Question

What are the equilibrium concentrations of all the solute species in a 0.92 M solution of phenol, HC₆H₅O?
(a) [H₃O⁺], M;
(b) [OH^-], M;
(c) [HC₆H₅O], M;
(d) What is the pH of the solution? For HC₆H₅O, K_a = 1.3 x 10^-10.

Answer 1

HC6H5O   ---------------> C6H5O+   +    H+

0.92                                     0                 0

0.92 - x                                x                  x

Ka = x^2 / 0.92 - x

1.3 x 10^-10 = x^2 / 0.92 - x

x = 1.09 x 10^-5

[H3O+] = 1.1 x 10^-5 M

[OH-] = 9.1 x 10^-10 M

[HC6H5O] = 0.92 M

pH = -log (1.1 x 10^-5)

pH = 4.96