JEE Main 2019 — Ionic Equilibrium Question with Solution

0.1 M Formic acid(HCOOH) (A),
0.1 M Acetic acid(CH₃COOH) (B),
0.1 M Benzoic acid(C₆H₅COOH) (C)

Conductivity (K) depends on no of ions present in unit volume of solution. When no of ions increases conductivity also increases.
Also no of ions depends on degree of dissociation($$\alpha$$).
We know, $$\alpha$$ = $$\sqrt{\frac{K_a}{C}}$$

For all solutions C = 0.1 M
So, $$\alpha$$ depends on only K_$$a$$ here.
K_$$a$$ of HCOOH = 10^-4
K_$$a$$ of CH₃COOH = 1.8 $$\times$$ 10^-5
and K_$$a$$ of C₆H₅COOH = 6 $$\times$$ 10^-5

$$\therefore$$ $$\alpha$$(HCOOH) > $$\alpha$$(C₆H₅COOH) > $$\alpha$$(CH₃COOH)

Therefore conductivity order = A > C > B