# JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Salt Hydrolysis

Salt Hydrolysis

Category : JEE Main & Advanced

It is the reaction of the cation or the anion or both the ions of the salt with water to produce either acidic or basic solution. Hydrolysis is the reverse of neutralization. (1) Hydrolysis constant : The general equation for the hydrolysis of a salt (BA),$\underset{\text{salt}}{\mathop{BA}}\,+{{H}_{2}}O$ $\rightleftharpoons$  $\underset{\text{acid}}{\mathop{HA}}\,+\underset{\text{base}}{\mathop{BOH}}\,$

Applying the law of chemical equilibrium, we get

$\frac{[HA][BOH]}{[BA][{{H}_{2}}O]}=K$,  where K is the equilibrium constant.

Since water is present in very large excess in the aqueous solution, its concentration $[{{H}_{2}}O]$ may be regarded as constant so,

$\frac{[HA][BOH]}{[BA]}=K[{{H}_{2}}O]={{K}_{h}}$

where ${{K}_{h}}$ is called the hydrolysis constant.

(2) Degree of hydrolysis : It is defined as the fraction (or percentage) of the total salt which is hydrolysed at equilibrium. For example, if 90% of a salt solution is hydrolysed, its degree of hydrolysis is 0.90 or as 90%. It is generally represented by ‘$h$’.

$h=\frac{\text{Number of moles of the salt hydrolysed}}{\text{Total number of moles of the salt taken}}$

 Types of salt Exp. for Kh Exp. for h Exp. for pH (i) Salt of weak acid and strong base Kh=Kw / Ka $h=\sqrt{\left( \frac{{{K}_{h}}}{C} \right)}$ pH=–$\frac{1}{2}$[log Kw+log Ka– log C] (ii) Salt of strong acid and weak base Kh=Kw / Kb $h=\sqrt{\left( \frac{{{K}_{h}}}{C} \right)}$ pH=–$\frac{1}{2}$[log Kw– log Kb+ log C] (iii) Salt of weak acid and weak base ${{K}_{h}}=\frac{{{K}_{w}}}{{{K}_{a}}{{K}_{b}}}$ $h=\sqrt{({{K}_{h}})}$ pH=–$\frac{1}{2}$[log Ka+ log Kw– log Kb]

(iv) Salts of strong acids and strong bases do not undergo hydrolysis (they undergo only ionization) hence the resulting aqueous solution is neutral.

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