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Topic: Isosbestic point  (Read 14649 times)

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Offline Babcock_Hall

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Re: Isosbestic point
« Reply #15 on: April 22, 2014, 11:45:06 AM »
Can you post the problem with a little bit of explanation?  I went to your link, but I did not find it easy to follow.

Offline Rutherford

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Re: Isosbestic point
« Reply #16 on: April 22, 2014, 12:41:17 PM »
In the pretext they explain the Beer-Lambert law basics. Then they say:
The change of an indicator color occurs due to its transformation from one colored form into another:
HIn ::equil:: H++In-
That can be used for the spectrophotometric determination of the acidity constant of the indicator. In the table below are presented the results of measurements of the absorbances A of a methyl orange solution (c0=1.8·10-5M, l=1.25cm) at different pH values (λ=610nm) Abosrption spectra of different light-absorbing forms of the indicator are shown in the figure:

...
Now here are the table and the spectrum.
And then they ask to calculate pKa of the indicator, ε, and pH interval of color transition.
The fourth question that gives me trouble is to determine the concentration of the indicator in its aqueous solution with pH=2.25, if at λгр A=2.213, ε(HIn)=2.2·104 and l=2.08cm.

Offline Babcock_Hall

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Re: Isosbestic point
« Reply #17 on: April 22, 2014, 01:24:39 PM »
The following theory was developed for an experiment using solutions of 4-NP in a series of buffers at different pH values.  I think it would also work for a situation where one titrated the chromophore with acid or base, as long as one made corrections for the change in volume during the titration, but perhaps someone else can weigh in on that.

The observed absorbance of any weak acid such as 4-NP at any pH is the sum of the absorbances of the neutral species (HB) and the anionic species (B-).  As you will see in this lab, the observed absorbance depends upon pH because the two species have different spectra.  Each form has its own molar absorptivity, which is pH-independent.  We use Beer's law to write:

Aobsd = AHB + AB- = εHB[HB] + εB-[B-]     (1)

By conservation of mass, we can write:

[Bt] = [HB] + [B-]     (2)

It is convenient to define a pH-dependent quantity, the apparent or observed molar absorptivity:

εH = (εHB[HB] + εB-[B-])/[Bt]     (3)

The observed absorbance is a function of the apparent molar absorptivity and the total concentration of conjugate acid and conjugate base:

Aobsd = εH[Bt]     (4)

The relative concentration of each species depends upon the [H+] and the acidity constant, Ka.  Therefore we can write:

εH = (εHB[H+] + εB-Ka)/(Ka + [H+])     (5)

An alternate form of equation (5) is

εH = (εHB + εB-•10(pH-pKa))/(10(pH-pKa) + 1)      (6)

One may fit equation (5) or (6) using nonlinear regression, which will estimate values of Ka, εHB and εB-.

With respect to 4-NP and equation 6, if one measures absorbance values at 400 nm and plots εH vs. pH, one will see a sigmoidal curve that rises with increasing pH.  The asymptotes are εHB and εB-, and the inflection point is pKa.  If one measures absorbance values at 317 nm and plots εH, then one will still get a sigmoidal curve, but it will fall with rising pH.

If one measures absorbance values at the isosbestic point (348 nm) one will get an uninformative straight line.  Putting it another way, suppose that I have a solution of a known total concentration of 4-NP at an unknown pH.  If I measure absorbance values at 348 nm, then I don't get any information on pH.  If I measure absorbance values at 400 nm or 317 nm, then I do get information on pH.  I'll try to write something more this evening if I have time.
« Last Edit: April 22, 2014, 01:38:14 PM by Babcock_Hall »

Offline Rutherford

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Re: Isosbestic point
« Reply #18 on: April 22, 2014, 02:04:07 PM »
That's okay and I saw different plot with ε against λ (I asked a problem earlier about one). In this problem absorbance is plotted against λ and three lines are represented: absorbance of HIn, absorbance of In- and absorbance of HIn+In-. In one point these three lines intercept (at λгр) and it can be clearly seen from the plot that the absorbance is the same for all three lines. And then I get back to:
A(HIn)=A(In-). They said that pH=2.25 which is different from pKa, meaning that the concentrations of HIn and In- are different and if A(HIn)=A(In-) the molar absorptivity for both species should be different. The only explanation would be that the point where the three lines intercept doesn't represent the isosbestic point. Isosbestic point is represented in plots where ε is plotted against λ. Is my reasoning correct?

Offline Big-Daddy

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Re: Isosbestic point
« Reply #19 on: April 22, 2014, 05:02:20 PM »
I'm a bit busy so I cannot try and address the problem properly (don't know the answer yet so would have to research - though I definitely want to know and I see your problem, it is non-trivial) but I can say that - this is the same question we discussed before from the Mendeleev book, where I said they had written one of the absorbance numbers wrong in the book. You don't need to define isosbestic point to solve this problem. Just assume equilibrium (after all acid-base equilibria are usually pretty quick and an indicator equilibrium has to be quicker than usual) and then apply Beer's Law. Exact solution is possible but obviously not what they are looking for - what approximations can you make?

For the "fourth question that gives (you) trouble", approximate that at such low pH only HIn is around; use A, l and ε to find [HIn] and then this is approximately identical to analytical concentration. pH range for colour transition of indicator is covered in one of Borek's online chemistry pages on indicators, not sure which but it's there.

Offline Rutherford

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Re: Isosbestic point
« Reply #20 on: April 23, 2014, 02:33:41 AM »
It is the same problem. I can't approximate that only HIn is present as the pH (2.25) doesn't differ so much from pKa (3.37). I got that [HIn]=4.84·10-5M (they got that this number represents the total concentration) and [In-]=3.67·10-6M, then co=5.2·10-4M. I think that their answer is wrong and that the interception point of the graph doesn't represent the isosbestic point, so ε(HIn)≠ε(In-)

Offline Borek

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Re: Isosbestic point
« Reply #21 on: April 23, 2014, 02:41:49 AM »
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline Big-Daddy

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Re: Isosbestic point
« Reply #22 on: April 23, 2014, 08:58:59 AM »
It is the same problem. I can't approximate that only HIn is present as the pH (2.25) doesn't differ so much from pKa (3.37). I got that [HIn]=4.84·10-5M (they got that this number represents the total concentration) and [In-]=3.67·10-6M, then co=5.2·10-4M. I think that their answer is wrong and that the interception point of the graph doesn't represent the isosbestic point, so ε(HIn)≠ε(In-)

Ok if pKInd=3.37 then this is too close. I had remembered it as 4.37.

I think you have overestimated the total concentration by a factor of 2? The total absorbance (=2.213) has to be the sum of the individual absorbances, and at this particular point we know the two absorbances are equal.

Offline Rutherford

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Re: Isosbestic point
« Reply #23 on: April 23, 2014, 09:33:02 AM »
You are right, it has to be the total absorbance. One line on the plot represents the absorbance of HIn at concentration co, the second line represents the absorbance of In- at concentration co. The third line on the plot represents the total absorbance of HIn and In- whose total concentration is co. So that point really is the isosbestic point. Too much headache from a single problem :D.

Offline Big-Daddy

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Re: Isosbestic point
« Reply #24 on: April 23, 2014, 10:41:48 AM »
side-point: you wrote co=5.2·10-4M before, I'm sure it was just a typo but in case anyone ever uses this for reference (  ::) ) I think it was meant to be co=2.6·10-5 M to 2sf.

You are right, it has to be the total absorbance. One line on the plot represents the absorbance of HIn at concentration co, the second line represents the absorbance of In- at concentration co. The third line on the plot represents the total absorbance of HIn and In- whose total concentration is co. So that point really is the isosbestic point. Too much headache from a single problem :D.

As I said I haven't done the research yet to define isosbestic point (I will try soon) but I can say that ε(HIn)≠ε(In-) at the wavelength when A(HIn)=A(In-).

You can calculate ε(In-) at this wavelength without much trouble, I think it was about 2.9·105 M-1·cm-1. That is not ε(HIn). In fact ε(In-) = ε(HIn) · [H+]/KInd. If the definition of isosbestic point is ε(HIn)≠ε(In-) then this is not the point.

BTW: did they get the answer wrong in the actual exam too or just the book?

Offline Rutherford

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Re: Isosbestic point
« Reply #25 on: April 23, 2014, 11:11:46 AM »
Actually their answer is right. I got confused. co=4.8·10-4M.
A(HIn)=A(In-) is true if the concentrations are the same.
They plotted the absorbances of both species separately at concentration co, and we can see that they intercept at a the isosbestic point where the molar absortivities are equal. The third line represents the absorbance of a solution containing both HIn and In-, where the sum of their concentrations is co. This line intercepts at the isosbestic point, too, but neither the concentrations, nor the absorbances of HIn and In- separately are equal for the solution of pH=2.25. The sum of their absorbances is equal to the absorbance of an equally molar solution containing only HIn or only In-.

Offline Big-Daddy

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Re: Isosbestic point
« Reply #26 on: April 23, 2014, 11:30:19 AM »
Actually their answer is right. I got confused. co=4.8·10-4M.

So you're saying that the graph really shows ε against λ rather than A against λ? Or, that it shows A against λ using the same concentration and length for each species? I can't see that suggested in the question...

Offline Rutherford

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Re: Isosbestic point
« Reply #27 on: April 23, 2014, 12:26:22 PM »
I mean the latter. It isn't suggested and that made me go the wrong path.

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