Reactions of Metal Ions with Bases Revision Notes | A Level AQA Chemistry

1. Why These Reactions Matter

Picture four test tubes, each containing a colourless solution of a different metal salt. Add a few drops of sodium hydroxide and suddenly you have a rainbow of precipitates — pale blue, rust-brown, white, and green. These colour changes are not decorative: they are the basis of qualitative analysis for metal ions at A-level.

The AQA specification requires you to predict, describe, and explain what happens when $\text{Fe}^{2+}$, $\text{Cu}^{2+}$, $\text{Al}^{3+}$, and $\text{Fe}^{3+}$ ions meet three different bases: $\text{OH}^-$, $\text{NH}_3$, and $\text{CO}_3^{2-}$. You also need to handle the special case of aluminium hydroxide, which dissolves in both acids and excess alkali — a property called amphoteric character.

2. The Core Pattern: Precipitation of Metal Hydroxides

When any of these metal ions meets hydroxide ions, the first thing that happens is the same: a metal hydroxide precipitate forms. The general ionic equation is straightforward — the metal cation combines with $\text{OH}^-$ ions to produce an insoluble hydroxide.

The colour of the precipitate is the key observation that distinguishes the ions. Each metal hydroxide has a characteristic appearance, and the examiner will expect you to state it precisely — not write "a precipitate forms."

Notice that the M2+ ions need two $\text{OH}^-$ ions and the M3+ ions need three — this follows directly from charge balance. Balancing these ionic equations is a common source of dropped marks, so always check that the total charge on each side is equal.

3. Adding Excess NaOH — Where Al3+ Stands Apart

Adding a limited amount of NaOH gives the precipitates described above. The interesting chemistry begins when you add excess NaOH. Three of the four precipitates remain unchanged — but aluminium hydroxide dissolves completely.

This is the amphoteric character of $\text{Al(OH)}_3$. In excess NaOH, the aluminium hydroxide reacts with $\text{OH}^-$ ions to form the soluble tetrahydroxoaluminate ion, $[\text{Al(OH)}_4]^-$. AQA accepts the description that $\text{Al(OH)}_3$ acts as an acid in this context (donating a proton to $\text{OH}^-$ in a Brønsted–Lowry sense), though the key requirement is knowing the equation and that the precipitate dissolves to give a colourless solution.

The tetrahydroxoaluminate ion $[\text{Al(OH)}_4]^-$ is sometimes written as $\text{Al(OH)}_4^-$ or as $\text{AlO}_2^- + 2\text{H}_2\text{O}$ in older texts — AQA uses the $[\text{Al(OH)}_4]^-$ form, so use that in your answers.

The other three hydroxides — $\text{Fe(OH)}_2$, $\text{Cu(OH)}_2$, and $\text{Fe(OH)}_3$ — do not dissolve in excess NaOH. Their precipitates persist. This contrast is exactly what makes the Al3+ test distinctive and examinable.

4. Reactions with Excess Ammonia Solution

Ammonia solution ($\text{NH}_3$(aq)) is a weak base. When added in limited amounts to each metal ion solution, it initially behaves like NaOH — the same hydroxide precipitates form, because $\text{NH}_3$ partially ionises to produce $\text{OH}^-$ ions in solution.

The difference emerges when excess $\text{NH}_3$ is added. Ammonia is also a ligand, and it can replace water molecules in the coordination sphere of certain metal ions. This ligand substitution dissolves the precipitate and forms a new, deeply coloured complex ion.

The deep blue solution formed with $\text{Cu}^{2+}$ and excess $\text{NH}_3$ is the tetraamminecopper(II) complex, $[\text{Cu(NH}_3)_4(\text{H}_2\text{O})_2]^{2+}$. The ionic equation for the dissolution step is:

For Fe2+, Fe3+, and Al3+, excess ammonia does not dissolve the precipitate. This is a key point of contrast with the NaOH test: Al(OH)3 dissolves in excess NaOH but not in excess NH3, while Cu(OH)2 dissolves in excess NH3 but not in excess NaOH.

5. Reactions with Sodium Carbonate Solution

The third base to consider is carbonate, $\text{CO}_3^{2-}$. When sodium carbonate solution is added to solutions of these metal ions, precipitates form again — but the chemistry is slightly different depending on whether you have an M2+ or M3+ ion.

For $\text{Cu}^{2+}$ and $\text{Fe}^{2+}$, the carbonate ion acts as a straightforward base and a precipitating anion. A metal carbonate or basic carbonate precipitate forms. The key ionic equations are:

• $\text{Cu}^{2+}\text{(aq)} + \text{CO}_3^{2-}\text{(aq)} \rightarrow \text{CuCO}_3\text{(s)}$ — green precipitate
• $\text{Fe}^{2+}\text{(aq)} + \text{CO}_3^{2-}\text{(aq)} \rightarrow \text{FeCO}_3\text{(s)}$ — green precipitate

For $\text{Al}^{3+}$ and $\text{Fe}^{3+}$, the situation is more interesting. These M3+ ions hydrolyse water so strongly that the solution is quite acidic. When carbonate is added, the carbonate ion is protonated by the acidic metal ion solution — $\text{CO}_3^{2-}$ reacts with the $\text{H}^+$ produced by hydrolysis and releases $\text{CO}_2$ gas. The result is that a metal hydroxide precipitate forms rather than a metal carbonate.

• $2\text{Al}^{3+}\text{(aq)} + 3\text{CO}_3^{2-}\text{(aq)} + 3\text{H}_2\text{O(l)} \rightarrow 2\text{Al(OH)}_3\text{(s)} + 3\text{CO}_2\text{(g)}$ — white precipitate with effervescence
• $2\text{Fe}^{3+}\text{(aq)} + 3\text{CO}_3^{2-}\text{(aq)} + 3\text{H}_2\text{O(l)} \rightarrow 2\text{Fe(OH)}_3\text{(s)} + 3\text{CO}_2\text{(g)}$ — rust-brown precipitate with effervescence

The effervescence of $\text{CO}_2$ is a distinctive observation that distinguishes M3+ ions from M2+ ions when carbonate is the reagent. If you see bubbles alongside a precipitate, you are almost certainly dealing with a 3+ metal ion.

6. Worked Identification: Putting the Tests Together

Exam questions often give you a series of observations and ask you to identify the ion, or give you an ion and ask you to predict all observations. The step-by-step walkthrough below models the analytical reasoning the examiner expects.

Worked example: Identifying an Unknown Metal Ion from Test-Tube Observations

An aqueous solution X gives the following results: (i) Addition of a few drops of NaOH(aq) gives a rust-brown precipitate. (ii) Addition of excess NaOH(aq) — the precipitate remains. (iii) Addition of Na2CO3(aq) gives a rust-brown precipitate and colourless gas. Identify the ion present and write the ionic equation for observation (iii).

1. Rust-brown precipitate with limited NaOH — From the colour table: rust-brown precipitate = Fe(OH)3(s). This points to Fe3+(aq). The ionic equation is: Fe3+(aq) + 3OH–(aq) → Fe(OH)3(s).
2. Precipitate remains in excess NaOH — Al3+ would dissolve in excess NaOH (amphoteric). Fe3+ does not dissolve — the rust-brown precipitate persists. This rules out Al3+ and confirms Fe3+.
3. Rust-brown precipitate + colourless gas with Na2CO3 — Fe3+ is an M3+ ion. It hydrolyses water strongly, so CO32– is protonated and CO2 gas is released. The precipitate is Fe(OH)3(s), not Fe2(CO3)3(s).
4. $2\text{Fe}^{3+}\text{(aq)} + 3\text{CO}_3^{2-}\text{(aq)} + 3\text{H}_2\text{O(l)} \rightarrow 2\text{Fe(OH)}_3\text{(s)} + 3\text{CO}_2\text{(g)}$ — Check balance: Fe: 2 = 2 ✓. C: 3 = 3 ✓. O: 9 + 3 = 9 + 6 → left: 9 from CO3 + 3 from H2O = 12; right: 6 from Fe(OH)3 + 6 from CO2 = 12 ✓. H: 6 = 6 ✓. Charge: left = 6+ + 6– = 0; right = 0 ✓.
5. Ion present = $\text{Fe}^{3+}$(aq) — All three observations are consistent with Fe3+: rust-brown Fe(OH)3 precipitate, no dissolution in excess NaOH, and CO2 effervescence with carbonate.

Answer: The ion present is Fe3+(aq). Ionic equation for observation (iii): 2Fe3+(aq) + 3CO32–(aq) + 3H2O(l) → 2Fe(OH)3(s) + 3CO2(g).

7. Amphoteric Character of Al(OH)3 in Depth

The amphoteric behaviour of $\text{Al(OH)}_3$ deserves its own focused treatment because it is a favourite topic for 4–6 mark "explain" questions. The key is understanding that $\text{Al(OH)}_3$ can act as both a Brønsted–Lowry acid and a Brønsted–Lowry base.

When $\text{Al(OH)}_3$ reacts with a strong acid such as $\text{HCl}$, the hydroxide acts as a base — it accepts protons from $\text{H}^+$ ions, and the precipitate dissolves to give $\text{Al}^{3+}$(aq) and water. When it reacts with excess $\text{NaOH}$, the hydroxide acts as an acid — it donates a proton to $\text{OH}^-$, and the precipitate dissolves to give the tetrahydroxoaluminate ion.

A useful way to remember this: the precipitate dissolves in both directions — add acid, it dissolves; add excess alkali, it dissolves. Only aluminium hydroxide behaves this way among the four ions on the AQA specification.

8. Summary Comparison of All Four Ions

Before moving to exam technique, it helps to see all the reactions side by side. The table below consolidates every observation you need for the four ions across all three bases.

Two patterns stand out from this table. First, only $\text{Al}^{3+}$ gives a precipitate that dissolves in excess NaOH. Second, only $\text{Cu}^{2+}$ gives a precipitate that dissolves in excess $\text{NH}_3$. These two contrasts are the most frequently tested distinctions in this topic.

9. Exam Technique and Common Mistakes

This topic rewards precision in language and equation writing. The observations are straightforward to learn, but marks are frequently lost through vague descriptions, unbalanced equations, and missing state symbols. The guidance below targets the most common errors seen in AQA mark schemes.