Question

The Ka of a monoprotic weak acid is 4.8 x 10^-3. What is the percent ionization of a .198M solution of this acid?
(I got 16.8% and it's wrong)

Answer 1

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

initial conc.            c               0         0

change                -cα            +cα      +cα

Equb. conc.         c(1-α)          cα    cα

Dissociation constant , K_a = cα x cα / ( c(1-α)

                                         = c α² / (1-α)

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

So K_a = cα²

==> α = √ ( K_a / c )

Given K_a = 4.8x10^-3

          c = concentration = 0.198 M

Plug the values we get α = 0.156

∴ % dissociation = 0.156 x 100 = 15.6 %