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 got the answer 5.44 but I think it's wrong. Do I have to subtract it from 14?

Answer 1

My answer was slightly different for pOH than yours (5.11) but yes you need to subtract from 14 since I suspect you solved for OH^-.
Call phenolate ion P^-, then
.......P^- + HOH ==> HP + OH^-
I....0.1..............0....0
C.....-x..............x.....x
E....0.1-x...........x.......x

Kb for phenolate = (Kw/Ka for phenol) = (x)(x)/(0.1-x) and solve for x = (OH^-) then convert to pOH and pH.

Answer 2

I reworked the problem and realized I didn't find Kb at first and plugged in Ka instead. Thanks!