What is the hydronium ion concentration of a 0.500 mol L^-1 acetic acid solution with K_a = 1.8 × 10^-5? The equation for the dissociation of acetic acid is below:

Question

Quizplus · Accepted Answer

The equation for the dissociation of acetic acid is:
CH3COOH + H2O ⇌ H3O+ + CH3COO-
So, the equilibrium constant expression (Ka) for this reaction would be:
Ka = [H3O+][CH3COO-]/[CH3COOH]
Given that Ka = 1.8 × 10^-5 for acetic acid,
we can write Ka equation as:
1.8 × 10^-5 = [H3O+][CH3COO-]/[CH3COOH]
Since we're given [CH3COOH] = 0.500 mol/L, we can take the x as the hydronium ion concentration.
1.8 × 10^-5 = x^2/0.500
x^2 = 0.500 × 1.8 × 10^-5
x = √(0.500 × 1.8 × 10^-5)
x = 2.12 × 10^-3 mol/L
Therefore, the hydronium ion concentration in the solution is 2.12 × 10^-3 mol/L or 3.0 × 10^-3 mol/L when rounded to two significant figures. 
The closest option to this answer is (C) 3.0 × 10^-3 mol/L.

What Is the Hydronium Ion Concentration of a 0