Nitric acid (hno3) is a strong acid that is completely ionized in aqueous solutions of concentrations ranging from 1% to 10% (1.50 m ). however, in more concentrated solutions, part of the nitric acid is present as un-ionized molecules of hno3. for example, in a 50% solution (7.50 m ) at 25°c, only 33% of the molecules of hno3 dissociate into h+ and no3–. what is the value of ka for hno3?

[HNO3] [H+] % ionization 
10.0 7.77 77.7% 
7.00 5.77 82.4% 
3.00 2.72 90.7% 
1.00 0.965 96.5% 
0.100 0.0996 99.6% 
0.0100 0.00999.. 99.9% 
0.00100 0.00100..100.0% 

HNO3(aq) <==> H+ + NO3^- ... Ka = ??? 
7.50M 0 0 initial 
3.02M 2.48...2.48 equilibrium, assuming that there is 33% ionization 

Ka = [H+][NO3-] / [HNO3] 
Ka = (2.48)² / 3.02 
Ka = 2.04 

your answer would be “dry”

the answer would be b i believe