The Ka of HClO is 3.0x 10^-8 at 25°c. What is the percent ionization of HClO in a 0.015M aqueous solution of HCLO at 25°c?

Answer 1

............HClO ==> H^+ + ClO^-
I..........0.015.....0......0
C...........-x.......x......x
E.........0.015-x....x......x

Ka = (H^+)(ClO^-)/(HClO)
Plug in the E line into Ka expession, solve for x, then %ionization = [(x)/0.015]*100 = ?

Answer 2

0.14

Answer 3

percent ionization=concentration of ionized acid/initial concentration of acid*100
[H3O+]equil/HA]init*100%