The molar conductivity of 0.007 M acetic acid is 20 S cm2 mol-1. What is the dissociation constant of acetic acid? Choose the correct option. [ H+ = 350 S cm2 mol-1 CH3COO- = 50 S cm2 mol-1]
- 2.50 10-5 mol L-1
- 1.75 10-4 mol L-1
- 2.50 10-4 mol L-1
- 1.75 10-5 mol L-1
Answer
(4) 1.75 10-5 mol L-1
Solution
By Kohlrausch's law:
m(CH3COOH) = H+ + CH3COO-
= 350 + 50 = 400 S cm2 mol-1
Degree of dissociation:
= m m = 20400 = 0.05
As per NCERT, Ka = c 21- c 2 (α is small)
⇒ Ka = 0.007 (0.05)2
= 0.007 2.5 10-3
= 1.75 10-5 mol L-1
Answer: (4)