The dissociation constant of aniline ( C 6 H 5 NH 2 ) as a base is 5 . 93 × 10 - 10 . The ionic product of water at 25 ° C is 1 . 02 × 10 - 14 . Calculate the percentage hydrolysis of aniline hydrochloride in 1 . 0 N solution at 25 ° C . Also calculate the pH of the solution.

Dissociation constant K b = 5 . 93 × 10 - 10 K w at 25 ° C = 1 . 02 × 10 - 14 Initial conc. ( C ) = 1 . 0 N Aniline hydrochloride C 6 H 5 N H + 3 Cl - is a salt of strong acid ( HCl ) and a weak base C 6 H 5 NH 2 , it will undergo cationic hydrolysis. ∴ K h = K w K b = 1 . 02 × 10 - 14 5 . 93 × 10 - 10 = 1 . 72 × 10 - 5 C 6 H 5 NH 3 + + H 2 O ⇌ C 6 H 5 NH 2 + H 3 + O Degree of hydrolysis ( h ) = K h C = 1 . 72 × 10 - 5 1 . 0 = 4 . 15 × 10 - 3 or 0 . 415 % H 3 O + = Ch = C K h C = 1 × 4 . 15 × 10 - 3 = 4 . 15 × 10 - 3 M Hence, pH = - log H 3 O + = - log 4 . 15 × 10 - 3 = - log ( 4 . 15 ) + 3 = - 0 . 6180 + 3 = 2 . 382

At 25 ° C , the ionisation constant of anilinium hydroxide is 4 . 6 × 10 - 10 . Taking ionic product of water as 1 × 10 - 14 . Calculate hydrolysis constant of anilinium chloride, the degree of hydrolysis and pH value of 0 . 2 molar solution of the salt.

K b = 4 . 6 × 10 - 10 C = 0 . 2 M K w = 1 × 10 - 14 Anilinium chloride is a salt of a weak base and a strong acid, ∴ K h = K w K b = 10 - 14 4 . 6 × 10 - 10 = 2 . 17 × 10 - 5 Degree of hydrolysis, h = K h C = 2 . 17 × 10 - 5 0 . 2 = 1 . 085 × 10 - 4 = 1 . 04 × 10 - 2 The given salt will undergo cationic hydrolysis as C 6 H 5 NH 3 + + H 2 O ⇌ C 6 H 5 NH 2 + H 3 O + Initial conc. 0 . 2 M At. eq. 0 . 2 - 0 . 2 × 1 . 04 × 10 - 2 2 . 08 × 10 - 3 2 . 08 × 10 - 3 ∴ pH = - log H 3 O + = - log 2 . 08 × 10 - 3 = - log ( 2 . 08 ) + 3 = - 0 . 318 + 3 = 2 . 682