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LABORATORY MANUAL CHEMISTRY 121 2013 - Napa Valley

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LABORATORY MANUAL
CHEMISTRY 121
EIGHTH EDITION
2013
Dr. Steven Fawl
Science, Mathematics, and Engineering Division
Napa Valley College
Napa, California
1
Table of Contents
Page
3
Preface
Laboratory Safety Rules
4
Lab Reports
5
Sample Lab Report………………………………………………………………………………. 6
EXPERIMENT ONE ................................................................................................................... 11
Kinetics of the acid hydrolysis of trans-[Co(en)2Cl2]Cl
EXPERIMENT TWO .................................................................................................................. 15
Determination of the tВЅ of a Radioactive Isotope
EXPERIMENT THREE.............................................................................................................. 18
Chemical equilibria and Le Chatelier's principle
EXPERIMENT FOUR................................................................................................................. 23
Hydrogen ion concentration and pH of aqueous solutions
EXPERIMENT FIVE ...................................................................................................................31
Neutralization and hydrolysis
EXPERIMENT SIX ..................................................................................................................... 36
Carbonic acid and its salts
EXPERIMENT SEVEN............................................................................................................... 41
Titration Curve for KHP and Determination of pKa1 and pKa2
EXPERIMENT EIGHT............................................................................................................... 46
Determination of The Equilibrium Constant For FeSCN2+
EXPERIMENT NINE...................................................................................................................50
Determination of the Heat of Reaction
EXPERIMENT TEN
.............................................................................................55
Hess’ Law - Heats of Solution
EXPERIMENT ELEVEN ............................................................................................................60
Electrochemical cells
EXPERIMENT TWELVE .......................................................................................................... 66
Mystery Experiment
2
PREFACE
Chemistry is an experimental science. Thus, it is important that students of chemistry do
experiments in the laboratory to more fully understand applications of the theories they study in
lecture and how to critically evaluate experimental data. The laboratory can also aid the student in
the study of the science by clearly illustrating the principles and concepts involved. Finally,
laboratory experimentation allows students the opportunity to develop techniques and other
manipulative skills that students of science must master.
The faculty of the Napa Valley College clearly understands the importance of laboratory work in
the study of chemistry. The Department is committed to this component of your education and
hopes that you will take full advantage of this opportunity to explore the science of chemistry.
A unique aspect of this laboratory program is that a concerted effort has been made to use
environmentally less toxic or non-toxic materials in these experiments. This was not only done to
protect students but also to lessen the impact of this program upon the environment. This
commitment to the environment has presented an enormous challenge, as many traditional
experiments could not be used due to the negative impact of the chemicals involved. Some
experiments are completely environmentally safe and in these the products can be disposed of by
placing solids in the wastebasket and solutions down the drain. Others contain a very limited
amount of hazardous waste and in these cases the waste must be collected in the proper container
for treatment and disposal. The Department is committed to the further development of
environmentally safe experiments which still clearly illustrate the important principles and
techniques.
The sequence of experiments in this Laboratory Manual is designed to follow the lecture
curriculum. However, instructors will sometimes vary the order of material covered in lecture and
thus certain experiments may come before the concepts illustrated are covered in lecture or after
the material has been covered. Some instructors strongly feel that the lecture should lead the
laboratory while other instructors just as strongly believe that the laboratory experiments should
lead the lecture, and still a third group feel that they should be done concurrently. While there is no
"best" way, it is important that you carefully prepare for each experiment by reading the related
text material before coming to the laboratory. In this way you can maximize the laboratory
experience.
In conclusion, we view this manual as one of continual modification and improvement. Over the
past few years many improvements have come from student comments and criticisms. We
encourage you to discuss ideas for improvements or suggestions for new experiments with your
instructor. Finally, we hope you find this laboratory manual helpful in your study of chemistry.
3
LABORATORY SAFETY RULES
Your participation in this laboratory requires that you follow safe laboratory practices. You are required to
adhere to the safety guidelines listed below, as well as any other safety procedures given by your instructor(s) in
charge of the course. You will be asked to sign this form certifying that you were informed of the safety
guidelines and emergency procedures for this laboratory. Violations of these rules are grounds for expulsion
from the laboratory.
Note: You have the right to ask questions regarding your safety in this laboratory, either directly or
anonymously, without fear of reprisal.
п‚·
Goggles must be worn at all times while in lab. You must purchase a pair of goggle for yourself and
you may store them in your locker. You will be advised of the appropriate goggles to be purchased.
п‚·
Locate the emergency evacuation plan posted by the door. Know your exit routes!
п‚·
Locate emergency shower, eyewash station, fire extinguisher, fire alarm, and fire blanket.
п‚·
Dispose of all broken glassware in the proper receptacle. Never put broken glass in the trashcan.
п‚·
Notify you instructor immediately if you are injured in the laboratory; no matter how slight.
п‚·
Never pipette fluids by mouth. Check odors cautiously (i.e. wafting). Never taste a chemical.
п‚·
Shoes must be worn in the laboratory. These shoes must fully enclose your foot.
п‚·
Long hair must be tied up in a bun during lab work. Loose long sleeves should be avoided in the lab.
п‚·
Children and pets are not allowed in the laboratory.
п‚·
Eating or drinking in the lab is prohibited. Do not drink from the laboratory taps.
п‚·
Wash your hands before and after working in the lab.
п‚·
Turn off the Bunsen burner when you are not using it.
п‚·
If any reagents are spilled, notify your instructor at once.
п‚·
Follow the instructor’s directions for disposal of chemicals.
п‚·
Only perform the assigned experiment. No unauthorized experiments are allowed.
п‚·
Every chemical in a laboratory must be properly labeled. If a label is unclear, notify your instructor.
п‚·
Use the proper instrument (eye-dropper, scoopula, etc.) to remove reagents from bottles. Never return
unused chemicals to the original container. Do not cross contaminate reagents by using the same
instrument for 2 different reagents. (e.g. don’t use the mustard knife in the mayonnaise jar)
п‚·
Material Safety Data Sheets (MSDS) are available for your reference. These contain all known health
hazards of the chemicals used in this course. In addition, there is information concerning protocols for
accidental exposure to the chemical. You are advised to inspect this binder.
4
LAB REPORTS
Every lab has a lab report worksheet for you to fill in and give to your instructor. A more detailed
explanation of what is expected in a lab report will be given to you by your instructor but some
general considerations are given below.
п‚·
Only turn the lab worksheets into your instructor, not the entire lab.
п‚·
Make sure that your name and date is on every page of the lab.
п‚·
Write in pen and never erase or white-out a result. A simple line through bad data is
sufficient.
п‚·
Do not write in pencil and then copy the information into your lab in pen.
п‚·
Always write every digit given to you by a balance. DO NOT ROUND.
п‚·
Pay attention to significant figures. It is usually safe to report an answer to 4 sig.figs.
п‚·
When drawing graphs, always use Guggenheim notation.
п‚·
Never connect the dots or label points when drawing the line on a graph. A best fit line
should be drawn to indicate the trend in the data.
п‚·
Labs should be formatted with your name, date, title, objective, procedure, data,
calculations, results, conclusions, sources of error and any questions found at the end of a
lab must be answered.
п‚·
A more detailed view of what a lab report should look like follows on the next page.
5
NAME (on each page)
DATE (on each page)
TITLE
OBJECTIVES- One or two sentences about the purpose of the experiment. Cute remarks will not
be tolerated.
PROCEDURE- This is to be a CONCISE OUTLINE of the experimental procedure. It is not
necessary to restate your handout. You do not need to write out concentrations and amounts used
of reagents or types of glassware and other equipment used. The exception to this rule is whenever
you do something not written in the lab instructions then (and only then) you must give a complete
description of everything you did.
DATA- Must be written in non-erasable ink. It is especially important that your data be written
neatly and directly into the table. DO NOT write on scratch or any other paper with the intention of
transferring the information your data table later. Any mistakes you make should be corrected by
drawing a single line through the incorrect information and the correct information written above.
Put your data in tabular form (in tables). There is only one exception to this rule and that is when a
single piece of data is taken and it applies to all the rest of the data. It is very important that your
data be written neatly and logically and in a table, an example follows from you first experiment
Beaker Measurements:
Beaker
size/mL
Circumference/cm
(y axis)
Diameter/cm
(x axis)
Your data section should also include any Constants that you plan on using in your calculations
examples include R (the gas law constant) and atomic mass.
CALCULATIONS- This includes one set of every calculation done to go from your raw data to
your final result. The last number you get is usually your result. Every single calculation must be
shown even if all you do is a simple addition. Also be very careful with your units. They should
be written every time they apply. It is not necessary to show every repeated calculation.
Calculations must be neat and easy to follow; units must be used every time they apply. You may
add paper to your report, if you need additional room for your calculations. (Hint: do your
calculations on scratch paper first, and then copy to report.) It is very important that you do not
use a number in your calculations that is not explained in your data section or in an earlier
calculation.
6
GRAPHS- Graphs are frequently needed for a proper interpretation of your data and are often an
integral part of your calculations. There is a standard method for making graphs that will be
outlined here. Always use a separate 8 1/2 x 11 sheet of graph paper for each graph. Always draw
your graph such that the x-axis is along the 11 inch side of the paper (turn the paper side-ways).
Draw in the axes using a pen and mark the axes so that your graph will fill the page. It is not
necessary to start at zero on every axis. Label the axes accordingly giving the units that were used.
When labeling the axes use the following format (Guggenheim Notation),
Time/sec or Temperature/K or Distance/cm
This notation will be explained further in class. Always give a title to your graph., This title should
tell you what the axes are and which system was used. For example for your first lab you could
label your graph,
Circumference vs. Radius for Several Beakers
Always include your name and date on your graph. Perhaps the two most common problems when
dealing with graphs are in the drawing of the best line and measurement of the slope which
represents your data. NEVER CONNECT DOTS. Always draw a smooth line through the data
putting as many dots above as below the line. NEVER FIND A SLOPE USING YOUR DATA
POINTS. Always use points on your line, as far apart as they can be, to measure your line. Do not
indicate the slope and intercept right on your graph. These belong in your calculations section.
You may indicate the points along the line that you used to determine the slope on the graph, but
make sure they are also in the Calculations section. Look at the graph on the next page as an
example of what your graph should look like.
RESULTS- A result differs from a conclusion in that a result usually reports numbers, but it may
also require reporting the presence or absence of chemical compounds based on the outcome of
some qualitative reactions. Do not assume that your data table suffices as a report for your results.
Every lab should have a separate result section indicating any unknown number (if required), and
the final answers obtained from your data tables, calculations, or graphs. Most labs indicate what
you need to have in your results. A result should look like,
п‚·
п‚·
п‚·
The formula of unknown #5 was ZnS.
The tablets contained 85.5 В± 0.3% aspirin
The percentage of oxygen in the air was 20.7%
7
Circumference/cm
8
5
10
15
20
25
30
35
40
45
50
2.5
Name
4.5
6.5
Diameter/cm
8.5
10.5
12.5
Circumference vs. Diameter for Several Beakers
14.5
Date
Sample Graph
CONCLUSIONS – This is an extension of what you have learned. It is a general statement
related to your experiment but giving a large view of what happened.
"Models are used in science to propose workable, testable methods from which information
can be obtained. These models are considered valid until more information reveals errors
whereupon the original model is either modified or completely abandoned for a new model
which describes the system more fully."
Or,
"When two elements combine the product has properties which are different from either of the
reactants. This indicates that a new compound has been formed."
DO NOT repeat your results or data here. There is always a conclusion that can be formed from
your data. A conclusion describes what were you supposed to learn from your experiment? The
answer should not be, I learned that my unknown was CaCl2 (which is a result), rather, you might
have learned that compounds can be identified by determining how each of the ions reacts with
other known compounds.
SOURCES OF ERROR- In all experiments there are probable sources of error. In this portion of
the lab write up you may discuss the possible reasons why the result differs from the expected
result. Note- "Human Error" or "I did not weigh out enough sulfur" are not acceptable. It is
assumed that you did the experiment properly. If you can identify error that you could not control
it belongs in this section.
QUESTIONS- Most experiments will include some questions which must be answered. Not all of
the questions will be graded, but all of them must be done. All work must be shown and yes/no or
multiple choice answers must be explained.
OTHER CONSIDERATIONS- Neatness does not count, but neither will excessive eye-strain be
tolerated. In most cases suggestions will be made on how to improve the quality of the lab report
as the quarter progresses. Do not write on the back of your report, anything written there will not
be graded. When you are finished collecting your data, the I.A. or Instructor will sign it. Data must
not be altered.
9
LABORATORY MANUAL
GENERAL CHEMISTRY 121
EIGHTH EDITION
SPRING 2014
10
EXPERIMENT
KINETICS OF THE ACID HYDROLYSIS OF trans-[Co(en)2Cl2]Cl
1
INTRODUCTION
In aqueous solution the green complex trans-dichlorbis (ethylenediamine) cobalt(III) chloride
dissociates into chloride ion and trans-[Co(en)2Cl2]+. This cobalt complex ion then reacts in acid
solution to yield a mixture of the cis and trans forms of [Co(en)2(H2O)Cl]2+. The progress of this
reaction can be followed visually because the products are red while the reactant is green. The
purpose of this exercise is to determine the rate law for the hydrolysis of trans-[Co(en)2Cl2]+ and to
determine the activation energy for the hydrolysis reaction by carrying out the reaction at several
different temperatures from 45В°C to 85В°C.
The equation for the hydrolysis reaction is,
trans-[Co(en)2Cl2]+ + H2O в†’ cis/trans-[Co(en)2(H2O)Cl]+ + Cl(dark green)
(burgundy to red violet)
The reactant is green while the product is supposed to be burgundy red to violet, but because of
problems we have had in recent years, the product is sometimes yellow or orange. To determine
the rate law and rate constants for this reaction, we shall measure the half-life for each experiment.
When 50% of the reactant has been converted to product, the mixture (50% green and 50% red)
has a characteristic color best described as "gun-metal gray", but other colors are possible and your
instructor will inform you of them..
EXPERIMENT: You may work in pairs. Weigh out two samples (one about 0.12 g and one 0.02
g) of trans-[Co(en)2Cl2]Cl, and dissolve them in separate test tubes each containing 8.0 mL of icecold 1 M sulfuric acid, then place these and all following solutions in an ice bath. With a pipet
transfer exactly 2 mL (save the rest for the next part) from each of these solutions to separate test
tubes, and heat these in boiling water for 5 minutes to completely hydrolyze the complex to trans[Co(en)2(H2O)Cl]+. The solutions should be red, but may be yellow or orange. Cool both the
tubes to room temperature, and add 2 mL of the corresponding green unhydrolyzed complex. You
should now have two pairs of solutions that are gun metal gray, or some other color. The exact
color is not important as long as you have a color representing the half-way point in the reaction.
Keep these tubes in an ice bath so that they do not undergo additional reaction during the
laboratory period. Adjust the temperature of a large beaker of water to 55 +/- 1В°C. One student
should place the tubes which contain the remaining unhydrolyzed green complex into the hot water
while a second student records the time. As the solutions change color a comparison should be
made to our standards prepared previously. When the colors match, the time and temperature
should be recorded. From this you should be able to determine the order of the reaction and the
rate law.
Prepare a solution of 0.15 g of trans-[Co(en)2Cl2]Cl in 30 mL of 1 M H2SO4, transfer 2 mL to a
test tube and boil it for 5 minutes. As before, cool it and add 2 mL of unreacted solution to make 4
mL of half converted standard. Using 4 mL portions measure the half life of the reaction at 85, 75,
65, and 45В°C. Record the times and temperatures in your notebook. Be sure to record t1/2, k, ln k
and 1/T in your data table.
11
Name ______________________
Date _______________
EXPERIMENT ONE
KINETICS OF THE ACID HYDROLYSIS OF trans-[Co(en)2Cl2]Cl
OBJECTIVE:
PROCEDURE:
DATA
Concentration
Time
Rate Law =
Temperature
Time
Rate Const. k
ln k
1/Temp
45В°C = 318 K
55В°C = 328 K
from above
65В°C = 338 K
75В°C = 348 K
85В°C = 358 K
Attach Graph
RESULTS:
Slope =_____________________________________
Ea = __________________________
Intercept = _________________________________
A = ___________________________
12
CALCULATIONS
(You may use this sheet in your lab report)
1) Write down the rate law for the acid hydrolysis of trans- [Co(en)2Cl2]Cl. What is the order of
the reaction?
2) The relationship between the rate constant k and the half-life t1/2 for first and second order
reactions is given by,
k = 0.693/t1/2 (First order)
or
k = 1/([A]t1/2) (Second order)
Using the order obtained in (1) and the proper half-life equation calculate k at each temperature of
your reaction. Show one example here; include all the k's in your data table.
3) Make a plot of ln(k) vs. 1/T, where k is the rate constant at each temperature and T is the
temperature in Kelvin. The relationship between the rate constant and a quantity called the
activation energy is given by the Arrhenius equation,
ln(k) = -Ea/RT + ln(A)
Where R = 8.314 J/mol-K, A is called a pre-exponential term (a constant), Ea is the activation
energy, and k is the rate constant. This equation is of the general form
y = mx + b
Where m is the slope and b is the intercept of a line. Your plot of ln(k) vs. 1/T should yield a slope
of -Ea/R and an intercept of ln(A). From the slope and intercept calculate Ea and A. Ea and A are
your results.
13
PROBLEMS
1) In this experiment you assumed that the t1/2 was very slow at ice bath temperatures of about
0В°C. Use your data to calculate the t1/2 at this temperature. Was the assumption valid?
2) While doing a different experiment a student found that the half-life of his reaction (not this
experiment) decreased as the concentration of his reacting species increased. What if anything was
wrong?
3) It takes 3 minutes to soft-boil an egg at 100В°C. Assuming a soft boiled egg represents the halflife of the denaturation of egg albumin, comment on why most people add salt to the water when
boiling an egg. (What does the salt do to the water?)
14
EXPERIMENT
DETERMINING THE HALF-LIFE OF AN ISOTOPE
2
INTRODUCTION
One type of nuclear reaction is called radioactive decay, in which an unstable isotope of an
element changes spontaneously and emits radiation. The mathematical description of this process
is shown below.
Af = Aie-kt
In this equation, k is the decay constant, commonly measured in sec-1 (or another appropriate unit
of reciprocal time) similar to the rate law constant, k, in kinetics analyses. Ai is the activity (rate of
decay) at t = 0. The SI unit of activity is the bequerel (Bq), defined as one decay per second. This
equation shows that radioactive decay is a first-order kinetic process.
One important measure of the rate at which a radioactive substance decays is called half-life, or
t1/2. Half-life is the amount of time needed for one half of a given quantity of a substance to decay.
Half-lives as short as 10-6 second and as long as 109 years are known.
In this experiment, you will use a source called an isogenerator to produce a sample of radioactive
barium. The isogenerator contains cesium-137, which decays to produce barium-137. The newly
made barium nucleus is initially in a long-lived excited state, which eventually decays by emitting
a gamma photon and becomes stable. By measuring the decay of a sample of barium-137, you will
be able to calculate its half-life.
EXPERIMENT
Each group should have a computer interfaced to a radiation monitor and have the Vernier “Lifetime” program running. You will also be given a shallow aluminum cup that will hold your
radioactive sample. Place the radiation monitor on top of, or adjacent to, the cup to get a
maximum rate of sample detection.
Your instructor will deliver a small sample of radioactive Barium-137 to each group. As soon as
delivery is complete, click the Collect button in the upper right corner of the data window.
Collect data for thirty minutes. Do not move the radiation monitor or the cup during the data
collection. Be careful not to spill the solution in the cup.
When the data collection is complete, you may dispose of the barium solution by pouring it down
the sink.
15
CALCULATIONS
The solution you obtained from the isogenerator may contain a small amount of long-lived cesium
in addition to the barium. To account for the counts due to any cesium, as well as for counts due to
cosmic rays and other background radiation, you can determine the background count rate from
your data. By taking data for 30 minutes, the count rate should have gone down to a nearly
constant value, aside from normal statistical fluctuations. The counts during each interval in the
last five minutes should be nearly the same as for the 20 to 25 minute interval. If so, you can use
the average rate at the end of data collection to correct for the counts not due to barium.
BACKGROUND RADIATION
Select the data on the graph between 25 and 30 minutes by dragging across the region with your
mouse. Click on the statistics button on the toolbar. Read the average counts during the intervals
from the floating box, and record the value in your data table as the average background counts.
Use the corrected count rates to derive an exponential function for the first fifteen minutes of the
data collection.
DETERMINING VALUES FOR k and Ai
Click and drag the computer mouse across the region between 0 and 15 minutes. Make sure that all
of the data points in this region are in the shaded area. Click the curve-fit icon on the tool bar.
Select natural exponent from the equation list. In the "Coefficients" box, type in the background
radiation value as the "B" value. This will account for the background radiation inherent in the
experiment. Click Try Fit. The function will be shown and plotted along with your data. Click to
see a full graph of the function and data. Record the fit parameters Ai, k, and B in your data table.
From the fit parameters, determine the half-life t1/2 for 137Ba and place it in your data table
16
Name__________________
Date____________________
DETERMINING THE HALF-LIFE OF AN ISOTOPE
OBJECTIVE:
PROCEDURE:
DATA TABLE
Experimental Data Fit to Af = Ai exp(- kt) + B
Ai =
k=
B=
t1/2 =
CRC t1/2 for 137 Ba =
PROBLEMS
1) What fraction of the initial activity of your barium sample would remain after 25 minutes?
2) Using your results calculate the counts after 25 minutes. Compare this to the actual value you
measured. Was it a good assumption that the counts in the last five minutes would be due entirely
to non-barium sources?
3) Would any of your t1/2 or k values change if you had been given more or less sample? Explain.
17
EXPERIMENT
CHEMICAL EQUILIBRIA AND LE CHATELIER'S PRINCIPLE
3
INTRODUCTION
All chemical reactions proceed toward an equilibrium position, some more rapidly than others. In
this experiment we shall consider only reactions that occur fairly rapidly, reaching equilibria in a
few minutes or less. Such reactions are said to be rapid and reversible.
The equilibrium position varies for different reactions. For example acetic acid dissociates only a
small extent.
CH3COOH ↔ H+ + CH3COOAt equilibrium in 0.1 M acetic acid, only about 1% of the acetic acid molecules are dissociated into
hydrogen ions and acetate ions. We say that this equilibrium "lies far to the left."
The equilibrium position is the same whether one starts with reactants or products. For example,
when equivalent amounts of H+ and CH3CO2- are mixed, they combine to a large extent to form
CH3COOH molecules, and at equilibrium only about 1% of the H+ and CH3COO- remain
unreacted. The final concentrations of H+, CH3COO-, and CH3COOH are exactly the same as in a
solution of acetic acid of the same concentration.
LE CHATELIER'S PRINCIPLE
The equilibrium position may be shifted to the left or to the right by changing experimental
conditions such as concentration, pressure or temperature. A very useful chemical principle, called
Le Chatelier's Principle, enables one to predict in which direction the equilibrium will shift. The
principle states that;
"If a change is made in any of the factors influencing a system at equilibrium, reaction
will occur in the direction which tends to counteract the change made."
To illustrate this principle, we shall examine the effect of increasing the concentration of
CH3COO- on the equilibrium of the reaction above. If this change causes the equilibrium to shift
to the left, then the concentration of H+ will decrease; if the equilibrium shifts to the right, then the
concentration of H+ will increase.
EQUILIBRIUM SHIFT WHEN A CONCENTRATION IS CHANGED
EXPERIMENT: Place a 5 mL portion of 1M acetic acid in each of two test tubes, a 5 mL portion
of 0.1 M acetic acid in a third test tube, and 5 mL of de-ionized water in a fourth. Add 2 drops of
methyl orange indicator to each test tube and record in your laboratory notebook the color of the
contents of each test tube. To one of the test tubes containing 1 M acetic acid, add 2 mL of 4 M
sodium acetate (CH3COONa) dropwise. Observe and record the changes in color that occur.
18
EQUILIBRIUM BETWEEN A SOLID SALT AND A SOLUTION
As a second example of a rapid reversible reaction we shall investigate the equilibrium between a
solid salt and a saturated solution of the salt. The solubility of solid silver acetate (CH3COOAg) is
0.06 mole per liter at room temperature, and the solubility product is therefore (0.06)2 or 3.6 x10-3.
CH3COOAg (s) ↔ Ag+ + CH3COO- Ksp = 3.6 x 10-3
EXPERIMENT: Prepare some solid silver acetate as follows: Place 5 mL of 0.1 M silver nitrate
(AgNO3) in a centrifuge tube and add 2 mL of 4 M sodium acetate. Stir the mixture thoroughly
using a glass rod. Record the color of the precipitate that is formed. Place the centrifuge tube in
the centrifuge and balance the centrifuge by placing a second centrifuge tube containing 7 mL of
water opposite the first tube. Turn on the centrifuge for about 30 seconds. When the centrifuge has
stopped, remove the tubes then decant the liquid and discard it.
To the solid which remains, add 3 mL of de-ionized water. Place the tube in a beaker of boiling
water and note the amount of solid in the tube. Cool the tube to room temperature by placing it in
a beaker of tap water for 5 minutes. Again note the amount of solid. Centrifuge the mixture as
above and decant the liquid into a clean test tube for use in another experiment.
To the remaining solid add 2 mL 6 M HNO3. Stir with a glass rod until all of the solid dissolves.
THE COMMON ION EFFECT
EXPERIMENT: To the test tube containing the saturated silver acetate solution add 2 mL of 4 M
sodium acetate. Stir the mixture with a glass rod. Observe any reaction that occurs. Write a net
reaction to account for the observed change. In your conclusion state briefly how this experiment
illustrates Le Chatelier's Principle. This application of Le Chatelier's principle is called the
common ion effect.
19
Name ______________________
Date _______________
EXPERIMENT THREE
LeChatelier’s Principle
OBJECTIVE:
PROCEDURE:
DATA:
Equilibrium Shift When Concentration is Changed
Solution
Color w/ Methyl Orange
1.0 M Hac
0.1 M HAc
Pure Water
Volume of NaAc added to
HAc
Color
0.0 mL
0.5 mL
1.0 mL
1.5 mL
2.0 mL
Equilibrium between Solid and Solution & Common Ion Effect
Procedure
Result (amount)
AgNO3 + NaAc Before Boiling
AgNO3 + NaAc After Washing and Boiling
AgAc(s) + 2 mL of HNO3
AgAc solution + 2 mL NaAc
Net Ionic Reaction =
20
PROBLEMS
1) What happens to the H+ concentration in acetic acid when sodium acetate is added? What is the
reaction taking place to account for this? Will the [H+] be higher or lower after the sodium
acetate/acetic acid system has come to equilibrium?
2) What substance disappeared when HNO3 was added to the wet solid formed by adding AgNO3
to sodium acetate? Write the reaction that has occurred (net ionic equation).
3) How would the solubility of AgCl be affected by addition of 6 M HNO3? Why do AgCl and
AgAc behave differently with addition of H+?
4) How would the equilibrium have shifted in the common ion experiment upon addition of Silver
Nitrate? of Sodium Nitrate?
21
5) Excess solid Ca(OH)2 is in equilibrium with its ions,
Ca(OH)2 ↔ Ca2+ + 2 OHState whether the amount of Ca(OH)2 increases, decreases, or stays the same upon addition of;
NaOH
KNO3
HCl
Ca(OH)2
CaCl2
6) Does the solubility of solid Ca(OH)2 in water depend on the amount of solid in contact with the
saturated solution?
7) The solubility of Ca(OH)2 in water is 0.165 g per 100 mL of water. Calculate the solubility
product of Ca(OH)2.
22
EXPERIMENT
HYDROGEN ION CONCENTRATION AND pH OF AQUEOUS SOLUTIONS;
DISSOCIATION CONSTANTS OF ACETIC ACID AND AMMONIA SOLUTIONS
4
INTRODUCTION
Water is a weak electrolyte. It dissociates slightly to give H+ and OH-.
H2O ↔ H+(aq) + OH-(aq)
In pure water [H+] = [OH-] = 1.0 x 10-7 M and thus the amount of dissociation is extremely small.
The dissociation constant for water can be written,
Kw = [H+] [OH-] / [H2O] = 1.0 x 10-14 @ 25В°C
Chemists use activities in the formulation of Kw instead of concentrations. By convention the
activity of water is set equal to one and the activity of dilute solutions is set equal to the
concentrations of the ion. Therefore our dissociation constant can be written as,
Kw = [H+] [OH-] = 1.0 x 10-14
We can use the dissociation constant of water to calculate the concentration of H+ or of OH- in any
aqueous solution if the concentration of the other ion is known.
THE USE OF pH TO MEASURE ACIDITY
It is generally easier to express [H+] in terms of pH. We shall define pH = -log[H+]. Thus the pH of
pure water would be pH = -log[H+] = -log[10-7] = 7.0. Acidic solutions have a pH less than 7.0
and basic solutions have a pH larger than 7.0. THE LOWER THE pH THE HIGHER THE
ACIDITY.
DETERMINATION OF HYDROGEN ION CONCENTRATION AND pH
In this section you will determine the [H+] and pH of two unknown aqueous solutions. In the back
of your Lab Manual you will find a table that will tell you the relationship between color and pH
for several different indicators. While the data in this table are useful as a general guide, it should
be realized that different individuals respond differently to the same color, and they may also use
different words to describe a particular color. For the accurate determination of pH with an
indicator, one must always make a direct comparison of the color of the unknown solution with the
colors of solutions of known pH values. Furthermore, the solutions being compared should be as
similar as possible in factors that may affect the color such as size of test tube, volume of solution,
and number of drops of indicator. It is usually helpful to observe the colors by placing the test
tubes side by side on a piece of white paper, and then looking down through the solutions from the
top.
23
EXPERIMENT: Obtain your unknowns from the instructional assistant. Test 2mL your first
unknown with 2 drops of bromothymol blue to ascertain whether it is acidic or basic. Then,
proceed with the appropriate indicators to determine the approximate pH. Record the name of
each indicator. Use its color with your unknown and the pH range indicated into your data table.
For these tests use 2 mL of your unknown solution and 2 drops of indicator.
When you have decided the approximate pH, you will need to verify it. To do this you must
prepare a solution of this pH and test it with the appropriate indicators. If your solution is in the 1
M to 10-3 M H+ concentration range (0 to 3 pH), prepare it from 6 M HCl. Solutions in the range
from 10-4 to lO-11 M H+ (pH 4 to 11) are available as buffers from the instructional assistant.
Solutions in the 10-11 to 10-14 M range of H+ concentration range (pH 11 to 14 ) are prepared from
6 M NaOH. Use serial dilution to prepare solutions from HCl and NaOH. For these final critical
comparisons use 5 mL of both the standard and the unknown (in different test tubes) and 3-4 drops
of indicator. Compare the color of the standard solution to the color of your unknown with at least
2 and preferably 3 different indicators. If they are the same, then you know the pH and [H+] of
your unknown solution.
AMMONIA SOLUTIONS
Ammonia is a gas at room temperature that is very soluble in water. The resulting solution is basic
because of the reaction,
NH3(g) + H2O ↔ NH4+(aq) + OH-(aq)
Kb = [NH4+] [OH-] / [NH3]
EXPERIMENT : Prepare a 1 M solution of NH3 from the 6 M NH3 found in the lab. Determine
the [OH-] in the 1 M NH3 by using the indicators alizarin yellow and indigo carmine. Record the
colors in your data table. Using these data calculate the dissociation constant Kb for NH3. Include
this in your results.
REACTIONS OF NH4+ AND OHPredict what will happen when solutions of ammonium chloride (NH4Cl), and sodium hydroxide
are mixed.
EXPERIMENT: Prepare 12 mL of 1 M NaOH by adding 2 mL of 6 M NaOH to 10 mL of water.
Mix thoroughly and then place 5 mL of the 1 M NaOH in each of two test tubes. Add several drops
of indigo carmine to each test tube. Record the color in your data table. Slowly add 5 ml of 1 M
NH4Cl to one test tube and record the color changes that occur. Does the OH- concentration
increase or decrease as you add NH4Cl to the NaOH solution?
Write an equation for the net reaction that occurs. Include this reaction and in your data section
and report it in your result section.
24
DISSOCIATION CONSTANT OF ACETIC ACID
In the following experiment we shall determine the dissociation constant of acetic acid by mixing
solutions containing equal numbers of moles of acetic acid and sodium acetate. The resulting
buffer solution will contain approximately equal concentrations of HAc and NaAc.
[HAc] ≈ [Ac-]
It then follows from the law of chemical equilibrium that the concentration of H+ in the solution is
equal to the dissociation constant,
HAc ↔ H+ + Ac-
Ka = [ H+ ] [ Ac- ] = [ H+ ]
[HAc]
Furthermore, if we define pKa = -logKa, then in this solution -logKa = -log[H+], or pKa = pH.
EXPERIMENT: Add 10 mL of 1 M HAc to 10 mL of 1 M NaAc. Determine the pH and [H+] of
the resulting solution using the appropriate indicators. Record the name of each indicator and its
color in this solution in your notebook. Complete the table in your worksheet for this solution
(which does not contain 0.5 M HAc and 0.5 M NaAc).
Using these data, calculate the dissociation constant Ka and pKa for acetic acid. Include these
numbers in your results.
25
Name _______________
Date ________________
HYDROGEN ION CONCENTRATION AND pH OF AQUEOUS SOLUTIONS;
DISSOCIATION CONSTANTS OF ACETIC ACID AND AMMONIA SOLUTIONS
OBJECTIVE:
PROCEDURE:
DATA: pH Unknowns
Indicator
Color of Unk#
Malachite Green
Methyl Violet
Methyl Orange
Bromocresol Green
Methyl Red
Bromothymol Blue
Cresol Red
Thymol Blue
Phenolphthalein
Alizarin Yellow R
Indigo Carmine
pH of Unknown # _____ = ______
pH of Unknown # _____ = ______
26
pH
Color of Unk#
pH
For 1 M NH3: Complete the following table
Indicator
Color
pH
Alizarin Yellow R
Indigo Carmine
pH of 1 M NH3 = _____________
Substance
H+
OH-
NH3
Equilibrium
Concentration
Kb for NH3 = ________________
For 1 M NaOH: Complete the following table
Indigo Carmine Color
Before adding NH4Cl
After adding NH4Cl
The solution becomes (more less) acidic as NH4Cl is added to the NaOH (Circle one).
Net reaction between NaOH and NH4Cl:
For 1 M HAc and 1 M NaAc: Complete the following table
Indicator Used
Color
pH
pH of 10 mL 1 M HAc mixed with 10 mL 1 M NaAc = __________
27
NH4+
Substance
H+
OH-
HAc
Equilibrium
Concentration
Ka for HAc = ___________
pKa for HAc = __________
RESULTS:
The pH of Unknown #_______ is _________
The pH of Unknown #_______ is _________
The Kb for NH3 = ___________
The net reaction between NaOH and NH4Cl is,
The Ka for HAc = ___________ and the pKa for HAc is ____________
28
Ac-
Na+
PROBLEMS
1. A solution is prepared by dissolving 0.050 mole of HCl and 0.030 mole of NaOH in sufficient
water to give a final volume of 2.0 liters. Calculate the concentration of all the ionic (Na+, etc.)
species in the final solution.
2. Instead of determining the dissociation constant of ammonia by our experimental procedure we
could mix equal volumes of 1 M NH3, and 1 M NH4Cl. Show that for such a solution Kb = [OH-],
and pKb = 14 - pH.
3. Calculate the H+, and OH- of a 0.10 M NH3 solution. (Use Kb = 1.8 x 10
following problems.
29
-5
in this and the
4. A solution is prepared by mixing 0.050 mole of NH4Cl and 0.010 mole of NaOH in sufficient
water to give a final volume of 200 mL. Calculate the concentration of all the molecular and ionic
species in the resulting solution.
5. A solution is prepared by mixing 0.050 mole of NH3 and 0.20 mole of HCl in sufficient water
to give a final volume of 500 mL. Calculate the concentration of all the molecular and ionic
species in the resulting solution.
30
5
EXPERIMENT
NEUTRALIZATION AND HYDROLYSIS
INTRODUCTION
The neutralization of an acid by a base to form a salt and water proceeds to an equilibrium position
which varies depending on the strengths of the acid and base that are used. In this assignment we
shall investigate the equilibrium positions reached for the reaction of;
H+ + OH- ↔ H2O
A. a strong acid and a strong base,
B. a weak acid (HX) and a strong base, HX + OH- ↔ X- + H2O
H+ + BOH ↔ B+ + H2O
C. a strong acid and a weak base,
or, in the special case of ammonia,
H+ + NH3 ↔ NH4+
In each of the above cases the same equilibrium position may be reached by mixing equivalent
amounts of the reactants (as in a titration) or by preparing an aqueous solution of the product (the
salt of the acid and base) of equivalent concentration. Since it is much easier experimentally to
prepare the salt solutions, we shall prepare them and measure their pH and [H+], from which we
can infer the equilibrium position. We shall therefore study the above reactions in the reverse
direction. The reverse of the neutralization reaction is called the "hydrolysis" reaction, because in
the typical case a water molecule is split by the reaction.
HYDROLYSIS OF THE SALT OF A STRONG ACID AND STRONG BASE
EXPERIMENT: Test about 5mL of a 50 mL portion of distilled water with bromothymol blue
indicator. It should give a green color. If it is yellow, boil the water until it gives a green color
with bromothymol blue. If your solution is blue inform the instructional assistant. Prepare an
approximately 1 M NaCl solution by dissolving about 1.2 grams of sodium chloride (compare with
the sample in the laboratory) in 20 mL of your distilled water. Determine the pH and [H+] of your
NaCl solution by using appropriate indicators, including bromothymol blue. Use the same
procedure you used in the preceding assignment for determining an unknown pH. Record the
names and colors of these indicators in your laboratory notebook.
Because sodium ion is the ion of a strong base, there is no tendency for Na+ to combine with OH-,
nor for H+ to combine with Cl-, and therefore the following reactions DO NOT occur.
Na+ + H2O ↔ NaOH + H+
Cl- + H2O ↔ HCl + OHMore precisely, the equilibrium positions of these two reactions are so far to the left that [NaOH] =
0 and [HCl] = 0.
31
HYDROLYSIS OF THE SALT OF A WEAK ACID AND A STRONG BASE.
The neutralization reaction for acetic acid, CH3COOH, by a strong base is
CH3COOH + OH- ↔ CH3COO- + H2O
The hydrolysis of acetate ion, CH3COO-, is represented by the reverse of the above reaction:
CH3COO- + H2O ↔ CH3COOH + OHEXPERIMENT: Determine the pH and [H+] of 1 M CH3COONa by using appropriate indicators,
including thymol blue and phenolphthalein. Record the names and color of these indicators in
your laboratory notebook. From your measured value of the pH and Kw, calculate [H+] and [OH-]
in 1 M CH3COONa. Using this value for [OH-], calculate [CH3COO-].
HYDROLYSIS OF THE SALT OF A STRONG ACID AND A WEAK BASE
The neutralization reaction for ammonia, NH3, by a strong acid is
NH3 + H+ ↔ NH4+
The reverse of the above reaction represents the “hydrolysis” reaction of ammonium ion, NH4+;
NH4+ ↔ NH3 + H+
EXPERIMENT: Determine the pH and [H+] of 1 M ammonium chloride by using appropriate
indicators, including bromocresol green and bromothymol blue. Record the names and colors of
these indicators in your laboratory notebook. From your measured value of the pH and Kw,
calculate [H+] and [OH-] in 1 M NH4Cl. Using this value for [H+], calculate [NH3] and [NH4+].
HYDROLYSIS OF SODIUM CARBONATE SOLUTIONS
The neutralization reaction for bicarbonate ion, HCO3- , by a strong base is
HCO3- + OH- ↔ CO32- + H2O
The reverse of the above reaction represents the hydrolysis of carbonate ion, CO32CO32- + H2O ↔ HCO3- + OHEXPERIMENT: Determine the pH and [H+] of 1 M sodium carbonate by using appropriate
indicators, including alizarin yellow and indigo carmine. Record the names and colors of these
indicators in your laboratory notebook. From your measured value of the pH and Kw, calculate
[H+] and [OH-] in 1 M Na2CO3. Using this value for [OH-], calculate [HCO3-] and [CO32-].
32
Name ___________________
Date_________________
NEUTRALIZATION AND HYDROLYSIS WORKSHEET
OBJECTIVE:
PROCEDURE:
Data: Strong Acid and a Strong Base: Complete the following table for 1M NaCl
pH =
Substance
H+
OH-
Na+
Cl-
HCl
NaOH
Equilibrium
Concentration
Weak Acid and Strong Base: Complete the following table for 1 M CH3COONa.
pH =
Substance
H+
OH-
Ac-
HAc
Na+
NaOH
NH3
HCl
Equilibrium
Concentration
Weak Base and Strong Acid: Complete the following table for 1 M NH4Cl.
pH =
Substance
H+
OH-
NH4+
Equilibrium
Concentration
33
Cl-
Weak Diprotic Acid and Strong Base: Complete the following table for 1 M Na2CO3.
pH =
Substance
H+
OH-
Na+
CO32-
HCO3-
Equilibrium
Concentration
PROBLEMS
1) a) Is the [H+] of the NaCl solution greater than, less than or equal to that of your boiled distilled
water? b) Which indicator would you choose for the most exact endpoint for the titration of an HCl
solution with NaOH?
2a) Would you expect a solution of sodium acetate to be acidic, basic or neutral? b) What is the
pH of sodium acetate? c) Which indicator would you choose for the titration of CH3COOH with
NaOH?
3a) Would you expect a solution of ammonium chloride to be acidic, basic or neutral? b) What is
the pH of 1 M NH4Cl? c) Which indicator would you choose for the titration of NH3 with HCl?
4) Would you expect a solution of sodium carbonate to be acidic, basic or neutral?
5) The Ka for acetic acid is 1.8 x 10-5, and the Kb for ammonia is also 1.8 x 10-5. Calculate the
equilibrium constant for the
(a) neutralization of a strong acid by a strong base,
(b) neutralization of acetic acid by sodium hydroxide,
(c) neutralization of ammonia by hydrochloric acid.
34
6) Ammonium acetate, NH4CH3COO, is a strong electrolyte. When a 1 M solution of ammonium
acetate is tested with bromothymol blue, the solution is green. Including all ions and molecules,
what substances are present in a 1 M NH4CH3COO solution? Write an equation for the net
reaction for the hydrolysis of NH4CH3COO. Calculate the equilibrium constant for this reaction,
using the information given in problem #5.
35
6
EXPERIMENT
CARBONIC ACID AND ITS SALTS
INTRODUCTION
Carbon dioxide is a colorless gas at room temperature that is somewhat soluble in water. Water in
equilibrium with gaseous CO2 at 1 atm pressure contains 0.034 M H2CO3.
CO2(1 atm) + H2O ↔ H2CO3 K = 0.034
Solid carbon dioxide is a convenient source of carbon dioxide (dry ice), but in this lab you will use
a carbon dioxide generator to make saturated solutions of H2CO3. Carbonic acid is a typical
diprotic acid in that it dissociates stepwise as shown below,
H2CO3 ↔ H+ + HCO3-
Ka1 = 4.3 X 10-7
HCO3- ↔ H+ + CO32-
Ka2 = 4.7 X 10-11
Thus we see that there are two possible salts, NaHCO3 (sodium bicarbonate) and Na2CO3 (sodium
carbonate). In the following experiments we will investigate the properties of H2CO3, NaHCO3,
and Na2CO3.
CARBONIC ACID SOLUTIONS
EXPERIMENT: Prepare about 100 mL of a saturated solution of carbon dioxide by adding a few
mL of concentrated H2SO4 to some solid sodium bicarbonate inside a CO2 generator found in the
lab. Allow the resulting gas to bubble through 100 mL of distilled water for about 5 minutes. Test
the resulting solution with bromocresol green and methyl orange indicators. Record the pH and
label the solution 0.034 M H2CO3. Note- the pH of the solution is due almost entirely to the first
acid dissociation.
EXPERIMENT: Bubble some CO2 through a 10 mL sample of 1M CaCl2. Does a precipitate
form? If we assume that the solution is saturated with CO2, what is the concentration of H2CO3 in
the solution? Calcium carbonate is very slightly soluble in water.
CaCO3 ↔ Ca2+ + CO32- Ksp = 4.7 x 10-9
Calculate the maximum concentration of CO32- that could exist in a 1 M CaCl2 solution. Is the
concentration of CO32- in your saturated H2CO3 solution larger or smaller than this value? Using
these data complete the table in your lab report. Use the value of Ka2 to calculate [CO32-].
36
SODIUM CARBONATE SOLUTIONS
Sodium carbonate, Na2CO3 is a solid that is very soluble in water and it is a strong electrolyte
which dissociates to give Na+ and CO32- in aqueous solutions. Sodium carbonate solutions can
hydrolyze stepwise in the following manner,
CO32- + H2O ↔ HCO3- + OHHCO3- + H2O ↔ H2CO3 + OHEXPERIMENT: Place 5 mL of 1 M Na2CO3 into a test tube and add 5 mL of 1 M CaCl2.
Centrifuge the mixture and note the amount of solid CaCO3 produced. Record your results. Write
an equation for the net reaction.
SODIUM BICARBONATE SOLUTIONS
Sodium bicarbonate, NaHCO3, is a solid that is very soluble in water and it is a strong electrolyte
which dissociates to give Na+ and HCO3- in aqueous solutions. Sodium bicarbonate solutions may
also be prepared by the reaction of solutions containing 1 mole of H2CO3 and 1 mole NaOH. The
equation for the net reaction is,
H2CO3 + OH- ↔ HCO3- + H2O
Since HCO3- is a weak acid, it should dissociate to give H+ and CO32-. However the H+ produced
would react with another HCO3- to form the weak acid H2CO3. As a result we predict that the
principal equilibrium in a NaHCO3 solution would be,
2 HCO3- ↔ H2CO3 + CO32- *
The position of this equilibrium is far to the left, the concentrations of H2CO3 and CO32- being
much smaller than the concentration of HCO3-.
EXPERIMENT: Place 5 mL of 1 M NaHCO3 and 5 mL of 1 M CaCl2 in a centrifuge tube.
Centrifuge the mixture and note the amount of precipitate obtained. Decant the solution into a
second centrifuge tube. Now, place the tube containing the decanted solution in a beaker of
boiling water. Keep it in the boiling water for 10 minutes, and then cool it by placing it in a beaker
of water at room temperature. After it has cooled to room temperature centrifuge it. Note the
amount of precipitate obtained. Are the results in accord with what you would predict from a shift
in equilibrium above?* Compare the amounts of CaCO3 obtained before and after boiling the
mixture with that obtained when you used the Na2CO3 solution. Record your observations.
EXPERIMENT: Determine the pH and H+ concentration of 1 M NaHCO3 by testing 5 mL
portions with the following indicators: bromothymol blue, cresol red, and phenolphthalein. Record
your observations. Calculate [OH-]. What would happen to the [OH-] if the 1 M NaHCO3 solution
is boiled for a few minutes and then cooled?
EXPERIMENT: Place 5 mL of 1 M NaHCO3 in a test tube, add 2 drops of phenolphthalein, and
place the test tube in a beaker of boiling water for 5 minutes. Record the color changes. Does [OH] increase or decrease?
37
RESULTS: In the result section of your lab report please list all the ions and molecules (except for
water) for the following solutions a) the carbonic acid solution (made by bubbling CO2 through
water); b) the 1 M Na2CO3 and c) the 1 M NaHCO3 (without the CaCl2 and before heating). Now
for each of the 3 solutions list the ions in order of decreasing concentration. It will be much
simpler if you group them in pairs.
38
Name ___________________
Date_________________
CARBONIC ACID AND ITS SALTS
OBJECTIVE:
PROCEDURE:
Data: Carbonic Acid Solutions
Bromocresol Green Color __________ Methyl Orange Color ____________ pH = _______
Substance
H2CO3
H+
HCO3-
OH-
CO32-
Equilibrium
Concentration
Sodium Carbonate Solutions
Amount of CaCO3 produced (mm) = _____________________
Net ionic reaction =
Sodium Bicarbonate Solutions
Amount of CaCO3 initially __________ Amount of CaCO3 after heating________________
pH of NaHCO3
Bromothymol Blue Color ___________ Cresol Red Color ______________
Phenolphthalein Color _____________
pH = ________
[OH-] = __________________
Phenolphthalein Color After Boiling _____________________ [OH-] increase decrease
Ion and Molecules Present: From highest to lowest concentration:
0.034 M H2CO3
1 M Na2CO3
1 M NaHCO3
_________
_________
_________
_________
_________
_________
_________
_________
_________
_________
_________
_________
_________
_________
_________
_________
_________
_________
39
PROBLEMS
1) In some of the following cases the two substances cannot coexist in the same solution because
they will react together. For example, an acid will react with a base, and some combination of ions
will form precipitates. For these cases write a net equation for the reaction that would occur. In
other cases write "No Reaction".
(a) OH- and HCO3-
(e) Ca2+ and HCO3-
(b) OH- and CO32-
(f) H2CO3 and CO32-
(c) OH- and H2CO3
(g) H+ and HCO3-
(d) H+ and H2CO3
(h) H+ and CO32-
2a) Using the equilibrium constants given in the lab calculate the value of the equilibrium constant
for 2 HCO3- ↔ H2CO3 + CO32-.
2b) Calculate the concentration of H2CO3 and CO32- in a 1 M solution of NaHCO3.
3a) Calculate the value of the equilibrium constant for H2CO3 ↔ 2 H+ + CO32-
3b) What is your experimental value for the pH of a solution of 1 M NaHCO3?
3c) Using your results from questions 2 and 3, show that for HCO3- that pH = (pK1 + pK2)/2
40
EXPERIMENT
TITRATION CURVE FOR POTASSIUM ACID PHTHALATE
7
INTRODUCTION
In this experiment a pH meter equipped with a combination electrode (a glass electrode and a
calomel reference electrode in the same housing) and drop counter will be used to construct the
titration curve for KHP titrated with NaOH. From the titration curve and the concentration of the
base you can determine the concentration of the unknown KHP solution and both the first and the
second dissociation constants of phthalic acid.
When molar concentration units are used, the concentration quotient
(1)
Kc2 = [H+][P2-]
[HP-]
is only approximately constant over the whole accessible range of concentrations. When activities
are used, however, the ionization constant defined by
(2)
Ka2 = (aH+)(aP2-)
(aHP-)
is truly constant at all concentrations of the ions.
The activity may be regarded, for our purposes, as an effective concentration. In ionic solutions
the nearest neighbor shell around a cation will contain a (slight) predominance of anions, and vice
versa. This partial screening of the ions by their counter ions makes the effective concentration of
an ionic species somewhat smaller than its stoichiometric concentration. The activity and
concentration of an ion i are related by
(3) ai = Cifi
where fi is an empirical parameter, called the activity coefficient, which depends on the size of the
ion and the total ionic strength of the solution.
Electrochemical experiments, such as the measurement of pH, yield ion activities rather than ion
concentrations. At the ionic-.strengths to be used in this experiment the activity coefficients of the
phthalate anionic species differ appreciably from unity and a serious misestimate of Ka results if
they are ignored,
The appropriate equations are
(4) Ka2 = (aH+)(aP2-) = (aH+)[P2-](fP2-)
(aHP-)
[HP-](fHP-)
41
Taking the log of both sides,
(5) log Ka2 = log (aH+) + log [P2-]/[HP-] + log (fP2-)/(fHP-)
Now since
(6) pH = - log(aH+) and pKa2 = -log Ka2, it follows that,
(7) pH = pKa2 + log [P2-]/[HP-] + log (fP2-)/(fHP-)
Thus when exactly half the KHP has been neutralized (when [P2-] = [HP-]) the measured pH and
pKa are roughly the same, except for the activity coefficient term.
At the equivalence point on the titration curve of a weak acid with strong base, the slope of the
curve is a maximum. We can use this fact to find the equivalence point very precisely. Using the
utilities available in Vernier LabPro, the program will calculate the first and second derivative of
our data. The first derivative produces a plot with a peak in the center. The top of this peak is
your final endpoint. If using the second derivative, the final endpoint is where the plot crosses the
X axis.
PROCEDURE:
A. Supply your instructor with a clean, dry, 125 mL Erlenmeyer flask labeled with your name
and locker number. You will receive 55 mL of your unknown KHP solution in this flask.
B. Set up a titration assembly, consisting of a pH meter, a magnetic stirrer, and a buret filled with
your standard 0.1 M NaOH..
C. Just before you are ready to prepare a titration curve standardize the pH meter according to the
directions given in a separate handout.
D. Using a pipet deliver 25 mL of the unknown KHP solution into a clean, dry 100 mL beaker
which already contains a clean magnetic stirring bar. Add 20 mL of distilled water and 2
drops of phenolphthalein indicator. Put the beaker onto the magnetic stirrer so that the bar
rotates freely, but there is still room alongside it for the electrode to be inserted near the
bottom of the beaker. Mix the solution thoroughly with the stirrer.
E.
Wash the standardized electrode with your wash bottle, blot it dry, and position it through the
large hole of the drop counter. Read the pH meter and the buret, and record both readings.
F.
Turn on the magnetic stirrer so it is rotating slowly. You are now ready to begin collecting
data. Click the Collect button. No data will be collected until the first drop goes through the
drop counter slot. Adjust the drop rate to about 1 drop every 2 seconds. When the first drop
passes through, check to see that the data has begun recording.
G. Continue watching your graph to see when the large increase in pH takes place - this will be
the equivalence point. When this jump in pH occurs, add about 3 more milliliters of NaOH,
click STOP, and stop the titration. Remove the electrode, wash it thoroughly, and put it back
into its storage bottle. Remove the stir bar from the beaker and pour the solution down the
sink. Clean and dry the stir bar.
42
H. Find the equivalence point. The best method for determining the equivalence point is to take
the second derivative of the pH vs. volume data. Do the following,
a. Open Page 3 by clicking on the Page window of the menu bar.
b. Analyze the second derivative plot and record the volume of NaOH at the equivalence
point.
I.
From this curve determine the pH at half and three-fourths neutralization of the KHP. Go to
the worksheet and use it to compute an average value for pKa2, taking into account the
activity coefficients. α’s (Na+ = 0.4, P2- = 0.6, HP- = 0.6, K+ = 0.3). The Debye-Hückel
equation is below.
0.51Z 2 пЃ­
пЂ­ log f a пЂЅ
1 пЂ« 3.329пЃЎ пЃ­
J.
Having Ka2 in hand you can obtain Ka1 from the initial pH of the diluted unknown. Starting
with the known equilibrium expressions,
(8)
K a1 пЂЅ
[H пЂ« ][HP пЂ­ ] f H пЂ« f HP пЂ­
п‚ґ
[H 2 P]
f H2P
K a2 пЂЅ
[H пЂ« ][P 2пЂ­ ] f H пЂ« f P 2пЂ­
п‚ґ
[HP - ]
f HP -
We can solve each of these equations for [H+] and multiply them together we get;
(9)
[H пЂ« ]2 пЂЅ K a1K a2
пЃ›H 2 PпЃќпЃ›HP пЂ­ пЃќ f H P f HP
пЃ›HP пЃќпЃ›P пЃќ f
пЂ­
2
2пЂ­
2
HпЂ«
пЂ­
f HP пЂ­ f P 2пЂ­
Cancelling terms that appear on the top and bottom of this equation we obtain,
(10)
[H пЂ« ]2 пЂЅ K a1K a2
пЃ›H 2 PпЃќ f H P
пЃ›P пЃќ f
2
2пЂ­
2
HпЂ«
f P2пЂ­
For the hydrolysis of pure KHP, the [H2P] = [P2-] so they cancel. In the Debye-HГјckel equation
the charge for H2P = 0 and fH2P = 1, so canceling the [H2P] and [P2-], setting fH2P = 1, and
remembering that the activity of H+ = aH+ = [H+] fH+we get;
(11)
[H пЂ« ]2 пЂЅ
K a1K a2
2
f H пЂ« f P 2пЂ­
and [aH пЂ« ]2 пЂЅ
K a1K a2
f P 2пЂ­
Since we have earlier defined pH as -log(aH+), Eq. 11 reduces to the simple form
(12)
2pH = pKa1 + pKa2 + log fP2-
The ionic strength of your diluted starting solution is easily calculated from the concentration of
your unknown. The activity coefficient of P2- at this ionic strength can be calculated from the
Debye-HГјckel equation. With this value, the measured pH, and average pKa2 calculated earlier,
pKa1 is in hand.
43
REPORT: Fill out the attached report sheet and turn it in. You should include your titration
curve, the concentration of the undiluted KHP unknown, and the calculation of pKa1 and pKa2
using equations 7 and 12.
44
45
______________
Vol. KHP used
For Initial Solution
Calculation of pKa1
3/4 Neutralization
1/2 Neutralization
Calculation of PKa2
Вѕ Neut
ВЅ Neut
Initial
Vol.
NaOH
Ој
Ој
Total
Vol.
pH
pH
Mol.
Na+
fHP-
fHP-
Conc.
Na+
fP2-
____
fP2-
Mol.
K+
pKa1
____
pKa2
Conc.
K+
Avg.pKa2
Mol.
HP-
Conc.
HP-
Mol.
P2-
Conc.
P2-
Ionic
Strength
Date _________________________
EXPERIMENT SEVEN : Titration Curve of KHP
not needed
Conc. of Unknown. ______________
______________
Vol. NaOH used
Conc. NaOH used
Name
EXPERIMENT
DETERMINATION OF THE EQUILIBRIUM CONSTANT FOR FeSCN2+
8
INTRODUCTION
When 2 reactants are mixed, the reaction typically does not go to completion. Rather, they will
react to form products until a state is reached whereby the concentrations of the reactants and
products remain constant. This is a dynamic state in which the rate of formation of the products is
equal to the rate of formation of the reactants. The reactants and products are in chemical
equilibrium and will remain so until affected by some external force. The equilibrium constant Kc
for the reaction relates the concentration of the reactants and products.
In this experiment we will study the equilibrium properties of the reaction between iron (III) ion
and thiocyanate ion:
Fe3+ (aq) + SCN– (aq) → FeSCN2+ (aq)
When solutions containing Fe3+ ion and thiocyanate ion are mixed, the deep red thiocyanatoiron
(III) ion (FeSCN2+) is formed. As a result of the reaction, the starting concentrations of Fe3+ and
SCN- will decrease: so for every mole of FeSCN2+ that is formed, one mole of Fe3+ and one mole
of SCN- will react. The equilibrium constant expression Kc, according to the Law of Chemical
Equilibrium, for this reaction is formulated as follows:
[FeSCN2+] / [Fe3+][ SCN- ] = Kc
Square brackets ([]) are used to indicate concentration in mols/liter, i.e., molarity (M). The value
of Kc is constant at a given temperature. This means that mixtures containing Fe3+ and SCN– will
react until the above equation is satisfied, so that the same value of the Kc will be obtained no
matter what initial amounts of Fe3+ and SCN– were used. Our purpose in this experiment will be to
find Kc for this reaction for several mixtures made up in different ways, and to show that Kc indeed
has the same value in each of the mixtures.
The reaction is a particularly good one to study because Kc is of a convenient magnitude and the
red color of the FeSCN2+ ion makes for an easy analysis of the equilibrium mixture using a
spectrophotometer. The amount of light absorbed by the red complex is measured at 447 nm, the
wavelength at which the complex most strongly absorbs. The absorbance, A, of the complex is
proportional to its concentration, M, and can be measured directly on the spectrophotometer using
the Beer-Lambert law:
A = пЃҐcпЃ¬
Where пЃҐ = extinction coefficient, c = concentration (Molarity), and пЃ¬ = path length (1cm)
46
EQUIPMENT
5 test tubes
10 ml graduated cylinder
10 ml graduated measuring pipette (0.1 ml graduations)
stirring rod
small labels or markers
5 cuvettes for the spectrophotometers
250 ml bottle acetone for rinsing
HAZARD: As always wear Safety glasses while performing this experiment
CONTAIMINATION NOTES: If your flask is wet before you prepare your standard/sample
solutions ensure that the flask is wet with diluant (in this case it is the 0.0200M Fe(NO3)3 in 1.0 M
HNO3 ).
EXPERIMENT: Beer-Lambert Data
A solution of 0.0200M Fe(NO3)3 in 0.5 M HNO3 has been prepared for you.
Dilute 5.0, 10.0 and 15.0 ml portions of 2.00 x10-4 M KSCN to 100 ml with the 0.0200M
Fe(NO3)3 in 0.5 M HNO3. This will give you 3 solutions that can be assumed to be 1.0x10-5,
2.0x10-5 and 3.0x10-5 Min FeSCN2+.
Measure the absorbance of these solutions at 447nm, using a solution of 0.0200M Fe(NO3)3 in 0.5
M HNO3 as the reference solution. Measure the absorbance of these solutions at this wavelength.
Plot Absorbance vs [FeSCN2+] using Excel. Find the slope and intercept for this plot.
EXPERIMENT: Determination of Kc
The mixtures will be prepared by mixing solutions containing known concentrations of iron (III)
nitrate, Fe(NO3)3, and potassium thiocyanate, KSCN. The color of the FeSCN2+ ion formed will
allow us to determine its equilibrium concentration. Knowing the initial composition of a mixture
and the equilibrium concentration of FeSCN2+, we can calculate the equilibrium concentrations of
the rest of the pertinent species and then determine Kc.
Label five regular test tubes 1 to 5, with labels or by noting their positions in your beakers. Pour
about 30 mL 0.02 M Fe(NO3)3 in 0.5 M HNO3 into a dry 100 mL beaker. Pipet 5.00 mL of this
solution into each test tube. Then add about 20 mL 2.00 x 10-4 M KSCN to another dry 100-mL
beaker. Pipet 1,2,3,4, and 5 mL from the KSCN beaker into each of the corresponding test tubes
labeled 1 to 5, then pipet the proper number of milliliters of 0.5 M HNO3 into each test tube to
bring the total volume in each tube to 10.00 mL.
47
The volumes of reagents to be added to each tube are summarized in the table below.
Reagents/mLs)
Fe(NO3)3
KSCN
HNO3
1
5.00
1.00
4.00
Test Tube #
2
3
4
5.00
5.00
5.00
2.00
3.00
4.00
3.00
2.00
1.00
5
5.00
5.00
0.00
Mix each solution thoroughly with a glass stirring rod. Be sure to dry the stirring rod after mixing
each solution to prevent cross-contamination.
Measure the absorbance of each mixture at 447 nm as demonstrated by your instructor and put the
data in the worksheet provided.
Determine the concentration of FeSCN2+ from your calibration curve. Record the value on your
Report form. Repeat the measurement using the mixtures in each of the other test tubes.
48
Name _______________________
Date____________________
Determination of the Equilibrium Constant for FeSCN2+
OBJECTIVE:
PROCEDURE:
Beer-Lambert Data
Data For Beer-Lambert Plot
5 mL
10 mL
Data
15 mL
Absorbance
Data for Equilibrium Calculation
Data
Test Tube #
1
2
3
Absorbance
Conc of FeSCN2+
Conc of Fe3+
Conc of SCN-
Value of Keq
RESULTS:
Average Keq =
Staple Beer-Lambert plot to the back of this sheet.
49
4
5
EXPERIMENT
DETERMINATION OF THE HEAT OF REACTION
9
INTRODUCTION
The purpose of the experiment is to determine the quantity of heat liberated in the reaction,
Mg(s) + 2 HCl (aq) в†’ H2 (g) + MgCl2 (aq)
The heat liberated in this reaction can be trapped and utilized to melt ice. Ice is less dense than
water and as a consequence there is a volume change when ice melts. This volume change can be
measured and used to calculate how much energy was transferred to the ice. This in turn is used to
measure the amount of energy released in the reaction (the enthalpy).
EXPERIMENT: Students may work in pairs with one student calling out the time and the other
student reading the volume and recording both the time and volume. The calculations are to be
made individually. Fill a Styrofoam container with crushed ice from the ice provided in the lab
and place in it a small flask containing 3 mL of 6 M HCl and 6 mL of water. While this solution is
cooling, assemble the bottle and stopper as shown in the lab discussion. Fill the bottle with tap
water and push the stopper in tightly. Water will spurt out the top of the pipet. Watch the water
level in the pipet over a 5 minute period. If it does not drop, the apparatus is free of leaks and you
may continue with the experiment. If the level drops, try to repair the leak with the aid of your
instructor or instructional assistant. (Do not use grease on the stopper.) When your assembly is
leak-proof, fill the bottle with crushed ice to the brim. Add ice water from the Styrofoam container
to fill the bottle completely and insert the stopper. Pour the 9 mL of HCl solution that you cooled
to 0В°C in the reaction test tube. Immediately place the bottle in the Styrofoam container and
surround the bottle with crushed ice and water to just below the rim of the test tube. Insert the cork
in the test tube loosely. Allow the apparatus to stand for 15 minutes during which time the
temperature should become constant at 0В°C.
During this time interval weigh a 0.1 g sample of magnesium ribbon to the nearest milligram.
Record the mass in your notebook.
After the apparatus has been allowed to stand for 15 minutes, adjust the water level in the pipet to
a reading of 0.8 mL or greater by placing a small piece of latex tubing on the end of the pipet and
filling it with water. By pushing on the stopper it is possible to raise the level of the water inside
of the pipet until it meets the water in the tubing. Releasing the stopper will pull the water from
the tubing into the pipet thus filling it. This technique may require practice. Begin reading the
pipet volume once the water level has reached some convenient mark on the pipet. Read the pipet
volume each minute for 5 minutes and record these readings in your notebook. If the volume
change in the first 5 minutes is greater than 0.05 mL then you have a leak and must readjust the
stopper and start again at the beginning of this paragraph. If the volume change during this 5minute period is 0.05 mL or less, roll the magnesium ribbon into a loose coil, and drop it into the
HCl solution. Place the cork LOOSELY over the test tube and continue to record the volume each
minute for a period of 15-20 minutes.
50
Plot your data using a computer-graphing program, or ask for a sheet of graph paper. The results
should be similar to those shown in the figure below.
CALCULATIONS: Draw parallel lines through the initial and final points. The volume change
due to the heat of reaction of your sample of magnesium is the vertical distance between these
parallel lines.
The density of ice at 0В°C is 0.9167 g/mL and for water it is 0.9998 g/mL. Inverting each of these
values we see that the volume of ice is 1.09087 mL/gram and for water it is 1.0001605 mL/gram at
0В°C. This means that for every gram of ice that melts the volume will change by,
1.09087 mL/gram ice - 1.00016 mL/gram water = 0.09071 mL/g of ice melted
Since it require 6.01 kJ to melt one mole of ice, we can calculate how much energy is needed to
melt one gram of ice,
6010 J
x 1 mole ice = 333.89 J/gram ice
1 mole ice
18 grams
Therefore, we can calculate how many Joules of energy must be added to our system to change the
volume by one mL,
333.89 J/gram ice = 3680.9 J/mL of volume change
0.09071 mL/gram ice
51
So, to calculate the energy required to melt your volume of ice, multiply your measured volume by
3680.9 J/mL. This amount of energy was released when approximately 0.1 gram of magnesium
reacted with HCl. To calculate the amount of energy released when one mole of magnesium
reacts, you must calculate the number of moles of magnesium used in your experiment. Calculate
the number of moles of magnesium used in your experiment (the atomic mass of magnesium is
24.3 g/mole). By dividing the number of calories calculated earlier and dividing this by the
number of moles of magnesium used one obtains the enthalpy for the following reaction,
Mg (s) + 2 HCl (aq) в†’ MgCl2 (aq) + H2 (g) + Enthalpy
Check your calculations and report the result of this calculation in your worksheet.
52
Name_________________
Date______________
THE DETERMINATION OF A HEAT OF REACTION
OBJECTIVE:
PROCEDURE:
DATA:
Time and Volume Measurements for the Reaction between Mg and HCl
Time
Volume
Time
Volume
Time
Volume
CALCULATIONS: (Include graph)
53
RESULTS:
SOURCES OF ERROR:
PROBLEM
The standard heat of formation of HCl (aq) is –167.5 kJ/mole. From this, using Hess’ law,
calculate the standard heat of formation of MgCl2 (aq). (Hint: What is meant by the standard heat
of formation?)
54
EXPERIMENT
DETERMINING THE ENTHALPY OF A CHEMICAL REACTION USING HESS’S LAW
10
INTRODUCTION
All chemical reactions involve an exchange of heat energy; therefore, it is tempting to plan to
follow a reaction by measuring the enthalpy change (О”H). However, it is often not possible to
directly measure the heat energy change of the reactants and products (the system). We can
measure the heat change that occurs in the surroundings by monitoring temperature changes. If we
conduct a reaction between two substances in aqueous solution, then the enthalpy of the reaction
can be indirectly calculated with the following equation.
пЃ„H = Cp Г— m Г— О”T
The term пЃ„H represents the heat energy that is gained or lost. Cp is the specific heat of water, m is
the mass of water, and О”T is the temperature change of the reaction mixture. The specific heat and
mass of water are used because water will either gain or lose heat energy in a reaction that occurs
in aqueous solution. Furthermore, according to a principle known as Hess’s law, the enthalpy
changes of a series of reactions can be combined to calculate the enthalpy change of a reaction that
is the sum of the components of the series.
In this experiment, you will measure the temperature change of two reactions, and use Hess’s law
to determine the enthalpy change, пЃ„H of a third reaction. You will use a Styrofoam cup nested in a
beaker as a calorimeter, as shown in Figure 1.
For purposes of this experiment, you may assume that the heat loss to the calorimeter and the
surrounding air is negligible.
MATERIALS
Vernier computer interface
2.0 M hydrochloric acid solution
2.0 M sodium hydroxide solution
2.0 M acetic acid solution
2.0 M sodium acetate solution
250 mL beaker
50 mL or 100 mL graduated cylinders
Utility clamp
Glass stirring rod
Temperature Probe
Computer
Styrofoam cup calorimeter
Ring stand
Figure 1: Calorimetry Setup
55
PROCEDURE
Obtain and wear goggles. Connect a Temperature Probe to Channel 1 of the Vernier computer
interface. Connect the interface to the computer with the proper cable. Use a utility clamp to
suspend the Temperature Probe from a ring stand, as shown in Figure 1.
Start the Logger Pro program on your computer. Open the file “13 Enthalpy” from the Advanced
Chemistry with Vernier folder.
Part I - Reaction Between Solutions of NaOH and HCl
Nest a Styrofoam cup in a beaker (see Figure 1). Measure 50.0 mL of 2.0 M HCl solution into the
cup. Lower the tip of the Temperature Probe into the HCl solution. CAUTION: Handle the
hydrochloric acid with care. It can cause painful burns if it comes in contact with the skin.
Measure out 50.0 mL of NaOH solution, but do not add it to the HCl solution yet. CAUTION:
Handle the sodium hydroxide solution with care.
Click to begin the data collection and obtain the initial temperature of the HCl solution. After
three or four readings have been recorded at the same temperature, add the 50.0 mL of NaOH
solution to the Styrofoam cup all at once. Stir the mixture throughout the reaction. Data collection
will end after three minutes. If the temperature readings are no longer changing, you may terminate
the trial early by clicking . Click the Statistics button, . The minimum and maximum temperatures
are listed in the statistics box on the graph. If the lowest temperature is not a suitable initial
temperature, examine the graph and determine the initial temperature.
Record the initial and maximum temperatures in your data table. Rinse and dry the Temperature
Probe, Styrofoam cup, and the stirring rod. Dispose of the solution as directed.
Part II - Reaction Between Solutions of NaOH and HAc
Measure out 50.0 mL of 2.0 M NaOH solution into a nested Styrofoam cup (see Figure 1). Lower
the tip of the Temperature Probe into the cup of NaOH solution. Measure out 50.0 mL of 2.0 M
HAc solution, but do not add it to the NaOH solution yet. Click to begin the data collection. After
three or four readings have been recorded at the same temperature, add the 50.0 mL of HAc
solution to the Styrofoam cup all at once. Stir the mixture throughout the reaction. Data collection
will end after three minutes. If the temperature readings are no longer changing, you may terminate
the trial early by clicking .
Examine the graph as before to determine and record the initial and maximum temperatures of the
reaction. Rinse and dry the Temperature Probe, Styrofoam cup, and the stirring rod. Dispose of
the solution as directed.
56
Part III - Reaction Between Solutions of HCl and NaAc
Measure out 50.0 mL of 2.0 M HCl solution into a nested Styrofoam cup (see Figure 1). Lower
the tip of the Temperature Probe into the cup of HCl solution. Measure out 50.0 mL of 2.0 M
NaAc solution, but do not add it to the HCl solution yet.
Click to begin the data collection. After three or four readings have been recorded at the same
temperature, add the 50.0 mL of NaAc solution to the Styrofoam cup all at once. Stir the mixture
throughout the reaction. Data collection will end after three minutes. If the temperature readings
are no longer changing, you may terminate the trial early by clicking .
Examine the graph as before to determine and record the initial and maximum temperatures of the
reaction. Rinse and dry the Temperature Probe, Styrofoam cup, and the stirring rod. Dispose of
the solution as directed.
57
Name _____________________
Date__________________________
Determining the Enthalpy of a
Chemical Reaction using Hess’s Law
OBJECTIVE:
PROCEDURE
DATA TABLE
Reaction 1
Reaction 2
Reaction 3
Maximum Temp (В°C)
Initial Temp (В°C)
Temperature Change
CALCULATIONS
1. Calculate the amount of heat energy, q, produced in each reaction. Use 1.03 g/mL for the
density of all solutions. Use the specific heat of water, 4.18 J/(gВ°C), for all solutions.
2. Calculate the enthalpy change, пЃ„H, for each reaction in terms of kJ/mol of each reactant.
58
3. Using Hess’ law and your experimentally determined enthalpies for reactions 2 and 3,
determine the experimental molar enthalpy for reaction 1.
4. How does your value compare to the accepted value of -56.4 kJ/mol?
5. Does this experimental process support Hess’s law? Suggest ways of improving your results.
SOURCES OF ERROR
59
11
EXPERIMENT
ELECTROCHEMICAL CELLS
INTRODUCTION
In this experiment you will investigate reactions in which electrons are transferred from one
reactant to another. Since electrons are being transferred we can force them to travel through an
electrical connection and thereby be measured. We call these set-ups electrochemical cells. We
can observe the effect of electron transfer in other ways. In the first experiment we will observe
the formation of elements by electron transfer.
EXPERIMENT: For each of the following experiments, record your observations and write an net
equation for each reaction. Also determine which of the two substances is the better oxidizer
(oxidizers get reduced and are usually called oxidizing agents).
Place a small piece of;
a) Metallic lead in 5 mL of 1 M Cu(NO3)2.
b) Metallic zinc in 5 mL of 1 M Pb(NO3)2.
c) Metallic copper in 5 mL of 0.10 M AgNO3.
d) Metallic zinc in 5 mL of 1 M HCl.
e) Metallic lead in 5 mL of 1 M HCl.
f) Metallic copper in HOT 6M HNO3. Heat in a water bath in the hood. (Silver also reacts
with HNO3).
g) 1 mL of CCl4 and 3 mL of 1 M KI and 3 mL of Bromine water.
h) Metallic copper in 5 mL of Bromine water. Heat in a water bath in the hood to remove
any excess bromine then add 2 mL of 6 M NH3 to see the blue color of the copper ions.
i) 1 mL of CCl4 and 3 mL of 1 M KI and 3 mL of 6 M HNO3. Heat in a water bath.
Some of these reactions may be slow, allow them to stand for 15 minutes and look carefully at the
surface of metal of some of the slower reactions.
In your result section, write all of the half reactions as reductions. Now arrange these reactions in
such a way that a reaction will occur when you take a reaction and reverse any of the reactions
above it (reversing a reaction makes it an oxidation). In other words, reverse the oxidation reaction
and put it on top of the reduction reaction. It is sometimes difficult to see the lead reaction with
H+, a reaction does occur so place the lead appropriately. What additional experiments would you
need to perform to place bromine and iodine in their proper position?
60
ELECTROCHEMICAL CELLS
In this portion of the experiment you will be measuring the voltage produced by three
electrochemical cells. The electrodes used in electrochemical cells are usually metals and
occasionally carbon rods. The oxidation reaction occurs at the anode and reduction at the cathode
so that in electrochemical cells electrons flow from the anode to the cathode and the anode will be
negatively charged compared to the cathode. You may work in pairs.
EXPERIMENT: Put about 40 mL of 1 M CuSO4 into a 250 mL beaker and about 40 mL of 1 M
ZnSO4 into a porous cup. Place the porous cup into the 250 mL beaker containing the CuSO4. Put
a copper strip into the copper sulfate solution and a zinc strip into the zinc solution. Attach either a
voltmeter to the zinc and copper electrodes and measure the voltage. Draw the cell and determine
which electrode is acting as the anode and cathode. Write the reaction occurring at each electrode.
EXPERIMENT: Repeat the experiment above using Bromine water and a carbon electrode to
replace the copper sulfate solution and the copper electrode. Measure the voltage. Draw the cell
and determine which electrode is acting as the anode and cathode and write the reaction occurring
at each electrode.
MEASUREMENT OF A SOLUBILITY PRODUCT
Electrochemical cells can be used to determine the equilibrium constant of chemical reactions. In
this lab you will calculate the Ksp of Cu(OH)2 by the following set of half-reactions,
Cu(s) + 2 OH- ↔ Cu(OH)2(s) + 2eCu2+(1 M) + 2e- ↔ Cu(s)
Cu2+(1 M) + 2 OH- ↔ Cu(OH)2 o = ?
From the measured value of пѓµo one could evaluate the Ksp for Cu(OH)2(s) using,
пѓµВ° = (0.0592/n) log (1/Ksp)
where n is the number of electrons transferred in the reaction.
EXPERIMENT: Clean the electrochemical cell used above. Place 40 mL of 1 M CuSO4 in the
beaker and 40 mL of 1 M NaOH in the cup. Add 4 drops of 1 M CuSO4 to the 40 mL of NaOH.
Place a copper strip into each cell and measure the voltage as previously. Note the negative
terminal. Which electrode is the anode? Calculate the Ksp for Cu(OH)2(s).
61
Name__________________
Date________________
Electrochemical Cell Worksheet
OBJECTIVE:
PROCEDURE:
DATA:
Ingredients
Half-Reactions
ox:
Pb + Cu(NO3)2
red:
ox:
Zn + Pb(NO3)2
red:
ox:
Cu + AgNO3
red:
ox:
Zn + HCl
red:
ox:
Pb + HCl
red:
ox:
Cu + HNO3 (hot)
red:
ox:
Br2 + KI
red:
ox:
Cu + Br2
red:
ox:
KI + HNO3 (warm)
red:
62
Observations
Electrochemical Cell Data:
Cu/Zn Cell Voltage =
Zn/Br2 Cell Voltage = ___________
Calculation of Ksp for Cu(OH)2
пѓµВ° = (0.0592/n) log (1/Ksp)
Cu(OH)2 Cell Voltage = _____________
63
RESULTS:
Ordered Half-Reactions
Cu/Zn Cell Voltage =
Zn/Br2 Cell Voltage = ___________
Cu(OH)2 Cell Voltage = _____________ Ksp of Cu(OH)2 = _________________
CONCLUSIONS:
64
PROBLEMS
1) Use your table of oxidizing and reducing agents to predict whether the following reactions will
occur as written, or will occur in the opposite direction.
a) I2 + Zn(s) = Zn2+ + 2 Ib) 2 NO3- + 8 H+ + 6 Cl- = 2 NO(g) + 4 H2O + 3 Cl2
c) Pb(s) + Cu2+ = Pb2+ + Cu(s)
d) Pb2+ + 2 Br- = Pb(s) + Br2
2) You are given the half-reaction for pure water,
2 H+ (10-7M) + 2e- = H2 (g, 1atm)
пѓµВ° = -0.41 eV
On a separate piece of paper, rank the following metals according their standard electrode
potentials as found in the CRC Handbook, (Na, K, Mg, Al, Zn, Fe, Sn, Pb, Cu, and Ag) by placing
the most positive one at the top of the list and then determine which of the following metals
should,
a) React with water evolving gas.
b) React with 1 M HCl but not with water to produce hydrogen gas.
c) What reagent and conditions would you use to dissolve the metals that cannot be dissolved in 1
M HCl?
3) Calculate the voltage of an electrochemical cell which uses the reaction,
Cu2+(1 M) = Cu2+(0.001 M)
4) Would your results have changed if you had added more drops of CuSO4 to your Ksp cell? Why
or why not?
65
12
EXPERIMENT
Name ____________________
Date______________
Title:
Objective:
Procedure:
Diagram of Apparatus:
Data:
Calculations:
Results:
Conclusion:
Sources of Error:
66
Exam I
Rate Laws
Activation Energies
Mechanisms
Radioactive Decay
Kinetics and Activation Energy
1) Rate information was obtained for the following reaction at 25В°C and 33В°C;
Cr(H2O)63+ + SCN- в†’ Cr(H2O)5NCS2+ + H2O(l)
[Cr(H2O)63+]
[SCN-]
2.0x10-11
2.0x10-10
9.0x10-10
2.4x10-9
1.0x10-4
1.0x10-3
2.0x10-3
3.0x10-3
0.10
0.10
0.15
0.20
1.4x10-10
1.0x10-4
0.20
Initial Rate
@ 33МЉC
a) Write a rate law consistent with the experimental data.
b) What is the value of the rate constant at 25МЉC?
c) What is the value of the rate constant at 33МЉC?
d) What is the activation energy for this reaction?
e) What is the reverse rate law?
2) The mechanism for the decomposition of phosgene
COCl2(g) в†’ CO(g) + Cl2(g)
is thought to be,
fast eq.
slow
fast eq.
Cl2(g) ⇆ 2 Cl(g)
COCl2(g) + Cl(g) в†’ COCl(g) + Cl2(g)
COCl(g) ⇆ CO(g) + Cl(g)
Based on this mechanism, what is the rate law for this reaction?
3) The following experimental data were obtained for the reaction at 250 K,
F2 + 2 ClO2 в†’ 2 FClO2
[F2]/M
[ClO2]/M
0.10
0.10
0.20
0.010
0.040
0.010
Rate/(M/sec)
1.2x10-3
4.8x10-3
4.8x10-3
3a) Write a rate law consistent with this data.
Rate =
3b) What is the value and units of the rate constant?
4) The bromination of acetone is acid catalyzed;
CH3COCH3 + Br2 в†’ CH3COCH2Br + H+ + BrThe rate of disappearance of bromine was measure for several different concentrations of
acetone, bromine and hydrogen ions;
Rate/Msec-1
6x10-5
6x10-5
1.2x10-4
3.2x10-4
8x10-5
[Acetone]/M
0.30
0.30
0.30
0.40
0.40
[Br2]/M
0.050
0.100
0.050
0.050
0.050
[H+]/M
0.050
0.050
0.100
0.200
0.050
a) What is the forward rate law for this reaction?
b) What is the value and units of the forward rate constant?
c) What is the reverse rate law?
d) If the equilibrium constant is 1.3x103, what is the value of the reverse rate constant?
5) At low temperatures, the rate law for the reaction,
CO(g) + NO2(g) в†’ CO2(g) + NO(g)
can be determined by the following data;
[NO]/10-3 M [CO]/10-3 M [NO2]/10-3 M Initial Rate/10-4
1.20
1.20
1.20
2.40
1.50
3.00
0.75
1.50
0.80
0.80
0.40
0.40
3.60
7.20
0.90
0.90
Write a rate law in agreement with the data.
Which of the following mechanisms is consistent with this rate
law? (Circle A, B, or C)
A) CO + NO2 в†’ CO2 + NO slow
B) NO2 ⇆ NO + O equil.
O + CO в†’ CO2 slow
C) 2 NO2 ⇆ 2 NO + O2 equil.
CO + O2 в†’ CO2 + O slow
O + NO в†’ NO2
fast
D) None of the above
6) The following experimental data were obtained for the reaction;
2 NO + 2 H2 в†’ N2 + 2 H2O
Initial Rate/10-5
0.60
2.40
0.30
[NO]/10-2 M
0.50
1.00
0.25
[H2]/10-2 M
0.20
0.20
0.40
a) Write a rate law in agreement with the data.
Rate =
b) What is the value of the rate constant?
c) What is the new rate when [NO] = 3x10-2 M and [H2] = 1.2x10-2 M?
d) Which of the following mechanisms are consistent with the rate law above?
i) 2 NO + H2 в†’ N2O + H2O (slow)
ii) 2 NO ⇆ N2O2 (equil)
N2O2 + H2 в†’ N2O + H2O (slow)
iii) NO + H2 ⇆ H2O + N (equil)
N + NO в†’ N2O (slow)
e) Platinum acts as a catalyst for this reaction. What term must be added to the rate law to
account for the presence of the catalyst?
f) Would platinum be a homogeneous or heterogeneous catalyst for this reaction?
g) If the temperature is increased by 6.8В°C the reaction rate increases 1.65 times. What is the
activation energy of the reaction?
7) The following experimental data were obtained for the reaction at 25В°C,
2 NO + H2 в†’ N2O + H2O
Initial [NO]
/10-3 M
6.40
12.8
6.40
Initial [H2]
/10-3 M
Initial Rate
/10-5 Msec-1
2.20
1.10
4.40
2.60
5.20
5.20
Which of the following mechanisms are consistent with the rate law for this reaction?
a)
2 NO + H2 в†’ N2O + H2O (slow)
b)
2 NO ⇆ N2O2 (equil)
N2O2 + H2 в†’ N2O + H2O (slow)
c)
NO + H2 ⇆ H2O + N (equil)
N + NO в†’ N2O (slow)
8) At low temperatures, the rate law for the reaction,
CO(g) + NO2(g) в†’ CO2(g) + NO(g)
is;
Rate = k [NO2]2
Which of the following mechanisms is consistent with this rate
law? (Circle A, B, or C)
a) CO + NO2 в†’ CO2 + NO
slow
b) 2 NO2 ⇆ N2O4
fast equil.
N2O4 + 2 CO в†’ 2 CO2 + 2 NO slow
c) 2 NO2 в†’ N2O4
N2O4 ⇆ 2 NO + O2
O2 + CO2 в†’ 2 CO2
slow
fast equil.
fast
9) Given the following rate data, calculate the rate law at 25В°C.
Rate/10-4
[I-]/M
[OCl-]/M
6.10
12.2
36.6
0.20
0.40
0.60
0.050
0.050
0.100
14.4
0.20
0.050
@ 25МЉC
@ 33МЉC
What is the rate law?
Rate =
What is the value of the rate constant at 25В°C and at 33В°C?
What is the activation energy for this reaction?
What is the half-life of this reaction at 45В°C if [I-] = [OCl-] = 0.25 M?
10) Thiosulfate (S2O32-) can react with triiodide (I3-) according to the following reaction at 25В°C,
2 S2O32- + I3- ⇆ S4O62- + 3 I- Keq = 3.75x105
Experimentally the forward rate law can be determined from the following data,
Rxn Rate/Msec-1
#1
#2
#3
2.56x10-4
1.28x10-4
1.92x10-4
[S2O32-]/M
[I3-]/M
0.040
0.020
0.060
0.12
0.12
0.06
a) What is the rate law?
b) What is the value of the rate constant? (Include units)
c) What are the minimum number of steps required in the mechanism of the forward rate law?
Circle one
1
2
3
d) If reaction #1 is heated to 35В°C the rate increases to 4.8x10-4 M/sec. What is the activation
energy of this reaction?
e) At what temperature would the reaction rate double from 25В°C?
11) A cook finds that it takes 30 minutes to boil potatoes at 100В°C in an open sauce pan and only
12 minutes to boil them in a pressure cooker at 110В°C. Estimate the activation energy for
cooking potatoes, which involves the conversion of cellulose into starch. Remember that there
is an inverse relationship between time and the rate constant.
12) When N2O4 decomposes it forms NO2,
N2O4 в†’ 2 NO2
If the half-life of this reaction is 1386 seconds, how much of a 10 gram sample would be left
after 1500 seconds? (Is the reaction first or second order?)
13) An archaeologist measured the amount of radioactivity of a piece of cloth used to wrap an
Egyptian mummy. The cloth was found to have a decay rate of 9.1 dpm. If the decay rate is 15.3
dpm in living tissue, how old is the mummy? t1/2 = 5730 years.
14) Uranium is radioactive and decays into lead. This process can be used to date rocks. A
piece of zirconium was dated using this process and it was found that this rock contained
3.2x10-3 grams of uranium and 2.2x10-5 grams of lead. If the half-life of uranium is 4.41x109
years how old is the rock?
15) Assuming that the loss of ability to recall learned material is a first-order process with a halflife of 35 days. Compute the number of days required to forget 90% of the material that you
learned in preparation for this exam.
16a) Under acidic conditions sucrose (table sugar) can be broken down into its individual sugars,
glucose and fructose. At 27В°C it takes 54.5 minutes to convert half the sucrose to glucose and
fructose and at 37В°C it takes 13.7 minutes. Estimate the activation energy for the breakdown of
sucrose.
16b) The above reaction is known to second order which means that the half-life is dependent on
the initial concentration. Will this fact effect your calculation of the activation energy above?
Why or why not. (Hint: What must you assume about the initial concentration of [A] when using
the Arrhenius equation?)
17) A 10 gram sample of 131I was sent from a pharmaceutical company to a hospital for use in
the treatment of hyperthyroidism. If the half life of 131I is 8.07 days, how much of the sample
would be left after a 2 day mail delivery?
18) The denaturation of the virus that causes the rabbit disease Myxomatosis can be followed by
heating the virus under a microscope. It is observed that the reaction is first-order and that it
takes 22.35 minutes at 50В°C and 0.35 minutes at 60В°C for the virus to denaturate. Estimate the
activation energy for the denaturation of the Myxomatosis virus.
Answer Key - Kinetics and Activation Energy
1a)
Rate = k[Cr(H2O)63+] [SCN-]2
1b)
2x10-11 = k [1x10-4] [0.1]2
1c)
1.4x10-10 = k [1x10-4] [0.2]2 k = 3.5x10-5 @ 33В°C
1d)
ln
1e)
kF [Cr(H2O)63+] [SCN-] = kR [X]
k = 2x10-5 @ 25В°C
2 x10пЂ­5
Ea пѓ¦ 298 пЂ­ 306 пѓ¶
пЂЅ
пѓ§
пѓ·
пЂ­5
3.5 x10
8.314 пѓЁ 298 п‚ґ 306 пѓё
Keq пЂЅ
kF
[X]
[Cr(H 2O)5 SCN 2 пЂ« ]
пЂЅ
пЂЅ
k R [Cr(H 2O)36пЂ« ][SCN пЂ­ ]2 [Cr(H 2O)36пЂ« ][SCN пЂ­ ]
[X] = [Cr(H2O)5SCN2+] [SCN-]
2)
Ea = 53,033 J/mol
п‚€
rateR = kR[Cr(H2O)5SCN2+] [SCN-]
Start with the slow step
rate = k [COCl2] [Cl]
Get rid of [Cl] using the equilibrium,
[Cl] 2
пЂЅ Keq пѓћ [Cl] пЂЅ Keq[Cl 2 ]
[Cl 2 ]
п‚€
1
1
1
rate пЂЅ k Keq[Cl 2 ][COCl 2 ] пЂЅ kKeq 2 [Cl 2 ] 2 [COCl 2 ] пЂЅ k ' [Cl 2 ] 2 [COCl 2 ]
3a)
rate = k[F2]2 [ClO2]
3b)
1.2x10-3 M/s = k [0.1 M]2 [0.01 M]
k = 12 1/M2s
4a)
rate = k [Acetone] [ H+]
4b)
6x10-5 M/s = k [0.3 M] [ 0.05 M]
k = 4x10-3 1/Ms
4c)
kF [Acetone] [H+] = kR [X]
kF
[X]
[AcetoneBr][H пЂ« ][Br пЂ­ ]
пЂЅ
пЂЅ
k R [Acetone] [H пЂ« ]
[Acetone] [Br2 ]
[X] пЂЅ
[AcetoneBr][H пЂ« ]2 [Br пЂ­ ]
[Br2 ]
so,
rate R пЂЅ k R
[AcetoneBr][H пЂ« ]2 [Br пЂ­ ]
[Br2 ]
[CO] [NO 2 ]
[NO]
5a)
rate пЂЅ k
5b)
A) rate = k [CO] [NO2]
No, doesn’t match
B) rate = k [O] [CO] and Keq пЂЅ
[NO] [O]
[NO 2 ]
so, [O] пЂЅ Keq
[NO 2 ]
[NO]
[CO] [NO 2 ]
[CO] [NO 2 ]
or rate пЂЅ k'
[NO]
[NO]
This matches our experimentally determined rate law so it is a possible mechanism.
Substituting, we get rate пЂЅ kKeq
C) rate = k [O2] [CO]
The only reason to use the equilibrium is to get rid of something that is not in the overall
reaction. Since the rate law does not contain anything that is not already in the overall
reaction (no intermediates) there is no reason to continue. This does not match.
6a)
rate = k [NO2]2 [H2]
6b)
2.4x10-5 = k [1x10-2 M]2 [0.2x10-2 M]2 so, K = 120 1/M2s
6c)
rate = (120 1/M2s) (3x10-2)2 (1.2x10-2) = 1.296x10-3 M/s
6d)
Both i & ii are consistent but i) is a single step 3rd order reaction so ii is the answer.
6e)
kcat [Pt] + 1
6f)
Heterogeneous catalyst. Platinum is a solid.
6g)
Assume k1 @ 298 K and 1.65 k1 @ 304.8 K
7)
1.65k1
Ea пѓ¦ 304.8 пЂ­ 298 пѓ¶
пЂЅ
пѓ§
пѓ·
k1
8.314 пѓЁ 304.8 п‚ґ 298 пѓё
rate = k [NO2]2 [H2]
ln
Ea = 55,6130 J/mol
A & B are consistent but A is a single step third order process so it doesn’t occur.
Therefore the answer is B.
8)
Only C matches the experimentally determined rate law.
9a)
rate = k [I-] [OCl-]
9b)
6.1x10-4 = k [0.20 M] [0.05 M] so, k = 0.061 1/Ms @ 25ВєC
14.4x10-4 = k [0.20 M] [0.05 M] so, k = 0.144 1/Ms @ 33ВєC
9c)
ln
0.144
Ea пѓ¦ 306 пЂ­ 298 пѓ¶
пЂЅ
пѓ§
пѓ·
0.061 8.314 пѓЁ 306 п‚ґ 298 пѓё
9d)
ln
81,399 пѓ¦ 318 пЂ­ 298 пѓ¶
пѓ§
пѓ·
0.061 8.314 пѓЁ 318 п‚ґ 298 пѓё
K 45 пЃЇ
пЂЅ
Ea = 81,399 J/mol
K45Вє = 0.4816
tВЅ = 1/[A]K for 2nd order reaction so, tВЅ = 1/[0.25](0.4816) = 8.305 seconds
10a)
rate = k [S2O32-] [I3-]
10b)
2.56x10-4 M/s = k [0.04 M] [ 0.12 M] so, k = 5.33x10-2 1/Ms
10c)
The reaction is 2nd order so it will take a minimum of 1 step.
10d)
2.56x10-4 M/s = k [0.04 M] [ 0.12 M] so, k1 = 5.33x10-2 1/Ms @ 25 C
4.80x10-4 M/s = k [0.04 M] [ 0.12 M] so, k2 = 0.10 1/Ms @ 35 C
ln
0.10
Ea пѓ¦ 308 пЂ­ 298 пѓ¶
пЂЅ
пѓ§
пѓ·
пЂ­2
5.33x10
8.314 пѓЁ 308 п‚ґ 298 пѓё
10e)
ln
2k1 47,969 пѓ¦ 1
1пѓ¶
пѓ§пѓ§
пЂЅ
пЂ­ пѓ·пѓ·
k1
8.314 пѓЁ 298 T2 пѓё
11)
t1 = 30 min @ 100ВєC and t2 = 12 min @ 110ВєC
30 min
Ea пѓ¦ 383 пЂ­ 373 пѓ¶
пЂЅ
пѓ§
пѓ·
12 min 8.314 пѓЁ 383 п‚ґ 373 пѓё
10
0.693
ln пЂЅ
(1500 sec)
X 1386 sec
ln
12)
13)
ln
15.3dpm
0.693
пЂЅ
пЃ„t
9.1dpm 5730yrs
Ea = 47,969 J/mol
T2 = 309.6 K
Ea = 108,830 J/mol
X = 4.72 grams
пЃ„t = 4296 yrs
14)
moles U + moles Pb = moles U initially in the sample (Parent)
3.2x10пЂ­3 g U
2.2x10пЂ­5 g Pb
пЂЅ 1.3445x10пЂ­ 5 mol U (Parent) and
пЂЅ 1.062x10пЂ­ 5 mol Pb (daughter)
238g/mol
207.2 g/mol
Total Parent = 1.3445x10-5 mol U + 1.062x10-7 mol Pb = 1.355x10-5 mol U
ln
1.355x10-5 mol U
0.693
пЂЅ
пЃ„t
-5
1.3445x10 mol U 4.41x109 yrs
15)
ln
100
0.693
пЂЅ
пЃ„t
10 35 days
16a)
54.5 min @ 27ВєC and 13.7 min @ 37ВєC
ln
16b)
пЃ„t = 5.01x107 yrs
пЃ„t = 116.3 days = summer vacation!
54.5 min
Ea пѓ¦ 310 пЂ­ 300 пѓ¶
пЂЅ
пѓ§
пѓ·
13.7 min 8.314 пѓЁ 310 п‚ґ 300 пѓё
Ea = 106, 764 J/mol
No, it makes no difference.
For 1st order reaction t 1 пЂЅ
2
For 2nd order reaction t 1
2
0.693
k
For both of these tВЅ пЂҐ 1/k
1
пЂЅ
[A] k
As long as [A] is the same for all runs, you can directly substitute tВЅ for 1/k
10
0.693
пЂЅ
2 days
X 8.07 days
17)
ln
18)
22.35 min @ 50ВєC and 0.35 min @ 60ВєC
ln
22.35 min
Ea пѓ¦ 333 пЂ­ 323 пѓ¶
пЂЅ
пѓ§
пѓ·
0.35 min 8.314 пѓЁ 333 п‚ґ 323 пѓё
X = 8.42 grams
Ea = 371,706 J/mol
Nuclear Chemistry - Radioactive Decay
1.
What particle is emitted when a Fr-210 nucleus decays to At-206?
(a) alpha
(b) beta
(c) neutron
(d) positron
(e) proton
2.
What particle is emitted when a Ra-221 nucleus decays to Rn-217?
(a) alpha
(b) beta
(c) neutron
(d) positron
(e) proton
3.
What particle is emitted when a Th-228 nucleus decays to Ra-224?
(a) alpha
(b) beta
(c) neutron
(d) positron
(e) proton
4.
What particle is emitted when a F-20 nucleus decays to Ne-20?
(a) alpha
(b) beta
(c) neutron
(d) positron
(e) proton
5.
What particle is emitted when an Ar-39 nucleus decays to K-39?
(a) alpha
(b) beta
(c) neutron
(d) positron
(e) proton
6.
What particle is emitted when a Sr-90 nucleus decays radioactively to Y-90?
(a) alpha
(b) beta
(c) neutron
(d) positron
(e) proton
7.
What particle is emitted when a carbon-11 nucleus decays to boron-11?
(a) alpha
(b) beta
(c) neutron
(d) positron
(e) proton
8.
What particle is emitted when a fluorine-17 nucleus decays to oxygen-17?
(a) alpha
(b) beta
(c) neutron
(d) positron
(e) proton
9.
What particle is emitted when a neon-19 nucleus decays to fluorine-19?
(a) alpha
(b) beta
(c) neutron
(d) positron
(e) proton
10. What nuclide is produced when Pt-175 decays by alpha emission?
(a)
(b)
(c)
(d)
(e)
171
76
175
76
171
78
171
79
175
79
Os
Os
Pt
Au
Au
11. What nuclide is produced when U-235 decays by alpha emission?
(a)
(b)
(c)
(d)
(e)
231
90
235
90
231
92
235
93
231
93
Th
Th
U
Np
Np
12. What nuclide is produced when Ra-223 decays by alpha and gamma emission?
(a)
(b)
(c)
(d)
(e)
219
86
227
86
227
88
219
90
227
90
Rn
Rn
Ra
Th
Th
13. What radionuclide decays to Pb-210 by alpha emission?
(a)
(b)
(c)
(d)
(e)
206
80
214
80
206
82
206
84
214
84
Hg
Hg
Pb
Po
Po
14. What nuclide is produced when K-43 decays by beta emission?
43
(a) 18 Ar
(b)
(c)
(d)
(e)
42
19
42
20
43
20
44
20
K
Ca
Ca
Ca
15. What nuclide is produced when Pb-210 decays by beta emission?
(a)
(b)
(c)
(d)
(e)
210
81
212
81
211
82
210
83
211
83
Tl
Tl
Pb
Bi
Bi
16. What nuclide is produced when Ar-39 decays by beta and gamma emission?
(a)
(b)
(c)
(d)
(e)
39
17
40
17
40
18
39
19
40
19
Cl
Cl
Ar
K
K
17. What radionuclide decays to Fe-56 by beta emission?
(a)
(b)
(c)
(d)
(e)
55
25
56
25
55
26
56
27
57
27
Mn
Mn
Fe
Co
Co
18. What nuclide is produced when N-13 decays by positron emission?
12
(a) 6 C
(b)
(c)
(d)
(e)
13
6
14
6
14
7
13
8
C
C
N
O
19. What nuclide is produced when O-15 decays by positron emission?
14
(a) 7 N
(b)
(c)
(d)
(e)
15
7
14
8
15
9
16
9
N
O
F
F
20. What nuclide is produced when K-40 decays by positron emission?
39
(a) 18 Ar
(b)
(c)
(d)
(e)
40
18
41
18
40
19
40
20
Ar
Ar
K
Ca
21. What radionuclide decays to Br-73 by positron emission?
(a)
(b)
(c)
(d)
(e)
72
34
74
34
72
35
74
35
73
36
Se
Se
Br
Br
Kr
22. What nuclide is produced when a Cs-129 nucleus decays by electron capture?
128
(a) 54 Xe
(b)
(c)
(d)
(e)
129
54
128
55
128
56
129
56
Xe
Cs
Ba
Ba
23. What nuclide is produced when a W-181 nucleus decays by electron capture?
180
(a) 73 Ta
(b)
(c)
(d)
(e)
181
73
180
74
180
75
181
75
Ta
W
Re
Re
24. What nuclide is produced when a Mn-52 nucleus decays by electron capture?
52
(a) 24 Cr
(b)
(c)
(d)
(e)
53
24
53
25
52
26
53
26
Cr
Mn
Fe
Fe
25. What radionuclide decays to Cs-133 by electron capture?
(a)
(b)
(c)
(d)
(e)
132
54
133
54
134
55
133
56
134
56
Xe
Xe
Cs
Ba
Ba
Radioactive Decay Series
26. In the final step of the uranium-238 disintegration series, the parent nuclide
decays to lead-206 and an alpha particle. What is the parent nuclide?
202
(a) 80 Hg
(b)
(c)
(d)
(e)
210
83
206
84
210
84
Bi
Po
Po
none of the above
27. In the final step of the uranium-235 disintegration series, the parent nuclide decays to lead207 and a beta particle. What is the parent nuclide?
207
(a) 81 Tl
(b)
(c)
206
82
208
82
207
83
Pb
Pb
(d)
Bi
(e) none of the above
28. In the final step of the thorium-232 disintegration series, the parent nuclide decays to lead208 and an alpha particle. What is the parent nuclide?
208
(a) 83 Bi
(b)
(c)
(d)
(e)
212
83
208
84
212
84
Bi
Po
Po
none of the above
29. The uranium-238 decay series begins with the emission of an alpha particle. If the daughter
decays by beta emission, what is the resulting nuclide?
234
(a) 89 Ac
(b)
(c)
(d)
(e)
233
90
234
90
233
91
234
91
Th
Th
Pa
Pa
30. The uranium-235 decay series begins with the emission of an alpha particle. If the daughter
decays by beta emission, what is the resulting nuclide?
231
(a) 89 Ac
(b)
(c)
(d)
(e)
230
90
231
90
230
91
231
91
Th
Th
Pa
Pa
31. The thorium-232 decay series begins with the emission of an alpha particle. If the daughter
decays by beta emission, what is the resulting nuclide?
228
(a) 87 Fr
(b)
(c)
(d)
(e)
227
88
228
88
227
89
228
89
Ra
Ra
Ac
Ac
Answer Key
1A, 2A, 3A, 4B, 5B, 6B, 7D, 8D, 9D, 10A, 11A, 12A, 13E, 14D, 15D, 16D, 17B, , 18B, 19B,
20B, 21E, 22B, , 23B, 24A, 25D, 26D, 27A, 28D, 29E, 30E, 31E
Chemistry 121
First Exam
Name______________________
February 17, 2011
CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full
credit. You may use a calculator.
Question
Credit
1 (20)
2 (25)
3 (15)
4 (15)
5(25)
TOTAL
Important Equations ln(Ai/Af) = kt
tВЅ = 0.693/k
ln(k1/k2) = Ea/R (1/T2 - 1/T1)
1/Af - 1/Ai = kt
tВЅ = 1/[A]k
R = 8.314 J/mol-K
1) As you know, ozone (O3) protects us from ultraviolet light in our upper atmosphere. Ozone
can break up to form oxygen by the following reaction,
2 O3 в†’ 3 O2
The rate law for this reaction is known to be rate = k [O3]2/[O2]. Which of the following is a
possible mechanism for this reaction? Show your work
a)
2 O3 в†’ 3 O2 slow
b)
O3 ⇆ O2 + O fast eq.
O3 + O в†’ 2 O2 slow
c)
O2 ⇆ 2 O fast eq.
O3 + O в†’ 2 O2 slow
2) The following experimental data was obtained for the bromination of acetone at 30В°C.
2 I- + 2 H+ + H2O2 в†’ I2 + 2 H2O
Rate/M/sec
1.64x10-5
3.28x10-5
7.38x10-5
1.64x10-4
[I-]/M
0.15
0.30
0.45
0.20
[H+]/M
0.10
0.10
0.15
0.25
[H2O2]/M
0.10
0.10
0.10
0.30
2a) Write a rate law consistent with this data.
Rate =
2b) What is the value and units of the rate constant?
2c) Copper ions (Cu2+) are known to be a catalyst for this reaction. Please write down the
catalyzed rate law for this reaction.
3) Cryosurgical procedures involve lowering the body temperature of the patient prior to surgery.
The activation energy for the beating of the heart muscle is about 30 kJ/mol (30,000 J/mol) A
persons normal body temperature is 98.6ВєF (or 37ВєC) and most people have a pulse rate of about
75 beats/min. Using this information estimate a person’s pulse if their body temperature is
dropped to 72ВєF (22ВєC).
4) Technetium is a relatively light element that has never been found on earth. As a
consequence, the only technetium we have is man-made and radioactive. If a sample of Tc has a
decay rate of 983 dpm and exactly 25 minutes later it has a decay rate of 937 dpm, what is halflife of Technetium?
5a) A reaction was set up using concentrations of 1.2 M and 0.30 M. When the half-life was
measured one took 3.1 minutes and the other took 3.6 minutes respectively. Was the reaction
first or second order? Please explain your choice.
5b) Can a 5th order reaction occur? Please explain.
5c) Please give two reasons why a reaction might be zero order.
5d) Why do reactions tend to speed up when you heat them?
Chemistry 121
First Exam
Name ______________________
February 17, 2011
CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full
credit. You may use a calculator.
Question
Credit
1 (20)
2 (25)
3 (15)
4 (15)
5(25)
TOTAL
ln(Ai/Af) = kt
tВЅ = 0.693/k
ln(k1/k2) = Ea/R (1/T2 - 1/T1)
1/Af - 1/Ai = kt
R = 8.314 J/mol-K
tВЅ = 1/[A]k
1) As you know, ozone (O3) protects us from ultraviolet light in our upper atmosphere. Ozone
can break up to form oxygen by the following reaction,
2 O3 –> 3 O2
The rate law for this reaction is known to be rate = k [O3]2/[O2]. Which of the following is a
possible mechanism for this reaction? Show your work
a)
2 O3 в†’ 3 O2 slow
rate = k [O3]2 does not match
b)
O3 ⇆ O2 + O fast eq.
rate = k [O3] [O]
[O ] [O]
[O ]
Keq пЂЅ 2
пѓћ [O] пЂЅ Keq 3
[O3 ]
[O 2 ]
O3 + O в†’ 2 O2 slow
Rate = k Keq
c)
O2 ⇆ 2 O fast eq.
O3 + O в†’ 2 O2 slow
[O3 ]2
[O ]2
= k’ 3
matches
[O2 ]
[O2 ]
rate = k [O3] [O]
[O]2
Keq пЂЅ
пѓћ [O] пЂЅ Keq [O 2 ]
[O 2 ]
Rate = k Keq½ [O3] [O2]½ = k’ [O3] [O2]½
does not match
2) The following experimental data was obtained for the bromination of acetone at 30В°C.
2 I- + 2 H+ + H2O2 в†’ I2 + 2 H2O
Rate/M/sec
1.64x10-5
3.28x10-5
7.38x10-5
1.64x10-4
[I-]/M
0.15
0.30
0.45
0.20
[H+]/M
0.10
0.10
0.15
0.25
[H2O2]/M
0.10
0.10
0.10
0.30
2a) Write a rate law consistent with this data.
Rate = k [I-] [H+] [H2O2]
2b) What is the value and units of the rate constant?
1.64x10-5 M/s = k [0.15 M] [0.10 M] [0.10 M]
k = 0.01093 1/M2s
2c) Copper ions (Cu2+) are known to be a catalyst for this reaction. Please write down the
catalyzed rate law for this reaction.
Rate = (kcat[Cu2+] + 1) k [I-] [H+] [H2O2]
3) Cryosurgical procedures involve lowering the body temperature of the patient prior to surgery.
The activation energy for the beating of the heart muscle is about 30 kJ/mol (30,000 J/mol). A
person’s normal body temperature is 98.6ºF (or 37ºC) and most people have a pulse rate of about
75 beats/min. Using this information estimate a persons pulse if their body temperature is
dropped to 72ВєF (22ВєC).
ln
75 bpm 30,000 пѓ¦ 310 пЂ­ 295 пѓ¶
пЂЅ
пѓ§
пѓ·
X bpm
8.314 пѓЁ 310 п‚ґ 295 пѓё
X bpm = 41.50 bpm
4) Technetium is a relatively light element that has never been found on earth. As a
consequence, the only technetium we have is man-made and radioactive. If a sample of Tc has a
decay rate of 983 dpm and exactly 25 minutes later it has a decay rate of 937 dpm, what is halflife of Technetium?
ln
983 dpm 0.693
пЂЅ
25 min
937 dpm
t1/2
tВЅ = 361.5 min
5a) A reaction was set up using concentrations of 1.2 M and 0.30 M. When the half life was
measured one took 3.1 minutes and the other took 3.6 minutes respectively. Was the reaction
first or second order? Please explain your choice.
The reaction is first order. First order reactions have half-lives that are independent of
concentration. Although the concentration changed by a factor of four, the time stayed
essentially constant. The small difference in time is experimental error. With a four-fold
change in concentration, a 2nd order reaction would have changed by a factor four.
5b) Can a 5th order reaction occur? Please explain.
Yes, but it would take at least four steps for a 5th order reaction to occur.
5c) Please give two reasons why a reaction might be zero order.
1. Though required for the reaction to occur, the compound may not be involved in the slow
step of the reaction.
2. The component may be a spectator ion and not involved with the reaction at all.
3. Adding more reactant to a saturated catalyst will not cause the reaction to speed up and
will appear to be zero order.
5d) Why do reactions tend to speed up when you heat them?
Reactions speed up because they are hitting each other more often so more of the
molecules have enough energy to exceed the activation energy.
Chemistry 121
First Exam
Name_______________________
February 21, 2008
CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full
credit. You may use a calculator.
Question
Credit
1 (20)
2 (35)
3 (35)
4 (10)
TOTAL
Important Equations ln(Ai/Af) = kО”t
t1/2 = 0.693/k
1/Af - 1/Ai = kО”t
t1/2 = 1/[A]k
ln(k1/k2) = Ea/R (1/T2 - 1/T1)
R = 8.314 J/mol-K
1a) A sample of 131I was purified and then analyzed for iodine content. It was found that the
sample was 99.8% pure. 25 hours later the sample was tested again and it was found that the
sample was now just 78.2% pure. What is the half-life of 131I?
1b) How long would it take for 90% of a sample of 131I to decay?
2) The following experimental data were obtained for the reaction at 295 K,
H2 + Br2 в†’ 2 HBr
[H2]/M
[Br2]/M
0.10
0.40
0.10
0.010
0.010
0.040
Rate/(M/sec)
1.2x10-3
4.8x10-3
2.4x10-3
2a) Write a rate law consistent with this data.
Rate =
2b) What is the value and units of the rate constant?
2c) What is the rate of the reaction when [H2] = 0.30 M and [Br2] = 0.025 M?
2d) Which of the following are a possible mechanism for the above reaction? Show your work.
a)
H2 + Br2 в†’ 2 HBr slow
b)
Br2 ⇆ 2 Br
Br + H2 в†’ HBr + H slow
H + Br в†’ HBr fast
c)
H2 + ВЅ Br2 в†’ HBr + H slow
H + ВЅ Br2 в†’ HBr fast
3) If a reaction has an activation energy of 112.4 kJ, how hot must you get the reaction to
increase the rate of the reaction 1.5 times? Assume an initial temperature of 25ВєC.
4a) Why do reactions tend to speed up when you heat them?
4b) What term must be added to a reaction to account for the presence of a catalyst?
5) Assume that the cooking of an egg is a first order process. Assume further that it takes 3
minutes to half-cook (soft boil) an egg at 100ВєC. If the activation energy of the cooking of an
egg is 75 kJ/mole, how long would it take to soft boil an egg at the top of Mt. Everest where
water boils at 80ВєC?
Chemistry 121
First Exam
Name_______________________
February 21, 2008
CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full
credit. You may use a calculator.
Question
Credit
1 (20)
2 (35)
3 (35)
4 (10)
TOTAL
Important Equations ln(Ai/Af) = kО”t
t1/2 = 0.693/k
1/Af - 1/Ai = kО”t
t1/2 = 1/[A]k
ln(k1/k2) = Ea/R (1/T2 - 1/T1)
R = 8.314 J/mol-K
1a) A sample of 131I was purified and then analyzed for iodine content. It was found that the
sample was 99.8% pure. 25 hours later the sample was tested again and it was found that the
sample was now just 78.2% pure. What is the half-life of 131I?
ln
99.8% 0.693
пЂЅ
25 hours
78.2%
t1/2
tВЅ = 71.03 hours (2.96 days)
1b) How long would it take for 90% of a sample of 131I to decay?
ln
100% 0.693
пЂЅ
пЃ„t
10% 71.03
пЃ„t = 236 hours
2) The following experimental data were obtained for the reaction at 295 K,
H2 + Br2 в†’ 2 HBr
[H2]/M
[Br2]/M
0.10
0.40
0.10
0.010
0.010
0.040
Rate/(M/sec)
1.2x10-3
4.8x10-3
2.4x10-3
2a) Write a rate law consistent with this data.
Rate = k [H2] [Br2]ВЅ
2b) What is the value and units of the rate constant?
1.2x10-3 M/s = k (0.10 M) (0.010 M)
пЃ™
k = 1.2 1/Ms
2c) What is the rate of the reaction when [H2] = 0.30 M and [Br2] = 0.025 M?
rate = 1.2 1/Ms (0.30 M) (0.025 M) = 9x10-3 M/s
2d) Which of the following are a possible mechanism for the above reaction? Show your work.
a)
H2 + Br2 в†’ 2 HBr slow
rate = k [H2] [Br2] doesn’t match
b)
Br2 ⇆ 2 Br
Br + H2 в†’ HBr + H slow
H + Br в†’ HBr fast
rate = k [Br] [H2]
[Br]2
Keq пЂЅ
пѓћ [Br] пЂЅ Keq [Br2 ]
[Br2 ]
rate = k KeqВЅ [H2] [Br2]ВЅ
rate = k’[H2] [Br2]½
possible right answer
c)
H2 + ½ Br2 —> HBr + H slow
H + ½ Br2 —> HBr fast
rate = k [H2] [Br2]ВЅ
possible right answer
3) If a reaction has an activation energy of 112.4 kJ, how hot must you get the reaction to
increase the rate of the reaction 1.5 times? Assume an initial temperature of 25ВєC.
ln
1.5 k 2 112,400 пѓ¦ 1
1 пѓ¶
пѓ§пѓ§
пЂЅ
пЂ­ пѓ·пѓ·
k2
8.314 пѓЁ 298K T2 пѓё
T2 = 300.68 K
4a) What do catalysts do to speed up a reaction?
They provide an alternate route for the reaction to occur.
4b) What term must be added to a reaction to account for the presence of a catalyst?
[kcat [Cat] + 1]
5) Assume that the cooking of an egg is a first order process. Assume further that it takes 3
minutes to half-cook (soft boil) an egg at 100ВєC. If the activation energy of the cooking of an
egg is 75 kJ/mole, how long would it take to soft boil an egg at the top of Mt. Everest where
water boils at 80ВєC?
t1 = 3 min @ 100ВєC and t2 @ 80ВєC
ln
t2
75,000 пѓ¦ 373 пЂ­ 353 пѓ¶
пЂЅ
пѓ§
пѓ·
3 min
8.314 пѓЁ 373 п‚ґ 353 пѓё
t2 = 11.81 min
Exam II
Chemical Equlibria
LeChatlier’s Principle
Acid Dissociation
Base Hydrolysis
Titration Curves
Solubility Product
pH Controlled Solubility
Amphoterism
Distribution Coefficients
Equilibria and LeChatliers Principle
1) For each of the following sets of compounds write the equilibrium reaction that would occur
when the compounds are mixed together.
a) HNO3 and NaBenz
b) NH4Cl and KOH
c) NaCN and NaOH
2a) If H2 and Cl2 are added to a container, both at 2 atm, what will the pressure of HCl be after
the system reaches equilibrium?
H2 + Cl2 в†’ 2 HCl K = 150
2b) If the equilibrium pressure of HCl is 2 atm what pressures of H2 and Cl2 must have been
added to the container originally?
3) If 100 grams of NaF and 70 grams of KOH are added to 250 mL water what is the equilibrium
concentration of all ionic species. Ka = 6.76x10-4
4) What is the pH of 100 mL of 1.2 M HAc after 20 mL of 2 M NaOH is added? Ka = 1.8x10-5
5) What is the pH of a solution that is made from 0.10 M HBenz, 0.25 M NaBenz, and 0.25 M
KOH? Ka = 6.46x10-5
6) What is the pH of a solution where 100 mL of 0.50 M NH4OH is completely neutralized by 65
mL of HCl? Kb = 1.8x10-5
7) Ca(IO3)2 is a slightly soluble precipitate. What would the concentration of IO3- be in a
solution saturated with Ca(IO3)2? Ksp = 1.5 x 10-15
8) What is the pH of a 0.10 M solution of NaAc? Ka = 1.8x10-5
Ac- + H2O ↔ HAc + OH-
9) When ammonia is heated it decomposes to N2 and H2 according to the following reaction,
2 NH3 ↔ N2 + 3 H2
Given 3 atm of NH3 and an equilibrium constant of 3x10-3, what will the final pressure (total
pressure) be in the system?
10) What is the pH of a solution made by adding 0.2 mole of NaAc to 250 mL of 1 M acetic
acid? Ka = 1.8x10-5
11) A 0.10 M solution of calcium ions can be precipitated with 0.10 M NaOH but not in 0.10 M
NH4OH. The solubility product for Ca(OH)2 is 7.88x10-6. Please explain why 0.10 M Ca2+ will
not precipitate in 0.10 M NH4OH (Kb = 1.8x10-5).
12) What is the pH of the resulting solution when 0.25 mole of NaCH3CO2 and 0.15 mole HCl
are added to 200 mL water? The Ka for CH3COOH is 1.8x10-5M.
13) What is the pH of a solution made by adding 0.30 mole NH3(aq) to 0.50 mole NH4Cl and
0.25 mole KOH? Kb for NH3(aq) = 1.8x10-5M
14) What is the concentration of all ionic and molecular species when you add 30 mL of 0.5 M
NaOH to 120 mL of 0.75M HCN? Ka for HCN is 4.8x10-10.
15a) What is the pH of the following solutions when the following amounts of 1.20 M NaOH
are added to 160 mL of 3 M dl Aspartic acid?
H2Asp ↔ H+ + HAspHAsp- ↔ H+ + Asp2-
1.38x10-4
1.51x10-10
0 mL
200 mL
400 mL
700 mL
800 mL
900 mL
15b) An Aspartic acid buffer of pH 5 was made by adding some NaHAsp to 0.5 moles of
H2Asp. How much 1.2 M NaOH or 1.2M HCl must you add to this buffer in order to make a
new buffer of pH 4.5?
16) What is the pH of the solution when 75 mL of 0.8 M dl-Histidine is titrated with the
following volumes of 1.20 M NaOH?
H3His ↔ H+ + H2HisH2His- ↔ H+ + HHis2HHis2- ↔ H+ + His3-
pKa1 = 2.40
pKa2 = 6.04
pKa3 = 9.33
0 mL
30 mL
50 mL
75 mL
100 mL
125 mL
135 mL
150 mL
175 mL
17) How much NH4Cl and NH4OH must you add to 250 mL of water to make a buffer of pH =
8.0? Kb = 1.8x10-5
18) How many moles of sodium acetate must be added to 100 mL of 0.25 M acetic acid to make
a buffer of pH = 4.0? pKa = 4.74 for acetic acid.
19) Another way of making the buffer from 2 above would be to titrate the acid with a base.
What volume of 0.40 M NaOH must be added to 100 mL of 0.25 M acetic acid to make a buffer
of pH = 4.0? pKa = 4.74 for acetic acid.
20) How much 0.5 M NaOH must be added to 0.75 M H3PO4 to make a buffer of pH = 8? Ka1 =
2.12, Ka2 = 7.21, Ka3 = 12.67
Solubility Products
21) Are the following molecules acidic, basic or neutral in aqueous solution?
NaF
Cr(NO3)3
KCl
NH4CN
acid
acid
acid
acid
basic
basic
basic
basic
neutral
neutral
neutral
neutral
can't tell
can't tell
can't tell
can't tell
22) Are the following compounds soluble or insoluble in water?
NaIO3
Cr(OH)3
PbSO4
CaCl2
Soluble
Soluble
Soluble
Soluble
Insoluble
Insoluble
Insoluble
Insoluble
FeSO4
ScCl3
Na2O
PbSO4
Soluble
Soluble
Soluble
Soluble
Insoluble
Insoluble
Insoluble
Insoluble
23) What is the solubility of PbBr2 in pure water? Ksp = 4x10-5
24) What is the solubility of Ca(OH)2 in 0.05 M NaOH? Ksp = 5.5x10-6
25) What is the solubility of Ca3(PO4)2 in water? Ksp = 5.87x10-8
26) What is the solubility of Ca3(PO4)2 in 0.5 M Na3PO4? Ksp = 5.87x10-8
27) What is the solubility of AgCl in a solution of 1 M HCl? Ksp = 1.8x10-10
28) At what pH will the concentration of Cu2+ exceed 0.02 M given the following equilibrium?
Cu(OH)2 ↔ Cu2+ + 2 OH- Ksp = 2.2x10-20
29) How much NH3(aq) must you add to 100 grams of AgCl in order to dissolve all of the AgCl.
Assume a liter of solution and calculate the concentration of NH3(aq).
AgCl ↔ Ag+ + Cl- Ksp = 1.8x10-10
Ag(NH3)2+ ↔ Ag+ + 2 NH3(aq) Keq = 1.6x10-9
30) Copper(I) ions in aqueous solution react with NH3 according to,
Cu+ + 2 NH3 ↔ Cu(NH3)2+ Keq = 6.3x1010
Calculate the solubility of CuBr (Ksp = 5.3x10-9) in a solution in which the equilibrium
concentration of NH3 is 0.185 M.
31) What is the solubility of PbCl2 in a solution of 0.10 M H2S at a pH=0? Ksp =1.6x10-5 for
PbCl2, Ksp = 8x10-28 for PbS, Keq = 1x10-20 for H2S (to 2 H+ + S2-).
PbS ↔ Pb2+ + S2- Ksp = 8x10-28
H2S ↔ 2 H+ + S2- Keq = 1x10-20
32) Calculate the maximum concentration of Fe3+ in an 0.10 M H2S solution buffered at pH=6.
Ksp = 1.6x10-21 for Fe2S3 and Keq = 1x10-20 for H2S (to 2 H+ + S2-).
33) What is the solubility cadmium sulfide (CdS) in a saturated solution of H2S = 0.10 M at pH
= 3?
CdS ↔ Cd2+ + S2- Ksp = 8x10-27
H2S ↔ 2H+ + S2Keq = 1x10-20
34) What is the solubility (silver ion concentration) of silver sulfide (Ag2S) in a saturated
solution of H2S = 0.10 M at pH = 3?
Ag2S ↔ 2 Ag+ + S2H2S ↔ 2H+ + S2-
Ksp = 1.6x10-49
Keq = 1x10-20
35) How much of each precipitate will form and what will the concentration of lead be if
0.75 mole sodium oxalate (Na2C2O4 = Na2Ox) is added to 100 mL of solution containing 0.30 M
Mg(NO3)2 and 0.5 M Pb(NO3)2?
MgOx(s) ↔ Mg2+ + Ox2- Ksp = 8.6x10-5
PbOx(s) ↔ Pb2+ + Ox2- Ksp = 4.8x10-10
36) What is the solubility of silver carbonate (Ag2CO3) in a solution saturated with carbonic acid
(0.034M) and which is buffered at pH = 9?
Ag2CO3 ↔ 2 Ag+ + CO32H2CO3 ↔ 2 H+ + CO32-
Ksp = 8.1x10-12
Ka = 2.02x10-17
Equilibria and LeChatliers Principle
1) For each of the following sets of compounds write the equilibrium reaction that would occur
when the compounds are mixed together.
a) HNO3 and NaBenz
b) NH4Cl and KOH
c) NaCN and NaOH
HBenz ↔ H+ + BenzNH4OH ↔ NH4+ + OHCN- + H2O ↔ HCN + OH-
2a) If H2 and Cl2 are added to a container, both at 2 atm, what will the pressure of HCl be after
the system reaches equilibrium?
H2 + Cl2 ↔ 2 HCl K = 150
x
x
4 atm – 2x
Since the equilibrium constant is large (150) we must shift the ingredients to the right.
This means that 2 atm H2 and 2 atm Cl2 will produce 4 atm HCl. Using this value we can
now solve the equilibrium
[4 пЂ­ 2 x ]2
пЂЅ 150
x2
You do not have to assume that 2x is small. You can solve this exactly.
Take the square root of both sides and then solve for x. So, x = 0.2808
atm so the pressure of HCl = 4 atm – 2(0.2808 atm) = 3.44 atm
2b) If the equilibrium pressure of HCl is 2 atm what pressures of H2 and Cl2 must have been
added to the container originally?
[2atm]2
пЂЅ 150
x2
x = 0.1633 atm H2 and Cl2 left after equilibrium. To make 2 atm of HCl, 1
atm H2 and 1 atm Cl2 must have been consumed. Therefore there must
have been 1 atm + 0.1633 = 1.1633 atm H2 and Cl2 to start.
3) If 100 grams of NaF and 70 grams of KOH are added to 250 mL water what is the equilibrium
concentration of all ionic species. Ka = 6.76x10-4 Note: HF is not ionic but was included.
100g NaF/42 g/mol = 2.381 mol NaF and 70g KOH/56.1 g/mol = 1.248 mol KOH
[ HF ][OH пЂ­ ] [ HF ][4.99 пЂ« x]
1x10 пЂ­14
F- + H2O ↔ HF + OH
пЂЅ
пѓћ [ HF ] пЂЅ 2.822 x10 пЂ­11
пЂ­
пЂ­4
[F ]
[9.52 пЂ­ x]
6.76 x10
[H+] = 1x10-14/[OH-] = 1x10-14/[4.99 M] = 2x10-15 M H+
Ion
Moles Before Rxn
Moles After Rxn
Conc (250 mL total)
Na+
FK+
OHHF
H+
2.381
2.381
1.248
1.248
0
0
2.381
2.381
1.248
1.248
9.52 M
9.52 M
4.99 M
4.99 M
2.822x10-11
2x10-15
From equilibrium
From equilibrium
4) What is the pH of 100 mL of 1.2 M HAc after 20 mL of 2 M NaOH is added? Ka = 1.8x10-5
(1.2M HAc)(0.100 L) = 0.12 mol HAc and (2M)(0.020L) = 0.040 mol NaOH
HAc + NaOH ↔ H2O + NaAc
0.12
0.040
0
-0.04
-0.040
+0.040
0.08
0
0.040
HAc ↔ H+ + Ac0.08-x x 0.04+x
[ H пЂ« ][ Ac пЂ­ ] [ H пЂ« ][0.040 пЂ« x]
пЂЅ
пЂЅ 1.8 x10пЂ­5 пѓћ [ H ] пЂЅ 3.6 x10пЂ­5 pH = 4.44
[ HAc ]
[0.080 пЂ­ x]
5) What is the pH of a solution that is made from 0.10 M HBenz, 0.25 M NaBenz, and 0.25 M
KOH? Ka = 6.46x10-5
Benz- + H2O ↔ OH- +
0.25
0.25
+0.10
-0.10
0.35-x
0.15+x
HBenz
0.10
-0.10
0+x
If you ignore x, [OH-] = 0.15 M and [H+] = 1x10-14/0.15 = 6.66x10-14 so pH = 13.18
6) What is the pH of a solution where 100 mL of 0.50 M NH4OH is completely neutralized by 65
mL of HCl? Kb = 1.8x10-5
If the NH4OH is completely neutralized, then the only thing left in solution is NH4Cl (an acid).
The only question is, what is the concentration of the NH4Cl? Total volume = 100 mL + 65 mL
(0.50 M)(0.100 L) = 0.050 mol NH4OH = 0.050 mol NH4Cl
0.050 mol NH 4 Cl
пЂЅ 0.3030 M NH 4Cl
0.165 L
NH4+ ↔ H+ + NH3
0.3030-x x
x
[H пЂ« ][NH 3 ]
x2
пЂЅ
пЂЅ 5.55x10пЂ­10 пѓћ x пЂЅ 1.297 x10пЂ­5 M H пЂ« pH = 4.89
[NH пЂ«4 ]
0.3030 пЂ­ x
7) Ca(IO3)2 is a slightly soluble precipitate. What would the concentration of IO3- be in a
solution saturated with Ca(IO3)2? Ksp = 1.5 x 10-15
[Ca2+] [IO3-]2 = x(2x)2 = 4x3 = 1.5x10-15
x = 7.21x10-6
8) What is the pH of a 0.10 M solution of NaAc? Ka = 1.8x10-5
Ac- + H2O ↔ HAc + OH[OH  ][HAc]
x2
пЂЅ
пЂЅ 5.55x10пЂ­10 пѓћ x пЂЅ 7.45x10пЂ­6 M OH пЂ­ pH = 8.87
[Ac ]
0.1 пЂ­ x
9) When ammonia is heated it decomposes to N2 and H2 according to the following reaction,
2 NH3 ↔ N2 + 3 H2
Given 3 atm of NH3 and an equilibrium constant of 3x10-3, what will the final pressure (total
pressure) be in the system?
2 NH3 ↔ N2 + 3 H2
3 – 2x
x
3x
x(3 x)3
27 x 4
пЂЅ
пЂЅ 3x10пЂ­3
(3 пЂ­ 2 x ) 2 (3 пЂ­ 2 x ) 2
ignore the 2x on the bottom and solve. x = 0.1778 atm
Total pressure = 3-2x+x+3x = 3+2x = 3+2(0.1778) = 3.3556 atm
10) What is the pH of a solution made by adding 0.2 mole of NaAc to 250 mL of 1 M acetic
acid? Ka = 1.8x10-5
(1 M)(0.250 L) = 0.250 mol HAc
Use the HH equation pH = 4.74 + log 0.20/0.25
pH = 4.64
11) A 0.10 M solution of calcium ions can be precipitated with 0.10 M NaOH but not in 0.10 M
NH4OH. The solubility product for Ca(OH)2 is 7.88x10-6. Please explain why 0.10 M Ca2+ will
not precipitate in 0.10 M NH4OH (Kb = 1.8x10-5).
A 0.10 M Ca2+ solution will require.
[Ca2+] [OH-]2 = [0.10 M] [OH-]2 = 7.88x10-6 [OH-] = 0.00888 M OH- to precipitate
0.10 M NH4OH can only produce,
[OH пЂ­ ][NH пЂ«4 ]
x2
пЂЅ
пЂЅ 1.8x10пЂ­5 пѓћ x пЂЅ 0.001342 M OH пЂ­ this is not enough to ppt. Ca2+
[NH 4OH]
0.1 пЂ­ x
12) What is the pH of the resulting solution when 0.25 mole of NaCH3CO2 and 0.15 mole HCl
are added to 200 mL water? The Ka for CH3COOH is 1.8x10-5M.
Ac- + H+ в†’ HAc
[H пЂ« ][AcпЂ­ ] [H пЂ« ][0.10 пЂ« x]
0.25 0.15
пЂЅ
пЂЅ 1.8x10пЂ­5 пѓћ [H пЂ« ] пЂЅ 2.7 x10пЂ­5 pH = 4.57
[HAc]
[0.15
пЂ­
x]
-0.15 -0.15 +0.15
0.10
0
0.15
13) What is the pH of a solution made by adding 0.30 mole NH3(aq) to 0.50 mole NH4Cl and
0.25 mole KOH? Kb for NH3(aq) = 1.8x10-5M
NH4OH ↔ NH4+ + OH0.30
0.50
0.25
Shift
+0.25
-0.25
-0.25
0.55
0.25
0
в†’ trickle
0.55-x
0.25+x
0+x
[OH пЂ­ ][NH пЂ«4 ] x(0.25 пЂ« x)
пЂЅ
пЂЅ 1.8x10пЂ­5 пѓћ x пЂЅ 3.96 x10пЂ­5 M OH пЂ­
[NH 4OH]
0.55 пЂ­ x
pH = 9.60
14) What is the concentration of all ionic and molecular species when you add 30 mL of 0.5 M
NaOH to 120 mL of 0.75M HCN? Ka for HCN is 4.8x10-10.
(0.5 M)(0.030 L) = 0.015 mol NaOH and (0.75 M)(0.120 L) = 0.090 mol HCN
HCN + OH- ↔ H2O + CN0.09
0.015
-0.015 -0.015
+0.015
0.075
0
0.015
[H+] = 2.4x10-9
Ion
+
Na
OHHCN
H+
CN-
[ H пЂ« ][CN пЂ­ ] [ H пЂ« ][0.015 пЂ« x ]
пЂЅ
пЂЅ 4.8 x10 пЂ­10
[ HCN ]
[0.075 пЂ­ x]
[OH-] = 1x10-14/[2.4x10-9] = 4.17x10-6 M OHMoles Before Rxn
0.015
0.015
0.090
0
0
Moles After Rxn
Conc (150 mL total)
0.015
0.10 M
4.17x10-6
0.05 M
2.4x10-9
0.05 M
From equilibrium
0.075
From equilibrium
0.015
15a) What is the pH of the following solutions when the following amounts of 1.20 M NaOH
are added to 160 mL of 3 M dl Aspartic acid?
H2Asp ↔ H+ + HAspHAsp- ↔ H+ + Asp2-
0 mL
1.38x10-4
1.51x10-10
pKA1 = 3.86
pKA2 = 9.82
x2
пЂЅ 1.38x10пЂ­4 пѓћ x пЂЅ 0.002035 M H пЂ« pH = 1.69
3пЂ­ x
M1V1 = M2V2
(3 M)(0.160 L) = (1.20 M)(V2)
V2 = 400 mL
Also, (3 M)(0.160 L) = 0.48 mol H 2Asp
200 mL
400 mL
700 mL
pH = 3.86
pH пЂЅ
pK A1 пЂ« pK A2 3.86 пЂ« 9.82
пЂЅ
2
2
300 mL into 2nd Region (1.2 M)(0.300 L) = 0.36 mol OHpH пЂЅ 9.82 пЂ« log
800 mL
[Asp2- ] пЂЅ
0.36
0.48 пЂ­ 0.36
pH = 10.30
0.48 mol
1x10-14
пЂЅ 0.50 M and Kb пЂЅ
пЂЅ 6.62 x10 пЂ­5
0.960 L
1.51x10-10
x2
пЂЅ 6.62 x10пЂ­5
0.50 пЂ­ x
900 mL
pH = 6.84
x = 5.75x10-3 M OH-
(0.100L)(1.20M) = 0.12 mol OH0.12 mol OH пЂ­
пЂЅ 0.1132 M OH пЂ­
1.060 L
pH = 11.76
pH = 13.05
15b) An Aspartic acid buffer of pH 5 was made by adding some NaHAsp to 0.5 moles of
H2Asp. How much 1.2 M NaOH or 1.2M HCl must you add to this buffer in order to make a
new buffer of pH 4.5?
First, calculate the composition of the existing buffer, pH пЂЅ pKa пЂ« log
B
A
[ HAsp пЂ­ ]
[HAsp-] = 6.90 mol
0.5
Now find out how much acid must be added to make new buffer (pH is going down so add acid).
5 пЂЅ 3.86 пЂ« log
4.5 пЂЅ 3.86 пЂ« log
6.90 пЂ­ x
0. 5 пЂ« x
x = 0.8792 mol H+ = (1.2 M)(V)
V = 732.7 mL of 1.2 M HCl
16) What is the pH of the solution when 75 mL of 0.8 M dl-Histidine is titrated with the
following volumes of 1.20 M NaOH?
H3His ↔ H+ + H2HisH2His- ↔ H+ + HHis2HHis2- ↔ H+ + His3-
M1V1 = M2V2
(0.8 M)(0.075 L) = (1.20 M)(V2)
V2 = 50 mL
pKa1 = 2.40
pKa2 = 6.04
pKa3 = 9.33
Also, (0.8 M)(0.075 L) = 0.060 mol H 3His
0 mL
x2
пЂЅ 3.98x10пЂ­3 пѓћ x пЂЅ 0.0564 M H пЂ«
0. 8 пЂ­ x
30 mL
(1.2 M)(0.030 L) = 0.036 mol OHpH пЂЅ 2.40 пЂ« log
0.036
0.060 пЂ­ 0.036
pK A1 пЂ« pK A2 2.40 пЂ« 6.04
пЂЅ
2
2
pH = 1.248
pH = 2.58
50 mL
pH пЂЅ
75 mL
pH = 6.04
100 mL
pH пЂЅ
125 mL
pH = 9.33
135 mL
35 mL into 3rd Region (1.2 M)(0.035 L) = 0.042 mol OH-
pK A2 пЂ« pK A3 6.04 пЂ« 9.33
пЂЅ
2
2
pH пЂЅ 9.33 пЂ« log
150 mL
0.042
0.060 пЂ­ 0.042
pH = 7.69
pH = 9.70
0.060 mol
1x10-14
[His ] пЂЅ
пЂЅ 0.2667 M and Kb пЂЅ
пЂЅ 2.14 x10 пЂ­5
-10
0.225 L
4.68x10
3-
x2
пЂЅ 2.14 x10пЂ­5
0.2667 пЂ­ x
175 mL
pH = 4.22
x = 2,39x10-3 M OH-
(0.025 L)(1.20M) = 0.030 mol OH0.030 mol OH пЂ­
пЂЅ 0.120 M OH пЂ­
pH = 13.08
0.250 L
pH = 11.38
17) How much NH4Cl and NH4OH must you add to 250 mL of water to make a buffer of pH =
8.0? Kb = 1.8x10-5
Using the base version of the HH equation;
6 пЂЅ 4.74 пЂ« log
[NH 4Cl]
[NH 4Cl] 18.19 0.1819
пѓћ
пЂЅ
пЂЅ
[NH 4OH]
[NH 4 OH]
1
0.01
Adding 18.19 moles to 250 mL of water will make a solution that is 72.76M, so divide
top and bottom by 100. Add 0.1819 mol NH4Cl and 0.010 mole NH4OH to 250 mL of
water to make a buffer of pH = 8
18) How many moles of sodium acetate must be added to 100 mL of 0.25 M acetic acid to make
a buffer of pH = 4.0? pKa = 4.74 for acetic acid.
(0.25 M HAc)(0.100 L) = 0.025 mol HAc
(x)
4 пЂЅ 4.74 пЂ« log
пѓћ x пЂЅ 0.00455 mol NaAc
(0.025 mol HAc)
19) Another way of making the buffer from 18 above would be to titrate the acid with a base.
What volume of 0.40 M NaOH must be added to 100 mL of 0.25 M acetic acid to make a buffer
of pH = 4.0? pKa = 4.74 for acetic acid.
4 пЂЅ 4.74 пЂ« log
(x)
пѓћ x пЂЅ 0.003849 mol NaOH
(0.025 - x)
(0.40 M)(V) = 0.003849 mol NaOH
V = 9.62 mL
20) How much 0.5 M NaOH must be added to 80 mL of 0.75 M H3PO4 to make a buffer of pH =
8? Ka1 = 2.12, Ka2 = 7.21, Ka3 = 12.67
The pH is nearest the 2nd pKa so the buffer is in Region II. You must add,
(0.75 M)(0.080 L) = (0.50 M)(V) V = 120 mL so add 120 mL to get to the starting point.
8 пЂЅ 7.21 пЂ« log
(x)
пѓћ x пЂЅ 0.05163 mol NaOH = (0.50 M)(V)
(0.060 - x)
Total volume = 103.25 mL + 120 mL = 223.35 mL NaOH
V = 103.25 mL
Solubility Products
21) Are the following molecules acidic, basic or neutral in aqueous solution?
NaF
Cr(NO3)3
KCl
NH4CN
acid
acid
acid
acid
basic
basic
basic
basic
neutral
neutral
neutral
neutral
can't tell
can't tell
can't tell
can't tell
22) Are the following compounds soluble or insoluble in water?
NaIO3
Cr(OH)3
PbSO4
CaCl2
Soluble
Soluble
Soluble
Soluble
Insoluble
Insoluble
Insoluble
Insoluble
FeSO4
ScCl3
Na2O
Hg2SO4
Soluble
Soluble
Soluble
Soluble
Insoluble
Insoluble
Insoluble
Insoluble
23) What is the solubility of PbBr2 in pure water? Ksp = 4x10-5
[Pb2+] [Br-]2 = x(2x)2 = 4x3 = 4x10-5
x = 2.15x10-2 M
24) What is the solubility of Ca(OH)2 in 0.05 M NaOH? Ksp = 5.5x10-6
[Ca2+] [0.50 M]2 = 5.5x10-6
x = 2.20x10-5
25) What is the solubility of Ca3(PO4)2 in water? Ksp = 5.87x10-8
[Ca2+]3 [PO43-]2 = (3x)3(2x)2 = 108x5 = 5.87x10-8
x = 1.40x10-2 M
26) What is the solubility of Ca3(PO4)2 in 0.5 M Na3PO4? Ksp = 5.87x10-8
[Ca2+]3 [PO43-]2 = (3x)3(0.50)2 = 0.75x3 = 5.87x10-8
x = 4.278x10-3 M
27) What is the solubility of AgCl in a solution of 1 M HCl? Ksp = 1.8x10-10
[Ag+] [1 M] = 1.8x10-10
[Ag+] = 1.8x10-10
28) At what pH will the concentration of Cu2+ exceed 0.02 M given the following equilibrium?
Cu(OH)2 ↔ Cu2+ + 2 OH- Ksp = 2.2x10-20
[Cu2+] [OH-]2 = [0.02 M] [OH-]2 = 2.2x10-20 [OH-] = 1.05x10-9 pH = 6.02
29) How much NH3(aq) must you add to 100 grams of AgCl in order to dissolve all of the AgCl.
Assume a liter of solution and calculate the concentration of NH3(aq).
AgCl ↔ Ag+ + Cl- Ksp = 1.8x10-10
Ag(NH3)2+ ↔ Ag+ + 2 NH3(aq) Keq = 1.6x10-9
If you reverse the bottom reaction and add it to the top reaction you get the following (see
below).
AgCl ↔ Ag+ + ClKsp = 1.8x10-10
+
+
Ag + 2 NH3(aq) ↔ Ag(NH3)2
Keq = 6.25x10-8
AgCl + 2 NH3(aq) ↔ Ag(NH3)2+ + ClK = 0.1125
This means that all the silver in solution will end up being Ag(NH3)2+. If you dissolve all of the
AgCl in 1 liter of solution, the concentration of Ag(NH3)2+ must be,
100 g AgCl/143.35 g/mol = 0.6976 mole AgCl =>
0.6976 mole Ag+ becomes,
0.6976 mole Ag(NH3)2+ and 0.6976 mole Cl- in 1 liter of solution.
So, [Ag(NH3)2+] = 0.6976 M and [Cl-] = 0.6976 M
Solving for the concentration of NH3(aq),
[Ag(NH 3 ) 2пЂ« ] [Cl пЂ­ ] [0.6976 M] [0.6976 M]
пЂЅ
пЂЅ 0.1125 [NH3] = 2.08 M
[NH 3 ]2
[NH3 ]2
30) Copper(I) ions in aqueous solution react with NH3 according to,
Cu+ + 2 NH3 ↔ Cu(NH3)2+ Keq = 6.3x1010
Calculate the solubility of CuBr (Ksp = 5.3x10-9) in a solution in which the equilibrium
concentration of NH3 is 0.185 M.
CuBr ↔ Cu+ + BrCu+ + 2 NH3 ↔ Cu(NH3)2+
CuBr + 2 NH3 ↔ Cu(NH3)2+ + Br-
Ksp = 5.3x10-9
Keq = 6.3x1010
K = 333.9
For every Cu(NH3)2+ there must be one Br-, so [Cu(NH3)2+] = [Br-] = x
x2
x2
пЂЅ
пЂЅ 333.9 x = 3.38 M CuBr
[NH 3 ]2 [0.185]2
31) What is the solubility of PbCl2 in a solution of 0.10 M H2S at a pH=0? Ksp =1.6x10-5 for
PbCl2, Ksp = 8x10-28 for PbS, Keq = 1x10-20 for H2S (to 2 H+ + S2-).
PbS ↔ Pb2+ + S2H2S ↔ 2 H+ + S2-
Ksp = 8x10-28
Keq = 1x10-20
Note: the Pb2+ concentration can only be as high as the least soluble compound so the
solubility of PbS will determine the concentration of Pb2+
[H+]2 [S2-]/[H2S] = 1x10-20
[S2-] = 1x10-20 [H2S]/[H+]2 = 1x10-20 [0.1]/[1]2 = 1x10-21 M S2Substitute this result into the following equation,
[Pb2+] [S2-] = 8x10-28
[Pb2+] = 8x10-28/[S2-] = 8x10-28/[1x10-21] = 8x10-7 M
Now, all equilibriums must be satisfied simultaneously. Since this is the maximum
concentration of Pb2+ that can be in solution, we can use this value to calculate the Clconcentration from the PbCl2.
PbCl2(s) ↔ Pb2+ + 2 Cl[Pb2+] [Cl-]2 = 1.6x10-5 => [8x10-7 M] [Cl-]2 = 1.6x10-5 [Cl-] = 4.47 M
This concentration of Cl- is fairly high. It can only be this high if all of the PbCl2
dissolved. So, if you put H2S in a solution of PbCl2, all of the PbCl2 would dissolve and
become PbS.
32) Calculate the maximum concentration of Fe3+ in an 0.10 M H2S solution buffered at pH=6.
Ksp = 1.6x10-21 for Fe2S3 and Keq = 1x10-20 for H2S (to 2 H+ + S2-).
[H+]2 [S2-]/[H2S] = 1x10-20
[S2-] = 1x10-20 [H2S]/[H+]2 = 1x10-20 [0.1]/[1x10-6]2 = 1x10-9 M S2Substitute this result into the following equation,
Fe2S3(s) ↔ 2 Fe3+ + 3 S2[Fe3+]2 [S2-]3 = 1.6x10-21
[Fe3+] = (1.6x10-21/[S2-]3)ВЅ = (1.6x10-21/[1x10-9]3)ВЅ = 1264.9 M Fe3+ => 632.45 M Fe2S3
33) What is the solubility cadmium sulfide (CdS) in a saturated solution of H2S = 0.10 M at pH
= 3?
CdS ↔ Cd2+ + S2- Ksp = 8x10-27
H2S ↔ 2H+ + S2Keq = 1x10-20
[H+]2 [S2-]/[H2S] = 1x10-20
[S2-] = 1x10-20 [H2S]/[H+]2 = 1x10-20 [0.1]/[1x10-3]2 = 1x10-15 M S2Substitute this result into the following equation,
[Cd2+] [S2-] = 8x10-27
[Cd2+] = 8x10-27/[S2-] = 8x10-27/[1x10-17] = 8x10-12 M
34) What is the solubility (silver ion concentration) of silver sulfide (Ag2S) in a saturated
solution of H2S = 0.10 M at pH = 3?
Ag2S ↔ 2 Ag+ + S2- Ksp = 1.6x10-49
H2S ↔ 2H+ + S2Keq = 1.1x10-20
[H+]2 [S2-]/[H2S] = 1x10-20
[S2-] = 1x10-20 [H2S]/[H+]2 = 1x10-20 [0.1]/[1x10-3]2 = 1x10-15 M S2Substitute this result into the following equation,
[Ag+]2 [S2-] = 1.6x10-49
[Ag+] = 1.6x10-49/[S2-] = 1.6x10-49/[1x10-17] = 1.6x10-32 M Ag+
35) How much of each precipitate will form and what will the concentration of lead be if
0.075 mole sodium oxalate (Na2C2O4 = Na2Ox) is added to 100 mL of solution containing 0.30
M Mg(NO3)2 and 0.5 M Pb(NO3)2?
MgOx(s) ↔ Mg2+ + Ox2PbOx(s) ↔ Pb2+ + Ox2-
Ksp = 8.6x10-5
Ksp = 4.8x10-10
Neither compound is very soluble so you can begin by assuming that 100% of the Pb2+ and
Mg2+ will react to form solids.
(0.30 M)(0.100 L) = 0.03 mole Mg(NO3)2
(0.50 M)(0.100 L) = 0.05 mole Pb(NO3)2
Start with the least soluble compound (PbOx),
You have 0.075 mole Ox2- and 0.50 mole of Pb2+, so all of the Pb2+ will react to make 0,50
mole of PbOx and leave 0.025 mole of Ox2- in solution.
The remaining 0.025 mole of Ox2- will react with the 0.03 mole Mg(NO3)2 leaving 0.005
mole of Mg2+ in solution and no Ox2-. The [Mg2+] = (0.005 mol)(0.100 L) = 0.05 M Mg2+.
[Mg2+] [Ox2-] = [0.05 M] [Ox2-] = 8.6x10-5
[Ox2-] = 1.72x10-3 M
[Pb2+] [Ox2-] = [Pb2+] [1.72x10-3 M] = 4.8x10-10 [Pb2+] = 2.79x10-7 M
36) What is the solubility of silver carbonate (Ag2CO3) in a solution saturated with carbonic acid
(0.034M) and which is buffered at pH = 9?
Ag2CO3 ↔ 2 Ag+ + CO32H2CO3 ↔ 2 H+ + CO32-
Ksp = 8.1x10-12
Ka = 2.02x10-17
[H пЂ« ]2 [CO32пЂ­ ] (1x10пЂ­9 M) 2 (CO32пЂ­ )
пЂЅ
пЂЅ 2.02x10пЂ­17 пѓћ [CO32пЂ­ ] пЂЅ 0.6868 M
[H 2CO3 ]
0.034M
[Ag+]2 [CO32-] = [Ag+]2 [0.6868 M] = 8.1x10-12
[Ag+] = 3.43x10-6 M
Solubility of Ag2CO3 = (3.43x10-6 M)/2 = 1.72x10-6 M Ag2CO3
Distribution Coefficients
1. If 0.500 grams of caffeine is dissolved in 100 mL of water, what percentage of caffeine can be
separated from the water using a single 40 mL sample of methylene chloride. P = 4.6
2. Benzoic acid can be separated from water using octanol as the organic solvent. The
distribution coefficient for this water/octanol system is P = 1.87. Assuming that 1 gram of
benzoic acid has been dissolved in 100 mL of water, how many 20 mL extractions must be done
to extract 60+ percent of the benzoic acid from the water?
3. A 0.100 gram sample of phthalic acid was dissolved in 100 mL of water. When 25 mL of
diethyl ether was used to extract the phthalic acid, 0.042 grams of phthalic acid were recovered.
What is the distribution coefficient for this extraction?
4. Draw a graph (using Excel) that shows the percentage of solute (extractable compound)
extracted from water using increasingly large volumes of organic solvent. Express the amount
extracted as a percent of the total. Make the spreadsheet very general. Allow for the input of
various distribution coefficients, amount of dissolved compound, and volume of aqueous
solution.
‫‪݈2‬ܥ‪2‬ܪܥܮ ݉ ‪⁄40‬ݔ‬
‫ܱ‪2‬ܪܮ ݉ ‪)⁄100‬ݔ ‪(0.5 −‬‬
‫ݔ‬
‫ݔ ‪0.5 −‬‬
‫= ‪1.84‬‬
‫‪1.‬‬
‫= ‪4.6‬‬
‫‪0.92 – 1.84x = x‬‬
‫‪0.92 = 2.84x‬‬
‫‪0.92‬‬
‫݂݂ܽܥ݃‪= 0.3239‬‬
‫‪2.84‬‬
‫=ݔ‬
‫݃‪0.3239‬‬
‫݀݁ݐܿܽݎݐݔ݂݂݁݁݊݅݁ܽܥ‪× 100 = 64.78 %‬‬
‫݃‪0.500‬‬
‫݈݋݊ܽݐܱܿܮ ݉ ‪⁄20‬ݔ‬
‫ܱ‪2‬ܪܮ ݉ ‪)⁄100‬ݔ ‪(1 −‬‬
‫ݔ‬
‫ݔ‪1−‬‬
‫‪2. First extraction‬‬
‫= ‪1.87‬‬
‫= ‪0.374‬‬
‫‪0.374 – 0.374x = x‬‬
‫‪0.374 = 1.374x‬‬
‫‪0.374‬‬
‫ݏ݊݅ܽ ݉݁ݎ݃‪0.7278‬݋ݏ ‪݅ܿܽܿ݅݀,‬݋ݖ݊݁ܤ݃‪= 0.2722‬‬
‫‪1.374‬‬
‫݃‪0.2722‬‬
‫݀݁ݐܿܽݎݐݔ݁ܿ݅݋ݖ݊݁ܤ‪× 100 = 27.22 %‬‬
‫݃‪1‬‬
‫݈݋݊ܽݐܱܿܮ ݉ ‪⁄20‬ݔ‬
‫ܱ‪2‬ܪܮ ݉ ‪)⁄100‬ݔ ‪(0.7278 −‬‬
‫ݔ‬
‫ݔ ‪0.7278 −‬‬
‫=ݔ‬
‫‪Second extraction‬‬
‫= ‪1.87‬‬
‫= ‪0.374‬‬
‫‪0.2722 – 0.374x = x‬‬
‫‪0.2722 = 1.374x‬‬
‫‪0.2722‬‬
‫ݏ݊݅ܽ ݉݁ݎ݃‪0.5291‬݋ݏ ‪݅ܿܽܿ݅݀,‬݋ݖ݊݁ܤ݃‪= 0.1987‬‬
‫‪1.374‬‬
‫=ݔ‬
Third extraction
0.2722 + 0.1987Эѓ
× 100 = 47.09 %‫݀݁ݐܿܽݎݐݔ݁ܿ݅݋ݖ݊݁ܤ‬
1Эѓ
1.87 =
‫ݔ‬⁄20 ݉ ‫݈݋݊ܽݐܱܿܮ‬
(0.5291 − ‫)ݔ‬⁄100 ݉ ‫ܪܮ‬2ܱ
0.374 =
‫ݔ‬
0.5291 − ‫ݔ‬
0.1979 – 0.374x = x
0.1979 = 1.374x
‫=ݔ‬
3.
0.1979
= 0.1440݃‫݀݅ܿܽܿ݅݋ݖ݊݁ܤ‬, ‫݋ݏ‬0.3851݃‫ݏ݊݅ܽ ݉݁ݎ‬
1.374
0.2722 + 0.1987 + 0.1440Эѓ
× 100 = 61.49 %‫݀݁ݐܿܽݎݐݔ݁ܿ݅݋ݖ݊݁ܤ‬
1Эѓ
ЬІ=
0.042݃⁄25 ݉ ‫ݐ݁ܮ‬ℎ݁‫ݎ‬
= 2.897
(0.100 − 0.042݃)⁄100 ݉ ‫ܪܮ‬2ܱ
4. You should begin by solving the extraction equation so that the grams of extracted solute is
in terms of the volume of extracting solvent.
ЬІ=
݃݁‫݁ݐݑ݈݋ݏ݀݁ݐܿܽݎݐݔ‬⁄‫݈݋ݒ‬. ݁‫ݐ݊݁ݒ݈݋ݏ݃݊݅ݐܿܽݎݐݔ‬
(݃‫݁ݐݑ݈݋ݏ‬− ݃݁‫)݁ݐݑ݈݋ݏ݀݁ݐܿܽݎݐݔ‬⁄‫݈݋ݒ‬. ‫ݎ݁ݐܽݓ‬
It will be difficult to solve this equation using the long words to describe each part so let’s
simplify the variables,
g extracted solute = gx
g solute = gs
vol. extracting solvent = vx
vol. water = vw
Our equation is now,
ЬІ=
݃‫ݔ‬⁄‫ݔݒ‬
(݃‫ݏ‬− ݃‫)ݔ‬⁄‫ݓݒ‬
We want to rearrange this equation so that we have gx in terms of vx. So, rearranging,
ܲ(݃‫ݏ‬− ݃‫ݔ݃ )ݔ‬
=
‫ݓݒ‬
‫ݔݒ‬
ܲ ∙ ‫ݏ݃(ݔݒ‬− ݃‫ݓݒ ∙ ݔ݃ = )ݔ‬
ܲ ∙ ‫ݏ݃ ∙ ݔݒ‬− ܲ ∙ ‫ݓݒ ∙ ݔ݃ = ݔ݃ ∙ ݔݒ‬
ܲ ∙ ‫ ݓݒ ∙ ݔ݃ = ݏ݃ ∙ ݔݒ‬+ ܲ ∙ ‫ݔ݃ ∙ ݔݒ‬
ܲ ∙ ‫ ݓݒ( = ݏ݃ ∙ ݔݒ‬+ ܲ ∙ ‫ݔ݃ ∙ )ݔݒ‬
ܲ ∙ ‫ݏ݃ ∙ ݔݒ‬
= ݃‫ݔ‬
(‫ ݓݒ‬+ ܲ ∙ ‫)ݔݒ‬
This is our equation. We can test is by using the number given in problem #1.
P = 4.6
vx = 40 mL
gs = 0.500 gram sample
vw = 100 mL
4.6 в€™ 40 в€™ 0.500
= 0.3239݃݁‫݂݋ݐݑ݋݀݁ݐܿܽݎݐݔ‬0.5000 (64.78%)
(100 + 4.6 в€™ 40)
Finally, we can calculate the percentage directly by dividing our answer (gx) by the grams of
sample (gs) and multiplying by 100.
ܲ ∙ ‫ݏ݃ ∙ ݔݒ‬
(‫ ݓݒ‬+ ܲ ∙ ‫)ݔݒ‬
× 100 = %݁‫݀݁ݐܿܽݎݐݔ‬
݃‫ݏ‬
So now, we need to make an Excel spreadsheet to do this calculation for us for various volumes
of extracting solvent. The Excel file that does this for you is online with this problem set. You
will notice that the amount extracted increases with the volume of extracting solvent but it also
increases when you lower the volume of the water. So, it is easier to extract stuff out of a more
concentrated water solution. You will also get more extracted if the distribution coefficient
increases.
Chemistry 121
Second Exam
Name_____________
April 18, 2007
CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit.
You may use a calculator.
Question
Credit
1(10)
2(15)
3(45)
4(20)
5(10)
Total
1) Which of the following compounds is the most soluble in water? Circle one. Show your
work.
BaSO4
Ksp = 1.6x10-10
PbI2
Ksp = 1.2x10-13
Ag3PO4
Ksp = 1x10-22
2) I found a bottle marked “Acetic Acid” in our back room, but there was no indication of
concentration. I wanted to know the concentration so I measured the pH and found it to be pH =
2.87. What was the concentration of the acetic acid in the bottle? Ka = 1.8x10-5
3) Calculate the pH of 100 mL of 1.2 M Arsenous acid upon addition of the following amounts
of 0.8 M NaOH
H3AsO4 ↔ H2AsO4- + H+
H2AsO4- ↔ HAsO42- + H+
HAsO42- ↔ AsO43- + H+
Ka1= 5.62x10-5
Ka2= 1.7x10-8
Ka3= 3.95x10-12
a) 0 mL
b) 150 mL
c) 225 mL
d) 400 mL
e) 450 mL
f) 600 mL
f) If you wanted to titrate to the first endpoint which indicator would you use? Circle one.
Indicator
Methyl Red
Bromothymol Blue
Cresol Red
Phenolphthalein
Range
4.8 - 6.0
6.2 - 7.6
7.0 - 8.8
8.2 - 10
g) How much 0.8 M NaOH must you add to 100 mL of 1.2 M Arsenous acid to make a buffer of
pH = 11.6?
4) What is the solubility of magnesium carbonate (MgCO3) in a solution saturated with carbonic
acid (0.034M) and which is buffered at pH = 8?
MgCO3 ↔ Mg2+ + CO32H2CO3 ↔ 2 H+ + CO32-
Ksp = 8.1x10-12
Ka = 2.02x10-17
5) Define amphoterism. Draw a graph of pH vs. concentration as part of your example.
Chemistry 121
Second Exam
Name_____________
April 18, 2007
CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit.
You may use a calculator.
Question
Credit
1(10)
2(15)
3(45)
4(20)
5(10)
Total
1) Which of the following compounds is the most soluble in water? Circle one. Show your
work.
BaSO4
Ksp = 1.6x10-10
(x)(x) = x2 = 1.6x10-10
x = 1.26x10-5
PbI2
Ksp = 1.2x10-13
(x)(2x)2 = 4x3 = 1.2x10-13
x = 3.11x10-5
Ag3PO4
Ksp = 1x10-22
(3x)3(x) = 27x4 = 1x10-22
x = 1.39x10-6
2) I found a bottle marked “Acetic Acid” in our back room, but there was no indication of
concentration. I wanted to know the concentration so I measured the pH and found it to be pH =
2.87. What was the concentration of the acetic acid in the bottle? Ka = 1.8x10-5
-log[H+] = 2.87
[H+] = 1.35x10-3 M also, when HAc dissociates, [H+] = [Ac-]
HAc ↔ H+
+
Ac-3
x
1.35x10
1.35x10-3
пЂЁ1.35x10 пЂ©
-3 2
x
пЂЅ 1.6 x10 пЂ­5
x = 0.1013 M HAc
3) Calculate the pH of 100 mL of 1.2 M Arsenous acid upon addition of the following amounts
of 0.8 M NaOH
H3AsO4 ↔ H2AsO4- + H+
H2AsO4- ↔ HAsO42- + H+
HAsO42- ↔ AsO43- + H+
a) 0 mL
x2
пЂЅ 5.62 x10 пЂ­5
1.2 пЂ­ x
Ka1= 5.62x10-5 pKA1 = 4.25
Ka2= 1.7x10-8 pKA2 = 7.74
Ka3= 3.95x10-12 pKA3 = 11.40
x = 8.21x10-3 pH = 2.09
M1V1 = M2V2
(1.2M)((0.100L) = (0.08 M)(V2)
V2 = 150 mL for each region
Also (1.2M)(0.100L) = 0.12 mol H3AsO4
pK A1 пЂ« pK A2 4.25 пЂ« 7.74
пЂЅ
2
2
b) 150 mL
pH пЂЅ
c) 225 mL
pH = 7.74
d) 400 mL
100 mL into 3rd Region (0.8 M)(0.100 L) = 0.08 mol OHpH пЂЅ 11.40 пЂ« log
e) 450 mL
[AsO34пЂ­ ] пЂЅ
0.08
0.12 пЂ­ 0.08
pH = 11.70
0.12mol
1x10-14
пЂЅ 0.2182M and Kb пЂЅ
пЂЅ 2.53 x10 пЂ­3
0.550L
3.95x10-12
x2
пЂЅ 2.54 x10 пЂ­3
0.2182 пЂ­ x
f) 600 mL
pH = 5.995
x = 7.43x10-2 M OH-
(0.150L)((0.80M) = 0.12 mol OH0.12 mol OH пЂ­
пЂЅ 0.1714 M OH пЂ­
0.700 L
pH = 12.87
pH = 13.23
g) If you wanted to titrate to the first endpoint which indicator would you use? Circle one.
Indicator
Methyl Red
Bromothymol Blue
Cresol Red
Phenolphthalein
Range
4.8 - 6.0
6.2 - 7.6
7.0 - 8.8
8.2 – 10
since pH = 5.995 @ first endpoint
choose an indicator that changes
just after 5.995
g) How much 0.8 M NaOH must you add to 100 mL of 1.2 M Arsenous acid to make a buffer of
pH = 11.6?
The pH is in the 3rd region so you will have to add 300 mL of NaOH to your answer,
11.6 пЂЅ 11.4 пЂ« log
x
0.12 пЂ­ x
x = 0.7357 mol OH-
0.7357 mol OH- = (0.8 M)(V)
V = 91.97 mL + 300 mL = 391.97 mL total
4) What is the solubility of magnesium carbonate (MgCO3) in a solution saturated with carbonic
acid (0.034M) and which is buffered at pH = 8?
MgCO3 ↔ Mg2+ + CO32H2CO3 ↔ 2 H+ + CO32-
Ksp = 8.1x10-12
Ka = 2.02x10-17
[H пЂ« ][CO32пЂ­ ] (1x10пЂ­8 M)(CO32пЂ­ )
пЂЅ
пЂЅ 2.02x10пЂ­17 пѓћ [CO32пЂ­ ] пЂЅ 6.868x10пЂ­3 M
[H 2 CO3 ]
0.034M
[Mg2+] [CO32-] = [Mg2+] [6.868x10-3 M] = 8.1x10-12
[Mg2+] = 1.18x10-9 M
5) Define amphoterism. Draw a graph of pH vs. concentration as part of your example.
Concentration of Metal Ion
Amphoterism = the ability of a metal ion to be soluble in acidic and basic solutions
0
7
pH
14
Chemistry 121
Second Exam
Name ________________
April 10, 2008
CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit.
You may use a calculator.
Question
Credit
1(10)
2(15)
3(45)
4(20)
5(10)
Total
1) Which of the following compounds is the most soluble in water? Show your work.
BaF2
Ksp = 1.84x10-9
Mg3(PO4)2
Ksp = 1.04Г—10-15
Ag3PO4
Ksp = 1.2x10-15
2) Please write the equilibrium reaction that will occur when the following compounds are
mixed. Write all reactions as dissociations where appropriate.
NaOH + HAc
Na3PO4 + HCl
HCl + KOH
3) Calculate the pH of 80 mL of 1.5 M Carbonic acid upon addition of the following amounts of
0.6 M NaOH
H2CO3 ↔ H+ + HCO3HCO3- ↔ H+ + CO32-
Ka1= 4.3x10-7
Ka2= 4.7x10-11
a) 0 mL
b) 100 mL
c) 150 mL
d) 200 mL
e) 300 mL
f) 365 mL
g) 400 mL
h) 450 mL
i) How much 0.8 M NaOH must be added to 80 mL of 1.5 M H2CO3 to make a buffer of pH =
10?
4) What is the solubility of Ag2S in a saturated solution of H2S = 0.01 M at pH = 3?
Ag2S ↔ 2 Ag+ + S2- Ksp = 6.0x10-50
H2S ↔ 2 H+ + S2- Keq = 6.84x10-23
5a) Iron can be separated from copper using pH controlled solubility in H2S. CuS is not soluble
in acid solutions and FeS is soluble. Draw and approximate ion separation curve for CuS and
FeS.
5b) Define amphoterism and give an example.
Chemistry 121
Second Exam
Name
April 10, 2008
CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit.
You may use a calculator.
Question
Credit
1(10)
2(15)
3(45)
4(20)
5(10)
Total
1) Which of the following compounds is the most soluble in water? Show your work.
BaF2
Ksp = 1.84x10-9
(x)(2x)2 = 4x3 = 1.84x10-9
x = 7.72x10-4 most soluble
Mg3(PO4)2
Ksp = 1.04Г—10-15
(3x)3(2x)2 = 108x5 = 1.04x10-15 x = 3.95x10-4
Ag3PO4
Ksp = 1.2x10-15
(3x)3 (x) = 27x4 = 1.2x10-15 x = 8.16x10-5
2) Please write the equilibrium reaction that will occur when the following compounds are
mixed. Write all reactions as dissociations where appropriate.
NaOH + HAc
H2O + Ac- ↔ HAc + OH-
Na3PO4 + HCl
HPO42- ↔ H+ + PO43HCl + KOH
H2O ↔ H+ + OH-
3) Calculate the pH of 80 mL of 1.5 M Carbonic acid upon addition of the following amounts of
0.6 M NaOH
H2CO3 ↔ H+ + HCO3HCO3- ↔ H+ + CO32-
Ka1= 4.3x10-7
Ka2= 4.7x10-11
a) 0 mL
x2/(1.5-x) = 4.3x10-7
x = 8.03x10-4M H+
pH = 3.095
b) 100 mL
ВЅ way point so pH = 6.37
c) 150 mL - use H-H equation
(0.6M)(0.150L) = .09 mole NaOH and (1.5M)(0.080L) = 0.12 mole H2CO3
pH = 6.37 + log(0.09/(0.12-0.09))
pH= 6.847
d) 200 mL
first end point so pH = (pka1 + pka2)/2 в†’ pH = (6.37 + 10.33)/2 = 8.35
e) 300 mL
2nd midpoint so pH = pKa2 = 10.33
f) 365 mL - you are now 165 mL into the second region - use H-H equation
(0.06M)(0.165L) = 0.099 mole NaOH and you still have 0.12 mole of acid (HCO3-)
pH = 10.33 + log(0.099/(0.12-0.099)) pH = 11.00
g) 400 mL - at final endpoint pKb = Kw/Ka2 = 1x10-14/4.7x10-11 = 2.13x10-4
also [CO32-] = 0.12 mole/ (0.480L total) = 0.25M CO32x2/(0.25 - x) = 2.13x10-4 в†’ x = 0.007297 M OH- pH = 11.86
h) 450 mL - this is 50 mL past the final endpoint
(0.050L)(0.60M) = 0.030 mole excess OH- in a total of 530 mL of solution.
Therefore, [OH-] = 0.030 mole / 0.530 L = 0.0566 M OH- в†’ pH = 12.75
i) How much 0.8 M NaOH must be added to 80 mL of 1.5 M H2CO3 to make a buffer of pH =
10?
The pH is in the 2nd region so you must add 200 mL to start this titration. Use H-H,
10 = 10.33 + log(x/(0.12-x)) в†’ x = 0.0382 mole NaOH
0.0382 mole NaOH = MV = (0.80M)(V) в†’ V = 0.0478 L or 47.8 mL, therefore,
200 mL NaOH to get to first endpoint + 47.8 mL = 247.8 mL of 0.80 M NaOH
4) What is the solubility of Ag2S in a saturated solution of H2S = 0.01 M at pH = 3?
Ag2S ↔ 2 Ag+ + S2- Ksp = 6.0x10-50
H2S ↔ 2 H+ + S2- Keq = 6.84x10-23
[H+]2 [S2-]/[H2S] = 6.84x10-23
[S2-] = 6.84x10-23 [H2S]/[H+]2 = 6.84x10-23 [0.01]/[1x10-3]2 = 6.84x10-19 M S2Substitute this result into the following equation,
[Ag+]2 [S2-] = 6x10-50
[Ag+]2 = 6x10-50/[S2-] = 6x10-50/[6.84x10-19] = 8.77x10-32
taking the square root of this answer to find [Ag+],
[Ag+] = 2.96x10-16 M Ag+
5a) Iron can be separated from copper using pH controlled solubility in H2S. CuS is not soluble
in acid solutions and FeS is soluble. Draw and approximate ion separation curve for CuS and
FeS.
[Ions]
CuS FeS
0
At this pH FeS is highly soluble
and CuS is not very soluble.
7
pH
14
5b) Define amphoterism and give an example.
Amphoterism is the ability of an ion to be soluble in either an acid or a base but not in
between. This property is commonly exhibited with transition metals like Zinc,
Zn(OH)2(s) ↔ Zn2+ + 2 OHZn(OH)2(s) + 2 OH- ↔ Zn(OH)42-
soluble in acid
soluble in base
Exam III
Thermodynamics
1 , 2nd, and 3rd Laws
Enthalpy, Entropy, and Gibbs Energy
Heats of Formation
Hess’ Law
Spontaniety
Gibbs Equation
Electrochemical Cells
Balancing Redox Reactions
Nernst Equation
Batteries
Electroplating and Hydrolysis
st
Thermodynamics
1) A 100 gram block of copper was heated with a bunsen burner and then thrown into a cup of
water (250 mL) initially at 25В°C. When the system reached thermal equilibrium the final
temperature of the water and the block was 61В°C. What was the initial temperature of the block
of copper? Cp(Cu)=24.5 J/mol-K, Cp(H2O)=75.2 J/mol-K, FWT(Cu) = 63.55 g/mol, FWT(H2O)
= 18 g/mol, density of water = 1 g/ml
2) Given the following information calculate the heat of formation of C2H4.
C2H4 + 3 O2 в†’ 2 CO2 + 2 H2O О”HВ° = -414 kJ/mol
C + O2 в†’ CO2
О”HВ° = -393.5 kJ/mol
H2 + ВЅ O2 в†’ H2O
О”HВ° = -241.8 kJ/mol
3) When ethanol (C2H5OH) burns in oxygen it forms CO2 and water. If the heat of combustion
is -1237.7 kJ/mol, the heat of formation of water is -241.8 kJ/mol, and the heat of formation of
carbon dioxide is -393.5 kJ/mol, what is the heat of formation of C2H5OH?
4) Given a block of ice at 0В°C how much of the ice will melt if you put a 100 gram block of
aluminum on it that is initially at 100В°C? Cp(Al)=24.1 J/mol-K, Cp(H2O)=75.2 J/mol-K, Heat of
Fusion = 6.01 kJ/mol for ice, FWT(Al) = 26.98 g/mol, FWT (water) = 18 g/mol
5) 10 grams of diethyl ether was placed into a bomb calorimeter and then combusted in excess
oxygen. The balanced combustion reaction was as follows;
C4H10O + 6 O2 -----> 4 CO2 + 5 H2O
The heat released by this reaction heated 2500 mL of water 8.6В°C. What is the heat of formation
of the diethyl ether? Cp(H2O) = 75.2 J/mol-K, О”HВ°f(CO2) = -393.5 kJ/mol, О”HВ°f(H2O) = -241.8
kJ/mol, FWT (Diethyl ether) = 74 g/mol, FWT (water) = 18 g/mol.
6) If a 25 gram block of copper at 45В°C is added to 50 grams of liquid ammonia at -60В°C,
calculate the final temperature of the system. For NH3: FWT(NH3) = 17 g/mol, О”Hvap = 23.4
kJ/mol, Cp(liq)= 35.1 J/mol.K and for Cu : Cp = 24.4 J/mol.K, FWT(Cu) = 63.55 g/mol
7) You have 50 grams of ice at -2.5В°C sitting in a dixie cup. If you heat 5 nickels and throw
them into the cup and 8 grams of the ice melts, how hot were the nickels? Each nickel weighs 5
grams and has a heat capacity of 24.3 J/mol-K. Assume that nickels are pure nickel with an
atomic mass of 57.81 g/mol.
8) What is the О”G and О”S for the dilution of 4.5 M H2SO4 to 0.03 M H2SO4 at 25В°C? The
enthalpy for this reaction is - 210 kJ/mol.
9) Calculate the О”G for the making of 6M NaOH from solid NaOH. Saturated solutions of
NaOH have a concentration of 19.25 M.
10) Mathematically we write the first law of thermodynamics as U = q + w. The work, w, is
pressure-volume work. What is the sign on this work (+ or -) and why does it have this sign?
11) Please give me a definition of a heat capacity.
12) What is the О”GВ° for the following reaction at 25В°C?
BaBr2(aq) + CuSO4(aq) ----> BaSO4(s) + CuBr2(s)
О”HВ°
BaBr2
CuSO4
BaSO4
CuBr2
SВ°
-186.47 kJ/mol
-201.51 kJ/mol
-345.57 kJ/mol
-25.1 kJ/mol
42.0 J/mol K
19.5 J/mol K
7.0 J/mol K
21.9 J/mol K
Circle all that apply about this reaction,
spontaneous
exothermic
entropy increases
the reaction occurs quickly
not spontaneous
endothermic
entropy decreases
the reaction occurs slowly
What is the equilibrium constant for this reaction?
13) Given the following set of thermodynamic data calculate the О”GВ°, О”HВ°, О”SВ°, and Keq for the
following reaction at 25В°C (Enthalpies are in kJ/mol and entropies are in J/mol-k)
CoCl2(s) ⇆ Co2+ + 2 ClCoCl2
Co2+
Cl-
О”HВ°
-325.5
-67.36
-167.4
SВ°
106.3
-155.2
55.1
Electrochemical Cells
14) Please balance each of the following redox reactions;
MnO4- + Fe2+ в†’ Fe3+ + Mn2+
CrO42- + I- в†’ I2 + Cr3+
Co2+ + NO3- в†’ Co2O3 + NO
I3- + S2O32- в†’ I- + S4O62H2S + Cr2O72- в†’ Cr3+ + S
Fe2+ + NiOOH в†’ Fe3+ + Ni(OH)2
15) Consider the following electrochemical reaction,
PbI2 + 2 e- в†’ Pb + 2 IPb в†’ Pb2+ + 2 ePbI2 в†’ Pb2+ + 2 IHow would the voltage change if you,
a) Add KI(s) to the oxidation cell
b) Add AgNO3 to the reduction cell
c) Add heat
d) Add water to the reduction cell?
e) Add NaNO3(s) to the oxidation cell
f) Enlarge the oxidation electrode?
Inc.
Inc.
Inc.
Inc.
Inc.
Inc.
N.C.
N.C.
N.C.
N.C.
N.C.
N.C.
16) Consider the following electrochemical reaction,
Au в†’ Au3+ + 3eAuCl4- + 3e- в†’ Au + 4 ClAuCl4- в†’ Au3+ + 4 ClHow would the voltage change if you,
a) Enlarge the Au electrodes?
b) Add KCl(s)?
c) Add AgNO3?
d) Add water to the reduction cell?
e) Add NaNO3(s)
f) Cool the reaction vessel?
Inc.
Inc.
Inc.
Inc.
Inc.
Inc.
N.C.
N.C.
N.C.
N.C.
N.C.
N.C.
Dec.
Dec.
Dec.
Dec.
Dec.
Dec.
Dec.
Dec.
Dec.
Dec.
Dec.
Dec.
17) Given the following half-reactions draw an electrochemical cell, calculate the voltage of the
cell and lable the anode and cathode. You may use a carbon electrode for the I- solution.
Fe3+ + e- в†’ Fe2+
I2 в†’ 2 I-
+0.771 V
+0.535 V
18) What is the value of the half reaction,
Cu2+ + e- в†’ Cu+
given that,
Cu2+ + 2e- в†’ CuВ° пѓµВ° = +0.3402
Cu+ + e- в†’ CuВ° пѓµВ° = +0.522
19) Given the following half-reactions draw an electrochemical cell that would work. Calculate
the voltage of the cell and label the anode and cathode, tell which electrode is positive and which
is negative, and where the oxidation and reduction reactions are occurring. In addition indicate
the direction of electron flow, and the concentration of any ionic species in solution.
Co3+ + e- в†’ Co2+ +1.842 V
Pb2+ + 2e- в†’ Pb -0.126 V
20) Given the following half-reactions, properly set up a WORKING electrochemical cell.
Ag + I- в†’ AgI + ePbI2 + 2e-в†’ Pb + 2 I-
пѓµВ° = 0.1519 V
пѓµВ° = 0.3580 V
Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all
ionic species. Indicate what the electrodes are made of.
20b) Draw the cell diagram for the above cell.
20c) What is the Keq for the above cell?
20d) Look very carefully at the overall reaction, and explain whether your answer for 20c) is
reasonable or not.
21) Balance the following half-reactions in a BASE and then use these reactions to set up a
WORKING electrochemical cell.
Bi(s) в†’ Bi2O3(s)
Hg2O(s) в†’ Hg(l)
0.460 Volts
-0.123 Volts
Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all
ionic species. Indicate the composition of the electrodes.
21b) Draw the cell diagram for the above cell.
21c) What is the Keq for the above cell?
22) How long will it take to deposit 1 ounce of gold (31.1 g) at a constant current flow of 10
amps? MW (Au) = 197.8
23) How long will it take to deposit 1 lb. of Palladium (453.6 g) from a solution of PdCl4 at a
constant current of 25 amps? MW (Pd) = 106.4
Thermodynamics
1) A 100 gram block of copper was heated with a bunsen burner and then thrown into a cup of
water (250 mL) initially at 25В°C. When the system reached thermal equilibrium the final
temperature of the water and the block was 61В°C. What was the initial temperature of the block
of copper? Cp(Cu)=24.5 J/mol-K, Cp(H2O)=75.2 J/mol-K, FWT(Cu) = 63.55 g/mol, FWT(H2O)
= 18 g/mol, density of water = 1 g/ml
-nCpпЃ„TCu = nCpпЃ„TH2O
пЂ­
100g
J пѓ¶
250g пѓ¦
J пѓ¶
пѓ¦
пѓ§ 24.5
пѓ·пЂЁ61п‚°C - Ti пЂ© пЂЅ
пѓ§ 75.2
пѓ·пЂЁ61п‚°C - 25п‚°C пЂ©
63.55 g/mol пѓЁ
molK пѓё
18 g/mol пѓЁ
molK пѓё
Ti = 1036.3В°C
2) Given the following information calculate the heat of formation of C2H4.
C2H4 + 3 O2 в†’ 2 CO2 + 2 H2O О” HВ° = -414 kJ/mol
C + O2 в†’ CO2
О” HВ° = -393.5 kJ/mol
H2 + ВЅ O2 в†’ H2O
О” HВ° = -241.8 kJ/mol
2 CO2 + 2 H2O в†’ C2H4 + 3 O2 О”HВ° = 414 kJ/mol
2 (C + O2 в†’ CO2)
О”HВ° = -393.5 kJ/mol x2
2 (H2 + ВЅ O2 в†’ H2O)
О”HВ° = -241.8 kJ/mol x2
2 C + 2 H2 в†’ C2H4 О”HВ° = 414 kJ/mol + 2 (-393.5 kJ/mol )+ 2(-241.8 kJ/mol)
О”HВ° = -856.6 kJ/mol
3) When ethanol (C2H5OH) burns in oxygen it forms CO2 and water. If the heat of combustion
is -1237.7 kJ/mol, the heat of formation of water is -241.8 kJ/mol, and the heat of formation of
carbon dioxide is -393.5 kJ/mol, what is the heat of formation of C2H5OH?
C2H5OH + 3 O2 в†’ 2 CO2 + 3 H2O
О”HВ° = О”HВ°prod - О”HВ°react
-1237.7 kJ/mol= [2(-393.5kJ/mol) + 3(-241.8kJ/mol)] – [ΔH°C2H5OH + 3(0kJ/mol)]
О”HВ°C2H5OH = -274.7 kJ/mol
4) Given a block of ice at 0В°C how much of the ice will melt if you put a 100 gram block of
aluminum on it that is initially at 100В°C? Cp(Al)=24.1 J/mol-K, Cp(H2O)=75.2 J/mol-K, Heat of
Fusion = 6.01 kJ/mol for ice, FWT(Al) = 26.98 g/mol, FWT (water) = 18 g/mol
-nCpпЃ„TAl = nпЃ„пЃ€fus
100 g
(24.1 J/molK)(100п‚°C) пЂЅ n (6010 J/mol)
26.98 g/mol
n = 1.4863 mol ice x 18g/mol = 26.75 g ice will melt
5) 10 grams of diethyl ether was placed into a bomb calorimeter and then combusted in excess
oxygen. The balanced combustion reaction was as follows;
C4H10O + 6 O2 в†’ 4 CO2 + 5 H2O
The heat released by this reaction heated 2500 mL of water 8.6В°C. What is the heat of formation
of the diethyl ether? Cp(H2O) = 75.2 J/mol-K, О” HВ°f(CO2) = -393.5 kJ/mol, О” HВ°f(H2O) = 241.8 kJ/mol, FWT (Diethyl ether) = 74 g/mol, FWT (water) = 18 g/mol.
-nCpпЃ„TH2O = пЃ„пЃ€rxn note: a negative sign was added because energy was given off.
2500g
J
(75.2
)(8.6п‚°C) пЂЅ пЂ­89,822 J
18
molK
пЂ­ 89,822 J
пЂЅ пЂ­664,683 J/mol
пѓ¦ 10 g пѓ¶
пѓ§пѓ§
пѓ·пѓ·
пѓЁ 74 g/mol пѓё
пЂ­
О”HВ° = О”HВ°prod - О”HВ°react
-664,683 kJ/mol= [4(-393.5kJ/mol) + 5(-241.8kJ/mol)] – [ΔH°Ether + 6(0kJ/mol)]
О”HВ°Ether = -2118.3 kJ/mol
6) If a 25 gram block of copper at 45В°C is added to 50 grams of liquid ammonia at -60В°C,
calculate the final temperature of the system. For NH3: FWT(NH3) = 17 g/mol, О”Hvap = 23.4
kJ/mol, Cp(liq)= 35.1 J/mol.K and for Cu : Cp = 24.4 J/mol.K, FWT(Cu) = 63.55 g/mol
-nCpпЃ„TCu = nCpпЃ„TNH3
25g пѓ¦
J пѓ¶
50g пѓ¦
J пѓ¶
пѓ§ 24.4
пѓ·пЂЁTf пЂ­ 45п‚°C пЂ© пЂЅ
пѓ§ 35.1
пѓ·пЂЁTf - (-60п‚°C) пЂ©
63.55 пѓЁ
molK пѓё
17 пѓЁ
molK пѓё
-9.9587(Tf – 45°C) = 103.235(Tf + 60°C)
-Tf + 45 = 10.3663(Tf + 60) = 10.3663Tf + 621.98
-11.3663Tf = 666.98
Tf = -50.76В°C
пЂ­
7) You have 50 grams of ice at -2.5 C sitting in a dixie cup. If you heat 5 nickels and throw them
into the cup and 8 grams of the ice melts, how hot were the nickels? Each nickel weighs 5 grams
and has a heat capacity of 24.3 J/mol-K. Cpice = 37.7 J/molK. Assume that nickels are pure
nickel with an atomic mass of 57.81 g/mol.
Note: You must heat all of the ice but melt only 8 grams of it.
nCpпЃ„TNi = nCpпЃ„Tice + nпЃ„пЃ€fus
25g пѓ¦
J пѓ¶
8g пѓ¦
J пѓ¶ 50g пѓ¦
J пѓ¶
пѓ§ 24.3
пѓ·пЂЁTf пЂ­ 0п‚°C пЂ© пЂЅ
пѓ§ 6010
пѓ·пЂ«
пѓ§ 37.7
пѓ·пЂЁ2.5п‚°C пЂ©
57.81 пѓЁ
molK пѓё
18 пѓЁ
mol пѓё 18 пѓЁ
molK пѓё
10.508 Tf = 2932.92
Tf = 279.1В°C so Tf = 279.1В°C
8) What is the О”G and О”S for the dilution of 4.5 M H2SO4 to 0.03 M H2SO4 at 25В°C? The
enthalpy for this reaction is - 210 kJ/mol.
пЃ„G = пЃ„GВ° + RT ln Q = 0 + (8.314 J/molK)(298K) ln (0.03 M/ 4.5M)
пЃ„G = -12,414 J/mol
J
J пѓ¶
пѓ¦
- 12,414
пЂ« 210,000
пѓ§
пѓ·
mol
mol пѓ· пЂЅ пЂ­663 J
пЃ„G = пЃ„H - TпЃ„S => пЂ­ пѓ¦пѓ§ О”G пЂ­ О”H пѓ¶пѓ· пЂЅ О”S пЂЅ пЂ­ пѓ§
T
298
molK
пѓ§
пѓ·
пѓЁ
пѓё
пѓ§
пѓ·
пѓЁ
пѓё
9) Calculate the О”G for the making of 6M NaOH from solid NaOH. Saturated solutions of
NaOH have a concentration of 19.25 M.
There are two processes going on,
1. Solid NaOH ↔ Sat’d NaOH
2. Sat’d NaOH → 6 M NaOH
The first process is an equilibrium so пЃ„G = 0 for the first process.
For the second process, since all solutions have a standard state of 1M, the standard state
for Sat’d NaOH and 6 M NaOH are both 1 M and the G° = 0 since there is no difference
between the G°’s for these two solutions. So,
пЂ±пЂ®пЂ пЂ пЃ„G1 = 0
пЂІпЂ®пЂ пЂ пЃ„G2 = пЃ„GВ° + RT ln Q
пЂ пЂ пЂ пЂ пЂ пЃ„G2 = 0 + (8.314 J/molK)(298K) ln (6 M/ 19.25M) = -2888.2 J/mol
пЂ пЂ пЂ пЂ пЂ пЃ„G1+ пЃ„G2 = 0 + (- 2888.2 J/mol) = - 2888.2 J/mol
10) Mathematically we write the first law of thermodynamics as U = q + w. The work, w, is
pressure-volume work. What is the sign on this work (+ or -) and why does it have this sign?
Work = PпЃ„V
A chemist thinks of himself as being inside the system pushing out (like expanding a
balloon). To a chemist, pushing out (expanding the balloon) is negative work. Negative
work is work that you must do as opposed positive work that is done for you. In
mathematics, a change in volume is always the Vfinal – Vinitial. In the case of an
expanding balloon, the final volume will be larger than the initial volume so Vfinal –
Vinitial will produce a positive value. Since PпЃ„V = negative for an expanding balloon,
and both P and пЃ„V are positive, we must add a negative sign to the work expression to
make this work overall negative. So, chemists define pressure-volume work as,
Work = - PпЃ„V (Chemist)
It should be noted that engineers see the world opposite from chemists. Engineers view
themselves as being outside of the system pushing in. Work for an engineer is
compressing the balloon, making it smaller. Since V = Vfinal – V initial, and the Vfinal
is smaller than Vinitial, the пЃ„V is negative. So, for an engineer,
Work = PпЃ„V (Engineer)
This value is still negative, like the chemist, but we see the world differently from an
engineer.
11) Please give me a definition of a heat capacity.
It is the amount of energy needed to raise the temperature of one mole of substance 1В°C.
12) What is the О”GВ° for the following reaction at 25В°C?
BaBr2(aq) + CuSO4(aq) ----> BaSO4(s) + CuBr2(s)
О”HВ°
BaBr2
CuSO4
BaSO4
CuBr2
SВ°
-186.47 kJ/mol
-201.51 kJ/mol
-345.57 kJ/mol
-25.1 kJ/mol
42.0 J/mol K
19.5 J/mol K
7.0 J/mol K
21.9 J/mol K
О”HВ° = О”HВ°prod - О”HВ°react and О”SВ° = SВ°prod - SВ°react
О”HВ° = О”HВ°prod - О”HВ°react
ΔH° = [(-345.57 kJ/mol) + (-25.1 kJ/mol)] – [(-186.47 kJ/mol) + (-201.51 kj/mol)]
О”HВ° = 17.31 kJ/mol
О”SВ° = SВ°prod - SВ°react
ΔS° = [(7.0 J/molK) + (21.9 J/molK)] – [(42.0 J/molK) + (21.9 J/molK)]
О”SВ° = -32.6 kJ/mol
пЃ„GВ° = пЃ„HВ° - TпЃ„SВ°
пЃ„GВ° = 17,310 J/mol + (298 K) (-32.6 J/molK)
пЃ„GВ° = 27,024.8 J/mol
Circle all that apply about this reaction,
spontaneous
exothermic
entropy increases
the reaction occurs quickly
not spontaneous
endothermic
entropy decreases
the reaction occurs slowly
What is the equilibrium constant for this reaction?
пЃ„GВ° = - RT ln Keq
Keq пЂЅ e
пЂ­ О”G п‚°
RT
пЂЅe
пЂ­ ( пЂ­27 , 024 .8 )
(8.314)(29 8)
пЂЅ 1.832 x10 -5
13) Given the following set of thermodynamic data calculate the О”GВ°, О”HВ°, О”SВ°, and Keq for the
following reaction at 25C (Enthalpies are in kJ/mol and entropies are in J/mol-k)
CoCl2(s) ↔ Co2+ + 2 ClCoCl2
Co2+
Cl-
О”HВ°
-325.5
-67.36
-167.4
SВ°
106.3
-155.2
55.1
О”HВ° = О”HВ°prod - О”HВ°react and О”SВ° = SВ°prod - SВ°react
О”HВ° = О”HВ°prod - О”HВ°react
ΔH° = [(-67.36 kJ/mol) + 2(-167.4 kJ/mol)] – [(-325.5 kj/mol)]
О”HВ° = -76.66 kJ/mol (exothermic)
О”SВ° = SВ°prod - SВ°react
ΔS° = [(-155.2 J/molK) + 2(55.1 J/molK)] – [(106.3 J/molK)]
О”SВ° = -151.3 J/molK (entropy decreases)
пЃ„GВ° = пЃ„HВ° - TпЃ„SВ°
пЃ„GВ° = -76,660 J/mol - (298 K) (-151.3 J/molK)
пЃ„GВ° = -31,572.6 J/mol (reaction is spontaneous as written)
пЃ„GВ° = - RT ln Keq
Keq пЂЅ e
пЂ­ О”G п‚°
RT
пЂЅe
пЂ­ ( пЂ­31,572.6)
(8.314)(29 8)
пЂЅ 3.423x10 5
Electrochemical Cells
14) Please balance each of the following redox reactions;
MnO4- + Fe2+ в†’ Fe3+ + Mn2+
Fe2+ в†’ Fe3+ + e5e- + 8 H+ + MnO4- в†’ Mn2+ + 4 H2O
CrO42- + I- в†’ I2 + Cr3+
3 e- + 8 H+ + CrO42- в†’ Cr3+ + 4 H2O
2 I- в†’ I2 + 2 eCo2+ + NO3- в†’ Co2O3 + NO
3 H2O + 2 Co2+ в†’ Co2O3 + 6 H+ + 2e3 e- + 4 H+ + NO3- в†’ NO + 2 H2O
I3- + S2O32- в†’ I- + S4O622 e- + I3- в†’ 3 IS2O32- в†’ S4O62- + 2 eH2S + Cr2O72- в†’ Cr3+ + S
H2S в†’ S + 2 H+ + 2 e6 e- + 14 H+ + Cr2O72- в†’ 2 Cr3+ + 7 H2O
Fe2+ + NiOOH в†’ Fe3+ + Ni(OH)2
Fe2+ в†’ Fe3+ + ee- + H+ + NiOOH в†’ Ni(OH)2
15) Consider the following electrochemical reaction,
PbI2 + 2 e- в†’ Pb + 2 IPb в†’ Pb2+ + 2 ePbI2 в†’ Pb2+ + 2 IHow would the voltage change if you,
a) Add KI(s) to the oxidation cell
b) Add AgNO3 to the reduction cell
c) Add heat
d) Add water to the reduction cell?
e) Add NaNO3(s) to the oxidation cell
f) Enlarge the oxidation electrode?
Inc.
Inc.
Inc.
Inc.
Inc.
Inc.
N.C.
N.C.
N.C.
N.C.
N.C.
N.C.
Dec.
Dec.
Dec.
Dec.
Dec.
Dec.
In electrochemical cells all concentrations are 1 M. If all concentrations are 1 M, the reverse
reaction is going to occur. Therefore, the real reaction going on is,
Pb + 2 I- в†’ PbI2 + 2 e- (oxidation)
Pb2+ + 2 e- в†’ Pb
(reduction)
2+
Pb + 2 I в†’ PbI2
To understand what would happen, you need to write down the Nernst Equation,
пЃ„пѓµ = пЃ„пѓµВ° - RT/nпѓ¶ ln 1/[Pb2+][I-]2
a) When you put KI in the oxidation cell you are increasing the I- concentration. This
causes 1/[Pb2+][I-]2 to become smaller. The natural log of a smaller number is a negative
number. The negative of a negative is a positive so the voltage will increase.
b) Nothing happens. There is nothing for the silver to react with.
c) Adding heat increases T in the equation. If everything else remains the same, the
- RT/nпѓ¶ ln 1/ [Pb2+][I-]2 becomes more negative and the voltage decreases.
d) Adding water is a dilution. Diluting the reduction cell lowers the concentration of I-.
If I- gets smaller, the 1/[Pb2+][I-]2 gets larger. The natural log will become larger but
there is a negative so it becomes more negative, so the voltage goes down.
e) NaNO3 does not participate in this reaction. No change.
f) Changing the size of the electrodes does not change the voltage. The electrodes are
not a part of the reaction. They are not in the Nernst Equation.
16) Consider the following electrochemical reaction,
Au в†’ Au3+ + 3eAuCl4- + 3e- в†’ Au + 4 ClAuCl4- в†’ Au3+ + 4 ClHow would the voltage change if you,
a) Enlarge the Au electrodes?
b) Add KCl(s)?
c) Add AgNO3?
d) Add water to the reduction cell?
e) Add NaNO3(s)
f) Cool the reaction vessel?
Inc.
Inc.
Inc.
Inc.
Inc.
Inc.
N.C.
N.C.
N.C.
N.C.
N.C.
N.C.
Dec.
Dec.
Dec.
Dec.
Dec.
Dec.
Note: This is NOT a Ksp problem since AuCl4- is not a solid (solids are not charged) so
we will leave the reaction as is.
пЃ„пѓµ = пЃ„пѓµВ° - RT/nпѓ¶ ln [Au3+][Cl-]4/[ AuCl4-]
a) No change. Changing the size of the electrode makes no difference.
b) Adding Cl- makes the natural log large which decreases the voltage
c) The Ag+ will react with the Cl- to make AgCl. This removes Cl-, making the natural
log smaller (more negative). So you will be subtracting a smaller number so the voltage
will increase.
d) Diluting the reduction cell lowers the concentration of both AuCl4- and Cl- but the Clis raised to the 4th power so it reduces faster. Reducing Cl- makes the natural log smaller
(more negative so you are subtracting a smaller (more negative) number so voltage
increases.
e) No change.
f) If T is smaller, then you are subtracting a smaller number so voltage increases.
17) Given the following half-reactions draw an electrochemical cell, calculate the voltage of the
cell and label the anode and cathode. You may use a carbon electrode for the I- solution.
Fe3+ + e- в†’ Fe2+
+0.771 eV
2 I- в†’ I2 + 2 e-0.535 eV
2 Fe3+ + 2 I- в†’ 2 Fe2+ + I2 0.236 eV
e-
Anode
Negative
Oxidation
C
Pt
I2
1M FeCl 3
1M FeCl 2
1M KI
18) What is the value of the half reaction,
Cu2+ + e- ----> Cu+
given that,
Cu2+ + 2e- в†’ Cu пѓµВ° = +0.3402
Cu+ + e- в†’ Cu пѓµВ° = +0.522
Cu2+ + 2e- в†’ Cu пѓµВ° = +0.3402
Cu в†’ Cu+ + e- пѓµВ° = -0.522
Cu2+ + e- в†’ Cu+
пѓµВ° = 2(0.3402) + 1(-0.522) = 0.1584 volts
1
Cathode
Positive
Reduction
19) Given the following half-reactions draw an electrochemical cell that would work. Calculate
the voltage of the cell and label the anode and cathode, tell which electrode is positive and which
is negative, and where the oxidation and reduction reactions are occurring. In addition indicate
the direction of electron flow, and the concentration of any ionic species in solution.
Co3+ + e- в†’ Co2+ О”пѓµВ° = + 1.842 eV
Pb2+ + 2e- в†’ Pb О”пѓµВ° = - 0.126 eV
e-
Anode
Negative
Oxidation
Pb
Pt
1M
Pb(NO3)2
1M CoCl3
1M CoCl2
Cathode
Positive
Reduction
Co3+ + e- в†’ Co2+ О”пѓµВ° = + 1.842 eV
Pb в†’ Pb2+ + 2e- О”пѓµВ° = + 0.126 eV
3 Pb + 2 Co3+ в†’ 3 Pb2+ + 2 Co
О”пѓµВ° = 1.968 V
20) Given the following half-reactions, properly set up a WORKING electrochemical cell.
AgI + e----> Ag + IО”пѓµВ° = -0.1519 V
PbI2 + 2e ---> Pb + 2 I О”пѓµВ° = 0.3580 V
Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all
ionic species. Indicate the composition of the electrodes.
-
e
Ag
Anode
Negative
Oxidation
AgI(s)
Pb
1M KI
Cathode
Positive
Reduction
PbI2(s)
20b) Draw the cell diagram for the above cell.
Ag/AgI(s)/1 M KI/PbI2(s)/Pb
20c) What is the Keq for the above cell?
пѓµ = 0.5099 volts = 0.0592/2 log Keq => Keq = 1.684x1017
20d) Look very carefully at the overall reaction, and explain whether your answer for 20c) is
reasonable or not.
The problem is that everything is a solid except for the 1 M KI. The equilibrium constant
must be,
[I пЂ­ ]
Keq пЂЅ пЂ­ Pb
[ I ] Ag
So the voltage is completely dependent on the difference between the amount of I- in
equilibrium with the Pb and the I- in equilibrium with the Ag. This makes what is known
as a concentration cell, that is, a cell whose voltage is due only to differences in
concentration on the two sides of the cell.
21) Balance the following half-reactions in a BASE and then use these reactions to set up a
WORKING electrochemical cell.
6 OH- + 2 Bi(s) в†’ Bi2O3(s) + 3 H2O + 6 e2 e- + H2O + Hg2O(s) в†’ 2 Hg(l) + 2 OH-
0.460 Volts
-0.123 Volts
Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all
ionic species. Indicate the composition of the electrodes.
e
Anode
Negative
Oxidation
-
Bi
Cathode
Positive
Reduction
1M NaOH
Bi2O3(s)
Hg2O
Pt
21b)
Draw the cell diagram for the above cell.
Bi / Bi2O3(s) / 1 M NaOH / Hg2O(s) / Hg / Pt
21c) What is the Keq for the above cell?
пѓµ = 0.337 volts = 0.0592/6 log Keq
Keq = 1.43x1034
22) How long will it take to deposit 1 ounce of gold (31.1 g) at a constant current flow of 10
amps? MW (Au) = 197.8
31.1 g/197.8 g/mol = 0.1572 mol Au has a 3+ charge, so multiply by 3.
3 mole e-/mol Au x 0.15723 mol Au = 0.4717 mol e0.4717 mol e- x 96,487 C/mol e- = 45,511.8 C
Amp x sec = Coulombs (C)
(10 amps)(sec) = 45,511.9 C
sec = 4,551.18 sec => 75.85 minutes
23) How long will it take to deposit 1 lb. of Palladium (453.6 g) from a solution of PdCl4 at a
constant current of 25 amps? MW (Pd) = 106.4
453.6 g/106.4 g/mol = 4.263 mol Pd, but Pd has a 4+ charge, so multiply by 4.
4 mole e-/mol Pd x 4.263 mol Pd = 17.053 mol e17.053 mol e- x 96,487 C/mol e- = 1,645,357 C
Amp x sec = Coulombs (C)
(25 amps)(sec) = 1,645,357 C
sec = 65,814 sec => 18.28 hours
Chemistry 121
Third Exam
Name____________________
May 15, 2007
CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full
credit. You may use a calculator.
Question
Credit
1 ( 18)
2 (40)
3(12)
4(15)
5(15)
Total
R = 8.314 J/mol-K
пѓµ = пѓµпЃ… - 0.0592/n log Q
О”GпЃ… = -nпѓ¶пѓµпЃ…
пѓ¶ = 96486 C/mol eО”G = О”GпЃ… + RT ln Q
PV = nRT
C = amp x sec
О”GпЃ…= -RTlnKeq
R = 0.08205 L-atm/mol-K
1) Balance each of the following oxidation-reduction half-reactions;
IO3- в†’ I3- (in acid)
Mn(OH)2 в†’ Mn2O3 (in base)
2) Given the following set of half-reactions, set up a WORKING electrochemical cell.
Hg2Cl2(s) + 2 e- в†’ 2Hg + 2Cl2 H+ + 2 e- в†’ H2
пѓµпЃ… = 0.27 V
пѓµпЃ… = 0.000 V
Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all
ionic species. Indicate the composition of the electrodes.
2b) Draw the cell diagram for the above cell.
2c) What is the voltage of the cell when all ionic species have a concentration of 0.10 M?
3) How long will it take to galvanize (coat with Zinc) a nail with 2 grams of zinc using a current
(amperage) of 0.8 amps? FWT of Zn2+ = 65.41 g/mol
4) You are given 100 mL of water at 25В°C. You add a 20 gram ice cube at 0В°C and 50 mL of
boiling water at 100В°C. What is the final temperature of the system?
5) What is the О”G and О”S for the dilution of 4.5 M H2SO4 to 0.03 M H2SO4 at 25В°C? The О”H =
- 210 kJ/mol.for this reaction.
Chemistry 121
Third Exam
Name__Answer Key____
May 15, 2007
CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full
credit. You may use a calculator.
Question
Credit
1 ( 18)
2 (40)
3(12)
4(15)
5(15)
Total
R = 8.314 J/mol-K
пѓµ = пѓµпЃ… - 0.0592/n log Q
О”GпЃ… = -nпѓ¶пѓµпЃ…
пѓ¶ = 96486 C/mol eО”G = О”GпЃ… + RT ln Q
PV = nRT
C = amp x sec
О”GпЃ…= -RTlnKeq
R = 0.08205 L-atm/mol-K
1) Balance each of the following oxidation-reduction half-reactions;
IO3-в†’ I3- (in acid)
16 e- + 18 H+ + 3 IO3- в†’ I3- + 9 H2O
Mn(OH)2 в†’ Mn2O3 (in base)
2 Mn(OH)2 в†’ Mn2O3 + H2O + 2 H+ + 2e2 OH- в†’ 2 OH2 Mn(OH)2 + 2 OH- в†’ Mn2O3 + 3 H2O + 2e-
2) Given the following set of half-reactions, set up a WORKING electrochemical cell.
Hg2Cl2(s) + 2 e- в†’ 2Hg + 2Cl2 H+ + 2 e- в†’ H2
e-
H2 1atm
Anode
Negative
Oxidation
пѓµпЃ… = 0.27 V
пѓµпЃ… = 0.000 V
Pt
1M HCl
Cathode
Positive
Reduction
1M HCl
Hg2Cl2
Pt
Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all
ionic species. Indicate the composition of the electrodes.
2b) Draw the cell diagram for the above cell. (Note: The solution in both cells is HCl so no salt
bridge is required).
Pt/H2 (1atm) / 1M HCl / Hg2Cl2(s) / Hg(l)/ Pt
2c) What is the voltage of the cell when all ionic species have a concentration of 0.10 M?
The overall reaction is,
Hg2Cl2(s) + 2 e- → 2Hg + 2ClH2 → 2 H+ + 2 eHg2Cl2(s) + H2 –> 2 Hg + 2 H+ + 2 Cl-
пѓµпЃ… = 0.27 V
пѓµпЃ…= 0.000 V
пѓµпЃ…= 0.27 V
So the voltage is,
пѓµ = 0.27 - 0.0592/2 log [H+]2 [Cl-]2 = 0.27 - 0.0591/2 log [0.10]2 [0.10]2 = 0.3882 Volts
3) How long will it take to galvanize (coat with Zinc) a nail with 2 grams of zinc using a current
(amperage) of 0.8 amps? FWT of Zn2+ = 65.41 g/mol
2 grams/65.41g/mol = 0.03058 mole of Zn
0.03058 mole of Zn x 2 mole e-/mol Zn = 0.06115 mole e0.06115 mole e- x 96,486 C/mol e- = 5900.4 Coulombs
1 C = 1 amp x sec
5900.4 C = 0.8 amps x sec
5900.4 C/0.8 amps = 7375.47 seconds
4) You are given 100 mL of water at 25В°C. You add a 20 gram ice cube at 0В°C and 50 mL of
boiling water at 100В°C. What is the final temperature of the system? О”Hfus = 6.01 kJ/mol and
Cp(H2O) = 75.2 J/mol K, FWT H2O = 18g/mol (These were not given in the original problem).
The easiest way of doing this is by mixing the hot and cold water and then adding the ice cube.
First add 100 mL of 25В°C water to 50 mL of 100В°C water,
nCpО”T25В°C = - nCpО”T100В°C
100g/18g/mol (75.2 J/molK)(Tf - 25) = -50g/18g/mol(75.2 J/molK)(Tf-100)
The 18g/mol and the 75.2 J/molK cancel on both sides so,
100(Tf-25) = -50(Tf-100)
Tf = 50В°C
So now we have 150 mL of water at 50МЉC and we add the 20 gram ice cube. I assume that the ice
cube will melt and then waters will mix to get to some final temperature so,
nО”Hfus + nCpО”T0В°C = - nCpО”T50В°C
20g/18g/mol(6010 J/mol) + 20g/18g/mol (75.2)(Tf - 0) = - 150g/18g/mol(75.2)(Tf - 50)
All the 18 g/mol cancel so,
120,200 + 1504Tf = - 11280Tf + 564000
12784Tf = 443,800
Tf = 34.71В°C
5) What is the О”G and О”S for the dilution of 4.5 M H2SO4 to 0.03 M H2SO4 at 25В°C? The О”H =
- 210 kJ/mol.for this reaction.
О”G = О”GпЃ… + RT ln Q
But О”GпЃ… = 0 since both concentrations have the same standard state of 1M. So,
О”G = 0 + RT ln 0.03M/4.5M
О”G = -5391.42 J/mol
Now, since О”G = О”H - TО”S, we can solve for О”S,
-5391.42 J/mol = -210,000 J/mol - (298)О”S
О”S = (-5391.42 J/mol + 210,000 J/mol )/298 = 686.61 J/mol K
Chemistry 121
Third Exam
Name____________________
May 14, 2009
CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit.
You may use a calculator.
Question
Credit
1 (18)
2 (28)
3(10)
4(20)
Total
R = 8.314 J/mol-K
пѓµ = пѓµпЃ… - 0.0592/n log Q
О”GпЃ… = -nпѓ¶пѓµпЃ…
пѓ¶ = 96486 C/mol eО”G = О”GпЃ… + RT ln Q
PV = nRT
C = amp x sec
О”GпЃ…= -RTlnKeq
R = 0.08205 L-atm/mol-K
1a) Please balance the following half-reaction in a base.
BH3 в†’ B2O3
1b) What is the voltage of the following half-cell reactions when added together?
Cu2+ + e- в†’ Cu+
Cu+ + e- в†’ Cu
Cu2+ + 2 e- в†’ Cu
пѓµпЃ…МЉ = 1.29
пѓµпЃ…= 1.68
пѓµпЃ…= ?
2) In lab, you were asked to experimentally determine the Ksp of PbI2. Given the information
below (found in the CRC), what voltage should you have obtained for your cell, and what should
the Ksp have been?
PbI2(s) ↔ Pb2+ + 2 IΔH
SпЃ…
PbI2 -174.1 kJ/mol
176.98 J/mol-K
Pb2+
1.63 kJ/mol
21.34 J/mol-K
I-55.94 kJ/mol
109.37 J/mol-K
The reaction is (Circle all that apply).
Spontaneous
Exothermic
Entropy increases
Occur Fast
Not Spontaneous
Endothermic
Entropy Decreases
Occurs Slow
2c) Like you did in lab, draw the electrochemical cell that was used to determine the Ksp of PbI2.
Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all
ionic species. Indicate the composition of the electrodes.
2c) Draw the cell diagram for the above cell.
3) It took 35 minutes and 10 amps to electroplate 10.615 grams of platinum (FWT = 195.08
g/mol). What was the charge on the platinum?
4) A 20 gram block of ice initally at -10.0В°C was dropped into 250 mL of water at 25В°C. What is
the final temperature of the system? О”Hfus = 6.01 kJ/mol, Cp(H2O) = 75.2 J/mol-K, Cp(ice) =
35.2 J/mol-K, FWT (H2O)= 18 g/mol
Chemistry 121
Third Exam
Name__Answer Key___
May 14, 2009
CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit.
You may use a calculator.
Question
Credit
1 (18)
2 (28)
3(10)
4(20)
Total
R = 8.314 J/mol-K
пѓµ = пѓµпЃ… - 0.0592/n log Q
О”GпЃ… = -nпѓ¶пѓµпЃ…
пѓ¶ = 96486 C/mol eО”G = О”GпЃ… + RT ln Q
PV = nRT
C = amp x sec
О”GпЃ…= -RTlnKeq
R = 0.08205 L-atm/mol-K
1a) Please balance the following half-reaction in a base.
BH3 в†’ B2O3
3 H2O + 2 BH3 в†’ B2O3 + 12 H+ + 12 e12 OH- в†’ 12 OH12 OH- + 2 BH3 в†’ B2O3 + 9 H2O + 12 e-
1b) What is the voltage of the following half-cell reactions when added together?
Cu2+ + e- —> Cu+
Cu+ + e- —> Cu
Cu2+ + 2 e- —> Cu
пѓµпЃ… = 1.29
пѓµпЃ… = 1.68
пѓµпЃ… = (1.29(1) + 1.68 (1))/2 = 1.485 Volts
2) In lab, you were asked to experimentally determine the Ksp of PbI2. Given the information
below (found in the CRC), what voltage should you have obtained for your cell, and what should
the Ksp have been?
PbI2(s) ⇆ Pb2+ + 2 IPbI2
Pb2+
I-
О”HпЃ…
-174.1 kJ/mol
1.63 kJ/mol
-55.94 kJ/mol
SпЃ…
176.98 J/mol-K
21.34 J/mol-K
109.37 J/mol-K
О”HпЃ…МЉ = [1.63 + 2(-55.94)] - [-174.1] = 63.85 kJ/mol
О” SпЃ… = [21.34 + 2(109.37)] - [176.98] = 63.1 J/molK
О”GпЃ… = 63,850 - 298(63.1) = 45,046.2 J/mol
45,046.2 J/mol = -8.314(298)ln(Ksp)
Ksp = 1.27x10-8
The reaction is (Circle all that apply).
Spontaneous
Exothermic
Entropy increases
Occurs Fast
Not Spontaneous
Endothermic
Entropy Decreases
Occurs Slow
2c) Like you did in lab, draw the electrochemical cell that was used to determine the Ksp of PbI2.
Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all
ionic species. Indicate the composition of the electrodes.
e-
Cathode
Positive
Reduction
Pb
Pb
Anode
Negative
Oxidation
PbI2(s)
1M
Pb(NO3)2
1M KI
2c) Draw the cell diagram for the above cell.
Pb / PbI2(s) / 1 M KI // 1 M Pb(NO3)2 / Pb
3) It took 35 minutes and 10 amps to electroplate 10.615 grams of platinum (FWT = 195.08
g/mol). What was the charge on the platinum?
10.615 g/195.08 g/mol = 0.05441 mole of Pt
0.05441 mole of Pt x Charge on Pt = moles of eAlso,
amp x sec/ 96,486 C/mol e- = mole of eSo,
0.05441 mole of Pt x Charge on Pt = amp x sec/ 96,486 C/mol e0.05441 mole of Pt x Charge on Pt = 10 amps x 35 min x 60 sec/min / 96,486 C/mol eCharge on Pt = [10 amps x 35 min x 60 sec/min / 96,486 C/mol e-]/0.05441 mole of Pt
Charge on Pt = +4
4) A 20 gram block of ice initally at -10.0В°C was dropped into 250 mL of water at 25В°C. What is
the final temperature of the system? О”Hfus = 6.01 kJ/mol, Cp(H2O) = 75.2 J/mol-K, Cp(ice) =
35.2 J/mol-K, FWT (H2O)= 18 g/mol
nCpО”Tice + nО”Hfus + nCpО”T0В°C H2O = -nCpО”T25В°C H2O
(20/18)35.2(10) + (20/18)6010 + (20/18)75.2(Tf-0) = -(250/18)75.2(Tf-25)
All the 18's cancel so,
7040 + 120200 + 1504Tf = -18800Tf + 470000
20304Tf = 342760
Tf = 16.88В°C
Final Exam
Comprehensive Exam
~ 30% Cut and Paste from Old Exams
~30% Organic Chemistry
~40% New Questions on Old Topics
Mechanism of Free Radical Halogenation using Bromine
Chain Initiation
light
Br2
2 Br
Chain Propagation
Br
H
H
H
H
H
H
C
C
C
H
H
H
H
H
H
C
C
C
H
H
H
Br2
H
+
HBr
H
H
H
C
C
C
H
H
H
H
Br H
C
C
C
H
H
H
H
Br H
C
C
C
H
H
H
+ Br
H
Chain Termination
Br
Br
H
Br
H
Br2
H
C
C
C
H
H
H
H
H
H
H
H
C
H
H
C
H
C
H
H
H
H
C
H
H
C
H
C
H
H
H
H
H
Relative Reactivity
1п‚° 2п‚°
3п‚°
Chlorination 1
3.5 5
Bromination 1
97
пЂґ
C
H
C
H
H
H H
C
H
C
H
C
H
H
C
H
H
H
1) Please draw the rotational energy diagram for the rotation of 2,3-dimethyl butane around the
C2-C3 bond.
Energy
0
60
120
180
240
300
360
Draw the molecule below, convert it to a Newman projection and then rotate it about the C2-C3
bond. Based on your rotations, draw the energy diagram above.
2) Please draw the rotational energy diagram for the rotation of 3-methyl pentane around the C2C3 bond.
Energy
0
60
120
180
240
300
360
Draw the molecule below, convert it to a Newman projection and then rotate it about the C2-C3
bond. Based on your rotations, draw the energy diagram above.
1) Please draw the rotational energy diagram for the rotation of 2,3-dimethyl butane around the
C2-C3 bond.
Energy
0
60
120
180
240
300
360
Draw the molecule below, convert it to a Newman projection and then rotate it about the C2-C3
bond. Based on your rotations, draw the energy diagram above.
0
CH3 CH3
H3C
C
C
H
H
120
CH3
H3C
H3C
CH3
CH3
CH3
4
60
H3C
CH3
H
HH
CH3
H3C
CH3
CH3
CH3
H
2
CH3
H3C
H
H3C
H CH3
3
H
H3C
H3C
CH3
H
H3C
CH3
H3C
H
CH3
H
H CH3
H
300
240
180
2
3
1
2) Please draw the rotational energy diagram for the rotation of 3-methyl pentane around the C2C3 bond.
6
5
Energy
4
3
2
1
0
60
120
180
240
300
360
Draw the molecule below, convert it to a Newman projection and then rotate it about the C2-C3
bond. Based on your rotations, draw the energy diagram above.
0
H
H3C
CH3
C
C
H
H
120
CH3
H3C
H
CH3
C2H5
C2H5
6
60
H
CH3
H
HH
CH3
H3C
CH3
CH3
C2H5
H
3
C2H5
H
H
H
H C2H5
5
H
C2H5
H
CH3
H
H
CH3
C2H5
H
CH3
H
H CH3
H
300
240
180
2
4
1
3) What is the ratio of products in the free radical chlorination of bicyclo [2,2,0] hexane?
2В°
‫ݔ‬
8 3.5
ቆ °ቇ
= в€™
‫ = ݔ‬26.32%
3
100 − ‫ ݔ‬2
5
4) What is the ratio of products in the free radical chlorination of 2-methylpropane?
CH3
H3C
C
CH3
H
1В°
‫ݔ‬
9 1
ቆ °ቇ
= ∙ ‫ = ݔ‬64.29%
3
100 − ‫ ݔ‬1 5
5) What is the ratio of products in the free radical chlorination of 1,1,3,3tetramethylcyclopentane?
H3C
CH3
H3C
CH3
1В°
‫ݔ‬
12
1
ቆ °ቇ
=
в€™
‫ = ݔ‬36.36%
2
100 − ‫ݔ‬
6 3.5
6) Please name the following compounds,
Acids
O
O
C
C
C
C
HO
OH
Propanoic acid
O
O
C
C
H
C
C
OH
OH
butadioic acid (malic acid)
methanoic acid (formic acid)
COOH
OH
C
C
Br
Benzoic acid
O
C
C
OH
3-bromo-3-hydroxybutanoic acid
Aldehydes and Ketones
O
O
C
C
C
C
C
O
C
C
C
H
C
H
H
Propanal
2-pentanone
OH O
O
C
C
methanal (formaldehyde)
C
Propanone (acetone)
C
C
C
C
3-hydroxybutanone
Alcohols
H
OH
C
C
C
OH
C
C
C
C
C
H
C
OH
H
Propanol
2-pentanol
OH OH
OH
C
C
methanol
C
2-propanol
C
C
C
2,3-butadiol
C
Aromatics
NH2
NO2
CH3
HO
Cl
Toluene
CH3
2-chloro-4-nitrophenol
Br
para methylaniline
Br
H2N
COOH
Br
1,3,5-tribromobenzene
para aminobenzoic acid
Chemistry 121
Final Exam
Name______________________
May 27, 1998
CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit.
You may use a calculator.
Question
Credit
Question
1(15)
5(20)
2(20)
6(20)
3(20)
7(25)
4(40)
8(40)
Total
Total
Credit
TOTAL POINTS =
R = 8.314 J/mol-K
О”пѓµ = О”пѓµВ° - 0.0592/n log Q
О”GВє = -nпѓ¶О”пѓµВ°
ln(Ai/Af) = kО”t
пѓ¶ = 96486 C/mol eО”G = О”GВє + RT ln Q
PV = nRT
1/Af - 1/Ai = kО”t
C = amp x sec
О”GВє= -RTlnKeq
R = 0.08205 L-atm/mol-K
ln(k1/k2) = Ea/R (1/T2 - 1/T1)
1)Write down the equilibrium that occurs when the following compounds are mixed.
a) NaOH + NaBenz
b) NH4OH + NH4Cl
c) NaCN + HCl
2) Last night I poured myself a glass of water from the tap and took a drink. The water was not
cold and didn't taste good so I put a couple of ice cubes in it. The ice cubes melted and the water
tasted better. If my cup had 250 mL of water at 22ВєC, and I put two ice cubes into the water that
weighed 20 grams each, what was the final temperature of the water if the ice cubes were initially
at 0В°C? О”Hfus = 6.01 kJ/mol, Cp(H2O) = 75.2 J/molK
3) Assume that the cooking of an egg is a first order process. Assume further that it takes 3
minutes to half-cook (soft boil) an egg at 100ВєC. If the activation energy of the cooking of an egg
is 75 kJ/mole, how long would it take to soft boil an egg at the top of Mt. Everest where water
boils at 80ВєC?
4a) Calculate the pH at each of the following points for the titration of 100 mL of 2.00 M
Phthalic acid with 5 M NaOH
H2Phthal в†’ H+ + HPhthalHPhthal- в†’ H+ + Phthal2-
pKA1 = 2.89
pKA2 = 5.41
INITIAL pH
20 mL NaOH
40 mL NaOH
50 mL NaOH
80 mL NaOH
100 mL NaOH
4b) How many milliliters of 5 M NaOH must you add to 100 mL of 2 M Phthalic acid to make a
buffer of pH = 3.5?
5) At low temperatures, the rate law for the reaction,
CO(g) + NO2(g) в†’ CO2(g) + NO(g)
can be determined by the following data;
Initial Rate/10-4 [NO]/10-3 M
3.60
7.20
0.90
0.90
1.20
1.20
1.20
2.40
[CO]/10-3 M
1.50
3.00
0.75
1.50
[NO2]/10-3 M
0.80
0.80
0.40
0.40
Write a rate law in agreement with the data.
Work out the rate law for each of the following mechanisms (show your work) and then circle the
one that is consistent with the rate law determined above. (Circle A, B, or C)
A) CO + NO2 в†’ CO2 + NO slow
B) NO2 ↔ NO + O equil.
O + CO в†’ CO2 slow
C) 2 NO2 ↔ 2 NO + O2 equil.
CO + O2 в†’ CO2 + O slow
O + NO в†’ NO2
fast
6) What is the solubility of Barium Carbonate in a solution saturated with H2CO3 at a pH of 10?
[H2CO3]sat'd = 0.034M.
BaCO3 ↔ Ba2+ + CO32- Ksp = 8.1x10-9
H2CO3 ↔ 2 H+ + CO32- Keq = 2.02x10-17
7) A cell has the following cell diagram;
Ag / AgCl / 1M HCl // 1M AgNO3 / Ag
7a) What is the reduction half reaction?
7b) What is the overall reaction?
7c) Based on the cell diagram please draw the electrochemical cell.
7d) If the Ksp for this cell is 1.8x10-10 then what is the cell voltage when all the ionic species are
0.01 M?
8) Please name or draw the following compounds.
OH
CH3
OH OH
C
H2N
C
C
Cl
O
C
C
C
C
C
C
C
O
C
C
C
C
C
Cl
C
C
CH3
butanone
3,3-dichloro butanoic acid
trans 1,3-cyclopentadiol
o-methylaniline
Chemistry 121
Final Exam
Name Answer Key
May 27, 1998
CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit.
You may use a calculator.
Question
Credit
Question
1(15)
5(20)
2(20)
6(20)
3(20)
7(25)
4(40)
8(40)
Total
Total
Credit
TOTAL POINTS =
R = 8.314 J/mol-K
О”пѓµ = О”пѓµВ° - 0.0592/n log Q
О”GВє = -nпѓ¶О”пѓµ В°
ln(Ai/Af) = kО”t
пѓ¶ = 96486 C/mol eО”G = О”GВє + RT ln Q
PV = nRT
1/Af - 1/Ai = kО”t
C = amp x sec
О”GВє= -RTlnKeq
R = 0.08205 L-atm/mol-K
ln(k1/k2) = Ea/R (1/T2 - 1/T1)
1)Write down the equilibrium that occurs when the following compounds are mixed.
a) NaOH + NaBenz
Benz- + H2O ↔ HBenz + OHb) NH4OH + NH4Cl
NH4OH ↔ NH4+ + OHc) NaCN + HCl
HCN ↔ H+ + CN-
2) Last night I poured myself a glass of water from the tap and took a drink. The water was not
cold and didn't taste good so I put a couple of ice cubes in it. The ice cubes melted and the water
tasted better. If my cup had 250 mL of water at 22ВєC, and I put two ice cubes into the water that
weighed 20 grams each, what was the final temperature of the water if the ice cubes were initially
at 0В°C? О”Hfus = 6.01 kJ/mol, Cp(H2O) = 75.2 J/molK
First, use the 22C water to melt the ice. The result will be colder water and melted ice.
nCpпЃ„T = nпЃ„HFus
250 g/18 g/mol (75.2 J/molK)(TF - 22ВєC) = 40g/18 (6010 J/mol)
TF = 9.212ВєC
Now, mix this colder water (9.212ВєC) with the melted ice (0ВєC),
-nCpпЃ„T = nCpпЃ„T
-[250 g/18 g/mol (75.2 J/molK)(TF - 22ВєC)] = 40 g/18 g/mol (75.2 J/molK)(TF - 0ВєC)
TF = 7.94ВєC
3) Assume that the cooking of an egg is a first order process. Assume further that it takes 3
minutes to half-cook (soft boil) an egg at 100ВєC. If the activation energy of the cooking of an egg
is 75 kJ/mole, how long would it take to soft boil an egg at the top of Mt. Everest where water
boils at 80ВєC?
t1 = 3 min @ 100ВєC and t2 @ 80ВєC
ln
t2
75,000 пѓ¦ 373 пЂ­ 353 пѓ¶
пЂЅ
пѓ§
пѓ·
3 min
8.314 пѓЁ 373 п‚ґ 353 пѓё
t2 = 11.81 min
4a) Calculate the pH at each of the following points for the titration of 100 mL of 2.00 M
Phthalic acid with 5 M NaOH
H2Phthal в†’ H+ + HPhthalHPhthal- в†’ H+ + Phthal2INITIAL pH
x2
пЂЅ 2.88 x10 пЂ­3
2пЂ­x
pKA1 = 2.89
pKA2 = 5.41
pH = 1.29
KA1 = 2.88x10-3
KA2 = 3.89x10-6
M1V1 = M2V2
(2M)((0.100L) = (5 M)(V2)
V2 = 40 mL for each region
Also (2M)(0.100L) = 0.2 mol H2Phthal
20 mL NaOH
pH = 2.89
40 mL NaOH
pH пЂЅ
50 mL NaOH
10 mL into 2nd Region (5 M)(0.010 L) = 0.05 mol OH-
pK A1 пЂ« pK A2 2.89 пЂ« 5.41
пЂЅ
2
2
pH пЂЅ 5.41 пЂ« log
80 mL NaOH
[Phthal 2- ] пЂЅ
0.05
0.2 пЂ­ 0.05
pH = 4.933
0.2 mol
1x10 -14
пЂЅ 1.111M and Kb пЂЅ
пЂЅ 2.57 x10 пЂ­9
0.180 L
3.89x10 -6
x2
пЂЅ 2.57 x10 пЂ­9
1.111 пЂ­ x
100 mL NaOH
pH = 4.15
x = 5.3x10-5 M OH-
(0.020 L)(5 M) = 0.10 mol OH0.10 mol OH пЂ­
пЂЅ 0.5 M OH пЂ­
0.200 L
pH = 9.72
pH = 13.70
4b) How many milliliters of 5 M NaOH must you add to 100 mL of 2 M Phthalic acid to make a
buffer of pH = 3.5?
The pH is nearest the 1st pKa so the buffer is in Region I. You must add,
3.5 пЂЅ 2.89 пЂ« log
(x)
пѓћ x пЂЅ 0.1607 mol NaOH = (5 M)(V)
(0.20 - x)
V = 32.14 mL NaOH
5) At low temperatures, the rate law for the reaction,
CO(g) + NO2(g) в†’ CO2(g) + NO(g)
can be determined by the following data;
Initial Rate/10-4 [NO]/10-3 M
3.60
7.20
0.90
0.90
[CO]/10-3 M
1.20
1.20
1.20
2.40
1.50
3.00
0.75
1.50
[NO2]/10-3 M
0.80
0.80
0.40
0.40
Write a rate law in agreement with the data.
rate пЂЅ k
[CO] [NO 2 ]
[NO]
Work out the rate law for each of the following mechanisms (show your work) and then circle the
one that is consistent with the rate law determined above. (Circle A, B, or C)
A)
CO + NO2 в†’ CO2 + NO slow
rate = k [CO] [NO2]
B)
NO2 ↔ NO + O
O + CO в†’ CO2
C)
2 NO2 ↔ 2 NO + O2
CO + O2 в†’ CO2 + O
O + NO в†’ NO2
No, doesn’t match
equil.
slow
[NO] [O]
[NO 2 ]
rate = k [O] [CO] and Keq пЂЅ
so, [O] пЂЅ Keq
[NO2 ]
[NO]
[CO] [NO 2 ]
[CO] [NO 2 ]
Substituting, we get rate пЂЅ kKeq
or rate пЂЅ k'
[NO]
[NO]
This matches our experimentally determined rate law so it is a possible mechanism.
equil.
slow
fast
rate = k [O2] [CO]
The only reason to use the equilibrium is to get rid of something that is not
in the overall reaction. Since the rate law does not contain anything that is
not already in the overall reaction (no intermediates) there is no reason to
continue. This does not match.
6) What is the solubility of Barium Carbonate in a solution saturated with H2CO3 at a pH of 10?
[H2CO3]sat'd = 0.034M.
BaCO3 ↔ Ba2+ + CO32- Ksp = 8.1x10-9
H2CO3 ↔ 2 H+ + CO32- Keq = 2.02x10-17
[H пЂ« ]2 [CO32пЂ­ ] (1x10пЂ­10 M) 2 (CO32пЂ­ )
пЂЅ
пЂЅ 2.02x10пЂ­17 пѓћ [CO32пЂ­ ] пЂЅ 68.68 M
[H 2CO 3 ]
0.034M
[Ba2+] [CO32-] = [Ba2+] [68.68 M] = 8.1x10-9
[Ba2+] = 1.18x10-10 M
7) A cell has the following cell diagram;
Ag / AgCl / 1M HCl // 1M AgNO3 / Ag
7a) What is the reduction half reaction?
Ag+ + e- в†’ Ag
7b) What is the overall reaction?
Ag+ + Cl- ↔ AgCl
7c) Based on the cell diagram please draw the electrochemical cell.
e-
Cathode
Positive
Reduction
Ag
Ag
AgCl(s)
1M
AgNO3
1M HCl
Anode
Negative
Oxidation
7d) If the Ksp for this cell is 1.8x10-10 then what is the cell voltage when all the ionic species are
0.01 M?
О”пѓµВ° = 0.0592/n log Ksp = 0.0592/(1) log (1/1.8x10-10) = 0.5769 volts
О”пѓµ = О”пѓµВ° - 0.0592/n log Q = О”пѓµВ° - 0.0592/n log [1/([Ag+][Cl-])]
О”пѓµ = 0.5769 V- 0.0592/(1) log 1/[(0.01)(0.01)]
О”пѓµ = 0.3401 volts
8) Please name or draw the following compounds.
OH
CH3
OH OH
C
C
C
H2N
para methylaniline
phenol
1,2-propadiol
Cl
O
C
C
C
C
C
C
O
C
H3C
C
C
C
3-methyl-2-pentanone
butanone
C
C
Cl
C
H
CH3
cis,trans-2,4-dichloro-3-methyl--2,4-hexadiene
3,3-dichloro butanoic acid
O
C
cyclopropyl ethyl ether
CH3
Cl
O
C
C
C
C
HO
trans 1,3-cyclopentadiol
C
C
Cl
o-methylaniline
NH2
OH
CH3
OH
Chemistry 121
Final Exam
Name
May 26, 2000
CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit.
You may use a calculator.
Question
Credit
Question
1(15)
6(20)
2(15)
7(15)
3(15)
8(30)
4(15)
9(20)
5(25)
10(30 )
Total
Total
Credit
TOTAL POINTS =
R = 8.314 J/mol-K
пѓµ = пѓµпЃ… - 0.0592/n log Q
О”GпЃ… = -nпѓ¶пѓµпЃ…
ln(Ai/Af) = kО”t
пѓ¶= 96486 C/mol eО”G = О”GпЃ… + RT ln Q
PV = nRT
1/Af - 1/Ai = kО”t
C = amp x sec
О”GпЃ…= -RTlnKeq
R = 0.08205 L-atm/mol-K
1) Which of the following compounds is the most soluble in water? Show your work.
BaSO4
Fe(OH)2
Ag3PO4
Ksp = 1.98x10-10
Ksp = 1.64x10-14
Ksp = 1.2x10-15
2) Please write the equilibrium reaction that will occur when the following compounds are mixed.
Write all reactions as dissociations where appropriate.
NaOH + HF
Na3Asp + HCl
HCl + KOH
3) The bromination of acetone is acid catalyzed;
CH3COCH3 + Br2 в†’ CH3COCH2Br + H+ + BrThe rate of disappearance of bromine was measured for several different concentrations of
acetone, bromine and hydrogen ions;
Rate
6x10-5
2.4x10-5
1.2x10-4
3.2x10-4
8x10-5
[Acetone]
0.30
0.30
0.30
0.40
0.40
[Br2]
0.050
0.100
0.050
0.050
0.050
[H+]
0.050
0.050
0.100
0.200
0.050
a) What is the forward rate law for this reaction?
b) What is the value and units of the forward rate constant?
4) What is the solubility of Cu2S in a saturated solution of 0.10M H2S that is buffered at pH=10?
Cu2S ↔ 2 Cu+ + S2H2S ↔ 2H+ + S2-
Ksp = 2x10-47
Keq = 1.1x10-20
5a) Set up a WORKING electrochemical cell that will determine the solubility product for HgO.
Note that this reaction is spontaneous in the reverse direction. Your overall reaction should be;
HgO + H2O в†’ Hg2+ + 2 OH-
Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all
ionic species. Indicate the composition of the electrodes.
5b) Draw the cell diagram for the above cell.
5c) If the пѓµпЃ… = 1.056 volts, what is the voltage of the cell when concentration of the ionic species
in the reduction cell are 0.5 M and the ionic species in the oxidation cell are 0.2 M?
6a) Assume that the cooking of an egg is a first order process. Assume further that it takes 3
minutes to half-cook (soft boil) an egg at 100В°C. If the activation energy of the cooking of an egg
is 75 kJ/mole, how long would it take to soft boil an egg at the top of Mt. Everest where water
boils at 80В°C?
6b) If the rate of a reaction doubles with a 15В°C change in temperature, what is it's activation
energy? State your assumptions.
7) A block of copper intially at 500В°C was placed on top of a 100 gram block of ice at 0В°C. All
of the ice melted and turned into water. The water and the block of copper ended up at 20В°C.
How much did the block of copper weigh? О”Hfus = 6.01 kJ/mol, Cp(Cu) = 24.5 J/mol-K,
Cp(H2O) = 75.2 J/mol-K, FWT (Cu) = 63.54 g/mol, FWT (H2O)= 18 g/mol
8) Calculate the pH at each of the following points for the titration of 100 mL of 3 M Phthalic
acid with 5 M NaOH
H2Phthal в†’ H+ + HPhthal- pKA1 = 2.89
HPhthal- в†’ H+ + Phthal2- pKA2 = 5.41
INITIAL pH
20 mL NaOH
60 mL NaOH
75 mL NaOH
90 mL NaOH
8b) How much 5 M NaOH would you add to 100 mL of 3 M H2Phthal to make a buffer of pH =
5?
9) How much heat must be added to 100 g of ice at -5В°C to turn it into 60 grams of water and 40
grams of steam, both at 100В°C? Cp H2O = 75.7 J/mol-K, Cp Ice = 37.7 J/mol-K, Cp Steam = 35.1
J/mol-K, О”Hfus = 6.01 kJ/mol, О”Hvap = 40.7 kJ/mol
Chemistry 121
Final Exam
Name Answer Key
May 26, 2000
CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit.
You may use a calculator.
Question
Credit
Question
1(15)
6(20)
2(15)
7(15)
3(15)
8(30)
4(15)
9(20)
5(25)
10(30 )
Total
Total
Credit
TOTAL POINTS =
R = 8.314 J/mol-K
пѓµ = пѓµпЃ… - 0.0592/n log Q
О”GпЃ… = -nпѓ¶пѓµпЃ…
ln(Ai/Af) = kО”t
пѓ¶= 96486 C/mol eC = amp x sec
О”G = О”GпЃ… + RT ln Q
О”GпЃ…= -RTlnKeq
PV = nRT
R = 0.08205 L-atm/mol-K
1/Af - 1/Ai = kО”t
1) Which of the following compounds is the most soluble in water? Show your work.
BaSO4
Fe(OH)2
Ag3PO4
Ksp = 1.98x10-10
Ksp = 1.64x10-14
Ksp = 1.2x10-15
(x)(x) = x2 = 1.98x10-10
(x)(2x)2 = 4x3 = 1.64x10-14
(3x)3(x) = 27x4 = 1.2x10-15
x = 1.41x10-5
x = 1.6x10-5
x = 8.16x10-5
2) Please write the equilibrium reaction that will occur when the following compounds are mixed.
Write all reactions as dissociations where appropriate.
NaOH + HF
F- + H2O ↔ HF + OH-
Na3Asp + HCl
HAsp2- ↔ H+ + Asp3-
HCl + KOH
H2O ↔ H+ + OH-
3) The bromination of acetone is acid catalyzed;
CH3COCH3 + Br2 в†’ CH3COCH2Br + H+ + BrThe rate of disappearance of bromine was measured for several different concentrations of
acetone, bromine and hydrogen ions;
Rate
6x10-5
2.4x10-5
1.2x10-4
3.2x10-4
8x10-5
[Acetone]
0.30
0.30
0.30
0.40
0.40
[Br2]
0.050
0.100
0.050
0.050
0.050
[H+]
0.050
0.050
0.100
0.200
0.050
a) What is the forward rate law for this reaction?
rate = k [Acetone] [Br2]2 [H+]
b) What is the value and units of the forward rate constant?
6x10-5 M/s = k [0.30 M] [0.050 M]2 [0.05 M]
k = 1.6 1/M3s
4) What is the solubility of Cu2S in a saturated solution of 0.10M H2S that is buffered at pH=10?
Cu2S ↔ 2 Cu+ + S2H2S ↔ 2H+ + S2-
Ksp = 2x10-47
Keq = 1.1x10-20
[H+]2 [S2-]/[H2S] = 1x10-20
[S2-] = 1x10-20 [H2S]/[H+]2 = 1x10-20 [0.10]/[1x10-10]2 = 0.10 M S2Substitute this result into the following equation,
[Cu+]2 [S2-] = 2x10-47
[Cu пЂ« ] пЂЅ 2 xпЂ±пЂ°пЂ­пЂґ7 /[S2- ] пЂЅ 2 xпЂ±пЂ°пЂ­пЂґ7 /[0.10] пЂЅ 1.41x10пЂ­23 M Cu пЂ«
[Cu 2S] пЂЅ
1.41x10-23
пЂЅ 7.07 x10 пЂ­24 M Cu 2S
2
5a) Set up a WORKING electrochemical cell that will determine the solubility product for HgO.
Note that this reaction is spontaneous in the reverse direction. Your overall reaction should be;
HgO + H2O ↔ Hg2+ + 2 OHThe half-reactions;
Hg2+ + 2 e- в†’ Hg
Hg + 2 OH- ↔ HgO + H2O + 2 e-
Cathode
Positive
Reduction
1 M NaOH
1 M Hg(NO3) 2
Anode
Negative
Oxidation
Hg2O
Pt
e-
Pt
Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all
ionic species. Indicate the composition of the electrodes.
5b) Draw the cell diagram for the above cell.
Pt / Hg(l) / Hg2O(s) / 1M NaOH // 1M Hg(NO3)2 / Hg / Pt
5c) If the пѓµпЃ… = 1.056 volts, what is the voltage of the cell when concentration of the ionic species
in the reduction cell are 0.5 M and the ionic species in the oxidation cell are 0.2 M?
пѓµ = пѓµпЃ… - 0.0592/n log 1/[Hg2+][OH-]2
пѓµ = 1.056 V - 0.0592/2 log 1/[0.5M][0.2M]2 = 1.0057 Volts
6a) Assume that the cooking of an egg is a first order process. Assume further that it takes 3
minutes to half-cook (soft boil) an egg at 100В°C. If the activation energy of the cooking of an egg
is 75 kJ/mole, how long would it take to soft boil an egg at the top of Mt. Everest where water
boils at 80В°C?
t1 = 3 min @ 100ВєC and t2 @ 80ВєC
ln
t2
75,000 пѓ¦ 373 пЂ­ 353 пѓ¶
пЂЅ
пѓ§
пѓ·
3 min
8.314 пѓЁ 373 п‚ґ 353 пѓё
t2 = 11.81 min
6b) If the rate of a reaction doubles with a 15В°C change in temperature, what is its activation
energy? State your assumptions.
Assume k1 @ 298 K and 2 k1 @ 313 K
ln
2k1
Ea пѓ¦ 313 пЂ­ 298 пѓ¶
пЂЅ
пѓ§
пѓ·
k1
8.314 пѓЁ 313 п‚ґ 298 пѓё
Ea = 35,834 J/mol
7) A block of copper intially at 500В°C was placed on top of a 100 gram block of ice at 0В°C. All
of the ice melted and turned into water. The water and the block of copper ended up at 20В°C.
How much did the block of copper weigh? О”Hfus = 6.01 kJ/mol, Cp(Cu) = 24.5 J/mol-K,
Cp(H2O) = 75.2 J/mol-K, FWT (Cu) = 63.54 g/mol, FWT (H2O)= 18 g/mol
nCpпЃ„TCu = nпЃ„пЃ€fus + nCpпЃ„TH2O
xg пѓ¦
J пѓ¶
100 g пѓ¦
J пѓ¶ 100 g пѓ¦
J пѓ¶
пЂ­
пѓ§ 24.5
пѓ·пЂЁ20п‚°C пЂ­ 500п‚°C пЂ© пЂЅ
пѓ§ 6010
пѓ·пЂ«
пѓ§ 75.2
пѓ·пЂЁ20п‚°C пЂ©
63.54 пѓЁ
molK пѓё
18 пѓЁ
mol пѓё
18 пѓЁ
molK пѓё
x g = 225.55 g Cu
8) Calculate the pH at each of the following points for the titration of 100 mL of 3 M Phthalic
acid with 5 M NaOH
+
-
H2Phthal ↔ H + HPhthal pKA1 = 2.89
HPhthal- ↔ H+ + Phthal2- pKA2 = 5.41
M1V1 = M2V2
(3 M)(0.100 L) = (5 M)(V2)
V2 = 60 mL
Also, (3 M)(0.1-0 L) = 0.300 mol H2Phthal
INITIAL pH
x2
пЂЅ 1.288x10пЂ­3 пѓћ x пЂЅ 0.06217 M H пЂ«
3пЂ­ x
20 mL NaOH
pH пЂЅ 2.89 пЂ« log
60 mL NaOH
pH пЂЅ
75 mL NaOH
pH пЂЅ 5.41 пЂ« log
90 mL NaOH
0.10
0.30 пЂ­ 0.10
pK A1 пЂ« pK A2 2.89 пЂ« 5.41
пЂЅ
2
2
0.075
0.30 пЂ­ 0.075
pH = 1.206
pH = 2.59
pH = 4.15
pH = 4.93
pH = 5.41
8b) How much 5 M NaOH would you add to 100 mL of 3 M H2Phthal to make a buffer of pH =
5?
The pH is nearest the 2nd pKa so the buffer is in Region II. You must add,
(3 M)(0.100 L) = (5 M)(V) V = 60 mL to get to the starting point.
5 пЂЅ 5.41 пЂ« log
(x)
пѓћ x пЂЅ 0.0840 mol NaOH = (5 M)(V)
(0.30 - x)
Total volume = 16.8 mL + 60 mL = 76.8 mL NaOH
V = 16.8 mL
9) How much heat must be added to 100 g of ice at -5В°C to turn it into 60 grams of water and 40
grams of steam, both at 100В°C? Cp H2O = 75.7 J/mol-K, Cp Ice = 37.7 J/mol-K, Cp Steam = 35.1
J/mol-K, О”Hfus = 6.01 kJ/mol, О”Hvap = 40.7 kJ/mol
100g пѓ¦
J пѓ¶
100g пѓ¦
J пѓ¶ 100g пѓ¦
J пѓ¶
40g пѓ¦
J пѓ¶
пѓ§ 37.7
пѓ·пЂЁ5п‚°C пЂ© пЂ«
пѓ§ 6010
пѓ·пЂ«
пѓ§ 75.2
пѓ·пЂЁ100п‚°C пЂ© пЂ«
пѓ§ 40,700
пѓ· пЂЅ 166,658 J
18 пѓЁ
molK пѓё
18 пѓЁ
mol пѓё
18 пѓЁ
molK пѓё
18 пѓЁ
mol пѓё
Chem 121 Practice Problems
Final Exam
Nomenclature
Isomers
Free Radical Halogenation
Hybridization
Sigma and Pi bonds
Intermolecular Forces
Mechanism of Free Radical Halogenation using Bromine
Chain Initiation
light
Br2
2 Br
Chain Propagation
Br
H
H
H
H
H
H
C
C
C
H
H
H
H
H
H
C
C
C
H
H
H
Br2
H
+
HBr
H
H
H
C
C
C
H
H
H
H
Br H
C
C
C
H
H
H
H
Br H
C
C
C
H
H
H
+ Br
H
Chain Termination
Br
Br
H
Br
H
Br2
H
C
C
C
H
H
H
H
H
H
H
H
H
C
C
C
H
H
H
H
H
H
H
H
C
C
C
H
H
H
H
Relative Reactivity
1п‚° 2п‚°
3п‚°
Chlorination 1
3.5 5
Bromination 1
97
пЂґ
H
C
H H
H
C
H
H
C
H
H
H
C
C
H
H
H
C
H
H
Section I
3) Please draw all of the isomers of C4H7Cl. There are more than a dozen.
4) Please draw all of the isomers of C3H5Cl.
5) Please draw all of the isomers of C3H4Cl2.
6) Please draw all of the isomers of C3H8O
7) Please answer the following questions concerning the molecule given below.
a) How many pi and sigma bonds are in this compound?
O8
C
7
9
10
O
CH3
1
6
b) What is the hybridization on the following atoms;
1
2
5
7
8
9
3
5
4
8) Please answer the following questions concerning the molecule given below.
8
2
a) How many pi and sigma bonds are in this compound?
O
1
H b) What is the hybridization on the following atoms;
N3
C
C
4
Cl
C
6
5
Cl 7
2
3
4
5
7
8
9) Please answer the following questions concerning the molecule given below.
7
O
6C
N
8
3
9
C
a) How many pi and sigma bonds are in this compound?
10
b) What is the hybridization on the following atoms;
5
4
C
1
2
1
2
5
7
8
14) Please describe what happens to the melting point of a large alkene (C20 or larger) if the
double bond changes from cis to trans. Also explain why it does this.
15) Please explain what happens to the melting point of an alkane as it becomes more branched.
16) What kinds of compounds can hydrogen bond? What is hydrogen bonding?
17) What is the major intermolecular bonding type for each of the following compounds?
CH3CH2NH2
CH2Cl2
CH3COOH
18) Which compound in each pair has the highest boiling point?
CH3OCH2CH3
or CH3CH(OH)CH3
CH3CH2CH2CH2CH3
or CH3CH2CH2CH3
(CH3)2CHCH3 or CH3CH2CH2CH3
19) Please predict the primary type of bonding that will occur in each of the following liquids.
CCl4
C16H34
Propanone
H-bond
H-bond
H-bond
Dipole-Dipole
Dipole-Dipole
Dipole-Dipole
Dispersion
Dispersion
Dispersion
Section I – Answer Key
3) Please draw all of the isomers of C4H7Cl. There are more than a dozen.
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl Cl
Cl
Cl
Cl
4) Please draw all of the isomers of C3H5Cl.
Cl
Cl
Cl
Cl
5) Please draw all of the isomers of C3H4Cl2.
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
6) Please draw all of the isomers of C3H8O
HO
OH
O
Cl
7) Please answer the following questions concerning the molecule given below.
a) How many pi and sigma bonds are in this compound? 5 pi and 16 sigma
O8
C
7
9
10
O
CH3
1
6
b) What is the hybridization on the following atoms;
1 sp2
2
5 sp
7 sp2
8 sp2
9 sp3
3
5
4
8) Please answer the following questions concerning the molecule given below.
8
2
a) How many pi and sigma bonds are in this compound?
4 pi and 16 sigma
O
1
N3
H
C
4
Cl
C
5
C
6
Cl 7
b) What is the hybridization on the following atoms;
2 sp2 3 sp3 4 sp2
5 sp
7 sp3 8 sp3
9) Please answer the following questions concerning the molecule given below.
7
O
6C
N
C
C
8
9
10
a) How many pi and sigma bonds are in this compound?
3 pi and 21 sigma
1
b) What is the hybridization on the following atoms;
5
4
3
2
1 sp2
2 sp2
5 sp3
7 sp2
8 sp3
14) Please describe what happens to the melting point of a large alkene (C20 or larger) if the
double bond changes from cis to trans. Also explain why it does this.
The melting point increases. A cis bond puts a kink in the chain that makes it hard to
stack. By changing the cis to a trans the chain straightens out and makes it easier to
stack. This allows for more dispersion forces and increases the melting point.
15) Please explain what happens to the melting point of an alkane as it becomes more branched.
The more branched an alkane becomes the lower the melting point.
Branching makes it more difficult to stack the molecules and this reduces
the molecules ability to use dispersion forces for bonding.
16) What kinds of compounds can hydrogen bond? What is hydrogen bonding?
Hydrogen bonds are formed between molecules that have OH or NH bonds (also HF but
HF is not an organic compound).
A hydrogen bond is a sharing of protons between the oxygen or nitrogen molecules. The
hydrogen atoms freely move between the oxygens or nitrogens by sharing grabbing onto
the lone pair electrons found on these atoms. Each atom of oxygen or nitrogen can have
up to 4 atoms of hydrogen surrounding them, each of them sharing their electrons.
17) What is the major intermolecular bonding type for each of the following compounds?
CH3CH2NH2 = H bond
CH2Cl2 = Dipole-dipole
CH3COOH = H bond
18) Which compound in each pair has the highest boiling point?
CH3OCH2CH3
or CH3CH(OH)CH3(because of H bonds)
CH3CH2CH2CH2CH3 (because it is bigger) or
CH3CH2CH2CH3
(CH3)2CHCH3 or CH3CH2CH2CH3(because it is less branched)
19) Please predict the primary type of bonding that will occur in each of the following liquids.
CCl4
C16H34
Propanone
H-bond
H-bond
H-bond
Dipole-Dipole
Dipole-Dipole
Dipole-Dipole
Dispersion
Dispersion
Dispersion
Section II
1) Please name or draw the structure of the following compounds.
H
CH3
Cl
CH3
H3C
H
H3C
CH3
Cl
Br
H3C
Br
Cl
3) Please draw all of the isomers of C3H3Cl
C
CH3
CH3
4) Please answer the following questions concerning the molecule given below.
O
3
C
O
4
C
CH3
a) How many pi and sigma bonds are in this compound?
N2
1
N
C
b) What is the hybridization on the following atoms;
1
2
3
4
6a) Please write down all the steps in the mechanism of the free radical chlorination of 2methylbutane..
6b) What is the ratio of products formed in the free radical chlorination of bicyclo [3,2,0]
heptane?
7b) What happens to the boiling point of an alkane when it becomes branched?
7c) How do the intermolecular bonding forces change when acids become larger?
7d) What is the primary intermolecular bonding force in each of the following compounds?
Isopropyl Alcohol
H-Bond
Dipole-Dipole
Dispersion
1,3-dichloropropane
H-Bond
Dipole-Dipole
Dispersion
C12H25NH2
H-Bond
Dipole-Dipole
Dispersion
methylcyclohexane
H-Bond
Dipole-Dipole
Dispersion
Section II – Answer Key
1) Please name or draw the structure of the following compounds.
H
CH3
Cl
CH3
H3C
Bicyclo [4, 2, 1] nonane
H
S-3-chloro-2-methylbutane
H3C
CH3
cis-1,4-dimethylcyclohexane
2,2,4-trimethylpentane
Cl
Br
H3C
C
Br
CH3
CH3
Cl
Z-bromo-1,2-dichlorocyclobutane
2-bromo-2-methylpropane
3) Please draw all of the isomers of C3H3Cl
H
Cl
C
C
H
C
H
H
C
C
H
H
C
C
Cl
H
Cl
C
C
H
Cl
H
H
H
H
Cl
H
H
H
4) Please answer the following questions concerning the molecule given below.
O
3
C
N2
1
N
C
O
4
C
CH3
a) How many pi and sigma bonds are in this compound?
20 sigma and 6 pi
b) What is the hybridization on the following atoms;
1
2
3
4
sp
sp3
sp3
sp3
6a) Please write down all the steps in the mechanism of the free radical chlorination of 2methylbutane.
Chain Initiation
light
Cl2
2 Cl
Chain Propagation
H
Cl
H3C
C
C2H5
HCl
+
H3C
C
C2H5
CH3
CH3
Cl
H3C
C
Cl2
C2H5
H3C
C
CH3
C2H5
+ Cl
CH3
Chain Termination
Cl
Cl
Cl2
Cl
Cl
H3C
C
H3C
C2H5
C
C2H5
CH3
CH3
CH3
H3C
C
C2H5
H3C
C
CH3
C2H5
CH3
H3C
C
C2H5
H3C
C
C2H5
CH3
6b) What is the ratio of products formed in the free radical chlorination of bicyclo [3,2,0]
heptane?
secondary
tertiary
X
100-X
=
10 x 3.5
2 5
X = 77.78% secondary
22.22% tertiary
7b) What happens to the boiling point of an alkane when it becomes branched?
The boiling point goes down
7c) How do the intermolecular bonding forces change when acids become larger?
As acids get larger they get less soluble in water so the intermolecular bonding force
changes from being mostly H-bond to mostly dispersion forces
7d) What is the primary intermolecular bonding force in each of the following compounds?
Isopropyl Alcohol
H-Bond
Dipole-Dipole
Dispersion
1,3-dichloropropane
H-Bond
Dipole-Dipole
Dispersion
C12H25NH2
H-Bond
Dipole-Dipole
Dispersion
methylcyclohexane
H-Bond
Dipole-Dipole
Dispersion
Section III
1) Please name or draw the structure of the following compounds.
Cl
Cl
C
C
CH3
Cl
H3C CH C
C
CH3
C
C
CH3
3) Please draw all of the isomers of C3H4Cl2
H3C
Cl
H
H
C2H5
4) Please answer the following questions concerning the molecule given below.
O
2
a) How many pi and sigma bonds are in this compound?
3
N
O
C
OH
4
1
b) What is the hybridization on the following atoms;
5
C
1
N
2
3
4
5
7a) Hexane and cyclohexane both have six carbons but one has a higher boiling point than the
other. Which one has the higher boiling point and why?
7b) Rank the following compounds from highest to lowest boiling point (highest =4, lowest = 1)
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
OH
C
7d) What is the primary intermolecular bonding force in each of the following compounds (Hbond, Dipole, Dispersion)?
C2H5NH2
cis 2,3-dichlorobutene
CH3(CH2)4CH2Br
C12H25COH
Section III – Answer Key
1) Please name or draw the structure of the following compounds.
Cl
Cl
Cis-1,2-dichlorocyclpentane
C
C
Bicyclo [4,1,0] heptane
CH3
Cl
H3C CH C
C
CH3
C
C
CH3
H3C
Cl
H
H
4-ethyl-5-methyloctane
C2H5
2R,3R-2,3-dichloropentane
3) Please draw all of the isomers of C3H4Cl2
C Cl
Cl
C
C
C
C
C
C
Cl
C
C
Cl
C
Cl
Cl
Cl
C
C
C
Cl
Cl
Cl
C
C
C
C
Cl
Cl
C
Cl
C
C
C
Cl
Cl
Cl
Cl
Cl
Cl Cl
Cl
4) Please answer the following questions concerning the molecule given below.
O
a) How many pi and sigma bonds are in this compound?
3
2
N
18 sigma and 4 pi
O C OH
4
1
5
C
N
b) What is the hybridization on the following atoms;
1
2
3
4
5
sp2
sp3
sp3
sp2
sp
Cl
6) Please predict the percentage of products made by the free radical chlorination of bicycle
[4,2,0] octane.
secondary
tertiary
X
100-X
=
12 x 3.5
2
5
X = 80.77% secondary
19.23% tertiary
7a) Hexane and cyclohexane both have six carbons but one has a higher boiling point than the
other. Which one has the higher boiling point and why?
Hexane has the higher boiling point because it is longer and less compact than
cyclohexane so it has more opportunity to exert dispersion forces on its neighbors.
7b) Rank the following compounds from highest to lowest boiling point (highest =4, lowest = 1)
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
OH
C
1
3
2
4
7d) What is the primary intermolecular bonding force in each of the following compounds (Hbond, Dipole, Dispersion)?
C2H5NH2
H-bond
CH3(CH2)4CH2Br Dipole-Dipole (borderline)
cis 2,3-dichlorobutene Dispersion
C12H25COH Dispersion (too big)
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