Question

The Ka of a monoprotic weak acid is 5.28 × 10-3. What is the percent ionization of a 0.116 M solution of this acid?

Answer 1

Ka = 5.28 E-3, Concentration of acid = 0.116 M

Lets assume HA is weak acid.

Dissociation :

            HA(aq) + H2O (l) --- > A^-(aq) + H3O⁺(aq)

I           0.116               0                      0                      0

C         -x                                             +x                    +x

E          (0.116-x)                                 x                      x

Ka expression

Ka = [A^-][H3O+]/[HA]

5.28 E-3 = x²/ (0.116-x)

Lets solve for x

x = 0.0222

So x = [H3O=] = x = 0.0222 M

Percent dissociation = (x / 0.116 ) x 100

= (0.0222 / 0.116 ) x 100

= 19.1 %

So the percent ionization of this acid = 19.1 %