# Calculation of concentration at equilibrium for extraction

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!

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

## 1 Answer

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:

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).
• 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.. – confused Feb 19 at 12:00
• 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. – Mathew Mahindaratne Feb 19 at 16:06