Two buffers, X and Y of pH 4.0 and 6.0 respectively are prepared from acid HA and the salt NaA. Both — Ionic Equilibrium Chemistry Question
Question
Two buffers, X and Y of pH 4.0 and 6.0 respectively are prepared from acid HA and the salt NaA. Both the buffers are 0.50 M in HA. What would be the pH of the solution obtained by mixing equal volumes of the two buffers? Ka of HA = 1.0 × 10^-5. (log 5.05 = 0.7)
💡 Solution & Explanation
Buffer X (pH = 4.0): pH = pKa + log([A-]X / [HA]) => 4.0 = 5.0 + log([A-]X / 0.5) => log([A-]X / 0.5 = -1.0) => [A-]X / 0.5 = 0.1 => [A-]X = 0.05 M. Buffer Y (pH = 6.0): pH = pKa + log([A-]Y / [HA]) => 6.0 = 5.0 + log([A-]Y / 0.5) => log([A-]Y / 0.5 = 1.0) => [A-]Y / 0.5 = 10.0 => [A-]Y = 5.0 M. When equal volumes are mixed, the concentrations are halved: [HA]_final = 0.5 M, [A-]_final = (0.05 + 5.0) / 2 = 2.525 M. The pH of resulting solution is: pH = pKa + log([A-]_final / [HA]_final) = 5.0 + log(2.525 / 0.5) = 5.0 + log 5.05 = 5.0 + 0.7 = 5.7.