What is the percent ionization of a solution prepared by dissolving 0.0175 mol of chloroacetic acid,(HC₂H₂O₂Cl)in 1.30 L of water? For chloroacetic acid,K_a = 1.4 × 10^-3. A)1.3 % B)100 % C)27 % D)72 % E)0.36 %

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First, we need to write the equation for the dissociation of chloroacetic acid: HC₂H₂O₂Cl (aq) + H2O (l) ⇌ H3O+ (aq) + C₂H₂O- (aq) + Cl- (aq) The equilibrium constant expression is: K_a = [H3O+] [C₂H₂O-] / [HC₂H₂O₂Cl] We know the equilibrium constant (K_a = 1.4 × 10^-3) and the initial concentration of chloroacetic acid (0.0175 mol/1.30 L = 0.0135 M). We can assume that the percent ionization is small, so we can simplify the equilibrium expression to: K_a ≈ [H3O+] [C₂H₂O-] / 0.0135 We can also assume that the change in concentration of chloroacetic acid is negligible compared to its initial concentration, so we can write: [H3O+] ≈ [C₂H₂O-] Substituting these values into the equilibrium expression, we get: K_a ≈ [H3O+]2 / 0.0135 Solving for [H3O+], we get: [H3O+] ≈sqrt(K_a x 0.0135) = 6.97x10^-4 M The percent ionization is: % ionization = [H3O+] / 0.0135 x 100% ≈ 27% Therefore, the correct answer is C).

What Is the Percent Ionization of a Solution Prepared by Dissolving