Question

Part A Find the percent ionization of a 0.250 M solution of HC2H3O2. (Note: Ka = 1.8×10−5). Express your answer numerically to two significant figures.

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 = 1.8x10^-5

          c = concentration = 0.250 M

Plug the values we get α = 8.485x10^-3

∴ % dissociation = 8.485x10^-3x 100 = 0.848