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I am trying to grasp how internal standardization works in principle, and I am having some trouble wrapping my head around it. It is not the choosing of an IS I think about, orwhen it might benefit the analysis, but just the underlying principle.

I add a known amount of the IS to every sample: unknown samples, blank samples, calibration samples, and quality control samples. Each of these samples contain equal amounts of the internal standard (or as "equal" as our pipettes may give). Considering the calibration, when the calibration samples are analyzed, I see two peaks in the chromatogram: the analyte and the IS.

Let's say I ran calibration samples of 0, 1, 2, 5, 10, and 20 ppm, each one consiting 20 ppm internal standard. The analyte peaks will grow for each calibration sample, but the IS peaks remain more or less stationary. I construct the calibration curve by plotting the ratio of analyte peak area to IS peak area (or height), as a function of the analyte conentration in the calibration samples.

I have been told that there is a constant ratio involved, one that is used to correct analyte concentrations for loss of analyte, or varying sample introduction system. If the IS peak is lower, then the analyte concentration will be corrected upwards. This apparently is because this ratio I mentioned is constant, so when reducing the denominator, the nominator must increase. Mathematically, this is trivial; however, what is this mystical constant, and what is its value? Is that arbitrary? Can I choose this constant's value myself?

I am quite confused. Hopefully someone here can help my out!

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This ratio has nothing mystical, and will not work in every case.

So let's first comment on the reasons to use an internal standard. As you mention, when you inject multiple times solutions containing the same concentration of IS:

but the IS peaks remain more or less stationary

More or less is a major point here: this means that repeatability of the measurement process is an issue, when multiple samples with the same concentration lead to a variation in the response. If the accuracy required for the measurement is such that these small variations might not be an issue, there is no need for an IS. In chrmatography measurement, sources of variability include repeatability in the volume injected, detector sensitivity. If there is some sample preparation steps involved, loss of analyte (in transfer from one tube to the other, in the filtration steps etc...) can also lead to variations from one analysis to the other.

So how will an IS solve this variability issue? The basic point is the assulmption that these external variability issues will affect both the analyte and the IS in a similar way. So one has to look at the sources of errors. For instance, let's consider an error in the measurement of a sample volume (let's call it $\epsilon$ and assume it is close to 1) . We have:

$$m_{analyte} = V_{real} c_{analyte} = \epsilon V_{set} c_{analyte}$$ $$m_{IS} = V_{real} c_{IS} = \epsilon V_{set} c_{IS}$$

If the measurement system accurately measures $m_{analyte}$ and $m_{IS}$, it is quite obvious that by dividing both equations we can remove this variability parameter and we get:

$$R_{analyte/IS} = {m_{analyte} \over{m_{IS}}} = {c_{analyte} \over c_{IS}}$$

This is the point were the ratio comes out: taking a ratio allows to remove the variability parameter ($\epsilon$) in this case.

Now this is a bit too simplistic, and a measurement procedure is not as simple. The general concept will still be to try to factor out the variability parameter as a proportion of the measured value, and then to take a ratio which will remove this variability from the measured values.

If the IS is correctly chosen so that it behaves as the analyte through the entire analytical procedure (both in terms of physico-chemical properties, as well as in the response of the detector), and if the chromatographic properties as well as the detector are such that peak area is proportional to concentration for the analyte as for the IS, then one can write:

$$A_{analyte} = \epsilon(analyte) R_{method}(analyte) c_{analyte}$$ $$A_{IS} = \epsilon(IS) R_{method}(IS) c_{IS}$$

where $A_{analyte}$ and $A_{IS}$ are the measured peak areas for both analyte ans IS, $R_{method}(x)$ is the method response factor, which is dependent on each compound. In general the variability factor $\epsilon$ should depend on each compound, but under the above assumption of similar properties, one can assume that $\epsilon(analyte) \approx \epsilon(IS) \approx \epsilon$. Without IS, one would determine $R_{method}(analyte)$ through a calibration procedure. With an IS, one takes the ratio and the above equation can be written as:

$$R_{area} = {A_{analyte} \over A_{IS}} = {R_{method}(analyte) \over R_{method}(IS)} {c_{analyte} \over c_{IS}}$$

The ratio:

$$R_{analyte/IS} = {R_{method}(analyte) \over R_{method}}$$

is the ratio of the response factors of the method for analyte and IS. This is probably the ratio that is referred to. And it is not mystical at all, it is obtained through a standard calibration procedure with the IS.

One caution word to conclude this answer: I have tried to be quite precise in the answer. I have often times heard some people state that the use of an, internal standard is a general solution for analytical issues. Thus I want to insist on a warning: the solution is good as long as the assumptions stated above hold. If any one of these fails, then the whole analytical procedure will also fail. And of course, the use of an IS will only protect from errors after addition of the IS, and not from variability in the addition of the IS (you mention pipetting it in the analyte solution, which will limit the repeatability to that achieved by the pipettes used).

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  • $\begingroup$ I'll throw in one more point. Retention time is frequently used as a means to identify a particular component in various chromatographic techniques. Having a consistent retention time for the IS shows that the chromatography setup is consistent since there are a zillion things that can cause the retention time to change. $\endgroup$ – MaxW Feb 15 '16 at 18:50

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