Question

Acetic acid has a Ka = 1.78×10-5. Determine the pH of a 0.250 M solution of
acetic acid

Answer 1

The balanced equation of the reaction of the dissociation of acetic acid is:

HCOOH + H₂O $\Leftrightarrow$ H₃O⁺ + HCOO^-.

The pH is defined as:

$pH = -log[H_3O^+]$ .

The initial concentration of HCOOH is 0.250 M. After the establishment of the equilibrium, the concentration of HCOOH is 0.250 -x, and the concentrations of H₃O⁺ and HCOO^- are equal to x. Then, the equilibrium constant is expressed as:

$K_a = \frac{x·x}{0.250-x}$ .

The equilibrium concentration of H₃O⁺ is:

$x^2 + K_a·x - 0.250·K_a =0$

$x^2 + 1.78·10^{-5}·x - 0.250·1.78·10^{-5} =0$

$x = 0.00210$ M.

$[H_3O^+] = 0.00210$ M.

The pH of the solution is:

$pH = -log(0.00210) = 2.68$ .

Answer: the pH of a 0.250 M solution of acetic acid is 2.68.