Enthalpies of solution and hydration

Chemistry A2 · Chemical energetics. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

When an ionic solid dissolves, think of two imaginary steps. First the lattice is broken into separate gaseous ions, which needs energy. Then each gaseous ion is surrounded by water molecules, which releases energy. This second step is hydration.

The enthalpy change of solution is the small difference between these two large energies. It can be slightly positive or slightly negative.

Hydration is exothermic because ions attract the polar water molecules. A small ion or a highly charged ion has a high charge density, attracts water more strongly and has a more exothermic enthalpy change of hydration.

Rules to remember

Enthalpy change of hydration, ΔH_hyd: one mole of gaseous ions dissolves in water to form a very dilute solution, e.g. Na⁺(g) + aq → Na⁺(aq). Always negative.
Enthalpy change of solution, ΔH_sol: one mole of solid dissolves in water to form a very dilute solution, e.g. NaCl(s) + aq → Na⁺(aq) + Cl⁻(aq).
ΔH_sol = ΣΔH_hyd − ΔH_latt (with ΔH_latt as the negative value, gaseous ions → solid)
Count every ion: for MgCl₂, ΣΔH_hyd = ΔH_hyd(Mg²⁺) + 2 × ΔH_hyd(Cl⁻).
Smaller ionic radius or higher ionic charge → more exothermic ΔH_hyd.
ΔH_sol is negative when the hydration enthalpies together are more exothermic than the lattice energy.

Common mistake

Sign errors with lattice energy are very common. Lattice energy is negative, so subtracting it means adding a positive number: write the brackets, e.g. −770 − (−787).

Worked example

For CaCl₂, ΔH_sol = −83 kJ mol⁻¹, ΔH_latt = −2258 kJ mol⁻¹ and ΔH_hyd(Cl⁻) = −364 kJ mol⁻¹. Find ΔH_hyd(Ca²⁺).

ΔH_sol = ΔH_hyd(Ca²⁺) + 2 × ΔH_hyd(Cl⁻) − ΔH_latt.
−83 = ΔH_hyd(Ca²⁺) + 2 × (−364) − (−2258) = ΔH_hyd(Ca²⁺) − 728 + 2258.
ΔH_hyd(Ca²⁺) = −83 + 728 − 2258 = −1613 kJ mol⁻¹.

Answer: ΔH_hyd(Ca²⁺) = −1613 kJ mol⁻¹

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