Enthalpy of Hydration and Solution

When an ionic solid such as sodium chloride dissolves in water, two separate energy processes are happening at once: the rigid ionic lattice must be pulled apart, and the free ions must then be surrounded by water molecules. Chemists separate the overall energy change into two linked quantities: the enthalpy of hydration, \(\Delta H_{hyd}\), and the enthalpy of solution, \(\Delta H_{sol}\). Together with lattice enthalpy, these values let you predict and explain whether dissolving a particular salt in water releases heat or absorbs it.

The enthalpy of hydration is the enthalpy change when 1 mole of a gaseous ion dissolves in a large excess of water to give an infinitely dilute aqueous ion, under standard conditions:

\(X^{n+}(g) + \)water\( \)→\( X^{n+}(aq)\)

Water molecules cluster around each ion, with the negatively charged oxygen ends facing a cation and the positively charged hydrogen ends facing an anion. This ion–dipole attraction releases energy, so \(\Delta H_{hyd}\) is always negative (exothermic). The strength of this attraction, and therefore the size of \(\Delta H_{hyd}\), depends mainly on:

• Ionic charge — a more highly charged ion attracts water molecules more strongly, giving a more negative hydration enthalpy.
• Ionic radius — a smaller ion has a higher charge density, so water molecules pack closer and bind more tightly, again making \(\Delta H_{hyd}\) more negative.

This is the same charge-density idea that explains trends in polarisability, so if you have already looked at how ion size and charge distort electron clouds, the pattern here should feel familiar.

The enthalpy of solution (or enthalpy of dissolution) is the enthalpy change when 1 mole of an ionic solid dissolves completely in enough water to form an infinitely dilute solution, so that the ions no longer interact with each other:

\(MX(s) + \)water\( \)→\( M^{+}(aq) + X^{-}(aq)\)

Unlike hydration enthalpy, \(\Delta H_{sol}\) can be either exothermic or endothermic. Its sign and size depend on the balance between how much energy is needed to break apart the lattice and how much energy is released when the ions are hydrated.

Dissolving an ionic solid can be thought of as happening in two imaginary steps: first the solid lattice breaks apart into gaseous ions, then those gaseous ions are hydrated. This is exactly the kind of indirect route used in Born-Haber cycles, and it works here because enthalpy is a state function — the overall change only depends on the start and end points, not the path taken.

Reading the cycle gives the key relationship:

\(\Delta H_{sol} = -\Delta H_{lattice(formation)} + \sum \Delta H_{hyd}(\)ions\()\)

Here \(\Delta H_{lattice(formation)}\) is the (exothermic, negative) enthalpy of forming the solid lattice from gaseous ions, so \(-\Delta H_{lattice(formation)}\) is the energy needed to break it apart. If you need a refresher on where lattice enthalpy values come from, see lattice energy, atomisation and electron affinity.

Using typical literature values, the lattice enthalpy of formation of NaCl is \(-787\ \)kJ/mol\(\), the hydration enthalpy of \(Na^{+}\) is \(-406\ \)kJ/mol\(\), and the hydration enthalpy of \(Cl^{-}\) is \(-364\ \)kJ/mol\(\).

\(\Delta H_{sol} = -(-787) + (-406 - 364) = +787 - 770 = +17\ \)kJ/mol\(\)

Because the result is positive, dissolving NaCl in water is slightly endothermic overall — breaking the lattice takes a little more energy than hydration releases, which matches the gentle cooling you can feel when salt dissolves in water.

Sodium hydroxide behaves very differently from sodium chloride. The \(OH^{-}\) ion is small, so both it and the \(Na^{+}\) ion have high charge density and strongly negative hydration enthalpies. In this case the energy released on hydration far outweighs the energy needed to break apart the lattice, so \(\Delta H_{sol}\) for NaOH is strongly exothermic, which is why solid NaOH pellets heat up noticeably as they dissolve.

Down a group, ionic radius increases, charge density falls, and hydration enthalpies become less negative. This trend explains patterns often quoted for the entropy and free-energy treatment of solubility, since a smaller (less negative) hydration enthalpy is not automatically enough to guarantee a substance dissolves: the entropy change of the process must also be taken into account.

• Enthalpy of hydration: gaseous ion to aqueous ion, always exothermic.
• Enthalpy of solution: ionic solid to aqueous ions, can be exothermic or endothermic.
• \(\Delta H_{sol} = -\Delta H_{lattice(formation)} + \sum \Delta H_{hyd}\)
• Smaller, more highly charged ions have more negative hydration enthalpies.

Related lessons

• Uk A Level Chemistry