What is the pH of a 2.0 molar solution of acetic acid. Ka acetic acid = 1.810–5

1 answer

To find the pH of a 2.0 M solution of acetic acid, we need to use the equilibrium constant expression for the dissociation of acetic acid:

CH3COOH + H2O ⇄ CH3COO- + H3O+

Ka = [CH3COO-][H3O+] / [CH3COOH]

We can assume that the initial concentration of acetic acid is 2.0 M, and that at equilibrium, some of it will dissociate to form acetate ions and hydronium ions. Let x be the concentration of H3O+ and CH3COO- ions formed at equilibrium.

Then, we can write:

Ka = x^2 / (2.0 - x)

Since the Ka value is small compared to the initial concentration of acetic acid, we can simplify the expression by assuming that x is much smaller than 2.0. This means that we can ignore the amount of acetic acid that does not dissociate, and consider the concentration of CH3COO- and H3O+ ions to be equal to x.

So, we can write:

Ka = x^2 / 2.0

Rearranging and solving for x, we get:

x = sqrt(Ka * 2.0) = 0.00294 M

The pH of the solution can now be calculated using the definition of pH:

pH = -log[H3O+]

pH = -log(0.00294) = 2.53

Therefore, the pH of a 2.0 M solution of acetic acid is 2.53.