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 (X i ) divided by the number of measurements or results (n):
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
CHEMICAL PRINCIPLES II LAB
____________________________________ Exercise #1: Statistical Treatment of Experimental Data
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
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.