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.
What is the pH of a 2.0 molar solution of acetic acid. Ka acetic acid = 1.810–5
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