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 25â°c, only 33% of the molecules of hno3 dissociate into h+ and no3–. what is the value of ka for hno3?

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            7.50               0                 0

C          -2.48          +2.48              +2.48

E            5.02            2.48              2.48

Ka = [H+][NO3-]/HNO3 = (2.48)²/5.02 = 1.23

Ans: Ka for HNO3 = 1.23

the potential energy of the reactants is 200j.

from the energy diagram, the energy of the product formed is 350j; this means that, this reaction is an endothermic reaction, because it absorbs energy from its environment.