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CHEMICAL PRINCIPLES II LAB
____________________________________ Exercise #1: Statistical Treatment of Experimental Data
_________________________________________________________
Objective: To gain experience in the use of statistical analysis of experimental data, and to
apply the statistical methods to establish the precision and accuracy of measurements
employing laboratory glassware.
Background: Most experiments involve multiple measurements or determinations of the
desired result. Usually these repeated trials are performed by one person, but they could also
represent a grouping of results from the same experiment performed by several lab workers.
Once the results have been calculated (correctly, we hope!) and tabulated, they must
be reported in a manner that represents their reliability to the reader or prospective user of the
information. The most common treatment of experimental data uses the science of statistics.
What follows is a brief summary of applications from that area. [Anyone majoring in a
scientific discipline would benefit greatly from a course in basic statistics.]
Two definitions are important to consider when using scientific data: precision and
accuracy. These two words are often used interchangeably in everyday language; however, in
scientific work there is a distinct difference between them.
Precision is the proximity of measured results to each other. It is an indication of
how reproducibly a result can be determined (i.e., if a procedure is done again and again, will
the same answer be attained?).
Accuracy is the comparison of the measured result to the actual or accepted value.
Consider the following examples of quiz grades:
CASE I_ _CASE II_ _CASE III_
97 % 58 % 97 %
58 % 53 % 99 %
21 % 55 % 98 %
poor precision good precision good precision
poor accuracy poor accuracy good accuracy
Obviously, the desired goal would be to have good precision and good accuracy (as shown in
CASE III) for scientific measurements and results.
Let us consider the following results from an experiment to determine the atomic
weight (mass) of the element molybdenum (Mo): 97.58, 92.17, 95.36, 90.11, 94.94, and
96.83 g/mole.
The first calculation would involve finding the average (or mean). This is defined as
the sum of the individual results (Xi) divided by the number of measurements or results (n):
iXX=
n
Let us begin by setting up a data table.
TRIAL (n)
Xi
1
97.58
2
92.17
3
95.36
4
90.11
5
94.94
6
96.83
SUM
566.99
X = 94.50
So the average X (pronounced "X bar") will be (566.99 g/mole)/6 = 94.50 g/mole.
Next we will calculate the deviation which is the absolute difference between the
average ( X ) and the individual result (Xi), or |( X - Xi)|. By adding deviations and dividing by
the number of results, we obtain the average deviation (AD):
average deviation (AD) = | (X-X ) |
n
i
For Trial #1, the deviation would be | 94.50 - 97.58 | = 3.08 g/mole.
TRIAL (i)
Xi
| X - Xi|
1
97.58
3.08
2
92.17
2.33
3
95.36
0.86
4
90.11
4.39
5
94.94
0.44
6
96.83
2.33
SUM
566.99
13.43
X = 94.50
AD=2.24
So the average deviation (AD) will be (13.43)/6 = 2.238333 or 2.24 g/mole. Note
each result is given to the hundredth’s place; therefore the mean and average
deviation are given to the hundredth’s place.
The relative deviation is defined as the ratio of the average deviation to the mean
value. It is often given as a %.
average deviationrelative deviation = x 100%
average result
In this example, it would be 2 24
94 50
.
.
g / mol
g / mol x 100% = 2.37 %
Here the relative deviation is given with three significant figures. The rubric for
multiplication and division is to present results using the least number of significant figures
of the factors in the problem.
The relative deviation can also be reported in "parts per thousand", ppt, if the fraction
is multiplied by 1000 instead of 100. Here, the answer would be 23.7 ppt. You might also see
the relative deviation reported in ppm, "parts per million", which means the fraction is
multiplied by 1,000,000 or 106 (our answer would be 2.37x10
4 ppm).
The average deviation gives a measure of the precision of the experimental results. In
this experiment, we know the accepted or actual result for the atomic mass of Mo. From
various books and charts, it is given as 95.94 g/mole. We can therefore report the error of the
data. The error is the difference between our average answer and the actual value (often
called absolute error). When this is divided by the actual value, we get the relative error.
When this is multiplied by 100% we get the percent error.
percent error = | actual value - average experimental result | X 100%
actual value
Using our results:
|(95.94 - 94.50)|
95.94 x 100% = 1.50 % error
We could also report relative error as 15.0 ppt, had we multiplied by 1000 instead of 100.
Error and relative errors are measures of accuracy.
Let’s summarize.
Represent a series of data using the average, or mean value - X
Average or percent deviation represents the precision of the data.
Absolute or percent error represents the accuracy of the data.
_______________________________________________________________________
What happens when we don’t know what the actual, or true value is. After all, isn’t
that the goal of doing the measurement to begin with? We have two choices, do the
measurement many, many (did I say many) times until we are confident in the result, or rely
on statistics to help us estimate the accuracy of our result. In what follows we introduce the
statistical approach.
The statistical approach is based on the use of the standard deviation, n-1
(sometimes designated, s). For this, we need to square the individual deviations in the table,
obtain the sum, divide by (n-1), and then take the square root of the answer.
standard deviation n-1 =
2
i(X - X )
n-1
For Trial #1, the deviation squared would be (3.08)2 = 9.4864 (g/mole)
2. We will not worry
about the units here, because later when we take the square root, we will end with g/mol
again.
With our data, this becomes 4054956 1
2 85.( )
.
g / mol = n-1
[You may also see an answer reported with n when the above calculation uses n as the
divisor rather than n-1. This is done sometimes if there is a large amount of data so that n ≈n-
1.] The details of the calculation are given in the table below.
TRIAL (i)
Xi
X - Xi
( X - Xi)2
1
97.58
3.08
9.4864
2
92.17
2.33
5.4289
3
95.36
0.86
0.7396
4
90.11
4.39
19.2721
5
94.94
0.44
0.1936
6
96.83
2.33
5.4289
SUM
566.99
13.43
40.5495
X = 94.50
AD = 2.24
n-l = 2.85
The standard deviation can be reported in terms of relative standard deviation, often
using units of % or parts per thousand, ppt:
relative standard deviation = n1
X x 100% =
2.85 g / mol
94.50 g / mol x 100% = 3.02 %
or = n1
X x 1000 =
2.85 g / mol
94.50 g / mol x 1000 = 30.2 ppt
The standard deviation is reported to the same decimal place as the individual results.
Here we report the value of n-1 to the hundredth’s place, just as the experimental molar
masses. The relative standard deviation is given with three significant figures since this was
the least number of significant figures of the factors used in its calculation.
To give an estimate of error of the mean (proximity to the true value), the standard
deviation is often converted to an expression called a confidence interval. This is a range,
centered at X , in which we can expect to find the true value with a certain degree of
confidence. To determine the confidence interval we make use of the statistical value, t,
which can be obtained from the table below.
n-1
t50%
t80%
t90%
t95%
t99%
1
1
3.078
6.314
12.706
63.657
2
0.816
1.886
2.92
4.303
9,925
3
0.765
1.638
2.353
3.182
5.841
4
0.741
1.533
2.132
2.776
4.604
5
0.727
1.476
2.015
2.571
4.032
6
0.718
1.440
1.943
2.447
3.707
7
0.711
1.415
1.895
2.365
3.500
8
0.706
1.397
1.860
2.306
3.355
9
0.703
1.383
1.833
2.256
3.250
10
0.700
1.372
1.812
2.228
3.169
15
0.691
1.341
1.753
2.131
2.947
20
0.687
1.325
1.725
2.086
2.845
∞
0.674
1.282
1.645
1.960
2.576
We will calculate a value
( )( )t
n
n 1
Using the value of t90% = 2.015 from the table for n -1 = 5 (n = 6 results), this gives us
( . )
.2 015
62 34
(2.85)
Now we can report our answer as 94.50 ± 2.34 g/mole, meaning that we have a range
extending from
the lower confident limit 94.50 - 2.34 = 92.16 g/mol
to the upper confidence limit 94.50 + 2.34 = 96.84 g/mol
This means that there is a 10% chance (100% - 90%) that the true value lies outside the range
92.16 to 96.84 g/mol. The value of 2 represents the width of the confidence interval.
The range becomes wider if we want to decrease the odds of misrepresenting the
actual value. Prove to yourself that for 95% confidence, our answer would be expressed as
94.50 ± 2.99 g/mole or that there is a 5% chance of finding the true value outside the range of
91.51 to 97.49 g/mole.
______________________________________________________________________
NOTE: If you have a scientific calculator, you will probably notice that you have keys
with labels of , 2, n-1, n. Thus your calculator is already programmed to do
calculations involved in deviations and standard deviations. Because procedures vary
amongst different models and manufacturers, you should consult the instruction
booklet that was included with your calculator. [Oh no, where did I put that.....]
_______________________________________________________________________
When performing an experiment, we often have occasion to question a result.
Sometimes we know that an error has occurred in the procedure and we make an entry in our
notebook to indicate what happened. We might then eliminate that particular trial from our
final calculations with good justification, but we do not obliterate the entry in our records.
However, even if a noticeable error has not happened, we might wonder about the validity of
a result. For example, notice that the answer for Trial #4 in our data has a high deviation
(4.39 g/mole). Should this result be eliminated from our final calculations or must we retain
it? There are several statistical ways to evaluate this and one (the Q test) is described below.
First, arrange the results in numerical order (ascending or descending). Working with
our data set, we get
97.58
96.83
95.36
94.94
92.17
90.11* (the suspected value that we want to consider)
Take the difference between the highest (97.58) and lowest (90.11) values. This gives us the
range of 7.47. Now take the difference between the suspected value (90.11) and its nearest
neighbor (92.17). Ignore the sign (+ or -). This gives us 2.06.
Now we define experimental
difference of suspect and nearest neighborQ =
range
With our numbers
We now consult a statistical table of Q values with different confidence levels
depending on the number of results, n (below). For example, with n = 6, Q50% = 0.560, Q95%
= 0.625, and Q99% = 0.740.
In our case Qexp = 0.276. This is less than any Q from the table. Thus we are NOT
allowed to remove the suspected value of 90.11 from our data. If Qexp > Q from the table, we
may eliminate our suspected value. For example, if Qexp = 0.676, we could say that the value
90.11 could be removed from the data with 95% certainty. We would then go back and redo
all calculations based on the 5 remaining results.
We finish up this introduction to error analysis by discussing how average deviation,
standard deviation and confidence limits relate to the representation of precision and accuracy
of experimental results. Measurements can deviate from true results as an accumulation of
random and systematic errors. Random errors are equally likely to lead to too high or too
low results. By performing a measurement numerous times we can reduce the influence of
random errors. We use the confidence interval to indicate the uncertainty resulting from
random errors. We use significant figures to represent the limitation of the measurement
device. For instruments with digital displays, the precision is given by the right-most digit on
the display. (eg. a reading of 9.076 g is precise to +/- 0.001 g). For equipment with graduated
scales the precision is taken as +/- 1/10 the smallest division. When representing
experimental measurements, the number of digits (significant figures) used should represent
the precision of the measurement.
Q-Test values
N
Q50%
Q95%
Q99%
3
0.941
0.970
0.994
4
0.765
0.829
0.926
5
0.642
0.710
0.821
6
0.560
0.625
0.740
7
0.507
0.568
0.680
8
0.468
0.526
0.634
9
0.437
0.493
0.598
10
0.412
0.466
0.568
15
0.338
0.384
0.475
20
0.300
0.342
0.425
25 0.277 0.317 0.393
30 0.260 0.298 0.372
If we discover that the actual (true) result lies outside the confidence interval it is
likely that there are systematic errors at play. Systematic errors will lead to inaccurate results,
no matter how many data points are averaged together. Now the confidence interval can help
us ferret out potential sources of these errors. As an example, if the true result lies outside a
95% confidence interval, there is a 95% certainty that our technique has introduced a
systematic error. One of the main sources of systematic error is the use of an uncalibrated
instrument. The following exercises will help you review statistical analysis of data.
10-11-206, Chemical Principles II Lab Name: _______________________
Exercise #1 Major: ______________________
[1] How many significant figures are indicated in each of the following
measurements?
(a) 0.02670 g _____ significant figures
(b) 328.0 mL _____
(c) 7000.0 ng _____
(d) 0.00200cm _____
[2] A sample containing the known amount of 102 g/L of chloride was analyzed twice by a
student. Answers of 101 and 98 g/L were reported by the student.
(a) Calculate the student's average (mean) answer __________
(b) Calculate the absolute error __________
(c) Calculate the percent error __________
[3] The weights (masses) of nuclear pellets were found to be 127.2, 128.4, 127.1, 129.0, and
128.1 grams.
(a) Calculate the average (mean) __________
(b) What is the median (middle measurement)? __________
(c) What is the range of the measurements? __________
[4] An alloy was analyzed four times for silver and the results were 95.67%, 95.61%,
95.71%, and 95.60% Ag.
(a) Calculate the standard deviation, n-1 __________
(b) Calculate the relative standard deviation (in %) __________
[5] Four measurements of sodium levels in blood samples gave results of 139.2, 139.8, 140.1,
and 139.4 mmol/L of Na+
(a) Calculate the average value __________
(b) Calculate the standard deviation, n-1 __________
(c) Calculate the relative standard deviation in ppt __________
(d) Calculate the range for 90% confidence __________
(e) Calculate the range for 95% confidence __________
[6] The precision of a method is being established, and the following data are obtained:
22.23, 22.18, 22.25, 22.09, and 22.17%. Is the 22.09% value a valid measurement at
the 95% confidence level? (i.e., should it be included or omitted in the results?)
CIRCLE ONE: INCLUDE 22.09 or OMIT 22.09
Lab Exercise 1: Calibration of a Volumetric Pipet.
Volumetric pipets are used to deliver a specific volume of a liquid. In this exercise
you will determine the accuracy and precision of using this tool.
a) Pull the 10-mL volumetric pipet from your drawer.
b) Obtain approximately 100 mL of deionized water in a clean beaker and
allow the temperature to equilibrate with that of the lab.
c) Record the temperature and refer to the density chart posted in the lab to
find the density of water at that temperature.
d) Determine the mass of a dry wash bottle.
e) Discharge 10 mL of deionized water into the wash bottle using the
volumetric pipet. Allow the pipet to drain; don’t force the contents out with the pipet
bulb. Record the mass and determine the volume based on the density of water.
f) Repeat steps (d-e) four more times by adding to the contents of the bottle.
Data:
Temperature of Water: _____________
Density of Water: _____________
Mass of Wash Bottle: _____________
Trial Mass of Bottle + Water Volume of Pipet
1
2
3
4
5
Average Volume ________________
Average Deviation ________________
Standard Deviation ________________
95% Confidence Interval ________________
1) Does the confidence interval include 10 mL?
Lab Exercise 2: Precision and Accuracy of a Buret.
Burets are used to deliver a measured volume of a liquid during a titration. In this
exercise you will determine experimental accuracy and precision using this tool.
a) Obtain a 50-mL buret
b) Obtain approximately 100 mL of deionized water in a clean beaker and
allow the temperature to equilibrate with that of the lab.
c) Record the temperature and refer to the density chart posted in the lab to
find the density of water at that temperature.
d) Mount the buret to a buret clamp supported by a ring stand.
e) With the stopcock closed, fill the buret to the 0-mark with water. It’s not
important that you fill exactly to the 0-mark, only that you know what the initial
reading is. Record this value in the table below.
f) Determine the mass of a clean, dry 150-mL beaker.
g) Discharge ~ 8 mL of deionized water into the beaker from the buret. Try to
get close. You’ll need this skill in week 3 when you perform a titration. Record the
final volume in the buret and the mass of the beaker + water. Calculate the volume
discharged using the density of water.
h) Repeat step (g) four more times by adding to the contents of the beaker.
Data:
Temperature of Water: _____________
Density of Water: _____________
Mass of Beaker: _____________
Trial Mass of Beaker +
Water
Initial Buret
Reading
Final Buret
Reading
1
2
3
4
5
The volume discharged is simply the final buret reading – initial buret reading.
The volume based on mass of water collected is given by V = m/d, where d is the
density of water in g/mL.
1) Use these definitions to complete the subsequent data table.
A B
Trial Volume from
buret reading
Volume from
mass of H2O
Percent
Error
1
2
3
4
5
2) Use column B as the “true” value, determine the percent error for each row.
3) What is the average % error? What is the average absolute error?
4) How does the average absolute error compare to the magnitude of the smallest
graduation on the buret?
5) How does the average absolute error compare to the magnitude of the precision of
the buret?
Revised: A. Langner 2/22/09
Experiment #2: Decomposition of KClO3
Objective: Determine the value of the gas constant, R, by measuring the decomposition of potassium chlorate, KClO3, using a liquid displacement method.
Background: Most gases obey the ideal-gas equation, PV = nRT, quite well under
ordinary conditions, that is, at room temperature and atmospheric pressure. Here P
is pressure, V is volume, n is moles of gas and T is absolute temperature. The gas
constant, R, relates these quantities and is now known to have a value of 0.08206
L.atm/mol.K. Small deviations from this law are observed, however, because real-
gas molecules are finite in size and exhibit mutual attractive forces. The van der
Waals equation,
2
2
n aP V nb nRT
V
eq. 1
where a and b are constants characteristic of a given gas, takes into account these
two causes for deviation and is applicable over a much wider range of temperatures
and pressures than the ideal-gas equation. The term nb in the expression (V - nb) is
a correction for the finite volume of the molecules; the correction to the pressure by
the term n2a/V2 takes into account the inter-molecular attractions.
In this experiment you will test the validity of the ideal gas law by determining
the gas constant by the thermal decomposition of potassium chlorate, KClO3. The
evolution of O2 from the sample can be measured gravimetrically (mass difference)
and by measuring the volume of evolved gas by displacement of water. A
manganese(IV) oxide, MnO2, catalyst will be used to speed up the decomposition
reaction. The overall reaction is given by:
2KClO3(s) 2KCl(s) + 3O2(g) eq. 2
Analysis of the results will be performed using both the ideal-gas law and the
van der Waals equation. The experiment will be performed in triplicate to establish
the precision of the measured value.
You will work with mixtures of potassium chlorate and potassium chloride,
KCl. From eq. 2 you can see that KCl(s) is a product of the reaction and,
consequently, will not react. If a sample of the KClO3/KCl mixture is accurately
weighed before and after the oxygen has been driven off, the mass of the evolved
oxygen can be obtained by difference. From this moles of O2 can be determined.
The oxygen can be collected by displacing water from a bottle or closed flask, and
the volume of gas can be determined from the volume of water displaced. The
measurement of displaced water is performed in such a way that the initial and final
temperature and pressure are identical. This way the ratio of moles of gas to volume
of gas in the closed system must be constant.
Letting n1 and V1 represent initial moles and volume of gas (air and water
vapor), and n2 and V2 by final moles and volume of gas (air, water vapor and evolved
O2), we can write the relation;
2 1
2 2 1O
V V Pn n n
RT
eq. 3
In the case of the van der Waals equation the relationship is a bit more complicated;
however, to a good approximation;
2
2 1
2 2 1 1O
V V n nn n n P a b
RT V V
eq. 4
Where n = n2-n1 and V = V2-V1, respectively. Equations 3 and 4 can be used to
determine the gas constant R.
Precautions:
You MUST wear safety goggles at all times while in the laboratoiry.
When sealing the test tube containing the KClO3/MnO2 mixture make sure
none of the reactants come in contact with the rubber stopper. KClO3 is a
strong oxidant and will react violently with the stopper if forced into contact
with it.
Return unused KClO3 and MnO2 in separate vials. Remember, MnO2 will
accelerate the decomposition of potassium chlorate.
Make sure all glassware is secured with clamps before starting the reaction.
Do NOT remove hoses from the glass tubing.
Replace the cap on the syringe needle after use.
All usual safety precautions should be followed, including any special
precautions given by your instructor.
Procedure: Work in partners
1. Obtain a vial of potassium chlorate (KClO3) and a vial of manganese(IV) oxide (MnO2). Both reactants have been mixed with inert potassium chloride (KCl).
2. Assemble the experimental apparatus shown in Figure 1, but don’t attach the test tube. The tubes and stoppers have already been pre-assembled. You will use a 250-mL Aspirator Flask rather than a bottle. The aspirator flask should be clamped to a ring stand to secure it. Fill the flask with water to the neck. Fill glass tube A and the rubber tubing with water by loosening the pinch clamp and attaching a rubber bulb to the end of tube B and applying pressure through it. Close the clamp when the tube is filled.
3. Add approximately 0.20 g of the MnO2/KCl mix and 0.5 g of KClO3 to the provided test tube. Accurately determine the mass of the test tube and contents to the nearest 0.0001 g. Mix the solids in the test tube by rotating the tube, being certain that none of the mixture is lost from the tube. Cover the tube with the rubber septa. Clamp the tube to a ring stand and attach tube B by pushing the needle through the septa. (CAUTION: When you attach the test tube, be certain that none of the solid mixture contacts the septa. With added heat and pressure a violent reaction could result.)
Figure 1: Experimental Apparatus
4. Fill the beaker about half full of water, insert the end of tube A in it,
open the pinch clamp, and lift the beaker until the levels of water in the
flask and beaker are identical. Then close the clamp, discard the water
in the beaker, dry the beaker, and determine it’s mass. The purpose of
equalizing the levels is to produce atmospheric pressure inside the
flask and test tube.
5. Set the beaker with tube A in it on the bench and open the pinch
clamp. A little water will flow into the beaker, but if the system is airtight
and has no leaks, the flow will soon stop and tube A will remain filled
with water. If this is not the case, check the apparatus for leaks and
start over. Leave the small amount of water in beaker, at the end of the
experiment, the water levels will be adjusted and this water will flow
back into the bottle.
6. Heat the lower part of the test tube gently (be certain that the pinch
clamp is open) so that a slow but steady stream of gas is produced,
as evidenced by the flow of water into the beaker. When the rate of gas
evolution slows considerably, increase the rate of heating, and heat
until no more oxygen is evolved. Allow the apparatus to cool to room
temperature, making certain that the end of the tube in the beaker is
always below the surface of the water. Equalize the water levels in the
beaker and the bottle as before and close the clamp.
7. Determine the mass of the beaker with water. Measure the
temperature of the water and, using the density of water in Table 1,
calculate the volume of the water displaced. This is equal to the volume
of oxygen produced.
8. Remove the test tube from the apparatus and accurately weigh the
tube plus the contents. The difference in mass between this and the
original mass of the tube plus solids is the mass of the oxygen
produced.
9. Record the barometric pressure. Your instructor will have access to a
barometer.
10. Repeat the experiment 2 more times for a total of three trials using a
new test tube for each trial.
Waste Disposal Instructions KClO3 is a powerful oxidizing agent and should not be disposed of in a waste basket! Do not attempt to clean out the residue that remains in the test tube. Return the test tube to the instructor or follow his instructions for disposal of its contents.
TABLE 1 Density and Vapor Pressure of Pure Water at Various Temperatures
Temperature oC
Density
g/mL
Temperature oC
Vapor Pressure
mmHg
15
0.999099
15
12.8
16
0.998943
16
13.6
17
0.998774
17
14.5
18
0.998595
18
15.5
19
0.998405
19
16.5
20
0.998203
20
17.5
21
0.997992
21
18.6
22
0.997770
22
19.8
23
0.997538
23
21.1
24
0.997296
24
22.4
25
0.997044
25
23.8
26
0.996783
27
0.996512
28
0.996232
Data & Analysis
Table 2: Moles of O2 Evolved.
Trial 1 Trial 2 Trial 3
Mass of test tube + reactants (g)
Mass of test tube + products (g)
Mass of O2 evolved (g)
Moles of O2 evolved (mole)
Table 3: Volume of Water Displaced
Trial 1 Trial 2 Trial 3
Mass of dry beaker (g)
Mass of beaker + water (g)
Mass of water displaced (g)
Temperature of water (oC)
Density of water (g/mL)
Volume of water displaced, V2-V1 (L)
Table 4: Determination of Gas Constant (For eq. 4 use a = 1.360 L2atm/mol2, b = 31.83 cm3/mol)
R from eq. 3 R from eq. 4
Trial 1
Trial 2
Trial 3
Average, X
Standard Deviation, n-1
95 % Conf. Int., X ±
Questions:
1. Does your value of R based on the ideal gas law agree with the accepted value within
the confidence limits of the result?
2. Does your value of R based on the van der Waals equation agree with the accepted
value within the confidence limits of the result?
3. How does the solubility of oxygen in water affect the value of R you determined?
Explain your answer.
4. Use the van der Waals equation to calculate the pressure exerted by 1.000 mol of Cl2;
in 22.41 L at 0.0 °C. The van der Waals constants for Cl2; are: a = 6.49 L2 atm/mol
2 and
b = 0.0562 L/mol.
5. How much potassium chlorate is needed to produce 20.0 mL of oxygen gas at 670
mm Hg and 20°C?
6. A certain compound containing only carbon and hydrogen was found to have a
vapor density of 2.550 g/L at 100°C and 760 mm Hg. If the empirical formula of this
compound is CH, what is the molecular formula of this compound?
Experiment #3: Acid-Base Titrations
Objective: Determine the concentration of acid in an aqueous unkown by using the technique of acid-base titration with an indicator.
Background: There are many definitions or classifications of acids and bases. For this
experiment, the Arrhenius definitions will be most useful:
An acid is a substance that provides hydrogen ions (H+) in aqueous solution.
A base is a substance that provides hydroxide ions (OH-) in aqueous solution.
Neutralization is the reaction of the H+ from the acid with the OH
- of the base:
HX(aq) + MOH(aq) MX(aq) + H2O(l) eq. 1
Here X- is a monovalent anion (e.g. Cl
-) and M
+ is a monovalent cation (e.g. Na
+). The
product of a neutralization reaction is a salt and water.
Using compounds as examples of acids and bases, we get equations such as these:
H+Cl
- + Na
+OH
- Na
+Cl
- + H2O eq. 2
K+H
+C8H4O4
2- + Na
+OH
- K
+Na
+C8H4O4
2- + H2O eq. 3
Acid Base salt water
Equation 2 represents the reaction between hydrochloric acid and sodium hydroxide to produce sodium chloride and water. Equation 3 appears more complicated, but it still
represents an acid, potassium hydrogen phthalate (KHP) reacting with a base, sodium
hydroxide (Na+OH
-) to produce a salt, potassium sodium phthalate (K
+Na
+C8H4O4
2-) and
water. The molecular structure of the phthalate anion is given below.
O
-O
O
O-
Notice that both reactions 2 and 3 involve reacting one mole of acid with one mole of base; some reactions have ratios that are not one-to-one. When the reaction is complete, this is called the equivalence point. All of the acid will have reacted with all of the base and the flask or beaker will now contain only the salt produced with water.
Often a visual method is used to determine when the reaction is completed. We will use an acid-base indicator that changes color dramatically as the solution changes from acidic to slightly basic (or vice versa). The indicator will be phenolphthalein, often abbreviated as "phth" It is colorless in acid and pink in base. When the indicator changes color, this is called the end point of the reaction and the indicator is selected so that the end point and the
equivalence point will be close, consequently giving results. Titration is a technique that involves a controlled reaction between two substances. In our experiment, one substance (the base, NaOH) is placed in a buret and the other substance (the acid, KHP) is in a flask. The phenolphthalein indicator is added to the flask containing the acid, then NaOH is slowly added to this mixture until the indicator changes from colorless to pink. This signals the end of the reaction.
We will first make a dilute solution of NaOH from a more concentrated solution (often called a stock solution). The concentration of the dilute NaOH solution will be determined by titrating it using a known amount of acid (KHP). Through this procedure the NaOH solution is standardized. Now that the concentration (molarity) of the NaOH is known, it can be used to determine the amount of acid in any appropriate sample, such as acid rain, stomach fluid, urine, soda, or a laboratory unknown.
PRECAUTIONS:
* You MUST wear safety goggles at all times while in the laboratory.
* Although the acids and bases we will use are rather dilute,
- wash your hands well if acid or base gets on them
- don't rub your eyes with your fingers
- you may wish to wear gloves
- you may wish to wear an apron or lab coat
- wash your lab bench well to remove any acid or base spills
* All solutions may be rinsed into the sinks with running tap water.
* Pipets should be used only with bulbs or pumps. NO pipeting by mouth! * All normal safety rules must be obeyed, including any special precautions issued by
your instructor.
Procedure: (work in pairs)
1. PREPARATION of NaOH SOLUTION and the BURET
The preparation of the NaOH titrant solution should be done by one lab partner.
A. Using a 50- or 100-mL graduated cylinder, measure 250 mL of distilled water into a 500-mL flask. With a 10-mL graduated cylinder, obtain 10 mL of 6 M NaOH and add it to the flask. Mix very well. Pour into a plastic or glass storage bottle and cap the bottle (or leave in flask and cover with plastic film). Label this solution as "NaOH titrant" and set aside. Make sure the valve to the carboy containing the 6 M NaOH(aq) is closed after use.
Determination 1: Calculate the approximate molarity of the "NaOH titrant" solution and record it in the data table. (This is a dilution problem so use M1V1 = M2V2 where M represents molarity and V is volume.)
B. Obtain a 50-mL buret. Clean with soapy water and rinse several times with tap water, then rinse with distilled water 2 or 3 times. Pour some of the "NaOH titrant" into a small beaker and rinse the buret with 2 or 3 small (5-10 mL) portions of it. The buret does not have to be dry, but now any liquid in the buret is the "NaOH titrant".
C. Set up a ring stand with the buret clamp. Attach the buret and, using the small buret funnel, slowly pour "NaOH titrant" solution into the buret, filling it to above the zero-line. Place a "waste beaker" below the buret and allow some solution to run out, making sure that the buret tip is filled and has no air bubbles. Adjust the level of the liquid in the buret to between 0 and 5 mL. It DOES NOT have to be at 0.00 mL to begin!
II PREPARATION of KHP SAMPLE
The preparation of the KHP solution should be done by one lab partner.
A. Clean five 125- or 250-mL flasks. Rinse with distilled water. They do not need to be dry. With a wax pencil, label them as A, B, C, D, E or 1, 2, 3, 4, 5 or any way you wish.
B. Using a plastic weighing boat, measure approximately 0.6 grams of KHP (potassium hydrogen phthalate). The amount doesn't have to be exactly 0.6 grams, but you need to know and record exactly what amount you have (i.e., 0.5832 g or 0.6117 g) in the data table. Add the weighed KHP to the first flask and using the squirt bottle, rinse the plastic weighing boat with distilled water allowing the liquid to go into the flask.
C. Repeat this process for the other four flasks.
D. Using the 50- or 100-mL graduated cylinder, measure and add 30 mL of distilled water to each flask. Swirl to dissolve the KHP solid.
Determination 2: Calculate the moles of KHP you have and enter it in your data table.
Example: Suppose you weighed 0.6117 grams of KHP.
0.6117 g KHP x 1 mole KHP = 0.002995 moles KHP
204.23 g KHP
III. STANDARDIZATION of NaOH TITRANT
A. Record the initial reading of the "NaOH titrant" in the buret, using a card with a dark line as background to read the bottom of the meniscus. Estimate your readings to the nearest 0.01 mL. Have your instructor verify your first reading. Enter it in the data chart.
B. Place a piece of white paper on the base of the ring stand. Add 3 or 4 drops of phenolphthalein to the first flask of KHP solution. Slowly add "NaOH titrant" while constantly swirling the flask to mix. Have your instructor show you. You will see some pink that disappears quickly, but as you continue to add NaOH, the pink will stay longer. Slow down and start adding very small amounts until the pink color persists when swirled. You are aiming for the lightest pink possible. Read and record the final volume of "NaOH titrant" from the buret. [After the sample sits for a few minutes, the color may disappear, but that is OK.]
C. Refill the buret. Repeat the procedure for the next flask until all five have been titrated. Dispose of the waste in the sink with running tap water. Leave the buret ready for the next part.
Determination 3: What is the molarity of the "NaOH titrant" obtained from each titration?
Enter this in the data table.
Example: The titration required 14.65 mL of NaOH for the sample from Part II which contained 0.002995 moles KHP.
0.002995 moles KHP x 1 mole NaOH = 0.2044 M NaOH
0.01465 L NaOH sol'n 1 mole KHP
Is the molarity of the NaOH solution close to your estimate from Part IA? [If it seems really different from your estimate or prediction, check for potential errors.]
Determination 4: Calculate the average molarity of the NaOH titrant solution, the standard deviation, and the 95% confidence interval. Include this in your report.
IV. CONCENTRATION of UNKNOWN ACID SAMPLE
A. Obtain an unknown from your instructor and record its number NOW!
B. Using a pipet bulb or pipet pump and your 10-mL volumetric pipet, transfer 10-
mL of unknown to a clean 125- or 250-mL flask. Add 20 mL of distilled water, using
the graduated cylinder. Add 3 or 4 drops of phenolphthalein. Record the initial
volume of NaOH in the buret and titrate until the color becomes a persistent light
pink. Record the final volume in the buret. Refill the buret with NaOH titrant solution
and repeat the procedure for a total of 5 trials.
After you finished the first titration, you should be able to predict how many mL the second sample should require. For subsequent trials add about 2/3 of that very quickly, then slow down to reach the end point.
Determination 5: What is the molarity of the acid in the original unknown?
Example: 10-mL of unknown required 22.50 mL of 0.2044 M NaOH to reach the end point. Be sure to use average molarity of NaOH from Part III.
22.50 mL NaOH sol'n x 0.2044 moles NaOH x 1 mole unk acid = 0.4599 M
10.00 mLunk acid 1 L NaOH sol'n 1 mole NaOH unk acid
Determination 6: Calculate the average molarity of the unknown acid solution, the standard
deviation, and the 95% confidence interval. Include this in your report.
V. CLEAN-UP
A. All solutions, including the NaOH, can be washed down the drain with running tap water.
B. Return the KHP vial and unknown container to the instructor.
C. Clean your equipment and lab bench area.
D. Have your instructor initial your lab book.
E. LOCK your LOCKER!
F. WASH your HANDS!
VI. DETERMINATIONS & DATA TABLES
PART III: STANDARDIZATION of NaOH TITRANT
Determination 1: Approximate Molarity of NaOH Titrant ________________
Table 1: Molarity of NaOH Titrant (Determination 2 & 3)
mL NaOH
mL NaOH
mL NaOH
TRIAL
g KHP
moles
KHP
Final
initial
Used
M NaOH
Determination 4: Average Molarity of NaOH _______________
Standard Deviation _______________
95% Confidence Interval _______________
PART IV: DETERMINATION of UNKNOWN ACID MOLARITY
Table 2: Molarity of Acid Unknown (Determination 5)
Unk#
mL NaOH
mL NaOH
mL NaOH
M of Acid
TRIAL
final
Initial
used
Unknown
Determination 6: Average Molarity of Acid _______________
Standard Deviation _______________
95% Confidence Interval _______________
Questions:
Question 1: Suppose in step II-D you unknowingly added 35 mL of distilled water to one of the five flasks instead of 30 mL. Does this lead to a systematic error in your results? Why or why not?
Question 2: Suppose in procedure IV you and your lab partner got your signals crossed and you both added 20 mL of distilled water to one of the flasks containing 10-mL of unknown. Does this lead to a systematic error in your results? Why or why not?
Question 3: When phenolphthalein is used as an indicator the pink color at the end point of a titration tends to fade over time. What does this indicate about changes in the solution? Give a possible explanation for this change.
Question 4: Potassium hydrogen phthalate is a monoprotic acid. dihydrogen phthalate (H2C8H4O4) is a diprotic acid. An analyte sample contains 0.127 M KHP and 0.0678 M H2C8H4O4. What volume of a 0.205 M NaOH solution is required to neutralize 25.0 mL of the analyte solution? Show your work.
Revised: A. Langner 3-21-2009
Experiment #4: Heat of Neutralization
Objective: To measure, using a calorimeter, the energy changes accompanying
neutralization reactions.
Background: Every chemical change is accompanied by a change in energy,
usually in the form of heat. The energy change of a reaction that occurs at constant
pressure is termed the heat of reaction or the enthalpy change. The symbol H =
Hfinal - Hinitial is used to denote the enthalpy change. If heat is evolved, the reaction is
exothermic (H < 0); and if heat is absorbed, the reaction is endothermic (H > 0). In
this experiment, you will measure the heat of neutralization (or the enthalpy of
neutralization) when an acid and a base react to form water and a soluble salt.
This quantity of heat is measured experimentally by allowing the reaction to
take place in a thermally insulated vessel called a calorimeter. The heat liberated in
the neutralization will cause an increase in the temperature of the solution and of the
calorimeter. If the calorimeter were perfect, no heat would be radiated to the
laboratory. The calorimeter you will use in this experiment is constructed from
styrofoam coffee cups covered by a plastic lid. Some heat will be lost through the
top; therefore, the calorimeter will have to be calibrated.
Calibration of the calorimeter involves determining its heat capacity, Ccal. By
"heat capacity of the calorimeter" we mean the amount of heat (that is, the number of
joules) required to raise its temperature 1 Kelvin, which is the same as 1°C. Provided
the calibration is done in the temperature range encountered during the
neutralization reaction, Ccal incorporates both the heating up of the cups themselves
and the heat loss through the plastic top. Once calibrated, the calorimeter can be
used to determine the enthalpy change of a reaction. At constant pressure the heat
released by a reaction is equal to –Hrxn for the reaction. The released heat of
reaction is absorbed by the calorimeter contents, the calorimeter and is lost through
the top. Mathematically we write
qreleased = -Hrxn = mCsT + CcalT eq. 1
Here m and Cs are the mass and specific heat capacity of the reaction solution. The
temperature change T = Tf – Ti (final – initial).
The heat capacity of the calorimeter is determined by measuring the tem-
perature change that occurs when a known amount of hot water is added to a known
amount of cold water in the calorimeter. The heat lost by the warm water is equal to
the heat gained by the cold water and the heat absorbed by the calorimeter. For
example, if T1 equals the temperature of 50 mL of cold water and the calorimeter, T2
equals the temperature of 50 mL of warm water added to it, and Tf equals the
temperature after mixing, then the heat lost by the warm water is
heat lost by warm water = (T2 - Tf) x 50 g x 4.18 J/goC eq. 2
The specific heat of water is 4.18 J/K-g, and the density of water is 1.00 g/mL. The
heat gained by the cold water is
heat gained by cold water = (Tf – T1) x 50 g x 4.18 J/goC eq. 3
The heat lost to the calorimeter is the difference between heat lost by the warm
water and that gained by the cold water:
(heat lost by warm water) — (heat gained by cold water) =
heat gained by the calorimeter
Substituting equations 2 and 3 we have
[(T2 - Tf) x 50 g x 4.18J/goC] - [{Tf – T1) x 50 g x 4.18J/goC]
= Ccal x (Tf – T1) eq. 4
Note that the heat lost to the calorimeter equals its temperature change times its
heat capacity. Thus by measuring T1, T2, and Tf, the heat capacity of the calorimeter
can be calculated from equation 4. The following example illustrates this procedure.
Example 1
Given the following data, calculate the heat lost by the warm water, the heat gained
by the cold water, the heat lost to the calorimeter, and the heat capacity of the
calorimeter:
Temperature of 50.0 mL warm water: 37.92°C = T2
Temperature of 50.0 mL cold water: 20.91 °C = T1
Temperature after mixing: 29.11 °C = Tf
SOLUTION: The heat lost by the warm water, where T = 37.92 °C - 29.11 °C, is
8.81 K x 50 g x 4.18 J/goC = 1840 J
The heat gained by the cold water, where T = 29.11 °C - 20.91 °C, is
8.20 K x 50 g x 4.18J/goC = 1710 J
The heat gained by the calorimeter is
1840 J - 1710 J = 130 J
The heat capacity of the calorimeter is, therefore,
130J/8.20K = 16.0 J/K
Once the heat capacity of the calorimeter is determined. equation 1 can be used to
determine the H for the neutralization reaction. Example 2 illustrates such a calculation.
Example 2
Given the following data, calculate the heat gained by the solution, the heat
gained by the calorimeter, and the heat of reaction:
Temperature of 50.0 mL of acid before mixing: Ti = 21.02 °C Temperature of 50.0 mL of base before mixing: Ti = 21.02 °C Temperature of 100.0 mL of solution after mixing: Tf = 27.53 °C
SOLUTION: The heat gained by the solution, where T = 27.53 °C - 21.02 °C, is
6.51 K x 100 g x 4.18 J/K-g = 2720 J
The heat gained by the calorimeter, where T - 27.53 °C - 21.02 °C, is
6.51 K x 16.0 J/K = 104 J
The heat of reaction is therefore
2720 J + 104 J = 2824 J = 2.82 kJ
Precautions:
You MUST wear safety goggles at all times while in the laboratory.
Although the acids and bases used are dilute,
-- wash your hands well if acid or base gets on them
-- don’t get solutions in your eyes or on your clothes
-- you may wear gloves and/or a lab apron
-- clean acid or base spills with ample water
All solutions may be rinsed into the sinks with running water.
All usual safety precautions should be followed, including any special precautions given by your instructor.
___________________________________________________________________
____
Procedure: Work in partners
A. Heat Capacity of Calorimeter
1) Construct a calorimeter similar to the one shown in Figure1 by nesting two
Styrofoam cups together. Make a hole in the lid just big enough to admit the
thermometer. Place the cups in a 400-mL beaker to provide stability. Using a clamp
support the thermometer in the calorimeter, making sure the thermometer bulb
doesn’t touch the bottom of the inner cup. To protect the thermometer, wrap some
paper towels around the top as you gently clamp down.
2) Place exactly 50.0 mL of tap water in the calorimeter cup and replace the
cover and thermometer. Allow 5 to 10 min for the system to reach thermal
equilibrium; then record the temperature to the nearest 0.1 °C on the report sheet.
Figure 1: Schematic of Calorimeter
3) Place exactly 50.0 mL of water in a clean, dry 250-mL beaker and heat the
water on a hot plate until the temperature is approximately 15° to 20°C above room
temperature. Do not heat to boiling, or appreciable water will be lost, leading to an
erroneous result. Allow the heated water to stand for a minute or two, recording the
temperature to the nearest 0.1 oC as you wait. The largest error in the lab occurs
when the thermometer hasn’t had the chance to reach the same temperature as the
hot water, so be patient.
4) Immediately after taking a temperature measurement of the warm water
pour it as completely as possible into the calorimeter, making sure the contents are
thoroughly mixed. Replace the lid and the thermometer, then observe the
temperature for the next 3 min and record the temperature every 15 s on the
Clamp
Thermometer
Lid
Styrofoam Cups
temperature vs. time data sheet.
5) Determine T by taking the difference between the maximum temperature
on your data sheet and the initial temperature of the cold water.
6) Repeat Steps 2-4 two more times, interchanging the inner and outer
calorimeter cup each time. Be sure the cups are thoroughly dried between uses.
B. Heat of Neutralization of HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)
1) Dry the calorimeter and the thermometer with a towel. Carefully measure
50.0 mL of 1.0 M NaOH and add it to the calorimeter. Place the lid on the calorimeter
but leave the thermometer out.
2) Measure out exactly 50.0 mL of 1.0 M HCl into a dry beaker. Allow it to
stand near the calorimeter for 3 to 4 min. Measure the temperature of the acid, rinse
the thermometer with tap water, and wipe dry.
3) Insert the thermometer into the calorimeter and measure the temperature
of the NaOH solution. The temperatures of the NaOH and the HCl should not differ
by more than 0.5°C. If the difference is greater than 0.5°C, adjust the temperature of
the HCl by either warming it by holding the beaker in your hands or cooling the out-
side of the beaker with tap water until the temperature of the HCl is within 0.5 °C of
that of the NaOH, then record the temperature of both solutions in your data table.
4) Lift the lid of the calorimeter and carefully add the 1.0 M HC1 all at once.
Be careful not to splash any on the upper sides of the cup. Make sure the contents
are thoroughly mixed then replace the lid and thermometer of the calorimeter.
5) Record the temperature as a function of time every 15 s for the next 3 min.
6) Determine T by using the highest temperature in your data table.
Calculate the heat of neutralization per mole of water formed. You may assume that
the NaCl solution has the same density and specific heat as water.
7) Repeat Steps 1-6 two more times, interchanging the inner and outer
calorimeter cup each time. Be sure the cups are thoroughly dried between uses and
the thermometer is rinsed with water.
C. Heat of Neutralization of HC2H3O2(aq) + NaOH(aq) H2O(l) + NaC2H3O2(aq)
Follow the same procedure as in Part B, but substitute 1.0 M HC2H3O2 for 1.0 M HCl.
Calculate the heat of neutralization per mole of water formed.
Waste Disposal Instructions Handle the stock solutions carefully. You may use a
wet sponge or paper towel to clean up any spills. The reaction mixtures produced in
the Styrofoam cups contain harmless salts. They can be discarded in the sink with
running water. Do not discard the Styrofoam cups. They should be returned to
your instructor.
Data & Analysis:
A. Heat Capacity of Calorimeter
Table 1: Temperature vs. Time Data for Calorimeter Calibration
Trial 1
Temp., oC
Trial 2
Temp., oC
Trial 3
Temp., oC
Calorimeter & Cool Water, T1
Warm Water, T2
Combined
Time (s)
0
15
30
45
60
75
90
105
120
135
150 165 180
Table 2: Heat Capacity of Calorimeter. (see example 1)
Trial 1 Trial 2 Trial 3
Final Temperature, Tf
Heat lost by warm water (J)
Heat gained by cool water (J)
Heat absorbed by calorimeter (J)
Calorimeter heat capacity, Ccal in J/oC
Average value of Ccal ______________________
Standard Deviation of Ccal ______________________
of Ccal at 95% confidence ______________________
B. Heat of Neutralization for HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)
Table 3: Temperature vs. Time Data for HCl – NaOH Neutralization
Trial 1
Temp., oC
Trial 2
Temp., oC
Trial 3
Temp., oC
Initial Temp of NaOH(aq), Ti
Initial Temp of HCl(aq)
Combined
Time (s)
0
15
30
45
60
75
90
105
120
135
150 165 180
Table 4: Molar Heat of Neutralization for HCl - NaOH. (see example 2)
Trial 1 Trial 2 Trial 3
T for reaction from data
Heat gained by solution (J)
Heat absorbed by calorimeter (J)
Total heat released by reaction (J)
Moles of water produced (moles)
Molar heat of reaction, Hrxn in kJ
Average value of Hrxn ______________________
Standard Deviation of Hrxn ______________________
of Hrxn at 95% confidence ______________________
C. Heat of Neutralization of HC2H3O2(aq) + NaOH(aq) H2O(l) + NaC2H3O2(aq)
Table 5: Temperature vs. Time Data for HC2H3O2 - NaOH Neutralization
Trial 1
Temp., oC
Trial 2
Temp., oC
Trial 3
Temp., oC
Initial Temp of NaOH(aq), Ti
Initial Temp of HCl(aq)
Combined
Time (s)
0
15
30
45
60
75
90
105
120
135
150 165 180
Table 6: Molar Heat of Neutralization for HC2H3O2 - NaOH. (see example 2)
Trial 1 Trial 2 Trial 3
T for reaction from data
Heat gained by solution (J)
Heat absorbed by calorimeter (J)
Total heat released by reaction (J)
Moles of water produced (moles)
Molar heat of reaction, Hrxn in kJ
Average value of Hrxn ______________________
Standard Deviation of Hrxn ______________________
of Hrxn at 95% confidence ______________________
Questions:
1. The accepted value for the molar enthalpy of neutralization is Ho = -55.81
kJ/mol. {P.J. Cerutti et al., Can. J. Chem. 56 (1978) 3084 ) Does this result fall within
the confidence limits of your experimental value for the HCl – NaOH neutralization
reaction? If not, give sources of error that may explain the discrepancy.
2. Answer question 1 using your HC2H3O2 – NaOH results. Why would the value of
Ho be the same for the two reactions?
3. The experimental procedure has you wash your thermometer and dry it after you
measure the temperature of the NaOH solution and before you measure the
temperature of the acid solution. Why?
4. Describe how you could determine the specific heat capacity of a metal by using
the apparatus and techniques in this experiment.
5. When a 3.25-g sample of solid sodium hydroxide was dissolved in a calorimeter
in 100.0 g of water, the temperature rose from 23.9 °C to 32.0°C. Calculate H (in
kJ/mol) for the solution process:
NaOH(s) Na+(aq) + OH-{aq)
Use a calorimeter heat capacity of Ccal = 15.8 J/oC.
6. A 50.0-mL sample of a 1.00 M solution of CuSO4 is mixed with 50.0 mL of 2.00 M
KOH in a calorimeter. The temperature of both solutions was 20.2°C before mixing
and 26.3°C after mixing. The heat capacity of the calorimeter is 12.1 J/K. From these
data calculate H for the process
CuSO4(l M) + 2KOH(2 M) Cu(OH)2(s) + K2SO4(0.5 M)
Experiment #5: Kinetics I - The Rate Equation
Reaction between Oxalic Acid and Potassium Permanganate
Objective: Determine the rate law for the reaction of oxalic acid with potassium
permanganate using a colorimetric analysis.
Background: Chemical reactions occur at various rates. Some occur instantaneously when
the reactants are mixed using optimum conditions - explosions are an example of this kind of
reaction. Others require a long time to occur - rusting of iron on a car is a slow process.
In industry, controlling the rate of a reaction is of prime importance; it can often
influence the purity of a product, its final form (i.e., big clumps or small particles), and
several other characteristics. Conditions that commonly play a role in the rate of a reaction
are concentration of the reactants, pressure (especially if gaseous reactants are involved),
temperature, and the presence of a catalysts. In this experiment we will consider the
importance of concentration of oxalic acid and of potassium permanganate in determining the
overall rate of reaction between them.
The rate is an experimentally determined quantity. Most reactions occur through a
series of steps called a reaction mechanism. More often than not, the mechanism is not
known. Consequently the observed rate cannot be predicted from the stoichiometry of the
reaction. For right now, we will simply write that
KMnO4(aq) + H2C2O4(aq) Intermediate products
purple colorless red pale yellowish
We can observe the progress of this reaction by its color. When the reactants are mixed, we
will see the dark purple color of potassium permanganate change to a red reaction
intermediate, which gradually changes over to a pale yellow product solution. By measuring
the time from mixing to the appearance of the red intermediate, we can obtain the rate of
reaction.
The rate of a reaction is a measure of the change in concentration of reactants and
products with time. Like the speed of a car, a rate is always a positive number. Therefore, the
rate based on the potassium permanganate concentration would be written:
4[KMnO ] = rate
t
eq. 1
Here [KMnO4] represents the change in potassium permanganate concentration during the
time interval t. Since the concentration decreases, the minus sign is needed to make the rate
come out positive.
The rate law expression gives the dependence of the reaction rate on the concentration
of reactants. For the reaction under investigation, we write the rate law as
rate = k[KMnO4]X[H2C2O4]
Y eq. 2
where the [ ] indicates molarity (moles/L), and X and Y are called the orders with respect to
potassium permanganate (X) and oxalic acid (Y), respectively. The values of X and Y are not
always integers; they are often fractions or decimal numbers. The overall rate order would
be "X + Y". Finally the constant, k, is called the rate constant for the reaction and it is a
constant at any one temperature.
Suppose we do two experiments, keeping the potassium permanganate concentration
constant for both, [KMnO4l1, but varying the oxalic acid concentration,
[H2C2O4]1 and [H2C2O4]2. We could write two rate expressions now:
Experiment #1: rate1 = k[KMnO4]1x[H2C2O4]1
y
Experiment #2: rate2 = k[KMnO4]1x[H2C2O4]2
y
If we now take the ratio of these two experimental results
X Y
1 4 1 2 2 4 1
X Y
2 1 2 2 4 2
rate k[KMnO ] [H C O ] =
rate k[KMnO4] [H C O ]
By cancelling k [KMnO4]1X, we would get
Y
1 2 2 4 1
Y
2 2 2 4 2
rate [H C O ] =
rate [H C O ] eq. 3
If we now take the natural (or Naperian) logarithm of both sides of eq. 3 we would get
1 2 2 4 1
2 2 2 4 2
rate [H C O ]ln = Yln
rate [H C O ]
We could now solve for Y.
1 2
2 2 4 1 2 2 4 2
ln rate /rateY=
ln [H C O ] /[H C O ] eq. 4
By reversing the process (varying the concentration of KMnO4 and keeping the
concentration of H2C2O4 constant), we can find the value of X. Then we can solve for the
overall rate order, X + Y. The rate constant, k, can be determined by substituting X and Y
into the rate expression eq. 2.
Precautions:
● You MUST wear safety goggles at all times while in the laboratory.
● KMnO4 (potassium permanganate) stains skin and clothes, the purple color
changes to dark brown. It will wear off the skin with normal washing, but will cause a
permanent stain on clothing.
- you may wish to wear gloves and an apron or lab coat
● It is always a good idea to
- wash your hands well if in contact with chemicals
- don't rub your eyes with your fingers
- wash your tab bench well to remove any spills
● All solutions should be placed in the special containers in the hoods.
● Pipets should be used only with bulbs or pumps. NO pipeting by mouth!
● All normal safety rules must be obeyed, including any special precautions issued by
your instructor.
________________________________________________________________________
Procedure: (work in pairs)
A. Using clean beakers and a graduated cylinder, obtain the following:
● 80 mL of distilled water in a 150- or 200- or 250- mL beaker
● 25 mL of 0.130 M KMnO4 in a small (50- or 100-mL) beaker
● 120 mL of 0.755 M H2C2O4 in a 200- or 250-mL beaker
● One sealed tube (vile) of Eriochrome Black T solution.
Record the temperature in the lab in °C
These solutions will be used to prepare the concentration series for the experiments. A table
of the solution combinations is given below. The subsequent procedure describes how the
solutions are to be mixed.
Table 1: Solution Volumes for Three Experimental
Reactants
Experiment #1
Experiment #2
Experiment #3
distilled water
6,00 mL
1.00 mL
5.00 mL
0.130M KMnO4
1.00 mL
1.00 mL
2.00 mL
0.755 M H2C2O4
5.00 mL
10.00 mL
5.00 mL
Total Volume
12.00 mL
12.00 mL
12.00 mL
[KMnO4], M
.
[H2C2O4], M
Determination #1: Calculate the molarity of KMnO4 and H2C2O4 for each set-up.
Example: Experiment #1
4 4(0.130 M KMnO )(1.00 mL KMnO ) = 0.0108 M
12.00 mL solution
Use a similar approach for the oxalic acid concentration.
B. Using disposable pipets, measure 6.00 mL of distilled water and 5.00 mL of oxalic acid
into a test tube and stir well with a glass stirring rod.
C. One partner should measure 1.00 mL of potassium permanganate and prepare to add it to
the test tube. The other partner should be ready to start the timer when the permanganate is
added. Stir immediately. The solution will change color from purple to red to yellow. Record
the time when the solution color matches that of the tube of Eriochrome Black T.
D. Dispose of the test tube contents in the designated container located in the hood. Clean
and dry the test tube. Repeat steps (B) and (C), rotating duties of the partners, for a total of 5
trials.
E. Repeat the procedure for the amounts in Experiment #2 and then for the amounts in
Experiment #3.
________________________________________________________________________
Data and Calculations
Table 2: Data for Experiment #1 Conditions T = oC
[KMnO4] [H2C2O4] time (s) Rate1 = [KMnO4]/time
Trial 1
Trial 2
Trial 3
Trial 4
Trial 5
Average Rate ____________________
Standard dev. ____________________
95% conf., ____________________
Table 3: Data for Experiment #2 Conditions T = oC
[KMnO4] [H2C2O4] time (s) Rate2 = [KMnO4]/time
Trial 1
Trial 2
Trial 3
Trial 4
Trial 5
Average Rate ____________________
Standard dev. ____________________
95% conf., ____________________
Table 4: Data for Experiment #3 Conditions T = oC
[KMnO4] [H2C2O4] time (s) Rate3 = [KMnO4]/time
Trial 1
Trial 2
Trial 3
Trial 4
Trial 5
Average Rate ____________________
Standard dev. ____________________
95% conf., ____________________
Determination of Reaction Order:
The results from the various pairs of experiments will now be combined to find X and
Y. The conditions for Experiments #1 and #2 differ by the oxalic acid concentration,
[H2C2O4]. We can therefore use eq. 4 to give the reaction order with respect oxalic acid:
1 2
2 2 4 1 2 2 4 2
ln rate /rateY=
ln [H C O ] /[H C O ] eq. 4
Substitute your values of rate1 and rate2 to obtain Y; remember that “ln” is natural log not log
base 10.
Now we can use results from Experiments #1 and #3 to get a similar expression for the
reaction order with respect to potassium permanganate.
1 3
4 1 4 3
ln rate /rateX=
ln [KMnO ] /[KMnO ] eq. 5
Substitute your values of rate1 and rate3 to obtain X.
Now we can write the overall rate equation, substituting numbers for X and Y:
rate = k [KMnO4]X[H2C2O4]
Y
Then we can solve for the rate constant, k, using data from any of the three
experiments.
X Y
4 2 2 4
ratek =
[KMnO ] [H C O ] eq. 6
The units of k depend on the values of X and Y. the rate has units of mol/L.s. Therefore, the
rate constant has units of mol(1-X-Y)
/L(1-X-Y)
.s. As an example, if X = 1.5 and Y=0.5, the units
of k are mol(1-1.5-0.5)
/L(1-1.5-0.5)
.s = mol-1
/L-1
.s = L/mol.s.
Results:
Reaction Order X (rounded to nearest 10th
) _______________
Reaction Order Y (rounded to nearest 10th
) ________________
Overall Order of Reaction X+Y ________________
Rate constant, k (with proper units) ________________
_____________________________________________________________________
Questions:
Question #1: In this experiment you estimated the reaction rate by
rate = [KMnO4]/time
Derive this expression from the more general relation given in eq. 1 by accounting for the
fact that all the permanganate is converted to intermediate at time t? Based on your
experimental observations, suggest another way to estimate the rate.
Question #2: Using your values of X, Y, and k, what would the rate be if [KMnO4] =
0.0500 M and [H2C2O4] =0.123 M?
Question #3: Each reaction condition was repeated in triplicate to enable the
determination of the precision of the reaction rates. Recall, precision is represented by the
width of the 95% confidence interval, 2. You will now use the s to estimate the precision
of the reaction orders X and Y. Use the determination of X as an example. From eqn 5
1 3
4 1 4 3
ln rate /rateX=
ln [KMnO ] /[KMnO ]
The numerator contains the ratio of two rates. This ratio is too large if rate1 is too large and
rate3 is too small. Alternatively, if rate1 is too small and rate3 is too large, X will be too small.
To incorporate the confidence limits we will use (rate+/2) and (rate-/2) to give estimates of
the precision of X and Y. Here (rate+/2) represents the point half way to the upper
confidence limit and (rate-/2) is half way to the lower confidence limit. The upper and lower
bound of X (X+ and X-) become
1 1 3 3
+
4 1 4 3
ln (rate /2)/(rate / 2)X =
ln [KMnO ] /[KMnO ]
1 1 3 3
-
4 1 4 3
ln (rate /2)/(rate / 2)X =
ln [KMnO ] /[KMnO ]
Determine X+, X-, Y+ and Y- from your data.
Question #4: The precision of the rate constant determination can be established in a
similar manner to that employed in Q3 for the precision of X and Y. Defining
X+1/2 = (X+ + X)/2 X-1/2 = (X- + X)/2
Y+1/2 = (Y+ + Y)/2 Y-1/2 = (Y- + Y)/2
We can estimate the upper and lower bounds of k to be;
-1/2 -1/2
+1/2 +1/2
+ X Y
4 2 2 4
X Y
4 2 2 4
ratek =
KMnO H C O
ratek =
KMnO H C O
Determine the upper and lower limits of the rate constant k from your data.
Experiment #6: Kinetics II - The Arrhenius Equation
Reaction between Oxalic Acid and Potassium Permanganate
Objective: Determine the activation energy for the reaction of oxalic acid with
potassium permanganate using a colorimetric analysis.
Background: In the previous experiment you determined the parameters of the rate equation
for the reaction between potassium permanganate and oxalic acid.
KMnO4(aq) + H2C2O4(aq) => products
purple colorless yellowish to colorless
The rate law expression gives the dependence of the reaction rate on the
concentration of reactants. For the reaction under investigation, we wrote
rate = k[KMnO4]X[H2C2O4]
Y eq. 1
where the [ ] indicate molarity (moles/L) and X and Y are called the orders with respect to
potassium permanganate (X) and oxalic acid (Y), respectively. The values of X and Y are not
always integers; they are often fractions or decimal numbers. The overall rate order would
be "X + Y". Finally the constant, k, is called the rate constant for the reaction and it is a
constant at any one temperature.
You have already observed that an increase in concentration of either reactant resulted
in an increase in the rate of the reaction, reduction of the time required for the complete
conversion of KMnO4. Provided the reactions were all conducted at the same temperature,
the rate constant, k, likely was determined to be independent of reactant concentration.
However, had the temperature been changed you would have discovered a dramatic change in
the rate of the reaction and the rate constant. For many reactions it has been observed that the
rate constant doubles for each temperature increase of 10 oC. In this experiment you will
determine the temperature dependence of the rate constant for permanganate/oxalic acid
reaction.
When two molecules react, they must collide with each other with sufficient energy
and with the correct orientation. Think of carrying some packages while walking in a crowd.
If someone brushes against you, it might be annoying, but otherwise nothing happens. If,
however, someone is running and bumps into you (high kinetic energy), they might knock off
a package or two, or even knock you over completely. Conceptually it’s the same with
colliding molecules. Low energy collisions don’t loosen the bonds of the molecules.
Generally, there is a threshold energy that the collision must overcome to break and
subsequently rearrange bonds (knock off the package). This threshold energy is called the
activation energy.
Recall that molecules in a fluid are in constant motion. At any given temperature
some are moving slow and others are fast, have sufficient energy to overcome the activation
energy barrier. As the temperature increase, the fraction of fast moving molecules increases,
thus increasing the chance of reactive collisions. The reaction rate is proportional to the
fraction of collisions with energy in excess of the activation energy. This is mathematically
described by the Arrhenius equation.
aE /RTk = Ae
eqn. 2
Where k is the rate constant, A is referred to as a frequency factor, Ea is the activation energy
for the reaction in J, R is the gas constant (8.314 J/mol.K) and T is the absolute temperature
in degrees Kelvin.
In the previous experiment you measured the rate constant of the KMnO4/H2C2O4
reaction at room temperature for various concentrations of the reactants. Now you will
determine the rate for a single concentration, but at multiple temperatures. To determine the
activation energy from the rate vs. temperature data it is most convenient to take the natural
log of both sides of eqn. 2.
aE /RT
a
ln(k) = ln(A) + ln e
Eln(k) = ln(A) -
RT
eqn. 3
Inspection of eqn. 3 reveals that a plot of ln(k) vs. 1/T should reveal a straight line, with a
slope of –Ea/R and an intercept of ln(A).
Example: An experiment is run at 20.0, 30.0 and 40.0 oC using 6.00 mL of distilled water,
5.00 mL of 0.755 M H2C2O4(aq) and 1.00 mL of 0.130 M KMnO4(aq) for each run. The
reaction rate at the three temperatures is given in Table 1. The table also gives values for
1000/T and ln(rate) to be used to construct a linear graph.
Temp. oC rate (mol/L.s) 1000/T (1/K) ln(rate)
20.0 5.4x10-5
3.41 -9.83
30.0 1.1x10-4
3.30 -9.12
40.0 2.5x10-4
3.19 -8.29
The ln(rate) vs. 1000/T data is plotting in Figure 1. As can be seen from the graph, the data
points lie along a line. The slope of the line is -7.026, while the y-intercept is 14.12. Using
eqn. 3 we can relate the slope to the activation energy.
Slope = -7.026 K.kJ/J = -Ea/R. eqn. 4
Ea = (7.026 K.kJ/J)(8.314 J/mol.K) = 58.4 kJ/mol
The units of the slope account for the fact that the x-axis is 1000/T. The factor of 1000 comes
from the conversion of J to kJ.
The intercept can be used to obtain a value for the frequency factor A. To obtain an
explicit value for A requires the conversion of rate to rate constant k through the use of the
rate expression given in eqn. 1. As an example, for X = 1.3 and Y = 0.75
rate = k[KMnO4]1.3
[H2C2O4]0.75
[KMnO4] = M1V1/V2 = (0.130 M)(1.00 mL)/(12.0 mL) = 0.0108 M
[H2C2O4] = (0.755 M)(5.00 mL)/(12.0 mL) = 0.315 M
Then rate = k(0.0108)1.3
(0.315)0.75
= 1.17x10-3
k
or ln(rate) = ln(1.17x10-3
k) = ln(1.17x10-3
) + ln(k) = -6.75 + ln(k)
From the above analysis we can see that for the example data the y-intercept for
ln(rate) vs. 1000/T would be shifted down 6.75 units compared to the y-intercept for ln(k) vs.
1000/T. The frequency factor can be obtained from the y-intercept of the graph by
ln(A) -6.75 = 14.115
or A = e20.87
= 1.15x109 L
1.05/mol
1.05s
Note the frequency factor must have the units of the rate constants. Your values of X and Y
are likely to be different; therefore, the units of A will be different.
________________________________________________________________________
Precautions:
● You MUST wear safety goggles at all times while in the laboratory.
● KMnO4 (potassium permanganate) stains skin and clothes, the purple color
changes to dark brown. It will wear off the skin with normal washing, but will cause a
permanent stain on clothing.
- you may wish to wear gloves and an apron or lab coat
● It is always a good idea to
- wash your hands well if in contact with chemicals
- don't rub your eyes with your fingers
y = -7.026x + 14.115 R² = 0.9964
ln(r
ate
)
1000/T (T in K)
Figure 1: Arrhenius Plot
- wash your tab bench well to remove any spills
● All solutions should be placed in the special containers in the hoods.
● Pipets should be used only with bulbs or pumps. NO pipeting by mouth!
● All normal safety rules must be obeyed, including any special precautions issued by
your instructor.
________________________________________________________________________
Procedure: (work in pairs)
A. Using clean beakers and a graduated cylinder, obtain the following:
● 100 mL of distilled water in a 150- or 200- or 250- mL beaker
● 15 mL of 0.130 M KMnO4 in a small (50- or 100-mL) beaker
● 60 mL of 0.755 M H2C2O4 in a 200- or 250-mL beaker
Obtain a vial of Eriochrome black T in CaCl2(aq) from your instructor. This red
solution will be use as a color guide to determine the stop time for the reaction.
B. Using disposable pipets, measure 6.00 mL of distilled water and 5.00 mL of oxalic acid
into each of 5 test tubes and stir well with a glass stirring rod. In 5 additional test tubes pipet
1.00 mL of 0.130 M KMnO4(aq).
C. Place the ten test tubes in a 400- or 500-mL beaker with enough water to cover the level
of the liquid in the tubes. This will be used as a temperature-control bath to regulate the
temperature of the reaction. The bath will initially be heated to ~45 oC, at which point the
time required for the first reaction is measured. As the bath is allowed to cool to room
temperature, successive trials are performed. The goal is to get obtain a total of 10 rate vs.
temperature measured recorded over a 25 – 30 oC temperature range.
D. For each measurement, one partner should add the aqueous oxalic solution to the test
tube containing the 1.0 mL of KMnO4(aq) and shake with a gloved finger over the top of the
tube. The other partner should be ready to start the timer when the permanganate is added.
Once mixed and placed back into the bath record the temperature of the bath water. Use the
red Eriochrome black T solution as a colorimeteric guide for when to stop the timer.
E. Dispose of the test tube contents in the designated container in the hood. Clean and dry
the test tube.
F. Repeat the experiment for the remaining 4 pairs of tubes, allowing the bath water to cool
between trials to adequately cover the specified temperature range.
G. Repeat steps B-F for an additional 5 trials. Depending on the temperature range you
cover previously, you may start at a lower initial temperature and/or use ice to cool.
________________________________________________________________________
Data and Calculations
Table 2: Raw Data for Temperature Studies
[KMnO4] [H2C2O4] Temp. oC Time to red (s)
Trial 1
Trial 2
Trial 3
Trial 4
Trial 5
Trial 6
Trial 7
Trial 8
Trial 9
Trial 10
Table 3: Data for Rate vs. Temperature Graph
rate = [KMnO4]/time 1000/T (T in K) ln(rate)
Trial 1
Trial 2
Trial 3
Trial 4
Trial 5
Trial 6
Trial 7
Trial 8
Trial 9
Trial 10
Determination of Arrhenius parameters:
A. Using the data in Table 3, make a graph of ln(rate) vs. 1000/T (ln(rate) is the y-
variable and 1000./T is the x-variable). This should be done using a spreadsheet program
such as Excel or Numbers. Fit the best line through the data using the trendline option of the
spreadsheet. Place the equation of the line on the graph.
B. Following the example given in the background section to obtain the value of the activation energy, Ea for the permanganate/.oxalic acid reaction.
C. Include the rate equation for the permanganate / oxalic acid reaction you determined last week in your report.
rate = k [KMnO4]X[H2C2O4]
Y
Using your values of X and Y, along with the concentrations [KMnO4] and [H2C2O4]
determine the numerical relation between rate and rate constant. The example gave rate =
1.17x10-3k ; however, your values will be different if X and Y are different.
D. Based on the y-intercept for your ln(rate) vs. 1000/T graph and the relationship
between rate and k, determine the value of the frequency factor, A. Be sure to use the proper
units based on the overall order of the reaction.
Results:
Overall Order of Reaction X+Y ________________
Activation Energy, Ea (in kJ/mol) ________________
Give C for rate = C*k ________________
Frequency Factor, A (proper units) ________________
_____________________________________________________________________
Questions:
Question #1: Consider your ln(rate) vs. 1000/T graph. Was the data evenly
distributed around the line or does the low (or high) temperature end of the curve have more
scatter than the other. Is this result consistent with your observations? Explain.
Question #2: The introduction stated that often the rate doubles for each temperature
increase of 10 oC. What value of the activation energy would cause the rate to double
between 25 and 35 oC?
Question #3: In this experiment you used a red colorimetric guide to establish the
end of the reaction. Would you expect to get the same value of the activation energy if yellow
was chosen rather than red? Support your answer with your observations of the reaction.
Question #4: How would you answer to question 3 differ if the frequency factor,
rather than the activation energy were considered?
Experiment #7: Visible Spectroscopy
Objective: Determine the wavelength of maximum absorbance for a variety of solutions;
then determine the concentration of an unknown through use of the Beer-Lambert Law.
Background: Light is a form of electromagnetic radiation, which includes a radiation
stretching from cosmic rays at high energies to radio waves at low energies. One small
portion of this radiation is visible light, which contains all of the colors of the spectrum.
Ordinary white light is a mixture of all possible colors, each characterized by the wavelength
(lambda). When we see a solution that has a color, it is because the solution absorbs certain
portions of the visible light; what we see is the complementary or transmitted light. For
example, a solution which appears orange has absorbed radiation in the blue-green region
with wavelengths of light from 485 to 495 nm (1 nm = 1x10-9
m). The table below shows the
complementary relationships of visible light.
Table 1: Complementary Colors of Visible Light
Color - Light
Observed
(transmitted)
Complementary Color
Absorbed
Wavelength of light
absorbed, (nm)
Transparent Ultra-violet (UV) < 400
Yellow-Green Violet 400-435
Yellow Indigo 435-465
Orange Blue 465-485
Red-Orange Cyan 485-500
Red Green 500-540
Violet Yellow-Green 540-560
Indigo Yellow 560-590
Blue Orange 590-620
Green Red 620-700
Transparent Infra-red > 700
To make accurate measurements using colored solutions, it is important to determine
the wavelength at which the solutions absorb the most light. This is called lambda max, max.
Experimentally, this can be predicted by matching the color of the solution with the
transmitted color in the table above; but we usually want more than a range. So we would
take the solution and measure its absorbance of light at different wavelengths close to our
prediction to find the maximum point.
To make this measurement, we will use a visible spectrophotometer, the Spec 20™ or
Spec 21TM
. A wavelength is selected, the machine is calibrated with distilled water to give a
reading of 0.00 absorbance (a procedure known as zeroing the instrument), and then the
colored solution is measured at the same wavelength. The scale on the spectrometer gives
two readings: A, the absorbance of light at that wavelength and %T, the percent transmittance
of light at that wavelength. These two measurements are related by the equation below.
A = 2 - log10(%T) eqn. 1
For example, if the %T reading was 23.5%, we could solve for absorbance, A:
A = 2 – log10(23.5)
A = 2 - 1.37
A = 0.63
Usually the absorbance is reported with two decimal places because that is indicative of the
precision with which we can estimate the scale increments from the Spec 21. The Spec 20D
allows for 3-place precision. . Many people feel that measuring %T and then mathematically
converting it to A is a better procedure to use rather than reading the value of A directly from
the scale. The reasoning for this is that the %T scale is linear and the A scale is not. [Note
that when we calibrate the spectrometer to have a 0.00 A reading with distilled water, it also
had a reading of 100 %T, meaning that no light was absorbed and all of it went through the
solution.]
The intensity (or darkness) of the color of a solution is proportional to the amount of
colored solute dissolved in the solution. This makes sense if you think about making 3 cups
of instant coffee in glass mugs. If you measured 1 teaspoon of instant coffee into one cup, 1/2
teaspoon of instant coffee into the second cup, and 2 teaspoons of instant coffee into the third
cup and then added equal amounts of boiling water to each one and stirred, you could give
them to three friends and ask them to line them up in order of increasing "strengths". By
merely observing the color changes, this could be easily done. You used this approach to
establish the extent of the permanganate/oxalic acid reaction in the kinetics experiments.
We can make quantitative measurements of the concentration of a colored solute
using a spectrometer. This can be done using any concentration unit (i.e., ppm, density,
molarity, % by weight), but most commonly molarity (M) is used. The relationship is
described by Beer's Law (sometimes called the Beer-Lambert Law):
A = bc eqn. 2
where A = absorbance of the solution (no units)
= extinction coefficient; this is a constant for the solute and represents the
absorbance of a 1-M solution of the solute for a 1 cm wide cuvette.
b = the width of the cuvette; the path length of light going through the
cuvette (usually measured in cm, 1.16 cm for the Spec-21)
c = the concentration of solution, usually in molarity (M)
Experimentally, we would prepare several solutions of known molarity of the solute
in solvent (often, but not always, water). We would then measure the absorbance of each
solution at the max for that solute/solution and make a graph of A (y-axis) versus
M (x-axis). Once this Beer’s Law plot is prepared we could determine the concentration of an
unknown by locating its absorbance on the graph reading off the concentration. This
approach will be taken in the subsequent experiment.
_________________________________________________________________________
Precautions:
* You MUST wear safety goggles at all times while in the laboratory.
* It is always a good idea to
- wash your hands well if in contact with chemicals
- don't rub your eyes with your fingers
- you may wish to wear gloves and an apron or lab coat
- wash your lab bench well to remove any spills
* All solutions containing copper sulfate (blue) should be placed in the special
containers in the hoods.
* Ammonia (aqueous) has a strong smell, as you may have observed in your kitchen.
It will be located in the hoods.
* Pipets should be used only with bulbs or pumps. NO pipeting by mouth!
* All normal safety rules must be obeyed, including any special precautions issued by
your instructor.
________________________________________________________________________
Procedure: (work in pairs)
I. Determination of max
A. Obtain the following solutions:
1. cuvette with red food dye
2. cuvette with green food dye
3. cuvette yellow food dye
4. 50 mL of 0.1 M CuSO4 [copper(II) sulfate], add ~ 5 mL to a cuvette
5. 20 mL of distilled water, add ~ 5 mL to a cuvette
B. Mix 5 mL of 0.1 M CuSO4, 5 mL of 2.5 M NH3, and 10 mL distilled H2O (if
not previously prepared) in the hood, and add ~5 mL to a cuvette.
C. Using the previous table, estimate the range where you expect to find max for
each solution. For example, if you had a solution that appeared orange, you
would expect max to fall in the range 465-485 nm.
Solution
Color
max , in nm
(give range)
1
Red
2
Green
3
Yellow
4
0.1 M CuSO4
5
0.05 M CuSO4 + NH3
D. Set the wavelength at mid-point of the range for the first solution. [In our
example of orange, we expect max to be ~ 475 nm.] Using the cuvette with
distilled water, set the scale to A = 0.00 (%T=100). Insert the cuvette with
solution #1 and read the absorbance, A (Spec20D+) or % transmittance
(Spec21). Record it in Table 2. Now set the wavelength at 10 nm higher (i.e.,
485 in our example), zero the machine with distilled water and measure
absorbance of solution and record. Repeat the procedure, adjusting the
wavelength in 10 nm intervals in the direction of increasing absorbance
(decreasing transmittance) until the absorbance starts to diminish. Once max
has been reached, obtain two more measurement (still 10 nm intervals) beyond
it.
E. Repeat step D for the remaining 4 solutions, alternating lab partners as you go.
F. Reserve the remaining CuSO4 solution for part II.
G. Prepare a graph of Absorbance (y-axis) versus Wavelength (x-axis), including
data from all 5 solutions. You will have 5 small curves on the graph (one for
each solution); distinguish the data by using different colors or symbols. Label
the max for each solution.
II. Preparation of Beer’s Law Plot
A. In the hood, measure 25 mL of 2.5 M NH3(aq) using a graduated cylinder.
Transfer it to a 400-mL beaker and add 225 mL of distilled water. Mix well.
This solution will be used to prepare the subsequent solutions.
Determination 1: What is the approximate M of the new NH3(aq) solution?
B. Obtain about 25 mL of 0.1 M CuSO4 solution in a clean, dry 50-mL beaker.
Using a disposable graduated pipet and pipet pump, pipet 1 mL of the CuSO4
solution into a 25-mL volumetric flask. Fill about ½ way with the NH3(aq)
solution prepared in (A). Mix well. Fill to the line (use a dropper near the end)
with the NH3(aq) solution. Cap and mix well.
The reaction that you observe here is:
Cu(H2O)42+
+ 4 NH3 Cu(NH3)42+
+ 4 H2O
light blue dark blue-violet
Determination 2: Calculate the molarity of the Cu(NH3)42+
ion in this solution.
Hint: You are really calculating the molarity of Cu2+
(or CuSO4) in the solution
because
1 mole CuSO4 1 mole Cu(H2O)42+
1 mole Cu(NH3)42+
C. Set the wavelength on the spectrometer at the max of solution #5 from PART
I. Set the scale at A = 0.00 (100%T) with a cuvette filled with distilled water.
Using ~ 5 mL of solution in a cuvette, measure the %T or absorbance for the
Cu(NH3)42+
prepared in B.
D. Repeat procedure B & C using 2 mL of CuSO4 solution; then with 3 mL, 4
mL, and 5 mL.
Determination 3: Calculate the molarity of each Cu(NH3)42+
solution measured in
step D. (hint: If you have the M from using 1 mL of CuSO4 solution, what's the quick
and easy answer if you use 2 mL? 3 mL? 4 mL? 5 mL?)
E. Using data from solutions (1) through (5), prepare a graph of absorbance, A
(y-axis), versus molarity, M (x-axis). This should be a straight line and must
go through the point (0,0) because we forced the spectrometer to read zero
absorbance with distilled water (i.e., 0.00 M CuSO4).
Determination 4: Based on eqn. 2, the slope of the graph created in part D gives the
value for b. Obtain the slope from your graph.
III. Concentration of Analyte
A. Obtain an unknown from your instructor and write down its number __. Using
a pipet, transfer 5 mL of unknown into a 25-mL volumetric flask and dilute
with NH3(aq) as before. Prepare a total of 4 separate solution of the unknown
in this manner.
B. Set the wavelength on the spectrometer at the max of solution #5 from PART
I. Set the scale at A = 0.00 (100%T) with a cuvette filled with distilled water.
Place each solution in a cuvette (~ 5 mL) and measure the %T or absorbance
for each.
Determination 5: For each absorbance value determine the concentration of the
diluted unknown using A = bc (b was established in determination 4). Using
the fact that the absorbance samples were a five-fold dilution of the original
unknown analyte, determine the molarity of the original unknown.
IV. Clean-up
A. All solutions containing copper(II) [blue] should be placed in the special
container in the hoods. All others can be rinsed into the sink with running tap
water.
B. Return the unknown container to the instructor.
C. Clean your equipment and lab bench area.
D. Have your instructor initial your lab book.
E. LOCK your LOCKER!
F. WASH your HANDS!
Data and Analysis
The data tables for Parts I –III are included below along with spaces for the various
determinations called for in the procedure. These should be included in your report. In
addition you should include the two graphs (absorbance vs wavelength and
absorbance vs molarity) in your report.
Table 2: Part I: Absorbance vs. Wavelength
Wavelength, nm %T
Absorbance, A
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Part II: Determination 1: Molarity of NH3(aq) solution ______________
Table 3: Absorbance of Cu(NH3)42+
solutions (Determination 2 & 3)
SOL'N
mL
M of
Cu(NH3)42+
%T
Absorbance, A
1
1 mL CuS04
2
2 mL CuS04
3
3 mL CuS04
4
4 mL CuS04
5
5 mL CuS04
Part II: Determination 4:
Slope of absorbance vs concentration graph b = _________________
Table 4: Part III: Concentration of Analyte Unknown #_______________
Trial
%T
Absorbance
A
c = A/b
Mol/L
M of analyte
1 unknown
2 unknown
3 unknown
4 unknown
Average [Cu2+
] _________________
Std. Dev. [Cu2+
] _________________
95 % l for [Cu2+
] _________________
Questions:
Question 1: Pure colored substances mostly exhibit a single peak (i.e. max) in the
visible; therefore the simple rules of complementary colors apply. Mixtures are a little
more complicated. When you were in grade school you learned that by mixing yellow
(max = 450 nm) with blue (max = 605 nm) you get green. Describe why this is based
on the wavelength range that is transmitted. What color would you see if you mixed
the two solutions with max = 520 and max = 610?
Question 2: The determination of the Cu2+
analyte concentration was based on the
proportionality between concentration and absorbance. The accuracy of that
determination in turn depended on how well the experimental absorbance data was fit
by the line A = bc. If we let Ai be a measured absorbance value for the concentration
ci, then |Ai – ebci| gives the absolute difference between the data and the line we used
to fit it. Recall the definition for the standard deviation was
2
11
i
n
X X
n
Similarly, we can describe how well a line fits the data by defining the standard error.
For the specific example of the Beer’s Law plot, the relationship is;
2
standard error = 1
i iA bc
n
Determine the standard error for your Beer’s Law plot.
Question 3: Part II of the procedure required that you make up 5 25-mL solutions of
different concentrations to prepare the Beer’s Law plot. This generated 125 mL of
waste per group, or ~25 L for all the sections of the lab course during the year.
Describe how you would change the procedure to reduce the waste generated and still
get an accurate result for the slope of the A = bc curve.
Question 4: A student proposes to use visible spectroscopy to measure the kinetics of
the permanganate/oxalic acid reaction. As you recall, the initial permanganate solution
was very dark. For a 1-cm sample cell, the absorbance A = 2500[MnO4-]. As a
practical rule of thumb, Beer’s law is only accurate below an absorbance value of
A=1.00. This limits the initial permanganate concentration that can be used.
a) What is the maximum permanganate concentration that can be used and
still have A < 1.00?
b) If the detection limit for the spectrometer is A = 0.01, what is the
minimum permanganate concentration that can be detected?
c) The rate law for the reaction was determined to be
4 1.29 0.76
4 2 2 4
[ ]0.052[ ] [ ]
o
o o
MnOMnO H C O
t
If [H2C2O4]o = 0.300 mol/L and [MnO4-] is the maximum it can be for visible
spectroscopy, how long would it take for the permanganate to be used up?
Remember the rate law given above was for the average rate for the complete
conversion of MnO4- to Mn
2+.
Experiment #8: Spectroscopic Determination of the Equilibrium Constant
for Fe3+
(aq) + SCN-(aq) ↔ FeSCN
2+(aq)
Objective: Determine the room temperature equilibrium constant for the reaction between
Fe3+
, ferric or iron(III) ion, and SCN-, thiocyanate ion, to form the complex ion, FeSCN
2+,
ferric thiocyanate ion (or thiocyanatoiron(III) ion) in an acidic aqueous solution by using
visible spectroscopy.
Background: Many chemical reactions do not proceed to 100% completion. This means that
not all of the reactants are converted to products even when they are present in the correct
stoichiometric amounts. When the reaction appears to be finished, an analysis would show
the presence of reactants and products. The reaction has not stopped, however; reactants
continue to be converted to products and at the same time products are converted to reactants.
The process continues until there is no apparent change in concentrations of reactants and
products. This state is called dynamic equilibrium; "dynamic" because the reactants and
products continue to convert back and forth and "equilibrium" because the concentrations of
the reactants and products reach a constant value (not an equal value).
If we represent an equilibrium reaction with a general equation such as
aA + bB ↔ cC + dD
then we can define an equilibrium constant, Keq, as the multiplication of the concentration of
products (each raised to the power of its coefficient in the balanced equation) divided by the
multiplication of the concentration of reactants (each also raised to the power of its
coefficient in the balanced equation). We use square brackets, [A] to indicate the
concentration of A, in units of molarity (i.e., moles A/liter of solution).
c d
eq a b
[C] [D]K =
[A] [B] eqn. 1
In this experiment, we will be studying the reaction below;
Fe3+
(aq) + SCN-(aq) ↔ FeSCN
2+(aq)
pale yellow-green colorless blood red
The equilibrium expression will be
2+
eq 3+ -
[FeSCN ]K =
[Fe ][SCN ] eqn 2
In this experiment the blood red FeSCN2+
ion absorbs light best at max of
477 nm, which we can measure using a spectrophotometer as we did in the previous
experiment. We can also use the Beer-Lambert or Beer's Law (A = bc) to relate the
absorbance (A) to the concentration (c) of the FeSCN2+
ion. By varying the initial
concentrations of Fe3+
and SCN-, the equilibrium concentration of FeSCN
2+ can be varied.
Through the established Beer’s Law relationship [FeSCN2+
] can be measured by monitoring
the absorbance of the equilibrium mixture at a wavelength of 477 nm.
The absorbance data for different initial values of [Fe3+
] and [SCN-] can be analyzed
in several ways. We will take the approach given in the article by H.A. Frank and R.L.
Oswalt in the Journal of the American Chemical Society, 69, (1947) 1321. Using the
experimental measures: A – absorbance at 477 nm, - Beer’s Law coefficient, [Fe3+
] and
[SCN-] – initial concentrations of the Fe
3+ and SCN
- ions we get:
3+ -
o o
eq eq3+ - 3+ -
o o o o
A [Fe ] + [SCN ]A = -K + bK
[Fe ] [SCN ] [Fe ] [SCN ] eqn. 3
If you look closely at this equation it is in the form of a straight line, y = mx + b. So if we
make a graph with
y-axis y = 3+ -
o o
A
[Fe ] [SCN ] eqn. 4a
and x-axis x = 3+ -
o o
3+ -
o o
A [Fe ] + [SCN ]
[Fe ] [SCN ] eqn. 4b
then the slope of the line, m, will be -Keq and the y-intercept will be bKeq.
_______________________________________________________________________
PRECAUTIONS:
* You MUST wear safety goggles at all times while in the laboratory.
* It is always a good idea to
- wash your hands well if in contact with chemicals
- don't rub your eyes with your fingers
- you may wish to wear gloves and an apron or lab coat
- wash your lab bench well to remove any spills
* All waste solutions should be placed in the special containers in the hoods.
* Pipets should be used only with bulbs or pumps. NO pipeting by mouth!
* All normal safety rules must be obeyed, including any special precautions issued by
your instructor.
_______________________________________________________________________
PROCEDURE: (work in pairs)
I. Preparation of SCN- solution.
A. Clean a 100-mL volumetric flask. Rinse with distilled water. It does not need to
be dry.
B. Take the volumetric flask to the hood and dispense 10.0 mL of the 0.00200 M
KSCN(aq) solution into the flask.
C. Measure 25 mL of 2 M HNO3(aq) using a 100-mL graduated cylinder and add to
the volumetric flask containing the KSCN solution.
D. Add distilled water, with swirling, up to the 100-mL line. Mix very well by
swirling and inverting the flask. Pour the contents into a clean 250-mL beaker
and label as "SCN" dilution solution".
Determination #1: What is the molarity (M) of the SCN- in this solution? [Hint:
This is a dilution problem.]
II. Preparation of FeSCN2+
and measurement of absorbance.
A. Turn on the spectrophotometer and let it warm up for about 15 minutes. Set the
wavelength at 477 nm.
B. Using a 10.0 mL graduated cylinder, combine 2.5 mL of the 2 M HNO3(aq)
solution with 7.5 mL of deionized water to prepare a 0.5 M HNO3(aq) solution to
act as a spectral blank.
C. Obtain 2 cuvettes and clean them. Fill one to above the line with 0.5 M HNO3.
This will be the "blank" and will be used to set the absorbance, A, to zero (instead
of distilled water).
D. Using a graduated cylinder, pour about 20 mL of ferric nitrate/nitric acid solution
("0.100 M Fe3+
solution") into a small beaker (50- or 100-mL).
E. Place the beaker containing the "SCN' dilution solution" from Part I on a
magnetic stirrer and add a magnetic stirring bar. Using a 5-mL graduated pipet
and bulb (or pump), measure 5-mL of the 0.100 M Fe3+
solution and add to the
beaker.
Determination #2 Calculate the molartity of Fe3+
and SCN- in the solution prepared
in step II-E. Assume volumes are additive. Place the values into columns D and E of
Table I.
First trial: 3+ 3+
-3 3+(5 mL Fe )x(0.100 M Fe ) = 4.76x10 M Fe
105 ml sol'n
-4
-4(100 mL SCN )x(2.00x10 M SCN ) = 1.90x10 M SCN
105 ml sol'n
F. Using the cuvette with 0.5 M HNO3, blank the spectrometer at %T=100 (A = 0.0).
Pour out a small amount of the solution from the reaction mixture beaker into the
second cuvette and rinse; pour back into the beaker: DO NOT DISCARD! Do the
rinsing a second time, then pour out enough solution into the cuvette so that it is
above the line. Measure and record its %T or absorbance in column F (and/or G)
of Table I. Return this solution to the beaker: DO NOT DISCARD!
Determination 3: If you are using a Spec21 you need to convert the %T to A, using
A = 2 - log(%T) and enter in column G of Table I.
G. Using a 1-mL graduated pipet, add 1 mL of the 0.100 M Fe3+
solution to the
reaction mixture beaker and repeat step (F) and record the new molarities and
absorbance reading into Table I. Continue adding 1-mL portions until you have
completed the trials indicated in the data table.
___________________________________________________________________
III. CLEAN-UP
A. All solutions should be placed in the special container in the hoods. All glassware
can be rinsed with running tap water in the sinks.
B. The disposable pipets should be placed in the "GLASS DISPOSAL" boxes.
C. Clean your equipment and lab bench area.
D. Have your instructor initial your lab book.
E. LOCK your LOCKER!
F. WASH your HANDS!
________________________________________________________________
VI. DATA and ANALYSIS
Table I: Concentration and Absorbance Data
A B C D E F G
Trial mL Fe3+
mL SCN- [Fe
3+]o [SCN
-]o %T A
1 5 100
2 6 100
3 7 100
4 8 100
5 9 100
6 10 100
7 11 100
8 12 100
9 13 100
10 14 100
In addition to the normal portions of the report, you must include the graph to
determine the value of the equilibrium constant, Keq.
A Using the values of [Fe3+
]o, [SCN-]o and A from Table I determine the values for
x = 3+ -
o o
3+ -
o o
A [Fe ] + [SCN ]
[Fe ] [SCN ]
y = 3+ -
o o
A
[Fe ] [SCN ]
and place them into the appropriate columns of Table II.
Table II: x & y Data for the Determination of Keq
Trial x =
3+ -
o o
3+ -
o o
A [Fe ] + [SCN ]
[Fe ] [SCN ] y =
3+ -
o o
A
[Fe ] [SCN ]
1
2
3
4
5
6
7
8
9
10
B Using a spreadsheet program, prepare a plot of y vs. x using the results in Table
II. The plot should give a straight line with a slope equal to –Keq. Set a linear trend line
through the data to determine the slope.
C Determine the equilibrium constant from the slope of the trend line. What are the
units of Keq?
D Show sample calculations as requested throughout the report, using any Trial
EXCEPT Trial #1 which has been used as an example.
_________________________________________________________________
Questions:
Question 1: Look at the numerical value for your answer for Keq. What does it
tell you about the ratio of products to reactants? Which side is favored, the
product side or reactant side (i.e., which has the higher concentration)?
Question 2: A student performed an experiment similar to this with a different
set of reactants. The experimental result was Keq = 0. How would you interpret
this result?
Question 3: The determination Keq was based on the goodness of fit of the
line y = Keqx + bKeq. If we let yi and xi correspond to a particular trial, then
|yi + Keqx - bKeq | gives the absolute difference between the data and the line
we used to fit it. The value of bKeq is simply the intercept of the fit line.
Recall the definition for the standard deviation was
2
11
i
n
X X
n
Similarly, we can describe how well a line fits the data by defining the
standard error. For the specific example of the equilibrium plot, the
relationship is;
2
i eq eqy K x bKstandard error =
1n
Determine the standard error for your equilibrium plot.
Experiment #9: Determination of [KMnO4] by Oxidation-Reduction
Titration
Objective: To determine the amount of oxalate in an impure sample through the use of a
titration with potassium permanganate, after standardization with pure sodium oxalate. The
titration is based on the oxidation-reduction properties of the reactants.
Background: The technique of titration has been used previously to detect the concentration
of an acid (the analyte) using a known concentration of a strong base (the titrant). The
technique can also be used in situations in which the reaction involves oxidation and
reduction.
Oxidation is defined as the loss of electrons or increase in oxidation state of a
reactant. The oxidation reaction is often represented explicitly by writing an oxidation half-
reaction. The half-reaction for the oxidation of the oxalate ion is shown below.
C2O42-
2 CO2 + 2 e- oxidation half-reaction
Reduction is defined as the gain of electrons or decrease (reduction) of oxidation state of a
reactant. The reduction half-reaction for the permanganate ion is given below. The oxidation
state of manganese decreases from +7 to +2.
8 H+ + MnO4
- + 5 e
- Mn
2+ + 4 H2O reduction half-reaction
Oxidation and reduction must occur together and are often designated as redox processes to
emphasize this; the number of electrons lost by one substance must equal the number of
electrons gained by the other substance. Since oxalate loses 2 electrons while permanganate
gains 5, the two species must react in a 5-to-2 ratio to equalize the electron transfer. This is
represented by the net redox equation below.
16 H+ + 2MnO4
- + 5 C2O4
2- 2 Mn
2+ + 10 CO2 + 8 H2O
Because the materials we weigh and measure occur as compounds, it is often useful to have
the balanced molecular equation (sometimes called the balanced total equation).
8H2SO4 + 2KMnO4 + 5 Na2C2O4 2 MnSO4 + 10 CO2 + K2SO4 + 5 Na2SO4 + 8 H2O
Potassium permanganate is reduced because it contains the permanganate ion; we can
also say that it behaves as an oxidizing agent because it causes something else to become
oxidized (the oxalate). Sodium oxalate is oxidized because it has the oxalate ion and we could
also specify it as the reducing agent because it causes the permanganate to become reduced.
The kinetics of the reduction of permanganate by the oxalate ion was studied in
experiments #5 and #6. As you may recall, in these studies oxalic acid was used instead of
sodium oxalate. The goal of the present experiment is to use redox titration as a way of
quantifying the oxalate concentration. To achieve this we will need to speed up the reaction.
This can be done by running the reaction under highly acidic conditions and by heating up the
reaction mixture. You should notice that H+ (or H2SO4) appear explicitly in the balanced
equations. Since you have already performed a titration (Experiment #3) and are familiar with
the redox reaction from the kinetics experiments, you will conduct the present quantitative
analysis individually. A full lab report will not be required; however, data tables and
statistical analyses should be submitted with your result.
________________________________________________________________________
Precautions:
* You MUST wear safety goggles at all times while in the laboratory.
* KMnO4 (potassium permanganate) stains skin and clothes, the purple color changes
to dark brown. It will wear off the skin with normal washing, but will cause a
permanent stain on clothing.
- you may wish to wear gloves and an apron or lab coat
* It is always a good idea to
- wash your hands well if in contact with chemicals
- don't rub your eyes with your fingers
- wash your lab bench well to remove any spills
* All solutions should be placed in the special containers in the hoods.
* Pipets should be used only with bulbs or pumps. NO pipeting by mouth!
* All normal safety rules must be obeyed, including any special precautions issued by
your instructor.
___________________________________________________________________
Procedure: (work individually)
I. Preparation of KMnO4 Solution
A. Obtain ~100 mL of the 0.130 M KMnO4 solution in a clean 500-mL Erlenmeyer
flask; you do not need to be exact because the next part will involve
standardizing the molarity of this solution. Using a 100-mL graduated cylinder,
add 100 mL of distilled water to the flask; swirl and mix well. Continue with
another 100 mL of distilled water until you have a total of 300 mL of well mixed
solution.
Determination 1: We will titrate this solution later to find out its exact molarity, but
for now it might be a good idea to have a rough "ballpark" estimate of its molarity.
B. Obtain a 50-mL buret, clean it, and rinse a couple of times with distilled water.
Rinse twice with about 10-mL of your KMnO4 solution (discard the rinses in the
container in the hood) and then fill the buret with the KMnO4 solution, making
sure that the tip contains no air bubbles.
II. Standardization of the KMnO4 Solution
A. Obtain ~200 mL of 3 M sulfuric acid, H2SO4, in a clean beaker. Label the beaker
so it will not be confused with other solutions.
B. Weigh approximately 1.5 grams of pure sodium oxalate, Na2C2O4 (fw =133.96
g/mol) in a plastic weighing boat. Record the exact weight measured. Transfer
the solid to a clean 100- or 150-mL beaker. Rinse the boat into the beaker with a
small amount of 3 M H2SO4; then add about 50 mL of the 3 M H2SO4 and swirl
to dissolve. You may need to heat this gently (DO NOT BOIL) to aid in
dissolving the salt.
C. Transfer the solution to a clean 100-mL volumetric flask. Rinse the beaker with a
small amount of 3 M H2SO4 and add the liquid to the flask. Mix, then fill to the
100-mL line with 3 M H2SO4. Cap and mix well.
Determination #2: What is the molarity (M) of the Na2C2O4 solution?
D. Using a 10-mL graduated disposable pipet, transfer 20 mL of the sodium oxalate
known solution to each of 4 clean 125- or 250-mL flasks.
Determination #3: Based on your results to determinations #1 and #2, what is the
anticipated volume of KMnO4 solution required to titrate the sodium oxalate sample?
Example: For a 0.0430 M KMnO4 solution titrating 20.0 mL of a 0.112 M Na2C2O4:
moles Na2C2O4 = (0.112 mol/L)(0.020 L) = 0.00224 mol
moles KMnO4 = (0.00224 mol Na2C2O4)4
2 2 4
2 mol KMnO
5 mol Na C O
= 8.96x10-4
mol
volume = (8.96x10-4
mol)/(0.0430 mol/L) = 0.0208 L = 20.8 mL
E. Record the initial reading of KMnO4 in the buret. Warm the contents of the first
flask to 70-90°C (DO NOT BOIL!). Titrate to a very faint pink endpoint
(reminds you of the phenolphthalein endpoint). Record the final reading of
KMnO4 in the buret.
F. Refill the buret and repeat step (E) for the other 3 samples. All solutions in the
flasks may be placed in the containers in the hoods when the titrations are
finished. [Keep the KMnO4 in the buret, you need it for the next part.]
Determination #4: What is the molarity (M) of the KMnO4 solution?
III. Determination of %Na2C2O4 in an Unknown
A. Obtain an unknown from your instructor. Record the number # _______ Measure
about 0.6-0.7 g of unknown for each of 4 titrations. Record the masses of each.
Transfer to 4 clean 125- or 250-mL flasks, rinsing each weighing boat with a
small amount of 3 M H2SO4 and add to the flasks. Using a graduated cylinder,
add 20 mL of 3 M H2SO4 to each flask and swirl to dissolve.
B. Record the initial volume of KMnO4 in the buret. Heat the first flask to about 70-
90°C and titrate to a faint pink endpoint as before. Record the final volume of
KMnO4 in the buret. Repeat with the remaining 3 samples. The contents of the
flasks may be placed in the specially marked containers in the hoods.
Determination #5: What is the % (by weight) of Na2C2O4 in the sample?
Example: Trial #1 of the unknown contained 0.645 g of unknown and required 23.65
mL of a 0.0430 M KMnO4 (average value of molarity from the first part) to reach the
endpoint.
moles KMnO4 = (0.0430 mol/L)(0.02365 L) = 0.001017 mol
mol Na2C2O4 = (0.001017 mol KMnO4)2 2 4
4
5 mol Na C O
2 mol KMnO
= 0.002542 mol
mass Na2C2O4 = (0.02542 mol)(133.96 g/mol) = 0.3406 g Na2C2O4
weight % Na2C2O4 = 100%0.3406 g
0.645 g
= 52.8 %
IV. Clean-up
A. The oxalate (colorless) solution in the volumetric flask may be flushed down the
sink with running tap water. All solutions containing potassium permanganate
(purple) should be placed in the special containers in the hoods. All glassware
may be rinsed with tap water into the sinks.
B. The disposable pipets should be placed in the "GLASS DISPOSAL" boxes.
C. Clean your equipment and lab bench areas.
D. Have your instructor initial your lab book.
E. LOCK your LOCKER!
F. WASH your HANDS before leaving!
________________________________________________________________________
Report: As mentioned earlier, you will not be submitting a full lab report for this
experiment. Instead, you will be evaluated based on the accuracy and precision of your
results and answers to the follow-up questions. To enable your instructor to evaluate your
results adequately you should submit the following items:
a) Data Analysis & Results
b) One complete set of sample calculations
c) Answers to the questions given below.
d) Answers to the optional Statistical Analysis review sheet may be submitted to
replace your score on exercise #1.
Data Analysis & Results
II. Standardization of the KMnO4 Solution
Approximate molarity of KMnO4 (Determination 1) _________________ mol/L
Molarity of Na2C2O4 (Determination 2) _________________ mol/L
Approximate titrant volume of KMnO4 (Determination 3) ________________L
Table I: Standardization of KMnO4 Solution with _______ moles Na2C2O4
Trial initial mL Final mL mL used [KMnO4]
1
2
3
4
Average [KMnO4] = CX __________ mol/L
Stdev. [KMnO4] __________ mol/L
@ 95% conf. = C __________ mol/L
III. Determination of %Na2C2O4 in an Unknown
Table II: Data for Determination of % Na2C2O4 in Unknown # _______
Trial grams of
Unknown
initial mL
KMnO4(aq)
initial mL
KMnO4(aq)
initial mL
KMnO4(aq)
grams
Na2C2O4
wt %
Na2C2O4
1
2
3
4
Average wt% Na2C2O4 = & M FX X __________
Stdev. wt% Na2C2O4 __________
@ 95% conf. = M __________
68
_______________________________________________________________________
Questions:
Question #1: Based on your experience with the permanganate/oxalate reaction why
was the titration performed by the addition of permanganate solution to oxalate solution and
not the other way around? Why was the experiment performed at an elevated temperature?
Question #2: The determination of the wt% sodium oxalate is based on a two step
procedure: 1) a calibration step, and 2) a measurement step. We have seen this approach
several times throughout this quarter. The accuracy of the result is therefore subject to the
errors associated with both steps. To estimate the overall error, and therefore relay the quality
of a particular experimental result, requires adding the contributions from both steps. To do
this we must distinguish two cases: 1) the result of the calibration step is added to the result
of the measurement step, or 2) the result of the calibration step multiplies the result of the
measurement step. To represent these cases let CX , MX and FX be the average values of the
calibration, measurement and final result, with C, M and F being their respective 95%
confidence limits
Case 1: For addition & subtraction 2 2 = F C M
Case 2: For multiplication & division
2 2
M
= XX X
C MF F
C
The first case was encountered in the Heat of Neutralization lab. Recall the
calorimeter constant was added to the mCsT term to give the overall heat of neutralization.
The analysis of titration data requires the later case.
Use the average values and confidence limits given for Tables I and II of the Data and
Analysis section to determine the true confidence limit of your wt% Na2C2O4 result (F). It
may seem strange, but the Table II value of average wt% Na2C2O4 is both MX and FX .
However, the Table II value of the 95% confidence limit must be identified with M since
only the error of the second set of titrations is considered in determining it. The values to use
for CX , MX and FX are indicated in Table I and II.
69
Statistical Treatment of Experimental Data –take 2
[1] What is the difference between precision and accuracy?
[2] How many significant figures are indicated in each of the following
measurements?
(a) 2.02670 g _____ significant figures
(b) 328.0 mL _____
(c) 7700 ng _____
(d) 0.006 cm _____
(e) 6.7600x10-4
M
_____
[3] The formula weight of sodium oxalate, Na2C2O4, was measured three times as 135.2
g/mol, 132.3 g/mol and 134.2 g/mol.
(a) Calculate the average measured value _______
(b) Calculate the absolute error _______
(c) Calculate the percent error _______
[4] Five groups of students measured the molarity of a CuSO4 solution using
spectrophotometry. Their results were 0.105 M, 0.112 M, 0.0983 M,
0.123 M and 0.114 M.
(a) Calculate the average (mean) ______
(b) What is the median (middle measurement)? ______
(c) What is the range of the measurements? ______
(d) Calculate the standard deviation, n-1 ______
(e) Calculate the relative standard deviation (in %) ______
[5] The concentration of lead in a soil sample revealed 12.7 ppm, 15.4 ppm, 13.6 ppm, 14.9
ppm, 13.5 ppm and 11.9 ppm for six replicate measurements on the same sample.
(a) Calculate the average value ______
(b) Calculate the standard deviation, n-1 ______
(c) Calculate at 95% confidence ______
(d) Give the limits of the 95% confidence interval ______
[6] Performing the acid-base titration lab a student determined the molarity of the
standardized NaOH(aq) solution to be 0.1005 +/- 0.0034 @ 95% confidence. The volume of
this NaOH(aq) solution required to titrate 20.00 mL of an acetic acid sample was 17.86 +/-
0.28 mL @ 95% confidence. Determine the molarity of the acetic acid solution and give the
95% confidence limit of the result.
