The equation for equilibrium of solvent-solvent extraction is: $$\ce{PhOH(toluene) <=> PhOH (aq)}$$ An aqueous solution contain $\pu{1.5 \times 10^{-2} M}$ phenol and is shaken with the same volume of toluene. Determine the equilibrium concentration of $\ce{PhOH(toluene)}$. $K_\mathrm{D} = 14$
I have done lots of similar questions and each time I think I've gotten the method of how to do these types of questions it seems like the method doesn't apply anymore. In this question, I am really confused because I initially solved it and got the right answer. But when taking a closer look, I noticed that my calculations were wrong, and did it again in the correct way (or at least what I think is the correct way), but I got the wrong answer.
My attempt at solving the problem
Given that the aqueous solution had the concentration $1.5 \times 10^{-2}$, I made the following table, where $n_a$ is what I want to calculate.
$$\begin{array}{c|c|c|} & \text{PhOH(toluene} & ⇌ & \text{PhOH (aq)} \\ \hline \text{t=0} & - & &\ \pu{1.5 \times 10^{-2} M}\\ \hline \text{n equilibrium} & n_a & &n_0-n_a \\ \hline \text{c equilibrium} & \frac{n_a}{V} & & \frac{n_0-n_a}{V} \end{array}$$
My question is, when writing the formula for $K_\mathrm{D}$, is the organic phase always in the numerator? and the aqueous phase in the denominator? Meaning that if my reaction is written as above, do I still have the organic phase divided by the aqueous phase or do I follow the rule of products over reactants. If I follow the rule of products over reactants, I get the correct answer, but if I take org/aq which is the formula our professor refers to, then I get the wrong answer.
Which way is the correct way to solve a problem like this? All help is truly appreciated!