A solution is prepared in which 0.1 mole each of , CH3COOH and CHCl2COOH is present in a litre. If t — Ionic Equilibrium Chemistry Question
Question
A solution is prepared in which 0.1 mole each of $HCl$, CH3COOH and CHCl2COOH is present in a litre. If the ionization constant of CH3COOH is 10^-5 and that of Cl2CHCOOH is 0.15, the pH of solution is (log 2 = 0.3, log 3 = 0.48)
💡 Solution & Explanation
$HCl$ is a strong acid and dissociates completely to give 0.1 M H+. Due to the high H+ concentration (common ion effect), the dissociation of CH3COOH (Ka = 10^-5) is negligible. For CHCl2COOH (Ka2 = 0.15): let x be the concentration of dissociated H+. Cl2CHCOOH ⇌ Cl2CHCOO- + H+. Initial: [Cl2CHCOOH] = 0.1, [H+] = 0.1. Equilibrium: [Cl2CHCOOH] = 0.1 - x, [H+] = 0.1 + x, [Cl2CHCOO-] = x. Ka2 = [H+][Cl2CHCOO-] / [Cl2CHCOOH] => 0.15 = (0.1 + x)x / (0.1 - x) => 0.015 - 0.15x = 0.1x + x^2 => x^2 + 0.25x - 0.015 = 0. Solving this quadratic: x = 0.05 M. Total [H+] = 0.1 + x = 0.15 M. pH = -log(0.15) = 1 - log 1.5 = 1 - (log 3 - log 2) = 1 - (0.48 - 0.30) = 0.82.