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HONORS CHEMISTRY – FINAL EXAM REVIEW – SPRING 2014
STRATEGY: Start by reading through your notes (chapters 8 – 15, and 18) to refresh your memory on these topics. Then, use this
review sheet as a starting point to identify the areas on which you need to spend more study time. For those areas, go back to
homework assignments, and notes to practice more problems. These questions are only samples and do not include specific examples
of how vocabulary and other conceptual information might appear in a multiple-choice or other format.
FORMAT:
 Questions will include 65 multiple-choice and 2 Essay/Calculations.
 You will be given a periodic table and formula sheet.
Chapter 8: Balancing Equations
Decomposition (D), Synthesis (S), Single Replacement (SR), Double Replacement (DR), Combustion (C), Neutralization (N)
S
N
SR
S
C
DR
S
C
N
DR
1. 2 H2 + O2  2 H2O
2. H3PO4 + 3 KOH  K3PO4 + 3 H2O
3. 6 K + B2O3  3 K2O + 2 B
4. 6 Cs + N2  2 Cs3N
5. 2 C6H6 + 15 O2  12 CO2 + 6 H2O
6. BaCl2(aq) + K2(SO4)  2 KCl + BaSO4
7. N2 + 3 H2  2 NH3
8. 2 C10H22 + 31 O2  20 CO2 + 22 H2O
9. Al(OH)3 + 3 HBr  AlBr3 + 3 H2O
10. CaBr2 + Na2CO3  CaCO3(aq) + 2 NaBr(aq)
Chapter 9: Stoichiometry
1. Ideal combustion, resulting in only carbon dioxide and water, rarely happens. In general, at the very least, some
carbon monoxide is also produced. The real combustion of methane is more closely represented, then, by the
unbalanced equation CH4 + O2  H2O + CO2 + CO. If 0.100 mol CH4 is allowed to react with 3.20 g O2, what is
the limiting reagent, how many grams of carbon monoxide will be produced, and how many liters of the excess
reactant will be left over when the reaction has gone to completion, assuming the reaction occurs at STP?
3 CH4 + 5 O2  6 H2O + CO2 + 2 CO
0.100 mole CH4 │ 2 mole CO │ 28.01 g CO = 1.87 g CO
│ 3 mole CH4 │ 1 mole CO
3.20 g O2 │ 1 mole O2 │ 2 mole CO
│ 32.00 g O2 │ 5 mole O2
1.87 g CO
- 1.12 g CO
0.75 g CO
│ 28.01 g CO = 1.12 g CO
│ 1 mole CO
0.75 g CO │ 1 mole CO │ 3 mole CH4 │ 22.4 L CH4 = 0.89 L CH4
│ 28.01 g CO │ 2 mole CO │ 1 mole CH4
2. Elemental zinc, Zn, reacts violently with elemental sulfur, S8. If 1020. g of zinc react with 1.210 x 1024 atoms of
sulfur, what is the limiting reagent, how many moles of product are produced?
8 Zn + S8  8 ZnS
1020. g of Zn │ 1 mole Zn │ 8 mole ZnS = 15.60 mole ZnS
│ 65.39 g Zn │ 8 mole Zn
1.210 x 1024 atoms S8 │ 1 mole S8
│ 8 mole ZnS = 16.07 mole ZnS
│ 6.022 x 1023 atoms S8 │ 1 mole S8
Zn is the LR
3. 25.0 mL of 0.350 M NaOH are added to 45.0 mL of 0.125 M copper (II) sulfate. How many grams of precipitate
will be formed? (0.427 grams copper II hydroxide) What are the molar concentrations of all the species remaining
in the solution? (0.0179 M copper ions, 0.125 M sodium ions, and 0.0804 M sulfate ions)
Chapter 10: States of Matter - Solids Liquids and Gases
1. Which of the following is NOT a characteristic of ionic substances?
a. Are usually gases at room temperature.
c. Have high melting points.
b. Conduct electricity in solution form.
d. Usually dissolve in water.
2. Which of the following is NOT a characteristic of metallic substances?
a. Malleable and ductile.
c. Have low melting points.
b. Conduct electricity.
d. Are usually solids at room temperature.
3. Which of the following is NOT a characteristic of covalent substances?
a. Have low melting points.
c. Form individual molecules.
b. Sometimes dissolve in water.
d. Conduct electricity.
4. What is the difference between an intermolecular force and an intramolecular force?
- What are the three types of intramolecular forces?
Ionic bond, Covalent, bond, and Metallic bond
- What are the three types of intermolecular forces?
Hydrogen bonding, Dipole-dipole, and London Dispersion Forces
- Arrange the intermolecular and intramolecular forces from weakest to strongest.
LDF, D-D, H. Bonding, (ionic, covalent, and metallic) bonding
5. What causes dipole-dipole interactions?
The cause of the dipole moment in a compound is due to the difference in electron affinity of the atoms that are
bonded in the compound.
6. Why is hydrogen bonding only possible with hydrogen?
Hydrogen is the only element that has an exposed proton when an electron is lost. The exposure of the proton
and the fact that the other element that the hydrogen in bonded to has a very high electron affinity, the compound
ends up having a very strong dipole moment called hydrogen bonding.
7. What are the three possible elements that hydrogen atom must be attached to, in a compound, in order for the
compound to contain hydrogen bonding? N, O, and F
8. Explain how London dispersion forces arise. Although London dispersion forces exist among all molecules, for
what type of molecules are they the only major intermolecular forces? Are London dispersion forces relatively
strong or relatively weak? Explain. LDF are created by a momentary polarity due to a valence electron moving
out of its normal orbit. Nonpolar molecules are the only type of molecules in which LDF is the major
intermolecular force of attraction. LDF are the weakest force of attraction.
9. What causes surface tension? The forces of attraction from molecule to molecule (intermolecular force –
hydrogen bonding) are much stronger than the attraction for the objects that the liquid comes in contact with.
10. Why is ice less dense than water? The density of ice is less because when ice is formed, the water molecules
form rings that create more space between the molecules.
11. What is the reasoning behind Chlorine being in a gaseous state, bromine being in a liquid state, and iodine being
in a solid state? LDF increases as the size of the atom increases. Since the size of the atoms increase from Cl 2
to Br2 to I2, the forces of attraction (LDF) are strong enough for the states to go from gaseous state, to liquid state
to solid state for the respective molecules.
12. Specify the predominant force of attraction that would affect the boiling point of the compound and list it
immediately following the substance. Then in the last column, indicate which member of the pair you would
expect to have the higher boiling point. Possible forces of attraction are: metallic bonding, ionic bonding,
network covalent bonding, hydrogen bonding, dipole-dipole and London dispersion forces
Predominant
Intermolecular Force
Dipole-dipole
Substance #2
I2
Predominant
Intermolecular Force
LDF
Substance with
Higher Boiling Point
HCl(l)
b. CH3F
Dipole-dipole
CH3OH
Hydrogen bonding
CH3OH
c. H2O(l)
Hydrogen bonding
H2S
Dipole-dipole
H2O(l)
d. SiO2(s)
SO2
Dipole-dipole
SiO2(s)
e. Fe(s)
Network covalent
bonding
Metallic Bonding
Kr
LDF
Fe
f. CH3OH(l)
Hydrogen bonding
CuO(s)
Ionic bonding
CuO
g. NH3
Hydrogen bonding
CH4
LDF
NH3
Dipole-dipole
NaCl
Ionic bonding
NaCl
Network covalent
bonding
Cu(s)
Metallic Bonding
SiC
Substance #1
a. HCl(l)
h. HCl(g)
i.
SiC
13. Draw a Heating Curve Diagram and label where each of the following would either be located on the
curve or label on what part of the curve it would be used to complete calculations.
a. Solid
e. Liquid
i. Vapor
b. Specific Heat of Solid
f. Specific Heat of Liquid
j. Specific Heat of Vapor
c. Melting/Freezing
g. Vaporizing/Condensing
d. Heat of Fusion
h. Heat of Vaporization
14. What formula is used to calculate the amount of energy needed to change the phase of a substance?
Q = H x moles (heat/energy = enthalpy of ______ x mole of substance)
15. CCl2F2 has a boiling point of -30. C, and a heat of vaporization of 0.165 kJ/g. The vapor and the liquid have
specific heats of 0.61 J/gK and 0.97 J/gK respectively. How much heat must be evolved when 10.0 g of
CCl2F2 is cooled from +40. C to -40. C?
Q = mCpT = (10.0 g)( 0.61 J/gK)(70. K)
= 430 J
Q = H x mass = (0.165 kJ/g)(10.0 g) = 1.65 kJ
= 1650 J
Q = mCpT = (10.0g)(0.97 J/gK)(10 K)
= 97 J
– 2180 J
(It is negative because energy is given off.)
16. For the following Phase Change Diagrams, identify what states are in sections A, B, and C and what occurs when
each line is crossed (in each direction). What is the point D called? What is the point E called? Which of the
diagrams is the Change of State Diagram for water? How can you tell?
A – solid, B – liquid, C – gas, going across the line from: A to B melting occurs, B to A freezing occurs, B to C
vaporization occurs, C to B condensation occurs, A to C sublimation, C to A deposition. D is the triple point. E is the
critical point. The first diagram is the diagram for water because the line between A and B leans to the left indicating
that the density of the solid is less than the density of the liquid.
Chapter 11: Gas Laws
1. What does Boyle’s Law state, what is the formula, and what is held constant?
– the volume of a fixed mass of gas varies inversely with the pressure at constant temperature.
P1V1 = P2V2
(at constant temperature and moles)
2. What does Charles’ Law state, what is the formula, and what is held constant?
– the volume of a fixed mass of gas at constant pressure varies directly with the Kelvin temperature.
V1 = V2
(at constant pressure and moles)
T1
T2
3. What does Avogadro’s Law state, what is the formula, and what is held constant?
4. What is the formula for the Combined Gas Law and what is held constant?
– expresses the relationship between pressure, volume, and temperature of a fixed amount of gas.
P1 V1 = P2 V2
T1
T2
5. What is the formula for the Ideal Gas Law and what is the value and units on R?
PV = nRT
R = 0.0821 L ▪ atm / mole ▪ K
6. What is diffusion and effusion? What is the formula that is used for these calculations?
Diffusion – the random movement of gases from an area of higher concentration to an area of lower concentration
in open space.
Effusion – the random movement of gases from an area of higher concentration (in a closed container) to an area
of lower concentration (outside of the container) through a small hole or opening in the container.
Formula for calculating diffusion:
Distance A
Distance B
=
mB
mA
Formula for calculating effusion:
Velocity A
Velocity B
=
mB
mA
7. 1 mole of any gas at STP = 22.4 L
8. The conversion factors for Pressure are equal to:1.00 atm = 760. mm = 760. torr = 101.3 kPa
9. A 2.00 L sample of a gas originally at 25 oC and a pressure of 700. torr is allowed to expand to a volume of
5.00 L. If the final pressure of the gas is 585 torr, what is its final temperature in Celsius? 350. oC
10. A gas has a volume of 350. mL at 740. torr. How many milliliters will the gas occupy at 900. torr if the
temperature remains constant? 288 mL
10. Calculate the number of liters occupied by the following at STP:
a. 0.200 moles of oxygen 4.48 L
b. 12.4 grams of chlorine gas 3.92 L
c. a mixture of 0.100 moles of nitrogen gas and 0.500 moles of oxygen gas 13.4 L
11. From the volume, temperature and pressure listed below, calculate the number of moles for the gas:
750.0 cm3 at 27.0 oC and 99.0 kPa. 0.0298 moles
12. A balloon has a volume of 2.0 L indoors at a temperature of 25.0 oC. If it is taken outdoors on a very cold day
when the temperature is –28.9 oC, what will its volume be in liters? Assume a constant air pressure within the
balloon. 1.6 L
13. Calculate the volume of 6.00 g of hydrogen gas at 24.0 oC and 1.50 atm. 48.3 L
14. Calculate the mass of 24.5 mL of SO2 gas at STP. 0.0701 g
15. If excess hydrochloric acid is added to 13.5 grams of Al, what volume of hydrogen gas will be produced if
the gas is collected at a temperature of 80.0 oC and a pressure of 750. torr? 22.1 L
16. At a certain temperature, the velocity of chlorine molecules is 0.0410 m/s. What is the velocity of sulfur dioxide
molecules at the same temperature and pressure? 0.0431 m/sec
17. What is the ratio of the speed of carbon monoxide molecules to that of nitrogen monoxide molecules when both
gases are at the same temperature? 1.04 : 1 (rate CO : rate NO)
18. The rate of effusion of an unknown gas was determined to be 2.92 times faster than that of ammonia. What is the
approximate molecular weight of the unknown gas? 2.00 g/mole
19. In the reaction, N2 + H2  NH3, how many mL of nitrogen, measured at STP, are required to produce 400 mL of
NH3, measured at STP? How many mL of hydrogen are required at STP? a. 200 mL N2
b. 600 mL
Chapter 12: Solutions
1. What is molarity (M), molality (m), normality (N), and mass percent?
M = mole of solute / liter of solution, m = moles of solute / kg of solvent, N = M x number of equivalents,
mass % = mass of solute / mass of solution x 100 or part over whole x 100
2.
3.
4.
5.
6.
What term is used to describe a substance dissolving? solubility
What is a solid – solid solution of two or more metals called? alloy
A mixture in which the particles are so small that they will not reflect the “light” from a laser are called solution.
A solution that contains a large amount of solute per amount of solvent is called a concentrated solution.
What is the term that is used to describe a solid being formed when two or more liquids are mixed together?
Precipitation or precipitate
7. When a solution is made, what is the smaller portion of the solution called? solute
8. A solution that contains as much solute as it can hold at a given temperature is called a saturated solution.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
When a solution is made, what is the larger portion of the solution called? solvent
A mixture in which the particles settle out upon stand is called a suspension.
If a substance will NOT dissolve it is said to be insoluble.
A solution that will hold more solute is said to be unsaturated.
A solution that contains a small amount of solute per amount of solvent is called a dilute solution.
A mixture in which the particles can only be seen with a laser and do NOT settle out upon standing is called a
colloid.
The amount of solute that can be dissolved in a solvent at a given temperature is described as solubility.
Due to the fact that so many substances dissolve in water, water has been called the universal solvent.
Calculate the molarity of a solution prepared by dissolving 15.6 g of solid KBr in enough water to make 1.25 L of
solution. 0.105 M
Calculate the number of moles of Ca(OH)2 in 150.0 mL of a 0.500 M solution. How many moles of each ion are
in solution? a. 0.0750 moles b. 0.0750 moles and 0.150 M
A solution is prepared by adding 2.0 L of 6.0 M HCl to 500 mL of a 9.0 M HCl solution. What is the molarity of
the new solution? (Remember, the volumes are additive) 6.6 M
Convert the following Molarities to Normalities. a. 2.5 M HCl, b. 1.4 M H2SO4, c. 1.0 M NaOH, d. 0.5 M Ca(OH)2
a. 2.5 N, b. 2.8 N, c. 1.0 N, d. 1 N
25.0 mL of 0.339 M Magnesium nitrate and 35.0 mL of 0.238 M Sodium hydroxide are mixed, what is the total
concentration of all ions left over in the solution? 0.494 M of ions
If 63.8 grams of Aluminum sulfide are precipitated when aluminum nitrate and sodium sulfide are mixed, how
many milliliters of 2.50 M Aluminum nitrate is needed to form that mass of precipitate? (Assume that there is an
excess of sodium sulfide dissolved in solution.) 340. mL of Aluminum nitrate
A solution is made by dissolving 25 g of NaCl in enough water to make 1.0 L of solution. Assume the density of
the solution is 1.0 g/cm3. Calculate the molarity, and molality of the solution. 0.43 M, 0.44 m
How many grams of KBr are contained in 250. grams of a 6.25 % KBr solution? 16.6 g KBr
25. Answer the following questions based on the solubility graph below.
- What is the temperature (oC) at which HCl and KNO3 have the same solubility? 38 oC
- What is the maximum amount of KClO 3 that can be dissolved in 25 g of water at 100 oC? 15 g
- What is the solubility of KNO3 in 100 g of water at 60 oC? 112 g
- What is the order of decreasing solubility for the substances NaCl, NaNO 3, HCl, NH4Cl in 100 g of
water at 40 oC? NaNO3 > HCl > NH4Cl > NaCl
Chapter 13: Colligative Properties and Solubility Rules (Net Ionic Equations)
1. What is a Colligative property?
Properties that depend on the concentration of solute particles but not on their identity. (For example: Freezing-point
Depression, Boiling-point Elevation, and Osmotic Pressure)
2. What are freezing point depression, boiling point elevation, and osmotic pressure? List the
formulas that are used to calculate each.
Freezing Point Depression (Tf) – is the difference between the freezing points of the pure solvent and solution of a
nonelectrolyte in the solvent, and it is directly proportional to the molal
concentration of the solution.
For an electrolyte solution, the number of ions (i) must be taken into account.
(Tf = i Kf m)
Boiling-point Elevation (Tb) – is the difference between the boiling points of the pure solvent and a nonelectrolyte
solution of that solvent, and it is directly proportional to the molal concentration of the
solution.
For an electrolyte solution, the number of ions (i) must be taken unto account.
(Tb = i kb m)
Osmotic Pressure – is the external pressure that must be applied to stop osmosis. The number of ions (i) must also
be taken into account in osmotic pressure calculations.
 = iMRT
3. What is the molal freezing-point constant (Kf) for water?
Kf = 1.86 oC ▪ kg
mole
4. What is the molal boiling point constant (Kb) for water?
Kb = 0.52 °C
m
5. a. What is the molality of a solution of 1.25 moles of sugar dissolved in 0.750 kg of water?
b. What is the boiling point of the resulting solution?
c. What is the freezing point of the resulting solution?
a.
m = 1.25 mole/0.750 kg = 1.67 m
b.
Tb = i kb m

Tb = (1)(0.52 °C/m)( 1.67 m)
Tb = 0.87 °C
BP = 100.00 °C + 0.87 °C = 100.87 °C
c.
Tf = i kf m
Tf = (1)(1.86 oC ▪ kg/mole)( 1.67 m)
Tf = 3.11 oC
FP = 0.000 oC – 3.11 oC = – 3.11 oC
6. An aqueous solution contains 3.00 % phenylalanine (C9H11NO2) (MM = 165.21 g/mole) by mass at 25.0 oC. Assume
the phenylalanine is non-ionic and volatile and that the density of the solution is 1.00 g/mL. Calculate the following:
a.
m = (3.00 g)/(165.21 g/mole)/(0.097 kg) = 0.19 m
The freezing point of the solution (T f)
Tf = i kfm
Tf = (1)(1.86 oC ▪ kg/mole)(0.19 m)
Tf = 0.35 oC
FP = – 0.35 oC
b.
The boiling point of the solution (T b)
Tb = i kbm
Tb = (1)(0.52 °C/m)(0.19 m)
Tb = 0.099 °C
BP = 100.000 °C + 0.099 °C = 100.099 °C
c.
The osmotic pressure of the solution ()
M = (3.00 g)/(165.21 g/mole)/0.100 L = 0.182 M
 = iMRT
 = (1) (0.182 M) (0.0821 L ▪ atm / mole ▪ K) (298 K)
 = 4.45 atm
7. What is the molarity of a solution in which the osmotic pressure is 80.7 atm at 27 oC? Assume the solute is a
nonelectrolyte.
 = iMRT
M =  / iRT = (80.7 atm) / (1)( 0.0821 L ▪ atm / mole ▪ K) (300. K)
Molarity = 3.28 M
8. At what temperature, in Celsius, would the osmotic pressure have to be 347.5 kPa in order to stop osmosis if there
was 24.37 g of glucose (MM = 180.18 g/mole) dissolve in enough water to make a 1.00 L solution?
a. Calculate moles of glucose:
24.37 g / 180.18 g/mole = 0.135 mole
b. Calculate molarity of solution:
M = 0.135 mole / 1.00 L = 0.135 M
c. Calculate the osmotic pressure in atm:

.5 kPa / 101.3 kPa = 3.43 atm
d. Calculate the temperature:
T =  / iMR = (3.43 atm) / (1) (0.135 M) (0.0821 L ▪ atm / mole ▪ K)
T = 309 K
9. When 5.0 g of CaCl2 dissolves in 50.0 g of water, what is the boiling point of the solution?
a. Calculate moles of CaCl2:
5.0 g / 110.98 g/mole = 0.045 mole
b. Calculate molality of CaCl2:
m = 0.045 mole/0.0500 kg = 0.90 m
c. Use boiling point elevation constant:
T = i Kb m
T = (3) (0.52 °C/m ) (0.90 m) = 1.4 °C
The CaCl2 solution boils at 101.4 °C
10. Seawater is about 3.5% (by weight) dissolved solids, almost all of which is NaCl. Calculate the normal boiling point of
seawater.
a. Calculate moles of NaCl:
3.5 g / 58.5 g/mole = 0.0598 mole
b. Calculate molality of NaCl:
m = 0.0598 mole / 0.0965 kg = 0.612 m
c. Use boiling point elevation constant:
T = i Kb m
T = (2) (0.52 °C/m) (0.612 m) = 0.64 °C
The seawater boils at 100.64 °C
11. Calculate the freezing point of a solution of 5.00 g of diphenyl C12H10 and 7.50 g of naphthalene, C10H8 dissolved in
200.0 g of benzene (fp = 5.5 °C and Kf = 5.12 °C/m).
There is a tiny curve in this problem, but keep in mind that Colligative properties are all about how many particles
in solution and nothing else.
a. So, the key to this problem is to calculate moles of each substance and add then together:
(5.00 g / 154.2 g/mole) + (7.50 g / 128.2 g/mole) = 0.0909 mole
b. Now, we calculate the molality:
0.0909 mole / 0.200 kg = 0.455 m
c. The last step is to calculate the freezing point depression:
T = i Kf m
T = (1) (5.12 °C/m) (0.455 m) = 2.33 °C
The solution freezes at 5.5 – 2.33 = 3.2 °C
12. Salt is often used to remove ice from roads and sidewalks. Explain how this process works in terms of Colligative
properties.
When the salt dissolves into solution it ionizes. It is the production of ions that causes the resulting solution to have a
lower freezing point than the solvent (water/snow ice). The more ions that are produced in solution the lower the
resulting freezing point.
13. Which salt, NaCl or CaCl2, has a greater effect on freezing point? Explain.
Since CaCl2 produces more ions in solution, it will cause the resulting solution to have a lower freezing point. The
more particles dissolved in the solvent the lower the freezing point.
14. Determine the products (indicating the states) and write the net ionic equation for each of the following.
a. AgNO3(aq) + KI(aq) → KNO3(aq) + AgI(s)
b. Pb(NO3)2(aq)+ Na2S(aq) → 2 NaNO3(aq) + PbS(s)
c. 2 (NH4)3PO4(aq) + 3 Ca(NO3)2(aq) → 6 NH4NO3(aq) + Ca3(PO4)2(s)
d. Fe(NO3)3(aq) + 3 NaOH(aq) → 3 NaNO3(aq) + Fe(OH)3(s)
15. Write a balance equation and label the states of the reactants and products for each the reactions. After writing the
balanced equation, write the net ionic equations below the balanced equation. If there is no driving force for a
reaction, write no reaction in place of the net ionic equation.
a. Hydrogen sulfide is bubbled through a solution of strontium hydroxide.
H2S(aq) + Sr(OH)2(aq)  2 HOH() + SrS(aq)
H+1(aq) + (OH)-1(aq)  HOH()
b. Aluminum metal is added to a solution of copper (II) chloride.
2 Al(s) + 3 CuCl2(aq)  2 AlCl3(aq) + 3 Cu(s)
Cu+2(aq)  Cu(s)
c.
A manganese (II) nitrate solution is mixed with a sodium hydroxide solution.
Mn(NO3)2(aq) + 2 NaOH(aq)  2 NaNO3(aq) + Mn(OH)2(s)
Mn+2(aq) + (OH)-1(aq)  Mn(OH)2(s)
d. Solutions of sodium carbonate and lead (II) nitrate are mixed.
Na2CO3(aq) + Pb(NO3)2(aq)  PbCO3(s) + 2 NaNO3(aq)
Pb+2(aq) + (CO3)-2(aq)  PbCO3(s)
e. Aqueous solutions of sodium phosphate and barium chloride are mixed.
2 Na3PO4(aq) + 3 BaCl2(aq)  Ba3(PO4)2(s) + 6 NaCl(aq)
Ba+2(aq) + (PO4)-3(aq)  Ba3(PO4)2(s)
f.
Hydrogen sulfide gas is added to a solution of cadmium (II) nitrate.
H2S(aq) + Cd(NO3)2(aq)  CdS(s) + 2 HNO3(aq)
Cd+2(aq) + S-2(aq)  CdS(s)
g. A solution of sulfuric acid is added to a solution of barium hydroxide.
H2SO4(aq) + Ba(OH)2(aq)  BaSO4(s) + 2 HOH()
Ba+2(aq) + (SO4) -2(aq)  BaSO4(s)
H+1(aq) + (OH)-1(aq)  HOH()
h. Dilute sulfuric acid is added to a solution of calcium acetate.
H2SO4 (aq) + Ca(C2H3O2)2(aq)  CaSO4 (s) + 2 H(C2H3O2) (aq)
Ca+2(aq) + (SO4) -2(aq)  CaSO4(s)
i.
Ammonium chloride solution is added to a solution of sodium hydroxide.
NH4Cl(aq) + NaOH(aq)  NaCl(aq) + (NH4)(OH)(aq)
NO REACTION
j.
A precipitate is formed when solutions of sodium sulfite and calcium chloride are mixed.
Na2SO4(aq) + CaCl2(aq)  CaSO4(s) + 2 NaCl(aq)
Ca+2(aq) + (SO4) -2(aq)  CaSO4(s)
Acids and Bases
1. What is the Arrhenius definition of an acid and a base? An acid is a compound that contains and produces a
hydrogen or hydronium ion when dissolved in water and a base is a compound that contains and produces a
hydroxide when dissolved in water.
2. What is the Brønsted-Lowry definition of an acid and a base? An acid is a proton donor and a base is a proton
acceptor.
3. List the 6 strong acids and the 8 strong bases. SA – HCl, HBr, HI, HNO3, H2SO4, HClO4, SB – LiOH, NaOH,
KOH, RbOH, CsOH, Ca(OH)2, Sr(OH)2, Ba(OH)2
4. What is the difference between a strong acid or base and a weak acid or base? A strong acid “completely”
dissociates when dissolved in water where as a weak acid on partially dissociates when dissolved in water,
5. What are diprotic and polyprotic acids? A diprotic acid produces two protons (H+1) in solution and a polyprotic acid
produces two or more protons in solution.
6. What are conjugate acids and a conjugate bases? Conjugates are always found on the product side of the
equation.
7. In each of the following chemical equations, identify the conjugate acid/base pairs and label them as weak or
strong. (HSO4-1 + H2O
SO4-2 + H3O+1)
(NH3 + H2O
NH4+1 + OH-1)
WA
WB
SCB
SCA
WB
WA
SCA
SCB
8. Label each of the following as either strong conjugate acid, weak conjugate acid, strong conjugate base, or weak
conjugate base. (Na+1 – WCA, CH3COO-1 – SCB, Cl-1 – WCB, Al+3 – SCA, Mg+2– SCA)
9. Which of the following is a conjugate acid/base pair? (HCl/OCl-1, H3O+1/OH-1, H2SO4/SO4-2, HPO4-2/PO4-3)
10. Indicate whether the following will produce an acidic, basic, or neutral solution when dissolved in water.
NaNO3 – N, BeO – B, NH4ClO4 – A, NO2 – A, CaCO3 – B, Cs2O – B, CuHSO4 – A, and Cl2O3 – A
11. Calculate the pH, pOH and [OH-1] of a solution in which the [H+1] = 1.2 x 10-3. Calculate each significantly!
pH = 2.92, pOH = 11.08, [OH-1] = 8.3 x 10-12 M
12. Calculate the pH, pOH, and [H+1] of a solution in which the [OH-1] = 2.34 x 10-5. Calculate each significantly!
pH = 9.369, pOH = 4.631, [H+1] = 4.27 x 10-10 M
13. Calculate the pH of a 0.10 M solution of Ca(OH)2. pH = 13.3
14. Using the following Ka values, place the following acids in order of increasing acid strength.
HClO4 (ka = 1 x 107), HCN (Ka = 4.93 x 10-10), CH3OOH (ka = 1.76 x 10-5 ), HF (Ka = 3.53 x 10-4)
HCN < CH3OOH < HF < HClO4
15. Using the Ka values from above, calculate the Kb values and place the following bases in
order of increasing strength. CN-1, F-1, CH3OO-1, ClO4-1 (Kw = Ka x Kb)
ClO4-1 < F-1 < CH3OO-1 < CN-1
16. Draw titration curves for the following titrations and label the x-axis, y-axis, and equivalence point: SEE YOUR
NOTES FOR THE FOLLOWING
a. 0.100 M NaOH is titrated into a solution a solution of 0.200 M HNO3.
b. 1.0 M HCl is titrated into a solution of 0.50 M NaOH.
17. What volume of 0.050 M NaOH(aq) will completely neutralize 100.0 mL of 0.075 M HBr solution? 150 mL
18. What is the conjugate base for each of the following acids? (HSO3 -1 – SO3 -2, H2S – HS-1, HF – F-1, CH3COOH –
CH3COO-1)
19. Which of the following ions can react as a buffer? For each buffer, write a reaction for how it would react with a
strong acid and a reaction for how it would react with a strong base.(HSO4 -1, HS -1, HCO3 -1, C2H3O2 -1)
HS -1 + H+1  H2S and HS -1 + OH-1  S-2 + H2O
HCO3 -1 + H+1  H2CO3 and HCO3 -1 + OH-1  CO3-2 + H2O
20. Which of the following would be an example of a buffered solution? (0.30 M HF / 0.30 M HCl,
0.30 M HC2H3O2 / 0.30 M NaC2H3O2, 0.30 M NaCl / 0.30 M NaOH, 0.30 M HClO4 / 0.30 M NaClO4)
21. What is an electrolytic solution? What is the difference between a strong electrolyte and a weak electrolyte?
An electrolyte solution is a solution that contains ions and will conduct electricity. A strong electrolyte solution
contains a large amount of ions and conducts electricity very well. A weak electrolyte solution contains very few
ions and conducts electricity poorly.
22. Which of the following solutions would be the strongest electrolytic solution?
(3.0 M HCN, 2.8 M HBr, 2.75 M HNO3, 3.1 M Mg(OH)2, 3.0 M RbOH)
Equilibrium
1. What is Activation Energy? How does Activation Energy effect a reaction?
It is the threshold energy that must be overcome to produce a chemical reaction. The activation energy
indicates the amount energy needed for the reaction to occur. If activation energy for a reaction is high, then
energy must be added (typically in the form of heat) in order for the reaction to occur.
2. Draw an Activation Energy diagram for an endothermic and an exothermic reaction and label the x-axis, y-axis,
reactants, products, activation energy, and total energy change.
3. What is a Catalyst and how does a catalyst effect a reaction?
It speeds up the reaction and is NOT consumed in the process. It helps an equilibrium reaction
reach its equilibrium more quickly.
4. What occurs in a reaction that causes a Chemical equilibrium?
A forward reaction occurs at the same rate as the reverse reaction. In other words, reactants react to produce
products at the same rate that the products react to produce reactants.
5. Write an Equilibrium expression for the following reactions:
a. CaCO3(s)
CaO(s) + CO2(g)
K = [CO2]
b. Ni(s) + 4 CO(g)
Ni(CO)4(g)
K = [Ni(CO)4]
[CO]4
c. 5 CO(g) + I2O5(s)
I2(g) + 5 CO2(g)
K = [CO2]5[ I2]
[CO] 5
d. Ca(HCO3)2(aq)
CaCO3(s) + H2O(l) + CO2(g)
K=
[CO2]
[Ca(HCO3)2]
6. What is an equilibrium constant and what does the equilibrium constant indicate about the reaction?
The value obtained when equilibrium concentrations of the chemical species are substituted into the equilibrium
expression and raised to a power that is the same as the coefficient of each species. The equilibrium constant
indicates which side of the reaction is favored. A larger equilibrium indicates that the products side is favored
(product side has greater concentrations) and a smaller equilibrium constant indicates that the reactant side of
the reaction is favored (reactant side has a larger concentrations).
7. Calculate the value of the equilibrium constant for the following system, given the data shown: H 2(g) + CO2(g)
H2O(g) + CO(g) Concentrations at equilibrium: [H2] = 1.5 M, [CO2] = 2.5 M, [H2O] = 0.5 M, [CO] = 3.0 M K = 2.0
8. For the reaction 2 H2O(g)
2 H2(g) + O2(g) K = 2.4 x 10-3 at a given temperature. At equilibrium it is found
that [H2O] = 1.1 x 10-1 M and [H2] = 1.9 x 10-2 M. What is the concentration of O2(g) under these conditions?
[O2] = 0.080 M
9. What is the difference between a homogeneous equilibrium and a heterogeneous equilibrium?
In a homogeneous equilibrium all species (reactants and products) are in the same state and in a heterogeneous
equilibrium some or all species (reactants and products) are in different state.
10. What is LeChâtelier’s Principle?
If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends
to reduce the effect (or stress) of that change.
11. Consider the equilibrium PCl3(g) + Cl2(g)
PCl5(g). How would the following changes affect each gas at equilibrium?
PCl3(g) + Cl2(g)
PCl5(g)
a. addition of PCl3
+
– → +
b. removal of Cl2
+
– ← –
c. removal of PCl5
–
– → –
d. decrease in the volume of the container –
– → +
e. addition of He without change in volume
NO EFFECT