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`?

Options

A. `0.019%`
B. `1.9%`
C. `3.0%`
D. `4.74%`

Answer

B

Explanation

`{:(CH_(3)COOH hArr,CH_(3)COO^(-),H^(+)),(1,0,0),(,alpha,alpha):}`
`pK_(a) =- log K_(a) = 4.74`
`:. K_(a) = 1.82 xx 10^(-5)`
From, `K_(a) = (Calpha^(2))/((1-alpha)) = Calpha^(@) ( :' 1- alpha = 1)`
`alpha = sqrt((K_(a))/(C)) = sqrt((1.82 xx 10^(-5))/(0.05))`
`=0.019` or `1.9%`