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Water and Environmental Sciences
Faculty of Graduate Studies
Spectrophotometric Determination of Iron
Student Name:
Lina M.A Tamimi
Supervisor:
Dr. Jack Mustaklem
2014
Spectrophotometric Determination of Iron
✓ Introduction
Samples can be analyzed spectrophotometrically for iron by forming the reddish-orange tris complex of iron (II) and 1,10-phenanthroline, C12H8N2 (see below). This complex absorbs light in the visible region rather strongly with a maximum absorbance occurring around 510 nm.
The absorption of light by this complex follows the Beer-Lambert law over a wide concentration range. In this experiment the iron content of an unknown will be determined by comparing the absorbance at a specific wavelength of an unknown solution to that of standards.
Beer-Lambert law ( A = abc
Where: A ( is the absorbance (in absorbance units) at a specific wavelength;
a: is the absorptivity at a specific wavelength; when expressed in units of L/mol-cm
it is the molar absorptivity (ε);
c: is the concentration of the absorbing species, usually expressed in units of mol/L;
b: is the pathlength of the sample through which the light beam passes, usually expressed in units of cm.
✓ Procedure
a) Preparation of solution :
1) Prepare a standard iron solution containing about 0.0702 g of iron per 1000 mL by first calculating the required amount of pure ferrous ammonium sulfate [Fe(NH4)2(SO4)2C6H2O] (FW=392.14). Weigh out the sample using an analytical balance and dissolve this in about 200 mL of deionized water containing 2.5 mL of H2SO4 in a 1 L volumetric flask. Dilute with water to the mark and mix thoroughly by inverting the stoppered flask several times. Do not dry the ferrous ammonium sulfate before weighing. This solution will be used in a subsequent analysis.
Note: in our lab we weighing 0.0933 g of ferrous ammonium sulfate, so the concentration of our solution is approximately 13 ppm
When 0.0702 ==( 10 ppm
0.0933 ==( X ; ( X = 10 ppm x 0.0933/0.0702 = 13 ppm
Conc = moles/L of solution =( gm/f.w)/L =( 0.0933/392.14)/1L= 0.000237 mol/l
2) Prepare 1, 10-phenanthroline solution ( dissolve 100 mg in 100 ml water .
3) Prepare hydroxylamonium hydrochloride ( dissolve 10 g in 100 ml water.
4) Sodium acetate solution ( dissolve 10 g in 100 ml water.
b) Second solution
1) Use 100 ml volumetric flasks and pipets to prepare the following solutions :
|Solution # |Standard iron solution |NH2OH.HCl |Phenanthroline |Sodium acetate |
|1 |2.0 ml |1.0 ml |5.0 ml |8.0 ml |
|2 |5.0 ml |1.0 ml |5.0 ml |8.0 ml |
|3 |10.0 ml |1.0 ml |5.0 ml |8.0 ml |
|4 |25.0 ml |1.0 ml |5.0 ml |8.0 ml |
|Unknown(# C) |10.0 ml |1.0 ml |5.0 ml |8.0 ml |
|Blank |___ |1.0 ml |5.0 ml |ml |
2) Mix the reagents in each flask and wait at least 15 minutes, to develop complex color.
3) Dilute the solution to 100 ml.
4) Use the blank as reference at spectrum 510 nm.
5) Measure the absorbance of each solution against blank at 510 nm.
C) The report
Calculate the concentration of the complex from the molar concentration of the iron solution and the cell path length.
Plotting the calibration curve with the absorbance against concentration in ppm.
Determine the concentration of unknown from the curve.
Calculation :
A = abc
Concentration of the diluted solution is 2.5 ppm ; the original is 10 ppm
In our lab : 10 ppm ( 2.5
13 ppm ( X ( X = 2.5 x 13 /10 = 3.25
|Solution # |Absorbance |PPM |
|balnk |zero |zero |
|2 ml |0.054 |0.26 |
|5 ml |0.133 |0.65 |
|10 ml |0.299 |1.3 |
|25 ml |0.668 |3.25 |
|Unknown (#C) |0.177 |???? |
Concentration of diluted solution (
Solution 2ml = (2 /1000 L x 13 mg/L)/ 0.1 L = 0.26 ppm
Solution 5 ml = (5/1000 L x 13 mg/L)/ 0.1 L = 0.65 ppm
Solution 10 ml = (10/1000 L x 13 mg/L)/ 0.1 L = 1.3 ppm
Solution 25 ml = (25/1000 L x 13 mg/L) / 0.1 L = 3.25 ppm
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From the chart above the concentration of unknown which its absorbance (0.177) is approximately ( 0.8 ppm)
A = abc ( 0.177 = a x 1 x 0.8 ( a = 0.177/0.8 = 0.2212 mg/L
Concentration (absorptivity )= 0.2212 mg/L
A = ( b c
( = molar absorptivity
b = path length (cm)
c = concentration (mol/L)
To calculate the molar absorbtivity, the C unit should be on mol/L ; so we should converted tow point from the chart above from mg/L (ppm) to mol/L and so from the slope of the new equation between that tow point we can determine the concentration of unknown on (mol/L).
Point 1 ( 0.26 mg/L
Mol/L = (mg/L x 1/1000 mg) x (1/FW for Iron)
So; concentration of point 1 mol/L = (0.26 mg/L x 1/1000 mg) x (1/55.845)
= 0.465 x 10-5 mol/L
Point 2 ( 0.65 mg/L
So; concentration of point 2 mol/L =( 0.65 x 1/1000 mg ) x (1/55.845)
= 1.16 x 10-5 mol/L
|Absorbance (y) |Conc.mol/L (x ) |
| Point 1 0.054 |0.465 x 10-5 |
| Point 2 0.133 |1.16 x 10-5 |
|Unknown # C ( 0.177 |???? |
The tow point is: (X1 , Y1) ,,, ( X2 , Y2 )
(0.465 x 10-5 , 0.054 ) and (1.16 x 10-5 , 0.133 )
So the slop is : M = Y2 – Y1 / X2 – X1
= (0.133 – 0.054) / (1.16 x 10-5 - 0.465 x 10-5 )
= 0.079 / 0.695 x 10-5 = 0.114 x 105
So , M = 0.114 x 105 ( the slope of the curve
We should take any point with the point ( X , Y ) to determine the equation; so we take (1.16 x 10-5 , 0.133 ) with ( X , Y )
So ; M = Y – 0.133 / X – 1.16 x 10-5
0.114 x 105 = Y – 0.133 / X – 1.16 x 10-5
(X – 1.16 x 10-5 ) x ( 0.114 x 105 ) = Y – 0.133
0.114 x 105 X – 0.0174 = Y – 0.133
( Y = 0.114 x 105 X – 0.1156
We know that the absorbance of unknown is 0.177 which is Y so we can find X;
0.177 = 0.114 x 105 X – 0.1156 ( 0.114 x 105 X = 0.1156 + 0.177
So, X = 0.2926/0.114 x 105 ( X = 2.57 x 10-5 mol/L
A = ( b c
0.117 = ( x 1 x 2.57 x 10-5
( ( = 0.177 /2.57 x 10-5
Molar absorbtivity (( ) = 0.069 x 105 L/mol.cm
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C 