[1] ANIONIC SALT HYDROLYSIS:

When a salt is dissolved in a solvent, it first dissociates into its constituent ions. This process is called dissolution. Now, if these ions chemically react with water, the process is called hydrolysis.

Salt hydrolysis is may be consider as the reverse of process of neutralization

We can also say that “combination of any of the ion furnished by the salt with water molecules is called hydrolysis “

Cationic hydrolysis will make the solution acidic but anionic hydrolysis will make the solution basic

(1) NEUTRAL SALTS: (Salts of strong acids and strong bases)

  A salts formed by complete neutralization of strong acid and strong base are called neutral salt such salts will not undergoes hydrolysis so aqueous solution of such salts be must neutral.

The salts that undergo hydrolysis after dissolution are
(2) ALKALINE SALTS:(Salts of weak acids and strong bases)
(3) ACIDIC SALTS:(Salts of weak bases and strong acids)
(4)  Salts of weak acids and weak bases

(1) ANIONIC HYDROLYSIS OR ALKALINE SALTS HYDROLYSIS:

                  (Salt of a Weak Acid and Strong Base) 

      Let us take a certain amount of weak acid (CH₃COOH) and add to it the same amount of a strong base (NaOH). They will react to produce CH₃COONa. 

      CH₃COONa being a strong electrolyte, completely dissociates into its constituent ions.

      Now, the ions produced would react with H₂O. This process is called hydrolysis

We know that NaOH is a strong base and therefore it would be completely dissociated to give Na⁺ and OH^– ions.

      Canceling Na⁺ on both the sides,

We can note here that ions coming from strong bases do not get hydrolysed. We should note here that the solution will be basic. This is because the amount of CH₃COOH produced and OH^– produced are equal. But CH₃COOH will not completely dissociate to give H⁺ ions. Therefore [OH^–] ions will be greater than [H⁺] ions.

      Since the reaction is at equilibrium,

This equilibrium constant K_c is given a new symbol, K_h

If we multiply and divide the above equation by [H⁺] of the solution, then

CASE (1): If a is very much less than 1, then 1-a= 1 this approximation valid when C/Kh is greater than 100”

CASE (2): If C/Kh is lower than 100 than calculate h by formation of quadratic equation.

ILLUSTRATIVE EXAMPLE (1): A 0.0258 M solution of the sodium salt, NaH of the weak monoprotic acid, HA has a pH of 9.65. Calculate Ka of the acid AH.

ILLUSTRATIVE EXAMPLE (2): What is the pH of 0.10 M CH₃COONa solution. Hydrolysis constant of sodium acetate is 5.6 × 10^-10   ?

SOLUTION: Hydrolysis of the salt may be represented as

ILLUSTRATIVE EXAMPLE (3): Calculate pH of 1.0 x 10^-3 M Sodium phenolate (Na⁺O^-C₆H₅ ) Ka for C₆H₅OH is 1.0 x10^{-10 .}

SOLUTION:  (Ans-  pH=10.43)

(2) CATIONIC HYDROLYSIS    OR  ACIDIC SALTS HYDROLYSIS:

         (Salt of a Weak Base and a Strong Acid)    .......