Answer:
The pH of the solution is 2.873
Explanation:
Given:
- Pka = 3.89
- pH of a 0.0140 M
Let the solution of the uric acid be HA. We have to set up the ICE table as follows;
HA(aq) + H20(l) ⇄ H3O^+(aq) + A^-
I 0.0140 0 0
C -x +x +x
E 0.0140 - x x x
Ka = 10^-(pka)
= 10^-3.89
= 1.28 x 10^-4
Ka = [H+] * [A-]/[HA]
=> 1.28 x 10^-4 = [H+]^2 / 0.0140
=>[H+]^2 = 1.28 x 10^-4 x 0.014
=> [ H+] = 1.338656 x 10^-3
so pH = - log [ H+ ] = - log 1.338656 x 10^-3
= 3 - log 1.338656
= 2.87333101111
Learn more about acids here: brainly.com/question/12978582
