Given:
Concentration of HNO3 = 7.50 M
% dissociation of HNO3 = 33%
To determine:
The Ka of HNO3
Explanation:
Based on the given data
[H+] = [NO3-] = 33%[HNO3] = 0.33*7.50 = 2.48 M
The dissociation equilibrium is-
      HNO3  ↔   H+    +    NO3-
I Â Â Â Â Â Â 7.50 Â Â Â Â Â Â Â 0 Â Â Â Â Â Â Â Â 0
C Â Â Â Â Â -2.48 Â Â Â Â Â +2.48 Â Â Â Â Â Â Â +2.48
E Â Â Â Â Â Â 5.02 Â Â Â Â Â Â 2.48 Â Â Â Â Â Â Â 2.48
Ka = [H+][NO3-]/HNO3 = (2.48)²/5.02 = 1.23
Ans: Ka for HNO3 = 1.23