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) = 4.74, i.e., - log K_(a) = 4.74 or log K_(a) = - 4.74 = bar(5).26 :. K_(a) = 1.82xx10^(-5)`
`alpha=sqrt(K_(a)//C)=sqrt((1.82xx10^(-5))//(5xx10^(-2)))=1.908 xx 10^(-2)`
In presence of HCl, due to high concentration of `H^(+)` ion, dissociation equilibrium will shift backward, i.e., dissociation of acetic acid will decrease .
(a) In presence of 0.01 MHCl, if x is the amount dissociated, then
`{:(,CH_(3)CO OH ,hArr,CH_(3)CO O^(-) ,+,H^(+),,,),("Initial conc.",0.05 M,,,,,,,),("After disso.",0.05 - x ,,x,,0.01 + x,,,),(,~=0.05,,,,,~=0.01M,,),(,,,,,(0.01 M H^(+) "ions are obtained from 0.01 M HCl"),,,):}`
`:. K_(a) = (x(0.01))/(0.05) or (x)/(0.05)=(K_(a))/(0.01)=(1.82xx10^(-5))/(10^(-2))=1.82xx10^(-3) . "But" (x)/(0.05)=alpha`
Hence, `alpha=1.82xx10^(-3)` (`:' alpha = ("Amount dissociated")/("Amount taken")`)
(b) In the presence of 0.1 M HCl, if y is the amount of acetic acid dissociated, then at equilibrium,
`[CH_(3)CO OH]=0.05-y~=0.05 M`
`[CH_(3)CO O^(-)]=y, [H^(+)]=0.1M+y~=0.1M`
`K_(a)=(y(0.1))/(0.05) or (y)/(0.05)=(K_(a))/(0.1)=(1.82xx106(-5))/(10^(-1))=1.82xx10^(-4), i.e., alpha = 1.82xx10^(-4)`