Question

A certain weak acid, HA , has a Ka value of 2.2×10⁻⁷.

Part A

Calculate the percent ionization of HA in a 0.10 M solution.

Answer 1

HA is a mono protic acid so the dissociation equation is as follows

   HA ----------------------> H+   + A-

I 0.10                               0         0

C -x                                 +x       +x

E (0.10-x)                          x         x

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

2.2*10^-7 = [x][x] / [0.10-x]

2.2*10^-7 * [0.10-x] = [x][x]

=>solving for x

x => 0.000148 M

percent ionization = [H+] / [HA] initial *100

                                 = 0.000148 / 0.10 *100

                                 = 0.148 %