Related topics: Group 2: the alkaline earth metals
Where this sits
This short topic is the practical payoff of transition metals: it is how you identify a metal ion in a test tube. It rests on two ideas: metal ions in water are really aqua complexes [M(H2O)6]n+, and adding a base removes H+ from those complexes to give a coloured hydroxide precipitate. Learn the colours and the few equations, and the identification follows.
Metal-aqua ions & their acidity
When a metal salt dissolves, the metal ion is surrounded by six water molecules as an aqua complex, so the “Fe3+(aq)” or blue “Cu2+(aq)” you wrote in earlier work is really [Fe(H2O)6]3+ or [Cu(H2O)6]2+; the six water molecules were always there. Some of these solutions are acidic:
- [M(H2O)6]2+ for M = Fe and Cu;
- [M(H2O)6]3+ for M = Al and Fe.
The metal ion polarises the O–H bonds in its water ligands, so a water ligand can lose an H+ to a free water molecule. The complex acts as a Brønsted–Lowry acid (the same H3O+-producing idea you met in acids and bases):
[Fe(H2O)6]3+ + H2O ⇌ [Fe(H2O)5(OH)]2+ + H3O+
Why 3+ ions are more acidic than 2+ ions
A [M(H2O)6]3+ ion is more acidic (lower pH) than a [M(H2O)6]2+ ion because the 3+ ion has a higher charge/size ratio (charge density). It polarises (weakens) the O–H bonds in the water ligands more strongly, so H+ is released more readily and the position of the equilibrium above lies further to the right.
“Fe3+(aq)” is really a complex
[Fe(H2O)6]3+: the six water ligands were always there, each attached by a co-ordinate bond.
Why the 3+ ion is acidic
The high charge density polarises the O–H bonds of the water ligands, so an H+ transfers to a free water molecule: the equilibrium in the equation above.
Exam questions
Q1
[3 marks]
Explain why an aqueous solution containing [Fe(H2O)6]3+ ions has a lower pH than one containing [Fe(H2O)6]2+ ions.
Show answer
The Fe3+ ion has a higher charge (and similar/smaller size), so a higher charge density / charge-to-size ratio. 1 mark
It polarises / weakens the O–H bonds in the coordinated water more strongly. 1 mark
So H+ ions are released more readily (the equilibrium lies further right), giving a lower pH. 1 mark
Q2
[2 marks]
Explain, with an equation, why a solution containing [Al(H2O)6]3+ has a pH below 7.
Show answer
[Al(H2O)6]3+ + H2O ⇌ [Al(H2O)5(OH)]2+ + H3O+. 1 mark
The highly charged Al3+ polarises the O–H bonds of the water ligands, releasing H+ (H3O+), so the solution is acidic. 1 mark
Source: AQA A-Level Chemistry past papers.
Reactions with hydroxide, ammonia and carbonate
A base removes H+ from the aqua complex. Removing enough H+ leaves a neutral metal hydroxide, which is insoluble: a coloured precipitate. This grid is the whole topic:
| Aqua ion | + a little NaOH or NH3 | + excess NaOH | + excess NH3 | + Na2CO3 |
|---|---|---|---|---|
| [Fe(H2O)6]2+ | green ppt, Fe(OH)2 | ppt remains | ppt remains | green ppt (FeCO3) |
| [Cu(H2O)6]2+ | blue ppt, Cu(OH)2 | ppt remains | deep blue solution | blue-green ppt (CuCO3) |
| [Al(H2O)6]3+ | white ppt, Al(OH)3 | dissolves (colourless) | ppt remains | white ppt + CO2 |
| [Fe(H2O)6]3+ | red-brown ppt, Fe(OH)3 | ppt remains | ppt remains | red-brown ppt + CO2 |
Writing the precipitate formula
Each precipitate is a solid with no overall charge, so its formula simply balances the metal ion with enough hydroxide ions to cancel that charge: a 2+ ion takes two OH− (Fe(OH)2, Cu(OH)2) and a 3+ ion takes three (Al(OH)3, Fe(OH)3).
Fe(OH)2 and [Fe(H2O)4(OH)2] are the same substance: two ways of writing one precipitate. The simplified formula (Fe(OH)2) leaves out the water ligands that are still attached; the hydrated formula ([Fe(H2O)4(OH)2]) shows them. The question won’t usually tell you which to use: both are accepted. Just pick one and stay consistent within an equation, rather than mixing a hydrated species on one side with a simplified one on the other.
The precipitate forms as the base removes H+ ions from the aqua ion. Each H+ lost turns a water ligand into an OH−, until a neutral, insoluble hydroxide is left:
[Fe(H2O)6]2+ + 2OH− → [Fe(H2O)4(OH)2] + 2H2O
[Fe(H2O)6]3+ + 3OH− → [Fe(H2O)3(OH)3] + 3H2O
With ammonia the base is the NH3 molecule itself: it removes H+ to form NH4+ (it does not supply OH− directly), giving the same hydroxide precipitate, for example [Fe(H2O)6]3+ + 3NH3 → [Fe(H2O)3(OH)3] + 3NH4+.
Two behaviours stand out and are worth extra marks.
Al(III) is amphoteric: its white hydroxide dissolves in both excess base and excess acid (that two-way behaviour is what “amphoteric” means). In excess NaOH it gives a colourless aluminate; in excess acid it reverts to the hexaaqua ion:
[Al(H2O)3(OH)3] + OH− → [Al(OH)4]− + 3H2O
[Al(H2O)3(OH)3] + 3H+ → [Al(H2O)6]3+
Cu(II) redissolves in excess ammonia by ligand substitution to give a deep blue solution:
[Cu(H2O)6]2+ + 4NH3 → [Cu(NH3)4(H2O)2]2+ + 4H2O
Interactive: amphoteric means it reacts with both
With carbonate, the two charges behave differently. The 2+ ions are only weakly acidic, so they form carbonate precipitates (FeCO3, CuCO3). The 3+ ions are acidic enough to react with the carbonate, releasing CO2 (effervescence) and giving the hydroxide:
2[Fe(H2O)6]3+ + 3CO32− → 2[Fe(H2O)3(OH)3] + 3CO2 + 3H2O
The four hydroxide precipitates
A little NaOH(aq) or NH3(aq) gives the same four precipitates. The colour is the first identification step.
Amphoteric Al(OH)3: excess matters
Only the white Al(III) hydroxide redissolves in excess NaOH; excess NH3 leaves it untouched.
Cu(II) & excess NH3
The blue hydroxide dissolves by ligand substitution. The deep blue solution is the Cu(II) giveaway.
Interactive: every ion × every reagent, with the equation
1 · aqua ion
2 · add
Interactive: precipitation quiz, both directions
Exam questions
Q3
[1 mark]
Aqueous aluminium sulfate is added to aqueous sodium carbonate. What are the formulae of the precipitate and the gas formed? Tick (✓) one box.
The acidic [Al(H2O)6]3+ reacts with carbonate to give the white hydroxide precipitate Al(H2O)3(OH)3 and carbon dioxide gas, not the carbonate Al2(CO3)3, which does not form.
Q4
[3 marks]
Aqueous sodium carbonate is added to a solution containing [Fe(H2O)6]3+. Give the formula and colour of the precipitate, and an ionic equation for the reaction.
Show answer
Formula: [Fe(H2O)3(OH)3] (iron(III) hydroxide). 1 mark
Colour: red-brown / brown. 1 mark
2[Fe(H2O)6]3+ + 3CO32− → 2[Fe(H2O)3(OH)3] + 3CO2 + 3H2O. 1 mark
Q5
[2 marks]
Aqueous potassium hydroxide is added, until in excess, to a solution of [Al(H2O)6]3+. Give the formula of the final aluminium species and an equation for its formation from the white precipitate.
Show answer
Final species: [Al(OH)4]−. 1 mark
[Al(H2O)3(OH)3] + OH− → [Al(OH)4]− + 3H2O. 1 mark
The white precipitate redissolving in excess is the signature of amphoteric Al(III).
Source: AQA A-Level Chemistry past papers.
Identifying the ion & Required Practical 11
Put the grid to work as a decision process. Add the reagents in turn and read off the ion from the colours and what dissolves.
The identification flowchart
Colour first, then the two discriminating tests: what dissolves in excess, and whether carbonate fizzes.
Required practical 11: identifying transition metal ions
You carry out these test-tube reactions on unknown solutions and record the observations. Two habits win marks: give the exact precipitate colour (not just “a precipitate”), and always test with excess reagent so you can report whether the precipitate redissolves. This is the step that separates Al3+ and Cu2+ from the rest.
AQA ask you to record rather than identify, so precision in the wording is the skill:
- Every colour observation must say precipitate or solution. “Green” is not an observation; “a green precipitate” is. This is the single commonest lost mark in the practical.
- Add dropwise, then to excess, and record both. The aluminium precipitate only gives itself away when the excess redissolves it.
- Write “no visible change” rather than leaving a cell blank. It is a real result and it carries marks.
- Watch the iron(II) precipitate on standing. It darkens from green towards brown as air oxidises it, which is why a standing step is built into the method.
- Effervescence with carbonate means a 3+ ion. Only Fe3+ and Al3+ are acidic enough to release CO2; the 2+ ions simply precipitate their carbonate.
You do this hands-on in Required Practical 11; the full method and safety sit in the required practicals guide.
The identification routine
- Precipitate colour narrows it down: green → Fe(II); blue → Cu(II); white → Al(III) or Mg(II); red-brown → Fe(III).
- Excess NaOH: only the white Al(III) hydroxide redissolves (amphoteric); white Mg(OH)2 stays.
- Excess NH3: only the Cu(II) hydroxide redissolves, to a deep blue solution.
- Carbonate: effervescence (CO2) means a 3+ ion; a plain precipitate means a 2+ ion.
- Judge colours fresh. On standing in air the green Fe(OH)2 darkens at its surface as the Fe(II) is oxidised towards red-brown Fe(III).
Precision points
- Examiner reports flag stopping at “white precipitate” without mentioning it redissolves in excess for Al(III), and omitting the exact colour.
- Keep the equations in the hydrated form throughout (e.g. [Fe(H2O)3(OH)3], not Fe(OH)3) when the question asks for the complex species.
Interactive: build the mark-scheme answer
“A colourless solution gives a white precipitate with a few drops of NaOH(aq); the precipitate dissolves in excess NaOH to give a colourless solution. With Na2CO3(aq) the solution gives a white precipitate and effervescence. Identify the ion and explain the observations.” [5 marks] Select every statement that earns a mark, and nothing that doesn’t. Order doesn’t matter.
Capstone question
Q6
[4 marks]
Two iron solutions, L and M, are tested. With ammonia, L gives a red-brown precipitate insoluble in excess and M gives a green precipitate insoluble in excess. With sodium carbonate, L gives a red-brown precipitate with effervescence, while M gives a green precipitate with no gas. Identify L and M, and explain why effervescence is seen with L but not M.
Show answer
L contains [Fe(H2O)6]3+ (iron(III)); its hydroxide is red-brown. 1 mark
M contains [Fe(H2O)6]2+ (iron(II)); its hydroxide is green. 1 mark
The 3+ ion (L) has a higher charge density and is more acidic, so it reacts with carbonate to release CO2 (effervescence). 1 mark
The 2+ ion (M) is less acidic, so it simply forms the carbonate precipitate (FeCO3) with no gas. 1 mark
Source: AQA A-Level Chemistry past papers.
3.2.6 Reactions of ions in solution: quick-reference summary
- Aqua ions: [M(H2O)6]2+ (Fe, Cu) and [M(H2O)6]3+ (Al, Fe). The 3+ ions are more acidic (higher charge density polarises the O–H bonds, releasing H+).
- Hydroxide precipitates (with a little NaOH or NH3): Fe(OH)2 green; Cu(OH)2 blue; Al(OH)3 white; Fe(OH)3 red-brown.
- Excess NaOH: only the white Al(III) hydroxide redissolves (amphoteric) to [Al(OH)4]−; being amphoteric it also dissolves in excess acid back to [Al(H2O)6]3+. Excess NH3: only the Cu(II) hydroxide redissolves to the deep-blue [Cu(NH3)4(H2O)2]2+.
- Required Practical 11: identify these ions by test-tube reactions. Give the exact precipitate colour and always test with excess reagent.
- Carbonate: 2+ ions form carbonate precipitates (FeCO3, CuCO3); 3+ ions are acidic enough to release CO2, giving the hydroxide precipitate + effervescence.