∧ m o for NaCl ,   HCl and NaA are 126.4 ,   425.9 and 100 .5   S   cm 2 mol - 1 respectively. If the conductivity of 0 .001   M   HA is 5 × 10 - 5 S   cm - 1 , degree of dissociation of HA is

Step-by-step solution

Kohlrausch's law states that the equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the conductances of the anions and cations. λ m o HA = λ m o HCl + λ m o NaA - λ o NaCl = 425.9 + 100.5 - 126.4 = 400 λ m o = K × 1000 M = 5 × 10 - 5 × 10 3 10 - 3 = 50 α = 50 400 = 0 .125