What is the pH of a solution prepared by dissolving 0.0163 mol of chloroacetic acid,(HC₂H₂O₂Cl)in 1.30 L of water? For chloroacetic acid,K_a = 1.4 × 10^-3.

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Quizplus · Accepted Answer

First, we need to write the equation for the dissociation of chloroacetic acid, which is: HC₂H₂O₂Cl ⇌ H+ + C₂H₂O2- + Cl- Then, we can set up an ICE table to calculate the concentration of H+ ions: HC₂H₂O₂Cl ⇌ H+ + C₂H₂O2- + Cl- I 0.0163 mol 0 0 C - x + x + x E 0.0163 - x x x The equilibrium constant expression for this reaction is: K_a = [H+][C₂H₂O2-] / [HC₂H₂O₂Cl] We are given that K_a = 1.4 × 10^-3 and the concentration of the acid (HC₂H₂O₂Cl) is 0.0163 mol / 1.30 L = 0.0125 M. Substituting the equilibrium concentrations into the expression for K_a gives: 1.4 × 10^-3 = x^2 / (0.0125 - x) Since the value of x is expected to be small compared to 0.0125 (due to the small value of K_a), we can simplify the expression: 1.4 × 10^-3 ≈ x^2 / 0.0125 Solving for x gives: x = 3.3 × 10^-3 M Therefore, the concentration of H+ ions in the solution is: [H+] = 3.3 × 10^-3 M Taking the negative logarithm of this value gives the pH: pH = -log[H+] = -log(3.3 × 10^-3) ≈ 2.45 So the answer is (B).

What Is the PH of a Solution Prepared by Dissolving