AQA A-Level Chemistry 7405 · 3.2.6 Reactions of ions in aqueous solution

Part 1: Aqua-ion acidity and hydroxide reactions

All 3 parts available · worked answers and exam guidance included. Reviewed 2 October 2026.

Explain why metal aqua ions are acidic, write deprotonation equations and distinguish precipitation from amphoteric dissolution.

A dissolved metal ion has water ligands

For this topic, represent the specified ions as [Cu(H₂O)₆]²⁺, [Fe(H₂O)₆]²⁺, [Fe(H₂O)₆]³⁺ and [Al(H₂O)₆]³⁺. Each water donates a lone pair from oxygen, giving six coordinate bonds in an octahedral arrangement; copper(II) is distorted octahedral. Short formulae such as Fe²⁺(aq) are useful shorthand, but a question about water-ligand acidity requires the complete complex.

Typical dilute laboratory copper(II) solutions are pale blue, iron(II) pale green and aluminium colourless. Iron(III) solutions commonly look yellow/brown because hydrolysed species and other ligands can be present; an isolated hexaaquairon(III) complex is pale violet. Use the actual conditions and observations rather than assuming every iron(III) solution has one fixed colour.

Aluminium is included in this section even though it is not a transition metal. The chemistry here is about charge density, coordinated water and acid–base reactions, not a requirement that every central ion has d electrons.

The metal ion makes an O–H bond easier to deprotonate

A small, highly charged metal ion attracts electron density from coordinated water. This polarises and weakens the water ligand’s O–H bond, making transfer of H⁺ to a neighbouring water molecule easier. The metal centre itself is not releasing a proton. A water ligand loses H⁺ to become an OH⁻ ligand; the metal oxidation state does not change.

For comparable ions, the 3+ charge gives a greater charge-to-size ratio and more acidic aqua ions than the 2+ charge. Explain the link all the way to O–H polarisation and proton loss. Size matters as well as charge; “three plus is more acidic” on its own is an incomplete causal explanation.

[Fe(H₂O)₆]³⁺(aq) + H₂O(l) ⇌ [Fe(H₂O)₅(OH)]²⁺(aq) + H₃O⁺(aq)
[Al(H₂O)₆]³⁺(aq) ⇌ [Al(H₂O)₅(OH)]²⁺(aq) + H⁺(aq)
[Fe(H₂O)₆]²⁺(aq) ⇌ [Fe(H₂O)₅(OH)]⁺(aq) + H⁺(aq)

Hydroxide removes protons in stages

OH⁻ accepts a proton from a coordinated water molecule to form water. For a 3+ aqua ion, each deprotonation reduces the complex charge by one. After three deprotonations the neutral hydrated hydroxide precipitates. In this A-level representation, six O donor sites remain around the metal as water ligands become hydroxide ligands.

[Al(H₂O)₆]³⁺(aq) + OH⁻(aq) → [Al(H₂O)₅(OH)]²⁺(aq) + H₂O(l)
[Al(H₂O)₅(OH)]²⁺(aq) + OH⁻(aq) → [Al(H₂O)₄(OH)₂]⁺(aq) + H₂O(l)
[Al(H₂O)₄(OH)₂]⁺(aq) + OH⁻(aq) → Al(H₂O)₃(OH)₃(s) + H₂O(l)

Small additions of hydroxide produce four useful precipitates

Write the neutral solid without an overall ionic charge. The hydrated formulae below make the water-ligand chemistry explicit; the simpler forms Cu(OH)₂, Fe(OH)₂, Fe(OH)₃ and Al(OH)₃ are often used as shorthand. Do not mix the two representations within an equation without rebalancing the waters.

[Cu(H₂O)₆]²⁺(aq) + 2OH⁻(aq) → Cu(H₂O)₄(OH)₂(s) + 2H₂O(l)
[Fe(H₂O)₆]²⁺(aq) + 2OH⁻(aq) → Fe(H₂O)₄(OH)₂(s) + 2H₂O(l)
[Fe(H₂O)₆]³⁺(aq) + 3OH⁻(aq) → Fe(H₂O)₃(OH)₃(s) + 3H₂O(l)
[Al(H₂O)₆]³⁺(aq) + 3OH⁻(aq) → Al(H₂O)₃(OH)₃(s) + 3H₂O(l)
Adding NaOH dropwise, then in excess
Initial metal ionSmall amount of OH⁻Excess OH⁻ under ordinary test conditions
Cu²⁺Blue Cu(H₂O)₄(OH)₂ precipitatePrecipitate remains
Fe²⁺Green Fe(H₂O)₄(OH)₂ precipitatePrecipitate remains; browns on standing in air
Fe³⁺Brown Fe(H₂O)₃(OH)₃ precipitatePrecipitate remains
Al³⁺White Al(H₂O)₃(OH)₃ precipitateDissolves to a colourless aluminate solution

Aluminium hydroxide reacts with acid and excess base

Amphoteric aluminium hydroxide dissolves in excess OH⁻ to give [Al(OH)₄]⁻, a colourless solution. Its white precipitate also dissolves in acid as hydroxide ligands are protonated back to water ligands. Aluminium stays +3 throughout.

In the shorthand representation, Al(OH)₃ + OH⁻ → [Al(OH)₄]⁻. In the hydrated representation used below, three water molecules are additionally released. These describe the same stoichiometric aluminium conversion using different conventions. The precipitation-stage coordination number is six; the tetrahydroxoaluminate product has four donor bonds.

Al(H₂O)₃(OH)₃(s) + OH⁻(aq) → [Al(OH)₄]⁻(aq) + 3H₂O(l)
Al(H₂O)₃(OH)₃(s) + 3H⁺(aq) → [Al(H₂O)₆]³⁺(aq)

Worked example: precipitation is not the final excess-base product

Constructed stoichiometric model: a sample contains 0.500 mmol Al³⁺ aqua ions, with no additional acid consuming hydroxide. Forming the hydroxide precipitate requires 3 × 0.500 = 1.50 mmol OH⁻. Converting all that precipitate to [Al(OH)₄]⁻ requires a further 0.500 mmol OH⁻. The total is therefore 2.00 mmol.

With 0.200 mol dm⁻³ NaOH, the idealised precipitation requirement is 7.50 cm³ and the additional dissolution requirement is 2.50 cm³, giving 10.0 cm³ overall. Actual observed boundaries depend on equilibria, concentrations and other acid present; these numbers are stoichiometric quantities, not a universal practical titration curve.

[Al(H₂O)₆]³⁺(aq) + 4OH⁻(aq) → [Al(OH)₄]⁻(aq) + 6H₂O(l)

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official AQA mark allocations.

Q1. Explain why [Fe(H₂O)₆]³⁺ is more acidic than [Fe(H₂O)₆]²⁺.Show answer

Fe³⁺ has greater charge density and polarises the O–H bonds in its water ligands more strongly. Proton transfer to water is therefore easier, giving greater acidity under comparable conditions.

Q2. After one water ligand in [Al(H₂O)₆]³⁺ loses H⁺, what is the complex formula and charge?Show answer

[Al(H₂O)₅(OH)]²⁺. Al remains +3; five waters are neutral and one hydroxide ligand contributes −1, giving overall +2.

Q3. State the observations when NaOH is added dropwise then in excess to aluminium aqua ions.Show answer

A white precipitate forms first. It dissolves in excess NaOH to produce a colourless solution containing [Al(OH)₄]⁻.

Q4. How many moles of OH⁻ are needed to precipitate 0.00200 mol Cu²⁺, assuming no other acid consumes base?Show answer

The ratio is two OH⁻ per Cu²⁺, so 0.00400 mol OH⁻. The blue hydrated copper(II) hydroxide remains in excess OH⁻ under the ordinary test conditions.

Q5. Why is the formation of [Al(OH)₄]⁻ not a reduction of aluminium?Show answer

Four OH⁻ ligands contribute −4; the overall charge is −1, so aluminium is still +3. The negative complex charge does not imply a negative oxidation state for Al.

Sources

Sources and examiner guidance (reviewed 2 October 2026)

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