Question

A solution of 0.10 M of a weak electrolyte is found to be dissociated to the extent of 1.20% at 25°C. Find the dissociation constant of the acid.

Answer 1

Given that α = 1.20%  = \(\frac{1.20}{100}\) = 1.2 x 10^-2

K_a = α²c = (1.2 X 10^-2)2 (0.1) 

= 1.44 x 10^-4 x 10^-1 = 1.44 x 10^-5