Solubility and Solubility Product - Application of Solubility Product for JEE

Solubility and solubility products are the key terms used in understanding the concept of solution itself. Someone who can clearly define these two terms will definitely be able to handle any complex question regarding the dissolution of solute in a solvent.

These are the terms used to describe how much of a solute is dissolved in a solvent. A deep study of these two terms is very much important for solving the numerical problems related to solutions.

What is Solubility and Solubility Product?

Solubility is the amount of solute dissolved in a definite amount of solvent (say 100 g), in order to form a saturated solution (solution in which further dissolving is not possible) at a given temperature. Solubility of the substance can change with the nature of solute taken, solvent, temperature and with the presence of common ions in the solution.

Whereas, the solubility product Ksp is the product of concentration of ions, raised to a power equal to the stoichiometry of those ions in a balanced equation of dissociation of those solutes, present in a saturated solution at that particular time.

Relationship between Solubility and Solubility Product

Consider the dissociation of a solid with the general formula AxBy. The equilibrium condition for the same can be written as: 

$A_{x} B_{y} \rightleftarrows x A^{y+}+y B^{x-}$

In order to derive the relationship between solubility and solubility product, we need to apply the law of mass action in this. Thus, we will get:

$\dfrac{\left[A^{y-}\right]^{x}\left[B^{x-}\right]^{y}}{\left[A_{x} B_{y }\right]}=K$  

Where K stands for equilibrium constant.

Since we know that the solution here is saturated, there will not be any more variation in the amount of undissolved solute AxBy. Thus, $\left[A_{x} B_{y}\right]$ will be a constant. So, the above equation becomes:

$\left[A^{y-}\right]^{x}\left[B^{x-}\right]^{y}=K_{s p}$   

Where Ksp denotes the solubility product.

Now, consider the solubility of the salt (solute) S \dfrac{mol}{L}$. Then at saturation,

$A_{x} B_{y} \rightleftarrows \underset{x S}{x A^{y+}}+\underset{y S}{y B^{x-}}$

Therefore,

$\begin{align} K_{s p} &=[x S]^{x}[y S] \\ &=x^{x} y^{y} S^{x-y} \end{align}$

Thus, the relation between solubility and solubility product can be written as:

$S^{(x+y)}=\dfrac{K_{s p}}{x^{x} y^{z}}$

To make it clearer, consider the dissociation of CaF2. There may be a question to write the relation between solubility and solubility product of CaF2 or any other solute. We know that CaF2 will dissociate as Ca2+ and 2F-. Thus, in the above relation, x value will be one and y value will be two.

$\mathrm{CaF}_{2} \rightleftarrows \underset{\mathrm{S}}{\mathrm{Ca}^{2+}}+\underset{2 \mathrm{~S}}{2 \mathrm{~F}^{-}}$

So, the Ksp will be $\left(1^{1} \times 2^{2}\right) S^{2+1}=4 S^{3}$

Similarly, solubility products of ternary electrolytes like Ag2CO3, Ag2CrO4, and PbCl2 with general formula AB2 will be equal to 4S3. For binary electrolyte with general formula AB (examples are AgCl, AgI), the solubility product will be S2. The solubility product for quaternary electrolyte with general formula AB3 will be 27S4. Examples of such types of salts are Al(OH)3, Fe(OH)3, AlI3.

Where Can the Ksp Value be Used?

Now, using this calculated value of K_SP, we can predict whether a reaction will form a precipitate or not, by analyzing the relation between ionic product and solubility product. Note that the solubility product is the ionic product when the solution is saturated. 

If the solution has not reached the saturation point, then it will not form any precipitate, because there will be solutes that are yet to be dissolved. Precipitation will occur only if solutes are not able to dissolve any more. This means that in order to form a precipitate, the ionic product should be greater than the solubility product. Thus, there will be a relation connecting the ionic product, solubility product and the precipitation of the solute.

Table of Relationship between Solubility Product and Precipitation

Common-Ion Effect and Solubility 

If we add a salt to the system in equilibrium, the common ion present in the salt will increase the concentration of the product side, due to the dissociation of salt into its ions. This increase in concentration of products will shift the equilibrium backward.

Since the equilibrium is shifted backward, more reactants will form again from the products. Thus, precipitation will occur from these undissolved reactants and the solubility of the solute will decrease.

For example: if we mix NaCl and KCl together, both NaCl and KCl will dissociate into their respective ions. However, here the Cl- ion is common in both salts. So the concentration of chloride ions becomes more than usual. This increase will shift the equilibrium backward according to Le Chatelier’s principle and the solubility will decrease in such cases. 

Applications of Solubility Product

• Purification of common salt

• Salting out of soap

• Manufacture of baking soda (sodium bicarbonate)

• Estimation of silver and silver chloride

• Estimation of barium and barium sulphate

• Precipitation of the sulphides of group II and IV

• Precipitation of group III hydroxides

Conclusion

Solubility product tells us about the degree of solubility of a solute in a solution. It refers to the extent to which a solute can dissolve in a solvent. Higher the solubility product, higher will be the solubility of that particular solute in a solvent. Solubility of a solute will depend upon the nature of both solvent and solute. It also depends on other characteristics like temperature, pressure, common ion effect, composition to name a few.

Solubility product does not have any units since it is a type of equilibrium constant. It is calculated only by taking the ratios of product and reactant concentration concept. Thus, the solubility product is a unitless quantity.