Chemistry I21) The pH of a 0.50M HNO2 solution is 1.83. What is the Ka of this acid? 6.4 x 10^-4 M 4.6 x 10^-2 M 4.6 x 10^-4 M 6.4 x 10^-2 MPrevious Question Next QuestionDo not use the 'Forward' or 'Back' button on your browser.

To find the Ka of the acid, we first need to understand that the pH is related to the concentration of H+ ions in the solution, which is related to the acid dissociation constant (Ka) of the acid.

The formula for pH is pH = -log[H+], where [H+] is the concentration of H+ ions.

Given that the pH is 1.83, we can find [H+] by rearranging the formula to [H+] = 10^-pH.

So, [H+] = 10^-1.83 = 0.0148 M.

The formula for Ka is Ka = [H+][A-]/[HA], where [HA] is the initial concentration of the acid, [H+] is the concentration of H+ ions, and [A-] is the concentration of the conjugate base.

In this case, because HNO2 is a weak acid, we can assume that it dissociates completely, so [H+] = [A-] and [HA] = initial concentration - [H+].

So, [HA] = 0.50 M - 0.0148 M = 0.4852 M.

Substituting these values into the Ka formula gives Ka = (0.0148 M * 0.0148 M) / 0.4852 M = 4.5 x 10^-4 M.

So, the Ka of the acid is 4.5 x 10^-4 M. The closest answer to this is 4.6 x 10^-4 M.