Qualitative analysis of ions

In a nutshell

Qualitative analysis at A level means identifying both ions in an unknown compound from test-tube observations: the anion from the AS tests, and the cation from the AS ammonium test or from the coloured precipitates that transition metal ions give with NaOH(aq) and NH₃(aq).

The five transition metal ions you must identify are Cu²⁺, Fe²⁺, Fe³⁺, Mn²⁺ and Cr³⁺. For each you need the colours, what happens with a few drops and with excess of each reagent, and the ionic equations.

Assumed knowledge: Qualitative analysis (AS); Transition elements.

Core content

Tests for anions: carbonate, sulfate and halide

These are the AS tests from 3.1.4, unchanged. You must still know them, because an A level unknown has an anion as well as a cation.

The order still matters: carbonate, then sulfate, then halide. BaCO₃ and Ag₂SO₄ are both insoluble, so an ion left over from an earlier test gives a false positive in a later one.

• Use dilute nitric acid: HCl adds Cl^- and H₂SO₄ adds SO₄^2-.
• Use barium nitrate, not barium chloride, when a halide test follows.
• If sulfate is present, add excess Ba(NO₃)₂(aq) and filter before adding AgNO₃(aq).

Test for the ammonium cation

Add NaOH(aq) and warm gently. Ammonium ions release ammonia gas, which turns moist red litmus paper blue.

NH₄⁺(aq) + OH^-(aq) → NH₃(g) + H₂O(l)

• Warm, because ammonia is very soluble and stays dissolved in a cold solution.
• Moist litmus, because the gas must dissolve on the paper to make it alkaline.
• Hold the paper at the mouth of the tube, not touching the glass (the NaOH(aq) itself would turn it blue).

The same NaOH(aq) you use for the transition metal tests can do this job too. If a precipitate forms and warming then gives a gas that turns moist red litmus blue, the compound contains both a transition metal ion and NH₄⁺.

Tests for transition metal cations: the results table

Add the reagent dropwise to a solution of the unknown, then keep adding until it is in excess. Do one test with NaOH(aq) and a separate test, on a fresh sample, with NH₃(aq).

The precipitate colour with a few drops identifies most ions. The excess step separates the ones that look alike.

Two patterns carry most of the marks:

• With excess NaOH(aq), only Cr(OH)₃ reacts further.
• With excess NH₃(aq), only Cu(OH)₂ and Cr(OH)₃ react further.

Every other precipitate stays as it is in excess.

Two precipitates also change on standing in air, because oxygen oxidises them:

• Fe(OH)₂: the green precipitate slowly turns orange-brown at the surface, as iron(II) is oxidised to iron(III).
• Mn(OH)₂: the light brown precipitate darkens.

Report the colour you see straight after mixing; that is the identifying colour.

Precipitation: the equations with a few drops of reagent

With a few drops of either reagent, the metal ion forms its hydroxide, which is insoluble, so it precipitates.

The OCR spec allows either form of the formulae for these precipitation equations: simple ions, Cu²⁺(aq) and Cu(OH)₂(s), or complexed formulae, [Cu(H₂O)₆]²⁺(aq) and Cu(OH)₂(H₂O)₄(s).

The number of OH^- ions always equals the charge on the metal ion: 2 for a 2+ ion, 3 for a 3+ ion. The hydroxide is then neutral.

Why aqueous ammonia gives the same precipitates. Ammonia is a weak base, so its solution contains some OH^- ions:

NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH^-(aq)

Those OH^- ions precipitate the metal hydroxide exactly as NaOH(aq) does. Written with ammonia as the reactant, for example:

Cu²⁺(aq) + 2NH₃(aq) + 2H₂O(l) → Cu(OH)₂(s) + 2NH₄⁺(aq)

Complex formation: the equations with excess reagent

In excess, the precipitate only dissolves if it can form a soluble complex ion. For these five ions that happens in just three cases.

Excess NaOH(aq): chromium only. Chromium(III) hydroxide takes up more hydroxide ions and forms a charged, soluble complex:

Cr(OH)₃(s) + 3OH^-(aq) → [Cr(OH)₆]^3-(aq)

Observation: the grey-green precipitate dissolves to give a dark green solution.

Excess NH₃(aq): copper and chromium only. Ammonia molecules act as ligands, replacing the ligands on the metal:

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

Observation: the pale blue precipitate dissolves to give a deep blue solution.

Cr(OH)₃(s) + 6NH₃(aq) → [Cr(NH₃)₆]³⁺(aq) + 3OH^-(aq)

Observation: the grey-green precipitate dissolves to give a purple solution.

For complexes like these, use complexed formulae with square brackets and the overall charge outside. Note the copper complex keeps two water ligands: it is [Cu(NH₃)₄(H₂O)₂]²⁺, not [Cu(NH₃)₆]²⁺.

Written as ligand substitution from the aqua ions (5.3.1), the same two products are:

• [Cu(H₂O)₆]²⁺(aq) + 4NH₃(aq) → [Cu(NH₃)₄(H₂O)₂]²⁺(aq) + 4H₂O(l)
• [Cr(H₂O)₆]³⁺(aq) + 6NH₃(aq) → [Cr(NH₃)₆]³⁺(aq) + 6H₂O(l)

Still don't get it? · why "a few drops" and "excess" give different results

Imagine a crowd of people each holding a helium balloon, floating along. Hand each person a small weight and they all sink to the floor in a heap. That heap is easy to see. Now keep handing out more of the same thing. For most people nothing more happens: they stay on the floor. But a few of them can use the extra items to build something that lifts them back up, and they float away again.

Now rebuild it step by step:

1. In solution, each metal ion is surrounded by water molecules and carries a positive charge. Charged particles stay spread out in water, so the solution is clear (though coloured).
2. A few drops of NaOH(aq) or NH₃(aq) supply OH^- ions. Enough OH^- join each metal ion to cancel its charge (2 for Cu²⁺, 3 for Fe³⁺). A neutral metal hydroxide does not stay dissolved, so it comes out as a solid: a precipitate. Every one of the five ions does this, which is why both reagents give the same first result.
3. Now add excess. Only a precipitate that can turn back into a charged complex ion will dissolve again.
4. With excess OH^-, only chromium manages it: Cr(OH)₃ picks up three more OH^- to become [Cr(OH)₆]^3-, which is charged and dissolves.
5. With excess NH₃, the ammonia molecules themselves become ligands. Copper and chromium form ammine complexes, [Cu(NH₃)₄(H₂O)₂]²⁺ and [Cr(NH₃)₆]³⁺, which are charged and dissolve. The other three precipitates stay put.

In the exam, give two observations for an excess step: the precipitate dissolves, and the colour of the solution formed (dark green, deep blue or purple). For "no reaction in excess", write "no change" or "precipitate does not dissolve".

Identifying an unknown compound

To identify a whole compound, find the cation and the anion separately, each on a fresh sample of the solution.

1. Look first. Note the colour of the solid and its solution. A coloured solution suggests a transition metal ion, but colour alone is not a test.
2. Cation: add NaOH(aq) dropwise, then in excess. Note the precipitate colour and whether it dissolves. Warm and test for ammonia with moist red litmus, in case NH₄⁺ is present.
3. Confirm the cation where needed with NH₃(aq), a few drops then excess (this separates Cu²⁺ and Cr³⁺ from the rest).
4. Anion: on a fresh sample, test in the order carbonate, sulfate, halide.
5. Combine the ions into a formula with the charges balanced (e.g. Fe³⁺ and Cl^- give FeCl₃).

Some of these ions are easy to mix up, so know how to separate them:

PAG4: qualitative analysis of ions

OCR links this subtopic to PAG4, qualitative analysis of ions on a test-tube scale. The skills are using the apparatus for the tests and making and recording qualitative observations.

Good practice that turns into marks:

• Use small volumes and add reagents dropwise, so you see the first precipitate before it can redissolve in excess.
• Then add the reagent in excess (several cm³, with shaking) before deciding "no change".
• Record a test, observation, inference table. An observation is colour + state: "orange-brown precipitate", "deep blue solution". If nothing happens, write "no change" or "no visible change".
• Record precipitate colours immediately, before Fe(OH)₂ or Mn(OH)₂ darken in air.
• Use a fresh sample for each separate test.
• Concentrated ammonia has a choking vapour: use it in a fume cupboard. NaOH(aq) is corrosive to skin and eyes, so wear eye protection.

Worked examples

Example 1: identify an unknown from its results

A pale green solid, W, dissolves in water to give a pale green solution. A student records:

Model answer, one inference per observation:

1. Green precipitate with NaOH(aq), not dissolving in excess, so Fe²⁺: Fe²⁺(aq) + 2OH^-(aq) → Fe(OH)₂(s)
2. Alkaline gas on warming with NaOH(aq), so NH₄⁺: NH₄⁺(aq) + OH^-(aq) → NH₃(g) + H₂O(l)
3. No effervescence with acid, so no carbonate.
4. White precipitate with Ba²⁺, so SO₄^2-: Ba²⁺(aq) + SO₄^2-(aq) → BaSO₄(s)
5. No precipitate with Ag⁺, so no halide.
6. W contains NH₄⁺, Fe²⁺ and SO₄^2-. Balancing the charges, two NH₄⁺ (2+) and one Fe²⁺ (2+) need two SO₄^2- (4-): (NH₄)₂Fe(SO₄)₂, ammonium iron(II) sulfate.

Where marks go: each ion must be linked to the observation that shows it. Test 6 used the filtrate after excess Ba(NO₃)₂(aq), so no sulfate was left to precipitate as Ag₂SO₄ and fake a halide result.

Example 2: tell two green solutions apart

Two unlabelled bottles each contain a green solution. One is iron(II) sulfate, the other chromium(III) sulfate. Describe how to identify each, with observations and equations.

1. Add NaOH(aq) dropwise to a sample of each. Both give a green-looking precipitate, so this step alone does not decide it:
   • Fe²⁺(aq) + 2OH^-(aq) → Fe(OH)₂(s), green precipitate
   • Cr³⁺(aq) + 3OH^-(aq) → Cr(OH)₃(s), grey-green precipitate
2. Add excess NaOH(aq) to both.
   • Chromium(III): the precipitate dissolves to give a dark green solution: Cr(OH)₃(s) + 3OH^-(aq) → [Cr(OH)₆]^3-(aq)
   • Iron(II): no change; the precipitate stays (and slowly turns orange-brown at the surface on standing).
3. Optional confirmation: excess NH₃(aq) gives a purple solution with chromium(III), [Cr(NH₃)₆]³⁺, and no change with iron(II).

Where marks go: the deciding observation is the excess step, so it must be stated for both solutions. "The chromium one dissolves" earns less than "the precipitate dissolves to form a dark green solution".

Example 3: write the equations for copper(II) with aqueous ammonia

A student adds aqueous ammonia dropwise, then in excess, to copper(II) sulfate solution. Give the observations and an ionic equation for each stage.

1. A few drops: pale blue solution gives a pale blue precipitate of copper(II) hydroxide.
   • Cu²⁺(aq) + 2OH^-(aq) → Cu(OH)₂(s)
   • or, with ammonia shown as the reactant: Cu²⁺(aq) + 2NH₃(aq) + 2H₂O(l) → Cu(OH)₂(s) + 2NH₄⁺(aq)
2. Excess: the precipitate dissolves to give a deep blue solution.
   • Cu(OH)₂(s) + 4NH₃(aq) + 2H₂O(l) → [Cu(NH₃)₄(H₂O)₂]²⁺(aq) + 2OH^-(aq)

Check the excess equation balances:

• Atoms: 1 Cu, 4 N, 4 O and 18 H on each side.
• Charge: 0 on the left; (2+) + 2 × (1-) = 0 on the right.

Where marks go: the complex needs square brackets, four NH₃ and two H₂O ligands, and the 2+ charge outside the brackets.

Common exam mistakes

• Mixing up the iron precipitates. Fe(OH)₂ is green; Fe(OH)₃ is orange-brown. Writing "brown precipitate" for iron(II) loses the mark: the brown colour only appears on standing, as it oxidises.
• Saying more precipitates dissolve in excess than really do. With excess NaOH(aq) only Cr(OH)₃ dissolves; Cu(OH)₂ does not. With excess NH₃(aq) only Cu(OH)₂ and Cr(OH)₃ dissolve.
• Wrong colour for chromium in excess ammonia. [Cr(NH₃)₆]³⁺ is a purple solution. The dark green solution is [Cr(OH)₆]^3-, from excess NaOH(aq).
• Calling a solution a precipitate. Excess ammonia with copper gives a deep blue solution, not a "deep blue precipitate". The observation needs both parts: precipitate dissolves and the colour of the solution.
• Wrong formula for the copper-ammonia complex. It is [Cu(NH₃)₄(H₂O)₂]²⁺, with two water ligands left on. [Cu(NH₃)₆]²⁺ and [Cu(NH₃)₄]²⁺ are not the formula the spec gives.
• Wrong charge on a complex. [Cr(OH)₆]^3- is 3- (Cr³⁺ with six OH^-: 3 - 6 = -3). [Cr(NH₃)₆]³⁺ is 3+ (NH₃ is neutral).
• Unbalanced precipitation equations. A 3+ ion needs three OH^-: Fe³⁺(aq) + 3OH^-(aq) → Fe(OH)₃(s), not 2OH^-. Check charge as well as atoms.
• Writing "clear" for "colourless". Every solution here is clear; give the colour (deep blue, purple, dark green), or "colourless" if it has none.
• Using solution colour as the test. A pale green solution suggests Fe²⁺ but could be Cr³⁺. The identifying evidence is the precipitate and the excess step.
• Writing "ammonium hydroxide" or calling the reagent ammonium. The reagent is aqueous ammonia, NH₃(aq); it supplies OH^- because NH₃ + H₂O ⇌ NH₄⁺ + OH^-.
• Missing a detail in the ammonium test. NaOH(aq) must be warmed, the litmus must be moist and red, and it turns blue. A precipitate with NaOH(aq) does not rule out NH₄⁺.
• Forgetting the anion order or the reagent choice. Carbonate, then sulfate, then halide; dilute HNO₃ (not HCl); Ba(NO₃)₂ (not BaCl₂) before a halide test.

Key definitions

• Qualitative analysis: identifying which ions (or substances) are present in a sample, rather than how much of each.
• Precipitate: an insoluble solid formed when two solutions react.
• Precipitation reaction: a reaction in which two aqueous ions combine to form an insoluble solid (e.g. Cu²⁺(aq) + 2OH^-(aq) → Cu(OH)₂(s)).
• Ligand: a molecule or ion that donates a pair of electrons to a central metal ion to form a coordinate (dative covalent) bond.
• Complex ion: a central metal ion surrounded by ligands bonded to it by coordinate bonds.
• Ligand substitution: a reaction in which one ligand in a complex ion is replaced by another (e.g. H₂O replaced by NH₃ in forming [Cu(NH₃)₄(H₂O)₂]²⁺).

Specification

• I can describe the tests for CO₃^2-, SO₄^2-, Cl^-, Br^- and I^-, with observations and ionic equations, and explain why they are done in the order carbonate, sulfate, halide. (5.3.2 a i; see 3.1.4 a)
• I can describe the test for NH₄⁺ by warming with NaOH(aq) to release NH₃, which turns moist red litmus blue, with the ionic equation. (5.3.2 a ii; see 3.1.4 a ii)
• I can state the colour of the precipitate that Cu²⁺, Fe²⁺, Fe³⁺, Mn²⁺ and Cr³⁺ each give with a few drops of NaOH(aq) and of NH₃(aq), and write the ionic equations, in simple or complexed formulae. (5.3.2 a ii; see 5.3.1 j i)
• I can state that with excess NaOH(aq) only Cr(OH)₃ reacts further, forming [Cr(OH)₆]^3- (dark green solution), and write the equation. (5.3.2 a ii; see 5.3.1 j ii)
• I can state that with excess NH₃(aq) only Cu(OH)₂ and Cr(OH)₃ react further, forming [Cu(NH₃)₄(H₂O)₂]²⁺ (deep blue solution) and [Cr(NH₃)₆]³⁺ (purple solution), and write the equations. (5.3.2 a ii; see 5.3.1 j ii, h)
• I can identify the cation and anion in an unknown compound from a set of test-tube results, and plan the tests to do so. (5.3.2 a)
• I can carry out the tests on a test-tube scale and record clear qualitative observations. (5.3.2 a, PAG4)

Related notes

• Qualitative analysis (AS): the anion and ammonium tests in full, including why the order matters
• The halogens: the silver nitrate test for halide ions
• Transition elements: complex ions, ligand substitution, and the colour changes behind these tests
• Redox and electrode potentials: the half-equations behind Fe²⁺ and Fe³⁺ interconversion
• Acids, bases and buffers: why aqueous ammonia is a weak base that supplies OH^-
• Chromatography and qualitative analysis: the organic functional-group tests that complete A level qualitative analysis