Calculate the acid dissociation constant Ka of benzoic acid.

Question

The pH of a 0.66 M solution of benzoic acid (HC₆H₅CO₂) is measured to be 2.19. Calculate the acid dissociation constant K_a of benzoic acid. Round your answer to 

2 significant digits.

K_a=

Answer 1

The balanced equation is

C₆H₅CO₂H ⇌ C₆H₅CO₂^– + H⁺

The acid dissociation constant K_a is related to the concentrations

of C₆H₅CO₂H, C₆H₅CO₂^– and H⁺ by the equation

K_a = [C₆H₅CO₂^–][H⁺]/[C₆H₅CO₂H]

pH = 2.19 = –log[H⁺]

which means that [H⁺] = 10^–2.19 = 0.00646

which means that [C₆H₅CO₂^–] = 0.00646

since the two concentrations have to be equal (see the balanced equation)

[C₆H₅CO₂^–] = 0.66 – 0.00646 = 0.65354

Plugging these three concentration into the K_a equation gives the K_a of benzoic acid:

K_a = (0.00646)(0.00646)/(0.65354) = 6.38 × 10^-5