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O Levels Chemistry (5070)•5070/11/O/N/20
Question 6 from 5070/11/O/N/20

Explanation

Amphoteric hydroxide formation and dissolution

Steps:

  • Adding NaOH to the solution forms a white precipitate, indicating an insoluble metal hydroxide.
  • The precipitate dissolves in excess NaOH, characteristic of amphoteric hydroxides that form soluble complex ions.
  • Aluminum ions (Al³⁺) produce Al(OH)₃, a white precipitate that dissolves as [Al(OH)₄]⁻ in excess base.
  • Other ions either do not precipitate or fail to dissolve in excess NaOH.

Why A is correct:

  • Al³⁺ + 3OH⁻ → Al(OH)₃ (white ppt), then Al(OH)₃ + OH⁻ → [Al(OH)₄]⁻ (soluble), per amphoteric metal hydroxide behavior.

Why the others are wrong:

  • B: Ca²⁺ forms Ca(OH)₂ white ppt, but it remains insoluble in excess NaOH.
  • C: Cu²⁺ forms blue Cu(OH)₂ ppt, insoluble in excess NaOH.
  • D: Na⁺ does not react with NaOH to form any precipitate.

Final answer: A

Topic: Identification of ions and gases

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