Determination of Concentration Using the Spectronic 20

Topics: Concentration, Absorbance, Sulfuric acid Pages: 2 (476 words) Published: November 2, 2008
Abstract:

The purpose of this experiment was to determine the concentration of an unknown amount of copper sulfate using the spectronic 20. Four solutions of different concentrations of H₂O and CuSO₄ were put into the spectronic 20 to observe their absorption of light. Then, a graph was created to plot down the measurements. Using the calibration curve of the graph, the concentration of the unknown substance was found out to be .12M.

Background/Introduction:

Beer’s law (which states that by holding the pathlength of a particular substance- in this case, CUSO₄ - constant, the absorption of light is directly proportional to the concentration of the substance) was used in this experiment to find the absorption measurement of the concentration of an unknown amount of copper sulfate.

Procedure:

Four different concentrated solutions of CuSO₄ (0.13M, 0.11M, 0.090M, and 0.070M) were prepared by crushing the copper sulfate with a mortar and pestle. The copper sulfate was then added into different volumetric flasks with 10.0 mL of distilled water. Also, 10 drops of concentrated nitric acid were added to the solutions. By placing each solution in the spectronic 20 device, the absorption measurements were noted down and graphed.

Data Collection:

M = mol/vol CuSO₄ + 5H₂OCuSO₄ = 249.5896g

0.13M0.13M = mol/ 0.100L = 0.013 mol (249.5896g) = 3.245g
0.11M0.11M = mol/0.100L = 0.011 mol (249.5896g) = 2.745g
0.090M0.090M = mol/0.100L = 0.009 mol (249.5896g) = 2.246g 0.070M 0.070M = mol/0.100L = 0.007 mol (249.5896g) = 1.747g

Concentration (M)Wavelength Determination (Trial One)Wavelength Determination (Trial Two) 0.13M50%T49%T
0.11M55%T54%T
0.090M62%T59%T
0.070M68%T67%T

Concentration (M)Absorbance Measurements (Trial One) 2-log(%T)Absorbance Measurements (Trial Two) 2-log(%T) 0.13M0.30M0.30M
0.11M0.26M0.27M
0.090M0.21M0.23M
0.070M0.17M0.17M

Unknown amount of CuSO₄: 52.0%T
2-log (%T)
2-log (52.0) = .28...
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