Solution:
Normality = 0.02
degree of ionization = 0.023
[H+] =0.02x 0.023
=0.00046N
=4.6 X10-5
pH = - log [H+] = - log (4.6 x10-4)
= - log 4.6 -log 10-4
= - 0.662 + 4 = 3.338 3.34
If acetic acid is ionized to the extent of 2.3% in 0.02 M aqueous solution, the pH of the solution is :
If acetic acid is ionized to the extent of 2.3% in 0.02 M aqueous solution, the pH of the solution is :
