Essay on Chemistry
THEORY:
The foundations of quantitative chemical analysis can be traced back to the development of titrimetric analysis in which titration end-points depended on the change of color of either the species being analyzed or of that of a specially added chemical indicator. These color transitions arise due to the molecular and structural changes in the substance being examined, leading to corresponding changes in the ability to absorb light in the visible region of the electromagnetic spectrum. In various ways absorption spectroscopy in the visible region has long been an important tool to the analyst.
Many important and the sensitive color tests have been developed for the detection and determination of a wide range of chemical species, both inorganic and organic in nature, and were first used long before the development of ultraviolet and visible spectrometers.
The foundations of quantitative chemical analysis can be traced back to the development of titrimetric analysis in which titration end-points depended on the change of color of either the species being analyzed or of that of a specially added chemical indicator. These color transitions arise due to the molecular and structural changes in the substance being examined, leading to corresponding changes in the ability to absorb light in the visible region of the electromagnetic spectrum. In various ways absorption spectroscopy in the visible region has long been an important tool to the analyst.
Many important and the sensitive color tests have been developed for the detection and determination of a wide range of chemical species, both inorganic and organic in nature, and were first used long before the development of ultraviolet and visible spectrometers.
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The UV/ visible spectrometer is often referred to as the workhorse of the analytical laboratory, and is applied to many thousands of determinations, which have been developed over the years. UV/ visible spectrometry has proved particularly useful in biochemical analysis.
Color Tests and Qualitative Chemical Analysis:
For those responsible for the analysis of materials, simple color tests have often provided useful preliminary or confirmatory evidence for the presence of particular chemical species.
The electromagnetic Spectrum:
One of the first statements made in the introduction to this part was that the appearance of color arises from the property of the colored material to absorb selectively within then visible region of the electromagnetic spectrum.
The visible region of the electromagnetic spectrum is defined in terms of the wavelength range to which the human eye responds. The usual range of the wavelengths quoted is from 380 nm at violet / blue end to 750 nm at the long wavelengths red end of the spectrum. However, some sources quote wavelengths down to 360 nm as the short wavelength limit and up to 780 nm at the long wavelength limit.
The lower limit is determined mainly by instrumental factors such as the lack of detector sensitivity, and reduced transmittance of radiation by optical components and by the oxygen in the air. An additional limitation is the marked reduction in the transmittance of radiation by common solvents at low wavelengths.
For a color test to have high sensitivity it is important for the colored material produced in the test to absorb strongly in the visible region.
The Beer -Lambert Law and Calibration Curve:
Measurement of the absorption of ultraviolet and visible radiation by species in solution provides one of the most widely used methods of quantitative analysis available in the analytical laboratory .The combined Beer-Lambert Law can be written in the sample form:
A = e b c
In order to apply this sample law to the determinations of an analyte species of unknown concentration in solution, it is necessary to first construct a calibration graph of absorbance versus concentration using standard solutions of known concentration of the analyte species.
CALCULATIONS:
1g / L ґ 1 mol / 327.3 gr = 3.06 ґ 10-3 mol/L
5 ґ 3.06 ґ 10-3 = M2 ґ 100
M2 = 1.53 ґ 10-4
1.53 ґ 10 -4 ґ 5 = 100 ґ M3
M3 = 7.65 ґ 10 -6
pH = 1
CT = M3 = ( CMOH ) = 7.65 ґ 10-6
450 nm 0.102 = e1 ґ 1ґ 7.65 ґ 10-6
e1 = 13333.3
510 nm 0.351 = e2 ґ 1 ґ 7.65 ґ 10-6
e3 = 45882.4
pH = 13
0.197 = e2 ґ 1 ґ 7.65 ґ 10-6
e2 = 25751.6
0.100 = e4 ґ 1 ґ 7.65 ґ 10-6
e4 = 13071.9
pH = 2
A = AMO + AMOH
A450 = e2 CMO + e1 CMOH CMOH = 7.598 ґ 10-6
A510 = e4 CMO + e3 CMOH CMO = 2.6 ґ 10 -7
pH = 3
A = AMO + AMOH
A450 = e2 CMO + e1 CMOH CMOH =5.715 ґ 10-6
A510 = e4 CMO + e3 CMOH CMO = 1.973 ґ 10 -6
pH = 4
A = AMO + AMOH
A450 = e2 CMO + e1 CMOH CMOH = 1.51 ґ 10-6
A510 = e4 CMO + e3 CMOH CMO = 6.32 ґ 10 -6
pH = 5
A = AMO + AMOH
A450 = e2 CMO + e1 CMOH CMOH = 1.412 ґ 10-7
A510 = e4 CMO + e3 CMOH CMO = 7.46 ґ 10 -6
pH = 2
log ( CMO / CMOH ) = -1.466
pH = 3
log ( CMO / CMOH ) = -0.462
pH = 4
log ( CMO / CMOH ) = 0.622
pH = 5
log ( CMO / CMOH ) = 1.723
Equation of the pH versus log ( CMO / CMOH )
y = 0.9385 x + 3.402
pKIn = 3.402 KIn =2524.4
Unknown
0.1
A = AMO + AMOH
A450 = e2 CMO + e1 CMOH CMOH = 5.92 ґ 10-6
A510 = e4 CMO + e3 CMOH CMO = 1.867 ґ 10 -6
log ( CMO / CMOH ) = -0.501
pH = pKIn + log ( CMO / CMOH )
pH = 3.402 - 0.501 = 2.901
[H+ ] = 1.256 ґ 10-3
K1 = [H+ ] / (a - [H+ ] ) = 1.598 ґ 10-5
0.01
A = AMO + AMOH
A450 = e2 CMO + e1 CMOH CMOH = 3.92ґ 10-6
A510 = e4 CMO + e3 CMOH CMO = 3.756 ґ 10 -6
log ( CMO / CMOH ) = -0.0186
pH = pKIn + log ( CMO / CMOH )
pH = 3.402 - 0.0186 = 3.383
[H+ ] = 4.14 ґ 10-4
K2 = [H+ ] / (a - [H+ ] ) = 2.925 ґ 10-3
0.001
A = AMO + AMOH
A450 = e2 CMO + e1 CMOH CMOH = 1.83 ґ 10-6
A510 = e4 CMO + e3 CMOH CMO = 5.89 ґ 10 -6
log ( CMO / CMOH ) = 0.508
K3 = [H+ ] / (a - [H+ ] )=2.137 ґ 10-6
pH = pKIn + log ( CMO / CMOH )
pH = 3.402 - 0.508 = 2.89
[H+ ] = 1.288 ґ 10-3
Kavarage = 9.8 ґ 10 -4
DISCUSSION:
In this experiment we determined the pK of an unknown acid by using an indicator.In order to find the concentrations we used the Beer’s law. Since e values are depend only on the changes in wavelengths, we calculated for different wavelengths, 450 nm and 510 nm.We choose them because between 400-450nm we observe MO, and between 450-550 nm MOH is observed.
For finding the concentration of MOH and MO at pH =1 we assume that we have only MOH and at pH = 13 our assumption is to neglect the MOH concentration. Because at pH=1 our solution only consists of MOH and at pH = 13 we have only MO. But for other pH values our assumptions are not valid. For those the concentration of MOH and MO are found by solving two equations with two unknowns. And after those calculations we draw the graph of pH versus log ( CMO / CMOH ) and the y intercept of it gives the pKIn .
For unknowns we found the hydrogen ion concentration and K for each concentration. We expect that they are equal but because of experimental errors they aren’t. So we take their average.
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