Calculation of concentration at equilibrium for extraction The equation for equilibrium of solvent-solvent extraction is:
$$\ce{PhOH(toluene) &lt;=&gt; PhOH (aq)}$$
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.
$$\begin{array}{c|c|c|} 
 &amp; \text{PhOH(toluene} &amp; ⇌ &amp; \text{PhOH (aq)} \\ \hline
\text{t=0} &amp; - &amp; &amp;\ \pu{1.5 \times 10^{-2} M}\\ \hline
\text{n equilibrium} &amp; n_a &amp; &amp;n_0-n_a \\ \hline
\text{c equilibrium} &amp; \frac{n_a}{V} &amp; &amp; \frac{n_0-n_a}{V}
\end{array}$$
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!
I'd like first to answer your question:

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?

The answer is yes. The IUPAC Recommendations 1993 (Ref.1) defines Partition Ratio ($K_\mathrm{D}$) as follows (also see the Goldbook):

Partition Ratio ($K_\mathrm{D}$): The ratio of the concentration of a substance in a single definite form, $\mathrm{A}$, in the extract to its concentration in the same form in the other phase at equilibrium, e.g. for an aqueous/organic system.
$$(K_\mathrm{D})_\mathrm{A} = \frac{[\mathrm{A}]_\mathrm{org}}{[\mathrm{A}]_\mathrm{aq}}$$

Thus, if you have used equal volumes of organic and aqueous phases, the equation you can set up to solve is:
$$(K_\mathrm{D})_\mathrm{A} = \frac{[\mathrm{A}]_\mathrm{org}}{[\mathrm{A}]_\mathrm{aq}}= \frac{[\mathrm{PhOH}]_\mathrm{org,eq}}{[\mathrm{PhOH}]_\mathrm{aq, initial}-\mathrm{PhOH}]_\mathrm{org,eq}} \\ =\frac{[\mathrm{PhOH}]_\mathrm{org,eq}}{\pu{1.5 \times 10^{-2} M}-[\mathrm{PhOH}]_\mathrm{org,eq}}= 14 \tag{1}$$
When you solve the equation $(1)$ for $[\mathrm{PhOH}]_\mathrm{org,eq}$, you get your answer:
$$[\mathrm{PhOH}]_\mathrm{org,eq} =\frac{\pu{1.5 \times 10^{-2} M} \times 14}{15}= \pu{1.4 \times 10^{-2} M} $$
Hence, $[\mathrm{PhOH}]_\mathrm{aq,eq} = \pu{1.5 \times 10^{-2} M} - \pu{1.4 \times 10^{-2} M} =  \pu{1.0 \times 10^{-3} M}$
Note: If you perform extraction in different volumes of aqueous and organic phases, you should read this Wikipedia article to get help for solution.
References:

N. M. Rice, H. M. N. H. Irving, M. A. Leonard, &quot;Nomenclature for liquid-liquid distribution (solvent extraction) (IUPAC Recommendations 1993),&quot; Pure &amp; App. Chem. 1993, 65(11), 2373-2396 (DOI: https://doi.org/10.1351/pac199365112373).