Atomic absorption spectrometry was used with the method of standard additions to
ID: 1042166 • 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 uL of a 1000.0 ng/mL Cd standard was added to10.0 mL of solution. The following data were obtained: Absorbance of reagent blank 0.036 Absorbance of sample 0.498 Absorbance of sample plus addition - 0.779 What was the concentration of the cadmium in the waste stream sample? Number 5.311x 10 g/mL Later, the analyst learned that the blank was not truly a reagent blank, but water. The absorbance of the actual reagent blank, was 0.079. Calculate the cadmium concentration using the new information of the blank. Number 5.0121×11 ?/mL Calculate the percent error caused by using water instead of the reagent blank. Number 5.630Explanation / Answer
Ans. Step 1: Calculate [Cd2+] in Final aliquot
Given - 10.0 uL (V1) of 1000.0 ug/ mL (C1) was added to 10.0 mL original sample solution.
10.0 uL = 0.010 mL
Total volume of the aliquot after standard addition, V2 =
10.0 mL (original sample) + 0.010 mL (Cd2+ standard)
= 10.010 mL
Now, Using C1V1 = C2V2
C1= Concentration, and V1= volume of initial solution 1 ; Cd2+ Std. solution
C2= Concentration, and V2 = Volume of final solution 2 ; Final Aliquot
Putting the values in equation 1-
C2 = (1000.0 ug mL-1 x 0.010 mL) / 10.010 mL = 0.9990 ug mL-1
Hence, spiked [Cd2+] in final aliquot = 0.9990 ug mL-1 = 1.0 ?g/ mL.
That is, addition of std. Cd increase the [Cd] in spiked waste water sample by 1.0 ug/mL.
# Step 2. Calculate [Cd2+] in original sample:
Given,
Abs of reagent blank = 0.036
Abs of original sample = 0.498
Abs of sample plus addition = 0.779
Now,
Actual Abs of original sample = Abs of original sample – Abs of blank
= 0.498 – 0.036 = 0.462
Actual Abs of sample plus addition = 0.779 – 0.036 = 0.743
Actual increase in Abs for spiking = (Abs of sample & addition) – Abs of sample
= 0.743 – 0.462
= 0.281
# Note that the increase in absorbance of the (sample plus addition) with respected to original sample is due to increase in Cd2+ concentration by 1.0 ?g/ mL.
So, an absorbance of 0.281 is equivalent to 1.0 ?g/ mL
Or, an abs of 1.0 is - (1.0 / 0.281) ?g/ mL
Or, an abs of 0.462 is - (1.0 / 0.281) x 0.462 ?g/ mL
= 1.6441 ?g/ mL
Hence, [Cd] of original waste water sample = 1.6441 ?g/ mL
#3. Corrected values due to applying correct abs of blank
Corrected Abs of reagent blank = 0.079
Abs of original sample = 0.498
Abs of sample plus addition = 0.779
Now,
Corrected absorbance of sample = 0.498 – 0.079 = 0.419
Corrected absorbance of (sample + addition) = 0.779 – 0.079 = 0.700
Corrected increase in absorbance = 0.700 – 0.419 = 0.281
So, an absorbance of 0.281 is equivalent to 1.0 ?g/ mL
Or, an abs of 1.0 is - (1.0 / 0.281) ?g/ mL
Or, an abs of 0.419 is - (1.0 / 0.281) x 0.419 ?g/ mL
= 1.4911 ?g/ mL
Hence, Corrected [Cd] in wastewater sample = 1.4911 ?g/ mL
#Step 4. Error = Inaccurate [Cd] with water as blank - Actual [Cd] with sample blank
= 1.6441 ug/mL – 1.4911 ug/mL
= 0.1530 ug/ mL
% error = (error in [Cd] / Actual [Cd]) x 100
= (0.1530 ug mL-1 / 1.4911 ug mL-1) x 100
= 10.26 %
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