Lab: Kinetics of Crystal Violet + NaOH
Name ________________________
Background
The piece of equipment we will be using for this lab is usually called a spectrophotometer
or just a “Spec 20”. It provides a means to determine the %Transmittance of light
through a solution at various wavelengths in the visible range. We can then see which
wavelengths of light our compound of interest is most absorbent. Using a wavelength of
maximum absorbance, we can then measure the transmittance of different concentrations
of our compound, which will then be used to produce a calibration graph for our
compound. This calibration graph will be used to determine the concentration of a
solution containing our compound.
The Math
The Transmittance (T) is defined as the ratio of the transmitted light (I) to the incident
light (Io). The percent transmittance is the transmittance converted to a percent. The
absorbance (A) is defined as the log of the incident light divided by the transmitted light.
In equation form, we have:
T = I/Io %
T = I/Io X 100%
A = log(Io/I) = log(1/T)
Our Spec 20 measures the %T, which can then be converted to absorbance. For
quantitative work the absorbance is more useful than the transmittance because it is
directly proportional to concentration
A = 2 – log(%T)
For quantitative work the absorbance is more useful than the transmittance because it is
directly proportional to concentration. The relationship is known as Beer’s Law:
A = abc
where “a” is a constant of proportionality (called the molar absorptivity) which varies
from substance to substance, “b” is the path length through which the light travels, and
“c” is the concentration in mol/L. Thus, for a given substance at a specific wavelength in
a given sample holder of fixed path length, a plot of A versus concentration is a straight
line.
Calibrating the Spec 20
Since our solutions are taking place in an aqueous environment, we must account for the
fact that the water (and the cuvet itself) will absorb some of the light passing through it.
By calibrating the % Transmittance to be 100% with a blank sample (one containing just
distilled water), we will effectively factor out any absorbance due to the water. The Spec
20 will need to be re-calibrated every time you change wavelengths.
1
Part 1 Pre-Lab
Suppose the data below was collected for Part 1 of this experiment.
 (nm)
350
92.0
 (nm)
500
41.7
370
89.4
520
31.0
390
85.3
530
28.6
400
80.1
540
24.7
420
440
72.6
71.3
550
570
20.6
16.9
455
65.5
580
12.2
470
60.1
600
12.2
490
52.9
625
23.5
%T
A
%T
A
1) At what wavelength should the analysis for Parts 2 and 3 be done?
2) Calculate the absorbance for each % T.
Answers:
1) About 590 nm
2)
 (nm)
%T
350
92.0
370
89.4
390
85.3
400
80.1
420
72.6
440
71.3
455
65.5
470
60.1
490
52.9
A
 (nm)
%T
500
41.7
520
31.0
530
28.6
540
24.7
550
20.6
570
16.9
580
12.2
600
12.2
625
23.5
0.036212
0.048662
A
0.379864
0.508638
0.069051
0.096367
0.139063
0.14691
0.543634
0.607303
0.686133
0.772113
0.183759
0.221126
0.276544
2
0.91364
0.91364
0.628932
Part 1: Absorption Spectra for 1.5 X 10-5 M crystal violet
1) Dilute 15 mL of 1.0 X 10-4 M crystal violet solution to 100. mL to create a solution
that is 1.5 X 10-5 M.
2) Set the wavelength to the wavelength 350 nm. Make sure the filter in the lower lefthand area of the Spec 20 is set to the correct spot. You will eventually need to change
this filter.
3) Make sure the “Transmittance” selection is chosen. If not, hit the MODE button until
“Transmittance” is selected. With the chamber empty and the lid closed, adjust the
left meter to 0% transmittance. Place the “blank” cuvet (one 2/3 filled with just
distilled water) into the chamber and close the lid. Adjust the right control knob to
read 100 % transmittance. Place the cuvet containing the crystal violet solution into
the chamber, close the lid, and record the %Transmittance. Hit the Mode button to
select Absorbance, and record the absorbance in the data table.
4) Repeat steps 2 and 3 for the wavelengths specified in the chart.
 (nm)
350
%T
A
 (nm)
500
370
520
390
530
400
540
420
440
550
570
455
580
470
600*
490
625
%T
A
*switch filter on lower left side of Spec 20 at 600 nm
Data Analysis
Use Excel to construct a graph of %Transmittance vs. Wavelength and a graph of
Absorbance (A) versus Wavelength for crystal violet. The x-axis on this graph needs to
run from 350 nm up to 625 nm. You should have two separate graphs when you are
finished, preferably on the same page. Use those graphs to help answer the questions
below:
3
Follow-up Questions:
1) At what wavelength(s) does crystal violet absorb the maximum amount of radiation?
What can you say about the %T at these wavelengths?
2) When a solution is red, does it absorb or transmit red light wavelengths? Explain.
3) Suppose a student got some fingerprints on the cuvet prior to taking data. What
effects would this error have on the % Transmittance and the Absorbance data
collected? Explain.
4) The Spec 20 can determine the absorbance of solutions that absorb in the 350 nm -650
nm range. What range of the electromagnetic spectrum does this range correspond to?
5) Suppose a solution allows 50.0% of the light that hits it to pass through at a certain
wavelength. What would be the absorbance of the solution at this wavelength? Hint:
it’s not 0.500!
4
Part 2: Beer’s Law
Background:
The amount of light a substance absorbs is governed by Beer’s Law, which states that the
absorbance is proportional to the product of the molar absorbtivity constant, the path
length, and the concentration:
A = abc
where a = molar absorbtivity constant
b = path length (usually in cm)
c = concentration (usually in M)
The absorbance (A) is dimensionless, the path length (b) is usually given in cm, and the
concentration (c) is usually expressed in M. Therefore, the molar absorbtivity constant
(a) has units of cm-1M-1. If the absorbance is measured as a function of concentration, the
relationship will be linear because the molar absorbtivity constant and the path length are
both constant. A graph of Absorbance versus Molarity is shown below:
Absorbance vs. Concentration
1.2
1
Absorbance
0.8
0.6
0.4
0.2
0
Molarity (M)
If a linear trendline is added to the plot, the slope can be set equal the product of molar
absorbtivity times the path length. Mathematically, y = mx becomes A = abc. If the
slope is dived by the path length, the molar absorbtivity constant for that substance is
obtained:
Slope = ab
a = Slope/b
5
Part 2 Pre-Lab:
The data below represents some hypothetical data collected for Part 2 of this experiment.
mL of 1.50 X 10-5 M Crystal Violet
mL of distilled H2O
10.0
90.0
[crystal violet]
(M)
1.50 X 10-6
A
0.484
20.0
80.0
3.00 X 10-6
0.968
30.0
70.0
4.50 X 10-6
1.452
40.0
60.0
6.00 X 10-6
1.936
50.0
50.0
7.50 X 10-6
2.419
60.0
40.0
9.00 X 10-6
2.903
70.0
30.0
1.05 X 10-5
3.387
80.0
20.0
1.20 X 10-5
3.871
90.0
10.0
1.35 X 10-5
4.355
Make a graph of Absorbance vs [CV], and find the slope of the line. You do not need to
turn this graph in, you just need to use it for a few minutes!
1) Assuming a path length of 1.27 cm, what is the molar absorbtivity for crystal violet?
Answer: 254,000 M-1cm-1
6
Procedure:
1) Put on safety glasses.
2) The crystal violet you have at your disposal is too concentrated for practical use.
Using a graduated cylinder, measure 75.0 mL of the 1.00 X 10-4 M crystal violet into
a 500 mL volumetric flask, and add distilled water until the solution is diluted to
500.0 mL. The diluted solution will have a crystal violet concentration of 1.50 X 10-5
M.
3) Set the Spec 20 to the wavelength of maximum absorbance you determine in Part 1.
This should be around 590 nm or so. Make sure that the transmittance radio dial is
selected. If it is not, hit the MODE button until transmittance is selected. Once you
set the wavelength, you will not need to readjust it for this portion of the lab.
4) With the chamber empty, set the left knob to 0% T.
5) Place a blank (a cuvet 2/3 filled with distilled water) into the chamber, and set the
right knob to 100% T.
6) Combine 10.0 mL of the 1.50 X 10-5 M crystal violet with 90.0 ml of distilled water in
a beaker and mix thoroughly. Make sure to measure both volumes with dedicated
graduated cylinders.
7) Rinse the cuvet with several mL of this dilute crystal violet solution, and place the
cuvet into the chamber. Hit the MODE button until Absorbance is selected, and record
the absorbance in the data table below.
8) Repeat steps 6 and 7 with the mixtures shown below.
9) When all the data has been collected, remove the cuvet from the chamber and rinse
with several mL of distilled water.
mL of 1.50 X 10-5 M Crystal Violet
mL of distilled H2O
10.0
90.0
[crystal violet]
(M)
1.50 X 10-6
20.0
80.0
3.00 X 10-6
30.0
70.0
4.50 X 10-6
40.0
60.0
6.00 X 10-6
50.0
50.0
7.50 X 10-6
60.0
40.0
9.00 X 10-6
70.0
30.0
1.05 X 10-5
80.0
20.0
1.20 X 10-5
90.0
10.0
1.35 X 10-5
7
A
Calculations:
1) Make a graph of Absorbance vs. Molarity of Crystal Violet, and add a linear trendline.
Record the slope below (including units) and attach your graph to the end of this lab.
Slope = ____________
2) Using the slope of the graph, calculate the molar absorbtivity for crystal violet to three
significant figures. Make sure to include units. The path length of the cuvets used in
this lab is 1.27 cm (1/2 inch).
Follow-up Questions
1) Show that diluting 75.0 mL of 1.00 X 10-4 M crystal violet to 500.0 mL produces a
crystal violet solution with a concentration of 1.50 X 10-5 M.
2) For the third solution in the data table, I tell you to mix 30.0 mL of 1.50 X 10-5 M
crystal violet with 70.0 mL of distilled water. Show that the resulting concentration
of crystal violet is 4.50 X 10-6 M.
3) Why is the cuvet rinsed each time with a small amount of the test solution before the
solution’s absorbance is measured?
8
4) Consider why I have you mix large quantities of crystal violet and distilled water for
the dilutions. For example, mixing 1 mL of crystal violet with 9 mL of water should
have the same effect as mixing 10 ml of the crystal violet with 90 mL of water. You
didn’t need much solution for each cuvet, so why is it advantageous to use larger
quantities?
5) Why do I have you use the same cuvet each time?
6) Suppose a student gets fingerprints on the cuvet while they are adding the crystal
violet solution. What effect would this mistake have on the measured value of the
absorbance? Explain.
7) Why could we use the Spec 20 to make a calibration curve for the concentration of
CuSO4 or CoCl6, but we could not use it to make a calibration curve for NaCl or
ZnCl2? Hint: think about follow-up question #4 in Part 1.
9
Part 3: Determination of the rate law of the reaction
The reaction of crystal violet and sodium hydroxide can be symbolized as:
CV + NaOH → colorless products
The rate law is Rate = k [CV]x [NaOH]y
For both the kinetic runs today, the [NaOH] > > [CV], so that the [NaOH] remains
virtually constant throughout the experiment. Therefore, the rate law can be written as:
Rate = k’[CV], where k’ = k[NaOH]y
By running the experiment twice, with different concentrations of sodium hydroxide,
plots of [CV], ln[CV] and 1/[CV] can be made to determine the order with respect to
[CV], and the relative slopes of the appropriate graphs will give us k’ for each run, and
therefore the order of the reaction with respect to [NaOH].
The experiment will be run at a wavelength of 590 nm, where crystal violet had
maximum absorbance. We will be using Beer’s Law:
Absorbance = molar absorbtivity x path length x concentration (mol/L)
(A = a x b x c)
For today’s lab, the molar absorbtivity of crystal violet at the wavelength of maximum
absorbance (about 590 nm) has been determined from Part 2, and the path length through
the cuvet is 1.27 cm. So, knowing the absorbance allows one to find the concentration of
CV:
[CV] = _____A_______
a ∙ 1.27 cm
The starting materials will be 1.00 X 10-4 M CV and 0.100 M NaOH.
10
Part 3 Pre-Lab
Assume the data below was collected for part 3 of this lab.
Time
(minutes)
0
1
2
3
4
A
(first run)
0.454545
0.766284
0.416667
0.475059
0.384615
0.344234
0.357143
0.269906
0.333333
0.221976
0.3125
0.188501
0.294118
0.1638
0.277778
0.144823
0.263158
0.129786
5
6
7
8
A
(second run)
[CV]
(first run)
[CV]
(second run)
1) Calculate the missing information on the data table. You can assume that the results
from the pre-lab for part 2 apply: a = 254,000 M-1cm-1 and b = 1.27 cm.
2) Make appropriate graphs to determine the rate law for the data collected in part 3.
Make sure to include a numerical value for k, including units. You do not need to
include these graphs for this pre-lab with this packet!
Rate = _________________ k = ____________
Answers
1)
0
1
2
3
4
5
6
7
8
0.454545
0.766284
1.40909E-06
2.37549E-06
0.416667
0.475059
1.29167E-06
1.47269E-06
0.384615
0.344234
1.19231E-06
1.06713E-06
0.357143
0.269906
1.10715E-06
8.3671E-07
0.333333
0.221976
1.03333E-06
6.88127E-07
0.3125
0.188501
9.68752E-07
5.84354E-07
0.294118
0.1638
9.11768E-07
5.07781E-07
0.277778
0.144823
8.61114E-07
4.48952E-07
0.263158
0.129786
8.15791E-07
4.02337E-07
2) Rate = k [CV]2 [NaOH]2 k = 6.5 X 108 M-3min-1
11
Procedure:
(First Run)
1) Set the Spec 20 to the wavelength of maximum absorbance (around 590 nm or so),
and calibrate.
2) Place 80.0 mL of distilled water into a beaker, and add a stirring magnet. Place
beaker on a stirrer, and stir gently.
3) With stirring, add 10.0 mL of 0.10 M NaOH.
4) With continued stirring, add 10.0 mL of crystal violet. Let the mixture stir for a
couple of seconds, then quickly rinse the cuvet out with a small portion of the
mixture.
5) Still working quickly, fill the cuvet ¾ of the way with the solution, and place into the
Spec 20. Record the first absorbance reading you see as the absorbance at time zero.
6) Record the absorbance every 1.00 minute for 12 minutes.
7) Dump out the contents of the beaker, and rinse it with distilled water.
(Second Run)
9) Place 70.0 mL of distilled water into a beaker, and add a stirring magnet. Place
beaker on a stirrer, and stir gently.
10) With stirring, add 20.0 mL of 0.10 M NaOH.
11) With continued stirring, add 10.0 mL of crystal violet. Let the mixture stir for a
couple of seconds, then quickly rinse the cuvet out with a small portion of the
mixture.
12) Still working quickly, fill the cuvet ¾ of the way with the solution, and place into
the Spec 20. Record the first absorbance reading you see as the absorbance at time
zero.
13) Record the absorbance every 1.00 minute for 10-12 minutes.
14) Dump out the contents of the beaker, and rinse it with distilled water.
15) Rinse off all glassware when you are done taking data.
12
Data
Time
(minutes)
0
A
(first run)
A
(second run)
1
2
3
4
5
6
7
8
9
10
11
12
13
[CV]
(first run)
[CV]
(second run)
Graphs
For this portion of the lab, you will need to make graphs of [CV] vs. time, ln[CV] vs. time, and
1/[CV] vs. time for each run. You should have a total of six graphs when you are finished. Make
sure each graph is adequately labeled, and find the slope of the graph that is most linear for both
runs.
Data and Graph Analysis
For the reaction:
CV + NaOH → colorless products
The rate law is Rate = k [CV]x [NaOH]y
For both the kinetic runs today, the [NaOH] > > [CV], so that the [NaOH] remains
virtually constant throughout the experiment. Therefore, the rate law can be written as:
Rate = k’[CV], where k’ = k[NaOH]y
1) Looking at the various graphs and determining which are most linear, determine the
order of the reaction with respect to the crystal violet, [CV]. What do you think is the
value of “x” ?
2) Look at the graphs for Run 1 and Run 2 that pertain to the order of the reaction with
respect to [CV]. The slope of the graphs is equal to k’, which is itself equal to
k[NaOH]y. In the first run, [NaOH] = 0.010 M, and in the second run, [NaOH] =
0.020 M. Use this information to find the value for “y”. Keep in mind, this is
experimental data, so round “y” off to the closest whole number.
14
3) Using the fact that k’= k[NaOH], and referring to the two graphs used in calculation
2, determine the value for k for both runs, and find an average value for k, including
units.
4) Write the rate law, and give the average value for k, including units.
15
Follow-up Questions
1) A student studies the hypothetical reaction between an orange substance called
Crystal Orange and sodium hydroxide.
Crystal Orange + NaOH → colorless products
(a) The student uses a Spec 20 and gathers some %Transmittance data on crystal
orange, which is shown below.
(i) At what wavelength should the reaction be studied?
(ii) What is the absorbance of crystal violet at λ = 500 nm?
% Transmittance vs. Wavelength
%Transmittance
100
80
60
40
20
0
300
400
500
600
700
Wavelength (nm)
(b) The student makes a series of dilutions of Crystal Orange, and records the
absorbance at different concentrations using a cuvet with a path length of
1.27 cm. A graph of the data is shown below.
A vs [Crystal Orange]
0.6
Absorbance
0.5
0.4
0.3
0.2
0.1
0
0
0.02
0.04
0.06
0.08
0.1
0.12
[Crystal Orange] (M)
What is the molar absorbtivity constant for crystal according to the graph
above?
16
(c) In this part of the experiment, the student conducts two different runs.
In RUN 1, 10.0 mL of 5.00 M NaOH is combined with 10.0 mL of 0.010 M Crystal
Orange solution and 80.0 mL of water. The absorbance is monitored as a function of
time, and Beer’s law is used to convert this absorbance to a Crystal Orange
concentration. Graphs of [Crystal Orange] vs time, ln[Crystal Orange] vs time, and
1/[Crystal Orange] vs time are made.
In RUN 2, 20.0 mL of 5.00 M NaOH is combined with 10.0 mL of 0.010 M Crystal
Orange solution and 70.0 mL of water. The absorbance is monitored as a function of
time, and Beer’s law is used to convert this absorbance to a Crystal Orange
concentration. Only one graph of this data for RUN 2 is made: a Graph 1/[Crystal
Orange] vs time.
All three graphs for RUN 1, and 1/[Crystal Orange] vs time for RUN 2 are shown
below:
Ln[Crystal Orange] vs time (RUN 1)
[Crystal Orange] vs time (RUN 1)
y = -0.4463x - 7.3334
R2 = 0.893
[Crystal Orange] (M)
0.0012
0.001
Ln [Crystal Orange] (M)
y = -0.0002x + 0.0007
R2 = 0.6565
-7
0
2
6
-7.5
0.0008
0.0006
0.0004
-8
-8.5
0.0002
0
-0.0002 0
2
4
Time (hour)
6
-9
-9.5
1/[Crystal Orange] vs time (RUN 1)
Time (hour)
1/[Crystal Orange] vs time (RUN 2)
y = 2000x + 1000
R2 = 1
y = 4000x + 1000
R2 = 1
12000
1/[Crystal Orange] (M)
1/[Crystal Orange] (M)
4
10000
8000
6000
4000
2000
0
0
2
Time (hour)
4
6
25000
20000
15000
10000
5000
0
0
2
Time (hour)
4
6
Use this graphical data to determine the rate law for the reaction of Crystal Orange
with NaOH. Also include a value for k, with units.
Rate = _______________________
k = _____________
17