What is the concentration of the acetate ion in a 1.00 M CH₃COOH solution given that the K_a of CH₃COOH is 1.8 $\times$ 10^-5?

Question

Quizplus · Accepted Answer

The concentration of acetate ion in a 1.00 M CH3COOH solution can be calculated using the acid dissociation constant (Ka) and the initial concentration of the acid. The dissociation of acetic acid (CH3COOH) in water is a dynamic equilibrium process represented by the equation CH3COOH ⇌ H+ + CH3COO-. The Ka expression for this equilibrium is Ka = [H+][CH3COO-]/[CH3COOH]. Given that Ka = 1.8 × 10^-5, we can assume that the concentration of H+ and CH3COO- produced is equal and much smaller than the initial concentration of CH3COOH, thus [CH3COOH] ≈ 1.00 M. Solving for [CH3COO-] (which equals [H+] at equilibrium) involves the square root of (Ka × [CH3COOH]), which gives a concentration of acetate ion around 4.2 × 10^-3 M, matching choice A.

What Is the Concentration of the Acetate Ion in a 1.00