Calculate the degree of ionisation of `0.05 M` acetic acid if its `pK_(a)` value is `4.74`. How is the degree of dissociation affected when its solution also contains
a. `0.01 M`, b. `0.1 M` in `HCl`?

Explanation

`pK_(a)=-logK_(a)implies K_(a)=1.82xx10^(-5)`
`alpha=sqrt(K_9a)//C` or `alpha=sqrt((1.82xx10^(-5))/0.05)`(C = 0.05 given)
`alpha= sqrt((1.82xx10^(-5))/(5xx10^(-2)))=sqrt(0.364xx10^(-3))=sqrt(3.64xx10^( 4))impliesalpha=1.908xx10^(-2)`
In the presence of 0.01 M `H^(+)`
`CH(3)COOH to CH_(3)COO^(-) + H^(+)`
Initial conc. 0.05 M 0 0
Equili. conc. `0.05-Calpha` `Calpha` `(Calpha+0.01)`
`approx 0.05` ` approx 0.01`
`[CH_(3)COOH` is a weak acid and HCl is a strong acid, so we can assume that `(Calpha + 0.01)` `approx 0.01`]
`K_(a)=([CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH])implies 1.82xx10^(-5)=(Calphaxx0.01)/0.05 implies alpha=1.82xx10^(-3)`
In the presence of 0.1 M HCl: similarly`K_(a)=([CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH])`
`1.82xx10^( 5)=(Calphaxx0.1)/0.05 [H^(+)]=(Calpha+0.1) approx 0.1M]`
`implies alpha=(0.91xx10^(-5))/0.05 = 1.82xx10^(-4)`
Note : In the presence of strong acid, dissociation of week acid i.e., decreases due to common ion effect. Also observe that we neglect the contribution of common ion from the weak electrolyte in such cases.