Calculate the pH of 0.02 M methanoic acid where Ka = 1.6 x 10^-4 M.

pH = 2.75

Explanation:

Methanoic acid, HCOOH, is in acidic equilibrium:

HCOOH ⇄ HCOO⁻ + H⁺

And Ka is defined as:

Ka = 1.6x10⁻⁴ = [HCOO⁻] [H⁺] / [HCOOH]

As both HCOO⁻ and H⁺ ions comes from the same equilibrium, we can solve these concentrations assuming the HCOOH produce X of both HCOO⁻ and H⁺ ions:

1.6x10⁻⁴ = [HCOO⁻] [H⁺] / [HCOOH]

1.6x10⁻⁴ = [X] [X] / [0.02M]

3.2x10⁻⁶ = X²

X = 1.79x10⁻³M = [H⁺]

As pH = -log [H⁺]