Question

You are analyzing for Selenium (Se) in a mixture of peanuts (0.843 g). Any Se from the original mixture has been converted to a fluorescent derivative and extracted into 25.0 mL of dichloromethane (DCM). This sample was too concentrated for fluorescent spectroscopy to work properly, and a dilution was made: 1.00 mL of the original DCM extract was diluted to 100.0 mL (the diluted sample is labeled "sample A"). A 1.00 mL aliquot of sample A was diluted to 2.00 mL. This sample was then placed in a cuvet for fluorescence analysis. The concentration was now low enough to generate an accurate signal, giving a reading of 23.7 units.

Another sample was prepared by mixing 10.0 µL of a 2.00 µg/L Se standard with 1.00 mL of sample A and diluted to 2.00 mL (the Se is completely converted to its fluorescent derivative when mixed with the sample A aliquot). This sample was placed in a cuvet and gave a fluorescent signal of 37.6 units.

What type of calibration is this? How did you determine this?

What is the concentration of Se in the mixture of peanuts (in ppb, w/w)?

Answer 1

The type of calibration performed here is known as the method of standard additions. A standard sample of known concentration is added as an internal standard to an analyte to spike the signal due to the analyte. The standard addition method is followed when the analyte is too dilute or less concentrated to give an usable signal. At times, the standard addition method is used to compare the signals due to the analyte only and the spiked sample (as in this case).

Let the concentration of Se in the sample A used for measurement of fluorescence be x µg/L. Therefore, concentration of Se in the sample used for measurement of fluorescence is (1.0 mL)*(x µg/L)/(2.0 mL) = (x/2) µg/L

Assume a path length of l units and write down Beer’s law as

23.7 units = ε*(x/2 µg/L)*l where ε = absorption co-efficient for Se ……(1)

Micrograms of Se standard added to sample A = (10.0 µL)*(1 L/10⁶ µL)*(2.00 µg/L) = 2.0*10^-5 µg.

Micrograms of Se in 1.00 mL of sample A = (1.00 mL)*(1 L/1000 mL)*(x µg/L) = 0.001*x µg.

Total micrograms of Se in the new sample = (2.0*10^-5 + 0.001*x) µg.

The total volume of the solution used for the measurement is 2.00 mL = 0.002 L; the concentration of Se in the new sample = (2.0*10^-5 + 0.001*x)/0.002 µg/L.

Therefore,

37.6 units = ε*[(2.0*10^-5 + 0.001*x)/0.002 µg/L]*l (ε and l stays the same for both the cases)…..(2)

Therefore,

37.6/23.7 = [(2.0*10^-5 + 0.001*x)/0.002 µg/L]/(x/2 µg/L)

====> 1.5865 = (2.0*10^-5 + 0.001*x)/(x/2)

====> 0.79325*x = 2.0*10^-5 + 0.001x

====> 0.79425*x = 2.0*10^-5

====> x = 2.5181*10^-5

The original diluted sample contained 2.5181*10^-5 µg/L Se.

We took 1.0 mL of sample A and diluted to 2.0 mL; the dilution factor is DF = (2.0 mL)/(1.0 mL) = 2.

Concentration of Se in sample A = (2.5181*10^-5 µg/L)*2 = 5.0362*10^-5 µg/L.

1.0 mL of the concentrated sample was taken for preparing sample A. The dilution factor is (100.0 mL)/(1.0 mL) = 100; the concentration of Se in the concentrated sample = (5.0362*10^-5 µg/L)*100 = 5.0362*10^-3 µg/L.

25.0 mL of the original extracted sample was diluted to 100.0 mL; the dilution factor is (100.0 mL)/(25.0 mL) = 4 and the concentration of Se in the extract = (5.0362*10^-3 µg/L)*4 = 0.0201448 µg/L.

Since we had 25.0 mL of the solution, the amount of Se in the 25.0 mL solution is (25.0 mL)*(1 L/1000 mL)*(0.0201448 µg/L) = 5.0362*10^-4 µg.

The concentration of Se in the peanuts = (5.0362*10^-4 µg)/(0.843 g) = 5.9741*10^-4 µg/g.

We know that 1000 ppb = 1 µg/g; therefore, 5.9741*10^-4 µg/g = (5.9741*10^-4 µg/g)*(1000 ppb/1 µg/g) = 0.59741 ppb ≈ 0.60 ppb (ans).