What is the hydronium ion concentration of a 0.300 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

First, write the equilibrium expression for the dissociation reaction of acetic acid:

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

We are given the value of Ka as 1.8 × 10-5 and the initial concentration of acetic acid as 0.300 M. Let x be the concentration of H3O+ and CH3COO- at equilibrium. Then, the concentration of CH3COOH at equilibrium is (0.300 - x).

Plugging in these values into the equilibrium expression and solving for x:

1.8 × 10-5 = x^2 / (0.300 - x)
x = 1.22 × 10-3 M

Therefore, the hydronium ion concentration is 1.22 × 10-3 M or 2.3 × 10-3 M when rounded to two significant figures. The closest choice is C, which is the correct answer.

What Is the Hydronium Ion Concentration of a 0