A2 · Practice questions · Enthalpies of solution and hydration

Dissolve it on paper.

Six original Cambridge-style questions on enthalpies of solution and hydration.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

Keep these straight

Lattice out, hydration in.

Hydration: gaseous ion → aqueous ion, always negative.
Solution: solid → aqueous ions, small, either sign.
ΔH_sol = ΣΔH_hyd − LE, counting every ion.
High charge density (small, highly charged): more exothermic hydration.

Recall

[2 marks]

Define (a) enthalpy change of hydration and (b) enthalpy change of solution, with an equation for each using magnesium chloride.

(a) The enthalpy change when 1 mol of gaseous ions dissolves in water to give an infinitely dilute solution: Mg²⁺(g) → Mg²⁺(aq). ✓
(b) The enthalpy change when 1 mol of an ionic solid dissolves in water to give an infinitely dilute solution: MgCl₂(s) → Mg²⁺(aq) + 2Cl⁻(aq). ✓

Calculation

[2 marks]

Calculate ΔH_sol(LiF). LE(LiF) = −1037; ΔH_hyd(Li⁺) = −519; ΔH_hyd(F⁻) = −506 (all kJ mol⁻¹).

ΣΔH_hyd = −519 + (−506) = −1025. ✓
ΔH_sol = −1025 − (−1037) = +12 kJ mol⁻¹. ✓

Calculation

[3 marks]

Calculate ΔH_sol(MgCl₂). LE = −2521; ΔH_hyd(Mg²⁺) = −1920; ΔH_hyd(Cl⁻) = −377 (all kJ mol⁻¹).

Two chloride ions: 2 × (−377) = −754. ✓
ΣΔH_hyd = −1920 − 754 = −2674. ✓
ΔH_sol = −2674 + 2521 = −153 kJ mol⁻¹. ✓

Calculation

[2 marks]

NaF has ΔH_sol = +1 kJ mol⁻¹. ΔH_hyd(Na⁺) = −406 and ΔH_hyd(F⁻) = −506 kJ mol⁻¹. Calculate the lattice energy of NaF.

LE = ΣΔH_hyd − ΔH_sol. ✓
LE = (−406 − 506) − 1 = −913 kJ mol⁻¹. ✓

Explain

[3 marks]

Explain why ΔH_hyd(Mg²⁺) is much more exothermic than ΔH_hyd(Na⁺).

Mg²⁺ has twice the charge of Na⁺. ✓
Mg²⁺ also has a smaller ionic radius, so it has a much higher charge density. ✓
It attracts the δ− oxygen atoms of water molecules more strongly, so more energy is released. ✓

Explain

[3 marks]

Explain, in terms of lattice energy and hydration enthalpy, why BaSO₄ is much less soluble than MgSO₄.

SO₄²⁻ is a large ion, so the lattice energy becomes only slightly less exothermic from MgSO₄ to BaSO₄. ✓
Ba²⁺ is much larger than Mg²⁺, so its hydration enthalpy is much less exothermic. ✓
So ΔH_sol becomes less exothermic (more endothermic) down the group and BaSO₄ is less soluble. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.