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Offline Rutherford

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Isosbestic point
« on: April 21, 2014, 04:52:24 AM »
When looking at an absorbance graph of an indicator, where A is plotted against λ, at one point the absorbances are equal and this point I think is called isosbestic point. Now, I found a definition where it is said that at the isosbestic point the two species would have the same molar absorptivity. Same absorbance and same molar absorptivity imply same concentrations of the two species, but that doesn't seem right. What if the pH of the solution is very different from Ka of the indicator? Where am I wrong at?
I don't understand how do the molar absorptivity coefficients need to be equal.
« Last Edit: April 21, 2014, 06:14:53 AM by Raderford »

Offline Hunter2

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Re: Isosbestic point
« Reply #1 on: April 21, 2014, 07:42:06 AM »
Isobestic point is found as an intersection of two graphs given from the same substance (or reaction of it)  in different stage (pH, Neutral or Ion,etc.)

http://nzic.org.nz/CiNZ/articles/2012/CiNZ%20OCT%202012_Sanjeev.pdf

Offline Rutherford

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Re: Isosbestic point
« Reply #2 on: April 21, 2014, 08:13:13 AM »
So if HA and A- absorbances intercept at the isosbestic point it doesn't mean that their molar absorptivity coefficients are equal?

Offline Hunter2

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Re: Isosbestic point
« Reply #3 on: April 21, 2014, 08:21:07 AM »
You have a substance HA => H+ + A-  Equilibrium is 30% left and 70% on right side. Of course the concentrations cannot be the same. So the molar absorptivity coefficient must be different as well.

Offline Rutherford

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Re: Isosbestic point
« Reply #4 on: April 21, 2014, 08:42:19 AM »
Okay, thanks. I got confused because in one problem solution it was stated differently.

Offline Babcock_Hall

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Re: Isosbestic point
« Reply #5 on: April 21, 2014, 01:44:09 PM »
"Same absorbance and same molar absorptivity imply same concentrations of the two species..."  This is a misunderstanding of what is meant (I would delete the word "absorbance" from your sentence).  Consider 4-nitrophenol/4-nitrophenolate for convenience.  At low wavelengths, the molar absorptivity of the conjugate acid is greater than the conjugate base, and at higher wavelengths (close to 400 nm) the opposite is true.  There is one wavelength where the two molar absorptivities are equal.

Offline Rutherford

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Re: Isosbestic point
« Reply #6 on: April 21, 2014, 02:00:00 PM »
I said that A is plotted against λ and in one point the two curves (for HA and A-) intercept, so the absorbance is equal here, but that doesn't mean that molar absorptivities are equal, as then the concentration of HA and A- need to be equal, but pH is different from Ka.

Offline Babcock_Hall

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Re: Isosbestic point
« Reply #7 on: April 21, 2014, 03:29:10 PM »
isosbestic point
Wavelength, wavenumber or frequency at which the total absorbance of a sample does not change during a chemical reaction or a physical change of the sample.
Notes:
A simple example occurs when one molecular entity is converted into another that has the same molar absorption coefficient at a given wavelength. As long as the sum of the concentrations of the two molecular entities in the solution is held constant there will be no change in absorbance at this wavelength  as the ratio of the concentrations of the two entities is varied.  http://goldbook.iupac.org/I03310.html


Offline Rutherford

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Re: Isosbestic point
« Reply #8 on: April 21, 2014, 03:52:52 PM »
The conclusion of that definition would be that the molar absorptivities are the same for both species, but the concentrations and absorbances of those species aren't equal, right?

Offline Rutherford

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Re: Isosbestic point
« Reply #9 on: April 21, 2014, 04:04:43 PM »
Here is the problem that confused me: http://www.chem.msu.ru/rus/olimpiad/olimp2003/Itur/welcome.html (click on Задача 1). On the plot you can see that they put A against λ, but now I think that it should be ε against λ. Am I right?

Offline Babcock_Hall

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Re: Isosbestic point
« Reply #10 on: April 21, 2014, 09:01:40 PM »
The conclusion of that definition would be that the molar absorptivities are the same for both species, but the concentrations and absorbances of those species aren't equal, right?
I think whether or not the concentrations are equal is only answerable on a case-by-case basis.  Consider the following experiment:  Take a series of 10 buffers of various pH values between 4 and 10, and put an equal mass of 4-nitrophenol (NP) into each of ten samples (so all are at the same concentration).  Now take the spectra and overlay them.  They will all cross at the isosbestic point.  The graph at this link gives you some idea of what it would look like, even though they only overlay two spectra.  Wikipedia give the isosbestic point of 4-NP and its conjugate base as 348 nm, which sounds reasonable (I have not measured it).  http://chemwiki.ucdavis.edu/Physical_Chemistry/Kinetics/Reaction_Rates/Experimental_Determination_of_Kinetcs/Spectrophotometry

This link also gives some isosbestic points of related phenols and has one or two good graphs:  http://nvlpubs.nist.gov/nistpubs/jres/71A/jresv71An5p385_A1b.pdf

Offline Babcock_Hall

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Re: Isosbestic point
« Reply #11 on: April 21, 2014, 09:05:20 PM »
Here is the problem that confused me: http://www.chem.msu.ru/rus/olimpiad/olimp2003/Itur/welcome.html (click on Задача 1). On the plot you can see that they put A against λ, but now I think that it should be ε against λ. Am I right?
I am not certain.  If one makes up a series of solutions at the same concentration but different pH values, then I don't see why it matters which one is plotted.  In other words, the plots of A or ε are basically the same thing with a different y-axis.

Offline Rutherford

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Re: Isosbestic point
« Reply #12 on: April 22, 2014, 02:54:03 AM »
If they are the same, then:
A(HA)=A(A-) (as seen from the plot), so ε(HA)·c(HA)·l=ε(A-)·c(A-)·l, then if ε is equal for both:
ε(HA)·c(HA)·l=ε(HA)·c(A-)·l, and c(HA)=c(A-), which can't be true if the pH differs that Ka. Still confused.

Offline Babcock_Hall

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Re: Isosbestic point
« Reply #13 on: April 22, 2014, 09:24:59 AM »
If they are the same, then:
A(HA)=A(A-) (as seen from the plot), so ε(HA)·c(HA)·l=ε(A-)·c(A-)·l, then if ε is equal for both:
ε(HA)·c(HA)·l=ε(HA)·c(A-)·l, and c(HA)=c(A-), which can't be true if the pH differs that Ka. Still confused.
Your first equation is not true.  I will try to return this evening to say more.

Offline Rutherford

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Re: Isosbestic point
« Reply #14 on: April 22, 2014, 10:39:43 AM »
It has to be by looking at the plot in the problem I posted.

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