Harvard University Crystal Violet Decolorize Chemistry Lab Report

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Harvard University

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Lab Investigation 3 - How Fast Does the Crystal Violet Decolorize?

GUIDING QUESTION

What is the rate law for decolorization of crystal violet?

INTRODUCTION

Crystal violet is used to dye paper and as a component of navy blue and black inks for printing, ballpoint pens, and ink jet printers. It is also used to colorize diverse products such as fertilizers, antifreezes, detergents, and leather jackets. Crystal violet decolorizes in the presence of hydroxide ions, which can limit its usefulness as a dye or ink; on the other hand, it makes crystal violet a useful pH indicator. This process is illustrated with Equation 1.

Figure:Equation 1.

Equation 1 can be simplified to:

CV+ + OH- ? CVOH

Given that the order with respect to [OH?] is first order, you will design and carry out an experiment to determine the order of the decolorization reaction with respect to [CV+] and the value of the rate constant, k.

FUNDAMENTALS OF CHEMICAL KINETICS

The rate of a reaction is expected to depend on the concentrations of reactants and possibly other substances. A rate law represents the dependence of the rate on concentrations, for example

rate = k[A]x[B]y

where k is the rate constant and the exponents (x, y) are described as the order with respect to that particular substance.

In this investigation, we will derive the rate law for the decolorization of crystal violet by hydroxide. In order to determine the rate law, we need to design an experiment that measures the concentration of a species at a particular time during a reaction. It is exceedingly difficult to get an accurate measurement of a concentration at a known time because the techniques used to measure concentrations don't work instantaneously, but rather take time to perform.

One of the earliest methods used to measure concentrations at specified times was to quench the reaction either by flash freezing it or by adding a substance that severely inhibits the reaction. The order would then be determined using the ratio method of comparing the effect of initial concentration on initial rate. These techniques are problematic because one can't be sure that the reaction has completely stopped. The reaction may still be going on during the analysis. Additionally, the reaction mixture is destroyed for the purposes of kinetic experiments, so the chemist must make multiple trial runs and waste a large amount of reagents to observe the concentrations at multiple points in time.

A more modern technique to measure concentration is absorbance spectroscopy. This experimental technique may be used when a product or reactant has an absorbance frequency unique to those of other components of the reaction mixture. By measuring the absorbance of a particular product or reactant at a variety of known concentrations, you can construct a plot of absorbance versus concentration called a Beer's law plot. This calibration chart allows you to calculate the unknown concentration given the reaction solution's absorbance. The advantage of this method is that a large number of data points with well known times can be quickly collected using only one reaction mixture.

Once you have data on concentration versus time, it is possible to infer the order of a reaction. The strategy works by comparing concentration versus time information to the mathematical predictions made using integrated rate laws. This is usually accomplished by assuming that if a reaction has a certain order, a plot of the data is then made in a format that should be linear if the assumption about the rate law is correct. If we guess correctly, the graph is linear. If we are wrong, then the graph is curved and we need to choose another order against which to plot the data. By integrating the various differential rate laws, the integrated rate laws for a reaction can be derived. The forms of the integrated rate laws for reaction orders zero through two are presented in Table 1.


You may have noticed that we only list three integrated rate laws, ignoring rate laws like rate = k[A][B]. That doesn't mean we can't use integrated rate laws to determine the rate law for those types of reactions; we just have to separate the two components. In order to determine the order of each component of a two-component rate law, we can make the concentration of one component, for example B, so large as compared to A that the concentration of B is almost constant. Assuming that the concentration of B is constant, the reaction becomes pseudo-order, and the rate constant is termed a pseudo k.

GOALS

As you complete this investigation you will:

Design an experiment that can measure the rate of decolorization of crystal violet in basic solutions as a function of time.

Determine the rate law for this reaction including the actual value of k.

MATERIALS AVAILABLE FOR USE

1.0 × 10?4 M crystal violet solution

0.10 M NaOH(aq)

1.0 M HCl(aq) (for cleaning glassware)

Spot plate

Pipettors and tips

Kimwipes

Vernier LabPro system with colorimeter

Cuvettes

SAFETY PRECAUTIONS

Caution: Wear goggles at all times.
Caution: Neutralize any excess liquid solutions and wash down the sink with water.
Caution: HCl and NaOH are corrosive. They can attack the skin and cause permanent damage to the eyes. If one of these solutions splashes into your eyes, use the eyewash station immediately. Hold your eyes open and flush with water. If contact with skin or clothing occurs, flush the affected area with water.
Caution: Be careful handling the crystal violet solution. It can stain skin and clothing. Gloves are available for you to wear.

GETTING STARTED

Your first task is to prepare a Beer's law calibration curve for absorbance versus concentration of crystal violet, [CV+].

  1. Download worksheet Download Download worksheet
  2. Your second task will be to determine the order of the crystal violet decolorization reaction with respect to the crystal violet concentration, [CV+].

You may want to start by just observing the reaction. Place about 8 drops of water in a spot plate and add one drop of 1.0 x 10-4 M crystal violet solution. Now add a drop of 0.10 M NaOH(aq) (this represents a large excess of hydroxide ion).

PROCEDURE

You need to make sure that your decolorization reaction occurs within the linear range of your calibration curve and that you have a large excess of hydroxide ion, OH?. Based upon the fact that one or more data points were dropped from your calibration curve, two dilutions should place your experiment in the linear range.

  • Make the first dilution with distilled water and the second with 0.10 M NaOH. This will give you a large excess of NaOH.

Have the colorimeter calibrated and ready before you perform the hydroxide dilution.

  • Fill the cell; then click start on the computer.

Your experiment should monitor the reaction until the absorbance value of the reaction mixture is 0.20 or less.

Use the regression equation from your calibration curve to find the concentration at the selected time intervals over the entire range of data.

  • Plot the concentration and time data, for zero, first, and second order reactions. Based on the correlation coefficient, determine the order.

Once you decide on the order, determine pseudo k from the slope.

Determine actual k, using the initial concentration of NaOH.

  • INTERACTIVE POSTER SESSION
  • Once your group has completed your work, prepare a whiteboard that you can use to share and justify your ideas. See the handout provided for details on this process.
  • REPORT
  • Once you have completed your research, you will need to prepare an investigation report that consists of three sections. This report may require more than 2 pages with data tables. This report must be typed and any diagrams, figures, or tables should be embedded into the document.
  • Section 1: What concept were you investigating, and how does it relate to the guiding question? How fast does crystal violet decolorize? Describe how kinetics concepts will allow determination of rate law using graphical analysis of integrated rate laws.

Section 2: How did you go about your work and why did you conduct your investigation in this way? Specifically, what technique is used to monitor the reaction and how is the Beer-Lambert calibration curve used?

Section 3: What is your argument? Your report should include the linear regression equation from the Beer-Lambert calibration curve along with the correlation coefficient; one sample calculation of how concentration at a selected time was determined; a data table with selected time and concentration values; the results from integrated rate law graphical analysis to determine order; how the actual value of the rate constant, k, was determined; and the complete rate law with the actual value of k included in the rate law with the correct units.

This third section is where you not only present your data, but use the values you obtain as evidence in your reasoning. Statements like, "see data table for values" are not acceptable!

Remember:  An argument is not just an answer to the question. It is a claim or conclusion supported by evidence with a rationale for why the evidence supports the claim or conclusion.

Unformatted Attachment Preview

Order Plot 0 1 [CV +¹] vs. time In [CV +¹] vs. time 1 [CV +¹] vs. time Slope -0.0000000774 -0.00221 0902 Correlation 0 -0.759 -0.761 0 0.541Use your regression equation to calculate concentration at each of the absorbance values below: Time 0 100 200 300 400 500 600 700 800 900 B or B 8 Absorbance 1.007 0.894 OF 0.700 On 0.928 Or 0.893 0.850 OB 0.803 q 8€ 0.751 8€ 0.699 0.646 On > [CV +¹] 0.0000478 0.0000405 0.0000278 0.0000427 0.0000362 × 0.0000376 0.0000345 0.0000312 0.0000278 0.0000243 In [CV +¹] -9.95 -10.1 -10.5 OB -10.1 08 -10.2 Or -10.2 Or -10.3 Or -10.4 Or -10.5 Or -10.6 1 [CV +1] 20921 0 20,900 24691 24,700 39571 36,000 23419 0 23,400 27624 024,800 26596 0 26,600 28986 32051 0 29,000 32,100 X X X 041,200 35971 * 36,000 X X X X X 411523 XEnter the slope, intercept, and correlation that you calculated from your last assignment: Regression equation A = 15335 x [20000,60000] C + 0.2733 x [-0.0800,0.100] R = 0.907 o [0.900,1.00]
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Explanation & Answer:
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Explanation & Answer

View attached explanation and answer. Let me know if you have any questions.Hiii Buddy! There was a lot of deviation in your provided data. However, I have made the report in 3 ways than.1st is with almost corrected data. (I have attached excel file also with it)2nd is with your provided data but I included tables from your data in that file. (Graphs are also in excel file)3rd is with your provided data and I have included your provided pictures in that file. (Graphs are also in excel file)Excel file is attached so you can observe the differences in the provided data.According to provided data the reaction is zero order. but in actual the reaction is first order. (However, I used first order notation in all files).

t

Abs

CV

0
100
200
300
400
500
600
700
800
900

1.007
0.894
0.7
0.928
0.893
0.85
0.803
0.751
0.699
0.646

Ln(CV)
1/CV
1.99E-05 -10.8247908 50251.26
1.93E-05 -10.8554055 51813.47
1.67E-05 -11.0001018 59880.24
1.42E-05 -11.1622686 70422.54
1.21E-05 -11.3223051 82644.63
1.01E-05 -11.5029751 99009.9
8.64E-06 -11.659108 115740.7
7.19E-06 -11.8428194 139082.1
6.09E-06 -12.0088625 164203.6
5.02E-06 -12.2020806 199203.2

1/[CV+] vs. Time
250000

1/[CV+]

200000
150000
1/[CV+]

100000

Linear (1/[CV+])
50000

y = 16216x + 14038
R² = 0.9344

0
0

100

200

300

400

500

600

Time (second)

-1.77345E-08

-0.99089133

-0.001597247

-0.99561924

162.1592333

0.966620488

700

800

900

[CV+] vs. Time
2.50E-05

[CV+] (M)

2.00E-05

1.99E-05
1.93E-05
1.67E-05

1.50E-05

1.42E-05
1.21E-05
1.01E-05
8.64E-06
7.19E-06
6.09E-06
y = -2E-06x + 2E-05
5.02E-06
R² = 0.9819

1.00E-05
5.00E-06

[CV+]
Linear ([CV+])

0.00E+00
0

100 200 300 400 500 600 700 800 900

Time (second)

Ln[CV+] vs. Time
-9.5
0

100 200 300 400 500 600 700 800 900

-10

Ln[CV+]

-10.5
Ln[CV+]

-11

Linear (Ln[CV+])

-11.5
-12
-12.5

y = -0.1597x - 10.56
R² = 0.9913

Time (second)

t

Abs

CV

0
100
200
300
400
500
600
700
800
900

1.007
0.894
0.7
0.928
0.893
0.85
0.803
0.751
0.699
0.646

Ln(CV)
1/CV
4.78E-05 -9.94848492 20920.5
4.05E-05 -10.1142086 24691.36
2.78E-05 -10.4904745 35971.22
4.27E-05 -10.0613116 23419.2
3.62E-05 -10.2264514 27624.31
3.76E-05 -10.1885065 26595.74
3.45E-05 -10.2745512 28985.51
3.12E-05 -10.3750925 32051.28
2.78E-05 -10.4904745 35971.22
2.43E-05 -10.6250342 41152.26

1/[CV+] vs. Time
45000

40000
35000

1/[CV+]

30000
25000
20000

1/[CV+]

15000

y = 1558.3x + 21168
R² = 0.5394

10000
5000
0
0

100

200

300

400

500

600

Time (second)

-1.85818E-08

-0.7590413

-0.000530162

-0.74915454

15.58276045

0.734435384

700

800

900

Linear (1/[CV+])

[CV+] vs. Time
0.00006

[CV+] (M)

0.00005

0.0000478
0.0000405

0.00004

0.0000427

0.0000376
0.0000362
0.0000345
0.0000312
0.0000278
0.0000278
[CV+]
0.0000243
Linear ([CV+])
y = -2E-06x + 5E-05
R² = 0.5761

0.00003
0.00002
0.00001

0
0

100 200 300 400 500 600 700 800 900

Time (second)

Ln[CV+] vs. Time
-9.6
0

100 200 300 400 500 600 700 800 900

-9.8

Ln[CV+]

-10
Ln[CV+]

-10.2

Linear (Ln[CV+])

-10.4
-10.6
-10.8

y = -0.053x - 9.9879
R² = 0.5612

Time (second)


How Fast Does the Crystal Violet Decolorize?

Name group
Department
Instructor
Institution

June 28, 2022

How Fast Does the Crystal Violet Decolorize?

Section1:
In this lab experiment, we were assigned to investigate the concept of Crystal Violet
decolorization by using Kinetics. We were supposed to evaluate the rate constant (k) for this
particular chemical reaction. Chemical Kinetics is the study of reaction rates by using a set of
known orders (0, 1, and 2). It is used to determine the rate constant of different chemical
reactions. We have done that by selecting particular time spots in the set chemical reaction. We
measured the time in seconds, and plotted it against the concentration of the known solution
and the natural log of the known solution. Then we did the same process with the reciprocal of
the known solution. Once we plotted these results, the data created three graphs. Each was
displaying a characteristic trend for aforesaid order. Each graph displayed a coefficient of
determination, also known as the R^2. It tells the experimenter which order the reaction falls
under by looking for the R^2 closest to the numerical value of one. We found, then, rate
constant (k) by the slope of this graph and the concentration of the reactant in excess. The rate
law illustrates how the rate of the reaction depends upon the concentration of the reaction and
is labeled by orders. For the purpose of this experiment, the we observed and classified the rate
in terms of disappearance.

Section 2:
First, we began by diluting a solution of 5mL of Crystal violet and 5mL of water into 8
unique series, each containing a lower concentration than the solution before it. Using a concept
known as Spectroscopy, or the study of the absorption and radiation of light in terms on
concentration and absorption, we were able to come up with a linear regression equation from
the set graph particular to the Crystal violet solutions. This linear regression equation represents

the relationship between the concentrations and the absorptions of the solution in the form y =
mx + b and can be used to determine unknown concentrations (x), from given absorptions (y) or
vice versa. This is due to the Beer Lambert law that states that absorbance and concentration are
directly proportional to one another. It was important to analyze the R^2 value given for the graph
and determine if the value was close enough to 1 to continue on with the most accurate form of
experimentation. We found that the value was not close enough to one, and found it useful to
eliminate the first and second solution points in order to come up with a coefficient of
determination of R^2 = 0.9913. Once those two-point values were eliminated, we came up with
the following Linear Regression equation: y= -0.1597x - 10.56, where the x represents
concentration and the y represents absorption. It was than time to begin part two of
experimentation. The following lab day the groups began experimentation by again creating a
dilution of 5mL of Crystal Violet and 5mL of water, but this time followed by a second dilute
solution made of up of 5mL of solution 1 and 0.10 M of NaOH. It was than crucial to quickly put
the second soluti...


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