Calculate the pH of a solution (to 2 decimal places) which is 0.106 M in phenol, Ka = 1.0 x 10-10.

Answer 1

Let's let HPh equal phenol with H representing the hydrogen that ionizes and Ph representing the remainder of phenol.
HPh ==> H^+ + Ph^-

Write the Ka expression.
Ka = 1 x 10^-10 = [(H^+)(Ph^-)]/(HPh)

Before the ionization (HPh) = 0.106 M.
Before the ionization (H^+)=(Ph^-)=0
After ionization, (H^+)= x
(Ph^-) = x
(HPh) = 0.106 - x

Plug these variables into the Ka expression and solve for x = (H^+).
Then convert (H^+) to pH with pH = - log(H^+). Post your work if you get stuck.

Answer 2

pH= -log(1.0x10^-10) + log x/0.106-x.
how??