Nathan Bahn

Beer’s Law Study Lab

Introduction:

In this lab, we used a spectrometer to observe the transmittance of light at a certain wave length. We experimented to see if the molarity of a solution changes the transmittance of light and the absorbance of that light by the solution. By observing the percent transmittance and the amount of light absorbed, we can calculate the amount of color absorbing components in the solution. Through this process is how we are able to discover the amount of copper in the solution.

Experimental Procedure:

250 mL of the copper solution was made by creating 100 mL of the solution, reacting CuO with HNO3, and then diluting to the mark of 250 mL. Using this stock solution, different concentrations were made and placed in the sprectrometer for observation. The absorbances and transmittances were recorded for use when identifying the amount of the color-absorbing copper ions later. A graph was plotted of Absorbance v. Molar Concentration easily see the results of the experiment.

Pre-Lab Questions:

1. Absorbance = - Log10T = 2-Log10 (%T)

a) 2 – Log (89.95) = .0460 f) 2 – Log (28.18) = .5501

b) 2 – Log (80.91) = .0920 g) 2 – Log (20.51) = .6880

c) 2 – Log (59.02) = .2290 h) 2 – Log (17.38) = .7600

d) 2 – Log (47.75) = .3210 i) 2 – Log (14.45) = .8401

e) 2 – Log (34.83) = .4580

2. Molar Concentration

1.213 g Cu (1 mol Cu/ 63.546 g Cu) (1 mol [Cu (H2O6)2+]/ 1 mol Cu) (1/ .100 L) = .191 M [Cu (H2O6)2+]

a) M2 = (M1V1)/V2 M2 = (.191 M [Cu (H2O6)2+])( 1.0 mL)/ (50.0 mL) = .0038 M

b) (.191 M [Cu (H2O6)2+]) (2.0 mL)/ (50.0 mL) = .0076 M

c) (.191 M [Cu (H2O6)2+]) (5.0 mL)/ (50.0 mL) = .0191 M

d) (.191 M [Cu (H2O6)2+]) (7.0 mL)/ (50.0 mL) = .0267 M

e) (.191 M [Cu (H2O6)2+]) (10.0 mL)/ (50.0 mL) = .0382 M

f) (.191 M [Cu (H2O6)2+]) (12.0 mL)/ (50.0 mL) = .0458 M

g) (.191 M [Cu (H2O6)2+]) (15.0 mL)/ (50.0 mL) = .0573 M

h) (.191 M [Cu (H2O6)2+]) (17.0 mL)/ (50.0 mL) = .0649 M

i) (.191 M [Cu...