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
`-log K_a` =4.74
log `K_a` = -4.74
`therefore K_a`=Antilog (`bar5` + 0.26)
`=1.8xx10^(-5)`
If `alpha` is the degree of dissociation , then
`alpha = sqrt(K_a/c)=sqrt((1.8xx10^(-5))/0.05)`
`=1.90xx10^(-2)` =0.019
(i) When the solution is also 0.01 M in HCl . In this case, the `H^+` concentration arises from two sources i.e. from HCl and from ionization of acetic acid . Let the concentration of `H^+` from acetic acid be x, then `[H^+]` =0.01 (from HCl) + x (from `CH_3COOH`)
`[CH_3COO^-] =x , [CH_3COOH]`=0.05 -x
`K_a=([H^+][CH_3COO^-])/([CH_3COOH])=1.8xx10^(-5)`
`=((0.01+x)x)/(0.05-x)=1.8xx10^(-5)`
Since x is very small, 0.01 + x = 0.01 , 0.05 - x = 0.05
`(0.01x)/(0.05)=1.8xx10^(-5)`
or `x=(1.8xx10^(-5)xx0.05)/0.01 =9.0xx10^(-5)` M
Degree of ionization =`x/c =(9.0xx10^(-5))/0.05 =1.8xx10^(-3)`
=0.0018
(ii) When the solution is 0.1 M in HCl
If y is the concentration of `H+` from `CH_3COOH`
`[H^+]`=0.1+ y `approx` 0.1 M, `[CH_3COO^-]`=y, `[CH_3COOH]` =0.05-y=0.05 M
`K_a=([H^+][CH_3COO^-])/([CH_3COOH])=1.8xx10^(-5)`
`=(0.1xxy)/0.05 =1.8xx10^(-5)`
`y=(1.8xx10^(-5)xx0.05)/0.1 =9.0xx10^(-6)`
Degree of ionization =`y/c=(9.0xx10^(-6))/0.05 =1.8xx10^(-4)`
=0.00018