IGCSE Chemistry Paper 4 (Theory / Extended) Challenge

Topic 7: Acids, Bases and Salts -- Challenge Paper

Duration1 hour 15 minutes

Total Marks80

Questions7

Time Remaining75:00

Syllabus0620

Instructions -- Challenge Paper

This paper is designed to stretch A* candidates with multi-step reasoning, data analysis and method evaluation.
Answer all questions in the spaces provided.
Show all working for calculations -- method marks are available even if the final answer is incorrect.
Use appropriate scientific terminology.
Your answers will be automatically graded when you submit.

Question Navigation

Question 1 -- Deducing the Basicity of an Unknown Acid

Total: 12 marks

A researcher in Pune isolates a white crystalline acid, labelled X, from a plant extract. 1.26 g of X is dissolved and made up to 250 cm³ of solution. 25.0 cm³ portions of this solution require a mean titre of 20.0 cm³ of 0.100 mol/dm³ sodium hydroxide for neutralisation using phenolphthalein. Separately, the relative molecular mass of X is found to be 126.

(a) [5]

(i) Calculate the number of moles of sodium hydroxide used in one titration. [1]
(ii) Calculate the number of moles of X present in the 250 cm³ of solution. [2]
(iii) Deduce the number of moles of sodium hydroxide that react with one mole of X, and hence state the basicity of X. [2]

Model Answer -- 1(a)

(i) Moles NaOH = 0.100 × 20.0/1000 = 2.00 × 10⁻³ mol [1]

(ii) Moles of X in 250 cm³ = mass ÷ Mₕ = 1.26/126 = 1.00 × 10⁻² mol [1]

(ii) Moles of X in each 25.0 cm³ portion = 1.00 × 10⁻² ÷ 10 = 1.00 × 10⁻³ mol [1]

(iii) Ratio NaOH : X = 2.00 × 10⁻³ : 1.00 × 10⁻³ = 2 : 1 [1]

(iii) X is therefore dibasic — each molecule can release two hydrogen ions, like sulfuric acid [1]

⚠ If you missed marks here: The trap is comparing the moles of alkali in one titration with the moles of acid in the whole 250 cm³. You must scale one to the other; forgetting the factor of ten makes X appear twenty-basic, which should immediately look impossible. Basicity means replaceable hydrogen atoms per molecule, not the strength of the acid.

1 mark -- moles NaOH 0.100 0.0200 2.00 1 mark -- moles X 1.26 126 1.00 10 2 total 1 mark -- portion divide ten 1.00 10 3 1 mark -- ratio 2 1 NaOH acid 1 mark -- dibasic two hydrogen ions replaceable sulfuric

(b) [4]

(i) Calculate the concentration of the solution of X in mol/dm³ and in g/dm³. [2]

(ii) If X is represented by the formula H₂A, write the balanced equation for its reaction with sodium hydroxide and give the formula of the salt formed. [2]

Model Answer -- 1(b)

(i) Concentration = 1.00 × 10⁻² mol ÷ 0.250 dm³ = 0.0400 mol/dm³ [1]

(i) In g/dm³: 0.0400 × 126 = 5.04 g/dm³ (or directly, 1.26 g in 0.250 dm³) [1]

(ii) H₂A + 2NaOH → Na₂A + 2H₂O [1]

(ii) The salt formed is Na₂A, the disodium salt, because both replaceable hydrogen atoms have been substituted by sodium [1]

⚠ If you missed marks here: Converting between mol/dm³ and g/dm³ requires multiplying by the relative molecular mass; dividing gives a value smaller than the concentration in mol/dm³, which is a clear signal of an error. Writing NaA for the salt would only be correct for a monobasic acid.

1 mark -- concentration 0.0100 0.250 0.0400 mol 1 mark -- multiply 126 5.04 g dm3 1 mark -- H2A 2NaOH Na2A 2H2O balanced 1 mark -- salt Na2A disodium both hydrogen substituted

The researcher's assistant suggests that the basicity could have been found more quickly by measuring the pH of the solution of X. Evaluate this suggestion, and explain what additional experiment would be needed to decide whether X is a strong or a weak acid. [3]

Model Answer -- 1(c)

The suggestion is unsound: pH measures the hydrogen ion concentration actually present, which depends on both the concentration and the degree of dissociation, so it cannot on its own reveal how many replaceable hydrogen atoms each molecule has [1]

Only the titration, which counts the total moles of alkali needed to remove every acidic hydrogen, can establish basicity [1]

To decide on strength, compare the measured pH of a solution of known concentration with the value expected for full dissociation (or compare its electrical conductivity, or its initial rate of reaction with magnesium, against a strong acid of the same concentration) [1]

⚠ If you missed marks here: Do not confuse basicity with strength. A dibasic acid can be weak (as carbonic acid is) and a monobasic acid can be strong (as nitric acid is). The distinction the examiner wants is that titration counts total acidic hydrogens while pH reports only those released at that moment.

1 mark -- pH hydrogen ion concentration depends dissociation cannot reveal replaceable 1 mark -- titration counts total moles alkali every acidic hydrogen basicity 1 mark -- compare pH expected full dissociation conductivity rate magnesium strong

Question 2 -- Investigating Acid Strength Quantitatively

Total: 12 marks

A class in Nottingham investigates four acids, all at 0.100 mol/dm³, by adding an identical 0.120 g piece of magnesium ribbon to 150 cm³ of each and measuring the volume of hydrogen collected after 30 seconds and the total volume collected when the reaction stopped. (Aₕ: Mg 24; one mole of any gas occupies 24.0 dm³ at room conditions.)

Acid	pH	Volume of H₂ after 30 s / cm³	Final volume of H₂ / cm³
hydrochloric	1.0	84	120
nitric	1.0	82	120
ethanoic	2.9	19	120
methanoic	2.4	31	120

(a) [4]

(i) Explain why the volumes collected after 30 seconds differ so much, but the final volumes are identical. [3]

(ii) Use the data to place the four acids in order of increasing strength, and justify your order. [1]

Model Answer -- 2(a)

(i) The initial rate depends on the hydrogen ion concentration; the strong acids are fully dissociated so they have a hundred times more H⁺ than ethanoic acid at the same concentration, giving far more frequent successful collisions with the magnesium [1]

(i) The final volume depends only on the limiting reactant, and in every flask this is the 0.120 g of magnesium (0.00500 mol), which can produce at most 0.00500 mol of hydrogen [1]

(i) As the weak acids react, their hydrogen ions are replaced by further dissociation of the undissociated molecules, so eventually all the acid becomes available and the same total volume is released [1]

(ii) Increasing strength: ethanoic < methanoic < nitric ≈ hydrochloric, justified by the falling pH values and the rising 30-second volumes [1]

⚠ If you missed marks here: The most frequently lost mark is the explanation of why the weak acids catch up. Saying only that they are slower does not explain the identical final volume — you must state that dissociation continues as hydrogen ions are consumed. Note that the two strong acids are effectively equal in strength, so ranking them apart is not required.

1 mark -- initial rate hydrogen ion concentration strong fully dissociated frequent collisions 1 mark -- final volume limiting reactant magnesium 0.120 0.00500 moles 1 mark -- weak further dissociation undissociated molecules all acid available same total 1 mark -- order ethanoic methanoic nitric hydrochloric pH 30 second volumes

(b) [4]

(i) Calculate the number of moles of magnesium used, and hence the volume of hydrogen expected at room conditions. Compare this with the measured final volume of 120 cm³ and comment. [3]

(ii) The class assumed the acid was in excess. Show by calculation that this assumption is correct for the hydrochloric acid. [1]

Model Answer -- 2(b)

(i) Moles of Mg = 0.120/24 = 5.00 × 10⁻³ mol; the equation Mg + 2H⁺ → Mg²⁺ + H₂ gives the same number of moles of hydrogen [1]

(i) Expected volume = 5.00 × 10⁻³ × 24.0 dm³ = 0.120 dm³ = 120 cm³ [1]

(i) This agrees exactly with the measured value, which confirms that all the magnesium reacted and that no significant gas escaped or dissolved [1]

(ii) Moles of HCl available = 0.100 × 150/1000 = 1.50 × 10⁻² mol; the magnesium needs 2 × 5.00 × 10⁻³ = 1.00 × 10⁻² mol, so 5.00 × 10⁻³ mol of acid is left over and the acid is genuinely in excess [1]

⚠ If you missed marks here: Forgetting that each mole of magnesium needs two moles of hydrogen ions halves or doubles the answer. In part (ii) it is not enough to say there is plenty of acid; you must compare the moles available with the moles required and quote both figures.

1 mark -- moles Mg 0.120 24 5.00 equation one mole hydrogen 1 mark -- expected volume 24.0 0.120 dm3 120 cm3 1 mark -- agrees measured all magnesium reacted no gas escaped 1 mark -- moles HCl 0.0150 needs two times 0.0100 left over excess

Identify two significant weaknesses in this investigation and, for each, describe a specific improvement and explain how it would make the conclusion more reliable. [4]

Model Answer -- 2(c)

Weakness: gas escapes in the moments between adding the magnesium and fitting the bung, so early readings are too low. Improvement: use a flask with a divided or tipping side-arm so the metal can be added without opening the system [1]

Improvement rationale: this makes the 30-second volumes directly comparable between acids, which is what the strength conclusion rests on [1]

Weakness: only a single run of each acid was done and the temperature was not controlled, so an anomalous result would go undetected and warmer acid would appear stronger. Improvement: repeat each acid at least three times and hold all four flasks in the same thermostatic water bath [1]

Improvement rationale: repeats allow a mean to be taken and anomalies to be spotted, and a fixed temperature removes a variable that also changes the rate, so any remaining difference can be attributed to acid strength [1]

⚠ If you missed marks here: Vague answers such as be more accurate or use better equipment score nothing. Each weakness must be paired with a specific practical change and with a statement of why the conclusion becomes safer. Suggesting a larger piece of magnesium is not an improvement, because it does not address any source of error.

1 mark -- gas escapes bung fitted early readings low divided side arm flask 1 mark -- comparable 30 second volumes strength conclusion rests 1 mark -- single run temperature not controlled repeat three times water bath 1 mark -- mean anomalies spotted fixed temperature variable removed attributed strength

Question 3 -- Identifying Elements from the Behaviour of Their Oxides

Total: 12 marks

Three unlabelled oxides, D, E and F, are investigated by a student in Melbourne. Each is shaken with water, then tested with dilute hydrochloric acid and with hot concentrated sodium hydroxide solution. The results are shown in the table.

Oxide	Shaken with water	With dilute HCl	With hot concentrated NaOH
D	dissolves, pH 13	dissolves, salt formed	no reaction
E	insoluble, pH stays 7	dissolves, salt formed	dissolves, salt formed
F	dissolves, pH 2	no reaction	dissolves, salt formed

(a) [4]

Classify D, E and F, justify each classification using the table, and explain why the water test alone would have been insufficient to classify E. [4]

Model Answer -- 3(a)

D is basic: it gives an alkaline solution with water and forms a salt with the acid but does not react with the alkali [1]

E is amphoteric: it forms a salt with the acid and also forms a salt with the hot alkali [1]

F is acidic: it gives an acidic solution with water and forms a salt with the alkali but does not react with the acid [1]

The water test alone would have shown only that E is insoluble and leaves the pH at 7, which is exactly what a neutral oxide would do; the acid and alkali tests are what separate amphoteric from neutral [1]

⚠ If you missed marks here: Amphoteric and neutral oxides can look identical in the water test, and this catches many candidates out. Only the pair of reactions with an acid and an alkali distinguishes them: amphoteric reacts with both, neutral reacts with neither.

1 mark -- D basic alkaline solution salt acid no reaction alkali 1 mark -- E amphoteric salt acid salt hot alkali 1 mark -- F acidic acidic solution salt alkali no reaction acid 1 mark -- water test insoluble pH 7 same neutral acid alkali tests separate

(b) [4]

(i) Suggest one possible identity for each of D, E and F, giving the formula in each case. [3]

(ii) Write a balanced equation for the reaction of your chosen oxide F with sodium hydroxide. [1]

Model Answer -- 3(b)

(i) D could be sodium oxide, Na₂O (accept potassium oxide, K₂O, or calcium oxide, CaO) [1]

(i) E could be zinc oxide, ZnO (accept aluminium oxide, Al₂O₃, or lead(II) oxide, PbO) [1]

(i) F could be sulfur dioxide, SO₂ (accept sulfur trioxide, SO₃, or carbon dioxide, CO₂) [1]

(ii) For example SO₂ + 2NaOH → Na₂SO₃ + H₂O or CO₂ + 2NaOH → Na₂CO₃ + H₂O [1]

⚠ If you missed marks here: Magnesium oxide is a poor choice for D because it is only sparingly soluble and gives a pH nearer 10 than 13. For F, remember that the salt formed from sulfur dioxide is a sulfite, Na₂SO₃, not a sulfate; only sulfur trioxide or sulfuric acid gives sulfates.

1 mark -- D sodium oxide Na2O potassium calcium 1 mark -- E zinc oxide ZnO aluminium lead 1 mark -- F sulfur dioxide SO2 carbon dioxide 1 mark -- equation SO2 2NaOH Na2SO3 H2O balanced

A fourth oxide, G, is the oxide of an element in Group IV, Period 3. Predict its acid–base character, predict what it would do in each of the three tests in the table, and explain your prediction in terms of the position of the element in the Periodic Table. [4]

Model Answer -- 3(c)

The element is silicon, so G is silicon(IV) oxide, SiO₂, and it is an acidic oxide [1]

With water: no visible reaction and the pH stays at 7, because silicon(IV) oxide has a giant covalent structure and is insoluble [1]

With dilute hydrochloric acid: no reaction; with hot concentrated sodium hydroxide: it slowly dissolves to form sodium silicate and water [1]

The prediction follows because silicon is a non-metal (a metalloid on the borderline), and across a period the oxides change from basic through amphoteric at aluminium to acidic, so the element immediately after aluminium gives a weakly acidic oxide [1]

⚠ If you missed marks here: Many candidates classify silicon(IV) oxide as neutral because it is insoluble in water and does not affect the pH. Solubility is not the criterion — the reaction with hot alkali is what establishes that it is acidic. This mirrors the trap in part (a) about oxide E.

1 mark -- silicon oxide SiO2 acidic Group IV Period 3 1 mark -- water no reaction pH 7 giant covalent insoluble 1 mark -- no reaction acid dissolves hot alkali sodium silicate 1 mark -- non metal metalloid across period basic amphoteric aluminium acidic

Question 4 -- Designing and Evaluating a Salt Preparation

Total: 12 marks

A student in Toronto must prepare 5.00 g of pure dry hydrated nickel(II) sulfate crystals, NiSO₄·7H₂O (Mₕ = 281). Nickel lies just above hydrogen in the reactivity series but reacts only very slowly with dilute acids at room temperature. Nickel(II) carbonate and nickel(II) oxide are both insoluble green solids. (Mₕ: NiCO₃ 119; NiO 75)

(a) [4]

Three routes are possible: (A) nickel metal with warm dilute sulfuric acid; (B) excess nickel(II) carbonate with warm dilute sulfuric acid; (C) mixing nickel(II) chloride solution with sodium sulfate solution. Evaluate all three and justify the route you would choose. [4]

Model Answer -- 4(a)

Route A would work in principle because nickel is above hydrogen, but the reaction is very slow, so it would take a long time and it is hard to judge when the acid has all been used [1]

Route C fails completely: nickel(II) sulfate is soluble, so no precipitate forms and the product could not be separated from the sodium chloride also present [1]

Route B is the best choice: the carbonate is insoluble so an excess can be added and filtered off, and the fizzing stops visibly when all the acid has reacted [1]

Route B also gives only carbon dioxide as a by-product, which escapes, so nothing is left to contaminate the salt solution [1]

⚠ If you missed marks here: Rejecting route A outright on the grounds that nickel is below hydrogen is a factual error — it is above hydrogen but kinetically slow, so the objection is practical, not thermodynamic. Route C is the one that is chemically impossible, and it must be rejected for the right reason.

1 mark -- route A above hydrogen very slow long time hard judge acid used 1 mark -- route C nickel sulfate soluble no precipitate sodium chloride separated 1 mark -- route B carbonate insoluble excess filtered fizzing stops 1 mark -- carbon dioxide by product escapes nothing contaminate

(b) [5]

The student uses 100 cm³ of 0.200 mol/dm³ sulfuric acid.

(i) Calculate the moles of acid used and the maximum mass of NiSO₄·7H₂O that could be obtained. [3]
(ii) Calculate the minimum mass of nickel(II) carbonate needed to react with all the acid, and explain why the student should weigh out more than this. [2]

Model Answer -- 4(b)

(i) Moles of H₂SO₄ = 0.200 × 100/1000 = 2.00 × 10⁻² mol [1]

(i) The equation NiCO₃ + H₂SO₄ → NiSO₄ + H₂O + CO₂ is 1 : 1, so 2.00 × 10⁻² mol of the salt can form [1]

(i) Maximum mass = 2.00 × 10⁻² × 281 = 5.62 g, which is more than the 5.00 g required, so the quantities are adequate [1]

(ii) Minimum mass of NiCO₃ = 2.00 × 10⁻² × 119 = 2.38 g [1]

(ii) More than this must be weighed out because the method depends on having a visible excess left over; only then can the student be sure that every last trace of acid has been neutralised before filtering [1]

⚠ If you missed marks here: Using the relative formula mass of the anhydrous salt (155) instead of the hydrated salt (281) is the most frequent error here and understates the yield badly. Read the formula asked for in the question. Note also that the excess is deliberate, so calculating only the exact stoichiometric mass misses the second mark.

1 mark -- moles H2SO4 0.200 0.100 2.00 10 2 1 mark -- equation 1 1 ratio same moles salt 1 mark -- 281 multiply 5.62 g more than 5.00 adequate 1 mark -- 119 multiply 2.38 g minimum carbonate 1 mark -- visible excess left over certain all acid neutralised filtering

The student obtains 4.21 g of crystals. Calculate the percentage yield based on your answer to (b)(i), and give two specific practical reasons why the yield falls short. State clearly why loss of mass in the chemical reaction itself is not one of them. [3]

Model Answer -- 4(c)

Percentage yield = 4.21/5.62 × 100 = 74.9 % (accept 75 %) [1]

Two practical reasons: some of the salt stays dissolved in the mother liquor that is poured off after crystallisation, and some solution is left wetting the filter paper, beaker and evaporating basin (accept: crystals lost when scraped out, or crystallisation stopped before complete) [1]

Mass cannot be lost in the reaction because atoms are only rearranged, not destroyed; the total mass of reactants equals the total mass of products, so every shortfall is a physical transfer loss [1]

⚠ If you missed marks here: Answers that say the reaction did not go to completion are weak unless justified — here the acid was fully neutralised by design. The examiner is looking for named transfer losses. Claiming mass was lost as carbon dioxide is also wrong: that gas comes from the carbonate, not from the salt being weighed.

1 mark -- 4.21 5.62 100 74.9 75 percentage yield 1 mark -- mother liquor poured off solution wetting filter paper beaker basin 1 mark -- atoms rearranged not destroyed conservation physical transfer loss

Question 5 -- Back Titration of an Impure Carbonate

Total: 10 marks

A limestone sample from a quarry near Buxton is analysed. A 2.50 g portion of the powdered limestone is added to 100 cm³ of 1.00 mol/dm³ hydrochloric acid, which is more than enough to dissolve all the calcium carbonate present. When the fizzing has stopped, the mixture is filtered to remove insoluble impurities and the filtrate is made up to 250 cm³. A 25.0 cm³ portion of this solution needs 12.0 cm³ of 0.500 mol/dm³ sodium hydroxide to neutralise the leftover acid. (Mₕ(CaCO₃) = 100)

(a) [4]

(i) Calculate the moles of sodium hydroxide used in one titration and hence the moles of unreacted hydrochloric acid in the 25.0 cm³ portion. [2]
(ii) Calculate the total moles of unreacted acid in the whole 250 cm³. [1]
(iii) Calculate the moles of hydrochloric acid added at the start. [1]

Model Answer -- 5(a)

(i) Moles NaOH = 0.500 × 12.0/1000 = 6.00 × 10⁻³ mol [1]

(i) NaOH + HCl react 1 : 1, so unreacted HCl in the 25.0 cm³ portion = 6.00 × 10⁻³ mol [1]

(ii) Total unreacted HCl in 250 cm³ = 6.00 × 10⁻³ × 10 = 6.00 × 10⁻² mol [1]

(iii) Moles of HCl added originally = 1.00 × 100/1000 = 0.100 mol [1]

⚠ If you missed marks here: The scaling factor of ten between the 25.0 cm³ portion and the full 250 cm³ is essential and is missed by many candidates. Watch the wording too: the titration measures the acid that is left over, not the acid that reacted with the carbonate.

1 mark -- moles NaOH 0.500 0.0120 6.00 1 mark -- 1 1 ratio unreacted HCl portion 6.00 1 mark -- scale ten 250 total 6.00 10 2 0.0600 1 mark -- added 1.00 0.100 dm3 0.100 mol

(b) [4]

(i) Calculate the moles of hydrochloric acid that reacted with the calcium carbonate. [1]
(ii) Using CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂, calculate the moles and mass of calcium carbonate in the 2.50 g sample. [2]
(iii) Calculate the percentage purity of the limestone. [1]

Model Answer -- 5(b)

(i) Moles reacted = 0.100 − 0.0600 = 0.0400 mol [1]

(ii) The ratio is 1 CaCO₃ : 2 HCl, so moles of CaCO₃ = 0.0400/2 = 0.0200 mol [1]

(ii) Mass of CaCO₃ = 0.0200 × 100 = 2.00 g [1]

(iii) Percentage purity = (mass of CaCO₃ ÷ mass of sample) × 100 = (2.00/2.50) × 100 = 80.0 % [1]

⚠ If you missed marks here: The 1 : 2 mole ratio must be applied when converting from acid to carbonate; leaving it out doubles the mass and gives an impossible purity above 100 %. Always finish a purity calculation by checking that the answer lies between 0 and 100 %; anything outside that range means a ratio or a scaling factor has gone astray.

1 mark -- moles reacted 0.100 0.0600 0.0400 1 mark -- ratio 1 2 divide two 0.0200 moles carbonate 1 mark -- mass 100 multiply 2.00 g 1 mark -- purity 2.00 2.50 80.0 per cent

Explain why the analyst used a back titration rather than titrating the limestone directly with acid, and state one precaution needed to make the result valid. [2]

Model Answer -- 5(c)

Limestone is an insoluble solid that reacts slowly and cannot be added from a burette, so no sharp end point could be seen in a direct titration; adding a known excess of acid and measuring what is left avoids the problem [1]

Precaution: make certain the acid really is in excess and that the reaction is complete (fizzing has entirely stopped, powder finely ground and mixture swirled or warmed) before filtering, otherwise unreacted carbonate would make the leftover acid appear smaller and the purity too high [1]

⚠ If you missed marks here: It is not enough to say a back titration is easier. The mark depends on identifying that the solid is insoluble and reacts too slowly for a sharp end point. For the precaution, generic answers such as be accurate are not credited — name the specific risk of incomplete reaction.

1 mark -- insoluble solid slow cannot burette no sharp end point known excess measure left 1 mark -- acid excess reaction complete fizzing stopped ground swirled before filtering

Question 6 -- Gravimetric Analysis by Precipitation

Total: 12 marks

A fertiliser blend sold near Ludhiana is claimed to contain ammonium sulfate as its only sulfate. To check this, a technician dissolves 4.00 g of the fertiliser in distilled water, adds excess aqueous barium chloride, filters, washes and dries the precipitate, and weighs it. The dry precipitate has a mass of 4.66 g. (Mₕ: BaSO₄ 233; (NH₄)₂SO₄ 132)

(a) [4]

(i) Name the precipitate and write the ionic equation with state symbols. [2]

(ii) Explain why barium chloride is added in excess, and why the precipitate must be washed before it is dried and weighed. [2]

Model Answer -- 6(a)

(i) The white precipitate is barium sulfate [1]

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

(ii) Excess barium chloride guarantees that every sulfate ion in the sample is precipitated, so the mass weighed corresponds to all the sulfate present and the result is not an underestimate [1]

(ii) Washing with distilled water removes the soluble ions (barium, chloride, ammonium) clinging to the solid; if they dried on the precipitate the mass would be too high and the calculated sulfate content would be overstated [1]

⚠ If you missed marks here: Note the direction of each error: too little barium chloride gives a mass that is too low, while failure to wash gives a mass that is too high. Examiners award the mark for the direction as well as for the fact. Using dilute sulfuric acid instead of barium chloride would of course add sulfate to the sample.

1 mark -- barium sulfate white precipitate 1 mark -- Ba2 SO4 BaSO4 aq s ionic state symbols 1 mark -- excess every sulfate ion precipitated not underestimate 1 mark -- washing removes soluble barium chloride ammonium mass too high overstated

(b) [5]

(i) Calculate the moles of barium sulfate obtained and hence the moles of sulfate ions in the sample. [2]
(ii) Calculate the mass of ammonium sulfate this corresponds to. [2]
(iii) Calculate the percentage of ammonium sulfate in the fertiliser. [1]

Model Answer -- 6(b)

(i) Moles of BaSO₄ = 4.66/233 = 0.0200 mol [1]

(i) Each formula unit of barium sulfate contains one sulfate ion, so moles of sulfate in the sample = 0.0200 mol [1]

(ii) Each formula unit of (NH₄)₂SO₄ also contains one sulfate ion, so moles of ammonium sulfate = 0.0200 mol [1]

(ii) Mass = 0.0200 × 132 = 2.64 g [1]

(iii) Percentage = (2.64/4.00) × 100 = 66.0 % [1]

⚠ If you missed marks here: The two 1 : 1 relationships are easy to state but must be stated: one sulfate per barium sulfate and one sulfate per ammonium sulfate. Candidates who introduce a factor of two because the formula contains two ammonium ions halve or double the final percentage.

1 mark -- moles BaSO4 4.66 233 0.0200 1 mark -- one sulfate per formula unit moles sulfate same 1 mark -- one sulfate ammonium sulfate 0.0200 moles 1 mark -- mass 132 multiply 2.64 g 1 mark -- percentage 2.64 4.00 100 66.0

The manufacturer claims the fertiliser is 70 % ammonium sulfate. Comment on whether the result supports the claim, suggest one chemical reason other than deliberate mislabelling why the measured figure might be lower, and describe a further test that would confirm that ammonium ions really are present. [3]

Model Answer -- 6(c)

The measured 66.0 % is close to but below the claimed 70 %, so the claim is not fully supported; the difference of about 4 percentage points is larger than would be expected from weighing errors alone [1]

A chemical reason: the fertiliser may have absorbed moisture from the humid air (ammonium sulfate is slightly hygroscopic), so part of the 4.00 g weighed was water and the sulfate content per gram appears lower [1]

Confirmatory test: warm a portion of the fertiliser with aqueous sodium hydroxide and test the gas with damp red litmus paper; ammonia is released and turns the paper blue, confirming ammonium ions [1]

⚠ If you missed marks here: The comment mark requires a numerical comparison, not just the word close. For the confirmatory test, the gas must be tested with damp litmus — dry paper will not change colour because the ammonia needs to dissolve first. Simply smelling the gas is not an acceptable school test.

1 mark -- 66.0 below claimed 70 not fully supported difference weighing 1 mark -- absorbed moisture humid air hygroscopic water content per gram lower 1 mark -- warm sodium hydroxide damp red litmus ammonia blue ammonium ions

Question 7 -- Determining a Formula from Combined Evidence

Total: 10 marks

A student in Accra is given a sample of a hydrated metal sulfate, MSO₄·yH₂O, where M is a divalent metal. She carries out two experiments. In experiment 1, 4.99 g of the crystals are heated to constant mass and 3.19 g of anhydrous solid remains. In experiment 2, a separate 4.99 g portion is dissolved and treated with excess barium chloride solution, giving 4.66 g of dry barium sulfate. (Mₕ: H₂O 18; BaSO₄ 233)

(a) [5]

(i) Use experiment 2 to find the moles of sulfate, and hence the moles of MSO₄, in the 4.99 g sample. [2]
(ii) Use experiment 1 to find the moles of water of crystallisation. [2]
(iii) Deduce the value of y. [1]

Model Answer -- 7(a)

(i) Moles of BaSO₄ = 4.66/233 = 0.0200 mol [1]

(i) Every sulfate ion is precipitated once, so the sample contains 0.0200 mol of sulfate and therefore 0.0200 mol of MSO₄ [1]

(ii) Mass of water lost = 4.99 − 3.19 = 1.80 g [1]

(ii) Moles of water = 1.80/18 = 0.100 mol [1]

(iii) Ratio MSO₄ : H₂O = 0.0200 : 0.100 = 1 : 5, so y = 5 [1]

⚠ If you missed marks here: The two experiments must be linked: experiment 2 gives the moles of the salt itself, experiment 1 gives the moles of water. Trying to answer using only the heating data is impossible without knowing the identity of M, which is exactly why the second experiment is provided.

1 mark -- moles BaSO4 4.66 233 0.0200 1 mark -- every sulfate precipitated 0.0200 moles MSO4 1 mark -- mass water 4.99 3.19 1.80 1 mark -- moles water 1.80 18 0.100 1 mark -- ratio 1 5 y equals five

(b) [3]

(i) Calculate the relative formula mass of the anhydrous salt MSO₄ and hence the relative atomic mass of M. [2]
(ii) Suggest the identity of M, and state one observation from the heating experiment that would support your suggestion. [1]

Model Answer -- 7(b)

(i) Mₕ(MSO₄) = mass ÷ moles = 3.19/0.0200 = 159.5, which rounds to 160 [1]

(i) The sulfate group SO₄ has a mass of 32 + 64 = 96, so Aₕ(M) = 160 − 96 = 64 [1]

(ii) M is copper; supporting evidence is that the hydrated crystals are blue and turn white on heating, which is characteristic of copper(II) sulfate [1]

⚠ If you missed marks here: The mass of the sulfate group is 96, not 80 — count all four oxygens. Subtracting an incorrect value gives an Aₕ that matches no real metal, which should prompt a recheck. Note that the value 64 is also consistent with the colour change, so the two lines of evidence agree.

1 mark -- Mr 3.19 0.0200 159.5 160 1 mark -- sulfate 96 subtract Ar M 64 1 mark -- copper blue crystals white heating characteristic

The student's teacher points out that experiment 1 alone would have given a wrong answer if the crucible had not been heated to constant mass, and that experiment 2 alone would have given a wrong answer if the barium chloride had not been in excess. State the direction of the error in each case and explain your reasoning. [2]

Model Answer -- 7(c)

If heating stopped too soon, some water of crystallisation would remain, so the mass loss recorded would be too small; the calculated moles of water and therefore the value of y would be too low [1]

If the barium chloride were not in excess, some sulfate would stay in solution, so the mass of barium sulfate and the deduced moles of MSO₄ would be too small; that would make the water-to-salt ratio, and therefore y, too large [1]

⚠ If you missed marks here: Stating only that the answer would be inaccurate earns nothing — the direction of the error is what is being tested. Notice that the two errors push y in opposite directions, which is a useful check: a value of y that is not a whole number usually signals one of these faults.

1 mark -- heating stopped soon water remains mass loss small moles water y too low 1 mark -- barium not excess sulfate stays solution moles salt small ratio y too large

Self-Assessment

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