3
$\begingroup$

When building a calibration curve for quantification of some analyte using any spectroscopic technique, what is the purpose of subtracting the intensity of the blank from the intensity of each point of the curve?

I can build two calibration curves using the same measurements, one plotting the total intensity recorded at each point (F) and the other plotting the total intensity recorded minus the intensity when the concentration of the analyte is zero (F-F0), both against analyte concentration.

When making a linear regression for each curve, the slope will be the same and the difference between the intercepts will be the intensity of the blank (F0). If I try to quantify an analyte in a given sample with a different matrix than the one used in the standard (different solvent, for instance) maybe the "blank of the sample" so to speak would be different than the blank of the standards (F0). In that case, the result would be equally wrong, using either of the curves, because I would either be comparing the response of the sample-F0 to analytic curve based on F-F0 or the total response of the sample to analytic curve based on F.

It would make sense to me if I could, when analyzing an unknown sample, measure an exact blank of the sample, and use this measurement to compare with the analytical curve built using F-F0. Then I would be comparing two intensities associated only with the analyte, but that is't possible.

Am I missing something? Does subtracting the blank helps in any way when the objective is to quantify the analyte in samples?

I've checked two different undergraduate level analytical chemistry books and found no answer. They just say it is done this way.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ (+1) for a perceptive question! You do not have to subtract the blank for each point on the curve. The analytical blank, and how it should be estimated and dealt with, is an unexpectedly subtle issue that has eluded the determined harmonization efforts of ISO and IUPAC for years. I recommend looking at these four publications: 1. L.A. Currie, “Detection: International update, and some emerging di-lemmas involving calibration, the blank, and multiple detection decisions”, Chemom. Intell. Lab. Syst. 37 (1997) 151-181. Continued next comment. $\endgroup$
    – Ed V
    Jul 9 '20 at 2:24
  • 1
    $\begingroup$ 2. L.A. Currie, for IUPAC, “Nomenclature in evaluation of analytical methods including detection and quantification capabilities”, Pure Appl. Chem. 67 (1995) 1699-1723. IUPAC © 1995. 3. K. Danzer, L.A. Currie, for IUPAC, “Guidelines for Calibration in Analytical Chemistry”, Pure Appl. Chem. 70 (1998) 993-1014. IUPAC ©1998. 4. ISO 11843-2, “Capability of Detection - Part 2: Methodology in the linear calibration case” ISO, Genève, 2000. $\endgroup$
    – Ed V
    Jul 9 '20 at 2:24
  • $\begingroup$ Thank you very much for your comment, @EdV. I will be checking these publications as soon as possible. $\endgroup$
    – Blg90
    Jul 9 '20 at 4:03
2
$\begingroup$

Good to see interesting analytical chemistry questions after a long time.

I can give a qualitative example. Suppose you wish to collect the spectrum of a star. What time would be the best to carry out the measurement? You would agree that night time would be the best, because constant unwanted sunlight background is missing at the time of measurements. Subtraction of the blank serves a similar purpose, you wish to remove the background signal from your analytical signal so that you can assess the magnitude of your analytical signal. The blank also helps to serve to establish the minimum concentration that can be detected reliably. This is why blank subtraction is carried out

One of the commentators on your question has written a whole book on this subject.

You did not mention the analytical instrument you are using, it would have helped to discuss more. Anyway, calibration curve is another story. In this case, you certainly can make two calibration curves, and you are very correct, their slopes are the same. The sample concentration will turn out to be the same from both.

With blank correction, your equation should have a form y= mx with negligible intercept. Without blank correction, your equation has the form y=mx+c, where c is the intercept, the average blank reading.

I may ask another thought provoking question which is certainly not discussed anywhere. For example, what if I square all the signal intensity and then plot a calibration curve as a function of concentration. The calibration curve will still remain linear.

a) What will happen to the magnitude of slope? b) Will the calibration curve sensitivity increase? "Apparently, yes!" c) If the sensitivity increases by this simple mathematical manipulation, why shouldn't we do it?

Coming to the second part of the question:

If I try to quantify an analyte in a given sample with a different matrix than the one used in the standard (different solvent, for instance) maybe the "blank of the sample" so to speak would be different than the blank of the standards (F0).

In this case, you should not/ perhaps cannot use a simple calibration curve. One of the key assumptions of a calibration curve is that the matrix of the sample is the same as the standard. If you violate this rule, then we cannot apply this technique.

Consequently, you will have to use a technique called Standard Addition Method. It produces very accurate results, when the sample matrix is significantly different from the standards.

Improve this answer
$\endgroup$
2
  • $\begingroup$ "if I square all the signal intensity and then plot a calibration curve as a function of concentration. The calibration curve will still remain linear" - hm.. did you mean a function of concentration$^2$? "night time would be the best, because constant unwanted sunlight background is missing" - but what OP is asking is about "measure the signal w/o star, then measuring star signal during day, then subtract one from another" - which removes the systematic error, but it also increases sd(noise). So while it allows replacing $mx+c$ with $mx$ it seem to also decrease the precision. $\endgroup$
    – Stanislav Bashkyrtsev
    Aug 3 '20 at 9:14
  • $\begingroup$ Could you please elaborate on how blank subtraction helps establishing the minimum detectable concentration? $\endgroup$
    – Stanislav Bashkyrtsev
    Aug 3 '20 at 9:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.