Question

6. Using the ionization Constants in the Appendices, calculate the hydrogen ion concentration and the percent ionization for 0.0092 M HClO

Answer 1

We know that HClO will ionize as

                                    HClO --> H+ + ClO-

Initial                             0.0092     0        0

Change                       -x             x         x

Equilibrium            0.0092-x         x        x

So Ka= [H+] [ ClO-] / [HClO]

From standard values

Ka = 3.5 X 10^-8 = x^2 / (0.0092-x)

As Ka <<1 we can ignore value of x in denominaotr

So

3.5 X 10^-8 = x^2 / 0.0092

x^2 = 3.22 X 10^-10

So x = 1.79 X 10^-5

Hence

[H+] = 1.79 X 10^-5

Percentage ionization = x X 100/ initial concentration = 1.79 X 10^-5 X 100 / 0.0092 = 0.194 %