Question

At equilibrium, the value of [H+] in a 0.230M solution of an unknown acid is 0.00413M . Determine the degree of ionization and the Ka of this acid.

Part 1: Degree of ionization

Part 2: K_a

Answer 1

HA dissociates as:

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

0.23 0 0

0.23-x x x

1)

x = [H+] = 4.13*10^-3 M

Use:

Degree of ionisation = x/c

= (4.13*10^-3)/0.23

= 0.0180

Answer: 0.0180

2)

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

Ka = x*x/(c-x)

Ka = 4.13*10^-3*4.13*10^-3/(0.23-4.13*10^-3)

Ka = 7.552*10^-5

Answer: 7.55*10^-5