Question

The ionization constant of propanoic acid is 1.32 × 10^–5. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?

Answer 1

Let the degree of ionization of propanoic acid be α.

Then, representing propionic acid as HA, we have:

\[\ce{\underset{(0.05 - 0.0\alpha ≈ 0.05)}{HA +}H2O <=> \underset{0.05 \alpha}{H3O+} + \underset{0.05 \alpha}{A-}}\]

`"K"_"a" = (["H"_3"O"^+]["A"^-])/["HA"]`

`= ((0.05alpha)(0.05alpha))/0.05 = 0.05 alpha^2`

`alpha = sqrt("K"_"a"/0.05) = 1.63 xx 10^(-2)`

Then, `["H"_3"O"^+]  = 0.05 alpha = 0.05 xx 1.63 xx 10^(-2)`

`= "K"_"b"0.15 xx 10^(-4) "M"`

`therefore "pH" = 3.09`

In the presence of 0.1M of HCl, let α´ be the degree of ionization.

Then `["H"_3"O"^+] = 0.01`

`["A"^(-)] = 005  alpha'`

[HA]  = .05

`"K"_"a" = (0.01 xx 0.5alpha')/.05`

`1.32 xx 10^(-5) = .01 xx alpha^'`

`alpha' = 1.32 xx 10^(-3)`