Enthalpy of solution, hydration and solubilityEdexcel International A Level Chemistry: Revision notes

• Enthalpy change of solution (ΔsolH): the enthalpy change when 1 mol of an ionic solid dissolves in water to form an infinitely dilute solution. Example: NaCl(s) + aq → Na⁺(aq) + Cl⁻(aq).
• Enthalpy change of hydration (ΔhydH) of an ion: the enthalpy change when 1 mol of gaseous ions dissolves in water to form 1 mol of aqueous ions at infinite dilution. Example: Cl⁻(g) + aq → Cl⁻(aq). It is always exothermic, because bonds form between the ion and polar water molecules.

"Infinitely dilute" means that adding more water causes no further enthalpy change.

Dissolving can be split into two stages: the lattice is broken into gaseous ions (the reverse of the lattice energy), then the gaseous ions are hydrated.

ΔsolH = −(lattice energy) + ΔhydH(cation) + ΔhydH(anion)

The lattice energy is defined for formation of the solid, so its sign is reversed. Multiply a hydration enthalpy by the number of ions in the formula.

NaCl: +780 − 406 − 364 = +10 kJ mol⁻¹. MgCl₂: +2524 − 1920 − 2(364) = −124 kJ mol⁻¹.

Charge density is charge divided by size. A smaller ion with a higher charge has a higher charge density.

• Hydration enthalpy: a higher charge density attracts water molecules more strongly, so hydration is more exothermic. Mg²⁺ (−1920) is more exothermic than Na⁺ (−406).
• Lattice energy: a higher charge and a smaller radius mean a stronger electrostatic attraction between the ions, so the lattice energy is more exothermic. MgCl₂ (−2524) is more exothermic than NaCl (−780).

Down a group the cation radius increases, so both values become less exothermic.

Solubility depends on the total entropy change, not just the enthalpy.

ΔStotal = ΔSsystem + (−ΔsolH/T)

Dissolving normally increases the entropy of the system (a solid becomes freely moving ions). If ΔsolH is exothermic, ΔSsurroundings is positive and the salt is very likely soluble. If ΔsolH is endothermic, ΔSsurroundings is negative and the salt dissolves only if ΔSsystem outweighs it.

NaCl: ΔsolH +10, ΔSsystem +43, ΔSsurroundings −33.6, ΔStotal = +9.4 J K⁻¹ mol⁻¹, so soluble. AgCl: ΔsolH +88, ΔSsystem +33, ΔSsurroundings −295.3, ΔStotal = −262.3 J K⁻¹ mol⁻¹, so insoluble.

The balance between the lattice energy and the hydration enthalpies explains the trends.

Sulfates become less soluble down the group. The sulfate ion is large, so the spacing of ions changes little as the cation grows: the lattice energy hardly changes. The cation hydration enthalpy becomes noticeably less exothermic. So ΔsolH becomes less exothermic (more endothermic). MgSO₄ −91, BaSO₄ +19 kJ mol⁻¹.

Hydroxides become more soluble down the group. The hydroxide ion is small, so the cation radius strongly affects the spacing: the lattice energy falls by more than the hydration enthalpy. So ΔsolH becomes more exothermic (less endothermic).