Tests for cations (Edexcel GCSE Chemistry): Revision Notes

Cations are positively charged ions. They form when atoms lose electrons.

Purpose: The flame test is used to identify certain metal cations based on the colour they produce when heated in a flame. Method:

1. Clean a nichrome or platinum wire by dipping it in hydrochloric acid and then holding it in a blue Bunsen burner flame until no colour is seen.
2. Dip the clean wire into a sample of the cation compound.
3. Hold the wire in the edge of a Bunsen burner flame and observe the colour. Observations:

• Lithium (Li⁺): Crimson red flame
• Sodium (Na⁺): Yellow flame
• Potassium (K⁺): Lilac flame
• Calcium (Ca²⁺): Orange-red flame
• Copper (Cu²⁺): Blue-green flame Sodium's yellow flame is very intense and can mask other colours. Always observe other colours first if sodium is present.

Purpose: When sodium hydroxide is added to solutions of metal cations, precipitates (solid particles) may form, allowing for the identification of specific cations. Method:

4. Add a few drops of sodium hydroxide solution (NaOH) to a solution of the cation.
5. Observe the colour of any precipitate formed.
6. Add excess sodium hydroxide, if needed, to see if the precipitate dissolves or remains. Observations and Results:

Calcium (Ca²⁺): White precipitate (insoluble in excess NaOH)

• $Ca²⁺ (aq) + 2OH⁻ (aq) → Ca(OH)₂ (s)$

Magnesium (Mg²⁺): White precipitate (insoluble in excess NaOH)

• $Mg²⁺ (aq) + 2OH⁻ (aq) → Mg(OH)₂ (s)$

Aluminium (Al³⁺): White precipitate (soluble in excess NaOH, forming a colourless solution)

• $Al³⁺ (aq) + 3OH⁻ (aq) → Al(OH)₃ (s)$
• $Al(OH)₃ (s) + OH⁻ (aq) → [Al(OH)₄]⁻ (aq)$

Copper (Cu²⁺): Blue precipitate

• $Cu²⁺ (aq) + 2OH⁻ (aq) → Cu(OH)₂ (s)$

Iron (II) (Fe²⁺): Green precipitate (turns brown on standing)

• $Fe²⁺ (aq) + 2OH⁻ (aq) → Fe(OH)₂ (s)$

• $Fe(OH)₂$ oxidises to Fe(OH)₃ in air, giving a brown colour. Iron (III) (Fe³⁺): Brown precipitate

• $Fe³⁺ (aq) + 3OH⁻ (aq) → Fe(OH)₃ (s)$

Zinc (Zn²⁺): White precipitate (soluble in excess NaOH, forming a colourless solution)

• $Zn²⁺ (aq) + 2OH⁻ (aq) → Zn(OH)₂ (s)$
• $Zn(OH)₂ (s) + 2OH⁻ (aq) → [Zn(OH)₄]²⁻ (aq)$