Introduction: Molarity (M) is used to determine concentration. Molarity is found by dividing the number of moles of a solute by the volume of the solution in liters. Multiple series of solutions with different concentrations can be used by diluting the concentration. The dilution technique is: Number Moles Concentrated Solution = Number Moles Dilute Solution.
An instrument called a spectrophotometer detects the amount of light that passes through the sample and the percent transmittance can be recorded from the meter. In the lab, multiple homogeneous solutions are made. There was not a way to determine the differences in concentrations, but the Spec 20 made it possible to measure the difference. The Beer-Lambert Law is a graph used to record data of admittance versus concentration. The relationship between percent transmittance and absorbance can be found by using the following equations: A= – logT or A=2 – log(%T).
The relation of this data can be used for quantitative analysis. When creating a hypothesis for this experiment, it was chosen that lower concentration makes light transmit through the solution easier. Procedure: The Spec 20 instrument warmed up at least 15 minutes before use. The wavelength was set to absorbance of 600 nm, which is the max for Cu2+. Ten small test tubes were cleaned and a piece of label tape was placed around the top of each tube. A vertical straight line was drawn on the tape to help align with the mark on the instrument sample holder. The transmittance of the Spec 20 was set to 0% using the left-hand adjustment knob while the well was empty and closed. One test tube was filled 3/4 full of a blank solution (distilled water).
A Kimwipe was used to wipe off any solution on the outside of the test tube. The test tube was then placed into the empty well with both the mark on the test tube and the mark on the well aligned. The lid of the well was closed. The right-hand adjustment knob was used for setting the transmittance to 100%. The percent transmittance of the standard solutions, 0.500 M, 0.200M, 0.100M, and 0.050M CuSO4 were measured on wavelength 600nm. The absorbance was calculated and recorded onto a table. Solid CuSO4 5H2O was used to create 20.0 mL of a 0.500 M solution. A 100 mL beaker was used to weigh the solid and 15mL of water was added to the beaker.
Water and solid were stirred to mix, then transferred back and forth between the beaker and graduated cylinder to ensure all the crystals had dissolved. Once dissolved, the solution was kept in the cylinder. The beaker was rinsed with small amounts of distilled water and then poured into the cylinder. The cylinder was less than 20 mL after rinsed solutions of the beaker were poured in. The pipette was then used to add water into the cylinder until the meniscus reached 20 mL. The mixture was then transferred back and forth between cylinder and beaker to ensure a homogeneous mixture.
Some amount was then poured into a test tube to check the percent transmittance in the Spec 20. This was considered the first solution. From the first solution, 20.0 mL of a 0.200 M CuSO4 solution was made. From an equation learned, an amount of water was added to dilute the sample. This was considered the second solution and the percent transmittance was checked. The same process was used to create 20.0 mL of a 0.100 M CuSO4 solution from the 0.200 M solution. 20.0 mL of a 0.050 M CuSO4 solution from the 0.100 M solution was then created. This were considered the third and fourth solutions. Percent transmittance was checked.
A clean test tube was filled with CuSO4 solution of an unknown concentration. The percent transmittance was recorded. Results: To find the percent transmittance of the concentration, a spectrophotometer was used. Once recorded, it was possible to find the absorbance of the solution by using the equation: A=2-log(%T). The Beer-Lambert Law was used to provide a calibration plot on the relationship between absorbance and concentration. From Figure 1, the calibration curve shows the absorbance vs. concentration of the standard solutions of Cu2+. Figure 2 shows absorbance vs. concentration of concentrations made.
The solutions in figure two were found my using the formula MC * VC = MD * VD. Concentration of molarity and volume were given from previous information in the lab. Dilution of molarity was also used from previous information form the lab. Dilution of volume was found using basic algebra. The VD plus water must equal 20 mL to create each made solution. Concentration is defined as molarity. It is found by dividing the number of moles of solute by the volume of solution in liters. Absorbance is defined as the amount of wavelengths taken in by the solution.
Figure 1: Calibration curve of standard Cu2+ solutions.
Figure 2: Calibration curve of made Cu2+ solutions. Discussion: Figure one gives a general idea of how the spectrophotometer worked. The spectrophotometer gave the percent transmittance, which made it possible to find the absorbance through a given formula. Figure one gives results of standard concentrations of Cu2+. Figure two was determined by made solutions in the lab. Solutions were made by adding water to dilute solutions. The Spec 20 gave the percent transmittance, and again made it possible to find the absorbance through a given formula. When completing the lab, different spectrophotometers were used because of an shortage. A different use of instruments can cause data to be less accurate. Different instruments are likely to measure differently, meaning data can’t be completely accurate.
Also, a spectrophotometer used in the lab had a broken lid on the well. It is possible that the lid wasn’t placed on exactly the way it should have been, and the transmittance wasn’t accurately shown on the meter. As the unknown data was recorded, it was obvious that the unknown was a concentration 0.2M. It was possible that the unknown sample, along with the other samples got contaminated from using the same pipet for each concentration. The calculated errors of concentration, were less than the calculated errors of absorbance. For solution one, the concentration error was 0%, meaning the spectrophotometer did an extremely good job.
Solutions two, three, and four had a higher concentration error of 11%, 11%, and 13%. Errors were very low for this part of the lab. As for the absorbance error,the instrument seemed to be more accurate when the absorbance level was higher. Solution one had an error of 28%, solution two had 22% error, solution three had a 5.8% error, and solution four had a 1.6% error. The experiment could have been improved if the same spectrophotometer had been used and the lid of the well wouldn’t have been broken. It is also possible that test tubes were slightly contaminated and should have been cleaned better before using them.
When placing the test tube into the well, there is a chance that some fingerprints and contamination on the outside of the tube, affected the accuracy of the spectrophotometer. As many other conflicts could have also been a reason, the ones listed are the most obvious in this lab. The conflicts could be fixed by allowing each lab group to have new test tubes, spectrophotometer for the lab, and for everyone to use gloves to keep fingerprints off the test tubes. Conclusion: The hypothesis was correct that the less that concentration is, the more light can be transmitted through the solution and less light is absorbed.
The hypothesis was supported by the data given by the spectrophotometer for each solution. Percent error was very low for the experiment meaning that there weren’t many mistakes throughout the experiment. If light passes through solution without any absorption, the absorbance is zero. If all light is absorbed, then the transmittance is zero. The technique of the spectrophotometer was accurate, but it was not as accurate on recording the absorbance level of lower concentration solutions. The instrument was better at reading the concentration of solutions. The experiment was good for learning exactly how dilutions work, and how to create new solutions from previous ones. Calculations: %T/100=T 92.9/100=0.929
A= -logT A= -log(0.929) A=0.032
MCxVC=MDxVD (0.2M)(0.1V)=(0.2M)(V) V=0.08L 0.08L=8.0mL of 0.5M
A=2-log(%T) A= 2-log(90.1) A=0.453
y=mx+b y=0.8573x+0.0091 y=0.8573(0.05)+0.0091 y=0.05
%error=(|measured value – actual value|/actual value)*100 %error= (|.1-.09 |/.09)*100 %error=11%