The ionization constant of propionic acid is `1.32xx10^(-5)`. Calculate the degree of ionization of the acid in its `0.05`M solution and also its pH. What will be its degree of ionization in the solution of `0.01N HCI` ?

Answer

`alpha = 1.63 xx 10^(-2), pH = 3.09`, In presence of `0.01 M HCl, alpha = 1.32 xx 10^(-3)`

Explanation

Let the degree of ionization of propanoic acid be `alpha`.
Then, representing propionic acid as HA, we have:
`{:(HA+,H_(2)O,harr,H_(3)O^(+),+,A^(-)),((.05-0.0alpha)~~.05,,,.05alpha,,.05alpha):}`
`K_(a)=([H_(3)O^(+)][A^(-)])/([HA])`
`= ((.05alpha)(.05alpha))/(0.05)=.05alpha^(2)`
`alpha=sqrt((K_(a))/(.05))=1.63xx10^(-2)`
Then, `[H_(3)O^(+)] = .05alpha = .05 xx 1.63 xx 10^(-2) = K_(b).15 xx 10^(-4) M`
`:. pH = 3.09`
In the presence of 0.1M of HCl, let `alpha´` be the degree of ionization.
Then, `[H_(3)O^(+)] = 0.01`
`[A^(-)] = 005alpha'`
`[HA] = .05`
`K_(a) = (0.01 xx .05alpha')/(.05)`
`1.32 xx 10^(-5) = .01 xx alpha'`
`alpha' = 1.32 xx 10^(-3)`