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    • #8959
      Chemhalp
      Participant

      I desperatly need help in relation to working out these questions so I know how to do them in my exam tomorrow.

      A 2mM solution of tyrosine was diluted by adding 5ml to 15ml of water. GIven that the molar absorbance coefficient of tyrosine at 294.5nm is 2375, what absorbance would this diluted solution give in the spectrophotometer on a 1 cm path-length cuvette when read at 294.5nm?

      All I have worked out is that when 5ml of the 2mM solution was added to the 15ml of water you get 20ml with 2 mM

      So it should be 100 mM per litre or 0.1M per litre (0.1 M L-1)

      ————–

      Any help on working this question out so I can answer similar ones tomorrow will be dearly appreciated

    • #80835
      blcr11
      Participant

      You can either convert the molar extinction coefficient to a mM extinction, or work with everything in M, in any event, you’ve miscalculated your dilution unless you misspoke.

      You diluted 5 ml of a 0.002M solution with 15 ml of water. This gives a final concentration of 0.0005 M (5×0.002/20). The molar extinction is in units of A(294.5 nm) cm^(-1)/M.

      The expected absorbance of the solution in the cuvette (assuming a 1 cm path length) is:

      0.0005 M x 2375 A/M = 1.188 A(294.5 nm)

      This is a bit on the high side. You typically want to read the absorbance between 0.1-1 A, so a better dilution would be maybe 5 ml of a 2 mM solution + 45 ml water—or even 95 ml water would work.

    • #80841
      Chemhalp
      Participant

      Thankyou 🙂

      I have another similar question.

      A solution of tryptophan was diluted by adding 2 ml to 18 ml of water. When the absorbance of this duluted solution was measured in a spectrophotometer in a 1cm light path cuvette at 280 nm , it gave a value of 0.94.

      Given than the molar absorbance coefficient of tryptophan at 280 nm is 5225 what was the tryptophan concentration in the original undiluted solution.

      —-

      This is what I tried.

      0.94/5225 = 0.00018

      (x*2ml)/20) = 0.00018

      (x*2ml) – 0.00018 * 20

      x = 0.00018 * 10

      0.0018 M?

      Is that the correct answer?

    • #80848
      blcr11
      Participant

      Looks good to me.

      You don’t have to do it this way, but it’s faster to realize that your first calculation tells you the concentration of tryptophan in the cuvette. You made a ten-fold dilution of the stock solution to get to the solution in the cuvette, so the stock concentration must be 10 x the concentration in the cuvette. The way you worked the problem is fine and shows all the details, so you should stick with whichever way is most comfortable for you.

    • #80852
      Chemhalp
      Participant

      Good point, If i’d noticed I’d just have to go from

      (0.94/5225)*10

      right?

    • #80853
      blcr11
      Participant

      That’s right. You just need to understand where the dilution factor comes from. It will be different for different dilutions. You should get the same result if you do the "full" calculation, so if you’re not sure with the shortcut, by all means, do the full calculation. The more problems you’ve seen, the faster you’ll get with dilution factors until it almost becomes second nature.

    • #80856
      mith
      Participant

      another great tutorial from blcr *huzzah*

    • #80867
      Chemhalp
      Participant

      Ok so I just sat the exam and this is the question as best I can remember (I can’t remember the words, just the values sorry)

      1ml of tyrosine diluted with 9ml water
      0.45A
      expected is 2375.
      work out the concentration in the original undiluted solution.

      so this is what I wrote.

      (0.45/2375)*10 = 0.00189 M

      Is that correct? If so I’m over the moon with happiness and I owe it all to you bclr 😀

    • #80869
      blcr11
      Participant

      Looks like you’ve got the idea now. Congrats.

    • #80871
      blcr11
      Participant

      Just out of curiosity–no obligation to answer if you don’t want to–how would your answer change if, given the same extinction coefficient and observed absorbance in the cuvette, the dilution instead was: 1 ml tryptophan stock solution + 19 ml water; or how about 1 ml tryptophan stock solution + 99 ml of water?

    • #80872
      Chemhalp
      Participant

      With the same extinction coefficient from which one?

      I’ll just do the working out (or try to)

      1ml tryptophan stock with 19ml water is a 20 fold change, so you divide the observed absorbance by the extinction coefficient and multiply by 20?

      or (Observed absorbance/Extinction Coefficient)*20

      and the same but *100 if you have it diluted with the 99ml of water?

      I think anyway 😀

    • #80873
      blcr11
      Participant

      Pretty good! I was intending you to use the same extinction they gave you for the exam, 2375, but it’s the same calculation no matter which wavelength/extinction you’re using.

      How about something a little more difficult—but also quite practical? You’ve prepared a solution of tryptophan that should be about 0.01 M, but you need to know the concentration precicely in order to use it to prepare a substrate solution for an enzyme assay. To do that you need to make a suitable dilution of the stock solution so that the final expected absorbance in the cuvette will fall in the range of 0.1 to 1 A. Suggest a suitable dilution. Use the wavelength with the extinction of 2375 for the calculation, but you might also consider how your answer would change if you were to use the wavelength with the extinction of 5225.

      There’s many possible correct answers here. I also realize that my “acceptable” range of absorbance is conservative by today’s standards. Most manufacturers these days claim their instrument responses are linear to at least 1.5 A and even a few claim linearity to 3 A, but I’m a skeptic.

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