What is the pH of 1000. mL of a 0.0033 M solution of phenol (C6H5OH)?

Question

Note: conjugate base, NaC₆H₅O  has a K_b = 9.43 x 10^-5 .

Answer 1

C6H5OH + H2O ==> C6H5O- + H3O where C5H5O- is the conjugate base (Kb = 9.43x10^-5)

From Kb, calculate Ka of phenol:

KaKb = Kw = 1x10^-14

Ka = 1x10^-14 / Kb = 1x10^-14 / 9.43x10^-5

Ka = 1.06x10^-10

Ka = 1.06x10^-10 = [C6H5O-][H3O+] / [C6H5OH]

1.06x10^-10 = (x)(x) / 0.0033

x² = 3.5x10^-13

x = 5.9x10^-7 = [H3O+]

pH = -log [H3O+]

pH = 6.23

NOTE: since the Ka is so small, we should include the auto ionization of H2O as contributing to the [H3O+], but I'm assuming that you need not do that if this is high school chem. If more advance, you would add 1x10^-7 M [H3O+] to the 5.9x10^-7 M H3O+ coming from the phenol.