Explain why sodium oxide dissolves in water to form an alkaline solution, whilst sulfur trioxide dissolves in water to form an acidic solution.

AQA A-Level Chemistry (7405) — 3.2.4 Properties of Period 3 elements and their oxides (A-Level only) · Explain · 5 marks · View as Markdown

Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).

Model answer (5 marks)

Sodium oxide is a metal oxide; it reacts with water to give
NaOH:
Na₂O + H₂O → 2 NaOH.
NaOH is a strong base and dissociates completely to give Na⁺ and OH⁻ ions, so the solution is alkaline.

Sulphur trioxide is a non‑metal oxide; it reacts with water to give
SO₃ + H₂O → H₂SO₄.
Sulphuric acid is a strong acid and releases H⁺ ions, making the solution acidic.

Examiner tips

  • Show the two reactions explicitly; include the ionisation step for NaOH.
  • Use the terms ‘metal oxide’ and ‘non‑metal oxide’ to justify the difference.
  • Mention that NaOH is a strong base and H₂SO₄ a strong acid to explain the pH.

Mark scheme (5 marks)

  1. Sodium oxide is a metal oxide / ionic oxide
  2. Sodium oxide reacts with water to produce sodium hydroxide (NaOH)
  3. Sodium hydroxide dissociates / ionises to release OH⁻ ions, making the solution alkaline
  4. Sulfur trioxide is a non-metal oxide / covalent oxide
  5. Sulfur trioxide reacts with water to produce sulfuric acid (H₂SO₄), which releases H⁺ ions making the solution acidic

Key terms in this question

alkaline · acidic

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