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DrBob222 Let's call phenol HP and it ionizes as
....................HP ==> H^+ + P^-
I...........2.55E-6........0..........0
C..................-x............x..........x
E..........2.55E-6-x.........x..........x
I don't know how advanced this course is but if you don't take into account the ionization of H2O you will make an error of approximately 5%. I will assume that such an error is OK and not include the water.
Then you need to look up the Ka for phenol and plug the E line and Ka values into the Ka expression. That will solve for x = (H^+). Then
%ionization = [(H^+)/2.55E-6)]*100 = ?
Post your work if you get stuck.