the unknown was prepared by dissolving 0.0598 gramos 5, Clif 0 ABS 100% T key to
ID: 564754 • Letter: T
Question
the unknown was prepared by dissolving 0.0598 gramos 5, Clif 0 ABS 100% T key to set the blank at 0.000 A. 6. Remove the blank cuvet and replace with standard solution/unknown solution cuvet. Read and record the absorbance of all your solutions. Note: Calibrations needs only be performed once per lab period. Questions Explain why it is necessary to add ortho-phenanthroline to the solutions. What is the purpose of preparing and analyzing standard iron solutions? Aeonvert 33.3 %T to A. 4. Consider the spectrophotometric determination of iron(il) using the 1,10 phenanthroline complex. Standards were prepared from a stock solution of iron(ll). All measurements were made in a 1.00 cm cell. Fe(ll) ppm 0.00 2.00 5.00 10.00 15.00 Transmittance 1.000 0. 896 0.759 0.541 0.399 Graph the following data as Absorbance vs. Concentration. Calculate the slope the y-intercept and the correlation coefficient for the linearly related data. S. The unknown was prepared by dissolving a o.0598 g sample containing iron and diluting this sample to 500.0 mL. A 10.00 mUaliquot of this solution was treated like the standards and diluted to 100.00 mL. This solution had an absorbance of 0.212) Calculate the percent iron in the unknown.Explanation / Answer
1) Iron (II) is pale yellow in color. The basic criterion for measuring the concentration of a species is that the species must be intensely colored and absorb light. Fe(II)-phenanthrolin complex is intensely colored and hence, ortho-phenanthrolin is added to Fe(II) solutions to output the complex for spectrophotometric determination.
2) The absorbance of a colored species is directly proportional to the concentration of the species by the Beer’s law:
A = *C*l where = molar absorptivity constant (constant for a particular species), C is the concentration of the species and l = path length of the solution.
A series of standard solutions of iron(II) with ortho-phenanthrolin are prepared and the concentrations of these solutions are accurately known. The absorbance of these solutions are measured and plotted against the concentration to prepare what is known as a calibration curve.
It is highly improbable that the concentration of iron(II) in an unknown solution is known. The said unknown solution is treated with ortho-phenanthrolin to convert the iron(II) to the colored iron(II)-phenanthrolin complex. The absorbance of this complex is then measured and inserted into the calibration curve to get the concentration of iron (II) in the unknown sample accurately.
3) We know that the absorbance and transmittance of a solution are related as
A = 2 – log (%T)
Put %T = 33.3 in the above expression and obtain
A = 2 – log (33.3) = 2 – 1.5224 = 0.4776 (ans).
4) Convert all the transmittance to absorbance using the formula above as shown in the table below.
Fe (II), ppm
Transmittance, T
%T = T*100
Absorbance, A = 2 – log (%T)
0.00
1.000
100.0
0.0000
2.00
0.896
89.6
0.0477
5.00
0.759
75.9
0.1197
10.00
0.541
54.1
0.2668
15.00
0.399
39.9
0.3990
Plot absorbance vs concentration of Fe(II) in ppm as below.
Plot of absorbance vs concentration of standard Fe(II) in ppm
The slope of the plot is m = 0.0269; the y-intercept is c = -0.0056 and the correlation co-efficient is R2 = 0.9988.
e) Put the absorbance as y = 0.212 in the regression equation above and obtain the concentration of Fe(II) in the unknown dilute sample.
0.212 = 0.0269x -0.0056
=====> 0.0269x = 0.212 + 0.0056 = 0.2176
=====> x = 0.2176/0.0269 = 8.0892
The concentration of Fe(II) in the diluted sample is 8.0892 ppm = 8.0892 mg/L (since 1 ppm = 1 mg/L).
10.00 mL sample was diluted to 100.00 mL to prepare the solution for measurement; hence, the concentration of Fe(II) in the 10.00 mL solution is (8.0892 mg/L)*(100.00 mL)/(10.00 mL) = 80.892 mg/L.
500.00 mL of the sample solution was prepared; hence, the amount of Fe(II) in the 500.00 mL solution = (500.00 mL)*(1 L/1000 mL)*(80.892 mg/L) = 40.446 mg = (40.446 mg)*(1 g/1000 mg) = 0.040446 g.
The above is the amount of Fe(II) in the unknown sample; hence, the percentage of Fe(II) in the unknown sample is (0.040446 g)/(0.0598 g)*100 = 67.6354% (ans).
Fe (II), ppm
Transmittance, T
%T = T*100
Absorbance, A = 2 – log (%T)
0.00
1.000
100.0
0.0000
2.00
0.896
89.6
0.0477
5.00
0.759
75.9
0.1197
10.00
0.541
54.1
0.2668
15.00
0.399
39.9
0.3990
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