from atmosphere. Others (such as lithium fluoride) have so little solubility that they are commonly termed as insoluble. The solubility depends on a number of factors important amongst which are the lattice enthalpy of the salt and the solvation enthalpy of the ions in a solution. For a salt to dissolve in a solvent the strong forces of attraction between its ions (lattice enthalpy) must be overcome by the ion-solvent interactions.
The solvation enthalpy of ions is referred to in terms of solvation which is always negative i.e. energy is released in the process of solvation. The amount of solvation enthalpy depends on the nature of the solvent. In case of a non- polar (covalent) solvent, solvation enthalpy is small and hence, not sufficient to overcome lattice enthalpy of the salt.
Consequently, the salt does not dissolve in non-polar solvent. As a general rule, for a salt to be able to dissolve in a particular solvent its solvation enthalpy must be greater than its lattice enthalpy so that the latter may be overcome by former. Each salt has its characteristic solubility which depends on temperature. We classify salts on the basis of their solubility in the following three categories.
Category I Soluble Solubility > .1M Category II Slightly .01M<Solubility< .1M Soluble Category III Sparingly Solubility < .01M Soluble We shall now consider the equilibrium between the sparingly soluble ionic salt and its saturated aqueous solution. . . Solubility Product Constant Let us now have a solid like barium sulphate in contact with its saturated aqueous solution.
The equilibrium between the undisolved solid and the ions in a saturated solution can be represented by the equation: BaSO (s) Ba + (aq) + SO – (aq), The equilibrium constant is given by the equation: K = {[Ba + ][SO – ]} / [BaSO ] For a pure solid substance the concentration remains constant and we can write K sp = K[BaSO ] = [Ba + ][SO – ] ( . ) We call K sp the solubility product constant or simply solubility product . The experimental value of