Question

What is the pH of a 0.11 M solution of C6H5OH (Ka = 1.3 x 10^-10)?

Answer 1

C6H5OH    ----------------------------> C6H5O- + H+

0.11                                                   0                0 ----------------> initial

0.11-x                                                x                x -------------------> equilibrium

Ka = [C6H5O-][H+]/[C6H5OH]

Ka = x^2 / 0.11-x

1.3 x 10^-10 = x^2 / 0.11-x

x^2 + 1.3 x 10^-10 x -1.43 x 10^-11 = 0

by solving this

x = 3.78 x 10^-6

[H+] = x = 3.78 x 10^-6 M

pH = -log [H+] = -log (3.78 x 10^-6)

pH = 5.42 -----------------------> answer