1. science
2. chemistry

-0-
Rules in My Class
1. Cheating on a test will result in a zero for a major exam.
Plagiarizing a lab report will result in a zero for that lab report.
2. Cutting class will result in a zero for any test or lab report expected
for that class day. In addition, I will require a note from your parents
informing me that they are aware you've cut class and have
received a zero for any test or lab report expected for that class day.
3. Going beyond the allotted time on a test will result in loss of all
credit for that question/problem.
4. Work missed due to an illness must be made up within a reasonable
period of time on returning to school.
5. Lab reports will not be accepted late, incomplete, or if they do not
meet guidelines set by me.
6. You are not allowed to eat in class.
-1-
Grading Policy
First Semester:
Tests comprise 54% of the semester grade. Tests consist of free response
which may require several concepts pieced together to provide a complete
response or math problems which require setups that lead to the answer.
Lab Work comprises 36% of the semester grade. Lab grades are built around
work in the lab, analyzing data, and interpreting what occurred during the lab.
Lab reports turned in late receive a zero.
Final Exam comprises 10% of the semester grade.
Second Semester:
60% for tests
40% for lab work
Assignments are usually begun in class. Those students needing additional time
finish the work outside of class.
The AP Chem Website provides an overview of each unit, assignments, handouts
along with accompanying answers to the text and handouts.
http://www.mro-chemweb.com
If you have questions and/or concerns, and you e-mail those to me BEFORE
8:00 P.M., I will try to get back to you the same evening.
[email protected]
-2-
Assignments
The Mole Concept & Stoichiometry
Readings:
Questions:
Experiments:
Chapter 2 & 3
2.5c, 78, 79, 84, 85; 3.119, 120, 123, 126 – 128, 135
(For Fun: 136, 140, 141)
Basic Laboratory Techniques
Identification of Substances by Physical Properties
Separation of the Components of a Mixture
Chemical Reactions of Copper & Percent Yield
Gravimetric Analysis of a Chloride Salt
Atomic Structure & The Periodic Table
Powerpoint:
Readings:
Questions:
Experiments:
AP Questions:
Atomic Theory & Structure
2.2, 2.3; Chapter 7
2.57
7.33, 36, 65, 66, 69, 71, 72, 85, 87, 88, 95, 97, 103, 104, 111,
112, 141, 142, 145 (For Fun: 138)
Beer's Law: Determining the Concentration of a Solution
2007: #2, #6; 2006: #5, #7; 2002: #6a,b
Bonding & Molecular Geometry
Powerpoint:
Readings:
Questions:
Handouts:
Video:
AP Questions:
Bonding & Molecular Geometry
Chapter 8 & 9
8.3 – 5, 8, 22, 24, 37, 46, 47, 51, 54, 69, 70, 82, 89 – 92,
94 – 100, 103, 113, 116, 118, 121
9.13, 14, 25, 27, 34, 39 – 43, 50 – 61, 77, 82, 86, 89, 93,
95 – 97, 99
Formal Charge Worksheet
Drawing Lewis Structures Worksheet #1, #2
Hybrid Orbital Identification Worksheet
Review of Lewis Structures, Hybrid Orbitals & Formal Charges
Bonding Worksheet
Signals From Within
2011: #6; 2006: #6; 2005: #8a – c; 2002: #6c
Gas Laws
Powerpoint:
Readings:
Gas Laws
Chapter 10
-3-
Questions:
Handouts:
Experiments:
AP Questions:
10.4, 19, 30, 35 – 37, 39, 41, 43, 47, 49, 51, 55, 57c, 59, 60,
63 – 65, 67 – 73, 75, 79 – 81, 97, 99
Gas Law Problems Worksheet
Dalton’s Law Worksheet
Behavior of Gases: Molar Mass of a Vapor
Determination of R: The Gas Law Constant
2011: #2; 2009: #3; 2005: # 6; 2004: #2; 2002: #3
Liquids, Solids & Changes of State
Powerpoint:
Readings:
Questions:
Handouts:
AP Questions:
Intermolecular Attractive Forces
Chapter 11
11.2, 5 – 7, 9 – 11, 16, 17, 20 – 24, 26, 29, 30, 32, 33, 36,
39 – 41, 43 – 44, 75 – 87, 102 – 110, 115
IMF Worksheet
2005: # 8d,c; 2002: #6d
Reactions in Aqueous Solutions & Physical Properties of Solutions
Powerpoint:
Readings:
Questions:
Handouts:
Experiments:
AP Questions:
Solution Chemistry
Chapter 4 & 12
4.11, 16, 17, 21, 33, 35, 40, 45, 49, 52, 55, 57 – 60, 63, – 66,
67a,b, 68a,c, 69, 71 – 79, 81b, 82b, 83b, 84a, 85, 87, 89, 92,
93, 95c,d, 97 – 99, 102, 104, 106, 109 – 114, 117, 120,
5.30, 49, 55, 56, 65, 67 – 70, 75, 76, 82
12.3, 6, 7, 9 – 12, 15, 16, 22, 23, 26, 33, 35, 36, 39, 44, 45,
47, 48, 50, 54 – 58, 61 – 63, 66, 72
Balancing Redox Reactions
Solution Chemistry Worksheet #1 – #3
Net Ionic Equations #1 – #3
Chemical Reactions
Solubility & Fractional Crystallization
Colligative Properties: Freezing Point Depression & Molar
Mass
20101: #3; 2008: #3, 5; 2005: #5
Thermochemistry, & The Spontaneity of Chemical Reactions
Powerpoint:
Readings:
Questions:
Chemical Thermodynamics
Chapter 6 & 18
6.3a-c, 11, 14, 15, 18, 19, 21 – 24, 27, 32 – 34, 38, 39, 45,
48 – 51, 55, 57, 59, 61 – 64, 67, 68, 71, 74, 76, 78, 84
11.45 – 49
18.4, 7, 9, 12, 13, 18 – 20, 23, 26, 27, 34 – 37, 44 – 46, 55,
56, 58, 60c-e, 66 – 68, 70, 74 – 76, 79, 88 – 90, 92, 94 – 96
-4-
Handouts:
Experiments:
AP Questions:
Thermochemical Worksheet
Heats of Reaction & Bond Enthalpies Worksheet
Heat of Fusion
Hess’s Law
2011: #3; 2009: #5; 2008: #6; 2006: #3; 2005: #7; 2004: #7;
2002:#5, #8
Kinetics
Powerpoint:
Readings:
Questions:
Handouts:
Experiments:
AP Questions:
Kinetics
Chapter 13
13.14, 15, 17 – 20, 25, 27, 29, 31, 33, 34, 37, 38, 40 – 42, 51,
53, 55, 56, 59, 63, 64, 67, 68, 71, 73 – 76, 79, 83, 84, 91, 93,
94, 95, 97
Kinetics Worksheet
Rates of Chemical Reactions: A Clock Reaction
2010: #6; 2009: #2; 2008: #2; 2006: #3; 2005: #3; 2004: #3;
2002:# 7
Chemical Equilibria, Solubility & Complex Ion Equilibria
Powerpoint:
Readings:
Questions:
Handouts:
Experiments:
AP Questions:
Chemical Equilibrium
Chapter 14 & 17
14.12, 13, 15, 18a,b, 21, 23d, 25, 26, 28, 29, 31, 35, 36, 39,
40, 43a,d, 44b,c, 54, 55, 57, 59, 61, 65
17.2, 3, 16a,c,d, 18 – 21, 32, 36, 38, 42, 53, 55
18.39 – 42, 80 – 82, 84, 87
Equilibrium Worksheet
Determination of the Ksp for a Sparingly Soluble Salt
2011: #1; 2010: #1; 2009: #1; 2008: #1; 2007: #1; 2006: #1;
2005: #1; 2004: #1, #8; 2002:# 1
Acid-Base Concepts Revisited & Acid-Base Equilibria
Powerpoint:
Readings:
Questions:
Handouts:
Experiments:
Acid-Base Chemistry
Chapter 15 & 16
15.4, 6, 9, 10, 14, 17, 19, 21, 25, 29, 31, 33, 39, 40, 43b,c,
44c,d, 59a, 60c, 63a, 64a, 67, 77, 78
16.2b, 5, 6, 8a, 9, 11, 12, 14, 15, 18, 20, 22, 24, 28, 29, 32,
36, 39, 40, 44 – 46, 49, 52, 56, 58, 62, 64, 88, 89
Acid Base Worksheet
Acid – Base Reactions Worksheet #1, #2
Buffers & Salts Worksheet
Ka for a Weak Acid
-5-
AP Questions:
Titration Curves of Polyprotic Acids
2011: #5; 2010: #5; 2007: #5; 2004: #5
Electron Transfer Reactions, Electricity & Chemical Change
Powerpoint:
Readings:
Questions:
Experiments:
AP Questions:
Electrochemistry
Chapter 5 & 19
19.2, 4, 5, 7, 8, 10, 11, 13, 15, 17 – 20, 24, 26, 28, 29, 50, 52,
55, 56a,d, 57a,d, 58, 59, 62, 63, 70, 78, 81, 83, 84, 87, 88
Activity Series of Metals
Electrolysis, the Faraday, & Avogadro's Number
Electrochemical Cells & Thermodynamics
2010: #2; 2009: #6; 2007: #3; 2006: #2; 2005: #2; 2004: #6;
2002: #2
-6-
Determining the Concentration of a Solution:
Beer’s Law
Introduction
The purpose of this lab is to become acquainted with how to use a colorimeter and
how it is used to determine the concentration of a solution.
Equipment & Materials
LabQuest
Vernier colorimeter
pipet pump
6 cuvettes
5 - 20 × 150 mm test tubes
1 - 10 mL pipet
2 - 100 mL beakers
0.40 M CuSO4(aq)
CuSO4(aq), unknown solution
distilled water
test tube rack
stirring rod
Kimwipe
Procedure
1.
Label five clean, dry, test tubes 1–5. Use pipets to prepare five standard
solutions according to the chart below. Thoroughly mix each solution with a
stirring rod. Clean and dry the stirring rod between uses.
Trial
1
2
3
4
5
CuSO4(aq)
(mL)
2
4
6
8
10
Distilled H2O
(mL)
8
6
4
2
0
Concentration
(M)
0.08
0.16
0.24
0.32
0.40
2.
Connect a colorimeter and your computer to the LabQuest.
3.
In Logger Pro, under the file menu, open the Advanced Chemistry Folder,
and then open file 17 - Colorimeter.
4.
Calibrate the colorimeter.
a)
Open the Experiment menu, click on Calibrate Colorimeter, click on
the Calibrate Tab, and click Calibrate Now.
-7-
b)
c)
d)
5.
Prepare a blank by filling an empty cuvette 3⁄4 full with distilled water.
Place the blank in the cuvette slot of the colorimeter and close the lid.
Turn the wavelength knob to the “0% T” position. When the voltage
reading stabilizes, type 0% in the first data box, and click
.
Move the knob to the Red LED position (635 nm), and when the reading
stabilizes, type 100% in the data box, and click
, then click Done.
To collect absorbance-concentration data for the five standard solutions:
a)
Remove the cuvette from the colorimeter and pour out the distilled
water. Using the solution in Test Tube 1, rinse the cuvette twice with ~1
mL amounts, and then fill it 3⁄4 full. Wipe the outside with a tissue, place
it in the colorimeter, and close the lid.
b)
Click
c)
When the absorbance readings stabilize, click
edit box, and press the ENTER key.
d)
Discard the cuvette contents, and repeat the procedure for test tubes 2,
3 and 4. Trial 5 is the original 0.40 M CuSO4 solution.
e)
When you finish testing the standard solutions, click
f)
Click the Linear Regression button, and a best-fit linear regression line
will be shown for your five data points.
.
, type 0.080 in the
.
6.
Record the absorbance values, for each of the five trials, in your data table.
7.
Determine the absorbance value of the unknown CuSO 4(aq).
a)
Rinse a cuvette with distilled water, followed by rinsing twice with the
unknown solution. Fill the cuvette about 3⁄4 full with the unknown. Wipe
the outside of the cuvette, place it into the colorimeter, and close the lid.
b)
Record the absorbance value in your data table.
c)
Dump the solutions into the sink, rinse the cuvettes, and turn these
upside down so they can drain.
-8-
DATA TABLE
Trial
1
2
3
4
5
Unknown #
Concentration
(M)
0.080
0.16
0.24
0.32
0.40
Absorbance
Questions
1.
For a given substance, the amount of light absorbed depends on what four
factors?
2.
How are percent transmittance and absorbance related algebraically?
3.
State the Beer-Lambert Law.
4.
What is the purpose of preparing a calibration curve?
5.
Why is the line on the cuvette always aligned with that of the sample holder?
-9-
- 10 -
- 11 -
- 12 -
- 13 -
- 14 -
Using Formal Charge to Predict Best Lewis Structure
Developing Possible Lewis Structures
●
●
●
●
Add up valence electrons
Add to or take away electrons based on the charge on a radical
Place a pair of electrons between central atom and surrounding atoms
With the exception of hydrogen, the remainder of the electrons are placed
around the surrounding atoms till four pairs of electrons surround them.
● Place any remaining electrons on central atom
Calculating Formal Charge (FC)
FC = Valence Electrons –Non-Bonding Electrons – Bonding Pairs
Determining The Best Lewis Structure
IF central atom has a FC of zero or a charge equal to its normal charge, the Lewis
dot structure is correct.
IF NOT, begin forming multiple bonds with the surrounding atom that has the most
negative FC; this will lead to a reduction in the FC, and to a lower potential energy
state.
● For a central atom in the second period, continue this process until the
central atom as an octet.
● For a central atom in the third period or lower, continue this process until
the FC on central atom is reduced to zero.
What if the choice is 2+ 2-
versus 1+ 1-
?
More potential energy is required to separate larger charges than smaller charges;
English . . . larger charges are less favored.
What other Considerations Are There?
Negative FC on atoms that are more electronegative, and Positive FC on atoms
that are less electronegative atoms are preferred.
- 15 -
Exercises Involving Formal Charge
http://www.cem.msu.edu/~reusch/VirtualText/Questions/General/resnce1.htm
http://www.usm.maine.edu/~newton/Chy251_253/Lectures/Formal%20Charge/FC
Exercises.html
- 16 -
Formal Charge Worksheet
Which structures satisfy the following conditions?
1.
No formally charged atoms are present in the structure.
2.
At least one nitrogen has a (+) formal charge
3.
At least one nitrogen has a (-) formal charge
4.
At least one oxygen has a (+) formal charge
5.
At least one oxygen has a (-) formal charge
6.
At least one carbon has a (+) formal charge
7.
At least one carbon has a (-) formal charge
8.
At least one sulfur has a (+) formal charge.
- 17 -
Drawing Lewis Structures Worksheet #1
PH4 +
CH2O
SiH4
SeF6
SeF4
ICl3
NO +
NO2 -
HCN
AsCl5
ICl2 -
XeF4
BF4 -
SCl2
PCl4 +
- 18 -
ClF3
SeCl4
PCl5
ClO3 –
SeO3 2-
OSCl2
Draw resonance structures for:
SO3
SO2
CH3COO -
N2O4
- 19 -
Drawing Lewis Structures Worksheet #2
Obey Octet Rule:
AsBr3
H3CCN
SeF2
O2SF2
TeF4
ClF5
CS2
NI3
H2CO3
PF6 -
SbCl6 -
XeF2
SiCl4
CN -
SnCl4
- 20 -
FCl2 +
PBr3
AsCl4 +
IO3 -
ICl4 -
BrF4 -
SeO3
C2O4 2-
Draw resonance structures for:
HN3
- 21 -
Hybrid Orbital Identification Worksheet
1.
For each structure shown below, identify the hybrid orbital on each atom, any
missing electrons, and the molecular geometry around each atom.
- 22 -
2. Draw the complete Lewis structures for the following molecules.
a) HC(NH2)HCOOH
c) CH3C(NH2)HCOOH
b) CH3CH(CH3)COOH
d) CH2CHCH2CH2Cl
- 23 -
Review of Lewis Structures, Hybrid Orbitals
& Formal Charges
1.
Draw Lewis structures for each of the following molecules/radicals, and then
identify:
a)
b)
c)
d)
the geometry
non-bonded electron pairs on each atom
formal charges on each atom
hybrid orbital on central atom
SO2
ClF5
O3
SnCl2
CH2O
SOF4
CO3 2-
NO2 -
SO3 2-
SnCl5 -
- 24 -
PH4 +
2.
In the following molecules
a)
which has the most polar bond?
b)
which has a net dipole greater than zero?
BH3
3.
CH4
H2O
HF
Write complete Lewis structures for each of the following compounds:
CH3C = N = O
CH2Cl2
CH2OHCHOHCH = O
H3CCN
- 25 -
CH3OCH3
CH3NH2
Bonding Worksheet
1.
Briefly discuss the relationship between the dipole moment of a molecule and
the polar character of the bonds within it. Account for the difference between
the dipole moments of CH2F2 and CF4.
2.
Suppose that a molecule has the formula AB 3. Sketch and name two
different molecular geometries this molecule may have. For each of the two
molecular geometries, give an example of a known molecule that has that
shape, and then identify the molecular geometry of each shape.
3.
For CO3 2-, CO2, and CO,
a)
draw the Lewis structures for all possible resonance structures.
b)
which of the three species has the shortest C-O bond length? Explain
the reason for your answer.
c)
predict the molecular geometry of the three species. Explain how you
arrived at your predictions.
- 26 -
4.
5.
Explain the following trends in lattice energy
a)
CaF2 > BaF2
b)
NaCl > RbBr
c)
BaO > KF
Calculate the lattice energy (and diagram the Born-Haber cycle) for CaH2
using the following information:
Electron affinity for H
Ionization energy for Ca
Ionization energy for Ca 1+
∆Hsubl for Ca
Bond dissociation energy for H2
∆Hform CaH2
- 27 -
- 72.8 kj/mol
+ 589.8 kJ/mol
+ 1145 kJ/mol
+ 178.2 kJ/mol
+ 435.9 kJ/mol
- 186.2 kJ/mol
6.
For each of the following molecules,
PH3
SF2
NI3
a)
draw the best Lewis structure
b)
identify the molecular geometry
c)
identify the hybridization on the central atom
d)
state whether the molecule is polar or nonpolar
- 28 -
OF2
Gas Law Problems Worksheet
1.
Air bags are activated when a severe impact causes a steel ball to compress
a spring and electrically ignite a detonator cap. This causes sodium azide,
NaN3, to decompose explosively according to the following unbalanced
equation:
NaN3(s)

Na(s) + N2(g)
What mass of NaN3(s) must react to inflate an air bag to 70.0 L at STP?
2.
Urea, NH2CONH2 is a nitrogen fertilizer that is manufactured from ammonia
and carbon dioxide, identified in the following unbalanced equation:
NH3(g) + CO2(g)  NH2CONH2(aq) + H2O(l)
0.908 g of ammonia reacts with 178 mL of CO 2 at a pressure of 35 C and
1.50 atm. How many grams of NH2CONH2 will be produced?
3.
When methane, CH4, reacts with molecular chlorine, hydrogen chloride and
carbon tetrachloride are produced. If 49.62 mL of methane reacts at a
temperature of 465.° C and at a pressure of 1400. torr, with 75.00 mL of
molecular chlorine, how many grams of carbon tetrachloride will be
produced?
- 29 -
4.
Calculate the volume of H2(g), measured at 26 C and 751 mm Hg, required
to react with 28.5 L CO2(g), measured at 0 C and 760 mm Hg, in the
following unbalanced equation:
3 CO(g) + 10 H2(g)
5.

C3H8(g) + 6 H2O(l)
Hydrogen cyanide is prepared commercially by the reaction of methane,
CH4(g), ammonia, and molecular oxygen, at high temperatures. The other
product is gaseous water. What volume of HCN(g) can be obtained from
20.0 L CH4(g), 20.0 L NH3(g), and 20.0 L O2(g)?
2 CH4(g) + 2 NH3(g) + 3 O2(g)  2 HCN(g) + 6 H2O(g)
6.
A compound has the empirical formula CHCl. A 256 mL flask, at 373 K and
750. torr, contains 0.800 g of the gaseous compound. Identify the molecular
formula.
- 30 -
7.
A gas consisting of only carbon and hydrogen has an empirical formula of
CH2. The gas has a density of 1.65 g/L at 27 C and 734 torr. Determine the
molar mass and molecular formula of the gas.
8.
The percent composition of a SFx compound is 29.69% S and 70.31% F. At
20 C, 0.100 g of the gaseous compound occupies a volume of 22.1 mL and
exerts a pressure of 1.02 atm. What is the molecular formula of this gas?
9.
An anesthetic contains 64.9% C, 13.5% H, and 21.6% O, by mass. At
120 C and 750 mm Hg, 1.00 L of the gaseous compound weighs 2.29 g.
What is the molecular formula of the compound?
- 31 -
Dalton’s Law Worksheet
1.
A mixture of 4.0 g of molecular hydrogen and 10.0 g of He(g) in a 4.3 L flask
is maintained at 0 C. What is the partial pressure of each gas? What is the
total pressure?
2.
A 2.00 L flask is filled with Ar(g) at 752 mm Hg and 35 C. A 0.728 g sample
of benzene, C6H6(g) is then added.
3.
a)
What is the total pressure in the flask?
b)
What is the mole fraction of each gas inside the flask?
A 1.65 g sample of Al reacts with excess HCl(aq), producing aluminum
chloride and H2(g). The molecular hydrogen is collected over water at 25 C
at a barometric pressure of 744 mm Hg. What volume of dry H 2(g), in L, is
collected?
(PH2O at 25 C = 23.8 mm Hg)
- 32 -
4.
Concentrated hydrogen peroxide solutions, H2O2(aq), are explosively
decomposed by traces of transition metal ions:
2 H2O2(aq)

2 H2O(l) + O2(g)
What volume of pure O2(g), collected at 27 C and 746 torr, would be
generated by decomposition of 125. g of hydrogen peroxide solution?
5.
On warming, formic acid, HCOOH, decomposes into:
HCOOH(l)  H2O(l) + CO(g)
If 3.85 L of carbon monoxide was collected over water at 25 C and
689 mm Hg, how many grams of formic acid reacted?
- 33 -
IMF Worksheet
Draw Lewis structures and identify all intermolecular and intramolecular attractive
forces.
C2H4
NCl3
HCN
H2S
N2H4
NH2Cl
SiH4
CO
(CH3)2CHCl
CH3F
CH3NH2
CH3CHO
NH2CH2COOH
CH3COCH3
HSCH2CH2SH
- 34 -
The Mystery of Excessive Perspiration
While Jogging
Ronald Delorenzo
Middle Georgia College
While jogging or running several miles, perspiration may be produced in such
large quantities that it flows down the body and drips onto the ground. Since the
purpose of perspiration is to produce a cooling effect by evaporation, why does the
human body not produce just enough perspiration to keep the skin surface moist?
The production of perspiration requires relatively large amounts of energy, yet
wasted effort is uncharacteristic of nature. In addition, excessive perspiration can
lead to dehydration and death. What is the advantage to the human body of
producing more perspiration than would be needed to keep the skin surface just
moist?
Perspiration contains solutes. If the human body were to produce just enough
perspiration to keep its skin moist, the solute concentration of the residual sweat
would increase as the perspiration evaporates. As the residual sweat increases,
the vapor pressure of this remaining concentrated solute solution decreases. This,
in turn, decreases the evaporation rate and consequently the cooling rate.
Therefore, it appears that one possible advantage of excessive sweating may be
that excess sweat rinses away concentrated solute solution from the skin surface
and thereby increases evaporation and cooling rates.
- 35 -
Fire Walking, Temperature & Heat
Ronald Delorenzo
Middle Georgia College
For thousands of years people have marveled at and been puzzled by the ability
of fire walkers to walk across beds of glowing coals without apparent harm. Many
scientists, having never observed fire walking firsthand, simplistically explained it
away with assumptions such as "fire walkers have toughened and calloused feet."
Today, tens of thousands of Americans without the benefits of toughened and
calloused feet engage in fire walking. How then do they do it?
In chemistry classes, teachers explain to students that there is a difference
between the amount of heat objects contain and the temperature of objects. For
example, sparks given off by a cigarette lighter are very hot (~ 1000° C), but when
sparks from a cigarette lighter touch one's hands or feet, one does not usually
detect any warmth. Likewise, there is a difference between the amount of heat
that glowing thousand-degree coals contain and their temperature. The dark
footprints that momentarily form after firewalkers step on previously red coals
indicate that the coals cool quickly. This cooling results in part by an ash
compression and oxygen depletion at the glow front and by the heat flow from
coals to feet. The specific heat of feet, which have a high water content, is much
greater than the specific heat of coals. If red-hot coals were replaced by aluminum
with a larger specific heat, the firewalkers would have problems!
A second factor to be considered is the conductivity of coal. Coal and coal ash
are insulators. Once human feet cool the surface of glowing coals, the heat
transfer from the internally hot coal to the cooler coal surface is relatively slow.
Again, if a heat conductor such as copper or aluminum replaced coals, firewalkers
would have additional problems.
A third factor to be considered is the Leidenfrost effect, which is used to explain
why water droplets spend such an unexpectedly long amount of time on hot
griddles before vaporizing. The bottoms of water droplets on hot griddles vaporize
very quickly, allowing the drops to float on insulating layers of vapor. Firewalkers
also dip there feet in ice water beforehand, or stand on damp grass at one end of
the coal bed before their walk.
Additional factors to be considered include: (1) coal beds are only 3 meters
long, (2) coal beds are traversed rapidly with only two quick steps, and (3) embers
are spread thinly.
- 36 -
Heat Calculations Involving Heating & Cooling Curves
Heat capacity is the amount of heat needed to change the temperature of a
substance by 1° C.
Specific Heat Capacity, s, is the amount of heat needed to change the
temperature of 1 g of a substance by 1° C.
The amount of heat lost or gained during a temperature change is:
∆H = (∆T)(m)(s)
The amount of heat lost or gained during a phase change is:
Heat of fusion, ∆Hfus is the amount of energy needed to melt, fuse, one gram
of a substance.
q = (∆Hfus)(m)
Heat of vaporization, ∆Hvap is the amount of energy needed to vaporize one
gram of a substance.
q = (∆Hvap)(m)
How much heat is needed to convert 46 g of ice at - 20° C to 138° C?
- 20° C to 0° C
fusion
0° C to 100° C
vaporization
100° C to 138° C
(2.092 J/g-°C)(20° C)(46 g)
(334.72 J/g)(46 g)
(4.184 J/g-°C)(100° C)(46 g)
(2259.36 J/g)(46 g)
(2.017 J/g-°C)(38° C)(46 g)
- 37 -
1.925 kJ
15.397 kJ
19.264 kJ
103.931 kJ
3.526 kJ
Thermochemical Worksheet
1. How much energy is needed to melt 86.8 g of HCl? (∆Hfus = 117.15 J/g)
2. How much energy is used to increase the temperature of 48.06 g of nickel
from - 14 °C to 1020 °C?
M.P. = 1200° C
B.P. = 1850° C
∆Hfus = 17.47 kJ/g
∆Hvap = 370.40 kJ/g
Cp(s) = 1.67 J/g-° C
Cp(l) = 2.47 J/g-° C
Cp(g) = 3.64 J/g-° C
3. How much energy is lost as 45 g of cesium is cooled from 880° C to 28° C?
M.P. = 28.59° C
B.P. = 690° C
∆Hfus = 2.09 kJ/g
∆Hvap = 67.74 kJ/g
- 38 -
Cp(s) = 30.04 J/g-° C
Cp(l) = 33.47 J/g-° C
Cp(g) = 20.79 J/g-° C
Heats of Reaction & Bond Enthalpies Worksheet
Based on average bond enthalpies, determine the heat of reaction for the following
chemical reactions:
C4H8 + O2

CH3CHCH2 + HCl
CH3COCH3 + HCN
CO2 + H2O


- 39 -
CH3CHClCH3
CH3C(OH)(CN)CH3
Heat of Fusion
Introduction
The purpose of this experiment is to identify the heat of fusion, ∆Hfus, of water.
Equipment & Materials
ice cubes
beaker, 150 or 250 mL
thermometer
calorimeter with lid
balance
crucible tongs
hot hands
Procedure
1.
Heat about 150 mL of water to around 50° C.
2.
Record the mass of a calorimeter and lid.
3.
Pour about 100 mL of the heated water into the calorimeter and then secure
with the lid.
4.
Record the mass of the calorimeter and its contents.
5.
Place a thermometer through the lid and record the temperature after
20 seconds.
6.
Put 4 – 6 ice cubes into the calorimeter, and stir the mixture until the
temperature is constant.
Note:
If the temperature does not drop to between 0 – 5° C, additional ice
cubes should be added.
7.
Remove any unmelted ice with crucible tongs, draining off any liquid on the
ice into the calorimeter.
8.
Measure the mass of the calorimeter and its contents along with the
temperature.
- 40 -
Processing the Data
1.
Determine the change in temperature.
2.
Calculate the number of joules transferred from the hot water, Htotal, to the
ice: (s = 4.184 J / g-° C)
3.
Calculate the number of joules that melted one gram of ice, ∆H fus.
4.
Percent Error: (Theoretical (∆Hfus is 334.72 J/g.)
Conclusions
1.
What does heat of fusion represent?
2.
In terms of potential and/or kinetic energy, what happens to ice at its melting
point?
3.
Discuss the type of attractions/bonds that exist between molecules of ice, and
what happens to these as melting occurs.
4.
What was your experimental heat of fusion?
5.
What was the percent error?
6.
Identify causes for this error.
- 41 -
Hess’s Law
Introduction
The purpose of this experiment is to use Hess's Law to determine the heat of
neutralization when an acid is reacted with a base.
Equipment & Materials
NaOH
electronic balance
0.5 M HCl
NaOH
Erlenmeyer flask, 250 or 500 mL
thermometer
Procedure
Part I. Solid NaOH + H2O
1.
Record the mass of a flask.
2.
Add 100.0 mL of tap water to the flask, and record the mass.
3.
Place a thermometer in the water, and after 20 seconds record the
temperature.
4.
Measure out 2.00 g of NaOH.
Caution: NaOH is caustic. If you get any on your hands or clothing wash it
off with soap and water.
5.
Add the NaOH to the flask, stirring gently with the thermometer. Record the
highest temperature.
Part II. Aqueous NaOH + Aqueous HCl
1.
Add 50 mL of HCl solution to a second flask.
2.
Record the temperature of the NaOH solution.
- 42 -
3.
Using the NaOH solution you prepared in Part I, pour 50. mL of NaOH into the
HCl solution. Record the highest temperature.
Part III. Solid NaOH + Aqueous HCl
1.
Add 50 mL of HCl solution to a flask, and record the temperature.
2.
Add 2.00 g of NaOH to the flask, gently stirring.
temperature.
Part I
Part II
Record the highest
Part III
ΔT
Heat Absorbed by Solution
Heat Absorbed by Flask
Mol NaOH
Total Heat/mol NaOH
Processing the Data
1.
Calculate the heat absorbed by the solution: (sH2O = 4.184 J/g-° C)
2.
Calculate the heat absorbed by the flask: (sflask = 0.92 J/g-° C)
3.
Total heat absorbed:
4.
Moles of NaOH:
Part I or Part III
Part II
- 43 -
5.
Total heat per mole of NaOH:
6.
Percent Error: (Part III = T.Y., Parts I + II = E.Y.)
Conclusions
1.
What does the heat of a reaction represent?
2.
Identify the net ionic equations that correspond to the reactions that took
place in Part I, II, and III.
3.
Using these net ionic equations, show how the changes that occurred in Part I
and Part II are the same as the changes that occurred in Part III.
4.
What is Hess’s Law?
5.
How do the results in this lab illustrate Hess's Law?
- 44 -
Balancing Redox Reactions Worksheet
Acidic Solutions
Mn 2+ + BiO3 -
ClO3 - + Cl -

MnO4 - + Bi 3+

Cl2 + ClO2
P + Cu 2+

Cu + H2PO4 -
PH3 + I2

H3PO2 + I -
NO2

NO3 - + NO
- 45 -
Basic Solutions
MnO4 - + C2O4 2-
ClO2

Zn

MnO2 + CO2
ClO2 - + ClO3 -

Cu(NH3)4 2+ + S2O4 2-
Zn + NO3 -


SO3 2- + Cu + NH3
Zn(OH)4 2- + NH3
Zn(OH)4 2- + H2
- 46 -
Solution Chemistry Worksheet #1
1.
How many grams of sodium hydroxide are required to neutralize 1.965 g of
H2S in each of the following unbalanced equations?
NaOH + H2S
NaOH + H2S

NaHS + H2O

Na2S + 2 H2O
2.
How many grams of lithium hydroxide are required to neutralize 75.0 g of
phosphoric acid if two its three hydrogens has been neutralized?
3.
What is the molar mass of a monoprotic acid (HX), 0.1248 g of which required
0.2475 g of potassium hydroxide for neutralization?
- 47 -
4. Determine the oxidation state for tin in the following tin oxide compounds:
oxide containing 88.12% Sn
oxide containing 78.77% Sn
5.
Balance the following redox equations:

MnO2 + Fe 2+ + H +
Cr2O7 2- + Sn 2+ + H +

- 48 -
Mn 2+ + Fe 3+ + H2O
Cr 3+ + Sn 4+ + H2O
Solution Chemistry Worksheet #2
1.
A chemist synthesizes a new organic acid. After dissolving 0.500 g of the
acid in water she finds that it requires 15.73 mL of 0.437 M sodium hydroxide
for neutralization to occur. If the acid contains three carboxylic acid groups,
what is the molar mass of this acid?
2.
500. mL containing 1.00 g of potassium permanganate is used in titrating a
reducing agent. During the reaction, KMnO4 is reduced to Mn 2+. What is
the molarity of the KMnO4 solution?
3.
31.25 mL of 0.100 M sodium oxalate (acidified) is titrated with 17.38 mL of
potassium permanganate solution of unknown strength. What is the molarity
of the potassium permanganate?
C2O4 2- + MnO4 - + H +

- 49 -
CO2 + Mn 2+ + H2O
4.
Titration of 1.500 g of an diprotic acid required 47.00 mL of 1.20 M
sodium hydroxide to reach the endpoint. What is the molar mass of the acid?
5.
A chemist synthesizes a substance which he believes is barbituric acid
(MM = 128.0), a precursor for many sleeping tablets. Barbituric acid has one
acidic hydrogen. To help with the identification, a 0.500 g crystalline sample
is titrated with 0.100 M sodium hydroxide. The equivalence point is reached
after 39.10 mL of base was added. Is this sample barbituric acid?
6.
What mass of sodium carbonate is required to neutralize 4.89 g of
hydrochloric acid in the following non-balanced equation?
HCl + Na2CO3

NaCl
- 50 -
+
H2O
+
CO2
Solution Chemistry Worksheet #3
1.
Hydrochloric acid reacts with zinc, resulting in the production of zinc chloride
and hydrogen gas. The mass of hydrogen gas that evolved was 0.0572 g.
How many grams of sodium hydroxide are required to neutralize this acid?
2.
3.245 g of cadmium was dissolved in a dilute nitric acid solution, producing
0.0532 g of hydrogen gas. Based on this data, identify the oxidation state of
cadmium.
3.
Solid potassium hydrogen phthalate, KHC8H4O4, is often used to standardize
a base. KHC8H4O4 has one ionizable H + per formula unit. How many moles
of acidic hydrogen are contained in 0.7325 g of this salt?
- 51 -
4. When FeCl3 oxidizes KI, I2, FeCl2 and KCl are produced. How many grams of
iodine will form when 13.20 g of iron (III) chloride reacts with an excess of
potassium iodide?
2 Fe 3+ + 2 I -
5.

I2 + 2 Fe 2+
Analysis shows that 16.20 g of an iron oxide compound contains 11.33 g of
iron.
a)
What is the formula for this iron oxide compound
b)
What is the oxidation state of iron in this oxide?
- 52 -
Reaction Type & Net Ionic Equations
Precipitation – the solubility rules need to be memorized, and applied.
Acid-Base Reactions fall into several categories:
(1)
Typical Acid-Base Reaction:

H + + OH –
CH3COOH + OH –
(2)

 Mg(OH)2
Non-Metal Oxides Added to Water:
SO2 + H2O
(4)

H2SO3
Formation of a Weak Acid or Base:
F-+ H+
(5)
CH3COO – + H2O
Metal Oxides Added to Water:
MgO + H2O
(3)
H2O

HF
Nonmetal oxide Added to a Basic Solution:

SO2 + 2 OH -
SO3 2- + H2O
SO2 dissolves in the solution, producing H2SO3,.which then reacts with
the base.
 H2SO3
 SO3 2- + 2
SO2 + H2O
H2SO3 + 2 OH (6)
H2O
Metal oxides Added to an Acidic Solution:
Na2O + 2 H +

- 53 -
2 Na + + H2O
Na2O dissolves in the solution, producing NaOH, which then reacts with
the acid.

Na2O + H2O
2 Na + + 2 OH -
Na + + OH - + H +  Na + + H2O
(7)
Salts Added to Water:
NaHSO4 + H2O
HSO4 -
↔
KC2H3O2 + H2O
C2H3O2 – + H2O

Na + + HSO4 –
H + + SO4 2-
↔ K + + C2H3O2 –
↔ HC2H3O2 + OH –
Redox Reactions
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/oxred_3.php
(1)
Free metal or non-metal undergoing a chemical change:
 NaOH
Br -  F - +
Na + H2O
F2 +
(2)
+ H2
Br2
An oxidizing agent grabs electrons from another source. The higher the
electronegativity, the greater its ability to absorb electrons. These tend to
include substances with higher oxidation states (MnO 4 -, CrO4 2-,
Cr2O7 2-, HNO3, H2SO4).
(3)
A reducing agent releases electrons. The lower the electronegativity, the
lower the ionization energy, the easier for valence electrons to be lost.
Formation of a Gas
You are expected to know common chemical reactions that produce CO 2, SO2,
and NH3.
- 54 -
Net Ionic Equations
Precipitate
NH4OH(aq) + FeCI3(aq)  NH4CI (aq) + Fe(OH)3(aq)
NH4 +(aq) + OH -(aq) + Fe 3+(aq) + CI -(aq)  Fe(OH)3(s) + NH4 +(aq) + Cl -(aq)
3 OH -(aq) + Fe 3+(aq)  Fe(OH)3(s)
NaI(aq) + Pb(NO3)2(aq)  PbI2(aq) + NaNO3(aq)
Na +(aq) + I -(aq) + Pb 2+(aq) + NO3 -(aq)  PbI2(s) + Na +(aq) + NO3-(aq)
2 I+ Pb 2+
 PbI
(aq)
(aq)
2(s)
Redox
Cl2(g) + NaBr(aq)  Br2(g) + NaCl(aq)
Cl2 + Na +(aq) + Br -(aq)
Cl + 2 Br 2
(aq)


Br2(l) + Na +(aq) + Cl -(aq)
Br
+ 2 Cl 2(l)
(aq)
Cu(s) + H2SO4(aq)  CuSO4(aq) + H2(g)
Cu(s) + 2 H +(aq) + SO4 2-(aq)  Cu 2+(aq) + SO4 2-(aq) + H2(g)
Cu
+ 2 H+
 Cu 2+
+ H
(s)
(aq)
(aq)
2(g)
Acid - Base
+ KOH(aq)  K2S(aq) + H2O(l)
H2S(aq) + K +(aq) + OH -(aq)  K +(aq) + S 2-(aq) + H2O(l)
H S
+ 2 OH  S 2+ 2 H O
H2S(aq)
2 (aq)
NaOH(aq)
(aq)
(aq)
+ NH4Cl(aq)  NH4OH(aq)

NH4 +(aq) 
Na +(aq) + OH -(aq) + NH4 +(aq) + Cl -(aq)
OH -(aq) +
- 55 -
2 (l)
+ NaCl(aq)
NH3(aq) + H2O(l) + Na +(aq) + Cl -(aq)
NH3(aq) + H2O(l)
HCl(aq)
+ K2CO3(aq)  KCl(aq)
+ H2CO3(aq)
H +(aq) + Cl -(aq) + K +(aq) + CO3 2-(aq)  K +(aq) + Cl -(aq) + H2O(l) + CO2(g)
2 H +(aq) + CO3 2-(aq)  H2O(l) + CO2(g)
- 56 -
Net Ionic Equations Worksheet #1
Precipitate
AgNO3(aq) + Na2CrO4(aq) 
H2SO4(aq) + BaCl2(aq) 
Na3PO4(aq) + CaCl2(aq) 
Redox
Bi(s) + O2(g) 
Ca(s) + H2O(l) 
Pb(s) + H2SO4(aq) 
Acid - Base
HC2H3O2(aq) + NaHCO3(s) 
(NH4)2SO4(aq) + KOH(aq) 
NaF(aq) + HCl(aq)
- 57 -

Net Ionic Equations Worksheet #2
Precipitate
H2S(aq) + Pb(NO3)2(aq) 
H2SO4(aq) + CaF2(aq) 
AgNO3(aq) + LiBr(aq) 
Redox
Mg(s) + FeCl3(aq) 
KClO3(s) 
CS2(l) + O2(g) 
Acid - Base
HCl(aq) + K2SO3(aq) 
CO2(g) + NaOH(aq) 
H2SO4(aq) + LiHCO3(s) 
- 58 -
Net Ionic Equations Worksheet #3
Sulfur dioxide gas is bubbled into distilled water.
Ammonia gas is mixed with hydrogen chloride gas.
Solid sodium bicarbonate is strongly heated.
A solution of potassium hydroxide is added to solid ammonium chloride.
Solutions of strontium nitrate and sodium sulfate are mixed.
Solid ammonium nitrate is heated to temperatures above 300° C
- 59 -
- 60 -
Fractional Crystallization
Fractional crystallization involves separating a mixture by using differences in
solubility between components of a mixture. As a heated solution cools, the less
soluble component of the mixture comes out of solution. The crystals are then
separated by filtration, and dried.
Suppose we have a mixture that consists of two amino acids, and the composition
of this mixture is 80% glycine (Gly) and 20% alanine (Ala). How much pure glycine
could be recovered from 100 g of this mixture?
The solubilities of Ala and Gly are as follows:
Alanine
Glycine
g solute/100 g H2O
0° C
100° C
13
37
14
67
The mixture contains 20 g of Ala, and the minimum amount of water required to
keep the Ala dissolved at 0° C is
20 g Ala x (100 g H2O/13 g Ala) ~ 154 g H2O
At 100° C the solubility of Gly is 67 g/100 g of H2O:
(67 g Gly/100 g H2O) x 154 g H2O ~ 103 g Gly
The 154 g of hot water needed to dissolve the Ala at 0° C is more than enough
to dissolve all of the Gly as well.
The amount of pure Gly that will separate from the solution after it has been
cooled to 0° C is the difference between the amount of Gly in the original
mixture, 80 g, and the amount that will remain in the solution at 0° C.
The solubility of Gly at 0° C is 14 g/100 g H2O and therefore in 154 g of water at
0° C we have
(14 g Gly / 100 g H2O) x 154 g H2O ~ 22 g Gly
The quantity of solid Gly that is precipitated is 80 g - 22 g = 58 g. This
represents the maximum amount of pure Gly that can be recovered from this
mixture.
- 61 -
Solubility & Fractional Crystallization
In this experiment we will be taking advantage of the differences in solubility of
solid substances in different kinds of liquid solvents to separate the components of
a mixture in essentially pure form.
This technique is called fractional
crystallization.
You will be given a sample containing silicon carbide, potassium nitrate, and
copper sulfate. Silicon carbide, SiC, is a black, very hard material; it is the classic
abrasive, and completely insoluble in water. Potassium nitrate, KNO 3, and copper
sulfate, CuSO4 • 5 H2O, are water-soluble ionic substances, with different solubility
at different temperatures. Copper sulfate is blue in its crystalline form and in
solution. The solubility of the hydrate increases fairly rapidly with temperature.
Potassium nitrate is a white solid, colorless in solution. Its solubility increases
about twenty fold between 0° C and 100° C.
The solid KNO3 that is recovered is contaminated by a small amount of copper
sulfate. Redissolving the solid in a minimum amount of hot water, cooling and
then recrystallizing the KNO3, increases its purity markedly. The purity of the
KNO3 can be established by the intensity of the color produced by the copper
impurity when treated with ammonia, NH3.
Procedure
1. Add the sample mixture to a beaker filled with 50-mL of distilled water.
2. Warm the solution, stirring continuously, until KNO3 and CuSO4 • 5 H2O are
in solution.
3. Pour the warm solution through a Büchner funnel with the aid of a rubber
police person, transferring as much of the mixture as is possible.
4. Set the SiC aside to dry, and transfer the filtrate to a beaker.
5. Rinse the filter flask and Büchner funnel with tap water.
6.
Place the Büchner funnel in the ice chest to chill for several minutes.
7. Add 15 drops of 6 M HNO3 to the filtrate, and heat the filtrate until white
crystals of KNO3 begin to form.
Note:
If the liquid begins bumping, gently move the flame back-and-forth
below the beaker to control the heating.
- 62 -
8.
Add 5-mL of distilled water to the solution to re-dissolve any crystals clinging
to the walls of the beaker.
9.
Cool the solution in an ice bath to force white KNO 3 crystals come out of
solution.
10. Reassemble the filtering apparatus, and filter the slurry.
11. Place a filter paper over the crystals and, with the aid of a rubber stopper,
press-dry the crystals.
12. Visually inspect the crystals. If CuSO4 is still present, spray the crystals with a
jet of ice-cold distilled water, and re-press the crystals.
Caution:
Using too much water will dissolve the KNO3 as well as the
CuSO4.
13. Allow the crystals to air dry.
- 63 -
Kinetics Worksheet
1.
Distinguish between average rate and instantaneous rate. Which of the two
rates gives us an unambiguous measurement of reaction rate? Why?
2.
Suggest experimental means by which the rates of the following reactions
could be followed:

a)
CaCO3(s)
CaO(s) + CO2(g)
b)
Cl2(g) + 2 Br –(aq)

Br2(aq) + 2 Cl –(aq)
3.
What are the units for the rate constants of zero order, first order, and second
order reactions?
4.
On which of the following properties does the rate constant of a reaction
depend? Briefly explain why.
a)
b)
c)
reactant concentration
nature of the reactants
temperature.
- 64 -
5.
Consider the reaction
X + Y

Z
From the following data, obtained at 360 K,
initial rate of
disappearance of X (M/s)
0.053
0.127
1.02
0.254
0.509
6.
[X]
[Y]
0.10
0.20
0.40
0.20
0.40
0.50
0.30
0.60
0.60
0.30
a)
determine the order of the reaction
b)
determine the initial rate of disappearance of X when the concentration
of X is 0.30 M and that of Y is 0.40 M
Consider the reaction
A

B
The rate of the reaction is 1.6 x10-2 M/s when the concentration of A is
0.35 M. Calculate the rate constant if the reaction is first order in A and
second order in A.
- 65 -
7.
The rate constant for the second order reaction
2 NOBr(g)

2 NO(g) + Br(g)
is 0.80 M-s at 10 C.
8.
a)
Starting with a concentration of 0.086 M, calculate the concentration of
NOBr after 22 s.
b)
Calculate the half-lives when [NOBr]0 = 0.072 M and [NOBr]0 = 0.054 M
The rate constant for the second order reaction
2 NO2(g)

2 NO(g) + O2 (g)
is 0.54 M-s at 300 C. How many seconds would it take for the concentration
of NO2 to decrease from 0.62 M to 0.28 M?
9.
The burning of methane in oxygen is a highly exothermic reaction. Yet a
mixture of methane and oxygen gas can be kept indefinitely without any
apparent change. Explain.
- 66 -
10. The rate law for the reaction
2 NO(g) + Cl2(g)

2 NOCl(g)
is given by rate = k [NO][Cl2].
a)
What is the order of the reaction?
b)
A mechanism involving the following steps has been proposed for the
reaction
NO(g) + Cl2(g)
NOCl2(g) + NO(g)


NOCl2(g)
2 NOCl(g)
If this mechanism is correct, what does it imply about the relative rates of
these two steps?
11. For the reaction
X2 + Y + Z

XY + XZ
it is found that doubling the concentration of X2 doubles the reaction rate,
tripling the concentration of Y triples the rate, and doubling the concentration
of Z has no effect.
a)
What is the rate law for this reaction?
b)
Why is it that the change in the concentration of Z has no effect on the
rate?
- 67 -
c)
Suggest a mechanism for the reaction that is consistent with the rate
law.
12. The decomposition of N2O to N2 and O2 is a first order reaction. At 730 C
the half-life of the reaction is 3.58 x103 min. If the initial pressure of N2O is
2.10 atm at 730 C, calculate the total gas pressure after one half-life.
Assume that the volume remains constant.
- 68 -
Equilibrium Worksheet
1.
What is the value of Kc at 225 C?
CO(g) + 2 H2(g)

CH3OH(g)
Kp = 6.3 x 10-3
2.
At 773 C the equilibrium shown above has a Kc of 0.40. What is the Kp, in
torr, at this temperature?
3.
If 0.100 mol of HCl and solid I2 are placed in a 3.00 L container, at 25 C, and
the following equilibrium occurred:
2 HCl(g) + I2(s)

Cl2(g) + 2 HI(g)
Kc = 1.6 x 10-34
What will be the equilibrium concentration of HI and Cl2 in the container?
- 69 -
4.
Based on the following equilibrium:
SO3(g) + NO(g)

NO2(g) + SO2(g)
Kc = 0.500
if 0.240 mol of SO3 and 0.240 mol of NO are placed in a 2.00 L vessel and
allowed to react, what will be the equilibrium concentration of each gas?
5.
Based on the equilibrium shown below:
2 HCl(g)

Cl2 (g) + H2(g)
Kc = 3.2 x 10-34
if a flask initially contains 0.0500 M of HCl which then reacts to reach
equilibrium, what will be the concentration of H2 and Cl2?
6.
The equilibrium system
PCl5

PCl3 + Cl2
Kp = 1.05
The system is found to have pressures for PCl5, PCl3, and Cl2 of 0.177 atm,
0.223 atm and 0.111 atm, respectively. Is the system at equilibrium? Prove
with a calculation and predict which way it will shift if it is not already at
equilibrium.
- 70 -
7.
When 1.05 moles of Br2 are put into a 0.980 L flask, 1.20% of the Br 2
dissociates into Br atoms. Calculate the Kc for the following reaction
Br2(g)
8.
Ammonium carbamate,
equilibrium:

2 Br(g)
NH4CO2NH2
NH4CO2NH2(s)

decomposes
into
the
following
2 NH3(g) + CO2(g)
Starting with only the solid, it is found at 40 C the total gas pressure is
0.363 atm. Calculate the Kp for the reaction.
9.
Consider the reaction
2 SO2(g) + O2(g)

2 SO3(g)
∆H = - 198.2 kJ
if the reaction has established equilibrium and then the following stresses
added, what would happen to the molarity of SO 3?
a)
increase the temperature
b)
remove some O2
- 71 -
c)
move the reaction mixture to a larger volume container
d)
add some SO2
e)
add a catalyst
- 72 -
Acid-Base Worksheet
1.
Determine the Molarity of H2S solution that produces a pH of 4.00?
(Ka = 1.10 x10-7)
H2S(g) + H2O(l)

H +(aq) + HS -(aq)
2.
Determine the [OH -] when 0.270 g of C5H5N is dissolved in 100. mL of
water? (Kb = 1.70 x10-9)
3.
Determine the pH and equilibrium concentrations of both products and
reactants of a 2.00 M NH3 solution. (Kb = 1.80 x10-4)
4.
Determine the pOH when 0.500 M CH3COOH ionizes 1.36%.
CH3COOH + H2O

- 73 -
H + + CH3COO -
5.
3.42 g of Mg(OH)2 is added to enough water to produce 350.0 mL of solution.
This solution is then diluted to 658. mL. What is the Molarity of the final
solution?
6. How many mL of 0.250 M H2SO4 is required to neutralize 0.0500 mol OH - in
Al(OH)3?
7. How many moles of Sr(OH)2 is required to neutralize 235. mL of 0.0560 M
H3PO4?
8. How many g of HC2H3O2 are needed to neutralize 50.0 mL of 0.100 M
Ca(OH)2?
- 74 -
Acid-Base Reactions Worksheet #1
1.
0.250 L of 0.100 M Mg(OH)2 is required to neutralize 4.63 g of a monoprotic
acid. What is the molar mass of this acid?
2.
25.0 mL of 0.100 M NaOH is reacted with 75.0 mL of 0.0200 M H2SO4. What
is the [OH -] of this solution?
3.
3.50 L of a 0.0400 M H3PO4 reacts with 0.500 L of a 0.160 M Ba(OH)2.
Show that the resulting concentration is 0.0217 M H3PO4.
- 75 -
4. 38.0 mL of 0.300 M Al(OH)3 reacts with 20.0 mL of 0.0300 M HClO4. What is
the [H3O +] of this solution?
5.
A dye mixture contains an unknown amount of H 2SO4. 60.0 g of this sample
neutralizes 36.50 mL of 0.250 M Ca(OH)2. Determine the percent, by mass,
of sulfuric acid in the sample.
6.
45.76 mL of 0.250 M hydroiodic acid is added to 2.09 g of barium hydroxide.
Is the solution acidic or basic? What is the concentration of hydronium and
hydroxide ions in this solution?
7.
How many grams of carbonic acid are required to neutralize 35.00 mL of an
aluminum hydroxide solution with a pH of 10.60?
- 76 -
8.
The empirical formula of ascorbic acid is HC3H3O3. A 0.200 g sample
required 17.48 mL of 0.0650 M Ba(OH)2 to reach neutralization. Assuming
the acid is diprotic, calculate its molar mass and its molar formula.
9.
If 45.0 mL of 0.490 M Ba(OH)2 is added to 36.00 mL of 0.330 M HNO 3, will
the resulting solution be acidic or basic? Identify the final [H +] and [OH -]
concentration?
10. Acetic acid, HC2H3O2, gives vinegar its sour taste. A 0.110 g sample of
vinegar requires 4.62 x10-3 L of 0.0150 M KOH to completely neutralize it.
What is the percent by mass of acetic acid in vinegar?
- 77 -
11. 6.80 mL of a monoprotic acid is diluted to 12.85 mL. 4.00 mL of this is reacted
with 5.86 mL of 0.160 M solution of NaOH. What is the original Molarity?
12. How much water must be added to 0.650 M solution of HNO 3 to produce
500.00 mL of 0.330 M HNO3?
- 78 -
Acid-Base Reactions Worksheet #2
1.
In water, hydrazoic acid, HN3, is a weak acid that has an equilibrium constant,
Ka, equal to 2.80 x10-5 at 25° C. A 0.300 L sample of 0.0500 M solution of
the acid is prepared.
a)
Write the expression for the equilibrium constant, K a, for hydrazoic acid.
b)
Calculate the pH of this solution at 25° C.
c)
To 0.150 L of this solution, 0.800 g of sodium azide, NaN 3, is added.
The salt dissolves completely. Calculate the pH of the resulting solution
at 25° C if the volume of the solution remains unchanged.
d)
To the remaining 0.150 L of the original solution, 0.0750 L of 0.100 M of
NaOH solution is added. Calculate the [OH -] for the resulting solution at
25° C.
- 79 -
2.
Ammonia is a weak base that dissociates in water.
dissociation constant, Kb, for NH3 is 1.80 x10-5.
NH3 + H2O

At 25° C, the base
NH4 + + OH -
a)
Determine the hydroxide ion concentration and the percentage
dissociation of a 0.150 M solution of ammonia at 25° C.
b)
Determine the pH of a solution prepared by adding 0.0500 mol of solid
ammonium chloride to 100. mL of a 0.150 M solution of ammonia.
c)
If 0.0800 mol of solid magnesium chloride, MgCl2, is dissolved in the
solution prepared in part (b) and the resulting solution is well stirred, will
a precipitate of Mg(OH)2 form? (Assume the volume of the solution is
unchanged). The solubility product constant for Mg(OH)2 is 1.50 x10-11.
- 80 -
Identifying pH in a Titration
A 50.0 mL sample of 0.100 M HCN, hydrogen cyanide, is titrated with 0.100 M
NaOH.
a)
0 mL of NaOH has been added:
HCN(aq)

H + (aq) + CN - (aq)
Ka = 6.2 x10-10
Ka = [H +][CN -]/[HCN]
6.2 x10-10 [X]2/[0.100]
[X] = 7.87 x10-6
pH = - log10 [7.87 x10-6] = 5.10
b)
8.00 mL of 0.100 M NaOH has been added:
[H +] = Ka [HCN]/[CN -]
[H +] = 6.2 x10-10 [7.24 x10-2]/[1.38 x10-2]
[H +] = 3.25 x10-9
c)
pH = 8.49
at the half-equivalence point:
[H +]/Ka = [HCN]/[CN -]
[H +] = Ka = 6.2 x10-10
d)
pH = 9.21
at the equivalence point:
CN - + H2O

HCN + OH -
Kb = Kw/6.2 x10-10 = 1.6 x10-5
Kb = [HCN][OH -]/[CN -] = 1.6 x10-5 = [X][ X]/[5.0 x10-2]
X = [OH -] = 8.9 x10-4
pOH = 3.05
pH = 10.95
- 81 -
Choosing an Indicator
http://www.psigate.ac.uk/newsite/reference/plambeck/chem1/p01172.htm
Indicator
pH range
methyl violet
thymol blue
methyl yellow
methyl orange
bromocresol green
methyl red
chlorophenol red
bromothymol blue
phenol red
cresol purple
thymol blue
phenolphthalein
thymolphthalein
alizarin yellow R
indigo carmine
0.0- 1.6
1.2- 2.8
2.9- 4.0
3.1- 4.4
3.8- 5.4
4.2- 6.2
4.8- 6.4
6.0- 7.6
6.4- 8.0
7.4- 9.0
8.0- 9.6
8.0- 9.8
9.3-10.5
10.1-12.0
11.4-13.0
pKa
Acid Form
0.8
1.6
3.3
4.2
4.7
5.0
6.0
7.1
7.4
8.3
8.9
9.7
9.9
11.0
12.2
yellow
red
red
red
yellow
red
yellow
yellow
yellow
yellow
yellow
colorless
colorless
yellow
blue
Base Form
blue
yellow
yellow
yellow
blue
yellow
red
blue
red
purple
blue
red
blue
red
yellow
A chemical indicator is a compound which changes color indicating when the
endpoint of a titration has been reached. The color of the indicator changes are
the result of the concentrations of ions in the solution. An acid-base indicator is an
weak acid-weak base conjugate pair, in which the two forms have different colors.
The indicator is added in low concentration to minimize any change to the pH of
the system.
Ka = [H3O +][In -]/[HIn]
When
[In -]/[HIn] = 1
[H3O +] = Ka
pH = pKa
In order to choose an indicator that will work for an acid – base reaction, you must
have an approximate idea where the equivalence point would be. Using the
expected pH range when neutralization occurs, the indicator with a pK a value
within ±1 pH unit of the equivalence point will be the correct indicator.
- 82 -
- 83 -
Buffers & Salts Worksheet
1.
How does a basic buffer differ from a base that dissolves in distilled water?
2.
Working with the buffer shown below:
0.350 M 0.200 M
NH3  NH4 + + OH –
a)
is the concentration of OH – be the same as NH4 +?
choice.
b)
what is the starting pH of this buffer?
c)
when 0.150 M HNO3 is added to this buffer, identify the balanced net
ionic equation showing how the buffer neutralizes the acid.
d)
calculate the equilibrium concentrations of the buffer after the acid has
been neutralized?
- 84 -
Explain your
3.
e)
calculate the pH of the solution when the acid has been neutralized?
f)
write the net ionic equation identifying what change occurs at the
equivalence point.
A 25.00 mL sample of a 0.500 M HN3 solution is titrated with a 0.350 M
solution of KOH. Ka = 2.80 x10-5
a)
What is the pH before any base is added?
b)
Is the [H +] and [N3 -] the same BEFORE any base is added? Explain
your choice.
c)
BEFORE base is added, is the HN3 solution a buffer? Explain your
choice.
d)
Calculate the pH after 15.25 mL of a 0.350 M KOH solution is added.
- 85 -
e)
Identify the pH at the ½ equivalence point.
f)
How does a ½ equivalence point differ from an equivalence point?
g)
Is a buffer is created when the ½ equivalence point is reached? Explain.
h)
How many mL of the KOH solution must be added to neutralize the HN 3
solution?
i)
When neutralization has occurred, what is the pH of the solution?
- 86 -
Essential Information for the AP Chemistry Exam
Strong Acids:
HCl, HBr, HI, HNO3, H2SO4, HClO4
Strong Bases:
LiOH, NaOH, KOH, Sr(OH)2, Ba(OH)2
Soluble Salts:
lithium, sodium, potassium, ammonium cations, nitrate, acetate;
chlorides except Ag, Pb, Hg(I), sulfates except Pb, Ca, Sr, Ba.
All other salts should be considered only slightly soluble
Complex Ion Formation
 Ag(NH3)2 +
 Al(OH)4- or
AgCl + NH3
Al(OH)3 + OH -
+ Cl Al(OH)6 3-
[Co(NH3)6] 2+
[CoCl4] 2-
[Ni(NH3)6] 2+
[NiCl4] 2-
[Zn(NH3)6] 2+
[Zn(OH)4] 2-
[Ag(NH3)] 2+
[Cu(NH3)4] 2+
[Cu(H2O)6] 2+
[Al(OH)4] –
[Al(OH)6] 3-
Hydrolysis Reactions
Three types of salts exist; acidic, basic, and neutral. When acidic and basic
salts are added to water the number of hydrogen and hydroxide ions change,
defining what the pH of the solution will be.
Neutral Salts
When neutral salts dissociate in water, both the cation and anion are spectator
ions; neither ion has the ability to tie up or contribute H + in the solution.
Ex: KNO3, NaCl
Acidic Salts
Acidic salts dissociate in water and increase the number of H + in the solution.
NH4Cl:
NH4 + + H2O

- 87 -
NH3 + H3O +
Basic Salts
Basic salts dissociate in water and decrease the number of H + in the solution.
C2H3O2 -
NaC2H3O2:

HC2H3O2 + OH -
Reactions Involving a Change in Oxidation State
Reactions between oxidizers and reducers can usually be predicted from the
following list of reagents:
Important Reducers
Product
halide ions
free metals
sulfite ions
nitrite ions
free halogens (dil, OH -), Cl2
free halogen
metal ions
sulfate ions
nitrate ions
hypohalite ions, ClO -
free halogens (conc, OH -), Cl2
metallous ions
halate ions, ClO3 metallic ions
Important oxidizers
Product
MnO4 - (H +)
Mn 2+
MnO2 (H +)
Mn 2+
MnO4 - (neutral/OH -)
Cr2O7
2-
(H
MnO2
+)
Cr 3+
Cr2O7 2- (OH -)
HNO3 (con)
HNO3 (dilute)
H2SO4 (hot, conc)
metallic ions
free halogens, X2, (F2, Cl2, etc)
Na2O2
CrO4 2NO2
NO
SO2
metallous ions
halide ions: F -, Cl -, I NaOH
HClO4
Cl -
- 88 -
Basic Organic Chemistry
Oxidation:
complete oxidation

CO2 + H2O
incomplete oxidation  CO + H2O
Addition:
involves the breaking of a multiple bond through the addition
of a mole of a reactant.
Addition reactions occur with H2 (catalyst Ni, Pt, or Pd),
halogens, hydrogen halide, water.
C2H4 + Br2
Substitution:
C2H4Br2
Hydrocarbons usually lacking multiple bonds. The hydrogen
atom leaves organic compound, becomes bonded to an
inorganic reactant, and is replaced by an atom/radical from
inorganic reactant producing two products
C2H6 + Br2
Esterification:


C2H5Br + HBr
Organic acid reacts with an alcohol to produce an ester and
water.
CH3COOH + CH3OH

CH3COO - CH3+
Being able to identify the names of organic compounds with 1-6 carbons is helpful.
meth
but
C1
C4
eth
pent
C2
C5
- 89 -
prop
hex
C3
C6
Functional Groups - Knowledge of functional groups is essential.
Isomers – Organic compounds with the same number of the same type of atoms,
set in different positions.
E.
Basics in Radioactivity
A basic knowledge of changes in the nucleus of a radioactive isotope is
required.
alpha particles

4
2
- 90 -
beta particles
0
1
gamma radiation
0
0


no change in nuclear charge or mass number
- 91 -
The AP Chemistry Exam
Section I:
You have 90 minutes to answer 75 multiple-choice questions;
72 seconds per question.
The first time through the multiple choice, focus only on questions that can be
answered within 20 - 30 seconds, then return to work on those questions that
require more time.
There are 10 questions you will not be able to answer. Accept that concept.
Total scores are based on the number of questions answered correctly.
Points are not deducted for incorrect answers.
The majority of students who get a 4 or 5 are able to correctly answer at least
50 of the 75 questions.
A simplified periodic table is provided.
Section II:
Part A:
You have 95 minutes to answer six questions.
You have 55 minutes to respond to 3 problems.
Calculators can ONLY be used in this part of the exam.
● One point is subtracted for arithmetic errors or not calculating the answer.
● Clearly define all unknowns, show reasoning and each formula used.
● Clearly show all work that leads to the numerical answer. Make sure the
grader can follow your work.
● If you do not have time to complete the arithmetic set-up, identify the
expected unit and number of significant figures in the answer. It ensures
you don’t lose points.
● If you are unsure how to respond to a question, move to the next question
and write, "Assuming the answer to the previous question was . . .", and
continue with the problem.
- 92 -
Part B
You have 40 minutes to answer 3 questions.
Question 4 requires balanced net ionic equations for three reactions,
along with an answer to a question about each reaction.
Questions 5 and 6 test your understanding of theory.
Answer questions 5 and 6 clearly, logically, and in a sequential order so the
grader can follow your work. Make sure you have addressed each point in
the question, but DO NOT write to write.
A simplified periodic table
and a list of equations and constants are provided.
Multiple choice and free response portions of the test are equally weighted.
- 93 -