AQA A-Level Chemistry Paper 1, 2023: Question 7

17 marks · Hard difficulty · State/Explain/Numerical

Answer questions on the chemistry of aluminium complexes, including electron configuration, acidity, ligand reactions, EDTA bonding, and a back-titration calculation to find the water of crystallisation.

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Question

Multi-part AQA A-Level Chemistry question about aluminium chemistry. Part 07.1 asks for the electron configuration of Al3+. 07.2 asks for the equation when anhydrous Al2(SO4)3 forms [Al(H2O)6]3+ in water. 07.3 asks why [Al(H2O)6]3+ solution is acidic. 07.4 asks why the concentration cannot be determined by colorimetry. 07.5 asks for the ionic equation and observation with excess aqueous ammonia. 07.6 asks to identify the species produced when excess dilute sulfuric acid is added to the product of 07.5. 07.7 shows the structure of the EDTA4- ion and asks to circle atoms of two different elements that coordinate to Al3+. 07.8 outlines a back-titration method where 1.036 g of hydrated aluminium sulfate is made up to 250 cm3, a 25.0 cm3 aliquot is reacted with 50.0 cm3 of 0.0100 mol dm-3 EDTA4-, and excess EDTA4- is titrated with 18.00 cm3 of 0.0105 mol dm-3 ZnSO4, requiring the calculation of the Mr and the integer value of x.
Question text

07 This question is about complexes containing the aluminium ion.

07.1 Give the electron configuration of the Al3+ ion.

[1 mark]

07.2 When anhydrous aluminium sulfate, Al2(SO4)3, is added to water a solution forms that

contains the complex aluminium ion, [Al(H O) ]3+

Give the equation for the reaction.

[1 mark]

07.3 Explain why the solution containing [Al(H O) ]3+ is acidic.

[2 marks]

07.4 State why the concentration of aluminium sulfate solution can not be determined by

colorimetry.

[1 mark]

07.5 An excess of aqueous ammonia is added to a solution containing [Al(H O) ]3+

Give an ionic equation for the reaction and state one observation.

[2 marks]

Equation

Observation

07.6 An excess of dilute sulfuric acid is added to the products of the reaction in

Question 07.5

Identify the aluminium species produced.

[1 mark]

07.7 Figure 3 shows the structure of the EDTA4 – ion.

Figure 3

Atoms of two different elements in EDTA4– can form co-ordinate bonds with an

aluminium ion.

On Figure 3, draw circles around the atoms of two different elements that would link

to an aluminium ion by a co-ordinate bond.

[2 marks]

07.8 Hydrated aluminium sulfate, Al2(SO4)3.xH2O, is soluble in water.

The relative formula mass and value of x can be found from a titration experiment.

Aqueous [Al(H O) ]3+ ions react to form a stable complex when treated with an excess

of EDTA4 – ions.

The excess of EDTA4 – ions is determined by titration with ZnSO solution.

*17* Method

• Dissolve 1.036 g of Al (SO ) .xH O in distilled water and make up to 250 cm3

24 3 2

• Add 25.0 cm3 of this solution to 50.0 cm3 of a solution containing

EDTA4– ions of concentration 0.0100 mol dm–3

• Determine the excess of EDTA4 – ions by titrating with ZnSO solution in the

presence of an indicator.

The excess of EDTA4 – ions requires 18.00 cm3 of 0.0105 mol dm–3 ZnSO solution to

react completely.

The equations for the reactions are

[Al(H O) ]3+ + EDTA4– → [AlEDTA]– + 6 H O

26 2

[Zn(H O) ]2+ + EDTA4– → [ZnEDTA]2– + 6 H O

26 2

For Al2(SO4)3 Mr = 342.3

Use the information given to calculate the Mr of Al2(SO4)3.xH2O

Calculate x

Give your answer as an integer.

[7 marks]

Mr

20 x

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Mark scheme

Show the mark scheme AQA mark scheme for Question 07. 07.1 gives 1s2 2s2 2p6 (1 mark). 07.2 gives Al2(SO4)3 + 12 H2O -> 2 [Al(H2O)6]3+ + 3 SO4 2- (1 mark). 07.3 gives M1: Al3+ has high charge and small size (or high charge density) and M2: weakens O-H bond in water ligands and donates H+ to water (2 marks). 07.4 accepts colourless solution / no d-d transitions / doesn't absorb visible light (1 mark). 07.5 requires [Al(H2O)6]3+ + 3 NH3 -> Al(H2O)3(OH)3 + 3 NH4+ and observation: white ppt or white solid (2 marks). 07.6 gives [Al(H2O)6]3+ (1 mark). 07.7 awards marks for circling one or more N and one or more O- (2 marks). 07.8 details 7 calculation marks: M1 moles EDTA added = 5 x 10^-4; M2 moles Zn2+ = 1.89 x 10^-4; M3 moles EDTA reacted in 25 cm3 = 3.11 x 10^-4; M4 moles reacted in 250 cm3 = 3.11 x 10^-3; M5 moles Al2(SO4)3.xH2O = 1.555 x 10^-3; M6 Mr = 666.2; M7 342.3 + 18x = 666.2 giving x = 18.

Question Answers Additional comments/Guidelines Mark

07.1 1s2 2s2 2p6 Allow [He] 2s22p6

(AO1)

07.2 Al (SO ) + 12 H O → 2 [Al(H O) ]3+ + 3 SO 2– Allow [Al(H O) ] (SO )

24 3 2 2 6 4 2 6 2 4 3

(AO1)

M1 Al3+ has a high charge and small size

OR

Al3+ has a high charge density 2

07.3 M2 Al3+ attracts electrons from the O-H bond (in

+ + (2 x AO1)

3+ + the ligand and releases H or H3O ions)

M2 Al weakens the O-H bond (in water ligands and donates H to

water or forms H O+ ions with water) OR

Al3+ polarises the O-H bond/water molecule

Allow no d-d transitions (as there are no d 1

electrons)

Colourless (solution) (AO3)

07.4 OR

Doesn’t absorb visible light

Question Answers Additional comments/Guidelines Mark 23

[Al(H O) ]3+ + 3 NH → Al(H O) (OH) + 3 NH + 2

26 3 2 3 3 4

07.5 Do not accept effervescence (1 x AO1,

White ppt or white solid Do not accept white ppt dissolves in excess NH3 1 x AO2)

Ignore state symbols

07.6 [Al(H O) ]3+

(AO3)

A circle around one or more N 2

07.7 A circle around one or more O – (2 x AO1)

M1 n (EDTA4–) added = 5 x 10 – 4 mol alternative methods will be allowed

M2 n (Zn2+) = 1.89 x 10– 4 mol

4– 3 3+ M3 = M1 – M2

M3 n (EDTA ) reacted with the 25 cm sample of Al

= 5 x 10 – 4 – 1.89 x 10– 4 = 3.11 x 10– 4 mol

M3 × 250

M4 n EDTA4– reacted with the 250 cm3 sample of Al3+ M4 =

– 4 – 3 25

= 3.11 x 10 x 10 = 3.11 x 10 mol

24 – 3 – 3 7

07.8 M5 n Al2(SO4)3 xH2O = 3.11 x 10 ÷ 2 = 1.555 x10 mol M5 = M4 ÷ 2

(7 x AO2)

M6 Mr Al2(SO4)3 xH2O = 1.036 ÷ 1.555 x10– 3 = 666.2 M6 = 1.036 ÷ M5

M6 – 342.3

M7 342.3 + 18 x = 666(.2) so x = 18 M7 = and answer as integer

How to answer it

Aluminium Chemistry: Complexes, Hydrolysis & Back Titrations

WHAT THIS QUESTION TESTS

This multi-step examination question tests core inorganic and analytical chemistry principles under the AQA A-Level specification:

  • Atomic Structure: Writing electron configurations for p-block metal cations.
  • Aqueous Ions & Acidity: Equation for hexaaqua ion formation and explaining acidity via high charge density and O–H bond polarisation.
  • Analytical Techniques: Understanding why colorimetry requires visible light absorption (d–d transitions).
  • Reactions with Bases & Acids: Stepwise deprotonation by NH₃ and re-protonation by strong acid.
  • Multidentate Ligands: Identifying donor atoms in hexadentate EDTA⁴⁻.
  • Quantitative Back Titration: A rigorous 7-mark multistep stoichiometric calculation involving ligand substitution, aliquots, and water of crystallisation.
QUESTION 07.1

Electron Configuration of the Al³⁺ Ion

1 Mark • Assessment Objective: AO1

✅ Correct Answer

1s² 2s² 2p⁶

Mark scheme: Allow [He] 2s² 2p⁶ . Full mark for the correct noble gas core configuration of neon.

❌ Common Errors

  • Writing the configuration of neutral aluminium: 1s² 2s² 2p⁶ 3s² 3p¹ (forgetting to remove the 3 valence electrons).
  • Writing simply [Ne] without showing sub-shell electron occupancy.
QUESTION 07.2

Formation of the Hexaaquaaluminium(III) Ion

1 Mark • Assessment Objective: AO1

✅ Correct Answer

Al₂(SO₄)₃ + 12 H₂O → 2 [Al(H₂O)₆]³⁺ + 3 SO₄²⁻

Mark scheme: Also allows full molecular complex notation: [Al(H₂O)₆]₂(SO₄)₃ .

🧠 Exam Technique & Balancing

  • Check atom balancing: 2 aluminium ions need 12 water molecules total (6 per Al³⁺ ion).
  • Make sure the charges balance: right side has 2 × (+3) + 3 × (-2) = 0.
QUESTION 07.3

Why Aqueous [Al(H₂O)₆]³⁺ is Acidic

2 Marks • Assessment Objective: 2 × AO1

✅ Correct Answer

  • M1: Al³⁺ has a high charge and small size (or high charge density).
  • M2: Al³⁺ polarises/attracts electrons from the O–H bonds in water ligands, weakening the O–H bond and releasing H⁺ / forming H₃O⁺ ions with water.

💡 Key Knowledge (Hydrolysis Mechanism)

[Al(H₂O)₆]³⁺ + H₂O ⇌ [Al(H₂O)₅(OH)]²⁺ + H₃O⁺

Because Al³⁺ has a +3 charge and very small ionic radius, its intense electric field pulls electron density away from the coordinated oxygen atom. This severely polarises the O–H bond, facilitating proton donation.

❌ Common Errors & Lost Marks

  • Stating only "Al³⁺ has a high charge" without mentioning its small size/radius. Both are required if "high charge density" is not explicitly stated.
  • Saying "Al³⁺ breaks the O–H bond directly" rather than stating it polarises/weakens the bond to release H⁺ into solution.
QUESTION 07.4

Limitation of Colorimetry with Aluminium Solutions

1 Mark • Assessment Objective: AO3

✅ Correct Answer

The solution is colourless (or it does not absorb visible light / has no d–d transitions because it has no partially filled d-orbitals).

💡 Key Knowledge

Colorimetry relies on a species absorbing specific wavelengths of visible light (Beer-Lambert law). Since Al³⁺ is a main-group ion with no d-electrons ( 2p⁶ ), it cannot undergo d–d electronic transitions in the visible spectrum.

QUESTION 07.5

Reaction of [Al(H₂O)₆]³⁺ with Excess Aqueous Ammonia

2 Marks • Assessment Objective: AO1 + AO2

✅ Correct Answers

Ionic Equation (1 Mark):
[Al(H₂O)₆]³⁺ + 3 NH₃ → Al(H₂O)₃(OH)₃ + 3 NH₄⁺

Observation (1 Mark):
White precipitate (or white solid).

❌ Common Errors & Examiner Warnings

  • DO NOT write effervescence / bubbles: Ammonia acts as a Brønsted–Lowry base, not a carbonate; no gas is evolved!
  • DO NOT state that the precipitate dissolves in excess: Although aluminium hydroxide is amphoteric and dissolves in excess NaOH, it does not dissolve in excess aqueous NH₃ because ammonia is a weak base.
  • Forgetting to deprotonate 3 times: the neutral complex precipitate is Al(H₂O)₃(OH)₃ or Al(OH)₃ .
QUESTION 07.6

Adding Excess Dilute Sulfuric Acid to the Product

1 Mark • Assessment Objective: AO3

✅ Correct Answer

[Al(H₂O)₆]³⁺

Mark scheme: Accepts [Al(H₂O)₆]³⁺ . The white precipitate re-dissolves as OH⁻ ligands are protonated back to neutral H₂O.

🧠 Exam Technique

Acid-base reactions of hexaaqua metal complexes are reversible:

Al(H₂O)₃(OH)₃ + 3 H⁺ → [Al(H₂O)₆]³⁺

The question asks to identify the aluminium species, so give the exact ionic complex formula.

QUESTION 07.7

Co-ordinate Bonding Atoms in EDTA⁴⁻

2 Marks • Assessment Objective: 2 × AO1

✅ Correct Answer

  • Mark 1: Circle around one or more Nitrogen (N) atoms.
  • Mark 2: Circle around one or more negatively charged Oxygen (O⁻) atoms.

🧠 Visualising the Figure 3 Diagram

EDTA⁴⁻ contains two amine groups and four carboxylate groups:
• Circle the central N atoms (each carries 1 lone pair).
• Circle the carboxylate oxygens with the negative charge (O⁻), NOT the carbonyl double-bonded oxygen (C=O).

EDTA⁴⁻ is a hexadentate ligand because it donates six electron pairs (2 from N, 4 from O⁻).

QUESTION 07.8

Back Titration: Determining Mr and x for Hydrated Aluminium Sulfate

7 Marks • Assessment Objective: 7 × AO2

📐 Step-by-Step Calculation

  1. Calculate moles of EDTA⁴⁻ added initially:
    Volume = 50.0 cm³ = 0.0500 dm³; Concentration = 0.0100 mol dm⁻³
    n(EDTA⁴⁻) = 0.0500 × 0.0100 = 5.00 × 10⁻⁴ mol
    Award M1
  2. Calculate moles of Zn²⁺ used to titrate unreacted excess EDTA⁴⁻:
    Volume = 18.00 cm³ = 0.01800 dm³; Concentration = 0.0105 mol dm⁻³
    n(Zn²⁺) = 0.01800 × 0.0105 = 1.89 × 10⁻⁴ mol
    Award M2
  3. Calculate moles of EDTA⁴⁻ reacted with Al³⁺ in the 25.0 cm³ sample:
    From equation: 1 mol Zn²⁺ reacts with 1 mol EDTA⁴⁻ → excess EDTA⁴⁻ = 1.89 × 10⁻⁴ mol
    n(EDTA⁴⁻ reacted) = (5.00 × 10⁻⁴) − (1.89 × 10⁻⁴) = 3.11 × 10⁻⁴ mol
    Award M3 (M1 − M2)
  4. Scale up to the full 250 cm³ volumetric flask:
    Scaling factor = 250 / 25.0 = 10
    n(Al³⁺ in 250 cm³) = 3.11 × 10⁻⁴ × 10 = 3.11 × 10⁻³ mol
    Award M4 (M3 × 10)
  5. Calculate moles of hydrated salt Al₂(SO₄)₃•xH₂O:
    Crucial 2:1 ratio: 1 mole of Al₂(SO₄)₃ contains 2 moles of Al³⁺.
    n(salt) = (3.11 × 10⁻³) / 2 = 1.555 × 10⁻³ mol
    Award M5 (M4 / 2)
  6. Calculate the relative formula mass (Mr):
    Mr = mass / moles = 1.036 g / (1.555 × 10⁻³ mol) = 666.24 ≈ 666.2
    Award M6 (1.036 / M5)
  7. Determine x (must be an integer):
    Anhydrous Al₂(SO₄)₃ has Mr = 342.3
    Mass of water of crystallisation = 666.24 − 342.3 = 323.94
    x = 323.94 / 18.0 = 17.997 → x = 18
    Award M7 (Integer answer required)

❌ Major Calculation Traps

  • Forgetting to divide by 2 for Al₂(SO₄)₃: This is where most students lose marks! 1 mole of Al₂(SO₄)₃ gives 2 moles of [Al(H₂O)₆]³⁺ ions. Forgetting this halves your Mr and ruins x.
  • Omitting the aliquot factor: 25.0 cm³ was taken from a 250 cm³ flask, so moles must be multiplied by 10.
  • Non-integer x: The question explicitly commands "Give your answer as an integer". Leaving 17.99 or 18.0 loses M7!

✅ Final Answers Summary

Mr: 666.2 (or 666)

x: 18

Full formula: Al₂(SO₄)₃•18H₂O

Topics

Physical Chemistry · Inorganic Chemistry · 3.1.1 Atomic Structure · 3.1.2 Amount of Substance · 3.1.3 Bonding · 3.2.5 Transition Metals · 3.2.6 Reactions of Ions in Aqueous Solution

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.