Calculate the degree of ionization of 0.05M acetic acid if its p*K*_{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?

#### Solution

c = 0.05 M

`pK_a = 4.74`

`pK_a = -log(K_a)`

`K_a = 1.82 xx 10^(-5)`

`K_a = calpha^2 alpha = sqrt((K_a)/c)`

`alpha = sqrt((1.82 xx 10^(-5))/(5xx10^(-2))) = 1.908 xx 10^(-2)`

When HCl is added to the solution, the concentration of H^{+} ions will increase. Therefore, the equilibrium will shift in the backward direction i.e., dissociation of acetic acid will decrease.

**Case I: **When 0.01 M HCl is taken.

Let *x* be the amount of acetic acid dissociated after the addition of HCl.

`CH_3COOH ↔ H^+ + CH_3COO^(-)`

Initial conc 0.05 M 0 0

After dissociation 0.05 - x 0.01 + x x

As the dissociation of a very small amount of acetic acid will take place, the values i.e., 0.05 – *x* and 0.01 + *x* can be taken as 0.05 and 0.01 respectively.

`K_a = ([CH_3COO^(-)][H^+])/[CH_3COOH]`

`:. K_a = ((0.01)x)/0.05`

`x = (1.82 xx 10^(-5) xx 0.05)/0.01`

`x = 1.82 xx 10^(-3) xx 0.05 M`

Now

`alpha = "Amount of acid disociuated"/"Amount of acid taken"`

`= (1.82 xx 10^(-3) xx 0.05)/0.05`

Case II When 0.1 MHCL is taken

Let the amount of acetic acid dissociated in this case be *X*. As we have done in the first case, the concentrations of various species involved in the reaction are

`[CH_3COOH] = 0.05 - X; 0.05 M`

`[CH_3COO^(-)] =- X`

`[H^+] = 0.1 + X ; 0.1 M`

`K_a = ([CH_3COO^(-)][H^+])/[CH_3COOH]`

`:. K_a =((0.1)X)/0.05`

`x = (1.82 xx 10^(-5) xx 0.05)/0.1`

`x = 1.82 xx 10^(-4)xx 0.05 M`

Now,

`alpha = "Amount of acid dissociated"/"Amount ofacid taken"`

`= (1.82 xx 10^(-4) xx 0.05)/00.5`

`= 1.82 xx 10^(-4)`