1. what is the ph of a 0.085 m solution of nitrous acid (hno2) that has a ka of 4.5 × 10-4? 2.21 5.33 3.35 4.42

The chemical reaction for nitrous acid in a solution with water is
                 HNO2 + H2O(l) ⇄ H3O^+ + NO2^-
Then
                 Ka = [H3O^+][NO2^-] / [HNO2]
If we let x represent the concentrations of [H3O^+] and [NO2^-] then
                 Ka = x^2 / [HNO2]
                    x = √(Ka * [HNO2])
                    x = √(4.5 × 10^-4 * 0.085)
                    x = 6.1846 x 10^-3 = [H3O+]
Therefore
                  pH = -log[H3O+]
                  pH = -log[6.1846 x 10^-3]
                  pH = 2.21
The answer to your question is 2.21.