# How can I prove ascorbic acid degradation with Uv-vis spectroscopy?

I have to prove scientifically why a sodium ascorbate solution in water turns yellowish - reddish over time. I know ascorbic acid is supposed to oxidize into dehydroascorbic acid, and the latter then oxidizes into more subproducts... But I have to prove this by using UV-vis spectroscopy and whenever I analyze the yellowish solution the only peak I see is the ascorbic acid peak (around 262 nm). Why don't I see the dehydroascorbic acid peak (210 nm) or even other peaks related to the other subproducts ? Update: I found this graph on a published article and I was expecting to obtain something like it:

Below are the peaks of a fresh sodium ascorbate 10% w/w solution (orange peak) and a solution with the same concentration that has been aged for 50 days and has acquired a deep yellow color (green peak). Both of the solutions have been diluted to 1:80.000 in water for the UV-vis analysis. The ascorbic acid in the aged solution has obviously degraded in some way since the peak is lower, but why are there no new peaks regarding the degradation products ?

• Since the solution turns “yellowish-reddish” over time, that means it is absorbing light in the visual region of the spectrum, i.e., between roughly 400 and 700 nm. So what happens when you obtain spectra from 200 nm to 700 nm? How about running experiments with fresh and aged solutions and put those into your edited question? – Ed V May 31 at 11:52
• Also something may diminish. – Alchimista May 31 at 13:08
• Thank you for your answers, I edited my question with your suggestions ! Any more thoughts ? – Yakata Jun 5 at 9:59
• How about not diluting the (deep yellow colored) aged solution: put 3 mL in a cuvette and run the visible spectrum from 400 to 900 nm? Use 3 mL of water as the blank. If no significant difference, then the spectrometer might be in need of repair or re-calibration. – Ed V Jun 5 at 11:57
• The way the baseline of the green sample increases with decreasing wavelength suggests you've got some light scattering going on. Possibly aggregation or precipitation. – Andrew Jun 5 at 12:54

Some of the experimental details are missing, but if you recall

$$\mathrm{Abs} = \varepsilon \cdot c \cdot d$$

about the absorption depending on the molar extinction coefficient $$\varepsilon$$, the analyte concentration $$c$$, and the optical path length $$d$$, it could be due to a low concentration of the analyte.

Independent of the former, given the $$\pi$$ electron systems of the carbonyl groups are not in conjugation with each other, nor with a $$\ce{C=C}$$ double bond (as in ascorbic acid) it is quite possible that $$\varepsilon$$ is low; equally attenuating the recorded signal.

(composite from here and here)

You mention to be able to record a signal at $$\lambda = 262\,\mathrm{nm}$$. Thirdly, given you now equally look for a signal around $$\lambda = 210\,\mathrm{nm}$$, either the material of your sample cell, or / and the solvent of your analyte no longer is optically transparent enough for the given path length $$d$$ for this wavelength. (Did you check that the spectrometer's second lamp, the deuterium lamp, equally is running and stabilized, too?)

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• Than you very much for your detailed answer. None of those things had occurred to me. Nevertheless I recently edited my post with more details so feel free to add a coment if you have some more thought on it, thank you again !! – Yakata Jun 6 at 8:08