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Author Topic: AAS  (Read 192 times)

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shafaifer

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AAS
« on: February 26, 2015, 07:27:49 AM »

In AAS, the standard calibration curve typically has a y-axis intercept value of 0.0x (where x goes from 1 to 9  :)), for example, a Na calibration standard curve of concentrations (0.1, 0.2, 0.4 and 1 mg/L) has the following equation:

A = 0.4384c + 0.0183,

I take the first value to be the extinction coefficient, but what should I do with the value of 0.0183? Can it be omitted to meet the Beer-Lambert law which tells: A = lcĪµ (l is in this case 1 cm).
Can I exclude it if I, for example, would like to calculate the unknown concentration of Na in a water sample?

Kind regards
« Last Edit: February 26, 2015, 07:52:37 AM by shafaifer »
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kriggy

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Re: AAS
« Reply #1 on: February 26, 2015, 08:16:29 AM »

You should deduct the 0,0183 from measured absorbance because the sample with 0 concentration of Na will have A=0,0183. So
c=(A-0,0183)/0,4384

shafaifer

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Re: AAS
« Reply #2 on: February 26, 2015, 10:27:20 AM »

You should deduct the 0,0183 from measured absorbance because the sample with 0 concentration of Na will have A=0,0183. So
c=(A-0,0183)/0,4384

A reasonable great answer - so thank you  :)
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Furanone

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Re: AAS
« Reply #3 on: February 26, 2015, 02:56:23 PM »

Another possibility if you know for sure a certain assay has a 0 intercept, then you could plot A=0.4834c+0.0183 using 0.1, 0.2, 0.3 all the way up to 1 in Excel, then add in 0, 0 to the data set & graph and then change the trendline to "set intercept = 0". This would give a new equation of A=0.4654c with still a respectable R^2 of 0.997. Changing the graph axes to switch the x an y values would then give c=2.1508A. The benefit of doing this way is increased sensitivity for very low absorbance values, where subtracting the 0.0158 (although very small) would give no detectable concentration below a 0.0158 absorbance threshold.
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shafaifer

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Re: AAS
« Reply #4 on: February 26, 2015, 10:20:08 PM »

Another possibility if you know for sure a certain assay has a 0 intercept, then you could plot A=0.4834c+0.0183 using 0.1, 0.2, 0.3 all the way up to 1 in Excel, then add in 0, 0 to the data set & graph and then change the trendline to "set intercept = 0". This would give a new equation of A=0.4654c with still a respectable R^2 of 0.997. Changing the graph axes to switch the x an y values would then give c=2.1508A. The benefit of doing this way is increased sensitivity for very low absorbance values, where subtracting the 0.0158 (although very small) would give no detectable concentration below a 0.0158 absorbance threshold.

Thanks for the advise.
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