AQA A-Level Chemistry Paper 1, 2018: Question 7

9 marks · Medium difficulty · State/Explain/Describe

Write equations and state observations for the reactions of aqueous hexaaquaaluminium(III) ions with water, sodium carbonate, and excess potassium hydroxide.

Practise this question

Question

Question 07 consists of four parts about the chemistry of aluminium ions in aqueous solution. Question 07.1 asks for an equation when anhydrous aluminium chloride reacts with water to form solution Y containing complex aluminium ion Z and chloride ions (1 mark). Question 07.2 asks for an equation showing how the complex ion Z acts as a Brønsted–Lowry acid with water (1 mark). Question 07.3 asks for two observations and a balanced equation when excess sodium carbonate solution is added to solution Y (3 marks). Question 07.4 asks for two observations and an equation for each reaction when aqueous potassium hydroxide is added until in excess to solution Y (4 marks).
Question text

07.1 When anhydrous aluminium chloride reacts with water, solution Y is formed that

contains a complex aluminium ion, Z, and chloride ions.

Give an equation for this reaction.

[1 mark]

07.2 Give an equation to show how the complex ion Z can act as a Brønsted–Lowry acid

with water.

[1 mark]

07.3 Describe two observations you would make when an excess of sodium carbonate

solution is added to solution Y.

Give an equation for the reaction. In your equation, include the formula of each

complex aluminium species.

[3 marks]

Observation 1

Observation 2

Equation

07.4 Aqueous potassium hydroxide is added, until in excess, to solution Y.

Describe two observations you would make.

For each observation give an equation for the reaction that occurs.

In your equations, include the formula of each complex aluminium species.

[4 marks]

Observation 1

Equation 1

Observation 2

Equation 2

Mark scheme

Show the mark scheme Mark scheme for Question 07 showing: 07.1 awards 1 mark for AlCl3 + 6H2O -> [Al(H2O)6]3+ + 3Cl- (or partial hydrolysis variants). 07.2 awards 1 mark for [Al(H2O)6]3+ + H2O -> [Al(H2O)5(OH)]2+ + H3O+. 07.3 awards 1 mark each for 'white ppt/solid', 'effervescence/bubbles/fizzing', and 1 mark for the equation 2[Al(H2O)6]3+ + 3CO3^2- -> 2[Al(H2O)3(OH)3] + 3CO2 + 3H2O. 07.4 awards 1 mark for 'White ppt/solid', 1 mark for equation [Al(H2O)6]3+ + 3OH- -> [Al(H2O)3(OH)3] + 3H2O, 1 mark for 'Colourless solution forms / ppt or solid dissolves', and 1 mark for equation [Al(H2O)3(OH)3] + OH- -> [Al(H2O)2(OH)4]- + H2O or [Al(OH)4]- + 3H2O. Total 9 marks.

Question Answers Additional Comments/Guidance Mark

AlCl + 6H O [Al(H O) ]3+ + 3Cl– Allow 1

32 2 6

AlCl + 6H O Al(H O) (OH)2+ + H+ + 3Cl-

07.1 3 2 2 5

Or equation to form Al(H O) (OH) +

24 2

3+ 2+ + allow equations to form [Al(H O) (OH) ]+ 1

07.2 [Al(H2O)6] + H2O [Al(H2O)5(OH)] + H3O 2 4 2

white ppt/solid M1 and M2 in either order 1

07.3 effervescence/bubbles/fizzing 1

2[Al(H O) ]3+ + 3CO 2– 2[Al(H O) (OH) 1

26 3 2 3 3] + 3CO2 + 3H2O accept multiples

only allow spectator ions in a balanced equation

White ppt/solid 1

[Al(H O) ]3+ +3OH– [Al(H O) (OH) ] + 3H O

26 2 3 3 2 only allow spectator ions in a balanced equation 1

07.4 Colourless solution forms / ppt or solid dissolves

[Al(H O) (OH) ] + OH- [Al(H O) (OH) ]– + H O only allow 6 or 4 co-ordination 1

23 3 2 2 4 2

OR

[Al(H O) (OH) ] + OH- [Al(OH) ]– + 3H O 3-

23 3 4 2 Allow [Al(OH)6] in a balanced equation

Total 9

How to answer it

Aqueous Aluminium(III) Complexes & Amphoteric Behaviour

📚 What this question tests

This question assesses your mastery of Topic 3.2.6 (Reactions of ions in aqueous solution), focusing on the aqueous chemistry of aluminium:

  • Formation of the hexaaqua complex ion [Al(H₂O)₆]³⁺ from anhydrous solid.
  • Explaining acidity using the Brønsted–Lowry acid-base definition with water.
  • Hydrolysis of 3+ aqua ions with weak bases like CO₃²⁻ forming a precipitate and CO₂ gas.
  • The amphoteric nature of aluminium hydroxide: precipitating with base and redissolving in excess base.
Part 07.1 • 1 Mark

Hydration of Anhydrous Aluminium Chloride

Writing the initial dissolving and complexation equation

✅ Mark Scheme Answer

AlCl₃ + 6H₂O → [Al(H₂O)₆]³⁺ + 3Cl⁻

Also allowed: Equations showing partial hydrolysis, such as:
AlCl₃ + 6H₂O → [Al(H₂O)₅(OH)]²⁺ + H⁺ + 3Cl⁻

🧠 Exam Technique

  • Identify complex Z: The prompt mentions "a complex aluminium ion, Z, and chloride ions". Complex Z is the hexaaqua ion [Al(H₂O)₆]³⁺ .
  • Balance chloride: Ensure that all 3 chloride atoms appear as separate aqueous ions ( 3Cl⁻ ), not as molecular Cl₃⁻ or Cl₂ .

❌ Common Errors

  • Writing AlCl₃ + 3H₂O → Al(OH)₃ + 3HCl : While vigorous hydrolysis can happen, the prompt explicitly states that solution Y forms containing a complex ion Z and chloride ions.
  • Forgetting square brackets or getting the complex charge wrong (aluminium is +3; water is neutral, so the charge is 3+).
Mark Breakdown: 1 mark for the correctly balanced equation showing [Al(H₂O)₆]³⁺ and 3Cl⁻ .
Part 07.2 • 1 Mark

Brønsted–Lowry Acidity in Aqueous Solution

Hydrolysis reaction of the hexaaqua ion acting as a proton donor

✅ Mark Scheme Answer

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

Note: A single forward arrow ( → ) or reversible arrows ( ⇌ ) are both accepted. Equations producing [Al(H₂O)₄(OH)₂]⁺ are also permitted.

💡 Key Knowledge

A Brønsted–Lowry acid is a proton ( H⁺ ) donor. When reacting with water, one water ligand loses an H⁺ to a solvent H₂O molecule, generating an oxonium ion ( H₃O⁺ ) and reducing the complex charge from 3+ to 2+.

❌ Common Errors

  • Writing free H⁺ : If the question asks for the reaction with water, water must be explicitly shown as a reactant, producing H₃O⁺ . Writing [Al(H₂O)₆]³⁺ → [Al(H₂O)₅(OH)]²⁺ + H⁺ fails to include water as a reactant.
  • Incorrect overall charge on the product complex: Remember that removing H⁺ leaves an OH⁻ ligand, so the charge drops from 3+ to 2+.
Mark Breakdown: 1 mark for balanced equation with correct formula/charges of complex, H₂O reactant, and H₃O⁺ product.
Part 07.3 • 3 Marks

Reaction with Sodium Carbonate

Observation and equation for the acidity-driven precipitation

✅ Mark Scheme Answer

Observation 1: White precipitate / solid

Observation 2: Effervescence / bubbles / fizzing

2[Al(H₂O)₆]³⁺ + 3CO₃²⁻ → 2[Al(H₂O)₃(OH)₃] + 3CO₂ + 3H₂O

(Spectator ions like Na⁺ permitted only if balanced on both sides.)

💡 Why Doesn't Aluminium Carbonate Form?

The Al³⁺ ion has a high charge density (small ionic radius, high 3+ charge), which strongly polarises the O–H bonds in attached water ligands. This makes the solution sufficiently acidic to react with carbonate ions ( CO₃²⁻ ) in an acid-base reaction, evolving CO₂ rather than forming Al₂(CO₃)₃ .

❌ Common Errors

  • Stating "carbon dioxide gas forms" as an observation. CO₂ is invisible! You must state what you actually see: effervescence / bubbles / fizzing.
  • Writing "white solution" instead of white precipitate.
  • Forgetting to balance the stoichiometric coefficients: 2 Al complex : 3 carbonate : 2 hydroxide ppt : 3 CO₂ : 3 H₂O.

🧠 Top Tip for Remembering the Stoichiometry

Think of it as neutralising 6 protons:
Each [Al(H₂O)₆]³⁺ must lose 3 H⁺ to become the neutral precipitate [Al(H₂O)₃(OH)₃] .
For 2 complexes, that's 6 H⁺ lost.
Each CO₃²⁻ accepts 2 H⁺ to make H₂O + CO₂ .
Therefore, exactly 3 CO₃²⁻ are required!

Mark Breakdown:
• M1: White ppt / solid (1 mark)
• M2: Effervescence / bubbles / fizzing (1 mark)
• M3: Balanced ionic equation with correct formulas for both aluminium complexes (1 mark)
Part 07.4 • 4 Marks

Reaction with Aqueous Potassium Hydroxide (Until in Excess)

Amphoteric behaviour of aluminium hydroxide

✅ Mark Scheme Answer

Observation 1: White precipitate / solid

[Al(H₂O)₆]³⁺ + 3OH⁻ → [Al(H₂O)₃(OH)₃] + 3H₂O

Observation 2: Precipitate / solid dissolves (OR colourless solution forms)

[Al(H₂O)₃(OH)₃] + OH⁻ → [Al(H₂O)₂(OH)₄]⁻ + H₂O

Accepted alternatives for Equation 2:
• [Al(H₂O)₃(OH)₃] + OH⁻ → [Al(OH)₄]⁻ + 3H₂O
• [Al(H₂O)₃(OH)₃] + 3OH⁻ → [Al(OH)₆]³⁻ + 3H₂O

💡 Understanding Amphoteric Hydroxides

Step 1 (Limited Base): OH⁻ deprotonates the hexaaqua complex until it is neutral: [Al(H₂O)₃(OH)₃] . Neutral complexes are insoluble in water, forming a white precipitate.

Step 2 (Excess Base): Unlike transition metal hydroxides like Fe(OH)₃ , aluminium hydroxide is amphoteric. It reacts further with excess OH⁻ acting as a Lewis acid / accepting another hydroxide to form a charged, soluble aluminate ion (e.g. [Al(OH)₄]⁻ ), so the precipitate redissolves.

❌ Common Errors

  • Mismatched observation 2: Saying "solution turns clear". The starting solution was already clear! The correct description is "precipitate dissolves" or "a colourless solution forms".
  • Direct reaction shortcut: Attempting to write one overall reaction for Equation 2 starting back at [Al(H₂O)₆]³⁺ . The question asks for the equation corresponding to each observation, so Equation 2 must start from the precipitate formed in Observation 1.
  • Incorrect co-ordination number: AQA accepts 4-coordinate tetrahydroxoaluminate [Al(OH)₄]⁻ or 6-coordinate [Al(H₂O)₂(OH)₄]⁻ / [Al(OH)₆]³⁻ . Do not mix up the charges and ligand counts!
Mark Breakdown:
• Observation 1: White ppt / solid (1 mark)
• Equation 1: Formation of neutral aluminium hydroxide (1 mark)
• Observation 2: Precipitate dissolves / colourless solution forms (1 mark)
• Equation 2: Reaction of precipitate with excess OH⁻ to form soluble complex ion (1 mark)

Topics

Inorganic Chemistry · Physical Chemistry · 3.2.6 Reactions of Ions in Aqueous Solution · 3.1.12 Acids and Bases

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.