Atomic absorption spectrometry was used with the method of standard additions to
ID: 525340 • 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 mu 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 = 0.362 Absorbance of sample plus addition = 0.850 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 0.084. 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
Ans. #1. Absorbance of reagent blank = 0.040
Absorbance of sample = 0.362
Actual absorbance of sample = Abs of sample – Abs of blank = 0.362 – 0.040 = 0.322
#2. Absorbance of sample + addition = 0.850
Actual absorbance of (sample + addition) = 0.850 – 0.040 = 0.810
Increase in absorbance = Abs of (sample + Addition) – Abs of sample alone
= 0.810 – 0.322
= 0.488
Now,
10.0 µL of 1000.0 µg/mL Cd standard contains 10.0 µg Cd.
That is, 10.0 µg (=10.0 µL) Cd was added to the sample.
Final volume of the sample + Addition = 10.0 mL
That is 10.0 µg standard Cd was diluted upto 10.0 mL
So, standard [Cd] in final solution = mass of Cd added / Final volume of solution
= 10.0 µg / 10.0 mL
= 1.0 µg/ mL
Therefore, the final concentration of standard Cd from addition µg/ mL. Or, the Cd concentration of the solution has increased by 1.0 µg/ mL due to addition.
Note that the increase in absorbance of the sample is due to increase in Cd concentration by 1.0 µg/ mL.
So, an absorbance of 0.488 is equivalent to 1.0 µg/ mL
Or, - 1.0 - (1.0 / 0.488) µg/ mL
Or, - 0.322 - (1.0 / 0.488) x 0.322 µg/ mL
= 0.9450 µg/ mL
Hence, [Cd] of wastewater sample = 0.6598 µg/ mL
Therefore, [Cd] in wastewater with Abs of 0.322 (un-corrected Abs was 0.362) = 0.6988 ug/mL
#2. Actual absorbance of sample = 0.362 – 0.084 = 0.278
Actual absorbance of (sample + addition) = 0.850 – 0.084 = 0.766
Increase in absorbance = Abs of (sample + Addition) – Abs of sample alone
= 0.766 – 0.278
= 0.488
So, an absorbance of 0.488 is equivalent to 1.0 µg/ mL
Or, - 1.0 - (1.0 / 0.488) µg/ mL
Or, - 0.278 - (1.0 / 0.488) x 0.278 µg/ mL
= 0.5697 µg/ mL
Hence, [Cd] in wastewater sample = 0.5697 µg/ mL
#3. Error = Inaccurate [Cd] with water as blank - Actual [Cd] with sample blank
= 0.6598 ug/mL – 0.5697 ug/mL
= 0.0901 ug/ mL
% error = (error in [Cd] / Actual [Cd]) x 100
= (0.0921 ug mL-1 / 0.5697 ug mL-1) x 100
= 15.815 %
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