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I have been taking many lab courses in my second year of university and I have come across the technique of creating a calibration curve quite a bit. One thing I noticed in a lot of my plots is that many of them do not intersect at the origin (0,0) when I make a linear regression.

This is one of my more recent plots from an MP-AES experiment we did, and the part that I cannot wrap my head around is that the blank solution for this calibration showed basically zero intensity. But we see a completely different spike in signal in the 2ppm trial than the normal sensitivity of the trial that the slope shows.

My questions are, what exactly does it mean that my calibration curve doesn't start at (0,0), and, why is it different from the sensitivity (normal slope of the points)?

I have additionally gotten solutions that get negative concentrations when I try to use external standards (plugging the solution intensity into the equation and solving backwards) and I am wondering what kinds of errors could account for something like this happening in terms of external standards.

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    $\begingroup$ Well, you made error(s), most likely. Who knows what exactly. $\endgroup$
    – Mithoron
    Nov 6, 2023 at 23:56
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    $\begingroup$ In the plot you've given, it looks like you might have saturated your detection method. If you do much lower concentrations (maybe 0-1 ppm), do you get a steeper slope? $\endgroup$
    – Andrew
    Nov 7, 2023 at 0:15
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    $\begingroup$ Am I correct in assuming you are doing microwave plasma atomic emission spectroscopy and detecting potassium by its emission at, say, 766.5 or 769.9 nm? So a decent blank may well produce negligible emission there. Your graph does not include the blank. Hate to say it, but the data are pretty bad: big error bars, not linear response, and not good R squared. @Andrew may be right, but those big error bars make it harder to test his hypothesis. $\endgroup$
    – Ed V
    Nov 7, 2023 at 0:51
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    $\begingroup$ By the way, it is almost always a bad idea to try to force a calibration curve to pass through the origin: chemistry.stackexchange.com/a/116224/79678. Doing so eliminates a possibility of learning the full story that the data are telling. $\endgroup$
    – Ed V
    Nov 7, 2023 at 0:58
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    $\begingroup$ For given data, using linear approximation does not seem to be numerically justified. It seems the data series is curved toward the origin. $\endgroup$
    – Poutnik
    Nov 7, 2023 at 2:48

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I saw a couple of reports about how forcing it was bad but I was just wondering if there was any other explanation for why my absorbance might be so high.

You are doing emission spectroscopy (AES= atomic emission spectroscopy) not atomic absorbance. Most well prepared calibration curves, or and carefully prepared standards have a very small intercept.

What was the composition of the blank?

My question is what exactly does it mean that my calibration curve doesnt start at 0,0 and why it is diffrent from the sensitivity (normal slope of the points). I have additionally gotten solutions that get negative concentrations when I try to use external standards (plugging the solution intensity into the equation and solving backwards) and I am wondering what kinds of errors could account for something like this happening in terms of external standards.

Calibration curves and most of the experimental analytical chemistry follows the GIGO principle, i.e., if the standards are not prepared with utmost care, the sample analysis is a useless exercise. The only absolute way to test the accuracy is to use a reference material with a known concentration of potassium. If the experimental calibration curve produces the correct result, it means everything is working alright. If it produces wrong numbers and they do not match the reference, it means the experiment has serious errors. Negative concentrations simply imply the standard preparation was erroneous. Recheck all calculations, purity of the potassium salt, and of course, the blank.

I also suspect ionization issues, the calibration curve is nowhere linear. Potassium ionizes very easily.

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  • $\begingroup$ Thank you for your comment, In that particular lab I think we were also comparing it to FAES and I suspected that it might have to do with the temperatures that an MP-AES can reach and how that may cause loss of signal because it might strip electrons. That was my first time using one of these machines so now I really have an appreciation for the precision that we need in sample preparation. Thank you again for the responce. $\endgroup$
    – user139281
    Nov 7, 2023 at 5:01

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