Van’t Hoff ’s factor for 0.01M aqueous solution of acetic acid is 1.04, the pH of that solution is

Molarity of CH₃COOH=0.01M

Normality of CH₃COOH=(0.01)1=0.01N

Van'thoff factor , i = 1.04

P^H = ?

   n=2

i=1-+n

i = 1+

1.04 = 1+

= 0.04

For weak acids , [H⁺] = c

[H⁺] =  (0.01)(0.04)

[H⁺] = 4X10^-4

P^H = - log [H⁺] = -log(4X10^-4)= 4-log4

        = 4-2 log2 = 4-2(0.3010)

         =3.398 3.4