Question

The pH of a 0.05 M solution of acetic acid is 3. What is the value of the equilibrium constant for the dissociation of acetic acid?

Answer 1

Answer:

k = 2,04x10⁻⁵

Explanation:

The equilibrium of acetic acid (CH₃COOH) in water is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺.

And the equilibrium constant is defined as:

k = [CH₃COO⁻] [H⁺] / [CH₃COOH] (1)

The equiibrium concentration of each specie if the solution of acetic acid is 0,05M is:

[CH₃COOH] = 0,05M - x

[CH₃COO⁻] = x

[H⁺] = x

-Where x is the degree of reaction progress-

As the pH is 3, [H⁺] = 1x10⁻³M. That means x =  1x10⁻³M

Replacing in (1):

k = (1x10⁻³)² / 0,05 - 1x10⁻³

k = 1x10⁻⁶ / 0,049

k = 2,04x10⁻⁵

I hope it helps!