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!

  • 4
    $\begingroup$ Why not to just take $$\frac{c}{c_0-c}=K_\mathrm{D}$$ ? $\endgroup$
    – Poutnik
    Feb 18, 2021 at 17:36

1 Answer 1


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.


  1. N. M. Rice, H. M. N. H. Irving, M. A. Leonard, "Nomenclature for liquid-liquid distribution (solvent extraction) (IUPAC Recommendations 1993)," Pure & App. Chem. 1993, 65(11), 2373-2396 (DOI: https://doi.org/10.1351/pac199365112373).
  • $\begingroup$ But I want to calculate $[PhOH]_{toluene,eq}$ which is why I am confused. Because $1 *10^{-3}$ is the correct answer but I get that answer when solving for the aqueous phase and not the organic phase.. $\endgroup$
    – katara
    Feb 19, 2021 at 12:00
  • $\begingroup$ That "correct answer" answer is incorrect, according to definition of $K_\mathrm{D}$, which is 14, indicating $\ce{PhOH}$ is more organic like. You should consult your teacher about it. $\endgroup$ Feb 19, 2021 at 16:06

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