I was looking through the literature when I found this source:


Which states that the equilibrium constant Khyd is:

Khyd = $\frac{[hydrate]}{[carbonyl]}$ = $\frac{\epsilon^{\mathrm{W}}_{\mathrm{0}} - \epsilon^{\mathrm{W}}}{\epsilon^{\mathrm{W}} }$

Where $\epsilon^{\mathrm{W}}$ is the molar extinction coefficient measured under conditions of hydration in water solvent and $\epsilon^{\mathrm{W}}_{\mathrm{0}}$ is the molar extinction coefficient of the carbonyl in the absence of hydration in water solvent, usually taken as the molar extinction coefficient in cyclohexane.

How did the researchers get to the above equation in terms of the molar absorption coefficients?

As [hydrate] is equal to [carbonyl]0 - [carbonyl], the equation turns to:

Khyd = $\frac{[carbonyl]_{\mathrm{0}} -[carbonyl]}{[carbonyl]}$

However the Beer-Lambert law states that A = c$\epsilon$l.

Shouldn't this mean that Hhyd is actually proportional to a fraction containing the $\frac{1}{\epsilon}$ values?

EDIT: My current working is shown in the picture below:

  • $\begingroup$ The assumption discussed by the paper (which they argue should be discarded because it is not a good one) is that the absorption of carbonyl species is independent of solvent. That is the same as saying the concentration of carbonyl is uniquely determined by absorbance. The other assumption is that the only thing that affects absorbance of a compound is the extent to which its hydrated: hydrated carbonyls are assumed not to absorb at all, and apparent per-molecule absorbance of carbonyl compounds in water is thus lowered by an amount proportional to the equilibrium constant. $\endgroup$ – Curt F. Apr 18 at 18:17
  • $\begingroup$ Under those assumptions, there is no real difference between $A$ and $\epsilon$, they are both direct measures of the concentration of carbonyl in a solution. They key is that the units of $\epsilon$ are L/mol/cm, where mol is mole of total carbonyl containing compound. $\endgroup$ – Curt F. Apr 18 at 18:19

The law is not as you state although it often gets reported as this. The 'A' you quote is the optical density. The Beer-Lambert law is $I_{tr}=I_0e^{-\epsilon_\lambda [C]L}$ where $I_{tr}$ is the intensity of transmitted light for a molecule at concentration [C] at wavelength $\lambda$ and cell path length L. $\epsilon_\lambda$ is the extinction coefficient at wavelength $\lambda$.

From the definition you can see where the expression you ask about comes from even with their different notation; it looks as if $w=-{\epsilon_\lambda [C]L}$ where ${\epsilon_\lambda [C]L}$ is the optical density.

  • $\begingroup$ I'm afraid I still don't follow. In your last equation, what is w, and why is it the negtative of the optical density? Additionally, doesn't it still imply that the extinction coefficient at a given wavelength is still inversely proportional to the concentration of the species? Working through the maths I get the relation: (e(water, measured) - e0) / e0, where e0 is the extinction coefficient for the initial concentration and e(water, measured) is equilivent to eW in my post. $\endgroup$ – K.P. Nov 19 '18 at 11:54
  • $\begingroup$ The equation you quote has $\epsilon^{+w}$ and from Beers law it should be $e^{-\epsilon [C]L}$ so $-w$, it may be my misunderstanding from what you wrote I has assume that it was a typo and $\epsilon^{w}$ should actually be $e^w$. $\endgroup$ – porphyrin Nov 19 '18 at 12:08
  • $\begingroup$ Ah that’s my fault - I didn’t define the terms in my equation because they were defined in the link. I’ve now edited it to make it more clear - the terms in the equation are all extinction coefficients; the W superscript says that they are the extinction coefficients in water solvent. This means when I wrote e in my first reply to you I meant epsilon, the extinction coefficients. $\endgroup$ – K.P. Nov 19 '18 at 13:34
  • $\begingroup$ I realized that in my calculations I erroneously cancelled Ao and A with each other. Now working through the mathematics I can't seem to get the desired equation at all. $\endgroup$ – K.P. Nov 19 '18 at 16:38

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.