Atomic absorption spectrometry was used with the method of standard additions to
ID: 892710 • Letter: A
Question
Atomic absorption spectrometry was used with the method of standard additions to determine the concentration of cadmium in a sample of an industrial waste stream. For the addition, 10.0 L of a 1000.0 g/mL Cd standard was added to 10.0 mL of solution. The following data were obtained: Absorbance of reagent blank = 0.040 Absorbance of sample = 0.381 Absorbance of sample plus addition = 0.785
What was the concentration of the cadmium in the waste stream sample?
Later, the analyst learned that the blank was not truly a reagent blank, but water. The absorbance of the actual reagent blank, was 0.080. Calculate the cadmium concentration using the new information of the blank?
Calculate the percent error caused by using water instead of the reagent blank?
Explanation / Answer
To do this, you need to use the following equation:
A = k C ----------> k = constant; C = concentration.
The blank is a value that we need to substract from all the absorbance values:
A1 = sample absorbance; A2 = Standard absorbance; At = Total absorbace = A1 + A2
A1 = 0.381 - 0.04 = 0.341
At = 0.785 - 0.04 = 0.745
The innitial concentration is 1000 mg/L or 1000 ppm and sample solution is 10 mL.
So:
A1 = k C1
0.341 = k C1 ------> k = 0.341/C1 (1)
At = A1 + A2
0.745 = k C1 + 1000k
0.745 - 1000k = k C1
0.745 - 1000(0.341/C1) = 0.341C1/C1
0.745 - 341/C1 = 0.341
0.745 - 0.341 = 341/C1
C1 = 341 / (0.745 - 0.341)
C1 = 844.05 ppm
Now, I don't know if the sample was originally diluted to to the volume of 10 mL, so, I'll assume that it was not diluted. This would be the concentration of Cd in the sample.
Now if the reagent was actually water, the concentration of Cd would be:
C1' = 301 / (0.705 - 0.301) = 745.05 ppm
% = (844.05 - 745.05 / 844.05) x 100 = 11.73 %
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