An original Thinka practice paper modelled on the structure and difficulty of the Jun 2022 CCEA A Level Chemistry 1110 paper. Not affiliated with or reproduced from CCEA.
Section Question 1: Qualitative Analysis of Transition Metal Cations & Functional Groups
Perform small-scale test-tube reactions on two unknown metal solutions and record detailed observations, pH measurements, and functional group deductions.
Solution A is a 0.1 mol/dm³ solution of a transition metal salt. Add aqueous sodium hydroxide, NaOH(aq), dropwise to 2 cm³ of Solution A until in excess. Note any observations. iron(II) salt: _________
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Worked solution
Adding NaOH(aq) to an Fe²⁺ solution precipitates iron(II) hydroxide: \( Fe^{2+}(aq) + 2OH^-(aq) \rightarrow Fe(OH)_2(s) \). This is a green precipitate, insoluble in excess NaOH(aq). On standing, the green precipitate darkens/turns brown at the surface, as the Fe(OH)2 is oxidised by oxygen in the air to Fe(OH)3. Final answer: green precipitate (insoluble in excess NaOH), which turns brown on standing in air.
Marking scheme
[2] 1 mark: green precipitate formed, insoluble in excess NaOH; 1 mark: precipitate darkens/turns brown (at least at the surface) on standing in air, due to oxidation. Accept 'green/grey-green' and 'brown/red-brown/rust-coloured'.
Solution B is a 0.1 mol/dm³ solution of a different transition metal salt. Add aqueous sodium hydroxide, NaOH(aq), dropwise to 2 cm³ of Solution B until in excess. Note any observations. iron(III) salt: _________
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Worked solution
Adding NaOH(aq) to an Fe³⁺ solution precipitates iron(III) hydroxide: \( Fe^{3+}(aq) + 3OH^-(aq) \rightarrow Fe(OH)_3(s) \). This is a red-brown precipitate, insoluble in excess NaOH(aq). Final answer: red-brown (orange-brown) precipitate, insoluble in excess NaOH.
Marking scheme
[2] 1 mark: red-brown/orange-brown precipitate formed; 1 mark: correctly noted as insoluble in excess NaOH(aq).
Using your observations from the addition of NaOH(aq) to Solutions A and B, identify which solution is the iron(II) salt and which is the iron(III) salt, giving a reason for each identification.
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Worked solution
Solution A is the iron(II) salt, because it formed a green precipitate that darkened/turned brown on standing in air (characteristic of Fe(OH)2 oxidising to Fe(OH)3). Solution B is the iron(III) salt, because it formed a red-brown precipitate immediately (characteristic of Fe(OH)3), with no colour change over time since the iron is already in the +3 oxidation state. Final answer: Solution A = iron(II) salt (green precipitate, darkens on standing); Solution B = iron(III) salt (red-brown precipitate).
Marking scheme
[3] 1 mark: Solution A correctly identified as the iron(II) salt; 1 mark: Solution B correctly identified as the iron(III) salt; 1 mark: valid reasoning given for at least one identification, referring to the precipitate colour/colour change observed.
The pH of Solution A (iron(II) salt) and Solution B (iron(III) salt) is measured using a pH meter. Both solutions are acidic. State which solution would have the LOWER pH, and give a reason for your answer in terms of the charge density of the metal ion.
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Worked solution
Solution B, containing Fe³⁺, would have the lower pH. This is because the Fe³⁺ ion has a higher charge than Fe²⁺, giving it a greater charge density; this more strongly polarises (weakens) the O–H bonds of the water molecules coordinated to it in the hexaaqua complex, \( [Fe(H_2O)_6]^{3+} \), making it easier for a proton to be released to a surrounding water molecule. This makes the hexaaqua iron(III) ion a stronger Brønsted–Lowry acid than the hexaaqua iron(II) ion, so Solution B is more acidic (lower pH). Final answer: Solution B (Fe³⁺) has the lower pH, because its higher charge density polarises the coordinated water molecules more strongly, making it release H⁺ ions more readily.
Marking scheme
[2] 1 mark: Solution B (iron(III)) correctly identified as having the lower pH; 1 mark: correct reasoning based on the higher charge/charge density of Fe³⁺ polarising coordinated water molecules and increasing acidity (hydrolysis).
A student is given a third unknown transition metal solution, Solution C, thought to contain either chromium(III) or copper(II) ions. Excess NaOH(aq) is added, and the grey-green precipitate initially formed dissolves in the excess NaOH(aq) to give a green solution. Identify the metal ion present in Solution C, and write an ionic equation for the reaction of its hydroxide precipitate with excess hydroxide ions.
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Worked solution
The metal ion is chromium(III), Cr³⁺, since Cr(OH)3 is amphoteric and dissolves in excess NaOH(aq) to give a green solution (unlike Cu(OH)2, which is insoluble in excess NaOH). The initial precipitate is \( Cr(OH)_3(s) \); this dissolves in excess hydroxide ions by acting as a Lewis acid, accepting three more OH⁻ ligands: \( Cr(OH)_3(s) + 3OH^-(aq) \rightarrow [Cr(OH)_6]^{3-}(aq) \). Final answer: chromium(III), Cr³⁺; \( Cr(OH)_3(s) + 3OH^-(aq) \rightarrow [Cr(OH)_6]^{3-}(aq) \).
Marking scheme
[3] 1 mark: chromium(III)/Cr³⁺ correctly identified; 1 mark: correct species Cr(OH)3 and [Cr(OH)6]3- shown; 1 mark: fully balanced ionic equation with state symbols. ECF for the ion identified applied consistently to the equation.
To a sample of copper(II) sulfate solution, Solution D, aqueous ammonia is added dropwise until in excess. Describe what would be observed.
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Worked solution
A pale blue precipitate of copper(II) hydroxide, Cu(OH)2, forms initially as ammonia is added. On continuing to add ammonia until it is in excess, this precipitate dissolves, giving a deep (royal) blue solution, as the hexaaquacopper(II) ion undergoes ligand exchange to form the tetraamminediaquacopper(II) complex, \( [Cu(NH_3)_4(H_2O)_2]^{2+} \). Final answer: a pale blue precipitate forms initially, which dissolves in excess ammonia to give a deep blue solution.
Marking scheme
[2] 1 mark: pale blue precipitate forms with limited ammonia; 1 mark: precipitate dissolves in excess ammonia to give a deep/royal blue solution.
To 2 cm³ of an unknown colourless organic liquid, X, a few drops of 2,4-dinitrophenylhydrazine (2,4-DNP) solution are added and an orange precipitate forms immediately. State what this observation shows about the functional group present in X, and name ONE other reagent that could be used to determine whether X is an aldehyde or a ketone.
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Worked solution
The formation of an orange precipitate with 2,4-DNP shows that X contains a carbonyl group, \( C=O \), meaning X is either an aldehyde or a ketone. To distinguish between the two, Tollens' reagent could be used: warming with Tollens' reagent gives a silver mirror with an aldehyde (which is oxidised), but no reaction with a ketone (which cannot be oxidised in this way). Fehling's solution would similarly distinguish them, giving a brick-red precipitate with an aldehyde but no change with a ketone. Final answer: X contains a carbonyl group (aldehyde or ketone); Tollens' reagent (silver mirror with aldehyde, no reaction with ketone) or Fehling's solution can distinguish between them.
Marking scheme
[2] 1 mark: correct identification of a carbonyl group/C=O (aldehyde or ketone present); 1 mark: a valid named reagent (Tollens' reagent or Fehling's solution) that distinguishes aldehydes from ketones, with correct differing results stated or implied.
Bromine water is added to a sample of an unknown organic liquid, Y, and the orange colour of the bromine water is decolourised immediately. State what type of functional group this observation indicates is present in Y, and explain, in terms of the bonding present, why this reaction occurs.
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Worked solution
The rapid decolourisation of bromine water indicates that Y contains a carbon–carbon double bond, \( C=C \) (an alkene). The region of high electron density in the π bond of the C=C induces a temporary dipole in the approaching (non-polar) bromine molecule, polarising it; the alkene then acts as a nucleophile, attacking the electrophilic (slightly positive) bromine atom, and an electrophilic addition reaction occurs across the double bond, forming a colourless dibromo-product and removing the orange colour of the bromine. Final answer: Y contains a C=C double bond (alkene); the electron-rich π bond induces a dipole in and reacts with the bromine molecule via electrophilic addition, decolourising the bromine water.
Marking scheme
[2] 1 mark: carbon–carbon double bond/alkene (C=C) correctly identified; 1 mark: valid explanation referring to the electron-rich π bond inducing a dipole in the bromine molecule and undergoing electrophilic addition.
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Section Question 2: Volumetric Preparation and Acid-Base Titration
Prepare a volumetric solution following reaction of a solid sample with standard acid, filter, dilute to mark, and perform repeated accurate titrations.
4 Question · 8 marks
Question 1 · Experimental Technique & Data Grid
2 marks
A 1.20 g sample of impure calcium carbonate is reacted with 100 cm³ of 0.500 mol/dm³ hydrochloric acid (an excess). The mixture is filtered to remove insoluble impurity, and the filtrate is made up to exactly 250 cm³ in a volumetric flask with distilled water. 25.0 cm³ portions of this solution are titrated against 0.100 mol/dm³ sodium hydroxide solution, using phenolphthalein indicator, to determine the excess (unreacted) hydrochloric acid. The table below shows the burette readings obtained. Titration / Rough / 1 / 2 / 3 Final burette reading / cm³ / 24.60 / 23.45 / 23.40 / 23.50 Initial burette reading / cm³ / 0.00 / 0.00 / 0.00 / 0.00
Complete the table by calculating the titre (final − initial burette reading) for each run, and identify which titre is the 'rough' titre.
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Worked solution
Each titre is calculated as final burette reading minus initial burette reading: Rough = 24.60 − 0.00 = 24.60 cm³; Run 1 = 23.45 − 0.00 = 23.45 cm³; Run 2 = 23.40 − 0.00 = 23.40 cm³; Run 3 = 23.50 − 0.00 = 23.50 cm³. The rough titre is the first, less precise run carried out quickly to give an approximate value before the accurate runs — here, 24.60 cm³, noticeably higher than the three accurate runs, as expected since the tap is opened quickly and the endpoint is easily overshot. Final answer: titres are 24.60 (rough), 23.45, 23.40 and 23.50 cm³; the rough titre is 24.60 cm³.
Marking scheme
[2] 1 mark: all four titres correctly calculated (24.60, 23.45, 23.40, 23.50 cm³), each to 2 decimal places matching the given readings; 1 mark: rough titre correctly identified as 24.60 cm³, with recognition that it is higher than the accurate runs.
Question 2 · Experimental Technique & Data Grid
2 marks
A 1.20 g sample of impure calcium carbonate is reacted with 100 cm³ of 0.500 mol/dm³ hydrochloric acid (an excess). The mixture is filtered to remove insoluble impurity, and the filtrate is made up to exactly 250 cm³ in a volumetric flask with distilled water. 25.0 cm³ portions of this solution are titrated against 0.100 mol/dm³ sodium hydroxide solution, using phenolphthalein indicator, to determine the excess (unreacted) hydrochloric acid. The table below shows the burette readings obtained. Titration / Rough / 1 / 2 / 3 Final burette reading / cm³ / 24.60 / 23.45 / 23.40 / 23.50 Initial burette reading / cm³ / 0.00 / 0.00 / 0.00 / 0.00
State which titres should be used to calculate a mean titre, and explain your choice.
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Worked solution
Concordant titres are those that agree closely with each other, conventionally within 0.10 cm³. Runs 1, 2 and 3 (23.45, 23.40 and 23.50 cm³) are all within 0.10 cm³ of one another, so they are concordant and should be used to calculate the mean. The rough titre (24.60 cm³) should be excluded, as it was obtained quickly and is not concordant with (differs by much more than 0.10 cm³ from) the accurate runs, making it unreliable. Final answer: use runs 1, 2 and 3 (23.45, 23.40, 23.50 cm³), as they are concordant (within 0.10 cm³ of each other); the rough titre is excluded as it is not concordant/reliable.
Marking scheme
[2] 1 mark: correctly identifies runs 1, 2 and 3 as the concordant titres to be used; 1 mark: valid explanation that concordant titres are within 0.10 cm³ of each other and that the rough titre is excluded for not meeting this/being less precise.
Question 3 · Experimental Technique & Data Grid
2 marks
A 1.20 g sample of impure calcium carbonate is reacted with 100 cm³ of 0.500 mol/dm³ hydrochloric acid (an excess). The mixture is filtered to remove insoluble impurity, and the filtrate is made up to exactly 250 cm³ in a volumetric flask with distilled water. 25.0 cm³ portions of this solution are titrated against 0.100 mol/dm³ sodium hydroxide solution, using phenolphthalein indicator, to determine the excess (unreacted) hydrochloric acid. The table below shows the burette readings obtained. Titration / Rough / 1 / 2 / 3 Final burette reading / cm³ / 24.60 / 23.45 / 23.40 / 23.50 Initial burette reading / cm³ / 0.00 / 0.00 / 0.00 / 0.00
Calculate the mean titre from the concordant results identified above.
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Worked solution
Mean titre = (23.45 + 23.40 + 23.50) ÷ 3 = 70.35 ÷ 3 = 23.45 cm³. Check by a second route: since 23.40 and 23.50 differ from 23.45 by equal and opposite amounts (−0.05 and +0.05), their average is exactly 23.45, and including the third value of 23.45 unchanged confirms the mean is 23.45 cm³. Final answer: mean titre = 23.45 cm³.
Marking scheme
[2] 1 mark: correct method shown (sum of the three concordant titres ÷ 3); 1 mark: correct final answer of 23.45 cm³, given to 2 decimal places.
Question 4 · Experimental Technique & Data Grid
2 marks
A 1.20 g sample of impure calcium carbonate is reacted with 100 cm³ of 0.500 mol/dm³ hydrochloric acid (an excess). The mixture is filtered to remove insoluble impurity, and the filtrate is made up to exactly 250 cm³ in a volumetric flask with distilled water. 25.0 cm³ portions of this solution are titrated against 0.100 mol/dm³ sodium hydroxide solution, using phenolphthalein indicator, to determine the excess (unreacted) hydrochloric acid. The table below shows the burette readings obtained. Titration / Rough / 1 / 2 / 3 Final burette reading / cm³ / 24.60 / 23.45 / 23.40 / 23.50 Initial burette reading / cm³ / 0.00 / 0.00 / 0.00 / 0.00
Using the mean titre, calculate the number of moles of sodium hydroxide used to neutralise the excess hydrochloric acid in the 25.0 cm³ sample titrated.
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Worked solution
Moles of NaOH = concentration × volume (in dm³) = \( 0.100 \times \frac{23.45}{1000} = 0.100 \times 0.02345 = 0.002345 \) mol. Check by a second route: \( 0.002345 \div 0.100 = 0.02345 \) dm³ = 23.45 cm³, matching the mean titre used, confirming the calculation. Final answer: moles of NaOH = \( 2.345 \times 10^{-3} \) mol.
Marking scheme
[2] 1 mark: correct method (moles = concentration × volume in dm³, using the mean titre of 23.45 cm³ converted to dm³); 1 mark: correct final answer \( 2.345 \times 10^{-3} \) mol (accept 0.002345 mol), with correct units/significant figures. ECF from part (c) applied if the mean titre used is consistent.
Section Question 3: Transition Metal Redox Observation
Carry out sequential redox additions on a transition metal salt and record distinct solution colour changes.
4 Question · 4 marks
Question 1 · Practical Observation
1 marks
A student is given a sample of acidified ammonium metavanadate(V) solution, containing the dioxovanadium(V) ion, \( VO_2^+ \). Zinc metal is added, and the solution is gently warmed and left to react, without shaking or exposure to air. As the reaction proceeds, the vanadium is progressively reduced through a sequence of lower oxidation states, each with a distinct colour. State the colour of the initial acidified ammonium metavanadate(V) solution, containing V in the +5 oxidation state as \( VO_2^+ \).
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Worked solution
The dioxovanadium(V) ion, \( VO_2^+ \), in acidified aqueous solution is yellow. Final answer: yellow.
Marking scheme
[1] yellow.
Question 2 · Practical Observation
1 marks
A student is given a sample of acidified ammonium metavanadate(V) solution, containing the dioxovanadium(V) ion, \( VO_2^+ \). Zinc metal is added, and the solution is gently warmed and left to react, without shaking or exposure to air. As the reaction proceeds, the vanadium is progressively reduced through a sequence of lower oxidation states, each with a distinct colour. State the colour of the solution once the vanadium has been reduced to the +4 oxidation state, as the oxovanadium(IV) ion, \( VO^{2+} \).
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Worked solution
The oxovanadium(IV) ion, \( VO^{2+} \), in aqueous solution is blue. Final answer: blue.
Marking scheme
[1] blue.
Question 3 · Practical Observation
1 marks
A student is given a sample of acidified ammonium metavanadate(V) solution, containing the dioxovanadium(V) ion, \( VO_2^+ \). Zinc metal is added, and the solution is gently warmed and left to react, without shaking or exposure to air. As the reaction proceeds, the vanadium is progressively reduced through a sequence of lower oxidation states, each with a distinct colour. State the colour of the solution once the vanadium has been reduced further to the +3 oxidation state, as \( V^{3+} \).
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Worked solution
The \( V^{3+} \) ion in aqueous solution is green. Final answer: green.
Marking scheme
[1] green.
Question 4 · Practical Observation
1 marks
A student is given a sample of acidified ammonium metavanadate(V) solution, containing the dioxovanadium(V) ion, \( VO_2^+ \). Zinc metal is added, and the solution is gently warmed and left to react, without shaking or exposure to air. As the reaction proceeds, the vanadium is progressively reduced through a sequence of lower oxidation states, each with a distinct colour. State the colour of the solution once the vanadium has been reduced fully to the +2 oxidation state, as \( V^{2+} \).
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Worked solution
The \( V^{2+} \) ion in aqueous solution is violet (also described as lilac/mauve). Final answer: violet (lilac).
Marking scheme
[1] violet/lilac/mauve (any one accepted).
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