AQA A-Level Chemistry 7405 · 3.2.4 Properties of Period 3 elements and their oxides

Part 2: Oxides in water and the structures of their acids

All 3 parts available · worked answers and exam guidance included. Reviewed 2 October 2026.

Predict oxide–water reactions and describe phosphoric, sulfurous and sulfuric acids and their anions.

Separate solubility, reaction and acidity

An oxide can be acidic without dissolving or reacting appreciably in water. Test the solution after mixing, and distinguish an unchanged suspension from a true dissolved product. These reactions use sufficient water; concentrated acid mixtures and different temperatures can behave differently.

Adding the specified oxides to water
OxideOutcomeSolution behaviour
Na₂OReacts to form dissolved NaOHStrongly alkaline; pH above 7
MgOSlow reaction to form sparingly soluble Mg(OH)₂Weakly alkaline suspension/solution, commonly around pH 9–10
Al₂O₃No appreciable reaction with waterWater approximately unchanged
SiO₂No appreciable reaction with waterWater approximately unchanged
P₄O₁₀Reacts strongly with sufficient water to form H₃PO₄Acidic; phosphoric acid is weak
SO₂Dissolves; establishes acid–base equilibria in waterAcidic; conventionally described using H₂SO₃
SO₃Reacts vigorously to form H₂SO₄Acidic; strongly acidic when appreciably concentrated

Write the species made by water

NaOH dissociates into Na⁺ and OH⁻. Magnesium hydroxide produces only a small dissolved concentration of OH⁻ because it is sparingly soluble. For sulfur dioxide, the H₂SO₃ equation is the conventional A-level representation of aqueous sulfurous acid; much dissolved sulfur dioxide remains as hydrated/dissolved SO₂. It is not converted to H₂SO₄ simply by adding water.

Na₂O(s) + H₂O(l) → 2NaOH(aq)
MgO(s) + H₂O(l) → Mg(OH)₂(s)
P₄O₁₀(s) + 6H₂O(l) → 4H₃PO₄(aq)
SO₂(g) + H₂O(l) ⇌ H₂SO₃(aq)
SO₃(g) + H₂O(l) → H₂SO₄(aq)

Why pH is not a label permanently attached to an oxide

A numerical pH depends on concentration, temperature, extent of reaction and solubility. Do not memorise “P₄O₁₀ always gives pH 0” or compare two acids at unspecified concentrations as though strength alone fixed pH. A strong acid ionises extensively; a concentrated acid contains many moles per unit volume. These are different descriptions.

Original worked example: 0.310 g Na₂O reacts completely and the solution is made up to 250.0 cm³. With Mᵣ(Na₂O) = 62.0, n(Na₂O) = 0.00500 mol, so n(OH⁻) = 0.0100 mol. [OH⁻] = 0.0400 mol dm⁻³. At 298 K, using Kw = 1.00 × 10⁻¹⁴, [H⁺] = 2.50 × 10⁻¹³ mol dm⁻³ and pH = 12.60. The volume and temperature are essential to the answer.

All the acidic hydrogens are bonded to oxygen

In the usual displayed structures, H₃PO₄ is O=P(OH)₃; H₂SO₃ is O=S(OH)₂ with a lone pair on sulfur; H₂SO₄ is O=S(=O)(OH)₂. Removing H⁺ from an O–H group leaves a negatively charged oxygen in a localised drawing. Never remove a whole OH group when forming the conjugate base. The O atoms around phosphorus in H₃PO₄ and around sulfur in H₂SO₄ are approximately tetrahedral; H₂SO₃ has a trigonal-pyramidal arrangement around sulfur because of its lone pair.

Phosphoric acid loses protons successively to H₂PO₄⁻, HPO₄²⁻ and PO₄³⁻. Sulfurous acid gives HSO₃⁻ then SO₃²⁻; sulfuric acid gives HSO₄⁻ then SO₄²⁻. Successive dissociations are not equally extensive. Resonance means the full anion is not best described by a unique, permanently localised double bond.

Fully deprotonated anions
AnionChargeArrangement around central atom
Phosphate, PO₄³⁻3−Four O atoms: tetrahedral, about 109.5°
Sulfite, SO₃²⁻2−Three O atoms and a sulfur lone pair: trigonal pyramidal
Sulfate, SO₄²⁻2−Four O atoms: tetrahedral, about 109.5°

Diagram placeholder

Acid and oxyanion displayed structures to add

Labels to include:

  • H₃PO₄: one P=O and three P–O–H groups
  • H₂SO₃: one S=O, two S–O–H groups and sulfur lone pair
  • H₂SO₄: two S=O and two S–O–H groups
  • PO₄³⁻: one P=O and three P–O⁻ in one resonance contributor
  • SO₃²⁻: one S=O, two S–O⁻ and sulfur lone pair
  • SO₄²⁻: two S=O and two S–O⁻ in one resonance contributor
  • Brackets and total anion charges

Pair each acid with its fully deprotonated anion; put H on O, label the bonds and localised oxygen charges, and show the central geometry with appropriate wedges. Indicate equivalent resonance contributors instead of implying one oxygen is permanently special.

Use charge conservation for partial dissociation

Each stage loses one H⁺ and lowers the remaining species charge by one unit. The first stage of phosphoric acid is a weak-acid equilibrium; a later neutralisation question may instead drive several stages to completion with a base.

H₃PO₄(aq) ⇌ H⁺(aq) + H₂PO₄⁻(aq)
H₂PO₄⁻(aq) ⇌ H⁺(aq) + HPO₄²⁻(aq)
HPO₄²⁻(aq) ⇌ H⁺(aq) + PO₄³⁻(aq)

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official AQA mark allocations.

Q1. Give the equation for P₄O₁₀ reacting with sufficient water.Show answer

P₄O₁₀ + 6H₂O → 4H₃PO₄. This makes phosphoric(V) acid, not an oxide containing four independent phosphorus atoms in the product.

Q2. Why can SiO₂ leave the water near pH 7 and still be an acidic oxide?Show answer

It does not react appreciably with water. Its acidic classification follows from reactions with bases, such as hot concentrated NaOH, rather than requiring an acidic aqueous solution.

Q3. Draw the localised sulfite structure in words and identify its shape.Show answer

Place sulfur centrally with one S=O, two S–O⁻ bonds and one lone pair on sulfur. Enclose the ion in brackets with overall 2− charge. The three oxygen positions are trigonal pyramidal; equivalent resonance forms can put the double bond on another oxygen.

Q4. A solution contains 0.00200 mol Na₂O fully reacted in a final volume of 0.500 dm³. Calculate its pH at 298 K; Kw = 1.00 × 10⁻¹⁴.Show answer

n(OH⁻) = 2 × 0.00200 = 0.00400 mol. [OH⁻] = 0.00800 mol dm⁻³. [H⁺] = 1.25 × 10⁻¹² mol dm⁻³, so pH = 11.90.

Q5. What changes when HSO₄⁻ loses its remaining acidic proton?Show answer

An O–H bond loses H⁺ and the ion becomes SO₄²⁻. The oxygen is retained. HSO₄⁻ ⇌ H⁺ + SO₄²⁻ has −1 total charge on both sides.

Sources

Sources and examiner guidance (reviewed 2 October 2026)

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