Reactions of ions in aqueous solution

In a nutshell

When a salt of iron, copper or aluminium dissolves in water, each metal ion is surrounded by six water molecules to form a metal-aqua ion, such as [Fe(H₂O)₆]³⁺.

The water ligands can lose H⁺ ions. That one idea explains why these solutions are acidic, and it explains every precipitate and colour change you see when you add hydroxide ions, ammonia or carbonate ions. Those test-tube reactions are how you identify the metal ion (Required practical 11).

Assumed knowledge: Bonding (co-ordinate bonds), Transition metals (ligands and complexes), Acids and bases (Brønsted-Lowry acids and bases).

Core content

Metal-aqua ions in solution

You need four metal-aqua ions. Each is octahedral: six water ligands, each forming a co-ordinate bond from a lone pair on its oxygen atom to the metal ion.

Write the charge once, outside the square bracket, for the whole complex. Never put charges on the individual ligands.

Acidity of metal-aqua ions: why 3+ beats 2+

A metal-aqua ion can act as a Brønsted-Lowry acid: one of its water ligands donates an H⁺ ion to a water molecule in the solution.

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

[M(H₂O)₆]²⁺(aq) + H₂O(l) ⇌ [M(H₂O)₅(OH)]⁺(aq) + H₃O⁺(aq)

Notice that only one H⁺ is lost, and the charge on the complex drops by one. Both are weak acids (the equilibria lie to the left), but:

• solutions of [M(H₂O)₆]³⁺ are noticeably acidic;
• solutions of [M(H₂O)₆]²⁺ are only very weakly acidic.

The acidity of [M(H₂O)₆]³⁺ is greater than that of [M(H₂O)₆]²⁺. The explanation, in the order that earns credit:

1. The M³⁺ ion has a higher charge and is smaller than the M²⁺ ion, so it has a higher charge-to-size ratio (higher charge density).
2. So the M³⁺ ion polarises the co-ordinated water molecules more strongly: it pulls electron density from the oxygen towards the metal.
3. This weakens the O-H bonds in the water ligands.
4. So H⁺ ions are released more easily: the equilibrium lies further to the right, [H₃O⁺] is higher and the pH is lower.

Iron is the cleanest comparison because it is the same metal: Fe³⁺ versus Fe²⁺.

Still don't get it? · why a 3+ ion makes the water more acidic

Picture a magnet holding a paperclip, and the paperclip is holding a small bead loosely on its far end. A stronger magnet pulls the paperclip harder towards itself, and the bead at the far end is left hanging on less firmly, so it drops off more easily.

Now rebuild it step by step:

• The metal ion is the magnet. A 3+ ion has more positive charge than a 2+ ion, and it is also smaller, so its charge is packed into less space. That is what "higher charge-to-size ratio" means.
• The water ligand is the paperclip. Its oxygen is bonded to the metal, and the metal pulls the electrons in that oxygen towards itself.
• The oxygen then pulls on the electrons in its own O-H bonds to make up for it. Those bonds get weaker, and the H end becomes more positive.
• The H⁺ is the bead. With a weaker O-H bond it is lost more easily to a water molecule in the solution, making H₃O⁺.
• More H₃O⁺ means a more acidic solution.

The exam version, in one chain: higher charge-to-size ratio, so the ion polarises the water ligands more, so the O-H bond is weakened, so H⁺ is released more easily. Write all four links; stopping at "higher charge density" leaves the explanation unfinished.

What a base does to a metal-aqua ion

Every reaction below is the same process: the base removes H⁺ ions from the water ligands, one at a time. Here OH^- and NH₃ act as Brønsted-Lowry bases (proton acceptors).

For example, with aluminium:

[Al(H₂O)₆]³⁺ + OH^- → [Al(H₂O)₅(OH)]²⁺ + H₂O

[Al(H₂O)₅(OH)]²⁺ + OH^- → [Al(H₂O)₄(OH)₂]⁺ + H₂O

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

The key point: once the complex has lost as many H⁺ as its original charge (two for M²⁺, three for M³⁺), it is uncharged, and the uncharged metal hydroxide is insoluble, so it forms a precipitate.

Reactions with hydroxide ions (sodium hydroxide solution)

A few drops of NaOH(aq) give the hydroxide precipitate. Adding excess NaOH(aq) changes nothing further, except for aluminium.

The equations:

• [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)

Iron(II) hydroxide darkens on standing. The green precipitate slowly turns brown at the surface because it is oxidised by oxygen in the air to iron(III) hydroxide:

4Fe(H₂O)₄(OH)₂(s) + O₂(g) → 4Fe(H₂O)₃(OH)₃(s) + 2H₂O(l)

Amphoteric aluminium hydroxide

Some metal hydroxides are amphoteric: they dissolve in both acids and bases. Aluminium hydroxide is the one you need.

With excess base, the precipitate loses another H⁺ and becomes a negative ion, which is soluble:

Al(H₂O)₃(OH)₃(s) + OH^-(aq) → [Al(OH)₄]^-(aq) + 3H₂O(l)

(The six-co-ordinate version, giving [Al(H₂O)₂(OH)₄]^- + H₂O, is equally correct.)

With acid, the precipitate gains H⁺ back and returns to the aqua ion:

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

So aluminium hydroxide acts as an acid (with OH^-) and as a base (with H⁺). Iron(II), copper(II) and iron(III) hydroxides do not dissolve in excess NaOH(aq), which is how you tell Al³⁺ apart.

Reactions with ammonia solution

A little ammonia behaves exactly like hydroxide: NH₃ is a Brønsted-Lowry base and takes H⁺ from the water ligands, so you get the same hydroxide precipitates (green, blue, brown, white).

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

You can also write this in two steps: NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH^-(aq), then the hydroxide equation. If you use OH^- in an ammonia question, you must show that first step.

Excess ammonia changes only one of the four: copper. The blue precipitate dissolves to give a deep blue solution. Now NH₃ is acting as a ligand (a Lewis base, an electron pair donor) and replaces four of the water ligands. The substitution is incomplete: two water ligands stay.

Cu(H₂O)₄(OH)₂(s) + 4NH₃(aq) → [Cu(NH₃)₄(H₂O)₂]²⁺(aq) + 2H₂O(l) + 2OH^-(aq)

Starting from the aqua ion, the overall change is:

[Cu(H₂O)₆]²⁺(aq) + 4NH₃(aq) → [Cu(NH₃)₄(H₂O)₂]²⁺(aq) + 4H₂O(l)

The iron(II), iron(III) and aluminium hydroxides do not dissolve in excess ammonia. In particular, aluminium hydroxide does not dissolve in excess ammonia, even though it dissolves in excess NaOH.

Reactions with carbonate ions (sodium carbonate solution)

Here the 2+ and 3+ ions behave differently, and this is the most examined contrast in the topic.

M²⁺ ions form the metal carbonate (a simple precipitation):

• [Fe(H₂O)₆]²⁺(aq) + CO₃^2-(aq) → FeCO₃(s) + 6H₂O(l)
• [Cu(H₂O)₆]²⁺(aq) + CO₃^2-(aq) → CuCO₃(s) + 6H₂O(l)

M³⁺ ions form the hydroxide and carbon dioxide (an acid-base reaction):

• 2[Fe(H₂O)₆]³⁺(aq) + 3CO₃^2-(aq) → 2Fe(H₂O)₃(OH)₃(s) + 3CO₂(g) + 3H₂O(l)
• 2[Al(H₂O)₆]³⁺(aq) + 3CO₃^2-(aq) → 2Al(H₂O)₃(OH)₃(s) + 3CO₂(g) + 3H₂O(l)

Why the difference:

• [M(H₂O)₆]³⁺ is acidic enough (more H₃O⁺ in solution) that the carbonate ions react with the H⁺ to give CO₂ and water. Removing H⁺ pulls the aqua ion all the way to the neutral hydroxide, which precipitates. No M₂(CO₃)₃ forms.
• [M(H₂O)₆]²⁺ is too weakly acidic to do this, so the carbonate ions simply combine with the metal ions to precipitate MCO₃.

Still don't get it? · why 3+ ions fizz with carbonate but 2+ ions do not

Think of carbonate as a shopper looking for a bargain. In the 3+ shop, H⁺ ions are on display everywhere, so carbonate grabs those. In the 2+ shop, there are hardly any H⁺ ions about, so carbonate takes the only thing on offer: the metal ion itself.

Step by step:

• From the acidity section, a 3+ aqua ion gives off enough H⁺ to make the solution noticeably acidic. A 2+ aqua ion gives off almost none.
• Carbonate ions react with H⁺: two H⁺ per carbonate gives H₂O and CO₂. That CO₂ is the fizzing.
• Every H⁺ removed lets the 3+ aqua ion lose another one, until it has lost three and is the neutral hydroxide, which is insoluble. So you see a hydroxide precipitate plus bubbles.
• With a 2+ ion there is not enough H⁺ for this, so carbonate ions just pair up with the M²⁺ ions, and the metal carbonate (insoluble) precipitates. No gas.

Exam wording: the 3+ aqua ion is more acidic (higher charge-to-size ratio), so carbonate reacts with H⁺ to give CO₂ (effervescence) and the hydroxide precipitates; the 2+ aqua ion is too weakly acidic, so the metal carbonate precipitates. For the observation write "effervescence" or "bubbles of gas", not just "CO₂".

Required practical 11: test-tube reactions to identify metal ions

Aim: identify the metal ion in an unknown solution from its colour and its reactions with three bases.

Method:

1. Note the colour of the original solution.
2. Put about 1 cm³ of the solution in a test tube and add sodium hydroxide solution dropwise, shaking after each drop. Record the colour and state (precipitate or solution).
3. Keep adding NaOH(aq) until it is in excess, and record any change (does the precipitate dissolve?).
4. Repeat steps 2 and 3 on a fresh sample using ammonia solution.
5. Repeat on a fresh sample using sodium carbonate solution, watching for effervescence.
6. Leave any green precipitate to stand in air and look for it turning brown at the surface.

Wear eye protection: sodium hydroxide and sodium carbonate solutions are irritants to the eyes.

Use the results like this:

An observation is only complete if it names the colour and the state together: "brown precipitate", "deep blue solution", never just "brown" or "deep blue".

Worked examples

Building the equation for iron(III) ions with carbonate ions

This is the equation examiners most often see unbalanced. Build it from what each species does.

Step 1, write the species you start and finish with:

$\text{[Fe(H}_2\text{O)}_6]^{3+} + \text{CO}_3^{2-} \rightarrow \text{Fe(H}_2\text{O)}_3\text{(OH)}_3 + \text{CO}_2 + \text{H}_2\text{O}$

Step 2, count H⁺ transferred. Each aqua ion gives 3 H⁺ to become neutral. Each carbonate takes 2 H⁺ to become CO₂ + H₂O:

$\text{per Fe: } 3\,\text{H}^+ \text{ given} \qquad \text{per CO}_3^{2-}\text{: } 2\,\text{H}^+ \text{ taken}$

Step 3, match the H⁺ (lowest common multiple 6): 2 aqua ions give 6 H⁺, and 3 carbonate ions take 6 H⁺:

$2\,\text{[Fe(H}_2\text{O)}_6]^{3+} + 3\,\text{CO}_3^{2-} \rightarrow 2\,\text{Fe(H}_2\text{O)}_3\text{(OH)}_3 + 3\,\text{CO}_2 + 3\,\text{H}_2\text{O}$

Step 4, check atoms and charge:

$\text{Fe: } 2 = 2 \quad \text{C: } 3 = 3 \quad \text{O: } 12 + 9 = 21 = 12 + 6 + 3 \quad \text{H: } 24 = 18 + 6$

$\text{charge: } 2(+3) + 3(-2) = 0 = \text{right-hand side}$

The aluminium equation is identical with Al in place of Fe. Where students lose credit: writing Fe₂(CO₃)₃ as the product, using 1 : 1 ratios so the charge does not balance, and leaving the H₂O off the right-hand side.

Model answer: explain why [Fe(H₂O)₆]³⁺ is more acidic than [Fe(H₂O)₆]²⁺

A full answer gives the equation and then the linked chain, in this order:

1. [Fe(H₂O)₆]³⁺ + H₂O ⇌ [Fe(H₂O)₅(OH)]²⁺ + H₃O⁺ (one H⁺ released).
2. Fe³⁺ has a higher charge-to-size ratio than Fe²⁺ (higher charge, smaller ion).
3. So Fe³⁺ polarises the co-ordinated water more, which weakens the O-H bond, so H⁺ is released more easily.

The word "higher" is what makes point 2 a comparison. Point 3 is the one examiners report students most often miss.

Identifying an unknown solution

Solution X is colourless. With NaOH(aq) it gives a white precipitate that dissolves in excess NaOH(aq) to a colourless solution. With Na₂CO₃(aq) it gives a white precipitate and bubbles of gas.

1. Colourless solution and white precipitate: rules out Fe²⁺, Cu²⁺ and Fe³⁺.
2. Precipitate dissolves in excess NaOH: an amphoteric hydroxide, so aluminium.
3. Effervescence with carbonate: a 3+ aqua ion. This confirms it.

X contains [Al(H₂O)₆]³⁺. The white precipitate is Al(H₂O)₃(OH)₃ and the colourless solution in excess NaOH contains [Al(OH)₄]^-.

Common exam mistakes

• Giving a colour without a state. "Deep blue" or "brown" alone is not an observation; write "deep blue solution", "brown precipitate". Examiners repeat this in almost every report on this topic.
• Writing [Cu(NH₃)₆]²⁺ for the deep blue complex. It is [Cu(NH₃)₄(H₂O)₂]²⁺: the substitution is incomplete. Getting the overall charge wrong also loses the species.
• Putting charges on the ligands inside a complex formula (e.g. writing OH^- inside the bracket). Only the overall charge is written, outside the square bracket.
• Stating "CO₂" as an observation. Carbon dioxide is a deduction; the observation is effervescence / bubbles / gas given off.
• Forgetting the effervescence for Fe³⁺ or Al³⁺ with carbonate, or adding effervescence for Fe²⁺ or Cu²⁺ with carbonate. Only the 3+ ions fizz; the 2+ ions just form MCO₃.
• Giving only one stage for aluminium with NaOH. The answer needs the white precipitate first, then the colourless solution in excess, with an equation for each.
• Starting an equation from the wrong species. If the question gives you the hydroxide precipitate (e.g. reacting it with excess ammonia), the equation must start from Cu(H₂O)₄(OH)₂, not from [Cu(H₂O)₆]²⁺.
• Writing the acidity equation as a full precipitation. The acidity (hydrolysis) equation releases one H⁺ and gives [M(H₂O)₅(OH)]²⁺; it does not make M(H₂O)₃(OH)₃.
• Stopping the acidity explanation at "higher charge density". You must go on to polarises the water / weakens the O-H bond / releases H⁺ more easily.
• Treating hexaaquairon(II) as a strong acid. Both aqua ions are weak acids; the 2+ ion is only very weakly acidic.
• Saying iron(II) hydroxide turns brown because "it reacts with air". Name the cause: it is oxidised by oxygen in the air to iron(III) hydroxide.
• Using an ion as a reagent. A reagent must be a compound: "sodium carbonate solution" or Na₂CO₃, not "CO₃^2-". NaCO₃ is a wrong formula.
• Describing aluminium hydroxide as grey or iron(II) hydroxide as blue-green. They are white and green.
• Confusing the two roles of ammonia. A little NH₃ acts as a Brønsted-Lowry base (proton acceptor) and gives the precipitate; excess NH₃ with copper acts as a ligand (electron pair donor). Swapping them loses the explanation.

Key definitions

• Metal-aqua ion: a complex ion in which water molecules act as ligands, each forming a co-ordinate bond to the central metal ion (e.g. [Fe(H₂O)₆]³⁺).
• Ligand: a molecule or ion that forms a co-ordinate bond with a (transition) metal by donating a pair of electrons.
• Brønsted-Lowry acid: a proton (H⁺) donor.
• Brønsted-Lowry base: a proton (H⁺) acceptor.
• Lewis base: an electron pair donor (a lone pair donor).
• Amphoteric: able to react with (dissolve in) both acids and bases.
• Why [M(H₂O)₆]³⁺ is more acidic than [M(H₂O)₆]²⁺: the M³⁺ ion has a higher charge-to-size ratio, so it polarises the co-ordinated water molecules more, weakening the O-H bonds, so H⁺ ions are released more easily.

Specification

• I can name and write the formulae of [Fe(H₂O)₆]²⁺, [Cu(H₂O)₆]²⁺, [Fe(H₂O)₆]³⁺ and [Al(H₂O)₆]³⁺ and state the colour of each solution.
• I can write the equation for a metal-aqua ion acting as an acid (losing one H⁺ to water).
• I can state that [M(H₂O)₆]³⁺ is more acidic than [M(H₂O)₆]²⁺.
• I can explain this difference in terms of the charge-to-size ratio of the metal ion, polarisation of the water ligands and weakening of the O-H bond.
• I can explain what amphoteric means and write equations for aluminium hydroxide dissolving in both an acid and excess hydroxide.
• I can describe (colour and state) and write equations for the reactions of Fe²⁺, Cu²⁺, Fe³⁺ and Al³⁺ aqua ions with a little and with excess sodium hydroxide.
• I can describe and write equations for the reactions of the four ions with a little and with excess ammonia, including the deep blue [Cu(NH₃)₄(H₂O)₂]²⁺.
• I can describe and write balanced equations for the reactions of the four ions with carbonate ions, and explain why the 2+ ions give MCO₃ but the 3+ ions give the hydroxide and CO₂.
• I can explain why iron(II) hydroxide turns brown on standing in air.
• I can carry out Required practical 11 and use the observations to identify an unknown metal ion.

Related notes

• Bonding
• Transition metals (3.2.5)
• Acids and bases (3.1.12)
• Group 2, the alkaline earth metals (Required practical 4, tests for other ions)
• Group 7(17), the halogens (tests for halide and other anions)
• Properties of Period 3 elements and their oxides (3.2.4)