What is the hydronium ion concentration of a 0.400 M acetic acid solution with K_a = 1.8 × 10^-5? The equation for the dissociation of acetic acid is: A) 2.7 × 10^-2 M B) 4.2 × 10^-2 M C) 2.7 × 10^-3 M D) 4.2 × 10^-3 M

Question

Quizplus · Accepted Answer

The dissociation equation is: CH3COOH + H2O ⇌ CH3COO- + H3O+ Where K_a represents the equilibrium constant for this reaction: K_a = [CH3COO-][H3O+] / [CH3COOH] At equilibrium, we can assume that x moles of CH3COOH dissociate into x moles of CH3COO- and x moles of H3O+. Therefore, the concentration of CH3COO- and H3O+ ions at equilibrium will be equal to x. The concentration of the remaining undissociated acetic acid will be (0.400 - x) M. Substituting these values into the equilibrium constant expression gives: 1.8 × 10^-5 = x * x / (0.400 - x) Solving for x gives x ≈ 2.7 × 10^-3 M Therefore, the hydronium ion concentration in the solution is equal to x, which is approximately 2.7 × 10^-3 M. The closest answer choice is C: 2.7 × 10^-3 M.

What Is the Hydronium Ion Concentration of a 0