You begin with a 0.050 M initial solution of HClO2 and when the reaction comes to equilibrium, the pH is measured to be 2.1. Use the following I-C-E table to find the Ka (which means when at equilibrium).
HClO2 (aq) ⇌ H+ (aq) + ClO2- (aq)
I: 0.050 M 0 M 0 M
C: -x +x +x
E: 0.050-x x x
The equilibrium expression for the dissociation of HClO2 is:
Ka = [H+][ClO2-] / [HClO2]
Ka = (x)(x) / (0.050 - x)
Since the pH is 2.1, we know that [H+] = 10^(-2.1) = 7.94 x 10^(-3)
So x = 7.94 x 10^(-3)
Now we can substitute x = 7.94 x 10^(-3) into the equilibrium expression:
Ka = (7.94 x 10^(-3))^2 / (0.050 - 7.94 x 10^(-3))
Ka ≈ 1.58 x 10^(-2)
Therefore, the Ka for HClO2 is approximately 1.58 x 10^(-2).