Question

Calculate the pH of 0.836 M anilinium hydrochloride (C6H5NH3Cl) solution in water, given that Kb for aniline is 3.83 ⋅ 10−4.

Answer 1

Ans :- pH = 5.33

Explanation :-

Given ,

K_b of C₆H₅NH₃Cl = 3.83 x 10^-4

So,

pK_b = - log K_b = - log 3.83 x 10^-4 = 3.4168 and

given, Concentration i.e. C = 0.83 M

C₆H₅NH₃Cl is a salt of weak base i.e. C₆H₅NH₂and strong acid i.e. HCl , and the pH of a salt of weak base and strong acid can be calculated by using the eqtauion :

pH = 1/2 [pK_w - pK_b - log C] , Here pK_w = 14

So,

pH = 1/2 [ 14 - pK_b - log C]

pH = 1/2 [ 14 - 3.4168 - log 0.836 ]

pH = 1/2 [ 14 - 3.4168 - ( - 0.07779) ]

pH = 1/2 [ 14 - 3.4168 + 0.07779 ]

pH = 1/2 [ 10.66099]

pH = 5.33