# What factors describe why ionic compounds are supposedly soluble in any polar solvent?

Dec 20, 2014

Ionic compounds are not always soluble in any polar solvent. It depends on the solvent (if it is water or another less polar solvent) whether they are soluble or not.

Also, ionic compounds constituted by small size ions, and/or ions with double or triple charge, and cations with similar dimensions to anion, are often insoluble in water.

When it happens that an ionic compounds is actually soluble in a polar solvent like water, this is worthy of explanation, because the electrostatic attraction between positive and negative ions is so strong that a simple ionic compound as table salt requires a temperature of 801 °C to get melted.

A high energy supply is necessary to dismount the ionic lattice, that is called lattice enthalpy. This energetic "payment" is partially compensated by the energy "gain" due to solvation enthalpy, resulting from the attraction between every ion and the many solvent molecules that can surround it with their opposite polarities.

A solvated ion can be surrounded by several shells of solvent molecules, depending by its charge and size (if the "naked ion" has a high charge and small size, it will carry a bigger "cloud" of solvent molecules).

The majority of ionic substances are dissolved in water endothermically, i.e. by spontaneously subtracting thermal energy from solvent and environment. This is an evidence that the lattice enthalpy is higher than the solvation enthalpy.

So, a second decisive factor is necessary to explain the solubility of ionic substances and to answer the question. This is a statistical or "entropic factor". By dissolving the substance there is an increase of entropy or "randomness" of motion, energies, positions, that is due to the passage from the very ordered structure of the solid lattice, to a disordered - gas type structure - of the solution. The structure of the mixture has higher statistical probability (measured by the number of equivalent configurations or "microstates" corresponding to the same "mixed" macrostate) than the unmixed macrostate.

There is always an increase in entropy, everytime a crystallline solid dissolves in a solvent, and it is the same kind of favoured process which happens with evaporation, sublimation or diffusion.

The ionic compound eventually dissolves in the solvent if the entropy contribution is sufficient to compensate the enthalpy loss that accompanies dissolution.

This can be translated quantitatively in a criterion for spontaneous dissolution: "${\Delta}_{s} G$, that is the variation of free energy, or Gibbs potential, G = (H-TS), for the disolution process, should be negative". In formulas:

${\Delta}_{s} G = {\Delta}_{l} H - T {\Delta}_{h} S < 0$

where ${\Delta}_{l} H$ is the lattice enthalpy, positive; ${\Delta}_{h} S$ is the solvation entropy difference, and it is converted in energy dimensions by multiplying the absolute temperature T. The entropy contribution $- T {\Delta}_{h} S$ is as favourable (negative) to dissolution as much the temperature is high. Thus the most usual behavior for ionic compounds is to become more soluble as the temperature increases.

Conversely, those compounds which dissolve themselves exothermically (${\Delta}_{l} H < 0$) are characterized by a solvation enthalpy that exceeds the lattice enthalpy, and are very soluble even at low temperature.