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Topic: Calibration curve linearity {Mod Edited title}  (Read 2384 times)

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Offline r-nik

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Calibration curve linearity {Mod Edited title}
« on: January 07, 2013, 12:14:21 PM »
Tghe reason you are getting 118.15 is that you are using the antilog function (ex) when entering 9E+07. You should be using the EXP function on your calculator.


using  1227456-61347/9e+07 (or 9 x 107), I get 0.012956 on my trusty casio


hello,i have a question,the calibration curve Is not necessary to be a linear? would is possible to be a poly or log or exp?

{MOD Edit: split new topic and edit title}
« Last Edit: January 07, 2013, 02:46:21 PM by Arkcon »

Offline curiouscat

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Re: Determining unknown from calibration curve??
« Reply #1 on: January 07, 2013, 12:26:46 PM »
would is possible to be a poly or log or exp?

Yes.

Offline JGK

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Re: Calibration curve linearity {Mod Edited title}
« Reply #2 on: January 08, 2013, 02:46:08 PM »
While it may well be possible to have calibrations which have a log or exponential relationship in a Calibration Line, the original problem was simply a case of mistyping the calculation into whatever instrument was being used to perform the calculation. The calibration range was 0 - 0.1 mg/L which yielded area responses in the range 0 - 8742401.

The unknown sample had an area of 1227456 which should yield a concentration within the calibration range (not 118.15 as the poster was calculating)

with the linear equation y= 9E+07x + 61347, 1227456-61347/9e+07=0.012956.

However, if you miscalculate and use 1227456-61347/9e+07=118.15
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