March 28, 2024, 07:14:39 PM
Forum Rules: Read This Before Posting


Topic: AAS  (Read 3150 times)

0 Members and 1 Guest are viewing this topic.

Offline shafaifer

  • Regular Member
  • ***
  • Posts: 95
  • Mole Snacks: +1/-2
  • Gender: Male
AAS
« on: February 26, 2015, 01:27:49 PM »
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, 01:52:37 PM by shafaifer »

Offline kriggy

  • Chemist
  • Sr. Member
  • *
  • Posts: 1519
  • Mole Snacks: +135/-16
Re: AAS
« Reply #1 on: February 26, 2015, 02:16:29 PM »
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

Offline shafaifer

  • Regular Member
  • ***
  • Posts: 95
  • Mole Snacks: +1/-2
  • Gender: Male
Re: AAS
« Reply #2 on: February 26, 2015, 04:27:20 PM »
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  :)

Offline Furanone

  • Chemist
  • Full Member
  • *
  • Posts: 211
  • Mole Snacks: +34/-2
  • Gender: Male
  • Actually more a Food Chemist
Re: AAS
« Reply #3 on: February 26, 2015, 08:56:23 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.
"The true worth of an experimenter consists in pursuing not only what he seeks in his experiment, but also what he did not seek."

--Sir William Bragg (1862 - 1942)

Offline shafaifer

  • Regular Member
  • ***
  • Posts: 95
  • Mole Snacks: +1/-2
  • Gender: Male
Re: AAS
« Reply #4 on: February 27, 2015, 04:20:08 AM »
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.

Sponsored Links