Hydrolysis of Salts


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Dissolution of different salts in water does not always result in neutral solutions. For example, aqueous solution of copper sulphate is acidic whereas aqueous solution of sodium acetate is basic and aqueous solution of sodium chloride forms neutral solution. This is due to the dissociation of the salt in water to form ions. This process of the reaction of anion or cation of the salt with water to produce an acidic or an alkaline solution is called hydrolysis. Thus, hydrolysis is reverse of neutralization.

The equilibrium constant of such a reaction is called hydrolysis constant.

Hydrolysis of anion (A-)

hydrolysis of anion

hydrolysis constant of anion

Hydrolysis of cation (B+)

Hydrolysis of cation

hydrolysis constant of cation

The fraction of the total salt that gets hydrolysed at equilibrium is called degree of hydrolysis. It is written as 'h'.

Relation of Hydrolysis Constant and Ka and Kb

Salt of weak acid and strong base

Here, the anion (A-) is a stronger base than OH-, hence it undergoes hydrolysis to give free OH- ions. Therefore the resulting solution will be basic in character having pH greater than 7. For example

Other examples of this type of salts are CH3COONa, Na2CO3, Na3PO4, etc.

The aqueous solution of a salt of weak acid and strong base is alkaline. For the general reaction:

aqueous solution of a salt of weak acid and strong base

aqueous solution of a salt of weak acid and strong base

anion hydrolysis

This is called anion hydrolysis.

Hydrolysis constant

The hydrolysis constant may be written as:

hydrolysis constant

Multiplying equation (i) and (ii) and dividing by equation (iii), we get

relation between hydrolysis constant and dissociation constant

The hydrolysis constant 'Kh', of the salt is inversely proportional to the dissociation constant, 'Ka' of the weak acid. Therefore, the weaker the acid, the greater is the hydrolysis constant of the salt.

Degree of hydrolysis

If the original concentration of the salt in the solution is 'c' mol/litre and 'h' is the degree of hydrolysis at that concentration, then

Intial concentrationc00
Concentration at eqilibriumc(1-h)chch

 

If 'h' is very small as compared to 1, we can assume 1 - h 1.

Kh = ch2

pH of the hydrolysed salt solution [OH-] = ch

Now, pH = -log [H+]

Now, -log Kw = pKw - log Ka = pKa

 

calculation of pH

 

Knowing the molar concentration 'c' of the solution, Ka and Kw, the pH of the solution can be calculated.

 


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