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 " or " - log K_(a) =4.74 , log K_(a) =-4.74 =bar(5).26`
`K_(a) = "Antilog " (bar(5).26) =1.82 xx 10^(-5)`
`alpha =(K_(a)//C)^(1//2) =[(1.82 xx 10^(-5))//(5xx10^(-2))]^(1//2) =1.908 xx 10^(-2)`
In the presence of HCI, more `H^(+)` ions will be formed in solution . the equilibrium will be shiffed to left .As a result `K_(a)` for acetic acid will decrease.
(ii) Calculation of degree of dissociation `(alpha)` in 0.01 M solution

`K_(a) =[[CH_(3)COO^(-)][H^(+)]]/[[CH_(3)COOH]] =(x(0.01))/((0.05)) " or " x = (K_(a) xx (0.05))/((0.01))`
`x=((1.82 xx10^(-5)) xx(0.05))/((0.01)) =(1.82 xx 10^(-5)) xx 5 = 9.1 xx 10^(-5)`
Degree of ionization `(alpha) =(x)/(C) =(9.1 xx10^(-5))/(5.0 xx10^(-2)) =1.82 xx10^(-3) = 0.0018`
(iii) Calculation of degree of dissociation `(alpha)` in 0.1 MHCI solution

`K_(a) [[CH_(3)COO^(-)][H^(+)]]/[[CH_(3)COOH]] =(x(0.01))/((0.05))`
`x=((1.82 xx 10^(-5))xx (0.05))/((0.1)) = 9.1 xx 10^(-6)`
Degree of ionization `(alpha) =(x)/(C) =(9.1 xx 10^(-6))/(5.0 xx 10^(-2)) =1.82 xx 10^(-4) =0.00018`