Atomic absorption spectrometry was used with the method of standard additions to
ID: 550420 • 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 = 0.043 Absorbance of sample = 0.320 Absorbance of sample plus addition = 0.840
What was the concentration of the cadmium in the waste stream sample?
_____ 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 0.099. Calculate the cadmium concentration using the new information of the blank.
_______g/ml
Calculate the percent error caused by using water instead of the reagent blank.
_____%
Explanation / Answer
Let the concentration of Cd in the actual waste stream be x g/mL. As per Beer’s law, we must have
0.320 = *x*l where = molar absorptivity of Cd (constant for a particular species) and l = path length of the solution (constant for a particular experiment) ……(1)
Again, we add 10.0 L of 1000.0 g/mL of Cd standard to 10 mL of the waste water sample; the mass of Cd in the spiked sample is [(10.0 mL)*(x g/mL) + (10.0 L)*(1 mL/1000 L)*(1000.0 g/mL)] = (10x + 10) g; the total volume of the solution is [10.0 mL + (10.0 L)*(1 mL/1000 L)] = (10.0 + 0.01) mL = 10.01 mL 10.0 mL.
The concentration of Cd in the spiked sample is [(10x + 10) g]/(10.0 mL) = (x + 1) g/mL. Write down Beer’s law as
0.840 = *(x + 1)*l …….(2)
The blank recorded an absorbance of 0.043; the corrected absorbances for the sample and the spiked samples are (0.320 – 0.043) = 0.277 and (0.840 – 0.043) = 0.797. Therefore, we must have,
0.277 = *x*l ……(3)
0.797 = *(x + 1)*l …….(4)
Divide (4) by (3) and obtain
0.797/0.277 = (x + 1)/x
====> 2.87726 = (x + 1)/x
====> 2.87726x = x + 1
====> 2.87726x – x = 1
====> 1.87726x = 1
====> x = 1/1.87726 = 0.53269 0.533
The calculated concentration of Cd in the waste stream is 0.533 /mL (ans, 1).
However, the reagent blank was actually water and the actual reagent blank had an absorbance of 0.099; therefore, the corrected absorbances for the sample and the spiked sample are (0.320 – 0.099) = 0.221 and (0.840 – 0.099) = 0.741.
Next, write down the modified Beer’s law as
0.221 = *x*l ……(5)
0.741 = *(x + 1)*l ……..(6)
Divide (6) by (5) and get
0.741/0.221 = (x + 1)/x
====> 3.35294 = (x + 1)/x
====> 3.35294x = x + 1
====> 3.35294x – x = 1
=====> 2.25294x = 1
=====> x = 1/2.25294 = 0.44386 0.444
The correct Cd concentration in the waste stream is 0.444 g/mL (ans, 2).
Percent error between the calculated and the correct concentrations are [(0.533 – 0.444) g/mL]/(0.444 g/mL)*100 = 20.0450% 20.045 (ans).
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