Enthalpies of solution and hydration

· Enthalpy change of solution, ΔHsol = enthalpy change when 1 mol of an ionic solid dissolves in enough water to form an infinitely dilute solution.
· Example: NaCl(s) → Na⁺(aq) + Cl⁻(aq).
· Enthalpy change of hydration, ΔHhyd = enthalpy change when 1 mol of gaseous ions dissolves in water to form aqueous ions in an infinitely dilute solution.
· Example: Na⁺(g) → Na⁺(aq) or Cl⁻(g) → Cl⁻(aq).
· ΔHhyd is always negative / exothermic because ion–dipole attractions form between ions and water molecules.
· ΔHsol can be positive or negative depending on the balance between lattice dissociation and hydration.

This diagram shows the required CIE energy cycle connecting ΔHsol, lattice energy, and ΔHhyd. It is useful for seeing that dissolving an ionic solid can be split into lattice breaking followed by hydration of gaseous ions. Source

Key energy changes in dissolving ionic solids

· Dissolving an ionic solid can be treated as two Hess’s law steps:
· Step 1: lattice dissociation = ionic solid → gaseous ions; this is endothermic.
· Step 2: hydration = gaseous ions → aqueous ions; this is exothermic.
· CIE defines lattice energy, ΔHlatt, as formation of the solid lattice from gaseous ions, so it is usually negative:
· M⁺(g) + X⁻(g) → MX(s)
· To break the lattice in a solution cycle, use –ΔHlatt.
· Overall equation: ΔHsol = –ΔHlatt + ΣΔHhyd.
· ΣΔHhyd means add the hydration enthalpy for all ions in the formula, using correct mole ratios.

The Chemguide cycle shows how Hess’s law links lattice enthalpy, hydration enthalpies, and enthalpy of solution. It helps explain why ΔHsol depends on the relative sizes of the endothermic and exothermic steps. Source

Using the solution–hydration cycle

· For MX(s):
· MX(s) → M⁺(aq) + X⁻(aq) = ΔHsol.
· Alternative route: MX(s) → M⁺(g) + X⁻(g) = –ΔHlatt.
· Then: M⁺(g) → M⁺(aq) = ΔHhyd(M⁺).
· And: X⁻(g) → X⁻(aq) = ΔHhyd(X⁻).
· Therefore: ΔHsol = –ΔHlatt + ΔHhyd(cation) + ΔHhyd(anion).
· For MX₂(s): ΔHsol = –ΔHlatt + ΔHhyd(M²⁺) + 2ΔHhyd(X⁻).
· For M₂X(s): ΔHsol = –ΔHlatt + 2ΔHhyd(M⁺) + ΔHhyd(X²⁻).
· Always check the formula ratio before substituting values.

Calculation method

· Write the correct equation for ΔHsol first, including state symbols.
· Draw or imagine the Hess cycle: solid ionic compound → gaseous ions → aqueous ions.
· Use: ΔHsol = –ΔHlatt + ΣΔHhyd.
· Rearrange carefully if asked for ΔHhyd or ΔHlatt.
· Apply stoichiometric multipliers to hydration enthalpies, e.g. 2 × ΔHhyd(Cl⁻) for MgCl₂.
· Keep signs: ΔHhyd is negative, lattice formation is negative, lattice dissociation is positive.
· Use units: kJ mol⁻¹.
· Quote answers to a suitable number of significant figures.

Interpreting signs of ΔHsol

· If ΣΔHhyd is more exothermic than lattice dissociation is endothermic, ΔHsol is negative.
· A negative ΔHsol means dissolving is exothermic and the solution tends to warm up.
· If lattice dissociation is larger than hydration energy released, ΔHsol is positive.
· A positive ΔHsol means dissolving is endothermic and the solution tends to cool down.
· Solubility is not determined by enthalpy alone; later topics may involve entropy and Gibbs free energy, but for 23.2 focus on enthalpy cycles.

This figure shows a thermodynamic cycle for dissolving NaCl. It is useful for practising how to combine lattice and hydration terms to calculate ΔHsol. Source

Effect of ionic charge and ionic radius on ΔHhyd

· Hydration enthalpy becomes more exothermic when ionic charge increases.
· Higher charge = stronger attraction between the ion and polar H₂O molecules.
· Hydration enthalpy becomes more exothermic when ionic radius decreases.
· Smaller radius = higher charge density, so stronger ion–dipole attractions form.
· Most exothermic ΔHhyd: small, highly charged ions such as Mg²⁺ or Al³⁺.
· Less exothermic ΔHhyd: large, singly charged ions such as K⁺, Br⁻, or I⁻.
· Exam phrase: greater charge density gives a more negative hydration enthalpy.

Common exam traps

· Do not use ΔHlatt directly if it is defined as lattice formation; use –ΔHlatt for lattice breaking.
· Do not forget multipliers: CaCl₂ needs 2 × ΔHhyd(Cl⁻).
· Do not say hydration enthalpy is positive; hydration is exothermic, so ΔHhyd is negative.
· Do not confuse enthalpy of solution with enthalpy of hydration.
· Do not ignore state symbols: (s), (g), (aq) are essential in definitions and equations.
· Do not compare hydration enthalpies using radius only; include charge and charge density.
· Do not write “smaller ion = larger hydration enthalpy” without stating more exothermic / more negative.