Explain the competing energy changes when an ionic solid dissolves and predict how charge and radius influence them.
Water stabilises ions through ion–dipole attraction
Hydration enthalpy is the enthalpy change when one mole of gaseous ions becomes hydrated in water. Water’s partially negative oxygen is attracted to cations; its partially positive hydrogens point towards anions. Forming these ion–dipole attractions releases energy, so hydration enthalpies are negative.
Write the physical-state change explicitly: Na⁺(g) → Na⁺(aq). Hydration does not mean the ion becomes a neutral atom or that an electron is transferred from water. The aqueous symbol describes an ion surrounded and stabilised by solvent molecules.
Smaller ions with the same charge attract water more strongly; ions with higher charge also generally have more negative hydration enthalpies, when other factors are comparable. Explain this in terms of charge density and attraction rather than just saying “smaller is stronger”.
Dissolution has an energy cost and an energy release
Enthalpy of solution is the enthalpy change when one mole of solute dissolves. To connect it to gaseous-ion data, first reverse lattice formation to separate the solid into gaseous ions, then hydrate all those ions. The gaseous-ion route is a Hess cycle, not a claim that dissolving salt literally releases free gaseous ions in the beaker.
For MX(s), ΔsolH = −ΔLEH + ΔhydH(M⁺) + ΔhydH(X⁻). Breaking the lattice costs energy and hydration releases energy. Their difference can be positive or negative; dissolving need not be exothermic. For MX₂ include two anion hydration terms.
Worked example: keep the lattice convention visible
For an original illustrative salt, ΔLEH = −780 kJ mol⁻¹ and the two hydration enthalpies are −390 and −370 kJ mol⁻¹. Dissolution gives +780 −390 −370 = +20 kJ mol⁻¹. The solution process is endothermic because separating the lattice requires more energy than hydration returns.
If a calorimetry experiment dissolves 0.0200 mol and measures +0.400 kJ absorbed by the dissolving system, the molar value is +0.400/0.0200 = +20.0 kJ mol⁻¹. The surrounding solution would cool in the idealised insulated experiment. Use the sign for the chemical system, not blindly the sign of the thermometer change.
To find an unknown hydration enthalpy, rearrange the same cycle. If ΔsolH = +15, ΔLEH = −800 and cation hydration is −405, the anion hydration is 15 −800 +405 = −380 kJ mol⁻¹. Substitute it back to verify the route.
Explain a lattice trend and its limitations
Comparing NaF with NaCl at the same ionic charges, the smaller fluoride ion gives a shorter separation between opposite charges and stronger electrostatic attraction, so NaF has the more negative lattice formation enthalpy. Comparing similarly sized ions, increasing charge also strengthens attraction.
Hydration changes with charge and size as well. A salt with a stronger lattice may also have more strongly hydrated ions, so lattice strength alone cannot establish solubility. Enthalpy of solution alone is also insufficient: entropy and temperature contribute to the free-energy change.
A useful comparison names the property held constant, the property changing and the resulting force. “MgO is more exothermic because magnesium is more reactive” does not identify the relevant charge and separation of the ions. Distinguish the enthalpy change of forming the lattice from the enthalpy change of forming the compound from elements.
Quick checks
Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official OCR A mark allocations.
Q1. For an illustrative MX₂ salt, ΔLEH = −2200, cation hydration = −1600 and each anion hydration = −320 kJ mol⁻¹. Find ΔsolH.Show answer
ΔsolH = +2200 −1600 + 2(−320) = −40 kJ mol⁻¹. Both anions must be hydrated; the result is exothermic.
Q2. Why is Mg²⁺ generally hydrated more exothermically than Na⁺?Show answer
Mg²⁺ has greater charge and smaller radius, giving greater charge density and stronger attraction to water’s dipoles. The stabilisation on hydration is larger, so the enthalpy is more negative.
Q3. Explain why a solution can cool while a salt dissolves.Show answer
If energy needed to separate the lattice exceeds energy released in hydration, dissolution absorbs energy from the surroundings. The measured temperature falls. Entropy can still make the overall process feasible.
Q4. Does a more negative lattice enthalpy alone prove lower solubility?Show answer
No. Hydration enthalpies also differ, and dissolution feasibility depends on entropy and temperature as well as enthalpy. Compare the complete process rather than one energy term.
Q5. A source gives lattice dissociation +900 and total hydration −850 kJ mol⁻¹. Should the +900 be reversed in the solution calculation?Show answer
No: dissociation already describes the solid-to-gaseous-ions direction needed. ΔsolH = +900 −850 = +50 kJ mol⁻¹. Reverse only a lattice formation value.
Sources
Sources and examiner guidance (reviewed 6 October 2026)
- OCR A H432 specification — version 3.1 — 5.2.1, printed pp. 47–48; outcomes and additional guidance, with relevant Module 1 practical skills.
- Chemrevise — OCR A 5.2.1 — Pages 1–3, 6–7; secondary coverage cross-check. Lesson explanations, data and questions are original Finesse material.
- OCR H432/01 mark scheme — June 2025 — Q3; printed pp. 10. Question-specific evidence, not universal marking rules.
- OCR H432/01 examiner report — June 2025 — Q3; printed pp. 8. Read with the corresponding question context.
- OCR H432/01 question paper — June 2025 — Q3; context for the assessment references, not reproduced questions.
Finesse Tuition is not endorsed by AQA or Chemrevise. All explanations and examples here are our own.