O Levels Chemistry (5070)•5070/12/M/J/23
Explanation
NaOH tests distinguish metal ions by precipitation and ammonium by gas evolution
Steps:
- Q's coloured solution forms a precipitate with NaOH drops, indicating Cu²⁺ as the ion producing insoluble Cu(OH)₂.
- R's colourless solution shows no precipitate with NaOH but evolves gas on heating, indicating NH₄⁺ releasing NH₃ via NH₄⁺ + OH⁻ → NH₃ + H₂O.
- Options B and C lack NH₄⁺ for R, so they fail the gas test.
- Option D has NH₄⁺ for R but Fe²⁺ for Q, which mismatches the specific ion behaviour.
Why A is correct:
- Cu²⁺ forms a characteristic precipitate with NaOH from its coloured (blue) solution, while NH₄⁺ confirms via ammonia gas evolution on heating, per standard qualitative analysis.
Why the others are wrong:
- B: CO₃²⁻ produces no gas with NaOH on heating; requires acid for CO₂.
- C: NO₃⁻ shows no reaction or gas with NaOH; needs reduction test.
- D: Fe²⁺ forms a green precipitate, not matching Q's described behaviour.
Final answer: A
Topic: Identification of ions and gases
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