Calculate the pH of an aqueous soln at 25 deg C that is 0.36 M in phenol (C6H5OH).

Question

Ka for phenol = 1.3X10^-10

Answer 1

phenol is a weak acid

[H+] = sqrt (Ka x [phenol]

pH = -lg [H+]

Answer 2

C6H5OH + H2O <- -> C6H5O- + H3O+

__0.36_____/_________0________0__(star...

_0.36-x_____/_________x________x_(end)

Ka = [C6H5O-]*[H3O+] / [C6H5OH] =

= 1.3*10^-10 = x^2 / (0.36-x) ≈

≈ x^2 / 0.36

then

x = (1.3*10^-10 * 0.36)^(1/2) = 6.84*10^-6 = [H3O+]

pH = -Log(6.84*10^-6) = 5.16