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)))=1.908xx10^(-2)`
In presence of `HCl`, due to high concentration of `H^(o+)` ion, dissociation equilibrium will shift backward, i.e, dissociation of acetic will decrease.
a. In presence of `0.01 M HCl`, if x is the amount dissociated then
`{:(,CH_(3)COOH,hArr,CH_(3)COO^(Θ),+,H^(o+)),("Initial",0.05 M,,,,),("Final",0.05-x,,x,,0.01+x),(,cong 0.05,,,,cong 0.01 M):}`

`(0.01 M H^(o+)` 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)`
or `alpha=1.82xx10^(-3)`
`( :' alpha=("Amount dissociated")/("Amount taken"))`
b. In the presence of `0.1 M HCl`, if `y` is amount of acetic acid dissociated, then at equilibrium
`[CH_(3)COOH]=0.05-y ~= 0.05 M`
`[CH_(3)COO^(Θ)]=y, [H^(o+)]=0.1 M+y ~= 0.1 M`
`K=(y(0.1))/0.05 or y/0.05=K_(a)/0.1=(1.82xx10^(-5))/10^(-1)`
`=1.82xx10^(-4), i.e., alpha=1.82xx10^(-4)`