Approximate equilibrium constant?

Question

9.What would be the approximate equilibrium constant, Keq, of the reaction
PhO- + C5H6 → PhOH + C5H5- ?

[PhOH is phenol, and C5H6 is cyclopentadiene.]

A. About 10e10
B. About 10e-10
C. About 10e6
D. About 10e-6
E. About 10e-16

As information i have these:

Approximate pKa values: RCOOH 4.8; ArOH 10; cyclopentadiene 16; ROH 16; RC≡CH 25; RNH2 38; RCH=CH2 44; RCH3 51.

How am i able to solve these example...?

Answer 1

This is an acid/base reaction. First you have to find your acid on each side. C5H6 is your acid on your reagent side, and Phenol is your acid on the product side.

Second you have to figure out the pKa of each acid. Reagent acid pKa~16; product acid Phenol~10 (ArOH)
The numbers tell you the Phenol wants to lose it's H more than the cyclopentadiene so the reaction will actually want to go to the left (reagents are favored). So you know you will have a - number for your Keq.
Then you take the difference between the pKa's (16-10 is 6) So your Keq will be approx 10 e-6 (d)

Answer 2

pKa for cyclopentadiene is 16, so Ka = 10^-16
pKa for phenol is 10, so Ka for phenolate GAINING a proton is 10+10

If you multiply those together you get about 10^-6 which should be the correct answer.