What is the concentration of H⁺ ions in a 2.0 M solution of acetic acid (HC₂H₃O₂) if the acid dissociation constant (Ka) is 1.8 x 10⁻⁵?

• A. 4.0 x 10⁻⁵ M
• B. 3.6 x 10⁻³ M
• C. 1.9 x 10⁻³ M
• D. 2.0 x 10⁻⁵ M

Answer: (C)

Verified step by step guidance

1. Start by writing the balanced chemical equation for the dissociation of acetic acid (HC₂H₃O₂) in water: HC₂H₃O₂ ⇌ H⁺ + C₂H₃O₂⁻.
2. Set up the expression for the acid dissociation constant (Ka) for acetic acid: \( K_a = \frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]} \).
3. Assume that the initial concentration of acetic acid is 2.0 M and that the change in concentration of H⁺ ions is x. Therefore, at equilibrium, [H⁺] = x, [C₂H₃O₂⁻] = x, and [HC₂H₃O₂] = 2.0 - x.
4. Substitute these equilibrium concentrations into the Ka expression: \( 1.8 \times 10^{-5} = \frac{x^2}{2.0 - x} \).
5. Assume that x is small compared to 2.0, so 2.0 - x ≈ 2.0. This simplifies the equation to \( 1.8 \times 10^{-5} = \frac{x^2}{2.0} \). Solve for x to find the concentration of H⁺ ions.