Acetic acid has a dissociation constant of 1.8*10^-5 . calculate the pH value of the decinormal solution of acetic acid ???????

Given:

 Normality of acetic acid = 0.1 N (decinormal solution)

Dissociation constant = Ka = 1.8x10^-5

Solution:

CH₃COOH + H₂O ? CH₃COO^- +H₃O⁺

Ka = [H3O⁺][CH3COO^-]/[CH3COOH]

1.8x10^-5 = x² / (0.1- x)

x is very negligible compare to initial concentration of acid.

Hence,

1.8x10^-5 = x² / 0.1

x² = 1.8 x 10^-5 x 0.1

x =[H₃O⁺] =  0.001342 N

0.001342 N acetic acid = 0.001342 M acetic acid

As [H+] = 0.001342 M

pH = -log₁₀[H⁺]

pH= - log₁₀(0.001342)

pH= 2.87