Question

The Ka of a monoprotic weak acid is 5.55x10-3. what is the percent ionization of a .182 M solution of this acid

Answer 1

Let a be the dissociation of the weak acid
                            HA <---> H ⁺ + A^-

initial conc.            c               0         0

change               -cα            +cα      +cα

Equb. conc.         c(1-a)          ca       ca

Dissociation constant , K_a = ca x ca / ( c(1-a)

                                         = c a² / (1-a)

In the case of weak acids α is very small so 1-a is taken as 1

So K_a = ca²

==> a = √ ( K_a / c )

Given K_a = 5.55x10^-3

          c = concentration = 0.182 M

Plug the values we get a = 0.175

∴ % ionization = 0.175x 100 = 17.5

Therefore the percent ionization is 17.5%