Choose the correct answer

Question

What is the POH of a 0.10 M solution of phenol C6H5OH. Ka=1.0*10^-10?

a)5.5

b)9.5

c)4.5

d)8.5

Answer 1

Hi Zuhra!

This is a pretty common style of question -- the easiest way is to visualize it in an ICE (Initial, Change, Equilibrium) table. Since we have a Ka value, we know phenol must be losing the H from it's OH group.

Phenol → H+ + Phenoxide (phenol minus the H)

I 0.1M - -

C -X +X +X

E 0.1-X X X

From our Ka expression - products over reactants, w(e can set up the following algebraic expression...

(X * X) / (0.1 - X) = 1.0*10^-10

Since the concentration of 0.1 is significantly greater than the overall Ka value, we can consider the X in the denominator negligible, and instead solve this much simpler approximation...

X² / 0.1 = 1.0*10^-10

Solving for X, that gives us 3.16*10^-6. This is our equilibrium concentration for H+. Taking the negative log of that number will give us the pH, from which we can calculate pOH.

-log(3.16*10^-6) = 5.5 = pH

14 - 5.5 = pOH = 8.5 (D)