Enthalpies of solution and hydration

Definitions (spec 12.16)

• Enthalpy change of solution, ΔsolH: the enthalpy change when one mole of an ionic solid dissolves in enough water that adding more causes no further heat change: NaCl(s) -> Na+(aq) + Cl-(aq).
• Enthalpy change of hydration, ΔhydH: the enthalpy change when one mole of gaseous ions dissolves in water to form one mole of aqueous ions: Na+(g) -> Na+(aq). Always exothermic: water's δ- oxygen is attracted to cations and its δ+ hydrogen to anions.

The cycle (spec 12.17)

Solid -> gaseous ions is the reverse of lattice energy (-LE, endothermic). Gaseous ions -> aqueous ions is the sum of the hydration enthalpies. So

ΔsolH = -LE + ΣΔhydH

For CaCl2 the hydration term is ΔhydH(Ca2+) + 2ΔhydH(Cl-).

Effect of charge and radius (spec 12.18)

Smaller, more highly charged ions have more exothermic hydration enthalpies and more exothermic lattice energies. Mg2+ is more strongly hydrated than Na+, which is more strongly hydrated than K+.

Predicting solubility (spec 12.19)

ΔsolH alone does not decide solubility; an endothermic ΔsolH can still give a positive ΔStotal if ΔSsystem is positive enough. Down Group 2, sulfates get less soluble because hydration enthalpy becomes less exothermic faster than lattice energy does (the large sulfate ion means lattice energy changes little as the cation grows). Hydroxides get more soluble because lattice energy weakens faster than hydration enthalpy.

Worked example

Calculate ΔsolH for NaCl. Lattice energy = -787 kJ mol^-1; ΔhydH(Na+) = -406 kJ mol^-1; ΔhydH(Cl-) = -378 kJ mol^-1.

1. ΔsolH = -LE + ΔhydH(Na+) + ΔhydH(Cl-)
2. = +787 - 406 - 378

Answer: ΔsolH = +3 kJ mol^-1 (slightly endothermic)

Check yourself

1. Calculate ΔsolH for KCl. Lattice energy = -711 kJ mol^-1; ΔhydH(K+) = -322 kJ mol^-1; ΔhydH(Cl-) = -378 kJ mol^-1.
   Show answer

   +11 kJ mol^-1. ΔsolH = +711 - 322 - 378 = +11 kJ mol^-1.

2. Calculate ΔsolH for CaCl2. Lattice energy = -2258 kJ mol^-1; ΔhydH(Ca2+) = -1650 kJ mol^-1; ΔhydH(Cl-) = -378 kJ mol^-1.
   Show answer

   -148 kJ mol^-1. ΔsolH = +2258 - 1650 - 2(378) = -148 kJ mol^-1.

3. Which equation represents the enthalpy change of hydration of the chloride ion?
   • Cl(g) + e- -> Cl-(g)
   • NaCl(s) -> Na+(aq) + Cl-(aq)
   • ½Cl2(g) -> Cl-(aq)
   • Cl-(g) -> Cl-(aq)
   Show answer

   Cl-(g) -> Cl-(aq). Hydration: gaseous ions become aqueous ions.

4. Which ion has the most exothermic enthalpy change of hydration?
   • Mg2+
   • Na+
   • K+
   • Ca2+
   Show answer

   Mg2+. Mg2+ is small and doubly charged, so it attracts water molecules most strongly.

5. Why is the hydration of a cation exothermic?
   • Water molecules lose electrons to the ion
   • Covalent bonds in water break
   • Bonds form between the ion and the δ- oxygen atoms of water molecules
   • The ion gains a full outer shell
   Show answer

   Bonds form between the ion and the δ- oxygen atoms of water molecules. Ion-dipole attractions form, releasing energy.

6. A salt has a positive ΔsolH but is still very soluble. Which statement explains this?
   • ΔStotal is positive because ΔSsystem for dissolving is large and positive
   • The lattice energy is endothermic
   • Hydration enthalpies are endothermic
   • The salt reacts with water
   Show answer

   ΔStotal is positive because ΔSsystem for dissolving is large and positive. Solubility depends on total entropy change, not on ΔH alone.

7. Why does the solubility of Group 2 sulfates decrease down the group?
   • Lattice energy becomes more exothermic down the group
   • The sulfate ion gets smaller
   • Hydration enthalpy becomes less exothermic faster than lattice energy does
   • The metals become less reactive
   Show answer

   Hydration enthalpy becomes less exothermic faster than lattice energy does. With the large sulfate ion, lattice energy changes little, while hydration of the larger cations falls a lot.

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Sources

• Pearson Edexcel IAL Chemistry (2018) specification
• Pearson Edexcel IAL Chemistry (2018) scheme of work

Last updated 7 November 2026. Spotted a mistake? Tell us.