Selenium from 0.108 8 of Brazil nuts was converted into the fluorescent product,

ID: 1046726 • Letter: S

Question

Selenium from 0.108 8 of Brazil nuts was converted into the fluorescent product, which was extracted into 2.00 mL of cyclohexane. Standard additions of fluorescent product containing 1.40 g Se/mL were then added and the associated intensities are given in the table. Construct a standard addition graph to find the concentration of Se in the 2.00-mL unknown solution. Note:you just need to provide the equation Fluorescence intensity (arbitrary units) 41.4 49.2 56.4 63.8 70.3 Volume of standard added (AL) 10.0 20.0 30.0 40.0

Explanation / Answer

We know that 1 mL = 1000 µL; therefore, we can assume that the volume of the solution remains unchanged at 2 mL; hence, we shall assume that the final volume of the solutions is 2 mL.

Use the dilution equation [take solution 2 as an example, 10 µL = (10 µL)*(1 mL/1000 µL) = 0.01 mL].

C1*V1 = C2*V2

====> (1.40 g Se/mL)*(0.01 mL) = C2*(2 mL)

====> C2 = (1.40 g Se/mL)*(0.01 mL)/(2 mL) = 0.007 g Se/mL.

Prepare the following table as below.

Solution #

Volume of 1.40 g Se/mL standard added (µL)

Concentration of added Se in the solution prepared (g/mL)

Fluorescence Intensity

41.4

10.0

0.007

49.2

20.0

0.014

56.4

30.0

0.021

63.8

40.0

0.028

70.3

Plot fluorescence intensity vs concentration of added Se in the solutions prepared.

Plot of fluorescence intensity vs concentration of added Se standard

It is assumed that the Se in the sample produces zero fluorescence; hence, put y = 0 and obtain

0 = 1034.3x + 41.74

=====> -1034.3x = 41.74

=====> x = -(41.74/1034.3) = -0.0403 ? -0.040

Ignore the negative sign and report the concentration of Se in the sample of Brazil nuts as 0.040 g/mL (ans).