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O Levels Chemistry (5070)•5070/12/O/N/22
Question 3 from 5070/12/O/N/22

Explanation

Ion identification via precipitation and gas tests

Steps:

  • Green precipitate with NaOH indicates Cr³⁺, as Cr(OH)₃ is green (Fe²⁺ also possible but tested further).
  • Ammonia gas (turns litmus blue) on heating excess NaOH confirms NH₄⁺, via NH₄⁺ + OH⁻ → NH₃ + H₂O.
  • White precipitate with Ba(NO₃)₂ after acidification indicates SO₄²⁻, as BaSO₄ is insoluble white.
  • Combination matches NH₄⁺, Cr³⁺, SO₄²⁻ in green crystals (e.g., (NH₄)₂Cr₂(SO₄)₃·12H₂O).

Why A is correct:

  • Tests align with NH₄⁺ (ammonia evolution), Cr³⁺ (green Cr(OH)₃), and SO₄²⁻ (BaSO₄ formation per solubility rules).

Why the others are wrong:

  • B: Fe²⁺ gives green Fe(OH)₂, SO₄²⁻ fits, but no NH₄⁺ for ammonia gas.
  • C: Cr³⁺ fits green ppt, SO₄²⁻ fits BaSO₄, but CO₃²⁻ yields CO₂ (not ammonia) with acid/heat.
  • D: Fe²⁺ fits green ppt, SO₄²⁻ fits, but NO₃⁻ produces no ammonia gas with NaOH.

Final answer: A

Topic: Identification of ions and gases

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