the same height. Drawing B finishing lower is the classic lost mark — and it is the same "rate versus amount" distinction you met in Topic 6.

Final AnswerA (HCl) is greater for: H+ concentration, hence lower pH, higher conductivity and faster initial rate — because it is fully dissociated. Identical for: total volume of hydrogen, volume of NaOH required, and (essentially) the temperature rise — because both beakers contain the same number of moles of monoprotic acid, and the weak acid dissociates further as the reaction proceeds.
Examiner's NoteThis question is designed so that half the answers are "different" and half are "the same". Candidates who have learned only "strong acids react more" answer "A is greater" six times and score 3 out of 8. The equilibrium-shift sentence in part (iv) is the top-band discriminator — examiners look specifically for the idea that removing H+ causes more acid to ionise. Part (vi) is rarely asked but always rewards a candidate who reasons from moles of water formed rather than guessing.
Walkthrough 4 — Identifying an Unknown Oxide from Experimental DataA white solid X is investigated. (1) Added to water it is insoluble and the pH is unchanged at 7. (2) Warmed with dilute hydrochloric acid it dissolves to give a colourless solution; no gas is given off. (3) Warmed with concentrated aqueous sodium hydroxide it also dissolves. (4) The colourless solution from (2) gives a white precipitate with a little aqueous ammonia, which dissolves in excess ammonia. Identify X, classify it, and write two equations. [7]
1

Insoluble, pH unchanged

It cannot be a soluble basic oxide such as Na2O or CaO, since those would dissolve and raise the pH. It cannot be a soluble acidic oxide either. But note carefully: insolubility tells you nothing about the acid–base classification. Do not conclude "neutral" here — that is the trap.

2

Two pieces of information, not one

It reacts with an acid ⇒ it is behaving as a base. And no gas is given off ⇒ it is not a carbonate and not a metal above hydrogen — so it is a metal oxide or hydroxide. The solution is colourless, so it is not a copper or iron compound.

3

Reacts with acid AND alkali

A substance that dissolves in both dilute acid and concentrated alkali is by definition amphoteric. The two amphoteric oxides on the syllabus are ZnO and Al2O3. Both are white and both give colourless solutions, so observation (3) alone cannot separate them.

4

The ammonia test separates zinc from aluminium

Both Zn2+ and Al3+ give a white precipitate of the hydroxide with a little aqueous ammonia. But only the zinc hydroxide redissolves in excess ammonia; aluminium hydroxide stays put. So X contains zinc, and X is zinc oxide, ZnO.

5

One with acid, one with alkali

With acid: ZnO(s) + 2HCl(aq) → ZnCl2(aq) + H2O(l) — here ZnO behaves as a base.
With alkali: ZnO(s) + 2NaOH(aq) → Na2ZnO2(aq) + H2O(l) — here ZnO behaves as an acidic oxide, forming sodium zincate.
Both produce a salt and water, which is exactly what the definition of amphoteric requires.

Final AnswerX is zinc oxide, ZnO. It is an amphoteric oxide, because it reacts with both acids and alkalis to form a salt and water. ZnO + 2HCl → ZnCl2 + H2O (basic behaviour); ZnO + 2NaOH → Na2ZnO2 + H2O (acidic behaviour, giving sodium zincate). The absence of gas in (2) rules out a carbonate, and the redissolving precipitate in excess ammonia identifies zinc rather than aluminium.
Examiner's NoteEvery clue in this question is load-bearing, which is typical of high-demand identification questions. The two most common errors are (a) concluding "neutral" from the pH-7 observation in (1), and (b) stopping at "amphoteric" without using clue (4) to distinguish ZnO from Al2O3. Note the phrase "no gas is given off" — examiners include it deliberately so that you can eliminate carbonates, and candidates who ignore it often propose ZnCO3 and lose two marks.
Walkthrough 5 — Water of Crystallisation from Experimental DataA student heats 6.25 g of hydrated copper(II) sulfate, CuSO4·xH2O, in a crucible until the mass is constant. The residue has a mass of 4.00 g. (a) Calculate x. (b) Explain the phrase "heated to constant mass". (c) The student's value came out as 4.6. Suggest two experimental reasons. (Ar: H = 1, O = 16, S = 32, Cu = 64) [7]
1

Subtract, and say where it went

mass of water driven off = 6.25 − 4.00 = 2.25 g. This water was the water of crystallisation held inside the crystal lattice; it escaped as vapour, which is why the blue crystals turned to a white powder.

2

Two separate Mr values

Mr of CuSO4 = 64 + 32 + (4 × 16) = 160. moles of CuSO4 = 4.00 ÷ 160 = 0.0250 mol.
Mr of H2O = 18. moles of H2O = 2.25 ÷ 18 = 0.125 mol.

3

Divide both by the smaller value

CuSO4: 0.0250 ÷ 0.0250 = 1. H2O: 0.125 ÷ 0.0250 = 5. So x = 5 and the formula is CuSO4·5H2O — the familiar blue crystals, which is a good sanity check.

4

Heat, cool, weigh, repeat

The crucible is heated, allowed to cool in a desiccator, and weighed; this cycle is repeated until two successive weighings agree. A constant mass proves that no more water is being driven off — that is, the dehydration is complete. Without it you cannot be sure you have the true anhydrous mass.

5

Reason from the direction of the error

An answer of 4.6 rather than 5 means too little water appeared to be lost, so the residue was heavier than it should have been. Two good reasons: (1) insufficient heating — the salt was not fully dehydrated, so some water remained in the residue; (2) the residue reabsorbed moisture from the air while cooling on the bench, because anhydrous copper(II) sulfate is hygroscopic (it should be cooled in a desiccator). A third acceptable answer: some solid spat out of the crucible — but note that would make the residue lighter and push x the other way, so it does not fit this data.

Final Answer(a) Water lost = 2.25 g = 0.125 mol; CuSO4 = 4.00 ÷ 160 = 0.0250 mol; ratio 1 : 5, so x = 5 and the salt is CuSO4·5H2O.
(b) Heating, cooling and weighing repeatedly until two successive masses agree, which shows all the water of crystallisation has been driven off.
(c) Either the salt was not heated for long enough so dehydration was incomplete, or the hygroscopic anhydrous salt absorbed moisture from the air while cooling — both leave the residue too heavy and x too small.
Examiner's NotePart (c) is the discriminator and it rewards directional reasoning: work out whether the error made the residue heavier or lighter, then suggest only causes that fit. Candidates who list every experimental error they can remember, including ones that push the answer the wrong way, score one mark at best. In (a), a common slip is dividing the water mass by 160 or using Mr = 250 for the hydrated salt when the question has already given you both masses.
Walkthrough 6 — Auditing a Flawed Salt-Preparation MethodA student writes this method for making pure dry crystals of copper(II) sulfate: "Add 20 cm3 of dilute sulfuric acid to a beaker. Add exactly 2 g of copper metal and stir until it dissolves. Add a few drops of methyl orange to check the acid has gone. Pour the mixture into an evaporating basin and heat strongly until all the water has boiled away, leaving the crystals." Identify and correct every error. [6]
1

Copper metal will not react

Copper is below hydrogen in the reactivity series, so it cannot displace hydrogen from a dilute acid. Nothing would dissolve and the method fails at the first step. Correction: use copper(II) oxide or copper(II) carbonate instead — the reactivity-series restriction applies only to acid + metal, not to acid + base or acid + carbonate.
CuO(s) + H2SO4(aq) → CuSO4(aq) + H2O(l)

2

Add it in excess, not by measurement

The whole point of this method is that you do not need to measure. Add the solid in excess, a spatula at a time, until no more dissolves and some remains at the bottom. That guarantees all the acid has reacted, so no acid is left to contaminate the crystals.

3

Two reasons it is wrong

First, it is unnecessary: undissolved solid is already the visual signal that the acid is used up. Second, it is harmful — methyl orange is a dye that would be evaporated into the crystals and colour them. Indicators belong in the titration method, and even there the preparation is repeated without indicator.

4

An entire step has been omitted

Once excess solid is present it must be removed: filter the mixture, keeping the blue filtrate (the copper(II) sulfate solution) and discarding the residue of unreacted copper(II) oxide. Without this, the crystals would be contaminated with black solid.

5

The most costly error of all

Copper(II) sulfate crystallises as CuSO4·5H2O. Boiling to dryness drives off the water of crystallisation and leaves a white anhydrous powder, not blue crystals. Correction: heat the filtrate gently until about half the water has evaporated and a hot saturated solution remains — the point of crystallisation, testable by dipping a cold glass rod in and watching crystals form on it. Then leave to cool slowly so that large, well-formed crystals grow, filter them off and dry them between filter papers or in a warm oven.

Final AnswerFive errors. (1) Copper metal does not react with dilute acid — use copper(II) oxide or carbonate. (2) Do not weigh out a fixed mass — add the solid in excess until no more dissolves, ensuring all the acid reacts. (3) Remove the methyl orange — it is unnecessary and would stain the crystals. (4) A filtration step is missing — filter off the excess solid and keep the filtrate. (5) Do not evaporate to dryness — evaporate only to the point of crystallisation, then cool slowly, filter and dry the crystals.
Examiner's Note"Identify the errors" questions are marked one mark per error with its correction — spotting an error without saying what to do instead usually scores nothing. Aim to write each point as a pair: "X is wrong because Y; instead do Z." Note also that examiners accept copper(II) hydroxide or copper(II) carbonate as alternatives in (1), but if you choose the carbonate you should mention the effervescence as your signal that the reaction is proceeding.

🔍 Spot the Difference

▼

Pairs of questions that look nearly identical but have different answers. Spot the key distinction.

Question A
Is 17 mol/dm3 ethanoic acid a strong acid?
No. It is a concentrated weak acid. Strength is about the proportion of molecules that dissociate, and ethanoic acid only partially ionises at any concentration.
Question B
Is 0.001 mol/dm3 hydrochloric acid a strong acid?
Yes. It is a dilute strong acid. Every HCl molecule present is fully dissociated, however few of them there are.
Key DifferenceStrength and concentration are independent axes. Strength = degree of dissociation, a fixed property of the substance. Concentration = moles per dm3, set by you and the measuring cylinder. All four combinations exist. If a question says "explain why the student is wrong", the answer is almost always that they have collapsed these two axes into one.
Question A
Is copper(II) oxide a base?
Yes. It neutralises acids to form a salt and water: CuO + H2SO4 → CuSO4 + H2O. That is the whole definition of a base.
Question B
Is copper(II) oxide an alkali?
No. An alkali must dissolve in water and release OH−(aq). Copper(II) oxide is insoluble, so it cannot form an alkaline solution.
Key DifferenceEvery alkali is a base; most bases are not alkalis. This is not pedantry — it decides your preparation method. An insoluble base can be added in excess and filtered off (Method 1). An alkali is a solution, so it must be titrated (Method 2). Get the word wrong and you will pick the wrong method.
Question A
50 cm3 of 0.1 mol/dm3 HCl versus 50 cm3 of 0.1 mol/dm3 CH3COOH with excess Mg. Which fizzes faster?
The hydrochloric acid. It is fully dissociated, so its H+ concentration is far higher, giving more frequent successful collisions.
Question B
Same two beakers. Which produces the greater total volume of hydrogen?
Neither — they are equal. Both contain the same moles of monoprotic acid, and as H+ is used up the weak acid dissociates further until every molecule has reacted.
Key DifferenceRate depends on the concentration of free H+ right now; total amount depends on the total moles of acid available. Ask yourself which of the two the question is really about. Steeper gradient, identical plateau — exactly the same shape rule you learned for rates in Topic 6.
Question A
A solution has pH 7. Must it be pure water?
No. Sodium chloride solution is pH 7. So is potassium nitrate solution. Neutral only means [H+] = [OH−].
Question B
A solution has pH 7 and does not conduct electricity. Could it be pure water?
Yes — now it is consistent. The absence of conductivity shows there are essentially no dissolved ions, which rules out a salt solution.
Key DifferenceOne measurement constrains an answer; two measurements can identify it. A pH reading alone never proves a substance's identity. Examiners set "comment on this conclusion" questions precisely to see whether you can say what a result does and does not establish.
Question A
What happens when copper metal is added to dilute sulfuric acid?
Nothing. Copper is below hydrogen in the reactivity series, so it cannot displace hydrogen from an acid.
Question B
What happens when copper(II) carbonate is added to dilute sulfuric acid?
Vigorous effervescence and a blue solution: CuCO3 + H2SO4 → CuSO4 + H2O + CO2.
Key DifferenceThe reactivity series rule applies only to acid + metal. Acid + oxide, acid + hydroxide and acid + carbonate are neutralisations and work for every metal, including copper, lead and silver. This is exactly why copper(II) sulfate is made from the oxide or carbonate rather than from the metal.
Question A
Which method for zinc sulfate?
Excess solid. The salt is soluble and zinc oxide (or the metal, or the carbonate) is insoluble, so add it in excess and filter off the surplus.
Question B
Which method for sodium sulfate?
Titration. Every sodium compound is soluble, so there is no insoluble solid to add in excess and nothing to filter off.
Key DifferenceBoth target salts are soluble, so solubility of the product does not separate them. The deciding question is whether an insoluble reactant exists. Sodium, potassium and ammonium salts can never use the excess-solid method — commit that sentence to memory.
Question A
How would you prepare lead(II) nitrate?
Excess solid. All nitrates are soluble, and lead(II) oxide or carbonate is insoluble — add it in excess to dilute nitric acid, filter, then crystallise.
Question B
How would you prepare lead(II) sulfate?
Precipitation. Lead(II) sulfate is insoluble, so mix solutions of lead(II) nitrate and sodium sulfate, then filter, wash and dry the residue.
Key DifferenceSame metal, completely different method — because the anion decides the solubility. Never memorise "lead salts are made this way". Look up the specific salt in the solubility table every time; the negative ion is doing the deciding.
Question A
Zinc oxide is added to (i) dilute HCl and (ii) hot NaOH(aq). What happens?
It dissolves in both. ZnO + 2HCl → ZnCl2 + H2O and ZnO + 2NaOH → Na2ZnO2 + H2O. This is amphoteric behaviour.
Question B
Carbon monoxide is bubbled through (i) dilute HCl and (ii) hot NaOH(aq). What happens?
Nothing in either. CO is a neutral oxide — it reacts with neither acids nor alkalis.
Key DifferenceAmphoteric = reacts with both. Neutral = reacts with neither. They are opposites, yet candidates routinely define one using the words of the other. Note too that CO2 and NO2 are acidic while CO and NO are neutral — one extra oxygen changes everything.
Question A
25.0 cm3 of 0.100 mol/dm3 NaOH needs 25.0 cm3 of HCl. Find the acid concentration.
0.100 mol/dm3. NaOH + HCl is 1 : 1, so equal volumes means equal concentrations.
Question B
25.0 cm3 of 0.100 mol/dm3 NaOH needs 25.0 cm3 of H2SO4. Find the acid concentration.
0.0500 mol/dm3. 2NaOH + H2SO4 is 2 : 1, so the acid needs only half as many moles.
Key DifferenceIdentical numbers, answers differing by a factor of two — and the only clue is the formula of the acid. Sulfuric acid is diprotic: one molecule supplies two H+. Write the balanced equation before any arithmetic, every single time. HCl and HNO3 are 1 : 1 with NaOH; H2SO4 is 1 : 2.
Question A
You want crystals of copper(II) sulfate. How far do you evaporate?
To the point of crystallisation only — about half the water — then cool slowly. The product is blue CuSO4·5H2O.
Question B
You want anhydrous copper(II) sulfate. How far do you evaporate?
To dryness, and then heat further until the blue solid turns white. The water of crystallisation is deliberately driven off.
Key Difference"Evaporate to dryness" is not always wrong — it is wrong when the question asks for crystals. Read the target product. If the words "pure dry crystals" appear, you must stop at the point of crystallisation, because the water of crystallisation is part of the product you were asked to make.

🔗 Acids, Bases and Salts Concept Map

▼

Click each node to see how the subtopics connect.

⭐ CORE FRAMEWORK 1
The H+ ion is the engine that drives all of 7.1
Definitions: Acid, Base, Alkali ▶
Acid: a source of H+ ions in aqueous solution. Supplement: a proton donor.
Base: a substance that neutralises an acid — metal oxide, metal hydroxide, metal carbonate or ammonia. Supplement: a proton acceptor.
Alkali: a base that is soluble in water, producing OH−(aq). All alkalis are bases; most bases are not alkalis.
Why proton definitions matter: they explain HCl(g) + NH3(g) → NH4Cl(s), an acid–base reaction with no water and no pH.
The Four Reaction Patterns ▶
acid + metal → salt + hydrogen. Only metals above hydrogen. Test: squeaky pop.
acid + base → salt + water. Works for any metal oxide or hydroxide. No gas.
acid + carbonate → salt + water + CO2. Works for every metal. Test: limewater turns milky.
base + ammonium salt → ammonia. Test for NH4+: damp red litmus turns blue.
Neutralisation and the pH Scale ▶
Ionic equation: H+(aq) + OH−(aq) → H2O(l). Everything else is a spectator ion. Always exothermic.
pH 0–14: below 7 acidic, 7 neutral, above 7 alkaline. Each unit is a ten-fold change in [H+].
Universal indicator: red / orange / yellow (acid) → green (neutral) → blue / indigo / violet (alkali). Compare against a colour chart.
Named indicators: litmus red→purple→blue; methyl orange red→orange→yellow; thymolphthalein colourless→colourless→blue.
Strong versus Weak (Supplement) ▶
Strong acid: fully dissociated, single arrow. HCl → H+ + Cl−. Also H2SO4, HNO3.
Weak acid: partially dissociated, reversible arrow. CH3COOH ⇌ H+ + CH3COO−. Also H2CO3, citric.
Differs: pH, conductivity, rate of reaction — all depend on free [H+].
Identical: total gas produced, titre required — both depend on total moles of acid.
Weak base: aqueous ammonia, NH3 + H2O ⇌ NH4+ + OH−, about pH 11 versus pH 14 for NaOH.
⭐ CORE FRAMEWORK 2
Oxides — two tests, four boxes
The Classification Rule ▶
Reacts with acid only → basic. Formed by metals. CuO, MgO, CaO, Na2O, Fe2O3.
Reacts with alkali only → acidic. Formed by non-metals. CO2, SO2, SO3, NO2, P4O10, SiO2.
Reacts with both → amphoteric. Only ZnO and Al2O3 at IGCSE.
Reacts with neither → neutral. CO, NO, N2O, H2O.
Links to Other Topics ▶
To the Periodic Table: across Period 3 the trend is basic → amphoteric → acidic, with Al2O3 exactly at the metal/non-metal border.
To air pollution: SO2 and NO2 are acidic oxides → acid rain. CO is neutral, so it damages people, not buildings.
To 7.1: soluble basic oxides (Na2O, CaO) form alkalis; soluble acidic oxides form acids. Insoluble ones are still classified the same way.
To metals: the amphoteric Al2O3 layer protects aluminium from acid and weather but dissolves in strong alkali.
Traps Built Into This Framework ▶
CO versus CO2, NO versus NO2: one neutral, one acidic. The commonest multiple-choice distractor in 7.2.
Insoluble ≠ neutral. CuO does not dissolve in water at all, yet it is firmly basic.
Amphoteric ≠ neutral. They are opposites: both versus neither.
Acidic oxide ≠ acid. SO2 is a gas; it becomes an acid only on meeting water.
⭐ CORE FRAMEWORK 3
Salt preparation — solubility decides everything
The Solubility Rules ▶
Always soluble, no exceptions: all nitrates; all sodium, potassium and ammonium salts.
Chlorides: soluble except silver and lead(II) chloride.
Sulfates: soluble except barium and lead(II) sulfate; calcium sulfate only slightly.
Carbonates and hydroxides: insoluble by default, except the sodium, potassium and ammonium versions.
The Two Deciding Questions ▶
Q1: Is the target salt insoluble? Yes → precipitation. Mix two soluble solutions, filter, wash, dry.
Q2: Is one reactant an insoluble solid? Yes → excess solid. Add in excess, filter off the surplus, evaporate to the point of crystallisation, cool, filter, dry.
Otherwise (both reactants are solutions) → titration. Find the titre with indicator, then repeat without indicator, evaporate and crystallise.
Shortcut: Na, K and NH4 salts are always titration — nothing of theirs is insoluble.
Naming, Formulae and Water of Crystallisation ▶
The acid names the salt: HCl → chloride; H2SO4 → sulfate; HNO3 → nitrate; CH3COOH → ethanoate.
Balance the charges: Mg2+ + SO42− → MgSO4; Ca2+ + 2NO3− → Ca(NO3)2.
Hydrated versus anhydrous: CuSO4·5H2O (blue) ⇌ CuSO4 (white) + 5H2O. Also CoCl2·6H2O pink ⇌ blue.
Why you never evaporate to dryness: it removes the water of crystallisation and destroys the crystals.
Titration Calculations ▶
Step order: mean titre → balanced equation → cm3 ÷ 1000 → moles of known → ratio → moles of unknown → concentration.
Ratios: NaOH : HCl = 1 : 1; NaOH : HNO3 = 1 : 1; NaOH : H2SO4 = 2 : 1.
Titres: discard the rough; average only results within 0.10 cm3.
Conversions: g/dm3 = mol/dm3 × Mr. Show every line so error carried forward can rescue you.

❌ "Why Is This Wrong?" Exercises

▼

Spot the error in each student's answer. Think before revealing.

Exercise 1: "Explain why 1 mol/dm3 ethanoic acid has a higher pH than 1 mol/dm3 hydrochloric acid. [2]"
Student's Answer"Because the ethanoic acid is more dilute, so it contains less acid and is weaker."
The FlawThe two solutions have the same concentration — the question says so. "More dilute" is factually contradicted by the stem. The student has fused concentration and strength into a single idea.
Correct Answer"Ethanoic acid is a weak acid, so it is only partially dissociated: CH3COOH ⇌ H+ + CH3COO− [1]. It therefore produces a much lower concentration of H+(aq) ions than the fully dissociated hydrochloric acid, and a lower [H+] means a higher pH [1]."
Key RuleWhen two solutions have the same concentration, any difference in pH, rate or conductivity must come from strength. Read the stem for the word "same" before choosing your explanation.
Exercise 2: "Excess magnesium is added to equal volumes of hydrochloric and ethanoic acid of the same concentration. Compare the total volume of hydrogen produced. [2]"
Student's Answer"The hydrochloric acid produces twice as much hydrogen, because it is a strong acid and reacts completely, whereas the ethanoic acid only partly reacts."
The Flaw"Partially dissociated" has been misread as "partially reacts". Dissociation is a dynamic equilibrium, not a limit on how much acid can react. The invented factor of two has no basis at all.
Correct Answer"The same volume of hydrogen is produced by each [1]. Both acids are monoprotic and present in the same number of moles. As the H+ ions are used up, the equilibrium CH3COOH ⇌ H+ + CH3COO− shifts to the right to replace them, so eventually every ethanoic acid molecule releases its proton [1]. Only the rate differs — the hydrochloric acid reacts faster."
Key RuleWeak means slower, not less. Strength controls the gradient of the curve; moles of acid control the height of the plateau.
Exercise 3: "Describe how to prepare pure dry crystals of copper(II) sulfate from copper and dilute sulfuric acid. [5]"
Student's Answer"Add copper to the sulfuric acid and warm it. Copper sulfate solution forms. Evaporate all the water to leave the crystals."
The FlawThree errors. (1) Copper metal does not react with dilute acid — it is below hydrogen in the reactivity series, so nothing happens. (2) No mention of excess or filtration. (3) "Evaporate all the water" destroys the water of crystallisation.
Correct Answer"Use copper(II) oxide (or carbonate) instead of the metal [1]. Warm the dilute sulfuric acid and add copper(II) oxide in excess, until no more dissolves, to ensure all the acid has reacted [1]. Filter to remove the unreacted excess oxide, keeping the blue filtrate [1]. Evaporate the filtrate to the point of crystallisation — not to dryness — then leave to cool slowly so crystals form [1]. Filter off the crystals and dry them between filter papers [1]."
Key RuleSalt preparation is marked step by step. Excess, filter, point of crystallisation, cool, dry — five words, five marks. And check the reactivity series before choosing a metal.
Exercise 4: "Describe how to prepare a pure dry sample of sodium nitrate. [4]"
Student's Answer"Add excess sodium hydroxide to dilute nitric acid, then filter off the excess sodium hydroxide and evaporate the filtrate."
The FlawSodium hydroxide is soluble. Excess alkali would simply dissolve, and there would be nothing on the filter paper. The student has applied the excess-solid method to two solutions.
Correct Answer"Use a titration. Pipette 25.0 cm3 of sodium hydroxide into a conical flask and add a few drops of methyl orange [1]. Run dilute nitric acid in from a burette, swirling, until the indicator just changes colour permanently; record the titre [1]. Repeat using the same volumes but with no indicator, so the product is not contaminated by the dye [1]. Evaporate the resulting solution to the point of crystallisation, cool, filter and dry the crystals [1]."
Key RuleSodium, potassium and ammonium salts must be made by titration, because every compound of those metals is soluble and there is nothing to filter off.
Exercise 5: "Describe how to prepare a pure dry sample of silver chloride. [3]"
Student's Answer"Mix silver nitrate solution with sodium chloride solution. A white precipitate forms. Filter it off and evaporate the solution to dryness to get the silver chloride."
The FlawThe first two sentences are correct, then it collapses. Evaporating the filtrate would give you sodium nitrate, not silver chloride — the product you want is already sitting in the filter paper as the residue. And the essential washing step has been omitted.
Correct Answer"Mix aqueous silver nitrate with aqueous sodium chloride; a white precipitate of silver chloride forms [1]. Filter, keeping the residue [1]. Wash the residue with distilled water to remove the soluble sodium nitrate, then dry it in a warm oven or between filter papers [1]. Ionic equation: Ag+(aq) + Cl−(aq) → AgCl(s)."
Key RuleIn precipitation, your product is the residue, never the filtrate. There is no evaporation and no crystallisation — but there is a washing step, and it carries a mark.
Exercise 6: "Explain what is meant by an amphoteric oxide. [2]"
Student's Answer"An amphoteric oxide is one that is neither acidic nor basic, so it does not affect the pH of water."
The FlawThat is the definition of a neutral oxide, which is the exact opposite. The student has also implied that solubility in water is the test, when the classification is about reaction with acids and alkalis.
Correct Answer"An amphoteric oxide reacts with both acids and alkalis, forming a salt and water in each case [1]. For example ZnO + 2HCl → ZnCl2 + H2O and ZnO + 2NaOH → Na2ZnO2 + H2O [1]. The IGCSE examples are zinc oxide and aluminium oxide."
Key RuleAmphoteric = both. Neutral = neither. If your definition contains only one of "acid" and "alkali", it cannot score.
Exercise 7: "Write the equation for the reaction of calcium carbonate with dilute sulfuric acid. [2]"
Student's Answer"CaCO3 + H2SO4 → CaSO3 + H2O + CO2"
The FlawThe salt is written as CaSO3 (calcium sulfite) rather than CaSO4 (calcium sulfate). Sulfuric acid, H2SO4, supplies SO42−. Sulfites come from sulfurous acid, H2SO3. Everything else is right, but the equation no longer balances for oxygen.
Correct Answer"CaCO3(s) + H2SO4(aq) → CaSO4(s) + H2O(l) + CO2(g) [2]. Note that calcium sulfate is only slightly soluble, so it forms as a solid coating — which is why hydrochloric acid, giving soluble calcium chloride, is preferred for this reaction in the laboratory."
Key RuleTake the anion straight from the acid formula: H2SO4 → SO42−, HNO3 → NO3−, HCl → Cl−. Never guess from the name alone.
Exercise 8: "A colourless solution turns universal indicator green and conducts electricity well. A student says it must be distilled water. Comment. [2]"
Student's Answer"The student is correct, because green means pH 7 and pH 7 means pure water."
The FlawThe student has ignored the second piece of evidence, which actually disproves the conclusion. Distilled water is a very poor conductor because it contains almost no ions.
Correct Answer"The student is incorrect [1]. pH 7 shows only that the solution is neutral, and many salt solutions are neutral. The good electrical conductivity shows the solution contains a high concentration of mobile ions, so it cannot be distilled water — it is more likely a solution of a neutral salt such as sodium chloride or potassium nitrate [1]."
Key RuleUse every observation the question gives you. Examiners include a second clue precisely to test whether you notice that it contradicts the obvious answer.
Exercise 9: "25.0 cm3 of 0.200 mol/dm3 NaOH is neutralised by 20.0 cm3 of H2SO4. Calculate the concentration of the acid. [3]"
Student's Answer"moles NaOH = 0.200 × 25.0 = 5.00 mol. So moles H2SO4 = 5.00 mol. Concentration = 5.00 ÷ 20.0 = 0.25 mol/dm3."
The FlawTwo independent errors that happen to partly cancel. (1) Volumes were never converted from cm3 to dm3, so "5.00 mol" of sodium hydroxide is out by a factor of 1000 — and physically absurd for 25 cm3 of dilute solution. (2) A 1 : 1 ratio was assumed, but sulfuric acid is diprotic.
Correct Answer"2NaOH + H2SO4 → Na2SO4 + 2H2O, a 2 : 1 ratio.
25.0 cm3 = 0.0250 dm3; 20.0 cm3 = 0.0200 dm3.
moles NaOH = 0.200 × 0.0250 = 5.00 × 10−3 mol [1]
moles H2SO4 = 5.00 × 10−3 ÷ 2 = 2.50 × 10−3 mol [1]
concentration = 2.50 × 10−3 ÷ 0.0200 = 0.125 mol/dm3 [1]"
Key RuleSanity-check every mole answer. A school-laboratory titration deals in millimoles — if your answer has no "× 10−3" in it, you have almost certainly forgotten to divide by 1000.
Exercise 10: "Explain the difference between a base and an alkali. [2]"
Student's Answer"A base is a strong alkali, and an alkali is a weak base. Both have a pH above 7."
The FlawCompletely invented, and it drags in "strong/weak" which is a different distinction altogether. Worse, "both have a pH above 7" is false — an insoluble base such as copper(II) oxide has no pH at all, because it does not form a solution.
Correct Answer"A base is any substance that neutralises an acid to form a salt and water — metal oxides, metal hydroxides, metal carbonates and ammonia [1]. An alkali is a base that is soluble in water, producing OH−(aq) ions; for example NaOH is an alkali, whereas CuO is a base but not an alkali [1]."
Key RuleBase versus alkali is about solubility. Strong versus weak is about dissociation. Two entirely separate ideas — never let one leak into the other.
Exercise 11: "Suggest why sodium carbonate solution can be used to precipitate calcium carbonate. [2]"
Student's Answer"It cannot be used, because all carbonates are insoluble, so sodium carbonate would not dissolve to make a solution."
The FlawAn over-generalised solubility rule, applied confidently in the wrong place. Carbonates are insoluble except those of sodium, potassium and ammonium — and those exceptions are exactly the reagents used to precipitate the others.
Correct Answer"Sodium carbonate is soluble, because all sodium salts are soluble, so it provides carbonate ions in solution [1]. Mixing it with a soluble calcium salt such as calcium chloride gives insoluble calcium carbonate as a white precipitate: Ca2+(aq) + CO32−(aq) → CaCO3(s) [1]."
Key RuleThe exceptions carry the chemistry. "All carbonates are insoluble" and "all sulfates are soluble" are both wrong, and both wrong in the exact place that examiners test.
Exercise 12: "A student wants to show that ethanoic acid is weaker than hydrochloric acid. Suggest a method. [3]"
Student's Answer"Add blue litmus paper to each acid. The stronger acid will turn the litmus a darker red than the weaker one."
The FlawLitmus is a switch, not a scale. It turns red in any acid, whether pH 6 or pH 1, and the shade is not a reliable measure of anything. The method could not distinguish the two acids at all.
Correct Answer"Use solutions of the same concentration and volume so that only strength varies [1]. Then either (a) measure the pH of each with a pH meter (or universal indicator against a colour chart) — the weaker acid gives the higher pH [1]; or (b) add identical pieces of magnesium and compare the rate of effervescence, or measure the electrical conductivity — the weaker acid reacts more slowly and conducts less well [1]."
Key RuleSingle indicators answer "acid or alkali?". Universal indicator and pH meters answer "how acidic?". And any comparison must control the concentration, or you cannot attribute the difference to strength.

✍️ Ultra-Detailed Practice Questions

▼

Ten Cambridge-style challenge questions. Write your answer, then reveal the model answer with mark scheme and examiner's notes.

Question 1
[7 marks]
Aditi in Chennai has two labelled bottles: 0.100 mol/dm3 hydrochloric acid and 0.100 mol/dm3 ethanoic acid. (a) State and explain which has the lower pH. [2] (b) She adds identical 3 cm strips of magnesium ribbon to 50 cm3 of each. Describe and explain the difference in the initial rate. [2] (c) The magnesium fully dissolves in both. State, with a reason, how the total volumes of hydrogen compare. [2] (d) State one further physical measurement that would distinguish the two acids. [1]
Model Answer(a) The hydrochloric acid has the lower pH [1]. It is a strong acid and is fully dissociated, so it has a much higher concentration of H+(aq) ions; ethanoic acid is weak and only partially dissociated, CH3COOH ⇌ H+ + CH3COO− [1]
(b) The hydrochloric acid reacts faster, with more vigorous effervescence [1]. Its higher concentration of H+ ions gives a greater frequency of successful collisions with the magnesium surface [1]
(c) The total volumes are the same [1]. Both acids are monoprotic and present in the same number of moles, and as H+ is used up the ethanoic acid equilibrium shifts to the right so that all of it eventually reacts [1]
(d) Measure the electrical conductivity — the hydrochloric acid conducts better because it contains more mobile ions [1] (accept: measure the temperature rise on neutralisation, or measure pH with a meter)
Examiner's NotesParts (a), (b) and (c) are deliberately arranged so that the answer flips from "different" to "the same". Candidates who have learned only "strong acids react more" answer "HCl is greater" three times and lose part (c) entirely. The equilibrium-shift sentence in (c) is the top-band discriminator — "they have the same number of moles" alone usually earns only one of the two marks. In (b) the word "surface" is worth including: the reaction happens at the magnesium surface, so it is collisions with that surface that matter.
Question 2
[8 marks]
A technician must prepare three salts: (a) magnesium sulfate, (b) potassium chloride, (c) barium sulfate. For each, name the method, state suitable starting materials, and give the single most important purification step with a reason. [6] (d) Explain why the same method cannot be used for all three. [2]
Model Answer(a) Magnesium sulfate — excess solid method. Add excess magnesium oxide (or carbonate, or the metal) to warm dilute sulfuric acid [1]. Key step: filter off the excess solid, because adding the solid in excess guarantees all the acid has reacted and none is left to contaminate the crystals [1]
(b) Potassium chloride — titration. Titrate potassium hydroxide solution against dilute hydrochloric acid [1]. Key step: repeat the titration with no indicator using the volumes found, so the crystals are not stained by the dye [1]
(c) Barium sulfate — precipitation. Mix barium nitrate solution with sodium sulfate solution (or dilute sulfuric acid) [1]. Key step: wash the residue with distilled water to remove the soluble sodium nitrate formed alongside it [1]
(d) Barium sulfate is insoluble, so it must be made by precipitation and collected as the residue [1]. Potassium chloride is soluble but every potassium compound is also soluble, so there is no insoluble reactant to add in excess and filter off — titration is the only option [1]
Examiner's NotesPart (d) is where the marks separate candidates. It asks for the reasoning behind method selection, not a description of the methods again. A full answer names the two deciding properties: solubility of the product, and availability of an insoluble reactant. Note that in (c) examiners accept dilute sulfuric acid as the sulfate source. A frequent error is offering "barium carbonate + sulfuric acid" for (c) — chemically it reacts, but the insoluble product coats the insoluble reactant and the two cannot be separated.
Question 3
[7 marks]
In a titration, 25.0 cm3 of sodium hydroxide solution required 22.50 cm3 of 0.120 mol/dm3 sulfuric acid for complete neutralisation. (a) Write the balanced equation. [1] (b) Calculate the concentration of the sodium hydroxide in mol/dm3. [3] (c) Calculate its concentration in g/dm3. [1] (d) Calculate the maximum mass of sodium sulfate obtainable. [2] (Ar: H = 1, O = 16, Na = 23, S = 32)
Model Answer(a) 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) [1]
(b) 22.50 cm3 = 0.02250 dm3; moles H2SO4 = 0.120 × 0.02250 = 2.70 × 10−3 mol [1]
ratio 1 H2SO4 : 2 NaOH, so moles NaOH = 2 × 2.70 × 10−3 = 5.40 × 10−3 mol [1]
concentration = 5.40 × 10−3 ÷ 0.0250 = 0.216 mol/dm3 [1]
(c) Mr of NaOH = 23 + 16 + 1 = 40; concentration = 0.216 × 40 = 8.64 g/dm3 [1]
(d) moles Na2SO4 = moles H2SO4 = 2.70 × 10−3 mol; Mr = 46 + 32 + 64 = 142 [1]
mass = 2.70 × 10−3 × 142 = 0.383 g [1]
Examiner's NotesNote the direction of travel here: the acid is the known substance, so you multiply by 2 to reach the alkali. In Walkthrough 2 the alkali was known, so you divided by 2. Candidates who memorise "divide by 2 for sulfuric acid" get this backwards and lose two marks. In (d), the ratio is 1 : 1 between the acid and the salt, which is easy to miss after concentrating so hard on the 2 : 1 in part (b) — read the equation again for every new part. Full working attracts error-carried-forward credit; a bare wrong answer attracts none.
Question 4
[6 marks]
Four oxides are labelled P, Q, R and S. P dissolves in water giving pH 13. Q is insoluble but dissolves in both dilute hydrochloric acid and hot aqueous sodium hydroxide. R dissolves in water giving pH 3. S is a gas that is insoluble in water and reacts with neither acids nor alkalis. (a) Classify each oxide. [4] (b) Suggest an identity for Q and S. [1] (c) State what can be deduced about the position in the Periodic Table of the element in R. [1]
Model Answer(a) P is basic (a soluble basic oxide, giving an alkaline solution) [1]. Q is amphoteric, because it reacts with both an acid and an alkali [1]. R is acidic, because it dissolves to give an acidic solution [1]. S is neutral, because it reacts with neither acids nor alkalis [1]
(b) Q could be zinc oxide (ZnO) or aluminium oxide (Al2O3); S could be carbon monoxide (CO), nitrogen monoxide (NO) or dinitrogen oxide (N2O) [1]
(c) The element in R is a non-metal, so it lies on the right-hand side of the Periodic Table [1]
Examiner's NotesThe trap in this question is Q. Because it is described as insoluble, weaker candidates classify it as neutral — but insolubility in water is irrelevant to the acid–base classification, which is decided by reaction with acids and alkalis. Note also that P is described only as "basic", not "alkaline": the oxide itself is basic and the solution it forms is alkaline, and examiners do distinguish these. In (b), any one valid example scores; naming a compound that is not on the standard lists (for example SiO2 for S) does not.
Question 5
[7 marks]
A student prepares hydrated copper(II) sulfate crystals from copper(II) carbonate and dilute sulfuric acid. (a) Write the balanced equation with state symbols. [2] (b) Describe how she would know when enough copper(II) carbonate had been added. [2] (c) Explain why she must not evaporate the filtrate to dryness. [2] (d) She obtains 4.20 g of crystals but the theoretical maximum is 6.25 g. Calculate the percentage yield. [1]
Model Answer(a) CuCO3(s) + H2SO4(aq) → CuSO4(aq) + H2O(l) + CO2(g) — correct products [1], correct state symbols [1]
(b) The effervescence stops, showing no more carbon dioxide is being produced [1], and solid remains undissolved at the bottom of the beaker, showing the acid has all been used up and the carbonate is now in excess [1]
(c) Copper(II) sulfate crystallises as the hydrated salt CuSO4·5H2O [1]. Evaporating to dryness would drive off the water of crystallisation, leaving a white anhydrous powder instead of the blue crystals required [1]
(d) percentage yield = (4.20 ÷ 6.25) × 100 = 67.2% [1]
Examiner's NotesPart (b) is worth two marks because there are two independent signals — the gas stopping and the solid remaining — and candidates who give only one score one. Part (c) requires the phrase "water of crystallisation" explicitly; "the crystals would be spoiled" is not credited. Expect a follow-up asking for reasons why the yield is below 100%: acceptable answers include crystals lost on the filter paper, some product remaining in solution, and losses during transfer between containers. Note that "the reaction is reversible" is not acceptable here.
Question 6
[6 marks]
A farmer near Birmingham finds his field has a soil pH of 4.5. He treats it with powdered limestone. (a) Name the type of reaction taking place and write an ionic equation. [2] (b) Explain why calcium carbonate is preferred to calcium oxide even though calcium oxide acts faster. [2] (c) His neighbour applies ammonium sulfate fertiliser to a field on the same day it is limed, and notices a strong smell. Explain the chemistry and the loss to the farmer. [2]
Model Answer(a) Neutralisation — an acid reacting with a base (a carbonate) [1]. Ionic equation: CaCO3(s) + 2H+(aq) → Ca2+(aq) + H2O(l) + CO2(g) [1]
(b) Calcium carbonate is insoluble, so it reacts only as fast as acid is available and cannot raise the pH far above neutral — it is self-limiting [1]. Calcium oxide reacts with water to give calcium hydroxide, a strong alkali, which can overshoot to a pH of 10 or more and damage crops and soil nutrient availability [1]
(c) A base reacts with an ammonium salt to release ammonia gas: NH4+ + OH− → NH3 + H2O, which is the pungent smell [1]. The nitrogen the farmer paid for escapes to the atmosphere instead of feeding the crop, so the fertiliser is wasted — lime and fertiliser should be applied weeks apart [1]
Examiner's NotesPart (c) is a genuine application question and it rewards candidates who recognise the laboratory test for the ammonium ion in an agricultural setting — same reaction, different scale. In (a), an ionic equation using H+ is expected because soil acidity is caused by hydrogen ions; writing the full equation with a named acid is usually accepted but the H+ version is the model. Part (b) must contrast both substances; describing only the limestone earns one mark. The word "self-limiting" is not required but the idea behind it is.
Question 7
[7 marks]
6.90 g of a hydrated sodium carbonate, Na2CO3·xH2O, is heated to constant mass, leaving 2.65 g of the anhydrous salt. (a) Calculate x. [4] (b) Explain what "heated to constant mass" means and why it is necessary. [2] (c) A second student obtained x = 8.7. Suggest, with reasoning, one cause. [1] (Ar: H = 1, C = 12, O = 16, Na = 23)
Model Answer(a) mass of water lost = 6.90 − 2.65 = 4.25 g [1]
Mr of Na2CO3 = 46 + 12 + 48 = 106; moles = 2.65 ÷ 106 = 0.0250 mol [1]
moles of H2O = 4.25 ÷ 18 = 0.2361 mol [1]
ratio = 0.2361 ÷ 0.0250 = 9.44 ≈ 10, so x = 10 and the salt is Na2CO3·10H2O (washing soda) [1]
(b) The sample is heated, cooled and weighed repeatedly until two successive weighings are the same [1]. This shows that no further water is being lost, so the dehydration is complete and the residue really is the anhydrous salt [1]
(c) A value of 8.7 is too low, meaning too little water appeared to be lost, so the residue was too heavy — most likely the salt was not heated for long enough and some water of crystallisation remained [1] (accept: the residue absorbed moisture from the air while cooling)
Examiner's NotesNotice that 9.44 rounds to 10, not to 9 — candidates sometimes truncate rather than round, or panic because the number is not clean. Real experimental data rarely gives exact integers, and examiners expect you to choose the nearest sensible whole number and, ideally, to recognise the compound. Part (c) requires directional reasoning: decide first whether the error made the residue heavier or lighter, then propose only a cause consistent with that direction. "Some solid spat out of the crucible" would make the residue lighter and give x too high, so it does not fit and would not be credited here.
Question 8
[6 marks]
A student is given four colourless solutions: dilute hydrochloric acid, dilute ethanoic acid, sodium hydroxide solution and sodium chloride solution. Using only universal indicator paper, a magnesium ribbon and a conductivity meter, describe how she could identify all four, giving expected results. [6]
Model AnswerTest 1 — universal indicator on each. The sodium hydroxide turns it blue/violet, pH about 13–14 [1]. The sodium chloride turns it green, pH 7 [1]. Both acids turn it red or orange, so a further test is needed to separate them [1]
Test 2 — distinguishing the two acids. The hydrochloric acid gives a lower pH (red, about 1) than the ethanoic acid (orange, about 3), because it is a strong acid and fully dissociated [1]
Test 3 — confirmation with magnesium. Both acids fizz, producing hydrogen, but the hydrochloric acid fizzes much more vigorously because of its higher H+ concentration; neither the sodium hydroxide nor the sodium chloride reacts [1]
Test 4 — conductivity. Sodium chloride and sodium hydroxide both conduct well (fully ionic in solution); hydrochloric acid conducts well; ethanoic acid conducts poorly, confirming it is only partially dissociated [1]
Examiner's NotesThe examiner is checking that you know the limits of each technique. Universal indicator alone cannot finish the job, and saying so explicitly is worth a mark. A very common error is claiming that sodium chloride solution will not conduct because it is neutral — neutrality says nothing about the presence of ions. Another is expecting magnesium to react with the alkali. Structure your answer as a sequence of tests with a conclusion after each; an unstructured list of observations rarely gains full credit.
Question 9
[7 marks]
Zinc oxide and magnesium oxide are both white insoluble powders. (a) Describe an experiment to distinguish them, with expected observations. [3] (b) Write equations for the reactions of zinc oxide with (i) dilute hydrochloric acid and (ii) hot aqueous sodium hydroxide, naming the salt in each case. [3] (c) State the term used to describe an oxide that behaves in this way and explain why aluminium cookware should not be cleaned with a strong alkali. [1]
Model Answer(a) Add a spatula of each powder to a separate test tube of hot concentrated aqueous sodium hydroxide and warm gently [1]. The zinc oxide dissolves to give a colourless solution [1]; the magnesium oxide remains undissolved as a white solid, because it is purely basic [1]
(b)(i) ZnO(s) + 2HCl(aq) → ZnCl2(aq) + H2O(l); the salt is zinc chloride [1]
(ii) ZnO(s) + 2NaOH(aq) → Na2ZnO2(aq) + H2O(l); the salt is sodium zincate [1]
Both reactions produce a salt and water [1]
(c) Amphoteric. The protective aluminium oxide layer on the cookware is also amphoteric and dissolves in strong alkali (Al2O3 + 2NaOH → 2NaAlO2 + H2O), exposing the metal beneath to attack [1]
Examiner's NotesIn (a) the test with acid is worthless as a discriminator, because both oxides are basic and both dissolve — candidates who propose it score zero for the experiment even if the observations are described well. Choose the test that exploits the difference. In (b), "sodium zincate" is on the mark scheme by name; "sodium zinc oxide" is not accepted. Part (c) is a classic application mark: examiners want the connection between an abstract classification and a practical consequence, so a bare "amphoteric" without the cookware explanation scores nothing.
Question 10
[8 marks]
A student writes: "To make pure dry crystals of sodium chloride, I will add excess sodium metal to 25 cm3 of dilute hydrochloric acid, add a few drops of litmus, filter off the unreacted sodium, and then heat the filtrate strongly until all the water has evaporated." (a) Identify and correct four errors in this method. [4] (b) Write the correct method, including one safety precaution. [3] (c) Explain why a titration is unavoidable for this particular salt. [1]
Model Answer(a) Error 1: sodium metal reacts explosively and dangerously with acid and is never used — use sodium hydroxide solution instead [1]. Error 2: excess cannot be used, because every sodium compound is soluble, so nothing would remain to filter [1]. Error 3: the filtration step is impossible for the same reason, and litmus is the wrong approach here — an indicator is needed, but as part of a titration, and the preparation must then be repeated without it [1]. Error 4: heating until all the water evaporates is wrong — evaporate only to the point of crystallisation, then cool [1]
(b) Pipette 25.0 cm3 of sodium hydroxide solution into a conical flask and add 2–3 drops of methyl orange; titrate with dilute hydrochloric acid from a burette until the indicator just changes colour permanently, and record the titre [1]. Repeat using exactly the same volumes but with no indicator, so the crystals are not contaminated [1]. Evaporate to the point of crystallisation, cool slowly, filter and dry the crystals between filter papers. Safety: wear eye protection — sodium hydroxide is corrosive [1]
(c) Both reactants are solutions and both the salt and every sodium compound are soluble, so there is no insoluble solid to add in excess and filter off; the exact volume must therefore be measured [1]
Examiner's NotesThis is a synoptic question and the highest-demand style in Topic 7. Marks in (a) require the error and its correction as a pair — simply listing what is wrong scores nothing. The safety mark in (b) must be relevant: "wear eye protection because sodium hydroxide is corrosive" scores, while "be careful" does not. Part (c) is testing whether you can articulate the principle behind method selection rather than reciting a recipe; answers that describe the titration procedure again, instead of explaining why it is forced, do not gain the mark.
IGCSE Chemistry 0620 — Topic 7 Challenge Prep Guide. Work through every section, and revisit the twelve traps the night before your exam.