Question

The Ka of hypochlorous acid (HCIO) is 3.0 x 10^-8. What is the % ionization of hypochlorous acid I'm a 0.015M aqueous solution of HCIO? Please show work

Answer 1

Given:

Ka of HClO4 = 3.0 E-8

[HClO4] = 0.015 M

By using reaction with HClO4 with H2O and ICE we can get ionized moles of acid.

ICE     

            HClO4 (aq)    + H2O---- > ClO4- (aq) + H3O+ (aq)

I           0.015                                       0                      0

C         -x                                             +x                    +x

E     (0.015-x)                                      x                      x

Now we use ka expression

ka = [H3O+][ClO4-] / [HClO4]

3.0 -8 = x² / (0.015-x)

Since the value of ka is very small that we can neglect the value of x in the bracket.

3.0 -8 = x² / 0.015

x = 2.12 E-5

we can see that x is the amount of mole ionized from acid.

So the percent ionization = (concentration of ionized ions / Initial concentration of acid )x 100

=(2.12E-5 / 0.015 ) x 100

=0.14 %

So the percent ionization = 0.14 %