Purpose
The purpose of this exercise is to get some experience with UV-VIS spectroscopy.
Method
Concentration in an unknown solution of jernalun determined from a standard curve in which absorbance is plotted as a function of concentration. Standard curve made by a spectrophotometer.
Theory
The concentration of colored substances in a solution can be determined with spectrophotometry, but when iron aqua-ion in our experiments is almost colorless, add potassium which are formed [Fe (SCN) n (H 2 O) 6-n] 3-n complexes. These complexes are reddish-brown color solution.
If the absorbance is measured at various wavelengths and is plotted in a coordinate system, obtained the substance absorption in the solvent. The wavelength giving the highest absorbance, is called the maximum absorption (l max) and it is at this wavelength to the standard curve is created.
If you send light with only one wavelength through a solution of a substance that absorbs light of this specific wavelength of light is weakened in intensity from 0 to I. The relationship between the solution absorbance and concentration is described by Lambert-Beers Law:
A = e (epsilon) * l * c
where A is the absorbance (dimensionless)
e is the molar absorption coefficient (L / (mol × cm))
l kuvettebredden (cm)
[A] is the current concentration of substance A (M)
In the intensity
Results
Measuring range:
We used a spectrophotometer to measure the spectrum.
l max read (the machine) to: 479.50 nm
This coincides very well with the color of the solution was orange-red. The blue light is absorbed as the wave field of 490-430 nm, and we can only see the complementary color to blue, orange (red).
Standard Curve:
We measured, using the spectrophotometer, the absorbance of the 6 samples and was signed following the standard curve.
It is evident that there is a linear relationship between absorbance and concentration, and are therefore proportional.
We chose the unknown iron solution No. II and measured using the spectrophotometer, the concentration of the "diluted version" to be: c = 4.9942 × 10 -- 5 M .
Since the measuring conditions are the same through the experiment, the slope of the curve e Liter One could therefore, if you were asking for example two absorbances and one concentration, calculate the other concentrations from Lambert-Beer's Law:
A1 = E * L * C1 and A2 = E * L * c2
It is available:
A1/c1 = A2/c2
But when spectrophotometer doing all the calculations for us, this calculation is unnecessary.
Calculations
Concentration of A:
The solution of is made by diluting 0025 M jernalun 100 times (10ml/1000ml), so the concentration of A must be:
0.25 * 10 ^-3M
Concentrations of standard solutions:
If we take 5 ml of A into flask, and a total of 100 ml of liquid A would have been diluted 20 times (5ml/100ml) and the concentration in the solution will be:
1.25 * 10 ^-5M
This calculation, we have created for all volumetric flask and the results are gathered in the table below:
V A
0 ml
5 ml
10 ml
20 ml
30 ml
40 ml
c
0.00 M
1.25 × 10 -- 5 M
2.5 × 10 -- 5 M
5.0 × 10 -- 5 M
7.5 × 10 -- 5 M
10 × 10 -- 5 M
Calculation of the concentration of our unknown iron solution:
10 ml of iron solution is diluted with water so there is 100 ml in total, it is then diluted 10 times.
5 ml of this mixture is diluted, so there again is 100 ml in total, it is now diluted 20 times.
In total, our unknown iron solution was diluted 200 times. We can therefore, from the measured diluted concentration, calculate the initial concentration (the iron solution II):
approx. 0.01 M
Discussion
As can be seen, is calculating the unknown concentration jernopløsnings incredibly close 0.01 M . This coincides very well with the iron solution No. II should have concentrations 0.01 M . So our attempt was overall very successful.
Exercise 5.b: Redox Reactions (tube tests)
Trials (a.)
Formed 2 phases in the test tube. The upper phase (water phase) had an orange-brown color and the lower phase (organic phase) had a purple color.
Reaction:
2 Fe 3 + (aq) + 2 I -- (aq) ® 2 Fe 2 + (aq) + I 2 (aq)
Iron (III) ion oxidizes iodide to iodine and is even reduced to iron (II) ion.
In 2 is miscible with dichloromethane, and the result is a phase with violet color.
Iodine may also react with iodide:
In 2 (aq) + I -- (aq) ⇌ I 3 -- (aq)
This has a brown color, and with iron yellow color provides the water phase an orange brown color.
Experiments (b.)
We reviewed two phases in the test tube. The upper phase (water phase) was yellow-brown and the lower (organic phase) was violet.
Reaction:
2 MnO 4 -- (aq) + 10 I -- (aq) + 16 H + (aq) ® 2 Mn 2 + (aq) + 5 I 2 (aq) + 8 H 2 O (l)
Permanganationen oxidizes iodide to iodine and is even reduced to manganese (II) ion.
In 2 is miscible with dichloromethane, and the result is that the organic phase turns violet.
Iodine may also react with iodide:
In 2 (aq) + I -- (aq) ⇌ I 3 -- (aq)
This has a brown color, and water phase will be a color combination of brown and slightly pink (from Mn 2 +).
Experiments (c.)
We saw that when we added a few drops of KMnO 4 to the clear mixture of sulfuric acid and iron (II) sulfate, was fluid in the area where the two solutions blended colors purple, but the color disappeared quickly and the liquid remained clear. But after 5-6 drops, there was a color change, and we now had a purple solution.
Reaction:
MnO 4 -- (aq) + 5 Fe 2 + (aq) + 8 H + (aq) ® 5 Fe 3 + (aq) + Mn 2 + (aq) + 4 H 2 O (l)
Permanganationen oxidizes iron (II) ions to iron (III) ions and is even reduced to manganese (II) ion.
At the beginning of the experiment, we have an excess of iron (II) ions from the dissolution of iron (II) sulfate. But as we add more and more KMnO 4 oxidizes Fe 2 + to Fe 3 + by permanganationen and when we reach 5-6 drops of KMnO 4 is an excess of MnO 4 - and Fe (II) ions are consumed. There is therefore no longer a reduction of MnO 4 - and the liquid will be purple color of MnO 4 -.
Trials (d.)
When heating was developed for a gas-dyed filter paper (with a solution of KI at) brown, which means that it is Cl 2 gas.
Reaction:
2 MnO 4 -- (aq) + 10 Cl -- (aq) + 16 H + (aq) ® 5 Cl 2 (g) + 2 Mn 2 + (aq) + 8 H 2 O (l)
Permanganationen oxidizes chloride to chlorine and is even reduced to manganese (II) ion.
We know also that Cl 2 can oxidize iodide to iodine in aqueous solution:
Cl 2 (g) + 2 I -- (aq) ® 2 Cl -- (aq) + I 2 (aq)
From some of the other test tube experiments, we know that iodine may also react with iodide:
In 2 (aq) + I -- (aq) ⇌ I 3 -- (aq)
And it is, then it (I 3 -) which colors the paper brown.
Experiments (e.)
We see in the experiment, how zinc powder decolourised copper (II) sulfate solution (blue ® clear).
Reaction:
Cu 2 + (aq) + Zn (s) ® Cu (s) + Zn 2 + (aq)
Copper can oxidize zinc because it is farther to the right in the voltage range and therefore are more electro negative than zinc.
Trials (f.)
We see in the experiment, the formation of solid silver around copper wire.
Reaction:
Cu (s) + 2 Ag + (aq) ® Cu 2 + (aq) + 2 Ag (s)
Copper is no longer left in the voltage range than silver and is therefore less electro negative. This means that silver can oxidize copper.
