When a solid substance dissolves and forms an aqueous solution, the phenomenon or the property is known as solubility. The solubility product constant is nothing but the equilibrium constant of the process of dissolution of the solid substance in an aqueous solution. The solubility formula or the solubility product formula is denoted by the symbol Ksp. Solubility can be properly defined as a property of the substance called the solute for getting dissolved in a solvent for the formation of a solution. It is a physical characteristic of the compounds. For example, the solubility of the ionic compound, which usually dissociates into cations and anions, in water varies to a great extent. Some compounds can be highly soluble such that they may even absorb moisture from the atmosphere while on the other hand, some are highly insoluble.
The Solubility Product
Solubility of a substance depends on a number of factors such as the lattice enthalpy of a salt, the solvation enthalpy of the ions in the solution, etc. Of those factors, the lattice and solvation enthalpy are the most important ones. Let us understand, the process of solubility below:
Whenever a salt is dissolved in a solvent, the ionic forces of attraction (the lattice enthalpy of the ions) within the solute are overcome by the interactive forces in-between the ions and the solvent for the solute to dissolve in the solvent.
Whenever the solute dissolves in the solvent, typically energy is released in the process and hence, the solvation enthalpy of the ions becomes negative.
The nature of the solvent is determined by the amount of energy that is released during the solvation giving the solvation enthalpy.
In the case of non-polar solvents, the energy is not sufficient to overcome the forces between the components of the compounds i.e. the lattice enthalpy, and hence in turn non-polar solvents have lower solvation enthalpy.
From the above point, it is clear that salts are not the ones that can be dissolved in non-polar solvents. Therefore, for the dissolution of the salt into the solvent, the solvation enthalpy has to be greater than the lattice enthalpy.
Solubility also depends on the factor of temperature. And hence owing to the different characteristics of different salts, the solubility of the salts varies depending on the temperature as well.
The salts can be classified into three groups depending on their solubility i.e. the formula of solubility product. They are:
Group I: It is the first category that is soluble and has solubility or the Ksp formula value of more than 0.1M.
Group II: It is the second category and includes the ones that are slightly soluble having solubility or in the range of 0.01M to 0.1M.
Group III: This is the third category and includes the ones that are sparingly soluble and have solubility less than 0.01M.
Let’s understand calculating solubility which also gives the Ksp formula, by taking an example.
Example: Consider Barium sulphate along with its saturated aqueous solution. The equation of the reaction that represents the equilibrium in-between the undissolved solids and ions is given below:
BaSO4 ⇌ Ba2+(aq) + SO4-(aq) (saturated solution in water)
For the given equilibrium, the equilibrium constant K or the solubility product formula can be given as,
K = ([Ba2+][SO4-])/[BaSO4]
Even when considering pure solid substances instead of liquid ones, the concentration remains constant and hence, the solubility product formula or the Ksp formula can be given as:
Ksp = Ksp[BaSO4] = [Ba2+][SO4-]; Ksp = solubility product constant
From this, we can understand that solid Barium sulphate when present in equilibrium with the saturated solution, the product of the concentration of the ions of both barium and sulphate is equal to the solubility product constant. Hence it will now be easier for calculating molar solubility from the solubility product as the molar solubility equation is nothing but the concentration of a given ion in a particular solution. Hence, if the solubility product is known, it can be used to determine the molar solubility i.e the concentrations of the ions from the above equation which can be in turn used as a molar solubility formula as well.