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 μL of a 1000.0 μg/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?

_____ μg/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.

_______μg/ml

Calculate the percent error caused by using water instead of the reagent blank.

_____%

Answer 1

Let the concentration of Cd in the actual waste stream be x μg/mL. As per Beer’s law, we must have

0.320 = ε*x*l where ε = 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 μL of 1000.0 μg/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 μg/mL) + (10.0 μL)*(1 mL/1000 μL)*(1000.0 μg/mL)] = (10x + 10) μg; the total volume of the solution is [10.0 mL + (10.0 μL)*(1 mL/1000 μL)] = (10.0 + 0.01) mL = 10.01 mL ≈ 10.0 mL.

The concentration of Cd in the spiked sample is [(10x + 10) μg]/(10.0 mL) = (x + 1) μg/mL. Write down Beer’s law as

0.840 = ε*(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 = ε*x*l ……(3)

0.797 = ε*(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 μ/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 = ε*x*l ……(5)

0.741 = ε*(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 μg/mL (ans, 2).      

Percent error between the calculated and the correct concentrations are [(0.533 – 0.444) μg/mL]/(0.444 μg/mL)*100 = 20.0450% ≈20.045 (ans).