I'm learning about the process of dissociation. I'm aware in dissolution that essentially three processes occur (I know I may be simplifying it, but I am in Year 11):

  1. Solute particles break apart (endothermic)
  2. Solvent particles break apart (endothermic)
  3. Solvent and solute particles bond/solvate (exothermic)

This much makes sense to me. However, when I get to dissociation my textbook only presented with two processes:

  1. Lattice energy - energy required to break apart ionic lattice (endothermic)
  2. Hydration energy - energy released by water molecules hydrating ions (exothermic)

What is confusing me is that the energy for the bonds (hydrogen bonds) between the water molecules to be broken apart doesn't seem to be accounted for?


The Lattice and hydration energies applies to cases of dissolution of ionic compounds like $\ce{NaCl}$.

For covalent liquid compounds with polar bonds like water, the lattice energy is replaced by the bond dissociation energy.

The value of the dissociation energy ( enthalpy ) -- involving hydration -- can be indirectly determined from the temperature dependence of the water auto-dissociation constant:

Water temperature Kw / 10−14 pKw[12]
0 °C 0.112 14.95
25 °C 1.023 13.99
50 °C 5.495 13.26
75 °C 19.95 12.70
100 °C 56.23 12.25

using van't Hoff equation

$$\frac{d}{dT} \ln K_\mathrm{eq} = \frac{\Delta_r H^\ominus}{RT^2}$$

  • $\begingroup$ My question is, is there no bond dissociation when salt is dissolved in water? Don't the water molecules also have to break their intermolecular bonds in order to hydrate the salt? $\endgroup$
    – scratch342
    Jul 30 at 10:45
  • 1
    $\begingroup$ Braking intermolecular hydrogen bonds to form hydration bonds is implicitly involved in the hydration enthalpy. Note that it is only partial breaking apart, not the full one, that can be derived from the evaporation enthalpy. $\endgroup$
    – Poutnik
    Jul 30 at 12:41
  • $\begingroup$ Ah that makes a lot of sense! Thanks! $\endgroup$
    – scratch342
    Jul 31 at 1:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.