6
$\begingroup$

Take aspirin (acetylsalicylic acid) and hydrolyze it with NaOH. This results in sodium acetylsalicylate and water. I think the sodium is held on with an ionic bond, so it is really salicylate dianions that are the important thing in the solution.

This is then diluted to several different concentrations. Using UV-VIS, the absorbance of these solutions are measured at 295 nm. Then a calibration line is plotted, that is, absorbance verse concentration. It will be a (pretty much) straight line according to the Beer–Lambert law. You can use this calibration line to calculate the concentration of a similar, but unknown, solution from its absorbance.

Why was 295 nm a good choice of wavelength to measure at?

The only idea I currently have is that at 295 nm there is a distinct difference between the absorbance of acetylsalicylic acid and salicylic acid. Apparently, the salicylic acid has a strong response at 295 nm, whereas the acetylsalicylic acid does not. However, I am not sure this is exactly relevant, since I'm not sure I have salicylic acid in my solution.

If I am on the right track here, then I have a follow-up question:

Apparently the lambda-max for salicylic acid is 302 nm. So, why would 295 nm be used in this experiment instead of 302 nm? Or, would anything between, say, 295 nm and 310 nm be just as good?

$\endgroup$
4
  • 2
    $\begingroup$ Searching for “aspirin absorbance spectrum” results in spectra showing 296 nm as the peak absorbance wavelength for salicylic acid and 229 nm for aspirin. Try it! And the peaks are broad, so being off a couple of nm just results in slightly lower sensitivity. $\endgroup$
    – Ed V
    Commented Apr 16, 2021 at 21:15
  • $\begingroup$ What is "UV-VIS"? Ultraviolet and visual [wavelengths]? Or the name of an instrument or a particular measurement method? $\endgroup$ Commented Apr 17, 2021 at 5:22
  • 1
    $\begingroup$ @PeterMortensen It is an analytical chemistry technique based on the absorption spectra of ultraviolet and visible light: en.wikipedia.org/wiki/Ultraviolet%E2%80%93visible_spectroscopy $\endgroup$
    – Cody Gray
    Commented Apr 17, 2021 at 5:57
  • 2
    $\begingroup$ Please consider giving the green checkmark (or upvote or both) to the most helpful of the posted answers. It encourages people to put some thought and time into crafting answers that are factually correct, relevant, understandable and likely to be of benefit to those, in future, who encounter the question and accepted answer. It is a small reward for those who volunteer their considerable time, effort and experience to aid others and they might well look favorably at future questions from the same person. Thanks for considering this! $\endgroup$
    – Ed V
    Commented Apr 17, 2021 at 15:08

1 Answer 1

7
$\begingroup$

I think this was basically an educational experiment. Your analytical method assumes that the hydrolysis was complete (this is the key requirement of a successful calibration curve). Your experiment did not include a test to assess the completion of hydrolysis. Acetylsalicylic acid begins to hydrolyze in alkaline solution to salicylic acid and basically it is an indirect method of detection.

You want to analyze acetylsalicylic acid but you are detecting salicylic acid. Guess where does this method fail? It will fail if you had a mixture of acetylsalicylic acid and salicylic acid in the beginning and you were asked to quantify both.

As EdV explained to you, being 1 nm off from the lambda max is not a big deal in UV-Vis spectra of solutions. Even the monochromator may be slightly off!

spectra

$\endgroup$
3
  • 2
    $\begingroup$ This is a good answer, but I had wanted to extend it with Woodward's rules. They seem to explain the aspirin absorption well enough, but unfortunately I didn't think of the reason why the salicylic acid absorption is so much higher. $\endgroup$ Commented Apr 17, 2021 at 1:19
  • 1
    $\begingroup$ Good point. I did not try to apply Woodward's rule. Your point could be a separate question. $\endgroup$
    – ACR
    Commented Apr 17, 2021 at 4:22
  • $\begingroup$ thank you for the great answers everyone! that helped a lot. $\endgroup$
    – fred_smith
    Commented Apr 17, 2021 at 15:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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