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### Topic: Beer-Lambert Law  (Read 4316 times)

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#### cyion

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##### Beer-Lambert Law
« on: January 21, 2024, 05:38:51 AM »
I have a question where I have absorbance as 1 and 0.7 when the concentration is 0.2 and 0.1M. What I would normally do is plot absorbance against concentration and then multiply the gradient by the path length (0.1dm) to get the extinction coefficient in mol^-1 dm^2. Instead I was asked to find the extinction coefficient without a graph. Would I be correct in using the equation (absorbance1 - absorbance2)/(conc1 - conc2) * 0.1 as I can combine the equations to get the difference between the 2 points as thats essentially what the gradient is?

FYI the equation is A = cεl  (A=absorbance, c=conc, ε=extinction coefficient, l =path length)
« Last Edit: January 21, 2024, 06:30:56 AM by cyion »

#### Borek

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##### Re: Beer-Lambert Law
« Reply #1 on: January 21, 2024, 08:20:16 AM »
What I would normally do is plot absorbance against concentration and then multiply the gradient

And where would you get the gradient from?
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#### cyion

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##### Re: Beer-Lambert Law
« Reply #2 on: January 21, 2024, 12:19:39 PM »
Sorry my mistake I would plot absorbance against concentration to find the gradient and then multiply by the path length

#### Hunter2

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##### Re: Beer-Lambert Law
« Reply #3 on: January 21, 2024, 12:26:30 PM »
That  is only repeatation, what is written above.

#### cyion

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##### Re: Beer-Lambert Law
« Reply #4 on: January 21, 2024, 12:56:08 PM »
Oh so absorbance is on the y axis and concentration is on the x so the gradient is just the change in absorbance over the change in concentration. (1-0.7)/(0.2-0.1) which gives a gradient of 3

#### Hunter2

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##### Re: Beer-Lambert Law
« Reply #5 on: January 21, 2024, 01:39:11 PM »
Yes, but if you do so you would have A = 3*l* c
Let say you have  l = 1 cm cuvette, then its A = 3*c
So if you put in c = 0,1 mol/l  then you get A = 0,3  and not 0,7 and for 0,2 mol/l you would get  for A 0,6 and not 1

The equation doesn't fulfill Lambert Beer. But if you calculate the straight line as
y = mx+ b you get a blank.
The result is y = 3x+ 0.4
y= 3*0.1 + 0.4 = 0.7
y =3*0.2 + 0.4 = 1
« Last Edit: January 21, 2024, 02:48:08 PM by Hunter2 »

#### cyion

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##### Re: Beer-Lambert Law
« Reply #6 on: January 21, 2024, 04:32:52 PM »
Sorry a little confused so if I need to include the y intercept (0.4) in order for the equation to work to help find either the concentration or absorbance. Does this mean it won't help me find the extinction coefficient. Also I'm not sure if it helps but in the solution there is another chromophore that has absorbance 1 and 0.7 when its concentration is 0.4 and 0.3. As it's a separate chromophore I thought that I would be able to find the extinction coefficient for one and then apply the same logic to the other.

#### Hunter2

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##### Re: Beer-Lambert Law
« Reply #7 on: January 21, 2024, 04:44:17 PM »
I am not sure what chemical you are measure. But to solve that problem you have to run the matrix of  your system to get rid of the blank. It must be Zero.
In your case the absorbance has to be 0.7 - 0.4 = 0.3 for 0.1 mol/l and 1-0.4 = 0.6 for 0.2 mol/l
Then the extinktions coefficent would be 3.

For the second chromophore  its the same way.
Also get a spectrum and find maybe different wavelength to measure.

In your second case you get
y = 3*x - 0.2
y = 3*0.4 - 0.2 = 1
y = 3*0.3 - 0.2 = 0.7

« Last Edit: January 21, 2024, 04:57:33 PM by Hunter2 »

#### cyion

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##### Re: Beer-Lambert Law
« Reply #8 on: January 21, 2024, 05:54:09 PM »
Oh I'm just a first year at uni and had an algebra worksheet so I didn't know about all the matrix stuff. thanks for your help I do appreciate it. In the answers it gave me extinction coefficients of 10 and 20 mol^-1 dm^2 for P and Q, but I'm sure they're wrong as your reasoning seems so much more logical and clearer. Below is the original question just incase I've missed something completely stupid, cheers tho )

A solution contains two chromophores, P and Q with extinction coefficient of εP and εQ, respectively. The pathlength l is constant at 0.01 m but the concentrations of P and Q vary between the two solutions. The absorbance is 1.0 when cP = 0.2 mol dm–3 and cQ = 0.4 mol dm–3 and drops to 0.7 when cP is 0.1 mol dm–3 and cQ = 0.3 mol dm–3. Determine values of εP and εQ

#### Borek

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##### Re: Beer-Lambert Law
« Reply #9 on: January 21, 2024, 06:10:14 PM »
Thread so far was seriously off, as you incorrectly stated the problem in your first post.

Absorbance of the mixture is a SUM of individual absorbances*. Write two equations - one for the first case (total absorbance of 1.) one for the second case (total absorbance of 0.7). You have two equations in two unknowns. Just solve.

*that's in general only an approximation, but definitely one you are expected to use.
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#### Hunter2

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##### Re: Beer-Lambert Law
« Reply #10 on: January 21, 2024, 06:18:52 PM »
Well thats true.
Its different exercise.

#### cyion

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##### Re: Beer-Lambert Law
« Reply #11 on: January 21, 2024, 06:26:37 PM »
Oh, sorry for wasting your time. if my 2 unknowns are the 2 extinction coefficients do you know how I would be able to set up a simultaneous equation to solve it. I was thinking of using substitution but since the extinction coefficients are different I'm not entirely sure how to work 1 equation into the other.

#### Hunter2

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##### Re: Beer-Lambert Law
« Reply #12 on: January 22, 2024, 01:26:52 AM »
Generally A = (εPcP + εQcQ)*l
This you have to solve with your values.

#### cyion

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##### Re: Beer-Lambert Law
« Reply #13 on: January 22, 2024, 05:20:08 PM »
yep I figured the total absorbance was the sum of the sum of chromophore P and Q absorbance. Once I had that I just has to solve the equation via elimination to get 1 of the chromophores values and then subbed it into one of the equations to get the other. Thanks for all your *delete me*