Calculate the pH of 0.1 M acetic acid solution if it's dissociation constant is 1.8×10^-5 if 1 litre of this solution is mixed with 0.05 Mole of HCl what will be the pH of the mixture

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

Regards