Question

What would be the pH of a 0.1 M aqueous solution of the phenolate ion, C6H5O�? (Ka for phenol, C6H5OH, is 1.3 x 10�10) I don't understand the steps to get this anwser

Answer 1

C6H5OH     ------------->   C6H5O-     +     H+
0.1 M                                           0                     0    (initial)
0.1 -x                                            x                    x    (at equilibrium)

Ka = [C6H5O-] [H+] / [C6H5OH]
1.3*10^-10 = x*x / (0.1-x)

since Ka is very small, X will be small and it can be ignored as compared to 0.1
So, above expression becomes,
1.3*10^-10 = x*x / 0.1
x = 3.61*10^-6 M

Thus, [H+] = 3.61*10^-6 M

pH = -log [H+]
      = -log (3.61*10^-6 )
      = 5.44

Answer: 5.44