What is the equilibrium concentration of chloroacetic acid,HC₂H₂O₂Cl,in a solution prepared by dissolving 0.0271 mol of HC₂H₂O₂Cl in 1.50 L of water? For chloroacetic acid,K_a = 1.4 × 10^-3.

Question

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The correct answer is A because the equilibrium concentration of chloroacetic acid can be calculated using the dissociation constant (K_a) and the initial concentration of the acid. The K_a expression for chloroacetic acid is:K_a = [H+][C₂H₂O₂Cl^-]/[HC₂H₂O₂Cl]At equilibrium, the concentration of the products ([H⁺] and [C₂H₂O₂Cl^-]) will be equal to x, and the concentration of the reactant ([HC₂H₂O₂Cl]) will be equal to (0.0271 mol/1.50 L) - x. Substituting these values into the K_a expression and solving for x gives:1.4 × 10^-3 = x²/(0.0271 mol/1.50 L) - xSolving for x gives x = 4.38 × 10^-3 M. Therefore, the equilibrium concentration of chloroacetic acid is (0.0271 mol/1.50 L) - x = 0.0227 mol/1.50 L = 4.38 × 10^-3 M.

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What Is the Equilibrium Concentration of Chloroacetic Acid,HC₂H₂O₂Cl,in a Solution