Chemistry 30 Solutions Manual
Unit A: Thermochemical Changes
Lesson: Analyzing Heat Transfer
1.
2.
Practice Problems: Analyzing Heat Transfer
1.
2.
3.
1
4.
5.
6.
7.
8.
9. Iron would undergo the greatest temperature change because it has the lowest specific heat
capacity—it takes less energy to raise the temperature of 1 g of iron by 1°C.
2
Lesson: Calorimetry
1.
2.
3.
4.
5.
3
6.
Practice Questions: Calorimetry
1. According to the law of conservation of energy, energy cannot be created or destroyed; rather, it
can only be converted from one form to another.
2. The calorimeter prevents energy and matter transfer between the inside of the calorimeter and the
outside environment. No calorimeter is a perfect isolated system and some energy is transferred.
3.
4.
5.
6. a.
4
b.
c.
d. The assumption of negligible heat transfer to the solid calorimeter materials is acceptable, since
the percent error introduced is very low and is likely less than the total experimental
uncertainties.
7. An enthalpy change or molar enthalpy with a negative sign indicates that the change is exothermic,
while a positive sign denotes an endothermic reaction.
8.
9.
10. a.
b.
11.
12.
13. a. 22.2%
5
b. 487 kg
14.
15.
16.
Lesson: Communicating Enthalpy Changes
1. –1036.0 kJ
2. C2H6(g) +136.4 kJ → C2H4(g) + H2(g)
3.
6
4.
Practice Questions: Communicating Enthalpy Changes
1. a. The standard molar enthalpy of combustion for methane
b. The chemical amount of propane multiplied by the standard molar enthalpy of formation for
propane
c. The change in quantity of thermal energy for water
2. a. ΔfHm° = +227.4 kJ/mol
C2H2
2 C(s) + H2(g) → C2H2(g)
ΔfH° = +227.4 kJ
2 C(s) + H2(g) + 227.4 kJ → C2H2(g)
b. ΔsdHm° = +1675.7 kJ/mol
7
Al2O3
Al2O3(s) → 2 Al(s) + O2(g)
ΔsdH° = +1675.7 kJ
Al2O3(s) + 1675.7 kJ → 2 Al(s) + O2(g)
c. ΔfHm° = −393.5 kJ/mol
C
C(s) + O2(g) → CO2(g)
ΔcH° = −393.5 kJ
C(s) + O2(g) → CO2(g) + 393.5 kJ
3. a. ΔcHm° = −241.8 kJ/mol
H2
b. ΔcHm° = −318.0 kJ/mol
NH3
c. ΔcHm° = +81.6 kJ/mol
N2
d. ΔcHm° = −372.8 kJ/mol
Fe
4. a. −114 kJ
b. H2SO4(aq) + 2 NaOH(aq) → Na2SO4(aq) + 2 H2O(l)
c. ΔnHm = −114 kJ/mol
ΔnH = −114 kJ
H2SO4
d. ΔnHm = −57 kJ/mol
NaOH
5. a. CH3OH(l) + O2(g) → CO2(g) + 2 H2O(g) + 725.9 kJ
b. C(s) + S8(s) + 89.0 kJ → CS2(l)
c. Zn(s) + O2(g) → ZnO(s) + SO2(g) + 441.3 kJ
d. Fe2O3 + 824.2 kJ → 2 Fe(s) + O2(g)
8
*Note the advantage of writing the chemical equation using one mole of the substance whose molar
enthalpy is given. All of these equations could be written with integer coefficients and the appropriate
enthalpy changes.*
6. a. C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) ΔcH° = −2043.9 kJ
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) + 2043.9 kJ
b. ½ N2(g) + ½ O2(g) → NO(g)
ΔfH° = +91.3 kJ/mol
½ N2(g) + ½ O2(g) + 91.3 kJ → NO(g)
c. C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g) ΔcH° = −1 234.8 kJ
C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g) + 1 234.8 kJ
7. a. H2(g) + ½ O2(g) → H2O(l)
ΔcH° = −285.8 kJ
H2O(l) → H2(g) + ½ O2(g)
ΔcH° = +285.8 kJ
b. The enthalpy changes are for the same magnitude and differ only in that one is negative
(exothermic) and the other is positive (endothermic). Reactions that are exactly the reverse of each
other have the same magnitude for the enthalpy change, but are of opposite sign.
In general, ΔrH = −ΔrH.
reverse
forward
Lesson: Hess’ Law
1. −110.5 kJ
2. −125.6 kJ
Practice Questions: Hess’ Law
1.
2.
3.
9
4.
5.
6.
7. a.
b.
c. This technique works because the addition of any chemical equations that yield the intended net
chemical reaction will provide the net enthalpy change. All of the reaction paths start and end with
the same chemicals and the same chemical potential energy (and thus the same change in
enthalpy).
8.
9.
10
10.
Lesson: Enthalpies of Formation
1. −64.5 kJ
2. −802.5 kJ
Practice Questions: Enthalpies of Formation
1.
3
2. a.
11
b.
c.
3. a.
b.
c.
4. a.
12
b.
5. a.
b.
6. a.
b.
7. a.
13
b.
c.
d.
e. They are the same.
8.
9.
10. a.
b.
14
c.
Lesson: Activation Energy
1.
Practice Questions: Activation Energy
1. Two conditions that must be met for two molecules to react are: the molecules must have sufficient
energy and they must have the correct orientation at the moment of collision.
2. a.
b.
c. The electrons or charge in the lightning are more effective in initiating the reaction because the
temperature and pressure are much higher during the strike. An alternative hypothesis is that the
electrons or charge in the lightning strike may act as a catalyst, or the electrons may cause other
nearby molecules to become ions and act as a catalyst in the reaction to increase the rate of the
reaction.
3. A successful reaction of two molecules involves a collision between molecules that have sufficient
kinetic energy to equal the activation energy. The molecules must also collide with the correct
orientation (positioning). Once activation energy is reached, the reactants form an activated
complex and the reaction begins. The activated complex then breaks apart into product molecules.
4. a. The reaction is endothermic as indicated by the chemical potential energy of the products being
greater than the chemical potential energy of the reactants.
b. i: the forward reaction activation energy; ii: the forward reaction enthalpy change
c. As the reactant entities, R, speed towards each other, some will combine to form an activated
complex, if they have the correct orientation and sufficient kinetic energy. To form the activated
complex, the kinetic energy of the reactant molecules must be converted to potential (bond) energy
within the activated complex. The activated complex breaks into the product entities, P, converting
15
the potential (bond) energy into kinetic energy. Since this reaction is endothermic, there will be less
kinetic energy and more potential energy in the products than in the reactants.
5. Enthalpy change and activation energy are similar in that they both measure the difference in
chemical potential energies between molecules at different stages in a reaction. They are different
in that activation energy measures the chemical potential energy difference between the reactants
and the activated complex, while enthalpy change measures the chemical potential energy
difference between the reactants and final products. [For the reverse reaction, the activation energy
changes but only the sign of the enthalpy change changes.]
6. a. For the reaction of hydrogen and oxygen to produce water, the hydrogen and water must collide
with the correct orientation, and they need sufficient kinetic energy (which could be supplied by a
spark or a flame) to overcome the activation energy for the reaction.
b.
c.
Practice Questions: Bond Energy and Reactions
1. According to bond energy theory, bond energy is the energy required to break a chemical bond.
Bond energy is also the energy released when chemical bonds are formed.
2. Our theoretical understanding of reactions includes the theory that bond energies are related to the
activation energy for a reaction in that the activation energy is the energy required to break the
bonds between reactant entities that can then create the activated complex.
3.
16
4. a. According to chemical energy theory, the decomposition of water is endothermic because the
quantity of energy required to break the O–H bonds within a water molecule is greater than the
quantity of energy released when the H–H and O=O bonds are formed.
b. Chemists explain that the combustion of hydrogen is exothermic because the quantity of energy
required to break the H–H and O=O bonds of the reactants is less than the quantity of energy
released when the O–H bonds are formed within water.
5. According to the table of molar enthalpies of formation, the molar enthalpy for the formation of
hydrogen bromide is −36 kJ/mol. On the basis of this evidence, the reac on is exothermic.
Therefore, the bond energy released from forming the product from the activated complex must be
greater than the bond energy absorbed as the reactants form the activated complex.
6. a.
4
5
6
.
7
k
J
b.
7. a. Cancelling entities common to both sides of the reaction equation gives the following net
equation.
Cl2(g) + H2(g) + light energy → 2 HCl(g)
b. Intermediate products are Cl(g) and H(g).
c. The dramatic increase in the speed of the reaction once light has produced atomic chlorine is
evidence that the activation energy for a successful collision between atomic chlorine and molecular
hydrogen is very much lower than for a collision between molecular chlorine and molecular
hydrogen.
Practice Questions: Catalysis and Reaction Rates
1. a. The reactants, products, and enthalpy change of reaction are the same for both catalyzed and
uncatalyzed reactions. The initial and final levels on a chemical potential energy diagram would
be
the same for both reactions.
b. The rate of the reaction, the activation energy, the activated complex(es), and the reaction
17
2.
3.
4.
5.
pathway are different in catalyzed compared to uncatalyzed reactions.
a. C and D because they are produced during one step and consumed during another; they are not
included in the net reaction.
b. A, B, and E because they are on the left side of the net reaction.
c. F is the only product; it is written on the right side of the net reaction.
d. 2 A + B + E → F
a. +60 kJ
b. +95 kJ
c. −35 kJ
d. +35 kJ
e. Forward
The reaction would not occur.
The reaction would not occur.
Unit A Review
1. C
2. 585
3. D
4. 98.9
5. C
6. D
7. 136
8. 245
9. 3421
10. 3214
11. 1243
12. C
13. A
14. D
15. 2678
16. D
17. The sun’s energy was initially trapped by photosynthesis in green plants to produce organic
compounds. The plant matter (either eaten by animals or left in the plant) eventually was converted
by high temperatures and pressures into hydrocarbons.
18. The enthalpy change, ΔrH, refers to the total energy change for a chemical system or substance
undergoing change, usually described by a chemical reaction equation. The molar enthalpy, Δ rHm,
change refers to the energy change per mole of the substance undergoing the change indicated;
e.g., combustion and formation.
19. a.
18
b.
20. a.
b.
c.
21.
22. a.
b.
c.
23.
19
24.
25. a.
b.
26.
20
27.
28. a. A successful collision between methane and chlorine requires that the molecules have sufficient
kinetic energy and the correct orientation to form an activated complex.
b. A catalyst would increase the rate of this reaction.
c. A catalyst increases reaction rates by providing an alternative reaction pathway. It does this by
allowing a different activated complex to be formed, which has a lower activation energy than
one without a catalyst. The catalyst is regenerated later during the reaction.
d. The bond energy is the energy required to break the reactant bonds to create an activated
complex. A catalyzed reaction forms an activated complex that requires less bond energy and
hence less kinetic energy of the reactant molecules. Since more reactant entities possess this
lower quantity of energy (at a given temperature), the catalyzed reaction can proceed more
quickly. Without a catalyst, more energy would be required to break reactant bonds to form the
activated complex, so fewer entities would possess this energy and the reaction would proceed
more slowly.
21
e. Bond energies significantly affect the net energy change of the reaction. The net energy
(enthalpy) change is the difference between the energy released by bond making and the
energy required for bond breaking. For example, if the reactant bonds are numerous and strong,
then the enthalpy change will be less exothermic or more endothermic.
29. a.
b.
30. a.
b.
31. Static electricity is a form of energy. During the refuelling of an automobile, static electricity may be
sufficient to provide the activation energy required to start the reaction of gasoline vapours with
oxygen in the air. Once the spark provides enough energy to ignite a few molecules of the fuel, the
energy thus produced can continue to provide the activation energy for the reaction of other,
nearby molecules, resulting in an explosion.
32. a. +50 kJ
b. +90 kJ
c. +20 kJ
d. +60 kJ
e. −40 kJ
f. +40 kJ
g. −40 kJ
h. +40 kJ
22
Unit B: Electrochemical Changes
Lesson: Oxidation and Reduction
1. Oxidation: Zn(s) → Zn2+(aq) + 2 e−
Reduction: Pb2+(aq) + 2 e− → Pb(s)
2. Oxidation: Al(s) → Al3+(aq) + 3 e−
Reduction: Fe2+(aq) + 2e− → Fe(s)
3. HNO2(aq) + H+(aq) + e− → NO(g) + H2O(l)
4. 8 H2O(l) + Cl2(aq) → 2 ClO4− + 16 H+(aq) + 14 e−
Oxidation
Practice Questions: Oxidation and Reduction
1. a.
b.
c.
d.
e.
2. a.
b.
c.
d.
e.
f.
3. a.
b.
4. a.
b.
c.
d.
e.
The gain of electrons by an entity.
The loss of electrons by an entity.
Causes oxidation by accepting (gaining) electrons from another substance in a redox reaction.
Causes reduction by donating (losing) electrons to another substance in a redox reaction.
The spontaneous redox reaction of metals, usually with oxygen, to form metallic compounds.
Oxidation of iron; oxygen is the oxidizing agent and iron is the reducing agent.
Reduction of lead(II) oxide; carbon is the reducing agent and lead(II) oxide is the oxidizing agent.
Reduction of nickel(II) oxide; hydrogen is the reducing agent and nickel(II) oxide is the oxidizing
agent.
Oxidation of tin; bromine is the oxidizing agent and tin is the reducing agent.
Reduction of iron(III) oxide; carbon monoxide is the reducing agent and iron(III) oxide is the
oxidizing agent.
Oxidation of copper; nitric acid is the oxidizing agent and copper is the reducing agent.
Zn(s) → Zn2+(aq) + 2 e−
Cu2+(aq) + 2 e− → Cu(s)
Mg(s) → Mg2+(aq) + 2 e−
2 H+(aq) + 2 e− → H2(g)
Ni(s) → Ni2+(aq) + 2 e− (oxidation)
Cu2+(aq) + 2 e− → Cu(s) (reduc on)
Pb(s) → Pb2+(aq) + 2 e− (oxidation)
Cu2+(aq) + 2e− → Cu(s) (reduc on)
2 H+(aq) + 2 e− → H2(g) (reduction)
Ca(s) → Ca2+(aq) + 2 e− (oxidation)
Fe3+(aq) + 3 e− → Fe(s) (reduc on)
Al(s) → Al3+(aq) + 3 e− (oxidation)
Cl2(aq) + 2 e− → 2 Cl−(aq) (reduction)
2 I−(aq) → I2(s) + 2 e− (oxidation)
23
5. The presence of the same ions in the reactants and products indicates that no electrons have been
transferred. Therefore, a redox reaction has not taken place.
6. a. Cl2(aq) + 2 e− → 2 Cl−(aq) (reduction)
2 Br−(aq) → Br2(l) + 2 e− (oxidation)
b. Ag+(aq) + e− → Ag(s) (reduc on)
Al(s) → Al3+(aq) + 3 e− (oxidation)
7. a. N2O(g) + 2 H+(aq) + 2 e− → N2(g) + H2O(l)
Reduction
b. NO3−(aq) + 3 H+(aq) + 2 e− → HNO2(aq) + H2O(l)
Reduction
c. NH3(aq) + 2 H2O(l) → NO2−(aq) + 7 H+(aq) + 6 e−
Oxidation
d. H2O2(aq) + 2 H+(aq) + 2e− → 2 H2O(l)
Reduction
8. a. 2 H2O(l) → O2(g) + 4 e− + 4 H+(aq)
Oxidation
b. 6 CO2(g) + 24 H+(aq) + 24 e− → C6H12O6(s) + 6 H2O(l)
Reduction
Lesson: Redox Tables
1.
Practice Questions: Redox Tables
1. A substance that is a very strong oxidizing agent has a very strong attraction for electrons.
2. A substance that is a very strong reducing agent has a weak attraction for its electrons, which are
easily removed.
3.
4.
24
5.
6.
7. According to atomic theory, nonmetal atoms have almost-filled valence energy levels and higher
electronegativity values (stronger affinity for bonding electrons) and tend to attract electrons to
attain stable, filled energy levels of the nearest noble gas. Metal atoms, however, have few
electrons in their valence energy levels, have low electronegativity values (weaker affinity for
bonding electrons), and tend to lose electrons to attain filled energy levels of their nearest noble
gas. This is consistent with the empirically determined table, which shows that nonmetals tend to
act as oxidizing agents (electron acceptors) and that metals tend to act as reducing agents (electron
donors).
Lesson: Predicting Redox Reactions
1. Yes, a redox reaction will occur.
spont.
MnO4−(aq) + 8 H+(aq) + 5 Fe2+(aq) → Mn2+(aq) + 4 H2O(l) + 5 Fe3+(aq)
pH test before and after the reaction should change from acidic to neutral. Purple colour of
permanganate ions should disappear after the solutions are mixed.
2. Yes, because the reaction is nonspontaneous.
nonspont.
Cu(s) + 2 H+(aq) → Cu2+(aq) + H2(g)
pH should show that the solution is acidic after the reactants are mixed. No evidence of hydrogen
gas production, copper(II) ions will appear blue.
nonspont.
2+
3. 3 Fe (aq) → 2 Fe3+(aq) + Fe(s)
Yes, it will be stable because the reaction was not spontaneous.
Practice Questions: Predicting Redox Reactions
1. a.
b.
c.
d.
e.
f.
Spontaneous
Nonspontaneous
Nonspontaneous
Spontaneous
Spontaneous
Spontaneous
25
2. a.
b.
c.
3. a.
b.
c. Both predictions cannot be correct; either iron(III) ions are formed or iron(II). The redox table
and rules are more likely to be correct because they are based on extensive observations of
relative strengths of oxidizing and reducing agents supported by the idea of electron transfer.
The single displacement rule is a generalization that has been useful in the past, but does not
have the same empirical basis or any theoretical justification.
d. Qualitatively, one could ensure complete reaction of the blue copper(II) solution and then
observe the colour of the solution. If the iron(III) ion is produced, the solution will be orangeyellow, but if the iron(II) ion is produced, the solution should be lime green. Quantitatively, one
could measure the mass of iron reacted and the mass of copper metal produced. Convert the
masses to chemical amounts and calculate the ratio of the chemical amount of copper produced
to the chemical amount of iron reacted. If the ratio is 3:2, then iron(III) is produced, but if the
ratio is 1:1, then iron(II) is produced.
4.
26
5. a.
b.
6.
7. a.
b.
8.
9.
10.
27
11.
Lesson: Balancing Redox Reactions with Half-Reactions
1.
2.
3.
4.
5.
2 K(s) + Cl2(g) → 2 KCl(s)
Cu(s) + 2 AgNO3(aq) → 2 Ag(s) + Cu(NO3)2(aq)
3 C2H5OH(aq) + 2 Cr2O72-(aq) + 16 H+(aq) → 3 CH3COOH(aq) + 4 Cr3+(aq) + 11 H2O(l)
4 H3PO3(aq) → 3 H3PO4(aq) + PH3(g)
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)
Practice Questions: Balancing Redox Reactions with Half-Reactions
1. a. Cu2+(aq) + 2 e− → Cu(s)
Co(s) → Co2+(aq) + 2e−(aq)
Cu2+(aq) + Co(s) → Co2+(aq) + Cu(s)
b. Zn2+(aq) + 2e− → Zn(s)
Cd(s) → Cd2+(aq) + 2e−
Zn2+(aq) + Cd(s) → Cs2+(aq) + Zn(s)
c. Br2(l) + 2 e−(aq) → 2 Br−(aq)
2 I−(aq) → I2(s) + 2 e−
Br2(l) + 2 I−(aq) → 2 Br−(aq) + I2(s)
2. a.
b.
28
3.
4.
Lesson: Oxidation Numbers
1. −4
2. +7
3. +6
4.
5.
6.
Practice Questions: Oxidation Numbers
1. An oxidation number is a positive or negative number corresponding to the oxidation state assigned
to an atom in an entity.
2. a. +4
b. +7
c. +6
d. +6
e. −1
f. −1
3. a. +1
b. +2
c. +4
d. −3
e. −2
29
f. +5
g. 0
h. −3
4. a. 0
b. 0
c. +4
d. +2
5.
6.
7. a. carbon: +2; oxygen: −2
b. oxygen: 0
c. nitrogen: −3; hydrogen: +1; chlorine: −1
d. hydrogen: +1; phosphorous: +5; oxygen: −2
e. sodium: +1; sulfur: +2; oxygen: −2
f. sodium: +1; phosphorous: +5; oxygen: −2
8. a. The oxidation number is calculated as being +8/3.
b. The fractional value of the answer is unusual because it would involve a fractional number of
electrons, which is not possible. The formula, Fe3O4, might represent a compound involving a
combination of iron(II) oxide and iron(III) oxide, which could be written FeO•Fe 2O3.
9. a.
b.
c.
10.
11. a.
b.
c.
d.
e.
30
f.
g.
h.
12. Double replacement reactions do not appear to be redox reactions.
Lesson: Balancing Redox Reactions Using Oxidation Numbers
1.
2.
3.
4.
2 H2S(g) + 3 O2(g) → 2 SO2(g) + 2 H2O(g)
3 H2O(l) + 5 ClO3−(aq) + 3 I2(s) → 5 Cl−(aq) + 6 IO3−(aq) + 6 H+(aq)
3 ClO−(aq) → 2 Cl−(aq) + ClO3−(aq)
16 H+(aq) + 3 C2H5OH(aq) + 2 Cr2O72−(aq) → 4 Cr3+(aq) + 3 CH3COOH(aq) + 11 H2O(l)
Practice Questions: Balancing Redox Reactions Using Oxidation Numbers
1. The oxygen atoms in hydrogen peroxide, H2O(l), have an oxidation number of –1, and can be either
oxidized to O2(g), (oxidation number of 0), or reduced to an oxidation state of –2 (as in H 2O(l) or
metallic oxides).
2. a.
b. Carbon is oxidized. Its oxidation number changes from +3 to +4.
c. Manganese is reduced. Its oxidation number changes from +7 to +2.
d. The oxidizing agent is MnO4– and the reducing agent is C2O42–.
3. a. CO2(g) + H2O(l) → H2CO3(aq)
b. This is not a redox reaction because the oxidation numbers of all atoms remain unchanged.
4. a.
b.
c.
31
5.
6. a.
b.
c.
7. a.
b.
c. The method presented to balance redox equations involves only one entity that is oxidized and
one entity that is reduced. However, some redox reactions may involve two entities gaining or
losing electrons.
Lesson: Redox Stoichiometry
1.
2. 5.25 mmol/L
32
Practice Questions: Redox Stoichiometry
1.
2.
33
3.
4.
34
5. a.
*Note: This answer was obtained using the average volume for all 4 trials. If you used the
average volume from the 3 most consistent trials, the answer is 0.0747 mol/L.*
b.
*Using the value of 0.0747 mol/L from the previous question, the percent difference is 6.67%.*
6.
35
7.
Lesson: Voltaic Cells
1.
Practice Questions: Voltaic Cells
1. a. A voltaic cell is an arrangement of two half-cells separated by a porous boundary that
spontaneously produces electricity.
b. A half-cell is an electrode–electrolyte combination that forms one-half of a complete cell.
c. A porous boundary is a barrier that separates two electrolytes while still permitting the
movement of ions.
d. A salt bridge is a U-shaped tube containing an inert electrolyte.
36
2.
3.
4.
5.
e. An electrolyte is a solute that forms a solution that conducts an electric current.
f. An external circuit is an electrical connection that completes the circuit outside of the cell,
allowing electrons to flow between the half-cells.
The cathode is the electrode where reduction occurs and the anode is the electrode where
oxidation occurs.
a. Cathode
b. Anode
c. Anode
d. Cathode
An inert electrode is used in a half-cell where no conducting solid is involved in the half-reaction
equation. A carbon, or graphite rod and platinum metal are two commonly used inert electrodes.
The solution in a salt bridge must be an inert (unreactive) electrolyte; that is, it must contain ions
that do not react during the operation of the cell. An example of an inert aqueous electrolyte is
sodium sulfate.
6. a.
b.
7. a. Cations move toward the cathode and anions move toward the anode.
b. Ions move to maintain electrical neutrality at each electrode. In a simple voltaic cell composed
of metals and metal ions, reduction removes positive ions (cations) from the solution around the
cathode, and oxidation adds positive ions (cations) to the solution around the anode. Anions
move toward the anode to balance the excess positive charge in the solution around the anode,
37
while cations move toward the cathode to replace the positive charge being removed from the
solution around the cathode.
c. Not only do the ions in the U-tube move, but the ions in the electrolyte solutions move as well.
Since the copper electrode is the anode, it loses electrons and is oxidized resulting in the
production of blue copper(II) ions. The blue colour is shown migrating through the salt bridge
and into the cathode compartment.
8.
[Alternatively, one half-cell could be placed
inside a porcelain cup set inside the beaker
containing the other half-cell.]
Lesson: Standard Cells and Cell Potentials
1.
2.
Practice Questions: Standard Cells and Cell Potentials
1. a. In a standard cell, each half-cell has all the entities shown in the half-reaction equation at
standard ambient temperature and pressure (SATP) conditions. Aqueous solutions have a
concentration of 1.0 mol/L.
b. The standard cell potential, E°cell, is the maximum electric potential difference (voltage) of the
cell at standard conditions. It represents the difference in ability of the two half-cells to gain
electrons. E°cell = E°r cathode − E°r anode.
2. a.
38
b.
3. a.
b.
c.
4.
5. Nonstandard conditions for concentrations, temperature, and pressure, the purity of the
substances, the presence of oxide coatings on metals, and the type and size of the porous boundary.
6. Changing the reference half-cell does not change the potential difference between any two halfreactions. Changing the lithium half-cell to 0.00 V means adding 3.04 V to each reduction potential
on the chart.
7. If a zinc-iron cell is run until the potential difference is zero, the concentration of reactants has
decreased and the concentration of products has increased to reach an equilibrium—the rate of the
forward reaction equals the rate of the reverse reaction.
8. A positive cell potential indicates a spontaneous reaction, while a negative cell potential indicates a
nonspontaneous reaction.
9. The reactions in standard voltaic cells are always spontaneous because voltaic cells are designed
with two different half-cells each containing the oxidized and reduced entities as shown in the halfreaction equation. Therefore, a spontaneous combination will always exist; the cell potential will
always be positive.
10. The standard hydrogen half-cell is an inert platinum electrode immersed in a 1.00 mol/L solution of
hydrogen ions, with hydrogen gas at 100 kPa bubbling over the electrode and a cell temperature of
25°C.
Pt(s) | H2(g) | H+(aq)
E°r = 0.00 V
39
11. E°cell = −0.28 V – (−0.76 V) E°cell = +0.48 V
The potential of a standard cobalt–zinc cell is +0.48 V. The theoretical interpretation of this cell
potential is that the cobalt(II) ions have a stronger attraction for electrons than do zinc ions. The
+0.48 V is a measure of the difference in their abilities to attract electrons.
12. a.
b.
13.
Practice Questions: Corrosion
1. For the corrosion of iron to occur, an oxidizing agent (most commonly oxygen and water) must be
present and in contact with the iron.
2. The presence of acidic solutions, electrolytes, and contact with less active metals accelerate the
corrosion of iron.
3.
40
4. a.
b.
c.
5. Since hydroxide ions (either alone or in combination with other ions) often act as reducing agents, a
basic solution might prevent or slow down the corrosion of iron.
6. A zinc coating on iron is better than a tin coating because zinc is more easily oxidized than iron and,
thus, the zinc will corrode first. Tin is less active than iron and will promote the oxidation of iron.
7. Two methods of cathodic protection are impressed current and sacrificial anode. These are similar in
that both methods force the iron to become the cathode by supplying it with electrons.
Lesson: Electrolytic Cells
1.
2. a.
b.
41
c.
Practice Questions: Electrolytic Cells
1. a.
b.
2. a.
b.
3. They consist of two electrodes and at least one electrolyte; the strongest oxidizing agent undergoes
a reduction at the cathode; the strongest reducing agent undergoes an oxidation at the anode;
electrons flow from the anode to the cathode through the external wire; and anions flow toward the
anode and cations flow toward the cathode.
4. The key difference between voltaic and electrolytic cells is that voltaic cells involve a spontaneous
reaction, while electrolytic cells involve a nonspontaneous reaction. (This is indicated by a positive
cell potential for a voltaic cell and a negative cell potential for an electrolytic cell.)
5. A power supply is required for an electrolytic cell because the reaction is nonspontaneous, which
means that the oxidizing agent cannot attract or remove electrons from the reducing agent. A
potential difference must be supplied to move electrons from one to the other.
42
6. a.
b.
7. a.
b.
8.
9. The chloride anomaly is an exception to the rule that the strongest reducing agent, (as identified
from a redox table), undergoes oxidation at the anode. This anomaly occurs during the electrolysis
of solutions that contain the chloride ion, where water is the strongest reducing agent present
43
according to the redox table, but chloride ions react preferentially. It can be recognized because
chlorine gas is produced instead of oxygen gas in situations where chloride and water are the only
reducing agents present.
10. a.
b.
c.
11. a.
b.
12.
13.
44
Lesson: Cell Stoichiometry
1.
2.
3.
4.
5.
45
6.
Practice Questions: Cell Stoichiometry
1.
2.
3.
4.
46
5.
6.
7.
47
8. a.
b.
*Significant digits = 1.6 x 103 kg*
*Significant digits = 4.8 x 103 kg*
9.
48
10.
11.
12.
49
13.
14.
Unit B Review
1.
2.
3.
4.
5.
6.
B
A
3142
C
A
B
50
7. C
8. 2134
9. D
10. 0.85
11. 87.4
12. A
13. D
14. C
15. B
16. C
17. D
18. 0.65
19.
20. a. Nonspontaneous
b. Spontaneous
c. Nonspontaneous
21. a.
b. In this table the strongest oxidizing agent is Co2+(aq) and the strongest reducing agent is Ti(s).
22. a.
b.
c.
d.
23.
51
24.
25. a.
b.
26.
52
27.
28. a. Electrolysis is the forcing of a possible but nonspontaneous oxidation-reduction reaction by the
application of a potential difference from an outside power source.
b. Oxidation (electron loss) occurs at the anode; reduction (electron gain) occurs at the cathode.
c. In electrolysis of molten compounds, water is not present, and there are fewer oxidizing agents
and reducing agents present. The temperature is usually very high, making container and
electrode selection more difficult, because they must be able to withstand the temperature
conditions. In all other aspects, the process is similar to electrolysis with aqueous electrolytes.
d. One important industrial application of electrolysis is the chloralkali process, which is the
production of chlorine gas and sodium hydroxide from the electrolysis of a sodium chloride
solution. Another important application is the HallHéroult process, which is the production of
aluminium metal from bauxite ore in a cryolite solvent. A third important industrial application
of electrolysis is the electroplating of metals such as chromium and nickel for everything from
oilfield equipment to jewellery.
29. a. nonspontaneous reaction (external voltage required)
b. spontaneous reaction (no external voltage required)
c. spontaneous reaction (no external voltage required)
d. nonspontaneous reaction (external voltage required)
30. a.
b.
c.
53
31. a.
b.
c.
32. a.
b.
54
33. a.
b.
34.
35.
55
Unit C: Chemical Changes of Organic Compounds
Lesson: Alkanes
1.
2.
3.
4.
2-methylbutane
3-ethyl-3,4-dimethylhexane
2,3-dimethylpentane
2,2-dimethylpropane (numbers not necessary for this one because if the methyl functional group
was on carbon 1 or 3, the root would be but).
5. 2,2,4,4-tetramethylhexane
6. heptane
7. 2,3,4-trimethylpentane
8. 2,2,4-trimethyl-4-propylheptane
9. 4-ethyl-2,3-dimethylhexane
10. 3-ethyl-2,5-dimethylheptane
11.
12.
13.
14.
15.
16.
56
17.
18.
19. a.
b.
20.
21. 1,2-dimethylcyclopentane
Practice Questions: Alkanes
1. a.
b.
c.
d.
e.
f.
g.
h.
inorganic
organic
inorganic
organic
organic
inorganic
inorganic
organic
2.
57
3. a.
b.
c.
d.
e.
f.
58
g.
h.
4. a.
b. These molecules are highly soluble in each other because they are all nonpolar and only have
London dispersion forces acting between molecules.
5. a. 2-methylheptane
b. 3,7-dimethylnonane
c. 2,4,6-trimethylheptane
d. 3-ethyl-6-methyloctane (preferred) or 6-ethyl-3-methyloctane
6. a.
59
b.
c.
d.
7. a.
b.
c.
d.
e.
Corrected name: 2,2-dimethylhexane. There should be a “2” for each methyl group.
Corrected name: 3,3-diethylpentane. The prefix “di” is required for two groups.
Correct as written.
Corrected name: 2,2-dimethylbutane. The longest continuous chain of carbon atoms is four.
Corrected name: 3-ethylhexane. An alkyl branch can never be on an end carbon. The longest
carbon chain is six.
8. a.
b.
c.
60
d.
e.
9. a.
b.
c.
d.
3-ethylpentane
2,2,3-trimethylbutane
4-ethyl-2,4-dimethylhexane
3-methyl-5-propyloctane
10.
11. a.
b.
12. a.
b.
c.
13. a. The boiling points of compounds in the alkane family increase as the number of carbon atoms
increases.
b. As the alkanes become larger, the strength of London forces between the molecules increases
because the number of electrons per molecule increases.
61
Lesson: Alkenes and Alkynes
1. 3-methylbut-1-ene
2. 5-methylhex-2-ene
3.
4.
Practice Questions: Alkenes and Alkynes
1. a.
b.
c.
d.
alkene (2n)
alkane (2n+2)
alkyne or cycloalkene (2n−2)
alkene or cycloalkane (2n)
2. a.
C3H8
b.
C3H6
c.
C3H4
d.
C3H6
3. a.
b.
c.
d.
4. Both propene and propyne have only one possible location for their double and triple bond,
respectively, so no number is needed to indicate the location.
62
5. a.
b.
c.
6.
7. a. alkanes (CnH2n+2); alkenes (CnH2n); alkynes (CnH2n−2)
b. Alkanes have only single carbon-carbon bonds, alkenes have at least one carbon-carbon double
bond, and alkynes have at least one carbon-carbon triple bond.
c. Every increase in the number of electrons shared between carbon atoms means two fewer
electrons available for sharing with two hydrogen atoms. Changing a single to a double or a
double to a triple means removing 2 hydrogen atoms at a time.
8. Unsaturated means it has at least one multiple bond and each carbon atom is bonded to less than
the maximum number of hydrogen atoms.
9. a.
b.
10. a. pentane
63
b.
c.
d.
e.
11. a.
b.
c.
d.
e.
f.
pent-2-ene
pent-1-yne
but-1-ene
4-methylpent-2-ene
but-2-ene
methylpropene
but-1-ene
pent-2-yne
1-ethyl-2-methylcyclopentane
4-methylcyclohexene
12.
13.
14. a.
b.
Lesson: Aromatics
1. 1-ethyl-2,4-dimethylbenzene
2. 3-phenyl-4-propyloctane
3.
Practice Questions: Aromatics
1. The simplest aromatic compound is benzene, C6H6. An unusual structural feature of benzene is the
circle in the centre of the 6-sided ring. This represents six “extra” electrons shared by all carbon
atoms in the ring. This is the best explanation that chemists have found for the chemical formula
and chemical reactions of benzene.
64
2.
3. a.
b.
c.
d.
4. a.
b.
65
c.
5. a.
b.
c.
d.
e.
f.
1-ethyl-3-methylbenzene
1-methyl-4-propylbenzene
3-methyl-2-phenylpentane
3-methyl-4-phenylhexane
2-phenylhept-3-ene
4-phenylpent-1-yne
6. a.
b.
c.
7. a.
b.
c.
d.
e.
f.
8. a.
b.
c.
d.
e.
f.
ethylbenzene → 1,2-dimethylbenzene
benzene + ethene → ethylbenzene
aliphatic
aromatic
aliphatic
aliphatic
aromatic
aromatic
unsaturated
saturated
not possible
unsaturated
saturated
not possible
Practice Questions: Crude Oil Refining
1. Fractionation is based on differences in boiling points.
2. Differences in the strength of the intermolecular bonds account for differences in boiling points.
3. Hydrogen bonding between water molecules and lack of hydrogen bonding with hydrocarbon
molecules prevent water from becoming a significant solute in a crude oil solution. Also, water
molecules are polar, whereas hydrocarbon molecules are nonpolar.
4. Chemical processes in oil refining are necessary to ensure that the resulting products have the
properties desired by the end market as well as to increase the yield of fractions that are in greatest
demand.
5. Hydrocracking is used to break down larger molecules into smaller molecules. Catalytic reforming
then converts some of the aliphatic molecules in this mixture into aromatic molecules. The use of
66
both processes helps to increase the yield of the gasoline fraction and improves its burning
characteristics.
6. Molecules in the gasoline fraction can be produced by either direct fractionation of crude oil or by
cracking larger hydrocarbon fractions. Increased branching is achieved during alkylation, while
catalytic reforming is used to produce more aromatic compounds in the gasoline fraction.
7. a.
b.
c.
8. a.
b.
c.
d.
Lesson: Organic Halides
1.
2. 2-bromo-4-chloroheptane
3.
Practice Questions: Organic Halides
1. a.
b.
c.
67
d.
2. a.
b.
c.
d.
e.
f.
3. a.
triiodomethane
3-chloro-2-methylpropene
dichloromethane
1,2,3-tribromopropane
chlorobenzene
phenylethene
Chloroethene will have a higher boiling point than ethene due to stronger London forces
(greater number of electrons) and dipole–dipole forces (polar). Chloroethene has 32 electrons
and is polar, due to the chlorine replacing a hydrogen atom, while ethene has only 16 electrons
and is nonpolar.
b. Chloroethene will be more soluble in water than ethene will. Chloroethene is polar because of
the chlorine, which has replaced a hydrogen atom and would be slightly soluble in the polar
water. Ethene is nonpolar and would essentially be insoluble in water.
4. a.
tetrachloromethane
b.
1,1,1-trichloroethane
c.
trichlorofluoromethane
d.
5. a.
b.
1,2,2-trichloro-1,1,2-trifluoroethane
chlorodifluoromethane
1,1-dichloro-1-fluoroethane
68
c.
trifluoromethane
d.
1,1,1,2-tetrafluoroethane
6. a.
b.
c.
d.
7. In this case, the number of electrons per molecule increases as the halogen in these compounds
increases in size. Since iodoethane contains the highest number of electrons, it has the strongest
London forces and hence the highest boiling point.
Lesson: Alcohols
1. 3-methylbutan-2-ol
2.
3. 2,3-dimethylbutan-2-ol
4. a.
b.
69
c.
d.
5.
6.
Practice Questions: Alcohols
1. To have greater volatility (ability to vaporize), the intermolecular forces between the molecules
must be lower. In other words, an order of increasing volatility is predicted to be the order of
decreasing intermolecular forces if molecular sizes are approximately the same. The order of
increasing volatility is therefore: alcohols, organic halides, and hydrocarbons.
2. a. butane
b. methylpropane
c. but-2-ene
d. methylpropene
e. butan-1-ol
f. butan-2-ol
g. methylpropan-1-ol
h. methylpropan-2-ol
3. a.
b.
c.
d.
4. a. butan-2-ol
b. pentane-1,3-diol
70
c. pentan-2-ol
d. pentane-2,3-diol
e. cyclohexane-1,3-diol
5. a.
b.
c.
d.
e.
f.
6.
7. The presence of a hydroxyl group in methanol makes the molecule and allows hydrogen bonding
between molecules. This increase in intermolecular forces results in a higher boiling point.
8. a. In order of their increasing boiling point: ethane, fluoromethane, methanol. Since the number of
electrons in these substances is the same, the strength of London dispersion forces are the
same. Ethane has only London forces among its molecules. Both fluoromethane and methanol
consist of polar molecules with dipole–dipole forces. Methanol has the highest boiling point
because it also has hydrogen bonding.
b. In order of increasing boiling point: pentane, 1-chlorobutane, butan-1-ol. All molecules have a
similar number of electrons. Pentane has the lowest boiling point because its molecules are
nonpolar and, as a result, have only London forces between them. The higher boiling point of 1chlorobutane is a result of dipole–dipole forces between its molecules, in addition to its London
forces. Butan-1-ol has the highest boiling point because its molecules attract each other with all
three types of intermolecular forces, in particular hydrogen bonding.
9. a. As the strength of intermolecular forces increases, the boiling point also increases, due to the
increase in attraction between the molecules. To vaporize, the molecules with higher forces
need more energy. Since ethane only has London forces and also has the fewest number of
electrons (18), it has the weakest intermolecular forces. Chloroethane has more electrons (34)
71
than ethane, and is polar due to the chlorine, so will rank next in strength of intermolecular
forces. Ethanol, due to its hydrogen bonding, will have the strongest intermolecular force even
though it has slightly fewer electrons (26) than chloroethane.
b. The solubility of molecules in water is dependent on the strength of attraction between the
molecule and the water. Ethanol is not only polar but is capable of hydrogen bonding with water
molecules. Ethanol should have the highest solubility in water. Chloroethane is the next most
soluble because it is polar due to the chlorine atom. Polar molecules tend to be at least partially
soluble in polar water. Ethane would be the least soluble in water because it is nonpolar.
Lesson: Carboxylic Acids, Esters, and Esterification Reactions
1.
2. butanoic acid
3. ethyl 2,3-dimethylpropanoate
4.
→
5. The ester is methyl ethanoate, and it can be prepared from methanol and ethanoic acid.
6.
Practice Questions: Carboxylic Acids, Esters, and Esterification Reactions
1. a.
b.
c.
2. a. methanoic acid
b. pentanoic acid
c. hexanoic acid
72
3. a.
b.
c.
d.
4. a. ethyl propanoate (from propanoic acid and ethanol)
b. methyl butanoate (from butanoic acid and methanol)
c. butyl methanoate (from methanoic acid and butan-1-ol)
d. propyl ethanoate (from ethanoic acid and propan-1-ol)
5. An ester contains an —OR group in place of the —OH in the carboxylic acid. The OH group is
responsible for the acidic properties of carboxylic acids, and also for hydrogen bonding; thus esters
have lower melting and boiling points, are less soluble in water, and are not acidic.
6. a.
b.
c.
Lesson: Organic Reactions
1.
73
2.
3.
4.
5.
6. 1) elimination
2) addition
3) substitution
4) esterification
Practice Questions: Organic Reactions
1. a.
b.
2.
74
a.
b.
c.
3. a. During the incomplete combustion of ethyne, the lack of oxygen prevents all the carbon in the
ethyne molecule from being converted completely into carbon dioxide. Consequently, some
soot (carbon) is observed. A chemical equation that describes this process is:
2 C2H2(g) + O2(g) → 4 C(s) + 2 H2O(g)
b. Burning the briquettes inside the tent results in a build-up of carbon monoxide gas. Once the
carbon monoxide gas concentration reaches a certain critical level, the campers inside the tent
succumb to carbon monoxide poisoning.
2 C(s) + O2(g) → 2 CO(g)
c. Incomplete combustion of gasoline produces carbon monoxide, which enters the back of a car
through the leaky muffler. The chemical equation for the incomplete combustion of octane is
given below to illustrate this process.
2 C8H18(l) + 17 O2(g) → 16 CO(g) + 18 H2O(g)
4. a.
b.
c.
d.
e.
f.
You are only required to predict one of the products.
g.
75
You are only required to predict one of the products.
5. a.
*coefficient of 2 in front of chlorine and hydrogen chloride*
b.
c.
d.
e.
f.
g.
h.
6. a.
b.
76
7. a.
(propan-1-ol) + Cl– (chloride)
b.
8. a.
b.
9.
10. a.
b.
c.
11.
12.
13.
14. a.
b.
c.
77
d.
15.
16. a.
b.
c.
d.
e.
f.
78
g.
Lesson: Polymers and Polymerization Reactions
1.
Practice Questions: Polymers and Polymerization Reactions
1.
2.
3. a. but-2-ene
b. 1-chloro-1,2-difluoropropene
4. The monomer must have a double or triple bond to form an addition polymer.
5.
6.
7. An example of a natural condensation polymer is a protein. An example of a synthetic condensation
polymer is nylon.
79
8.
9.
Unit C Review
1. C
2. B
3. A
4. B
5. D
6. C
7. A
8. D
9. B
10. 4, 9
11. 1, 8
12. 5, 7
13. 2, 3, 6
14. 2, 6
15. 2
16. 5, 7
17. 4
18.
19. a. octane
b. 2-methylheptane
c. 2-methylpentane
20. a.
b.
80
c.
21. a. benzene
b. methylbenzene
c. 1,2-dimethylbenzene
d. 1,3-dimethylbenzene
e. ethylbenzene
f. 1,4-dimethylbenzene
22. Alkylation (isomerization) is the process where a molecule transforms from one structural formula,
or shape, into another structural formula, or shape, without changing the number or type of atoms
in the molecule.
23. a. butane → methylpropane
b. 3-methylhexane → 2,2,3-trimethylbutane
c.
d.
24. Catalytic reforming
25. a.
b. From lowest to highest boiling points would be: butane, chloroethene, 1-propanol, then
ethanoic acid. Since they have approximately the same number of electrons, London forces will
be approximately the same strength for each. Since butane only has London forces, it has the
weakest forces and therefore the lowest boiling point. Chloroethene has dipole–dipole
intermolecular forces due to the chlorine, but no hydrogen bonding as in 1- propanol and
ethanoic acid, so chloroethene has the second-lowest boiling point. Finally, ethanoic acid has
the highest boiling point because it has an added carbonyl group to increase its dipole–dipole
bonding as compared to 1-propanol and this also increases the number of hydrogen bonds that
are possible. This makes the intermolecular forces the strongest in ethanoic acid and gives it the
highest boiling point.
c. Ethanoic acid and 1-propanol would have high solubility in water because they are both polar
and can form hydrogen bonds with water molecules. Chloroethene may be slightly soluble in
water as it is polar, like water. Butane would have negligible solubility in water due to the fact
that it is non-polar.
81
26.
27.
28. a.
b.
c.
d.
e.
29. a.
b.
c.
30. a.
b.
82
Unit D: Chemical Equilibrium Focussing on Acid-Base Systems
Lesson: Explaining Equilibrium Systems
1.
Practice Questions: Explaining Equilibrium Systems
1. a. For a chemical system at equilibrium, some directly observable characteristics include colour
and
volume of solid, liquid, or gas phases, size and shape of any solid particles, and odour of any
vapours. All of these will appear to remain constant when the system is at equilibrium.
b. At equilibrium, both the forward and reverse chemical processes are occurring, although
observation indicates that the reaction appears to have stopped.
c. The rates of the forward and reverse changes are assumed to be equal at equilibrium.
2. a.
b.
83
3. a.
b.
c.
d.
4. a.
Initial
Change
Equilibrium
[HI(g)] mol/L
[H2(g)] mol/L
[I2(g)] mol/L
0.100
-0.022
0.078
0
+0.011
0.011
0
+0.011
0.011
b. n = cV
nHI = 0.078 mol/L x 2.00 L = 0.16 mol
nH2 = 0.011 mol/L x 2.00 L = 0.022 mol
nI2 = 0.011 mol/L x 2.00 L = 0.022 mol
5. a.
b.
c. Initially, hydrogen is reacting quickly as a result of numerous favourable collisions with iodine
molecules. As the number of iodine and hydrogen molecules decreases, the frequency of
collisions producing product decreases. At the same time as the number of product hydrogen
iodide molecules increases, the rate at which they collide to break up back into hydrogen and
iodine molecules will steadily increase. An observer will see the rate of reaction of hydrogen and
iodine gradually “slowing down.” Eventually, the rate at which hydrogen molecules react to
form hydrogen iodide becomes equal to the rate at which hydrogen iodide molecules break
84
apart to form hydrogen and iodine molecules. An observer will see the reaction as having
“stopped,” because no macroscopic changes will be observable.
6. a.
b.
c.
d.
e.
>50%
Ca
f.
2+
(aq) + SO42-(aq) ⇆ CaSO4(s)
2%
CH3COOH(aq) + H2O(l) ⇆ H3O+(aq) + CH3COO–(aq)
Lesson: The Equilibrium Constant
1.
2.
3.
4.
5. 3.0 x 1011
6. 2.6 x 10-3 mol
7.
85
Practice Questions: The Equilibrium Constant
1. PCl5(g) ⇆ PCl3(g) + Cl2(g)
Kc = [PCl3(g)][Cl2(g)]
[PCl5(g)]
Kc = (0.014)(0.014)
4.3 x 10−4
Kc = 0.46
2. a.
86
b.
c.
d.
e.
f.
g.
3.
4.
5. a.
b.
c.
d.
e.
f.
87
g.
; products are favoured
6.
7.
8. a.
b.
9. a.
88
b.
Practice Questions: Le Châtelier’s Principle
1. Changes in concentration, temperature, and pressure (in gaseous systems).
2. a.
b.
c. The equilibrium shifts to the right, increasing the concentration of CO2 to replace the lost CO2.
89
d.
3. a.
b.
c.
d.
4. a.
b.
c. The equilibrium shifts to the right in an attempt to replace the removed nitrogen monoxide.
d. The equilibrium shifts to the left in an attempt to reduce the pressure by shifting to the side with fewer
molecules.
5.
6. a.
90
b.
7.
8.
9. a.
b.
c.
d.
10. a.
b.
c.
d.
Lesson: Water Ionization Constant
1.
91
2.
3.
4. pOH = 11.681
Practice Questions: Water Ionization Constant
1.
2.
92
3.
4.
5.
6.
7.
8.
9.
10.
11.
93
Lesson: Brønsted–Lowry Acids and Bases
1.
2.
3.
Practice Questions: Brønsted–Lowry Acids and Bases
1. a.
94
b.
c.
2. a.
b.
c.
d.
3. a.
b.
4.
5. a.
b.
6. a.
b.
c.
d.
7.
8. a.
b.
c.
95
d.
e.
f.
g.
h.
9.
10. Note that these solutions are not all in net ionic equation form. Recall that carbon dioxide is not on
the table, but using the modified Arrhenius theory, it reacts with water to produce carbonic acid
which then reacts with water to produce hydronium ions - the answer key is showing these two
steps combined. Recall that calcium oxide is not on the table, but metallic oxides react with water to
produce ionic hydroxides which then dissociate in water - the answer key is showing these two steps
combined. USe your logic about which entities can accept a proton and which entities can donate a
proton!!
a.
b.
c.
d.
e.
f.
96
g.
11.
12.
Lesson: Acid Strength and the Equilibrium Law
1.
97
2.
3.
4.
98
5.
Practice Questions: Acid Strength and the Equilibrium Law
1. a.
b.
99
c.
d.
e.
f.
2.
3. HBr(aq), HNO3(aq), HF(aq), HNO2(aq), CH3COOH(aq), and H2S(aq). HBr(aq) and HNO3(aq) have
essentially the same strength, due to the leveling effect with the water solvent.
4. a.
b.
100
c.
5. a.
b.
c.
6. a.
b.
101
7. a.
b.
8. a.
b.
9.
102
10.
11.
12.
103
Lesson: Base Strength and the Equilibrium Law
1.
2.
3.
104
4.
105
5.
106
6.
Practice Questions: Base Strength and the Equilibrium Law
1. a.
b.
2.
3.
107
4.
5.
6.
108
7.
8. a.
b.
109
c.
Practice Questions: Interpreting pH Curves
1. a.
b.
2. a.
b.
c.
d.
e.
3.
4. a.
110
b.
5. a.
b.
c.
d.
e.
yellow
red
green
colourless
yellow
6.
7. a.
b.
c.
8. a.
b.
9.
10. a.
b.
111
c.
11. a.
b.
c.
d.
e.
f.
12. a. As this is a strong acid-strong base reaction, the pH endpoint is 7, and bromothymol blue should be used.
b.
c.
As this is a strong acid-weak base reaction, the pH endpoint is less than 7, and methyl orange should be
used.
As this is a weak acid-strong base reaction, the pH endpoint is greater than 7, and phenolphthalein should
be used.
13. a.
b.
c.
d.
14.
Lesson: pH Curve Buffering Regions and Buffers
1.
Buffering region 1 is the first flat portion—buffer is H 3PO4(aq)—H2PO4-(aq)
Buffering region 2 is the second flat portion—buffer is H 2PO4-(aq)—HPO42-
112
Practice Questions: pH Curve Buffering Regions and Buffers
1.
2. a.
b.
3.
4. a.
b.
5. a.
b.
c.
d.
6.
113
7. a.
b.
8. a.
b.
c.
d.
e. A buffering region on the graph occurs where about 5 to 20 mL of HCl(aq) has been added. The entities
present before the equivalence point are Na+(aq), SO32-(aq), HSO3-(aq), Cl-(aq), and H2O(l).
Unit D Review
1.
2.
3.
4.
5.
6.
7.
C
B
A
3611
A
C
D
8.
9.
10.
11. a.
b.
114
12.
13.
14.
115
15.
16.
17.
116
117
18.
19.
20.
118
21.
22.
23.
119
24.
120