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I would like to draw a comparison of the sensitivity between several spectrometers. Therefor I measure the absorption of a serial dilution of potassium hydrogen phthalate (KHP). The idea is to compare the theoretical absorption with the measured absorption.

To calculate the theoretical absorption I need the molar absorption coefficient of KHP at 254 nm wavelength. Unfortunately I haven't found anything till now.

Does anyone know where I can find this coefficient?

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    $\begingroup$ Why use KDP? Benzene or naphthalene and many other stable compounds have well known extinction coefficients vs wavelength. $\endgroup$ – porphyrin May 17 at 10:20
  • $\begingroup$ @porphyrin Maybe the manufacturer has some documented arcane calibration procedure involving KHP that has to be followed for accreditation, but this just my speculation. $\endgroup$ – andselisk May 17 at 10:22
  • $\begingroup$ I use KHP because it's used for calibration. $\endgroup$ – Helen May 20 at 8:49
  • $\begingroup$ According to Burke et al. (R.W. Burke, E.R. Deardorff, O. Menis, Liquid Absorbance Standards, J. of Research of the National Bureau of Standards - A, Physics and Chemistry, 76A, Sept. - Oct. 1972, 469-482), solutions of KHP have "been used as a spectral standard in the comparative evaluation of spectrophotometers" (p. 478). But they say that absorbance is pH dependent and 262 nm is the shortest wavelength they apparently used. So I would do what @porphyrin suggested, assuming that degree of freedom. $\endgroup$ – Ed V Jul 18 at 20:17
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I would imagine that you could just use one of the spectrophotometers you have available to determine the molar absorptivity constant experimentally. After all, the molar absortivity constant is simply the constant which equates the terms (those being Absorbance and Concentration) which are already proportional. So, using something like the beer-lambert law...

$$A = \epsilon b C$$

...make a solution of a known concentration, measure its absorbance at your wavelength of interest, and solve for $\epsilon$.

$$\epsilon = \frac{A}{b C}$$

You might consider doing this a number of times to get an average value and thus minimize the random error associated with your constant.

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    $\begingroup$ This is a very good answer, but it can be even better with a little tweaking! From the reference I gave in my comment above, $\epsilon$ is roughly 1400 L/mole cm at 254 nm. A 0.5 mM KHP solution in water gives an absorbance of about 0.7. So how about suggesting that the OP carefully make a KHP solution of about 0.5 mM, measure the pH and then do as you say in your answer? Averaging a few absorbances is right. And then OP can experimentally get the actual $\epsilon$ for use in calibrations. They also need to measure pH to see if anything unexpected is happening. Upvoting now! $\endgroup$ – Ed V Jul 19 at 1:35
  • $\begingroup$ I get the logic behind noting the pH of your solutions, as Absorbance can change as a function of pH, but I don't know how necessary that piece of data is if the system we're investigating is the actual spectrophotometer instead of the chemistry of the sample. Actually, now that I'm thinking about it, I'm not even sure if you need the $\epsilon$ value if you're comparing machines; since $\epsilon$ is constant then just rearrange BL such that $\frac{A_1}{C_1} = \frac{A_2}{C_2}$ and use the same solution such that $C_2 = C_1$. That's all you need to compare results I would suspect. $\endgroup$ – Michael Green Jul 19 at 12:47
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    $\begingroup$ Well, the OP says she wants to compare the theoretical and experimental absorbances, so the theoretical one depends on just what the molar absorptivity happens to be. If this was just a comparison of two spectrometers, then no issue. $\endgroup$ – Ed V Jul 19 at 12:52
  • $\begingroup$ Ya, I guess I'm not entirely tracking what she means when she says 'theoretical absorbance'. Like, if $\epsilon$ is determined experimentally, then 'theoretical absorbance' just means the correlation associated with the machine from which you determined $\epsilon$, unless there's a first-principles method for determining $\epsilon$ that I'm not familiar with. And at that point, you don't need $\epsilon$ since you have $A_i$ of the spectrophotometer and ideally $C_i$ as well. $\endgroup$ – Michael Green Jul 19 at 12:57

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