The equilibrium constant for the reaction C₆H₅COOH(aq) + CH₃COO^-(aq) $\leftrightharpoons$ C₆H₅COO^-(aq) + CH₃COOH(aq)
Is 3.6 at 25^$\circ$C. If K_a for CH₃COOH is 1.8 * 10^-5, what is the acid dissociation constant for C₆H₅COOH

Question

The equilibrium constant for the reaction C₆H₅COOH(aq) + CH₃COO^-(aq)

\leftrightharpoons

C₆H₅COO^-(aq) + CH₃COOH(aq)
Is 3.6 at 25^$\circ$C. If K_a for CH₃COOH is 1.8 * 10^-5, what is the acid dissociation constant for C₆H₅COOH

Quizplus · Accepted Answer

Using the equilibrium constant and the expression for the reaction, Kc = ([CHCOOH][C16H15COOH])/([CH3COO-][C16H15COO-]) Substitute the equilibrium concentrations of the reactants and products and cancel out the solvent molecules to get: 3.6 = ([H+][C16H15COO-])/([CH3COO-][C16H15COOH]) Since K_a for CH₃COOH is given as 1.8 F1F1F1S1F1F1F10 10^-5, we can write the expression for its dissociation as: K_a = ([H+][CH3COO-])/([CH3COOH]) Substituting the value of K_a and solving for [H+], we get: [H+] = sqrt(K_a*[CH3COOH]) = 7.33E-5 M Substituting this value in the equilibrium constant expression, we get: 3.6 = (7.33E-5*[C16H15COO-])/([CH3COO-][C16H15COOH]) Solving for [C16H15COOH]/[C16H15COO-], we get: [C16H15COOH]/[C16H15COO-] = 6.5 F1F1F1S1F1F1F10 10^-5 Therefore, the acid dissociation constant for C₆H₅COOH is 6.5 F1F1F1S1F1F1F10 10^-5, which is option B.

The Equilibrium Constant for the Reaction C6H5COOH(aq) + CH3COO-(aq) ⇋\leftrightharpoons⇋ C6H5COO-(aq) + CH3COOH(aq) Is 3

The Equilibrium Constant for the Reaction C₆H₅COOH(aq) + CH₃COO^-(aq) $\leftrightharpoons$ C₆H₅COO^-(aq) + CH₃COOH(aq)
Is 3