Atomic absorption spectrometry was used with the method of standard additions to
ID: 525276 • 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 mu L of a 1000.0 mug/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 = 0.362 Absorbance of sample plus addition = 0.850 What was the concentration of the cadmium in the waste stream sample? Number mug/mL Later, the analyst learned that the blank was not truly a reagent blank, but water. The absorbance of the actual reagent blank, was 0.084. Calculate the cadmium concentration using the new information of the blank Number mug/mL Calculate the percent error caused by using water instead of the reagent blank. Number %Explanation / Answer
a) Let the sample contain x µg/mL Cd.
Mass of Cd in 10 mL of the industrial waste = (10 mL)*(x µg/mL) = 10x µg.
Mass of Cd added as standard addition = (10 µL)*(1 mL/1000 µL)*(1000 µg/mL) = 10 µg.
Total volume of the solution = (10 mL) + (10 µL)*(1 mL/1000 µL) = 10.01 mL 10.0 mL.
Total mass of Cd in the sample + standard = (10x + 10) µg; concentration of Cd in the sample + standard = (10x + 10) µg/(10.0 mL) = (x + 1) µg/mL.
Corrected absorbance for the sample, A1 = (absorbance of the sample) – (absorbance of the reagent blank) = (0.362 – 0.040) = 0.322.
Corrected absorbance for sample + standard, A2 = (absorbance of sample + standard) – (absorbance of the reagent blank) = (0.850 – 0.040) = 0.810.
Make an assumption that the absorbance varies linearly with the concentration of Cd as
A = k*C where k = constant of proportionality and C = concentration of Cd in µg/mL.
Take a ratio:
A2/A1 = k*(x+1)/kx = (x+1)/x
===> 0.810/0.322 = (x+1)/x
===> 2.5155 = (x+1)/x
===> 2.5155*x = x + 1
===> 1.5155x = 1
===> x = 1/1.5155 = 0.6598 0.66
The concentration of Cd in the waste water is 0.66 µg/mL (ans).
b) Again, determine A1 and A2.
A1 = (0.362 – 0.084) = 0.278
A2 = (0.850 – 0.084) = 0.766
Take ratio again,
A2/A1 = (x+1)/x
===> 0.766/0.278 = (x+1)/x
===> 2.7554 = (x+1)/x
===> 2.7554x = x + 1
===> 1.7554x = 1
===> x = 0.5697 0.57
The concentration of Cd in the waste water = 0.57 µg/mL (ans).
c) Percent error = (0.66 – 0.57)/(0.57)*100 = 15.789 15.80 (ans).
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