What is the hydronium ion concentration of a 0.500 M acetic acid solution with K_a = 1.8 × 10^-5? The equation for the dissociation of acetic acid is:
CH₃CO₂H(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃CO₂^-(aq) .

Question

Quizplus · Accepted Answer

Since acetic acid is a weak acid, we will use the equilibrium expression for its dissociation:
Ka = [H3O+][CH3COO-]/[CH3COOH]

We know that the initial concentration of acetic acid is 0.500 M. At equilibrium, some of it will have dissociated into H3O+ and CH3COO-. Let's say x M of acetic acid has dissociated. Then, at equilibrium:

[CH3COO-] = x M
[H3O+] = x M
[CH3COOH] = (0.500 - x) M

Using the Ka value given in the problem, we can set up an equation and solve for x:

1.8 × 10-5 = (x)(x)/(0.500 - x)

Solving this equation gives:
x = 3.0 × 10-3 M

So, the hydronium ion concentration at equilibrium is 3.0 × 10-3 M.

Answer choice C (3.0 × 10-3 M) is the best answer.

What Is the Hydronium Ion Concentration of a 0