Calculate the pH of a solution that is 0.250 M in acetic acid (Ka = 1.8 × 10^-5) and 0.125 M in HClO (Ka = 2.9 × 10^-8).

• A. pH = 2.87
• B. pH = 5.60
• C. pH = 4.25
• D. pH = 3.12

Answer: (D)

Verified step by step guidance

1. Identify the acids in the solution: acetic acid (CH₃COOH) and hypochlorous acid (HClO). Both are weak acids, and their dissociation in water can be represented by their respective equilibrium expressions.
2. Write the equilibrium expressions for each acid: For acetic acid, CH₃COOH ⇌ H⁺ + CH₃COO⁻ with Ka = 1.8 × 10⁻⁵. For hypochlorous acid, HClO ⇌ H⁺ + ClO⁻ with Ka = 2.9 × 10⁻⁸.
3. Set up the ICE (Initial, Change, Equilibrium) tables for both acids to determine the concentration of H⁺ ions at equilibrium. Assume initial concentrations are 0.250 M for acetic acid and 0.125 M for hypochlorous acid, and initial H⁺ concentration is 0.
4. Use the equilibrium expressions to solve for the concentration of H⁺ ions for each acid. For acetic acid, [H⁺] = √(Ka × [CH₃COOH]₀). For hypochlorous acid, [H⁺] = √(Ka × [HClO]₀).
5. Calculate the total [H⁺] in the solution by adding the contributions from both acids. Finally, use the formula pH = -log[H⁺] to find the pH of the solution.