OxfordAQA A-Level Chemistry: Reactions of ions in aqueous solution (9620)

This study guide teaches the aqueous-ions topic, section 3.2.6, from OxfordAQA’s International AS and A-level Chemistry specification (9620), Version 5.3, for exams from May/June 2020. The section has one subsection, 3.2.6.1, and it is International A2 content, so all of it is International A-level only. The specification places 3.2.4 to 3.2.6 in Unit 3 (Inorganic 2 and Physical 2), and Unit 5 (Practical and synoptic) can draw on any content. Required practical 9 belongs here.

Companion pages: the revision notes and the practice questions. Background: transition metals (ligands), acids and bases and Period 3 oxides and chlorides. Course links: hub, printable checklist, free diagnostics.

What 3.2.6.1 asks of you

Spec ref	Outcome	Stage
3.2.6.1	Know the aqua ions [M(H₂O)₆]²⁺ for M = Fe and Cu, and [M(H₂O)₆]³⁺ for M = Al and Fe	International A-level only
3.2.6.1	Use the charge/size ratio of the metal ion to explain why 3+ aqua ions are stronger acids than 2+ aqua ions	International A-level only
3.2.6.1	Know that some metal hydroxides are amphoteric, dissolving in acids and in bases (for example the hydroxide of Al³⁺)	International A-level only
3.2.6.1	Describe and explain what Fe²⁺, Cu²⁺, Fe³⁺ and Al³⁺ in solution do with the bases OH⁻, NH₃ and CO₃²⁻	International A-level only
Required practical 9	Identify transition metal ions in solution using simple test-tube reactions	International A-level only

Four ions and three bases give twelve combinations. Learn each as an observation, an equation and a reason.

Where the aqua ions come from

Dissolve a soluble salt of one of these metals in water and the cation picks up six water ligands. Each oxygen donates a lone pair to the metal, so the complex is octahedral with a co-ordination number of 6.

Aqua ion	What the solution looks like
[Fe(H₂O)₆]²⁺(aq)	pale green
[Cu(H₂O)₆]²⁺(aq)	blue
[Fe(H₂O)₆]³⁺(aq)	yellow to brown in a normal lab solution
[Al(H₂O)₆]³⁺(aq)	colourless (Al³⁺ has no d electrons)

Colour narrows the field but never proves an identity.

Acidity of the aqua ions

What happens

A water ligand on the metal can lose a proton to a solvent water molecule. The aqua ion is behaving as a Brønsted–Lowry acid:

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

One H₂O ligand becomes OH⁻, the complex keeps six ligands, and its charge drops by one. That pattern explains every precipitate below.

Why the 3+ ions are stronger acids

The specification asks for the explanation in terms of charge/size ratio. Build it in this order:

• An M³⁺ ion carries more positive charge than an M²⁺ ion and has a smaller ionic radius.
• So the M³⁺ ion has the larger charge/size ratio. It is more strongly polarising.
• It draws electron density from the O–H bonds of its water ligands towards the oxygen and the metal. Those O–H bonds are weakened.
• A proton is lost more readily, so at equal concentrations the M³⁺ solution has the higher [H₃O⁺] and the lower pH.

Worked example 1: putting numbers on charge/size

Ionic radii for high-spin octahedral iron ions are 78 pm for Fe²⁺ and 64.5 pm for Fe³⁺. Compare charge/radius for the two ions and say what this predicts.

Fe²⁺: 2 / 78   = 0.0256 per pm
Fe³⁺: 3 / 64.5 = 0.0465 per pm
ratio: 0.0465 / 0.0256 = 1.81

Iron(III) has about 1.8 times the charge/size ratio of iron(II). It polarises its water ligands far more, so iron(III) nitrate solution is the more acidic of the two at the same concentration. Both factors push the same way.

How a base turns an aqua ion into a precipitate

Every base in this section can pull protons off the water ligands. Follow iron(III) as each proton goes:

[Fe(H₂O)₆]³⁺ → [Fe(H₂O)₅(OH)]²⁺ → [Fe(H₂O)₄(OH)₂]⁺ → [Fe(H₂O)₃(OH)₃]

A charged complex stays dissolved. Once it reaches zero charge it comes out as a solid. So a 2+ ion precipitates after losing two protons, and a 3+ ion after losing three. The neutral products are written [M(H₂O)₄(OH)₂] and [M(H₂O)₃(OH)₃], or more simply M(OH)₂ and M(OH)₃.

With hydroxide ions (sodium hydroxide solution)

Ion	NaOH(aq) dropwise	NaOH(aq) in excess
Fe²⁺	green precipitate, turning brown at the surface on standing in air	no further change
Cu²⁺	blue precipitate	no further change
Fe³⁺	orange-brown precipitate	no further change
Al³⁺	white precipitate	dissolves, leaving a colourless solution

[Fe(H₂O)₆]²⁺(aq) + 2OH⁻(aq) → [Fe(H₂O)₄(OH)₂](s) + 2H₂O(l)
[Cu(H₂O)₆]²⁺(aq) + 2OH⁻(aq) → [Cu(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)

The brown surface on iron(II) hydroxide is air oxidising it to iron(III) hydroxide.

Amphoteric aluminium hydroxide

An amphoteric hydroxide dissolves in acids and in bases. In 3.2.6.1 the example is the hydroxide of Al³⁺.

• In excess hydroxide, the neutral solid loses one more proton and becomes a soluble anion.
• In acid, the hydroxide ligands are protonated back to water and the aqua ion returns.

[Al(H₂O)₃(OH)₃](s) + OH⁻(aq) → [Al(H₂O)₂(OH)₄]⁻(aq) + H₂O(l)
[Al(H₂O)₃(OH)₃](s) + 3H₃O⁺(aq) → [Al(H₂O)₆]³⁺(aq) + 3H₂O(l)

The anion can also be written [Al(OH)₄]⁻. The second equation is often written with 3H⁺ on the left and no water on the right; both are balanced.

Worked example 2: running the aluminium sequence backwards

A student has a colourless solution made by adding excess sodium hydroxide to aluminium nitrate solution. She adds dilute hydrochloric acid one drop at a time until the acid is in excess. Predict and explain what she sees.

1. The solution holds [Al(H₂O)₂(OH)₄]⁻. The first drops of acid protonate one OH⁻ ligand, giving back the neutral [Al(H₂O)₃(OH)₃]. A white precipitate appears.
2. More acid protonates the remaining OH⁻ ligands until [Al(H₂O)₆]³⁺ re-forms. The precipitate dissolves and the solution becomes colourless again.

With ammonia (aqueous ammonia)

Ammonia can do two different jobs, and you must say which one it is doing.

• As a base. A small amount of NH₃ removes protons, forming NH₄⁺. Each ion gives the same precipitate as it does with NaOH.
• As a ligand. With copper, excess NH₃ substitutes for ligands on the copper and the precipitate dissolves.

Ion	NH₃(aq) dropwise	NH₃(aq) in excess
Fe²⁺	green precipitate	no further change
Cu²⁺	blue precipitate	dissolves to give a deep blue solution
Fe³⁺	orange-brown precipitate	precipitate remains
Al³⁺	white precipitate	precipitate remains

[Fe(H₂O)₆]²⁺(aq) + 2NH₃(aq) → [Fe(H₂O)₄(OH)₂](s) + 2NH₄⁺(aq)
[Al(H₂O)₆]³⁺(aq) + 3NH₃(aq) → [Al(H₂O)₃(OH)₃](s) + 3NH₄⁺(aq)
[Cu(H₂O)₄(OH)₂](s) + 4NH₃(aq) → [Cu(NH₃)₄(H₂O)₂]²⁺(aq) + 2H₂O(l) + 2OH⁻(aq)

The deep blue complex keeps a co-ordination number of 6: four ammonia ligands and two water ligands. The transition metals guide explains why NH₃ and H₂O swap without a change in co-ordination number.

Ammonia does not redissolve aluminium hydroxide. It is a weak base, and the concentration of OH⁻ it provides is too low to take the fourth proton.

With carbonate ions (sodium carbonate solution)

Here 2+ and 3+ ions behave differently, because of their acidity.

Ion	Na₂CO₃(aq)	What forms
Fe²⁺	green precipitate, no gas	iron(II) carbonate, FeCO₃
Cu²⁺	blue-green precipitate, no gas	copper(II) carbonate, CuCO₃
Fe³⁺	orange-brown precipitate plus bubbles	[Fe(H₂O)₃(OH)₃] + CO₂ gas
Al³⁺	white precipitate plus bubbles	[Al(H₂O)₃(OH)₃] + CO₂ gas

A 2+ aqua ion is too weak an acid to protonate carbonate, so the insoluble metal carbonate precipitates:

[Cu(H₂O)₆]²⁺(aq) + CO₃²⁻(aq) → CuCO₃(s) + 6H₂O(l)

A 3+ aqua ion is acidic enough to give up protons. Carbonate takes them and breaks down to CO₂ and water, leaving the neutral hydroxide. No 3+ carbonate forms.

2[Fe(H₂O)₆]³⁺(aq) + 3CO₃²⁻(aq) → 2[Fe(H₂O)₃(OH)₃](s) + 3CO₂(g) + 3H₂O(l)

The balancing logic: two M³⁺ ions release six protons in total, and three carbonate ions take two protons each.

Worked example 3: volume of gas from the iron(III) reaction

30.0 cm³ of 0.140 mol dm⁻³ iron(III) nitrate solution reacts with excess sodium carbonate solution. Calculate the volume of CO₂ released, measured at 298 K and 100 kPa. Take R as 8.31 J K⁻¹ mol⁻¹.

n(Fe³⁺) = 0.140 × 30.0/1000          = 4.20 × 10⁻³ mol
n(CO₂)  = 4.20 × 10⁻³ × 3/2          = 6.30 × 10⁻³ mol
V = nRT/p = 6.30 × 10⁻³ × 8.31 × 298 / (1.00 × 10⁵)
          = 1.56 × 10⁻⁴ m³ = 156 cm³

Convert kPa to Pa before dividing, and m³ to cm³ at the end.

Required practical 9

The aim is to name a transition metal ion in solution from a short series of test-tube reactions.

A workable routine:

1. Put about 2 cm³ of the test solution in each of three clean test tubes. Note the colour.
2. To the first tube add dilute NaOH(aq) a drop at a time, shaking between drops. Note any precipitate, then keep going to excess and note whether it dissolves.
3. Repeat with dilute NH₃(aq) in the second tube, again to excess.
4. Add Na₂CO₃(aq) to the third tube and watch for bubbles as well as a precipitate.
5. Leave the NaOH tube for a few minutes and look again for any change at the surface.

Wear eye protection: NaOH(aq) is corrosive, and NH₃(aq) irritates eyes and lungs. Separate tubes stop one reagent’s products confusing the next result.

Worked example 4: a solution containing two ions

A pale blue solution may contain Cu²⁺ alone, or Cu²⁺ and Al³⁺ together. Plan a test that decides, with the expected results.

1. Add NaOH(aq) to excess, then filter.
2. Cu²⁺ is present either way, so a blue precipitate stays on the filter paper: copper(II) hydroxide does not dissolve in excess hydroxide.
3. If Al³⁺ is present, it has passed into the filtrate as [Al(H₂O)₂(OH)₄]⁻, which is colourless.
4. Add dilute acid dropwise to the filtrate. A white precipitate that then dissolves in more acid shows Al³⁺. No precipitate means Al³⁺ was absent.

A white precipitate on its own is not enough: Mg²⁺ from Group 2 also gives one. The dissolve-and-return behaviour is what pins down aluminium.

Common errors

• Saying Fe³⁺ is more acidic “because it is bigger”. It is smaller, and that is half of the reason.
• Putting a charge on a neutral hydroxide, or leaving the minus sign off [Al(H₂O)₂(OH)₄]⁻.
• Writing Fe₂(CO₃)₃ or Al₂(CO₃)₃ as a product. Neither forms with the aqua ions.
• Leaving out “bubbles” or “effervescence” for a 3+ ion with carbonate; it is a separate observation.
• Claiming excess ammonia dissolves aluminium hydroxide, or excess NaOH dissolves copper hydroxide.
• Recording only the final state after excess reagent and missing the precipitate seen first.

Official syllabus

Source: OxfordAQA’s International AS and A-level Chemistry specification (9620), Version 5.3, for International AS and A-level exams from May/June 2020. Section 3.2.6 (International A2), the topic on reactions of ions in aqueous solution, subsection 3.2.6.1, including required practical 9.