In the laboratory, a general chemistry student measured the pH of a 0.425 M aqueous solution of benzoic acid, C6H5COOH to be 2.270. Use the information she obtained to determine the Ka for this acid.

Ka(experiment) =

Ka = 6.87x10⁻⁵

Explanation:

The equilibrium of benzoic acid in water is:

C₆H₅COOH(aq) + H₂O(l) ⇄ C₆H₅COO⁻(aq) + H₃O⁺(aq)

The equilibrium constant, Ka, is:

Ka = [C₆H₅COO⁻] [H₃O⁺] / [C₆H₅COOH]

The initial concentration of benzoic acid is 0.425M. In equilibrium its concentration is 0.425M - X and [C₆H₅COO⁻] [H₃O⁺] = X.

X is the reaction coordinate. How many acid produce C₆H₅COO⁻ and H₃O⁺ until reach equilibrium.

Concentrations in equilibrium are:

pH is defined as -log [H₃O⁺]. As pH = 2.270

2.270 = -log [H₃O⁺]

10^-2.270 = [H₃O⁺]

5.37x10⁻³M = [H₃O⁺] = X.

Replacing, concentrations in equilibrium are:

[C₆H₅COOH] = 0.425M - 5.37x10⁻³M = 0.4196M

[C₆H₅COO⁻] = 5.37x10⁻³M

[H₃O⁺] = 5.37x10⁻³M

Ka = [5.37x10⁻³M] [5.37x10⁻³M] / [0.4196M]