Question

An aqueous solution of acetic acid is found to have the following equilibrium concentrations at 25 C:

[CH3COOH] = 1.65 x 10^-2 M; [H+] = 5.44 x 10^-4 M; and [CH3COO-] = 5.44 x 10^-4 M. Calculate the equilibrium constant Kc for the ionization of acetic acid at 25 C. The reaction is
CH3COOH -> H+ + CH3COO-

Answer 1

Answer: 1.79 x 10^-5

Explanation: The equilibrium constant of a reaction can be calculated from the quotient of the concentrations of the products over the concentrations of the reactants, with each termed raised to their respective stoichometric coefficients.

For acetic acid, this equilibrium expression is:

`Kc=([H+] [CH3COO-])/([CH3COOH])`

Replacing the equilibrium concentrations given by the exercise into the expression above, the equilibrium constant, Kc will be obtained and it is found to be equal to 1.79 x 10^-5.

Answer 2

The Kc for the ionization of acetic acid is 1.79x10^-5 M

Step-by-step explanation:

the reaction is as follows:

CH3COOH = CH3COO- + H+

the constant Kc is calculated with the following equation:

Kc = ([CH3COOH-] * [H+])/[CH3COOH]

Kc = (5.44x10^-4 * 5.44x10^-4)/(1.65x10^-2) = 1.79x10^-5 M