Kinetics of Crystal Violet Fading

March 10, 2018 | Author: w_kang | Category: Absorbance, Hydroxide, Chemical Reactions, Chemical Kinetics, Spectrophotometry
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This experiment used spectroscopy and graphical analysis to determine the rate law for the reaction of crystal violet wi...

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Walsh Kang Dr. Lowery AP Chemistry P1 26 November 2014 Kinetics of Crystal Violet Fading II. Objective: The purpose of this lab is to use spectroscopy and graphical analysis to determine the rate law for the reaction of crystal violet with sodium hydroxide. III. Background: Crystal violet is a violet-colored compound that loses its color when in a strong basic solution. The crystal violet cation (CV+) combines with hydroxide ions to form CVOH, a neutral colorless product. The first reaction is shown here: CV+ + OH-  CVOH The general rate law form of this reaction is shown here: Rate = k[CV+]n[OH-]m The variables n and m are the order of the reaction with respect to each reactant and k is the rate constant at a specific temperature. If there is a huge excess of OH-, then the concentration of OH- will fold into the rate constant, changing the rate law as so: Rate = k’[CV+]n, where k’ = k[OH-]m This is a pseudo rate law because it is a simpler version of the actual rate law. A Beer’s Law curve must be made in order to convert the absorbances measured from the spectrophotometer into concentrations. The general relationship between concentration and absorbance is this: A = abc A is the absorbance, a is the molar absorptivity coefficient, b is the path length in centimeters, and c is the concentration. Beer’s Law is used to track the concentration of CV+ as the reaction progresses in a test tube with the absorbance being measured.

After finding the concentrations at certain times of the reaction of CV+ and OH-, graphical analysis can be used, because zero, first, and second order derivative rate laws can be graphed: zero order ([CV+] vs. time), first order (ln[CV+] vs. time), or second order (1/[CV+] vs. time). If the data shows a straight line, then the reaction is that order. IV. Materials: Materials for part 1 (constructing Beer’s Law curve) -

Crystal violet solution, 25 µM (2.5 * 10-5 M), 50 mL Sodium hydroxide solution, 0.02 M, 30 mL Distilled water Beaker, borosilicate, 50 mL Spectrophotometer Test tubes Pipets, serological, 10-mL Pipet bulb filler 5 disposable pipets Stirring rod Permanent marker

Materials for part 2 (graphical analysis of rxn. of CV+ and OH-) -

Crystal violet solution, 25 µM (2.5 * 10-5 M), 50 mL Sodium hydroxide solution, 0.02 M, 30 mL Spectrophotometer 1 test tube, 13*100mm 2 pipets, serological, 10-mL Pipet bulb filler Stirring rod Stopwatch 1 Kimwipe Permanent marker

V. Safety Issues: For both parts of the experiment, sodium hydroxide solution is irritating to eyes and skin. Crystal violet will stain clothes and skin. Caution should be used with these solutions. Goggles, gloves, and aprons should be worn. Chemicals should be avoided. All lab safety rules should be adhered to. VI. Procedure: Procedure for obtaining a calibration curve for crystal violet 1. Turn on the spectrophotometer and allow it to warm up for 15-20 minutes before. Adjust the wavelength setting to 540 nm.

2. Obtain five test tubes and label them with a permanent marker: A, B, C, D, and E. Using one serological pipet and pipet bulb filler, fill test tube A with 9.0 mL of water, and using the other serological pipet, fill test tube A with 1.0 mL of the stock solution of crystal violet. 3. Making sure to use each pipet for its own respective solution, construct the other four solutions: 8.0 mL of water and 2.0 mL of stock solution in test tube B, 7.0 mL of water and 3.0 mL of stock solution in test tube C, 6.0 mL of water and 4.0 mL of stock solution in test tube D, and 5.0 ml of both water and stock solution in test tube E. Use the stirring rod to mix all of these solutions, making sure to dry the stirring rod after each stir. 4. Take one test tube, 13 by 100mm, and make a small line on the top of the tube, to mark where the tube will be lined up in the spectrophotometer. Fill the tube three quarters of the way full with distilled water. 5. Put the test tube into the spectrophotometer and adjust the percent transmittance to 100 percent. 6. Take the test tube out and fill it ¾ of the way full with mixture A using a disposable pipet. Wipe the exterior of the tube with the Kimwipe and place the test tube in the spectrophotometer, making sure to line up the drawn line with the mark in the spectrophotometer. 7. Set the spectrophotometer to absorbance and record the absorbance of mixture A. 8. Take the test tube out and pour out mixture A. With a different disposable pipet, take some of mixture B and rinse out the test tube. Then, pipet mixture B into the test tube. 9. Repeat steps 6-8 with all of the mixtures A-E, by rinsing with the next mixture and recording the absorbances.

Procedure for obtaining the graphs needed to find the order of the reaction with respect to CV+: 1. Using a serological pipet with a bulb filler, pipet 830 mL of .02 M naOH into the 13 by 100 mm test tube. Make a mark on the test tube with the permanent marker. Wipe the test tube with the Kimwipe. 2. Set the wavelength of the spectrophotometer to 540 nm and place the test tube with 8.0 mL of .02 M NaOH in the spectrophotometer set for transmittance by lining up the marks. 3. Adjust the percent transmittance to 100, then take the test tube out of the spectrophotometer. 4. Empty the test tube and pipet 4.0 mL of the 25 µM crystal violet into the test tube with the 4.0 mL of .02 M Naoh, starting the stopwatch as soon as the crystal violet is pipetted. 5. Use the stirring rod to mix the crystal violet and NaOH, wipe the test tube, and put the test tube into the spectrophotometer by lining up the marks.

6. Flip the switch of the spectrophotometer to absorbance. Steps 5-7 should have happened in less than 20 sec. 7. Record the absorbance value every 20 seconds for 20 minutes.

VII. Results: Data Table 1 (measured with a wavelength of 540 nm) Mixt ure

Absorba nce

A

Concentratio n of CV+ (µM) 2.5

B

5.0

.24

C

7.5

.34

D

10.0

.48

E

12.5

.58

.12

Graph 1

Abs. vs. [CV+] (µM) A b s o r b a n c e

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 2

4

6

8 [CV+] (µM)

Y = .0464x + .004, r = .9988

10

12

14

Data Table 2 (measured with a wavelength of 540 nm) ***attach data from Excel***

Graphs 2-4 ***attach data from Excel and write in r-values*** Calculation 1 (determination of the order with respect to CV+)

Calculation 2 (determination of the order with respect to OH-)

Final rate law

Guided inquiry design and procedure A. 1-4 and B. 1-4

VIII. Conclusion: The objective of this experiment was to find the rate law of the reaction between crystal violet and sodium hydroxide. This was done by first constructing a Beer’s law curve to have the ability to convert absorbance into concentration, then measuring the change in absorbance (inherently concentration) of crystal violet as the reaction commenced. The rate law was determined to be Rate = k[CV+][OH-] 3, an overall fourth order reaction. With a fourth order reaction being unlikely, there are three sources of error that could have contributed to this absurd order. The three errors are a lack of agitation in the test tube containing CV+ and OH-, having too little CV+ in the solutions used to construct the Beer’s law curve, and not pipetting enough CV+ into the test tube with sodium hydroxide. The first error is not having agitation in the test tube that contains the reaction of CV+ and OH-. This would change the order of the reaction. Without any agitation, the CV+ and OH- cannot react as much because the molecules of CV+ will not collide with as many OH- ions. This reduces the rate of the reaction because there are not as many effective collisions or frequency of collisions and so since the rate decreased, the concentration will be higher at a certain time because there is more CV+ left than there should be. The slope of [CV+] vs. time will be much shallower because the concentration won’t change as much. That means that the slope of ln[CV+] vs. time and 1/[CV+] vs. time will also be shallower. The amount the points change cannot be determined because the magnitude of the change in

slopes cannot be determined exactly. The r-values may or may not change, but they most likely will. This might force an order to be preferred over another order. So the lack of agitation in the test tube will change the order of the reaction. A way to prevent this error is to use the stirring rod and mix the solution well. The second error is having too little CV+ in the solutions used to construct the Beer’s law curve. This is a decrease in concentration. According to the equation A = abc, concentration and absorbance are directly proportional, so a decrease in concentration would lead to a decrease in absorbance. The slope and y-intercept of the Beer’s law curve will change, depending on how lacking the CV+ is. Yet if all of the solutions lack the same amount of CV+, only the y-intercept will decrease, with the slope staying the same. So now, for any absorbance, the corresponding concentration will be larger-than-normal because you are subtracting a smaller bvalue from the absorbance to calculate the concentration. There would be an increase in the values for ln[CV+] and a decrease in the values for 1/[CV+]. All of these changes in values would be constant and so the shapes of the graphs would not change. Therefore the order would be determined just the same. This error would not have an effect on the determination of the order of the reaction with respect to either reactant. A way to avoid this source of error would be to properly pipet the CV+ and to make sure that the dead space or lack of it is observed. The third error is not pipetting enough CV+ into the test tube with sodium hydroxide. According to the rate law, a decrease in concentration of CV+ (which is what happens when less CV+ is pipetted into the test tube), will lead to a decrease in rate. So now the absorbances do not correlate with the right times. Since the rate decreased, the concentration will be higher at a certain time because there is more CV+ left than there should be. The difference in CV+ remaining is more evident as time progresses. The slopes of the graphs of [CV+] vs. time, ln[CV+] vs. time, and 1/[CV+] vs. time will be shallower, possibly changing the r-values. This will possibly change the determination of the order. Depending on what the final graphs look like, the order might look like zero, first, or second. This is how not pipetting enough CV+ into the test tube with sodium hydroxide affects the determination of the order of the rate law. A way to avoid this error would be to properly pipet and to make sure that the dead space or lack of it is observed.

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