The Ka for benzoic acid is 6.5 * 10-5. Calculate the pH of a 0.12 M benzoic acid solution.

pH=4.05

Explanation:

C7H6O2 - > C7H6O - + H+

ka = [C7H6O-] [H+]/[C7H6O2]

During equilibrium

[C7H6O-] = [H+] = x^2

[C7H6O2]=0.12-x

Replace

ka = x^2/0.12-x

6.5 x10^-5 = x^2/0.12-x

7.8x10^-6 - 6.5 x10^-5x=x^2

x^2 + 6.5 x10^-5x - 7.8x10^-6

Solution of quadratic equation

x=8.8 x10^-5

pH = - log [H+] = - log 8.8 x10^-5=4.05