The Ka of a monoprotic weak acid is 429x103 What is the perc

The Ka of a monoprotic weak acid is 4.29x10^3. What is the percent ionization of a 0.152 M solution of this acid?

Solution

.....HA...<=>...H+...+...A-

I...0.152.......0........0

C....-a........+a.......+a

E..0.152-a......a........a


Ka = [H+][A-]/[HA]

= a^2/(0.152 - a) = 4.29 x 10^(-3)

a^2 + 4.29 x 10^(-3)a - 6.5208 x 10^(-4) = 0


The positive root of the quadratic equation is:

a = 0.02348


[H+] = a = 0.02348 M

Percent ionization = [H+]/[HA]initial x 100%

= 0.02348/0.152 x 100%

= 15.4%