|
|
g(Fe3+) [mg/L] |
Stock
solution of the
complex |
Diluent
solution |
Sample
solution |
Blank |
- |
0 |
450 |
- |
Sample |
- |
- |
- |
450 |
Calibration solution 1 |
12.1 |
50 |
400 |
- |
Calibration solution 2 |
24.3 |
100 |
350 |
- |
Calibration solution 3 |
36.4 |
150 |
300 |
- |
Calibration solution 4 |
48.6 |
200 |
250 |
- |
Calibration solution 5 |
60.7 |
250 |
200 |
- |
Calibration solution 6 |
72.9 |
300 |
150 |
- |
Calibration solution 7 |
85.0 |
350 |
100 |
- |
Calibration solution 8 |
97.2 |
400 |
50 |
- |
Calibration solution 9 |
109.3 |
450 |
0 |
- |
Insert a blister into the measuring chamber of a SpektraTM spectrometer so that a blank solution occupies the measuring site. Use blue LED at its maximal intensity and set the transmittance for the blank to 100.0. Take measurements for samples and calibration solutions and write the transmittances into the table below. Calculate absorbance and prepare a calibration graph as demonstrated below.
|
|
Calibration solutions |
||||||||
|
Sample |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
g(Fe3+) [mg/L] |
? |
12.1 |
24.3 |
36.4 |
48.6 |
60.7 |
72.9 |
85.0 |
97.2 |
109.3 |
T |
|
|
|
|
|
|
|
|
|
|
A |
|
|
|
|
|
|
|
|
|
|
Use the linear equation obtained for your calibration data and calculate the iron concentration in the sample solution.
Note
If the absorbance of the sample solution exceeds the absorbance of the final calibration solution, the sample solution must by no means be diluted with water. This would shift the equilibrium and consequently cause erroneous results. The sample can only be diluted with a diluent solution. Dilution must be taken into account while calculating the iron content in the condiment sample.
The iron content per 100 g of the condiment sample is then calculated as follows:
m(Fe3+) = |
g(Fe3+) · 0.013 L · 100 g |
msample |
Worked out example
In our case 1.0041 g of a dry basil sample was taken for analysis. The transmittance of a sample solution measured against the blank using the blue LED was 77.5.
A = -log(77.5/100) = 0.1107
0.1107 = 0.0037 x - 0.0361
g(Fe3+) = x = 39.67 mg/L
m(Fe3+) = |
39.67 mg · 0.013 L · 100 g |
= 51.4 mg |
L · 1.0041 g |
The iron content in a dry basil sample is 51.4 mg per 100 g of the condiment.
Developed
and prepared
by: Margareta
Vrtačnik*, Vida Mesec*, Nataša
Gros** and Domen Klančar**
*University of Ljubljana, Faculty of Natural Sciences
and Engineering
**University
of Ljubljana, Faculty of Chemistry
and Chemical Technology