Spectrophotometric Determination of Phosphate by Colorimetric Assay
Essay by jlpy888 • March 27, 2018 • Lab Report • 1,237 Words (5 Pages) • 1,139 Views
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Experiment 1 : Spectrophotometric determination of phosphate by colorimetric assay
Introduction
Spectrophotometry is a biochemical technique used to identify a biomolecule or for determination of concentration of a specific substance in a solution. It measures how much the chemical substance absorbs light (absorbance, A) when a beam of light passes through the sample solution at a particular wavelength (LibreTexts 2017). The relationship between light absorption and the concentration of the absorbing species can be described mathematically by Beer-Lambert Law (A= [pic 1]cl) where [pic 2] is the extinction coefficient. If the extinction coefficient of a substance is unknown, the absorbance and concentration of standard solutions can be used to construct a standard curve. Absorbance of the unknown samples can then be measured and their concentration can be read directly from the standard curve (Boyer 2012). However, the Beer-Lambert Law applies only for light of narrow wavelength, dilute solutions and solutions in which chemical reactions are not proceeding. The purpose of this experiment is to determine the concentration and mass of phosphate in two unknown samples through the calibration of a standard curve. The determination can either be done by tracing the Y-value (absorbance of the unknown sample) down to the X-value (mass of phosphate) or via calculation using the linear equation Y = mX.
Results
Table 1 : The concentration and volumes of solution added to produce the standard solutions.
Total mass of phosphorus (μg) | Volume of standard phosphate solution added (ml) | Volume of water added (ml) | Total volume of solution (ml) | Concentration (μg/ml) |
50.0 | 0.5 | 4.5 | 5.0 | 10.0 |
40.0 | 0.4 | 4.6 | 5.0 | 8.0 |
30.0 | 0.3 | 4.7 | 5.0 | 6.0 |
20.0 | 0.2 | 4.8 | 5.0 | 4.0 |
10.0 | 0.1 | 4.9 | 5.0 | 5.0 |
Sample Calculations
(C1V1 = C2V2)
For 50 μg of phosphorus:
(100μg)(1ml) = (50μg)(Xml)
X = 0.5 ml
Amount of water needed = 5.0 – 0.5 = 4.5 ml
0.5 ml of the standard phosphate solution should be added to 4.5 ml of water.
The same method is used to calculate the amount of standard phosphate solution and water needed to prepare the rest of standard solutions (5ml). Results are tabulated in Table 1.
Table 1.1 : Table of solutions and absorbance values at a Vmax of 660nm.
Solution (μg) | Absorbance (660nm) 1st reading | Absorbance (660nm) 2nd reading | Average Absorbance (660nm) |
0 | 0 | 0 | 0.000 |
10 | 0.013 | 0.015 | 0.014 |
20 | 0.043 | 0.047 | 0.045 |
30 | 0.081 | 0.083 | 0.082 |
40 | 0.123 | 0.125 | 0.124 |
50 | 0.176 | 0.181 | 0.179 |
X | 0.221 | 0.231 | 0.226 |
Y | 0.121 | 0.132 | 0.127 |
[pic 3]
Figure 1 : Standard curve of Absorbance at Vmax 660nm against Mass of phosphorus (μg)
Table 1.2 : Table of absorbance at 660nm, mass and concentration of unknown samples.
Sample | Absorbance 1st reading | Absorbance 2nd reading | Mass 1st reading (μg) | Mass 2nd reading (μg) | Concentration 1st reading (μg/ml) | Concentration 2nd reading (μg/ml) |
X | 0.221 | 0.231 | 69.063 | 72.1875 | 13.813 | 14.438 |
Y | 0.121 | 0.132 | 37.813 | 41.250 | 7.563 | 8.250 |
A | 0.043 | 0.033 | 13.438 | 10.313 | 2.688 | 2.063 |
A (5x diluted) | 0.013 | 0.011 | 3.438 | 3.438 | 0.688 | 0.6875 |
A (10x diluted) | 0.008 | 0.010 | 3.125 | 3.125 | 0.625 | 0.625 |
B | 0.347 | 0.345 | 108.438 | 107.813 | 21.688 | 21.563 |
B (5x diluted) | 0.070 | 0.068 | 21.875 | 21.250 | 4.375 | 4.250 |
B (10x diluted) | 0.047 | 0.049 | 15.313 | 15.313 | 3.063 | 3.063 |
C | 0.006 | 0.008 | 1.875 | 2.500 | 0.375 | 0.500 |
C (5x diluted) | 0.005 | 0.006 | 1.875 | 1.875 | 0.375 | 0.375 |
C (10x diluted) | 0.007 | 0.008 | 2.500 | 2.500 | 0.500 | 0.500 |
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