The dissociation constant of acetic acid at a given temperature is `1.69xx10^(-5)`.The degree of dissociation of 0.01 M acetic acid in presence of 0.01 M HCl is equal to :

Options

A. `0.41`
B. `0.13`
C. `0.169 xx 10^(-2)`
D. `0.013`

Answer

C

Explanation

Or `alpha = (K_(a))/(C) = (1.69 xx 10^(-5))/(0.01) = 0.169 xx 10^(-2)`