Distribution of acid in Organic and Aqueous Phases

The value for the distribution constant for an organic acid, HA, between chloroform and water is $10^2$. The dissociation constant of the acid in water is $10^{-5}$, and it can be assumed that it does not dissociate or dimerize in the chloroform phase. Increasing the pH of the aqueous phase from 4 to 9 has what effect on the fraction of HA extracted into the chloroform phase?


(a) The fraction is more than doubled.


(b) The fraction is increased but not doubled.


(c) The fraction is unaffected.


(d) The fraction is decreased but is not halved.


(e) The fraction is reduced to less than half.


The correct answer is (d). I understand why the concentration of HA in chloroform should decrease, but I'm not sure I'm completely confident as to why it decreases to less than half.


As the acid in the organic phase is in equilibrium with the acid in the aqueous phase, obviously, increasing the pH past the pK$\_a$ of the acid will drive it to completely dissociate, causing the HA in the organic layer to go into the aqueous layer. How can I quantitatively know that this process will decrease the $\frac{[HA]\_{CHCl\_3}}{[HA]\_{water}}$ to less than 50?


EDIT: So, I've taken your suggestion, but I seem to be getting stuck.


I'll use the notation HA$\_{org}$ to denote the acid in the chloroform layer and HA$\_{aq}$ to denote the acid in the water layer.


H$^+$ + A$^-$ $\rightleftharpoons$ HA$\_{aq}$


HA$\_{aq}$ $\rightleftharpoons$ HA$\_{org}$


K$\_a$ = $10^{-5}$ and the equilibrium constant for the bottom reaction is K$\_D$.


We know that: 


Our wanted ratio, D = $\frac {[HA]\_{org total}}{[HA]\_{aq total}}$ = $\frac {[HA]\_{org}}{[HA]\_{aq} + [A^-]} = \frac {[HA]\_{org}}{[HA]\_{aq} + 10^{-5}[HA]\_{aq}/[H^+]} = \frac{K\_D [H^+]}{[H^+] + 10^{-5}}$


But, I don't know what K$\_D$ is and it would also change with pH.


