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I am in the lab and trying to perform the molybdenum blue test for phosphorus using a photometer. It appears common speak in colorimetric lab sheets that

Absorbance of a mix of two concentrations is the sum of the absorbances of the individual concentrations

This is often exemplified by statements such that: “You can zero your photometer on the reagent and subtract the absorbance of the sample to just measure the absorbance due to the reaction”. I tried to test if this logic is correct, but I think it is false and performed the following experiment by preparing 4 curettes to find out:

  • I zeroed my instrument on a cuvette with distilled water.
  • I prepared 1ml of the molybdenum reagent
  • I prepared 1ml of the molybdenum reagent to which I add 10 ul phosphorus sample
  • I prepared 1ml of distilled water to which I add 10 ul of phosphorus sample.

I measured the absorbance of the four cuvettes within about 1350 milliseconds of mixing and they were 0, 0.172, 0.072, 0.402. I measured again after 10 seconds and got similar absorbances.

As 0.172 + 0.072 = 0.244 and not 0.402, I conclude that the logic cited above is false.

QUESTION: Am I correct in jumping to this conclusion? What is the reason that the mixed substance does not have the absorbance of the sum of the individual concentrations?

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  • $\begingroup$ As fas as I can imagine, your 4th sample is a simple dilution of a nearly colorless reagent. As a consequence, it does not have the described absorbance, because a simple dilution should produce an absorbance next to zero. How do you explain this ? Anyway it should not be the sum of anyhow previous solutions, because the molybdenum reagent does react with phosphorous producing a chemical reaction. I presume that instead of distilled water, the 4th sample was made with molybdenum reagent. Am I wrong ? $\endgroup$
    – Maurice
    Commented Aug 27, 2021 at 18:46
  • $\begingroup$ Thanks for having a look. “Anyway it should not be the sum of anyhow previous solutions, because the molybdenum reagent does react with phosphorous producing a chemical reaction.” How fast must I mix them to get the absorbance before the reaction really gets going? The fourth solution has the colour of the phosphorus solution but with colour intensity divided by 100. The phosphorus solution is blood serum, so the solution is yellow. The absorbance on number 4 is correct, I checked it unfortunately $\endgroup$
    – Mikkel Rev
    Commented Aug 27, 2021 at 18:56
  • $\begingroup$ What are the people in the link above talking about, if it is not this? $\endgroup$
    – Mikkel Rev
    Commented Aug 27, 2021 at 18:59
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    $\begingroup$ The statement quoted about absorbance is quite correct and can be shown to be so from the Beer Lambert law. If species react then of course things are different as new species are produced that may absorb where you measure or not absorb at all. $\endgroup$
    – porphyrin
    Commented Aug 27, 2021 at 19:11
  • $\begingroup$ The reaction between reagent and sample is rather fast. You cannot hope to mix the liquid and to measure the absorbance before the reaction. $\endgroup$
    – Maurice
    Commented Aug 27, 2021 at 19:22

2 Answers 2

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There is a fallacy in your experiment. Yes, absorbance is indeed additive tha low concentrations provided the reagents do not react. This sum property allows simultaneous determination or two or more components in a solution using the additive property of Beer's law.

Note: When you zero the photometer you are not taking absorbance into account but a lot of other things too. Reflection, refraction, scattering etc by cuvets.

For that do a simple experiment: Zero the photometer when you have nothing in the spectrophotometer (i.e., with air only). Now put an empty cuvet back. Do you still see zero absorbance? Mostly likely not, and you know very well that the cuvet is not absorbance any light at light. It should be transparent.

In order to test the absorbance idea, take two dilute dyes and then check the sum after mixing and taking dilution into account.

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Is the following statement true?

Absorbance of a mix of two concentrations is the sum of the absorbances of the individual concentrations.

My answer is need more information to give a correct answer. The statement is not clear about few things:

  1. It is not clear about the mixture of two concentrations are of the solutions of same reagents. If they are the solutions of same reagents, the sum of absorbance of each does not match the absorbance of the mixture of them in any ratio. For example, suppose that the concentration of one of the two reagents is $\pu{1.25E-05 M}$ of molybdenum reagent, which displays $0.172$ uv/vis absorbance at certain wavelength (similar to OP's second sample). Now, if the second sample has the concentration of $\pu{2.05E-05 M}$ of the same molybdenum reagent, it would show $0.282$ uv/vis absorbance at same wavelength. However, if you mix $\pu{5.00 mL}$ of first sample with $\pu{5.00 mL}$ of second sample, the uv/vis absorbance of the mixture at the same wavelength is not the sum of individual samples, which is $0.172 + 0.282 = 0.454$. Instead, it is the average of individual absorbance, which is $\frac{0.172 + 0.282}{2} = 0.227$. This is simply because when you add two solutions together, you are diluting each other. You can prove this by yourself using the Beer Lambert law: $A = \epsilon \ell c$. The important fact here is your solutions should be made of same compounds and hence have same $\epsilon$. The measurement should be done at same wavelength, preferably at $\lambda_\mathrm{max}$.
  2. If the mixture of two concentrations are of the solutions of different reagents, the given statement is not true at all. If they are the solutions of different reagents, each solution have a different $\lambda_\mathrm{max}$ and $\epsilon$. Thus, the sum of absorbance of each does not match the absorbance of the mixture of them in any ratio. This is the major reason why the absorbance of your sample 3 does not match the sum of absorbance of sample 2 and 4. One solution (sample 2) is the pure molybdenum reagent while the second solution (sample 4) is the pure phosphorus reagent. The sample 3, the mixture of samples 2 and 4, is different even if they did not react each other. The things become worst because these two reagent react each other to give the third reagent, which also have totally different $\lambda_\mathrm{max}$ and $\epsilon$!
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