Part 3: Hydration and enthalpies of solution

Combine lattice separation with hydration and distinguish a heat change from a prediction of solubility.

Gas ions become aqueous ions

Hydration enthalpy is the enthalpy change when one mole of gaseous ions becomes hydrated aqueous ions. Use the appropriate standard conditions for a standard value. It is exothermic because attractive interactions with water are established. Water’s oxygen end is δ− and points towards cations; its δ+ hydrogen ends point towards anions.

For ions of equal charge, smaller radius generally gives a more negative hydration enthalpy. Higher ionic charge can also increase its magnitude. Use ionic radius, not atomic radius, and refer to water’s partial charges rather than free H⁺ ions.

Two hypothetical steps give the actual change

The enthalpy of solution describes dissolving one mole of solute in enough solvent that further dilution produces negligible heat change. The cycle is a Hess’s law route, not a claim that free gaseous ions exist inside the beaker. Separate the solid lattice to gas ions, then hydrate every ion.

For MX₂, count two anions. If the data provide lattice formation, reverse it. The solution enthalpy may be positive or negative because it is the sum of competing contributions.

Worked example: keep the coefficients

Illustrative data for MX₂ are lattice formation −2100, cation hydration −1500 and anion hydration −340 kJ mol⁻¹. Dissociation is therefore +2100. ΔsolH° = 2100 − 1500 − 680 = −80 kJ mol⁻¹. In an insulated experiment this exothermic dissolution would raise the solution temperature.

If 0.0250 mol dissolves and losses are neglected, 2.00 kJ is transferred to the solution. For 100 g of solution with c = 4.18 J g⁻¹ K⁻¹, ΔT = 2000/(100 × 4.18) = 4.78 K. The reaction enthalpy is negative while the warming solution’s q is positive. Real calorimetry must account for heat loss and container heat capacity.

Enthalpy alone does not decide solubility

A solid can dissolve endothermically. An entropy contribution can offset the positive enthalpy in ΔG = ΔH − TΔS. Conversely, a favourable solution enthalpy does not imply unlimited solubility: equilibrium and concentration still matter.

Do not assume every dissolution has positive entropy. Dispersing ions increases possible arrangements, but strongly ordering nearby water can oppose that change. Use supplied entropy data where available. At saturation, dissolution and crystallisation continue at equal rates; adding more solid need not raise the dissolved concentration.

Quick checks

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

Sources