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Topic: Absorbance calculated from pH - need help  (Read 4727 times)

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Offline Borek

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Re: Absorbance calculated from pH - need help
« Reply #15 on: April 25, 2020, 04:45:11 PM »
1) So yes, the pKa is close to the pH (4,7) of the solution of the unknown absorbance.

What does it tell you about ratio of acid and conjugate base? Take a look at the Henderson-Hasselbalch equation.

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2) To be honest, im not certailny sure, but i would guess that both conjugate base and conjugate acid both absorb light.

Yes. Do they both have the same molar absorption coefficient?

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So the concentration of [A-] is the same as the concentration of [H+]

No. That would be true for a solution that contains only the acid, in this case pH is forced with something else. Hint: once again, take a look at the HH equation.
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Offline Adam_ko

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Re: Absorbance calculated from pH - need help
« Reply #16 on: April 25, 2020, 04:53:37 PM »
1) so the ratio is close to one, so the logarithm(1)=0
    and if the logarith=0 than pKa = pH

2) No, they dont have the same molar absorption coefficient

3) so i should  -> [A-] = (Ka - [H+])*c ? 

Offline Borek

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Re: Absorbance calculated from pH - need help
« Reply #17 on: April 25, 2020, 05:07:45 PM »
1) so the ratio is close to one, so the logarithm(1)=0

"Close" is perhaps an overstatement, but yes, definitely closer to 1 than at other pH values given.

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2) No, they dont have the same molar absorption coefficient

What does it tell you about the total absorption of a solution containing both forms? What do you need to calculate it?

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3) so i should  -> [A-] = (Ka - [H+])*c ?

You might be on the right track, but it is definitely wrong. Check your math.
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Offline Adam_ko

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Re: Absorbance calculated from pH - need help
« Reply #18 on: April 25, 2020, 05:19:47 PM »
2) The total absorption of solution is sum of partial absorptions of compounds in it? A = A1 + A2 + ...
     So i need to know the concentrations and molar absorption coefficient for both of the absorbances
3) pH = pKa + log(A-/HA) -> [H+]= Ka + [A-]/HA -> ([H+]-Ka) * HA = [A-]

Offline Borek

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Re: Absorbance calculated from pH - need help
« Reply #19 on: April 25, 2020, 05:28:08 PM »
2) The total absorption of solution is sum of partial absorptions of compounds in it? A = A1 + A2 + ...
     So i need to know the concentrations and molar absorption coefficient for both of the absorbances

Yes.

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3) pH = pKa + log(A-/HA) -> [H+]= Ka + [A-]/HA -> ([H+]-Ka) * HA = [A-]

No, that's not how logs work.

It can be easier for you to start with just Ka definition.
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Offline Adam_ko

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Re: Absorbance calculated from pH - need help
« Reply #20 on: April 25, 2020, 05:39:58 PM »
3)

Ka = [H+][A-]/[AH]

One question out of the topic, how can i make fractions?

Offline Borek

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Re: Absorbance calculated from pH - need help
« Reply #21 on: April 25, 2020, 05:43:11 PM »
With LaTeX (click About post formatting... link above the edit field for details).
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