4
$\begingroup$

The $K_\mathrm{D}$ (distribution coefficient) of coffee water at $25\,°\mathrm{C}$ is $7.8$ in a methylene chloride/water system. If there is $2.0\,\mathrm{g}$ of caffeine in $200\,\mathrm{mL}$ of water, calculate the grams and the % of caffeine that can be extracted with a single $100\,\mathrm{mL}$ portion of methylene chloride.

My attempt:

$$\frac{x}{\frac{2\,\mathrm{g}}{200\,\mathrm{mL}}}=7.8$$

$$\frac{x}{0.01\,\mathrm{g\,mL^{-1}}}=7.8$$

$$x=0.078\,\mathrm{g\,mL^{-1}}$$

$$x=0.078\,\mathrm{g\,mL^{-1}}$$

$$0.078\,\mathrm{g\,mL^{-1}}\cdot 100\,\mathrm{mL}=7.8\,\mathrm{g}$$

But I think I'm wrong. Am I supposed to maybe use the molar mass of caffeine ($194\,\mathrm{g\,mol^{-1}}$)?

$\endgroup$
4
$\begingroup$

You got it almost right ;-)

The distribution coefficient gives the ratio of equilibrium concentrations in a biphasic mixture.

It is crucial to realize that a caffeine concentration of $2.0\,\mathrm{g}$ in $200\,\mathrm{mL}$ of water is the initial concentration, but not the concentration after a single extraction step! The latter will be lower, since a part of the caffeine went into the organic layer.

It doens't matter whether you express the concentrations in $\mathrm{g\cdot L^{-1}}$ or $\mathrm{mol\cdot L^{-1}}$ since $K_D$ is a dimensionless constant.

$$K_D = \frac{{[\mathrm{caffeine}]_{\ce{CH2Cl2}}}}{{[\mathrm{caffeine}]_{\ce{H2O}}}} = \frac{x}{10-x}\ \mathrm{g\cdot L^{-1}} = 7.8$$

$$x = 7.8\cdot (10-x)\,\mathrm{g\cdot L^{-1}} = \mathbf{8.86} \,\mathrm{g\cdot L^{-1}}$$

Note that $x$ is the concentration in $\mathrm{g\cdot L^{-1}}$. In order to calculate the absolute amount of caffeine in the organic layer or the remaining amount of caffeine in the water phase, you have to consider the volumes.

By the way, there's an article by G.S. Klebanov, L.N. Mednikova, and A.D. Ovcharova on this topic: Extraction of caffeine from aqueous solutions, published in Pharmaceutical Chemistry Journal, 1967, 1, 221-223 (DOI).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.