Question

The Ka of a monoprotic weak acid is 7.23 × 10-3. What is the percent ionization of a 0.154 M solution of this acid?

must use quadratic formula.

Answer 1

Given

Ka = 7.23 * 10^-3

Molarity of Acid = 0.154 M = [HA]inital

we will represent the weak acid as HA and

HA <---> H+ + A-

                  HA           H+          A-

Intial         0.154          -             -

C                 -x           x             x

E          0.154-x         x             x

ka = [H+]eq * [A-]eq / [HA]eq

Ka = x.x / (0.154 - x) = 7.23 * 10^-3

x = 0.02995

percent ionization = ( [H+]eq / [HA]inital ) * 100 % = (0.02995 / 0.154) * 100 % = 19.45 %

Hence percent ionizaation is 19.45 %