Chemistry 12
Provincial Exam Review
Prescribed Learning Outcomes
with Selected Questions
from Past Provincial Exams
CHEMISTRY 12 PROVINCIAL EXAM REVIEW
UNIT I REACTION KINETICS
A:
REACTION KINETICS (Introduction)
A1.
A2.
A3.
A4.
A5.
A6.
A7.
give examples of reactions proceeding at different rates
describe rate in terms of some quantity (produced or consumed) per unit of time
experimentally determine rate of a reaction
identify properties that could be monitored in order to determine a reaction rate
recognize some of the factors that control reaction rates
compare and contrast factors affecting the rates of both homogeneous and heterogeneous reactions
discuss situations in which the rate of reaction must be controlled
1.
A1. DIFF RXN RATES 011
Which of the following reactions would have the greatest reaction rate at room temperature?
A.
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
B.
Ca(s) + 2H2O(l) → Ca(OH)2(aq) + H2(g)
C.
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
D.
Na2CO3(s) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g)
2.
A2. DESCRIBE RATE 015
Which of the following could be used to describe the rate of a reaction
change in time
change in mass
A.
B.
change in concentration
change in concentration
C.
3.
change in concentration
change in time
D.
change in concentration
change in mass
A3. DETERMINE RATE 032
Consider the following reaction:
Fe2O3(s) + 2Al(s) → Al2O3(s) + 2Fe(s)
If 0.50 mol of Fe is produced in 10.0 sec, what is the rate of consumption of Fe2O3 in mol/s ?
A.
5.0 x 10–2 mol/s
B.
2.5 x 10–2 mol/s
C.
1.0 x 10–1 mol/s
D.
5.0 mol/s
— Chemistry 12 Provincial Exam Review — Page 1
Chemistry 12
4.
Unit I Reaction Kinetics
A3. DETERMINE RATE 033
Consider the following reaction;
3Cu(s) + 8HNO3(aq) → 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l)
A piece of copper is added to a nitric acid solution in an open beaker, allowing the NO(g) to escape. The
following data was obtained:
5.
A.
Calculate the reaction rate for the time period 2.0 to 6.0 min. (2 marks)
B.
Calculate the mass of copper consumed in the first 5 minutes. (3 marks)
A4. MONITOR RATE 018
Consider the following reaction:
COCl2(g) → CO(g) + Cl2(g)
Which of the following could be used to determine reaction rate in a closed system?
A.
a decrease in gas pressure
B.
an increase in gas pressure
C.
a decrease in the mass of the system
D.
an increase in the mass of the system
6.
A6. HOMO/HETERO RXNS012
Which of the following does not affect both homogeneous and heterogeneous reaction rates?
A.
addition of a catalyst
B.
change in temperature
C.
change in surface area
D.
change in concentration
7.
A7. CONTROL RATE 001
Situations exist in everyday life in which chemical reaction rates must be decreased. Describe one specific
situation and state how the decrease could be attained; explain the principle involved. (2 marks)
— Chemistry 12 Provincial Exam Review — Page 2
Chemistry 12
B:
Unit I Reaction Kinetics
REACTION KINETICS (Collision Theory)
B1.
B2.
B3.
B4.
B5.
B6.
B7.
B8.
B9.
demonstrate an awareness of the following:
• reactions are the result of collisions between reactant particles
• not all collisions are successful
• sufficient kinetic energy (KE) and favourable geometry are required
• to increase the rate of a reaction one must increase the frequency of successful collisions
• energy changes are involved in reactions as bonds are broken and formed
describe the activated complex in terms of its potential energy (PE), stability and structure
define activation energy
describe the relationship between activation energy and rate of reaction
describe the changes in KE and PE as reactant molecules approach each other
draw and label PE diagrams for both exothermic and endothermic reactions, including ∆H, activation
energy and the energy of the activated complex
relate the sign of ∆H to whether the reaction is exothermic or endothermic
write a chemical equation including the energy term (given a ∆H value)and vice versa
describe the role of the following factors in reaction rate:
• nature of reactants
• concentration
• temperature
• surface area
8.
B1. COLLISION THEORY021
Which of the following would result in a successful collision between reactant particles?
A.
particles have sufficient KE
B.
particles convert all their PE into KE
C.
particles are in an excited state and are catalyzed
D.
particles have sufficient KE and proper molecular orientation
9.
B1. COLLISION THEORY023
Using collision theory, explain why reactions between two solutions occur more rapidly than reactions
between two solids. (2 marks)
10.
B2. ACTIVATED COMPLX006
An activated complex is a chemical species that is
A.
stable and has low PE.
B.
stable and has high PE.
C.
unstable and has low PE.
D.
unstable and has high PE.
11.
B3. ACTIVATION ENRGY005
The minimum amount of energy required to overcome the energy barrier in a chemical reaction is the
A.
heat of reaction.
B.
activation energy.
C.
KE of the reactants.
D.
enthalpy of the products.
— Chemistry 12 Provincial Exam Review — Page 3
Chemistry 12
12.
Unit I Reaction Kinetics
B4. EA & RATE
010
A certain reaction is able to proceed by various mechanisms. Each mechanism has a different Ea and
results in a different overall rate. Which of the following best describes the relationship between the Ea
values and the rates?
A.
B.
C.
D.
13.
B4. EA & RATE
011
What is the relationship between the activation energy and the rate of a reaction?
A.
When the activation energy is high, the rate of reaction is fast.
B.
When the activation energy is low, the rate of reaction is slow.
C.
When the activation energy is high, the rate of reaction is slow.
D.
There is no relationship between activation energy and rate of reaction.
14.
B5. PE/KE CHANGES 008
The following diagram shows reactant molecules approaching one another:
What is happening to the kinetic energy and the potential energy?
A.
B.
C.
D.
Kinetic Energy
decreasing
decreasing
increasing
increasing
Potential Energy
decreasing
increasing
increasing
decreasing
— Chemistry 12 Provincial Exam Review — Page 4
Chemistry 12
15.
Unit I Reaction Kinetics
B6. PE DIAGRAMS 032
Consider the following PE diagram:
The activation energy for the forward reaction is represented by
A.
I
B.
II
C.
III
D.
IV
16.
B7. ENTHALPY EXO/END004
Consider the reaction:
P2O3(g) + O2(g) → P2O5(g) + 114 kJ
This reaction may be described as
A.
exothermic and ∆H = –114 kJ.
B.
endothermic and ∆H = 114 kJ.
C.
exothermic and ∆H = 114 kJ.
D.
endothermic and ∆H = –114 kJ.
17.
B8. EQUATIONS/ENERGY005
Which of the following reactions is endothermic?
A.
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) + 890.3 kJ
B.
2Na2O2(s) + 2H2O(l) – 287.0 kJ → 4NaOH(aq) + O2(g)
∆H = – 65.2 kJ
C.
CaO(s) + H2O(l) → Ca(OH)2(aq)
D.
CaO(s) + 3C(s) → CaC2(s) + CO(g)
∆H = + 464.8 kJ
18.
B9. ROLE OF FACTORS 024
Consider the following KE distribution curve for colliding particles:
A.
On the diagram above, sketch a line for the distribution of collisions at a higher temperature.
(2 marks)
B.
Shade in the area representing the collisions that could result in forming an activated complex at
the lower temperature. (1 mark)
— Chemistry 12 Provincial Exam Review — Page 5
Chemistry 12
C:
REACTION KINETICS (Reaction Mechanisms and Catalysts)
C1.
C2.
C3.
C4.
C5.
C6.
19.
20.
Unit I Reaction Kinetics
use examples to demonstrate that most reactions involve more than one step
describe a reaction mechanism as the series of steps (collisions) that result in the overall reaction
define catalyst
compare and contrast the PE diagrams for a catalyzed and uncatalyzed reaction in terms of:
• reaction mechanism
• ∆H
• activation energy
identify reactant, product, reaction intermediate and catalyst from a given reaction mechanism
describe the uses of specific catalysts in a variety of situations
C2. DESC RXN MECH 018
Consider the following reaction mechanism:
A.
Determine the overall reaction. (2 marks)
B.
Identify a reaction intermediate. (1 mark)
C3. CATALYSTS
014
A substance that increases the rate of a reaction without appearing in the equation for the
overall reaction is a(n)
A.
product.
B.
catalyst.
C.
reactant.
D.
intermediate.
— Chemistry 12 Provincial Exam Review — Page 6
Chemistry 12
21.
Unit I Reaction Kinetics
C4. PE DIAGRAMS CAT 029
Consider the following potential energy diagram for a reaction:
Which of the following represents the correct activation energies?
Forward Catalyzed
Ea
40 kJ
80 kJ
100 kJ
100 kJ
A.
B.
C.
D.
22.
Reverse Uncatalyzed
Ea
140 kJ
40 kJ
80 kJ
160 kJ
C5. RXN MECHANISMS 037
Consider the following proposed reaction mechanism:
Step 1
Fe3+ + H2O2 → FeH2O23+
Step 2
FeH2O23+ → FeOH3+ + HO
Step 3
HO + H2O2 → H2O + HO2
Step 4
FeOH3+ + HO2 → Fe3+ + H2O + O2
A.
Write the overall reaction. (2 marks)
B.
Define the term catalyst and identify a catalyst in the above mechanism. (2 marks)
— Chemistry 12 Provincial Exam Review — Page 7
Chemistry 12
Unit II Dynamic Equilibrium
UNIT II DYNAMIC EQUILIBRIUM
D:
DYNAMIC EQUILIBRIUM (Introduction)
D1.
D2.
D3.
D4.
D5.
D6.
D7.
D8.
D9.
1.
describe the reversible nature of most chemical reactions
identify the reversible pathways of a chemical reaction on the PE diagram
relate the changes in rates of the forward and reverse reactions to the changing concentrations of the
reactants and products as equilibrium is established
describe chemical equilibrium as a closed system at constant temperature:
• whose macroscopic properties are constant
• where the forward and reverse reaction rates are equal
• that can be achieved from either direction
• where the concentrations of reactants and products are constant
describe the dynamic nature of chemical equilibrium
infer that a system not at equilibrium will tend to move toward a position of equilibrium
determine entropy and enthalpy changes from a chemical equation (qualitatively)
state that systems tend toward a position of minimum enthalpy and maximum randomness (entropy)
predict the result when enthalpy and entropy factors:
• both favour the products
• both favour the reactants
• oppose one another
D3.CHANGE IN RATE/[]029
Consider the following:
2NH3(g) →
← N2(g) + 3H2(g)
Initially, NH3 is added to an empty flask. How do the rates of the forward and reverse reactions change as
the system proceeds to equilibrium?
A.
B.
C.
D.
2.
Forward Rate
increases
increases
decreases
decreases
Reverse Rate
increases
decreases
increases
decreases
D4.CHARACTER OF EQ 021
Which of the following applies to a chemical equilibrium?
I. Forward and reverse reaction rates are equal
II. Equilibrium can be achieved from either direction
III. Macroscopic properties are constant
A.
B.
C.
D.
I only
I and II only
II and III only
I, II and III
— Chemistry 12 Provincial Exam Review — Page 8
Chemistry 12
Unit II Dynamic Equilibrium
3.
D5.DYNAMIC EQ
009
A chemical equilibrium is described as “dynamic” because
A.
maximum randomness has been achieved.
B.
the pressure and temperature do not change.
C.
both reactants and products continue to form.
D.
the concentrations of chemical species remain constant.
4.
D7.ENTROPY/ENTHAPLY 018
Consider the following:
2N2(g) + O2(g) + energy
→ 2N2O(g)
←
What positions do minimum enthalpy and maximum entropy tend toward?
A.
B.
C.
D.
Minimum
Enthalpy
products
products
reactants
reactants
Maximum
Entropy
products
reactants
products
reactants
5.
D8.DRIVING FORCES EQ002
Chemical systems tend to move toward positions of
A.
minimum enthalpy and maximum entropy.
B.
maximum enthalpy and minimum entropy.
C.
minimum enthalpy and minimum entropy.
D.
maximum enthalpy and maximum entropy.
6.
D9.SPONT/NONSPONT RX017
In which of the following will the driving forces of minimum enthalpy and maximum entropy oppose one
another?
A.
2C(s) + O2(g) → 2CO(g) ∆H = –221 kJ
B.
2N2(g) + O2(g) → 2N2O(l) ∆H = +164 kJ
C.
2CO(g) + O2(g) → 2CO2(g) ∆H = –566 kJ
D.
4CO2(g) + 6H2O(g) → 2C2H6(g) + 7O2(g)
∆H = +3122 kJ
— Chemistry 12 Provincial Exam Review — Page 9
Chemistry 12
E:
DYNAMIC EQUILIBRIUM (Le Châtelier’s Principle)
E1.
E2.
E3.
E4.
E5.
7.
Unit II Dynamic Equilibrium
describe the term shift as it applies to equilibria
apply Le Châtelier’s principle to the shifting of equilibrium involving the following:
• temperature change
• concentration change
• volume change of gaseous systems
explain the above shifts using the concepts of reaction kinetics
identify the effect of a catalyst on dynamic equilibrium
apply the concept of equilibrium to a commercial or industrial process
E2.LE CHATELIER 007
Consider the following equilibrium:
2NO(g) + Br2(g) + energy
→
← 2NOBr(g)
The equilibrium will shift to the left as a result of
A.
adding a catalyst.
B.
removing NOBr.
C.
increasing the volume.
D.
increasing the temperature.
8.
E2.LE CHATELIER 024
Consider the following graph for the reaction:
H2(g) + I2(g) →
← 2HI(g)
The temperature is increased at t1 and equilibrium is re–established at t2.
A.
On the above graph, sketch the line representing the [HI] between time t1 and t3. (1 mark)
B.
Calculate the value of Keq after t2. (2 marks)
— Chemistry 12 Provincial Exam Review — Page 10
Chemistry 12
9.
Unit II Dynamic Equilibrium
E3.SHIFTS & KINETICS009
Consider the following equilibrium:
PCl3(g) + 3NH3(g) →
← P(NH2)3(g) + 3HCl(g)
The volume of the equilibrium system is increased and a new equilibrium is established.
How have the rates been affected?
A.
B.
C.
D.
Rate (forward)
increased
decreased
decreased
did not change
Rate (reverse)
decreased
increased
decreased
did not change
10.
E4.CATALYST & EQ 007
A catalyst affects a reversible reaction by
A.
making the value of Keq larger.
B.
increasing the yield of products.
C.
decreasing the value of ∆H for the reaction.
D.
enabling equilibrium to achieved more rapidly.
11.
E5.APPLY LE CHAT 009
The graph below shows the amount of ammonia produced at various temperatures and pressures during the
Haber process. The reaction is N2(g) + 3H2(g)
→ 2NH3(g).
←
100
200 °C
300 °C
80
400 °C
60
industrial
operating
conditions
% Yield of NH 3
500 °C
40
600 °C
20
0
0
20
40
60
80
100
120
Pressure (x 10 3 kPa)
A.
The production of ammonia is an exothermic process. Use information from the graph to support
this statement. (1 mark)
B.
Use Le Châtelier’s principle to explain the relationship between increased pressure and the
percentage yield of ammonia. (2 marks)
C.
What theoretical conditions produce the greatest yield of ammonia? (1 mark)
D.
Industrial operating conditions are indicated on the graph. Explain why these low yield conditions
are used by industry. (2 marks)
— Chemistry 12 Provincial Exam Review — Page 11
Chemistry 12
F:
DYNAMIC EQUILIBRIUM (The Equilibrium Constant)
F1.
F2.
F3.
F4.
F5.
F6.
F7.
F8.
12.
Unit II Dynamic Equilibrium
gather and interpret data on the concentration of reactants and products of a system at equilibrium
write the expression for the equilibrium constant when given the equation for either a homogeneous or
heterogeneous equilibrium system
relate the equilibrium position to the value of Keq and vice versa
predict the effect (or lack of effect) on the value of Keq of changes in the following factors:
• temperature
• pressure
• concentration
• surface area
• catalyst
calculate the value of Keq given the equilibrium concentration of all species
calculate the value of Keq given the initial concentrations of all species and one equilibrium
concentration
calculate the equilibrium concentrations of all species given the value of Keq and the initial
concentrations
determine whether a system is at equilibrium and if not, in which direction it will shift to reach
equilibrium when given a set of concentrations for reactants and products
F1.CONC AT EQUIL'M 013
Consider the following:
2SO3(g) →
← 2SO2(g) + O2(g)
Initially, some SO3 is placed into a 3.0 L container. At equilibrium there is 0.030 mol SO2
present. What is the [O2] at equilibrium?
A.
0.0050 mol/L
B.
0.010 mol/L
C.
0.015 mol/L
D.
0.030 mol/L
13.
F2.KEQ EXPRESSIONS 039
Which reaction has the following equilibrium expression?
K eq =
A.
PCl3(g) + Cl2(g) →
← PCl5(g)
B.
PCl3(g) + Cl2(l) →
← PCl5(g)
C.
PCl5(g) →
← PCl3(g) + Cl2(g)
D.
PCl5(g) →
← PCl3(g) + Cl2(l)
[PCl5 ]
[PCl3 ][Cl 2 ]
— Chemistry 12 Provincial Exam Review — Page 12
Chemistry 12
14.
Unit II Dynamic Equilibrium
F3.POSITION & KEQ 022
An equal number of moles of I2(g) and Br2(g) are placed into a closed container and allowed to establish
equilibrium:
I2(g) + Br2(g) →
← 2IBr(g)
Keq = 280
Which one of the following relates [IBr] to [I2] at equilibrium?
A.
[I2] = [IBr]
B.
[I2] < [IBr]
C.
[I2] = 2 [IBr]
D.
[I2] = 280 [IBr]
15.
F4.LE CHAT & KEQ 031
Consider the following reaction:
C(s) + 2H2(g) →
← CH4(g)
∆H = –74.8 kJ
Which of the following will cause an increase in the value of Keq ?
A.
increasing [H2]
B.
decreasing the volume
C.
finely powdering the C(s)
D.
decreasing the temperature
16.
F5.KEQ FROM [EQ] 036
Consider the following equilibrium:
N2O4(g) →
← 2NO2(g)
An equilibrium mixture contains 4.0 x 10–2 mol N2O4 and 1.5 x 10–2 mol NO2 in a 1.0 L flask.
What is the value of Keq ?
A.
5.6 x 10–3
B.
3.8 x 10–1
C.
7.5 x 10–1
D.
1.8 x 102
17.
F6.KEQ FROM [I]/[EQ]031
Consider the following: (4 marks)
H2(g) + I2(g) →
← 2HI(g)
Initially, 0.200 mol H2 and 0.200 mol I2 are added to an empty 2.00 L container. At equilibrium,
the [I2] = 0.0200 mol/L. What is the value of Keq ?
18.
F7.[EQ]FROM KEQ/[I] 023
Consider the following: (4 marks)
H2(g) + Br2(g) →
← 2HBr(g)
Keq = 12.0
Initially, 0.080 mol H2 and 0.080 mol Br2 are placed into a 4.00 L container. What is the [HBr] at
equilibrium?
19.
F8.TRIAL KEQ
026
Consider the following: (4 marks)
N2O4(g) →
Keq = 9.5 x 10–3
← 2NO2(g)
Initially, 0.060 mol N2O4 and 0.020 mol NO2 are placed into a 2.00 L container. Use calculations to
determine the direction in which the reaction proceeds in order to reach equilibrium.
— Chemistry 12 Provincial Exam Review — Page 13
Chemistry 12
Unit III Solubility Equilibria
UNIT III SOLUBILITY EQUILIBRIA
G:
SOLUBILITY EQUILIBRIA (Concept of Solubility)
G1.
G2.
G3.
G4.
G5.
G6.
G7.
G8.
classify solutions as ionic or molecular given the formula of the solute
describe the conditions necessary to form a saturated solution
describe solubility as the concentration of a substance in a saturated solution
use appropriate units to represent the solubility of substances in aqueous solutions
measure the solubility of a compound in aqueous solution
describe the equilibrium that exists in a saturated aqueous solution
write a net ionic equation that describes a saturated solution
calculate the concentration of the positive and negative ions given the concentration of a solute in an
aqueous solution
1.
G1.IONIC/MOLEC SOL'N011
Which of the following will dissolve in water to form an ionic solution?
A.
O2
B.
CH4
C.
NH4Cl
D.
CH3OH
2.
G1.IONIC/MOLEC SOL'N009
Which of the following dissolves in water to form a molecular solution?
A.
KCl
B.
Na2O
C.
NH4Br
D.
C2H5OH
3.
G2.SATURATED SOL'NS 006
A saturated solution is formed by adding 10.0 g PbI2(s) to 10.0 mL of water in a beaker. Describe the
situation which exists in the beaker.
A.
[Pb2+] = [I–]
B.
moles PbI2(s) = moles Pb2+(aq)
C.
mass of PbI2(s) = mass of Pb2+(aq)
D.
rate of crystalization = rate of dissociation
4.
G3.DESCRIBE SOL 006
The solubility of SrCO3 is 2.4 x 10–5 M . How many moles of dissolved solute are present in 100.0 mL of
saturated SrCO3 solution?
A.
5.6 x 10–10 mol
B.
2.4 x 10–6 mol
C.
2.4 x 10–5 mol
D.
2.4 x 10–4 mol
5.
G4.UNITS OF SOL 005
Which of the following could be used to express solubility?
A.
mol
B.
M/s
C.
g/mL
D.
mL/min
— Chemistry 12 Provincial Exam Review — Page 14
Chemistry 12
Unit III Solubility Equilibria
6.
G5.MEASURE SOL
006
When 100.0 mL of a saturated solution of BaF2 is heated and all the water is evaporated, 3.6 x 10–4 mol of
solute remains. The solubility of BaF2 is
A.
1.9 x 10–10 M
B.
1.3 x 10–5 M
C.
3.6 x 10–4 M
D.
3.6 x 10–3 M
7.
G6.EQUIL & SATURAT'N017
The equation that describes the solubility equilibrium of Ca3(PO4)2 is
8.
9.
A.
6+
3–
Ca3(PO4)2(s) →
← Ca3 (aq) + 2PO4 (aq)
B.
3–
2+
Ca3(PO4)2(s) →
← 3Ca (aq) + 2PO4 (aq)
C.
3+
2–
Ca3(PO4)2(s) →
← 2Ca (aq) + 3PO4 (aq)
D.
2+
3–
Ca3(PO4)2(s) →
← (Ca )3(aq) + (PO4 )2(aq)
G7.NET IONIC EQN 002
The equation that describes the solubility equilibrium of Ag2CrO4 is
A.
2+
2–
Ag2CrO4(s) →
← Ag2 (aq) + CrO4 (aq)
B.
+
2–
Ag2CrO4(s) →
← 2Ag (aq) + CrO4 (aq)
C.
Ag2CrO4(s) →
← 2Ag(s) + Cr(s) + 2O2(g)
D.
+
6+
2–
Ag2CrO4(s) →
← 2Ag (aq) + Cr (aq) + 4O (aq)
G8.CALCULATE [ION] 021
What are the ion concentrations in 0.30 M CuCl2?
[Cu2+] [Cl–]
A.
0.10 M 0.20 M
B.
0.20 M 0.10 M
C.
0.30 M 0.30 M
D.
0.30 M 0.60 M
— Chemistry 12 Provincial Exam Review — Page 15
Chemistry 12
H:
Unit III Solubility Equilibria
SOLUBILITY EQUILIBRIA (Solubility and Precipitation)
H1.
H2.
H3.
H4.
H5.
H6.
H7.
describe a compound as having high or low solubility relative to 0.1 M by using a solubility chart
use a solubility chart to predict if a precipitate will form when two solutions are mixed and identify the
precipitate
write a formula equation, complete ionic equation and net ionic equation that represent a precipitation
reaction
use a solubility chart to predict if ions can be separated from solution through precipitation and outline
the process
predict qualitative changes in the solubility equilibrium upon the addition of a common ion
identify an unknown ion through experimentation involving a qualitative analysis scheme
devise a procedure by which the contaminating ions in hard or polluted water can be removed
10.
H1.CMPD SOLUBILITY 023
Which of the following has the lowest solubility?
A.
CaS
B.
CuS
C.
FeS
D.
MgS
11.
H2.PPT 0.1 M SOL'NS 019
When equal volumes of 0.2 M solutions are mixed, which of the following combinations forms a
precipitate?
A.
CaS and Sr(OH)2
B.
H2SO4 and MgCl2
C.
(NH4)2SO4 and K2CO3
D.
H2SO3 and NaCH3COO
12.
H3.PPT EQUATIONS 016
When equal volumes of 0.20 M Pb(NO3)2 and 0.20 M KCl are mixed, a precipitate of PbCl2 forms.
13.
A.
Write the formula equation for the above reaction. (1 mark)
B.
Write the complete ionic equation for the above reaction. (1 mark)
C.
Write the net ionic equation for the above reaction. (1 mark)
H4.SELECTIVE PPT 013
A solution contains both 0.2 M Mg2+(aq) and 0.2 M Sr2+(aq) . These ions can be removed separately through
precipitation by adding equal volumes of 0.2 M solutions of
A.
OH– and then S2–
B.
Cl– and then OH–
C.
CO32– and then SO32–
D.
SO42– and then PO43–
— Chemistry 12 Provincial Exam Review — Page 16
Chemistry 12
14.
Unit III Solubility Equilibria
H5.COMMON ION EFFECT022
Consider the following equilibrium:
2+
2–
CaSO4(s) →
← Ca (aq) + SO4 (aq)
Which of the following would shift the above equilibrium to the left?
A.
adding CaSO4(s)
B.
adding MgSO4(s)
C.
removing some Ca2+ (aq)
D.
removing some SO42–(aq)
15.
H6.QUAL ANALYSIS 008
Consider the following anions:
I. 10.0 mL of 0.20 M Cl–
II. 10.0 mL of 0.20 M OH–
III. 10.0 mL of 0.20 M SO32–
When 10.0 mL of 0.20 M Pb(NO3)2 are added to each of the above, precipitates form in
A.
I and II only.
B.
I and III only.
C.
II and III only.
D.
I, II and III.
16.
H7.HARD WATER
006
Which of the following could be added to a sample of hard water to remove
both 0.2 M Ca2+ and 0.2 M Mg2+?
A.
0.2 M S2–
B.
0.2 M Cl–
C.
0.2 M OH–
D.
0.2 M SO42–
— Chemistry 12 Provincial Exam Review — Page 17
Chemistry 12
I:
Unit III Solubility Equilibria
SOLUBILITY EQUILIBRIA (Quantitative Aspects)
I1.
I2.
I3.
I4.
I5.
I6.
I7.
describe the Ksp expression as a specialized Keq expression
write a Ksp expression for a solubility equilibrium
calculate the Ksp for AB and AB2 type compounds when given the solubility of a compound
calculate the solubility of AB and AB2 type compounds from the Ksp
predict the formation of a precipitate by comparing the trial ion product to the Ksp value using specific
data
calculate the maximum concentration of one ion given the Ksp and the concentration of the other ion
demonstrate and describe a method for determining the concentration of a specific ion
17.
I2.KSP EXPRESSION 012
The Ksp expression for a saturated solution of Ag2SO3 is
A.
Ksp = [2Ag+][SO32–]
B.
Ksp = [Ag+]2[SO32–]
C.
Ksp = [Ag22+][SO32–]
D.
Ksp = [2Ag+]2[SO32–]
18.
I3.CALC KSP FROM SOL029
The solubility of CaF2 is 3.3 x 10–4 M. Determine the Ksp value of CaF2.
A.
3.6 x 10–11
B.
1.4 x 10–10
C.
1.1 x 10–7
D.
3.3 x 10–4
19.
I3.CALC KSP FROM SOL028
The solubility of CdCO3 is 2.5 x 10–6 M . Calculate the Ksp value for CdCO3.
A.
6.3 x 10–12
B.
2.5 x 10–6
C.
5.0 x 10–6
D.
1.6 x 10–3
20.
I4.CALC SOL FROM KSP023
Calculate the solubility of CaC2O4.
A.
2.3 x 10–9 M
B.
1.2 x 10–5 M
C.
4.8 x 10–5 M
D.
8.3 x 10–4 M
21.
I4.CALC SOL FROM KSP013
The solubility of SrF2 is
A.
4.3 x 10–9 M
B.
6.6 x 10–5 M
C.
1.0 x 10–3 M
D.
1.6 x 10–3 M
— Chemistry 12 Provincial Exam Review — Page 18
Chemistry 12
Unit III Solubility Equilibria
22.
I5.TRIAL ION PRODUCT018
When a solution containing Ag+ is mixed with a solution containing BrO3– , the trial ion product is
determined to be 2.5 x 10–7. What would be observed?
A.
A precipitate would form since trial ion product < Ksp.
B.
A precipitate would form since trial ion product > Ksp.
C.
A precipitate would not form since trial ion product < Ksp.
D.
A precipitate would not form since trial ion product > Ksp.
23.
I6.MAX [ION] W/O PPT020
Determine the maximum [Na2CO3] that can exist in 1.0 L of 0.0010 M Ba(NO3)2 without forming a
precipitate.
A.
2.6 x 10–12 M
B.
2.6 x 10–9 M
C.
2.6 x 10–6 M
D.
5.1 x 10–5 M
24.
I7.TITRATIONS
006
Consider the following information and accompanying diagram:
In a titration experiment, AgNO3(aq) was used to determine the [Cl–] in a
water sample and the following data were recorded:
[AgNO3] = 0.125 M
Volume of water sample containing Cl– = 20.00 mL
Initial buret reading of AgNO3 = 5.15 mL
Final buret reading of AgNO3 = 37.15 mL
The equation for this reaction is
Ag+(aq) + Cl–(aq) → AgCl(s)
Using the above data, determine the [Cl–] in the water sample. (3 marks)
— Chemistry 12 Provincial Exam Review — Page 19
Chemistry 12
Unit IV Acids and Bases
UNIT IV ACIDS AND BASES
J:
ACIDS, BASES and SALTS (Properties and Definitions)
J1.
J2.
J3.
J4.
J5.
J6.
J7.
J8.
J9.
J10.
J11.
J12.
1.
identify acids and bases through experimentation
list general properties of acids and bases
write balanced equations representing the neutralization of acids by bases in solution
define Arrhenius acids and bases
write names and formulae of some common acids and bases and outline some of their common
properties, uses and commercial names
define Brönsted–Lowry acids and bases
identify Brönsted–Lowry acids and bases in an equation
write balanced equations representing the reaction of acids or bases with water
identify an H3O+ ion as a protonated H2O molecule that can be represented in shortened form as H+(aq)
define conjugate acid–base pair
identify the conjugate of a given acid or base
show that in any Brönsted–Lowry acid–base equation there are two conjugate pairs present
J1.IDENTIFY A/B 002
Which of the following tests could be used to distinguish between 1.0 M HCl and 1.0 M NaOH?
I.
II.
III.
A.
B.
C.
D.
electrical conductivity
reaction with zinc to produce hydrogen gas
colour of the indicator phenolphthalein
III only
I and II only
II and III only
I, II, and III
2.
J2.A/B PROPERTIES 031
Which of the following is a property of sodium hydroxide?
A.
feels slippery
B.
releases H3O+ in aqueous solution
C.
changes litmus paper from blue to red
D.
reacts with magnesium to produce hydrogen gas
3.
J3.NEUTRALIZAT'N EQN005
Which of the following represents the complete neutralization of H3PO4 by NaOH?
A.
H3PO4 + NaOH → NaH2PO4 + H2O
B.
H3PO4 + 3NaOH → Na3PO4 + 3H2O
C.
H3PO4 + 2NaOH → Na2HPO4 + 2H2O
D.
H3PO4 + NaOH → NaH + HPO4 + H2O
4.
J4.ARRHENIUS A/B 005
An Arrhenius base is defined as a compound that
A.
accepts OH– in solution.
B.
releases OH– in solution.
C.
accepts protons in solution.
D.
donates protons in solution.
— Chemistry 12 Provincial Exam Review — Page 20
Chemistry 12
Unit IV Acids and Bases
5.
J5.COMMON A/B
002
Caustic soda, NaOH, is found in
A.
fertilizers.
B.
beverages.
C.
toothpaste.
D.
oven cleaners.
6.
J6.DEFINE B–L A/B 005
A Brønsted–Lowry acid
A.
donates protons to a Brønsted–Lowry acid.
B.
donates protons to a Brønsted–Lowry base.
C.
accepts protons from a Brønsted–Lowry acid.
D.
accepts protons from a Brønsted–Lowry base.
7.
J7.IDENTIFY B–L A/B 033
Consider the following Brønsted–Lowry equilibrium:
–
+
C6H5NH2(aq) + H2O(l) →
← C6H5NH3 (aq) + OH (aq)
The substances acting as acids and bases from left to right are
A.
acid, base, acid, base.
B.
acid, base, base, acid.
C.
base, acid, acid, base.
D.
base, acid, base, acid.
8.
J8.EQNS W/ H2O
004
Water acts as an acid when it reacts with which of the following?
I. CN–
II. NH3
III. HClO4
IV. CH3COO–
A.
B.
C.
D.
I and IV only
II and III only
I, II and IV only
II, III and IV only
9.
J9.HYDRONIUM ION 001
A hydronium ion has the formula
A.
H 2+
B.
OH–
C.
H 2O +
D.
H 3O +
10.
J10.DEFINE CONJ A/B 003
A.
Define the term Brønsted–Lowry conjugate acid–base pair. (1 mark)
B.
Give an example of a conjugate acid–base pair. (1 mark)
— Chemistry 12 Provincial Exam Review — Page 21
Chemistry 12
11.
J11.CONJUGATE A/B 018
What is the conjugate acid and what is the conjugate base of HPO42– ?
A.
B.
C.
D.
12.
Unit IV Acids and Bases
Conjugate Acid Conjugate Base
PO43– H2PO4–
H2PO4– PO43–
H2PO4– H3PO4
H3PO4 PO43–
J12.B–L THEORY
006
Consider the following equilibrium:
HS– + H3BO3
–
→
← H2BO3 + H2S
The two species acting as Brønsted–Lowry bases in the above equilibrium are
A.
HS– and H2S
B.
H2BO3 and H2S
C.
HS– and H2BO3–
D.
H2BO3 and H2BO3–
— Chemistry 12 Provincial Exam Review — Page 22
Chemistry 12
K:
Unit IV Acids and Bases
ACIDS, BASES and SALTS (Strong and Weak Acids and Bases)
K1.
K2.
K3.
K4.
K5.
relate electrical conductivity in a solution to the concentration of ions
classify an acid or base in solution as either weak or strong by comparing conductivity
define a strong acid and a strong base
define a weak acid and a weak base
write equations to show what happens when strong and weak acids and bases are dissolved in water
(dissociation, ionization)
K6. compare the relative strengths of acids or bases by using a table of relative acid strengths
K7. identify and explain why the strongest acid in aqueous solutions is H3O+ and the strongest base in
aqueous solutions is OH–
K8. predict whether products or reactants are favoured in an acid–base equilibrium by comparing the
strength of the two acids (or two bases)
K9. compare the relative concentrations of H3O+ (or OH–) between two acids(or between two bases) using
their relative positions on an acid strength table
K10. define amphiprotic
K11. identify chemical species that are amphiprotic
K12. describe situations in which H2O would act as an acid or base
13.
K1.CONDUCTIVITY & []011
The electrical conductivities of 0.10 M solutions of NaCl, HCN, and HNO2 are measured. The order by
conductivity from highest to lowest is
A.
NaCl > HNO2 > HCN
B.
HCN > HNO2 > NaCl
C.
NaCl > HCN > HNO2
D.
HNO2 > HCN > NaCl
14.
K2.CLASSIFY A/B 015
When comparing equal volumes of 0.10 M HNO3 with 0.10 M HNO2, what would be observed?
A.
The pH values would be the same.
B.
The electrical conductivities would be different.
C.
The effects on blue litmus paper would be different.
D.
The volumes of 0.10 M NaOH needed for neutralization would be different.
15.
K3.DEFINE STRONG A/B002
Which of the following is a property of 1.0 M HCl but not a property of 1.0 M CH3COOH ?
A.
turns litmus red
B.
ionizes completely
C.
has a pH less than 7.0
D.
produces H3O+ in solution
16.
K4.DEFINE WEAK A/B 003
Which of the following statements applies to 1.0 M NH3(aq) but not to 1.0 M NaOH(aq)?
A.
partially ionizes
B.
neutralizes an acid
C.
has a pH greater than 7
D.
turns bromocresol green from yellow to blue
— Chemistry 12 Provincial Exam Review — Page 23
Chemistry 12
17.
Unit IV Acids and Bases
K5.A/B EQUATIONS 008
An equation representing the reaction of a weak acid with water is
A.
HCl + H2O → H3O+ + Cl–
B.
–
+
NH3 + H2O →
← NH4 + OH
C.
–
HCO3– + H2O →
← H2CO3 + OH
D.
+
–
HCOOH + H2O →
← H3O + HCOO
18.
K6.REL STRENGTH A/B 030
List the bases C2O42–, NH3, and PO43– in order from strongest to weakest.
A.
PO43– > NH3 > C2O42–
B.
C2O42– > NH3 > PO43–
C.
NH3 > PO43– > C2O42–
D.
PO43– > C2O42– > NH3
19.
K7.STRONGEST AQ A/B 003
In aqueous solutions, H3O+ is the strongest acid present. This phenomenon is called the levelling effect.
Explain why this occurs. (2 marks)
20.
K8.POSITION OF EQM 026
Consider the equilibrium:
C6H5COOH + NO2–
→ HNO2 + C6H5COO–
←
Identify the stronger acid and predict whether reactants or products are favoured.
A.
B.
C.
D.
21.
Stronger Acid
HNO2
HNO2
C6H5COOH
C6H5COOH
Side Favoured
reactants
products
reactants
products
K9.RELATIVE [H+] 009
A student records the pH of 0.1 M solutions of two acids:
Acids
X
Y
pH
4.0
2.0
Which of the following statements can be concluded from the above data?
A.
Acid X is stronger than acid Y.
B.
Acid X and Y are both weak.
C.
Acid X is diprotic while acid Y is monoprotic.
D.
Acid X is 100 times more concentrated than acid Y
22.
K10.DEF AMPHIPROTIC 005
The term amphiprotic describes a substance that can act as
A.
a proton donor and as a proton acceptor.
B.
a proton donor but not as a proton acceptor.
C.
a proton acceptor but not as a proton donor.
D.
neither a proton donor nor as a proton acceptor.
— Chemistry 12 Provincial Exam Review — Page 24
Chemistry 12
23.
Unit IV Acids and Bases
K11.ID AMPHIPROTIC 015
Which of the following chemical species are amphiprotic in aqueous solution?
I. F–
II. NH4+
III. HPO42–
A.
B.
C.
D.
I only.
II only.
III only.
II and III only.
— Chemistry 12 Provincial Exam Review — Page 25
Chemistry 12
L:
Unit IV Acids and Bases
ACIDS, BASES and SALTS (Kw , pH, pOH)
L1.
L2.
L3.
write equations representing the ionization of water using either H3O+ and OH– or H+ and OH–
write the equilibrium expression for the ion product constant of water, Kw
predict the effect of the addition of an acid or base to the equilibrium system:
+
–
2H2O(l) →
← H3O (aq) + OH (aq)
L4.
L5.
L6.
L7.
L8.
L9.
L10.
L11.
L12.
state the relative concentrations of H3O+ and OH–in acid, base and neutral solutions
state the value of Kw at 25°C
describe the variation of the value of Kw with temperature
calculate the concentration of H3O+ (or OH–) given the other, using Kw
describe the pH scale with reference to everyday solutions
define pH and pOH
define pKw, give its value at 25°C and its relation to pH and pOH
perform calculations relating pH, pOH, H3O+ and OH–
calculate H3O+ or OH– from pH and pOH
24.
L1.WATER IONIZ'N EQN006
The ionization of water can be represented by
A.
2H2O(l) → 2H2(g) + O2(g)
B.
H2O(l) → 2H+(aq) + O2–(aq)
C.
H2O(l) → H3O+(aq) + OH–(aq)
D.
2H2O(l) → H3O+(aq) + OH–(aq)
25.
L2.KW EXPRESSION 007
Which of the following describes the relationship between [H3O+] and [OH–] ?
A.
[H3O+][OH–] = 14.00
B.
[H3O+] + [OH–] = 14.00
C.
[H3O+][OH–] = 1.0 x 10–14
D.
[H3O+] + [OH–] = 1.0 x 10–14
26.
L3.SELF IONIZATION 013
Consider the following equilibrium at 25 °C :
+
–
2H2O(l) →
← H3O (aq) + OH (aq)
What happens to [OH–] and pH as 0.1 M HCl is added?
A.
[OH–] decreases and pH increases.
B.
[OH–] decreases and pH decreases.
C.
[OH–] increases and pH increases.
D.
[OH–] increases and pH decreases.
27.
L4.RELATIVE CONC 007
A basic solution can be defined as one in which
A.
[H3O+] is not present
B.
[H3O+] is equal to [OH–]
C.
[H3O+] is less than [OH–]
D.
[H3O+] is greater than [OH–]
— Chemistry 12 Provincial Exam Review — Page 26
Chemistry 12
Unit IV Acids and Bases
28.
L5.VALUE OF KW
003
What is the value of the ionization constant for water at 25 °C ?
A.
7.0
B.
14.0
C.
1.0 x 10–7
D.
1.0 x 10–14
29.
L6.KW & TEMP
013
Consider the following equilibrium:
2H2O + energy
→ H3O+ + OH–
←
Which of the following describes the result of decreasing the temperature?
A.
B.
C.
D.
30.
[H3O+]
increases
decreases
increases
decreases
[OH–]
increases
increases
decreases
decreases
Kw
increases
decreases
no change
decreases
L10.DEFINE PKW
002
Which of the following statements concerning pKw are true?
I. pKw = – log Kw
II. pKw = pH + pOH
III. pKw = [H3O+][OH–]
A.
B.
C.
D.
I and II only
I and III only
II and III only
I, II, and III
31.
L11.PH, POH, H+, OH–055
Calculate the pOH of a 0.050 M HBr solution.
A.
0.30
B.
1.30
C.
12.70
D.
13.70
32.
L12.CALC [H+] FRM PH015
Determine the pH of 3.0 M KOH .
A.
0.48
B.
11.00
C.
13.52
D.
14.48
— Chemistry 12 Provincial Exam Review — Page 27
Chemistry 12
M:
ACIDS, BASES and SALTS (Ka and Kb Problem Solving)
M1.
M2.
M3.
M4.
M5.
33.
Unit IV Acids and Bases
write Ka and Kb equilibrium expressions
relate the magnitude of Ka or Kb to the strength of the acid or base
given the Ka, Kb and initial concentration, calculate any of the following:
• H 3O +
• OH–
• pH
• pOH
calculate the value of Kb for a base given the value of Ka for its conjugate acid (or vice versa)
calculate the value of Ka or Kb given the pH and initial concentration
M1.KA & KB EXPRESS'N013
[H3 BO 3 ][OH− ]
The relationship [H 2BO 3
A.
Ka for H3BO3
B.
Kb for H3BO3
C.
Ka for H2BO3–
D.
Kb for H2BO3–
−
]
is the expression for
34.
M2.MAGNITUDE OF K 013
Four acids are analyzed and their Ka values are determined. Which of the following values represents the
strongest acid?
A.
Ka = 2.2 x 10–13
B.
Ka = 6.2 x 10–8
C.
Ka = 1.7 x 10–5
D.
Ka = 1.2 x 10–2
35.
M3.KA/KB CALCULAT'N 040
Calculate the pH of 0.35 M H2CO3. (4 marks)
36.
M3.KA/KB CALCULAT'N 036
A 0.0200 M solution of methylamine, CH3NH2, has a pH = 11.40. Calculate the Kb for methylamine. (4
marks)
37.
M4.CALC KB
026
Calculate the value of Kb for HPO42– .
A.
4.5 x 10–2
B.
1.6 x 10–7
C.
2.2 x 10–27
D.
6.2 x10–22
38.
M5.CALC KA/KB
012
At a particular temperature a 1.0 M H2S solution has a pH = 3.75. Calculate the value of Ka at this
temperature. (4 marks)
— Chemistry 12 Provincial Exam Review — Page 28
Chemistry 12
N:
Unit IV Acids and Bases
ACIDS, BASES and SALTS (Hydrolysis of Salts)
N1.
N2.
N3.
N4.
write a dissociation equation for a salt in water
write net ionic equations representing the hydrolysis of salts
predict qualitatively whether a salt solution would be acidic, basic, or neutral
determine whether an amphiprotic ion will act as a base or an acid in solution
39.
N1.SALT DISSOCIAT'N 006
The dissociation of NH4NO3 is represented by
A.
NH4NO3(s) → NH4+(aq) + NO3–(aq)
B.
NH4+(aq) + NO3–(aq) → NH4NO3(s)
C.
NH4+(aq) + H2O(l) → H3O+(aq) + NH3(aq)
D.
NO3–(aq) + H2O(l) → HNO3(aq) + OH–(aq)
40.
N2.HYDROLYSIS EQN 017
The equation for the predominant hydrolysis of NH4NO3 can be represented by
A.
+
–
NH4NO3(s) →
← NH4 (aq) + NO3 (aq)
B.
+
NH4+(aq) + H2O(l) →
← H3O (aq) + NH3(aq)
C.
–
NO3–(aq) + H2O(l) →
← HNO3(aq) + OH (aq)
D.
+
–
NH4NO3(aq) + H2O(l) →
← H3O (aq) + NH3NO3 (aq)
41.
N3.SALT HYDROLYSIS 042
Which of the following salt solutions will be neutral?
A.
1.0 M NH4Cl
B.
1.0 M LiClO4
C.
1.0 M K2C2O4
D.
1.0 M NaHCO3
42.
N4.AMPHIPROTIC IONS 013
A solution made from baking soda (NaHCO3) has an amphiprotic anion which is
A.
basic since Ka < Kb
B.
basic since Ka > Kb
C.
acidic since Ka < Kb
D.
acidic since Ka > Kb
— Chemistry 12 Provincial Exam Review — Page 29
Chemistry 12
O:
Unit IV Acids and Bases
ACIDS, BASES and SALTS (Indicators)
O1.
O2.
O3.
O4.
O5.
describe an indicator as a mixture of a weak acid and its conjugate base, each with distinguishing
colours
describe the term transition point of an indicator, including the conditions that exist in the equilibrium
system
describe the shift in equilibrium and resulting colour changes as an acid or a base is added to an
indicator
predict the approximate pH at the transition point using the Ka value of an indicator
predict the approximate Ka value for an indicator given the approximate pH range of the colour change
43.
O1.INDICATOR COMP 005
A chemical indicator in solution consists of
A.
a weak acid and its conjugate acid.
B.
a weak acid and its conjugate base.
C.
a strong acid and its conjugate acid.
D.
a strong acid and its conjugate base.
44.
O2.TRANSITION PT 013
Which of the following applies at the transition point for all indicators, HInd ?
A.
[HInd] = [Ind–]
B.
[Ind–] = [H3O+]
C.
[H3O+] = [OH–]
D.
[HInd] = [H3O+]
45.
O3.INDICATOR SHIFT 036
Consider the following equilibrium for the chemical indicator phenol red, HInd, at a pH = 7.3 (orange).
HInd (yellow) + H2O
+
–
→
← H3O + Ind (red)
When some NaOH is added, what stress is imposed on the equilibrium and what colour change occurs?
A.
B.
C.
D.
Stress Indicator
increased [H3O+]
decreased [H3O+]
increased [H3O+]
decreased [H3O+]
Colour Change
turns red
turns red
turns yellow
turns yellow
46.
O4.PH AT TRANSITION 012
A chemical indicator has a Ka = 2.5 x 10–5. Determine the pH at the transition point.
A.
2.30
B.
4.60
C.
7.00
D.
9.40
47.
O5.INDICATOR KA 015
A chemical indicator has a transition point at a pOH = 8.0. Calculate its Ka value and identify the indicator.
A.
Ka = 1 x 10–8, phenol red
B.
Ka = 1 x 10–6, methyl red
C.
Ka = 1 x 10–8, thymol blue
D.
Ka = 1 x 10–6, chlorophenol red
— Chemistry 12 Provincial Exam Review — Page 30
Chemistry 12
P:
Unit IV Acids and Bases
ACIDS, BASES and SALTS (Neutralizations of Acids and Bases)
P1.
P2.
P3.
P4.
P5.
P6.
demonstrate an ability to design and perform a neutralization experiment involving the following:
• primary standards
• standardized solutions
• titration curves
• indicators selected so the end point coincides with the equivalence point
calculate from titration data the concentration of an acid or base
calculate the volume of an acid or base of known molarity needed to neutralize a known volume of a
known molarity base or acid
write formula, complete ionic and net ionic neutralization equations for:
• a strong acid by a strong base
• a weak acid by a strong base
• a strong acid by a weak base
calculate the pH of a solution formed when a strong acid is mixed with a strong base
contrast the equivalence point (stoichiometric point) of a strong acid strong base titration with the
equivalence point of a titration involving a weak acid–strong base or strong acid–weak base
48.
P1.NEUTRALIZ'N EXPT 012
In acid–base titrations, the solution of known concentration is called a(n)
A.
basic solution.
B.
acidic solution.
C.
standard solution.
D.
indicating solution.
49.
P2.TITRATION
025
During a titration, 25.0 mL of H3PO4(aq) is completely neutralized by 42.6 mL of 0.20 M NaOH. Calculate
the concentration of the acid.
A.
0.11 M
B.
0.17 M
C.
0.34 M
D.
1.0 M
50.
P3.VOL TO END POINT 024
Calculate the volume of 0.300 M HNO3 needed to completely neutralize 25.0 mL of 0.250 M Sr(OH)2.
A.
10.4 mL
B.
15.0 mL
C.
20.8 mL
D.
41.7 mL
51.
P4.NEUTRALIZAT'N EQN013
Write the formula equation and the net ionic equation for the reaction between 0.10 M H2SO4 and 0.100 M
Sr(OH)2. (3 marks)
52.
P5.PH OF A/B MIXTURE027
Calculate the pH of a solution prepared by mixing 15.0 mL of 0.50 M HCl with 35.0 mL of 1.0 M NaOH.
53.
P6.TITRATION CURVES 045
A strong acid–strong base titration has a pH = 7.0 at the equivalence point. A weak acid–strong base
titration has a pH > 7.0 at the equivalence point.
A.
What causes the difference in these pH values? (2 marks)
B.
Select one indicator which could be used for both titrations. (1 mark)
— Chemistry 12 Provincial Exam Review — Page 31
Chemistry 12
Q:
Unit IV Acids and Bases
ACIDS, BASES and SALTS (Buffer Solutions)
Q1.
Q2.
Q3.
Q4.
Q5.
Q6.
describe the tendency of buffer solutions to resist changes in pH
describe the composition of an acidic buffer and a basic buffer
outline a procedure to prepare a buffer solution
identify the limitations in buffering action
describe qualitatively how the buffer equilibrium shifts as small quantities of acid or base are added to
the buffer
describe common buffer systems present in industrial, environmental, or biological systems
54.
Q1.PURPOSE OF BUFFER003
All buffer solutions are able to
A.
maintain pH at 7.00.
B.
neutralize acidic solutions only.
C.
maintain a relatively constant pH.
D.
keep the pH of a solution at a constant value.
55.
Q2.BUFFER COMPOSIT'N023
Equal moles of which of the following chemicals could be used to make a basic buffer solution?
A.
HF and NaOH
B.
HCl and NaCl
C.
KBr and NaNO3
D.
NH3 and NH4Cl
56.
Q3.PREPARE BUFFER 006
A buffer solution can be prepared by dissolving equal moles of
A.
a weak base and a strong base.
B.
a weak acid and its conjugate base.
C.
a strong base and its conjugate acid.
D.
a strong acid and its conjugate base.
57.
Q4.BUFFER LIMITAT'N 001
When 10 mL of 0.10 M Sr(OH)2 is added to 20 mL of a solution of 0.10 M CH3COOH and 0.10 M
NaCH3COO, the pH increases greatly. This result occurs because
A.
the solution is a buffer.
B.
Sr(OH)2 is a strong base.
C.
Sr(OH)2 contains a common ion.
D.
the amount of OH– exceeds the buffer’s capacity.
58.
Q5.BUFFER SHIFT QUAL011
A.
Write the net ionic equation that represents the equilibrium that exists in the buffer system
produced when equal volumes of 1.0 M NH3 and 1.0 M NH4Cl are mixed.
(1 mark)
B.
59.
Explain why the pH of this buffer system changes very little when a small amount of strong base
is added. (2 marks)
Q6.COMMON BUFFERS 001
Which of the following pairs of substances form a buffer system for human blood?
A.
HCl and Cl–
B.
NH3 and NH2–
C.
H2CO3 and HCO3–
D.
H3C6H5O7 and HC6H5O72–
— Chemistry 12 Provincial Exam Review — Page 32
Chemistry 12
R:
Unit IV Acids and Bases
ACIDS, BASES and SALTS (Acid Rain)
R1.
R2.
R3.
R4.
R5.
write equations representing the formation of acidic solutions or basic solutions from non–metal and
metal oxides
describe the pH conditions required for rain to be called acid rain
relate the pH of normal rain water to the presence of dissolved CO2
describe sources of NOx and SOx
discuss general environmental problems associated with acid rain
60.
R1.OXIDES
028
Which of the following equations describes the reaction that occurs when MgO is added to water?
A.
MgO + H2O → Mg(OH)2
B.
MgO + H2O → MgO2 + H2
C.
MgO + H2O → MgH2 + O2
D.
2MgO + 2H2O → 2MgOH + H2 + O2
61.
R2.PH OF ACID RAIN 003
The pH of acid rain could be
A.
5.0
B.
7.0
C.
9.0
D.
11.0
62.
R3.PH OF RAIN WATER 011
The pH of normal rainwater is
A.
less than 7.0 due to dissolved SO2(g)
B.
less than 7.0 due to dissolved CO2(g)
C.
greater than 7.0 due to dissolved CO2(g)
D.
equal to 7.0 due to dissolved N2 and O2
63.
R4.ACID RAIN
003
A common source of NO2 is
A.
a fuel cell.
B.
a lead smelter.
C.
an aluminum smelter.
D.
an automobile engine.
64.
R5.ACID RAIN
001
SO2 is a waste product in some industrial processes. State the environmental problem associated with
SO2(g), write the equation that accounts for this problem, and give one effect on the natural environment. (2
marks)
— Chemistry 12 Provincial Exam Review — Page 33
Chemistry 12
Unit V Electrochemistry
UNIT V ELECTROCHEMISTRY
S:
OXIDATION REDUCTION (Introduction)
S1.
S2.
S3.
S4.
S5.
S6.
define and apply the following:
• oxidation–reduction
• oxidizing agent
• reducing agent
• half–reaction
• redox reaction
determine the following:
• the oxidation number of an atom in a chemical species
• the change in oxidation number an atom undergoes when it is oxidized or reduced
• whether an atom has been oxidized or reduced by its change in oxidation number
relate change in oxidation number to gain or loss of electrons
from data for a series of simple redox reactions, create a simple table of reduction half–reactions
identify the relative strengths of oxidizing and reducing agents from their positions on a half–reaction
table
use a table of reduction half–reactions to predict whether a spontaneous redox reaction will occur
between any two species
1.
S1.DEFINITIONS
056
Which of the following describes a strong oxidizing agent?
A.
a substance which loses electrons readily
B.
a substance which gains electrons readily
C.
a substance which has a large increase in oxidation number
D.
a substance which has a small increase in oxidation number
2.
S1.DEFINITIONS
046
In which reaction is nitrogen reduced?
A.
2NO + O2 → 2NO2
B.
4NH3 + 5O2 → 4NO + 6H2O
C.
Cu2+ + 2NO2 + 2H2O → Cu + 4H+ + 2NO3–
D.
4Zn + 10H+ + NO3– → 4Zn2+ + NH4+ + 3H2O
3.
S1.DEFINITIONS
043
Consider the following redox reaction:
2MnO4– + 3ClO3– + H2O → 3ClO4– + 2MnO2+ 2OH–
The reducing agent is
A.
H 2O
B.
ClO3–
C.
MnO2
D.
MnO4–
4.
S2.OXIDATION NUMBER 072
What is the oxidation number of S in S2O62– ?
A.
+3
B.
+5
C.
+6
D.
+7
— Chemistry 12 Provincial Exam Review — Page 34
Chemistry 12
Unit V Electrochemistry
5.
S3.OX NUM & REDOX 013
The oxidation number of zinc in a reaction increases by 2. This indicates that
A.
zinc is reduced and loses 2 electrons.
B.
zinc is reduced and gains 2 electrons.
C.
zinc is oxidized and loses 2 electrons.
D.
zinc is oxidized and gains 2 electrons.
6.
S4.HALF–RXN SERIES 022
A solution containing Pd2+ reacts spontaneously with Ga to produce Pd and Ga3+ . However, a solution
containing Pd2+ does not react with Pt. The metals, in order of increasing strength as reducing agents, are
A.
Pt < Pd < Ga
B.
Pt < Ga < Pd
C.
Ga < Pt < Pd
D.
Ga < Pd < Pt
7.
S5.OA/RA STRENGTHS 034
Which of the following is the weakest oxidizing agent?
A.
Cl2
B.
Al3+
C.
Sn2+
D.
acidified Cr2O72–
8.
S6.SPONTANEOUS RXNS 042
Which of the following could react spontaneously with Ag metal?
A.
Cl–
B.
Fe2+
C.
acidified SO42–
D.
acidified NO3–
— Chemistry 12 Provincial Exam Review — Page 35
Chemistry 12
T:
OXIDATION REDUCTION (Balancing Redox Equations)
T1.
T2.
T3.
T4.
T5.
T6.
9.
Unit V Electrochemistry
balance a half–reaction in solution (acid, base, neutral)
balance a net ionic redox reaction in acid and base solution
write the equations for reduction and oxidation half–reactions given a redox reaction
identify reactants and products for several redox reactions performed in a laboratory and balance the
equations
select a suitable reagent to be used in a redox titration in order to determine the concentration of a
species
determine the concentration of a species by performing a redox titration
T1.BALANCE HALF–RXN 028
Which of the following is the balanced half–reaction for
N2O → NH3OH+
A.
B.
C.
D.
10.
(acidic)
N2O + 4H+ + 3e– → NH3OH+
N2O + 3H+ + H2O → NH3OH+ + 2e–
N2O + 6H+ + H2O → 2NH3OH+ + 4e–
N2O + 6H+ + H2O + 4e– → 2NH3OH+
T1.BALANCE HALF–RXN 026
Consider the following half–reaction in a basic solution:
Ag2O3 → AgO
(basic)
The balanced half–reaction is
A.
Ag2O3 + 4H+ + 4e– → AgO + 2H2O
B.
Ag2O3 + 2H+ + 2e– → 2AgO + H2O
C.
Ag2O3 + H2O + 2e– → 2AgO + 2OH–
D.
Ag2O3 + 2H2O + 4e– → AgO + 4OH–
11.
T2.BALANCE REDOX 024
Balance the following redox equation: (4 marks)
ClO3– + S2O32–
12.
S4O62– + Cl–
(acidic)
T2.BALANCE REDOX 019
Balance the following redox reaction in basic solution. (5 marks)
SO32–
13.
→
+
MnO4–
→
SO42–
+
MnO2
(basic)
T3.HALF–RXNS
013
Consider the following redox reaction which occurs in a lead–acid storage cell:
PbO2(s) + Pb(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)
The balanced reduction half–reaction is
A.
Pb → Pb2+ + 2e–
B.
Pb + SO42– → PbSO4 + 2e–
C.
2H2SO4 + 2Pb + 2e– → 2PbSO4 + 2H2O
D.
PbO2 + 4H+ + SO42– + 2e– → PbSO4 + 2H2O
— Chemistry 12 Provincial Exam Review — Page 36
Chemistry 12
Unit V Electrochemistry
14.
T4.LAB REDOX RXNS 010
What occurs when a piece of Zn is placed in 1.0 M Cu(NO3)2 ?
A.
[Cu2+] decreases
B.
[Zn2+] decreases
C.
[NO3–] increases
D.
no change occurs
15.
T5.REAGENT FOR TITRA015
Which of the following could be titrated using acidified MnO4– ions?
A.
Na+
B.
IO3–
C.
SO42–
D.
H 2O 2
16.
T6.[] REDOX TITRAT'N018
A titration is performed to determine the concentration of Fe2+ in 25.00 mL of an FeSO4 solution. It
requires 22.52 mL of 0.015 M KMnO4 to reach the equivalence point according to the following equation:
MnO4– + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O
Calculate the [Fe2+] . (4 marks)
— Chemistry 12 Provincial Exam Review — Page 37
Chemistry 12
U:
Unit V Electrochemistry
OXIDATION REDUCTION (Electrochemical Cells)
U1.
U2.
U3.
U4.
U5.
U6.
U7.
U8.
U9.
U10.
U11.
define, construct and label the parts of an electrochemical cell
identify the half–reactions that take place at each electrode
predict the direction of movement of each type of ion in the cell
predict the direction of flow of electrons in an external circuit
predict which electrode will increase in mass and which will decrease in mass as the cell operates
predict the voltage of the cell when equilibrium is reached
assign voltages to the reduction half–reactions of oxidizing agents by comparison of several cells
describe the significance of the E° of an electrochemical cell
predict the voltage (E° ) of an electrochemical cell using the table of standard reduction half–cells
predict the spontaneity of the forward or reverse reaction from the E° of a redox reaction
describe how electrochemical concepts can be used in various practical applications
U1.PARTS OF ECELL 041
Use the following cell to answer questions 17 and 18.
17.
U3.ION MIGRATION 012
Which of the following represents the relationship between [NO3–] and the mass of the Cu electrode in the
complete cell as it operates?
B.
A.
C.
D.
— Chemistry 12 Provincial Exam Review — Page 38
Chemistry 12
Unit V Electrochemistry
18.
U9.PREDICT E°
030
The E° for the above cell is
A.
–1.10 Volts
B.
–0.42 Volts
C.
+0.42 Volts
D.
+1.10 Volts
19.
U1.PARTS OF ECELL 039
Draw and label an electrochemical cell using a copper anode and having an E° value > 1.00 V. (2 marks)
U5.ELECTRODE CHANGE 022
Use the following diagram to answer questions 20 to 22.
20.
U5.ELECTRODE CHANGE 023
In the above electrochemical cell, how do the mass of the anode and the [Ag+] change as the cell operates?
A.
B.
C.
D.
Mass of the Anode
decreases
increases
decreases
no change
21.
U2.ELECTRODE RXNS 023
What is the overall cell reaction?
A.
2Ag + Sn2+ → Sn + 2Ag+
B.
2Ag + Sn → Sn2+ + 2Ag+
C.
2Ag+ + Sn2+ → Sn + 2Ag
D.
2Ag+ + Sn → Sn2+ + 2Ag
22.
U9.PREDICT E°
038
What is the value of E° for the cell?
A.
–0.94 V
B.
–0.66 V
C.
+0.66 V
D.
+0.94 V
[Ag+]
increases
increases
decreases
decreases
— Chemistry 12 Provincial Exam Review — Page 39
Chemistry 12
23.
Unit V Electrochemistry
U4.ELECTRON FLOW 006
The direction of electron flow in an electrochemical cell is from
A.
anode to cathode through the external wire.
B.
cathode to anode through the external wire.
C.
anode to cathode through the external wire and back through the salt bridge.
D.
cathode to anode through the external wire and back through the salt bridge.
U3.ION MIGRATION 009
Use the following diagram to answer questions 24 and 25.
24.
U3.ION MIGRATION 010
Which of the following statements apply to this electrochemical cell?
I. Electrons flow through the wire toward the copper electrode.
II. The copper electrode increases in mass.
III. Anions move toward the Zn half cell.
A.
B.
C.
D.
I and II only
I and III only
II and III only
I, II and III
25.
U6.EQUIL'M VOLTAGE 006
At equilibrium, the voltage of the cell above is
A.
–1.10 V
B.
0.00 V
C.
+0.42 V
D.
+1.10 V
26.
U7.REDUCTION POTENTI003
Which of the following statements would be correct if the zinc half–cell had been chosen as the standard
instead of the hydrogen half–cell?
A.
The reduction potentials of all half–cells would remain unchanged.
B.
The reduction potentials of all half–cells would increase by 0.76 V.
C.
The reduction potentials of all half–cells would have positive values.
D.
The reduction potential of the hydrogen half–cell would decrease by 0.76 V.
— Chemistry 12 Provincial Exam Review — Page 40
Chemistry 12
27.
Unit V Electrochemistry
U8. SIGNIFICANCE E° 004
Which of the following affects the potentials of electrochemical cells?
I. species used as oxidizing agent
II. temperature
III. concentration of reactants
A.
B.
C.
D.
28.
I and II only.
II and III only.
I and III only.
I, II and III.
U10.SPONTANEITY E° 008
Consider the following equation:
Cd2+ + 2I–
→
← Cd + I2
E°cell = –0.94 V
What is E° for the reduction of Cd2+ ?
A.
–0.40 V
B.
–0.14 V
C.
+0.14 V
D.
+0.40 V
29.
U11.APPLICATIONS 022
Consider the following:
I. electrolysis of water
II. electroplating of copper
III. rusting of iron
Which of the above involve non–spontaneous redox reactions?
A.
I and II only
B.
I and III only
C.
II and III only
D.
I, II and III
— Chemistry 12 Provincial Exam Review — Page 41
Chemistry 12
V:
OXIDATION REDUCTION (Corrosion)
V1.
V2.
V3.
V4.
30.
Unit V Electrochemistry
describe the conditions necessary for corrosion to occur
analyse the process of metal corrosion in electrochemical terms
suggest several methods of preventing or inhibiting corrosion of a metal
describe and explain the principle of cathodic protection
V1.CORROSION
007
Which of the following must be present to produce rust by the corrosion of iron?
I. water
II. oxygen
III. salt
A.
B.
C.
D.
I only
II only
I and II only
I, II and III
31.
V2.METAL CORROSION 005
What happens to iron as it corrodes?
A.
It loses electrons and is reduced.
B.
It gains electrons and is reduced.
C.
It loses electrons and is oxidized.
D.
It gains electrons and is oxidized.
32.
V3.PREVENT CORROSION012
Describe two chemically different methods of preventing the corrosion of iron. Explain how each method
works. (3 marks)
33.
V4.CATHODIC PROTECT 014
Which of the following metals could be used to cathodically protect iron?
A.
tin
B.
lead
C.
zinc
D.
copper
— Chemistry 12 Provincial Exam Review — Page 42
Chemistry 12
W:
Unit V Electrochemistry
OXIDATION REDUCTION (Electrolytic Cells)
W1.
W2.
W3.
W4.
W5.
W6.
W7.
W8.
define electrolysis and electrolytic cell
design and label the parts of an electrolytic cell capable of electrolyzing an aqueous salt (use of over–
potential effect not required)
predict the direction of flow of all ions in the cell
write the half–reaction occurring at each electrode
demonstrate the principles involved in simple electroplating
construct an electrolytic cell capable of electroplating an object
describe the electrolytic aspects of metal refining processes
draw and label the parts of an electrolytic cell used for electrolysis of a molten binary salt
34.
W1.DEFINE ELY
006
The process of applying an electric current through a cell to produce a chemical change is called
A.
corrosion.
B.
ionization.
C.
hydrolysis.
D.
electrolysis.
35.
W2.PARTS OF ELY–CELL010
Consider the following operating cell:
Which of the following describes the cell above?
A.
B.
C.
D.
Electrode #1
anode
anode
cathode
cathode
Electrode #2
cathode
cathode
anode
anode
Gas Produced
H2(g)
O2(g)
H2(g)
O2(g)
— Chemistry 12 Provincial Exam Review — Page 43
Chemistry 12
Unit V Electrochemistry
W4.ELECTRODE RXNS 045
Use the following diagram to answer questions 36 and 37.
36.
W4.ELECTRODE RXNS 046
What reaction occurs at the cathode?
A.
2I– → I2 + 2e–
B.
Cu2+ + 2e– → Cu
C.
H2O → 1/2O2 + 2H+ + 2e–
D.
2H2O + 2e– → H2 + 2OH–
37.
W3.ION MIGRATION 005
What happens to the [I–] in the operating cell?
A.
[I–] increases overall.
B.
[I–] decreases overall.
C.
[I–] remains constant overall.
D.
[I–] decreases near the anode and increases near the cathode.
38.
W5.ELECTROPLATING 010
Which of the following are necessary for electroplating to occur using an electrolytic cell?
I. Two electrodes
II. A metal ion being reduced
III. A direct current power source
A.
B.
C.
D.
I and II only.
I and III only.
II and III only.
I, II, and III.
39.
W6.ELECTROPLATE CELL008
Draw and label a simple electrolytic cell capable of electroplating an inert electrode with silver. (2 marks)
40.
W7.METAL REFINING 005
Draw a diagram of an operating electrolytic cell used to extract pure lead from an impure lead sample.
Identify the electrolyte and the material used for the anode. (3 marks)
41.
W8.ELY BINARY SALT 007
A.
Draw and label the parts of an electrolytic cell that can be used for the electrolysis of molten NaCl
(1.5 marks)
B.
Write the half–reaction that occurs at the cathode. (1 mark)
C.
Identify the product at the anode. (0.5 mark)
— Chemistry 12 Provincial Exam Review — Page 44
Chemistry 12
Answers to Questions
ANSWERS TO QUESTIONS
UNIT I REACTION KINETICS
Q
SOURCE
ANS
1.
January - June 2003
C
2.
August 2002
C
3.
January - June 2003
B
4.
January - June 2003
WR
5.
January - August 2001
B
6.
January - June 2003
C
7.
COMMENTS
A.
195.45 g - 188.15 g
= 1.83 g/min
6.0 min - 2.0 min
B.
3 Cu 63.5 g
200.00 – 189.90 g NO
x
x
= 32.1 g
2 NO
mol
30.0 g/mol
WR
Food spoilage → refrigerate
Body enzymes → catalysts
8.
January - June 2003
D
9.
January - June 2003
WR
10.
Jan - Aug 1999
D
11.
Jan - Aug 1999
B
12.
Jan - Aug 1999
C
13.
January - August 2001
C
14.
January - August 2001
B
15.
January - August 2001
B
16.
Solutions react more quickly than solid because the particles can mix
thoroughly (more surface area) and the particles have greater mobility
(greater number of collisions).
A
17.
January - August 2001
D
18.
Jan - Aug 1999
WR
19.
January - June 2003
WR
20.
January - August 2001
B
21.
January - June 2003
A
22.
January - August 2001
WR
A.
B.
2NO + 2H2 → N2 + 2H2O
N2O2 or N2O
A.
B.
2H2O2 → 2H2O + O2
A catalyst is a substance that speeds up a chemical reaction by
providing a lower energy pathway.
Catalyst in mechnism is Fe3+
— Chemistry 12 Provincial Exam Review — Page 45
Chemistry 12
Answers to Questions
UNIT II DYNAMIC EQUILIBRIUM
Q
SOURCE
ANS
1.
January - August 2002
C
2.
January - August 2000
D
3.
January - August 2000
C
4.
January - August 2002
D
5.
April 1997
A
6.
January - August 2000
C
7.
January - August 1999
C
8.
August 2001
9.
January - August 2001
10.
11.
COMMENTS
A.
[HI]2
(1.0)2
=
= 20
[H2][I2] (0.5)(0.1)
B.
Keq =
A.
C.
As the temperature increases the % yield of NH3 decreases. This
indicates a shift to the left. When energy is on the product side, a shift
to the left will occur when the temperature is increased. If energy is on
the product side the reaction is exothermic.
At 600 °C when pressure is increased, the % yield of NH3 also
increases. According to Le Chatelier’s principle, when the pressure on
an equilibrium is increased, there will be a shift to the side that
produces fewer gas molecules. In this equilibrium, there would be a
shift to produce more NH3 when the pressure is increased.
Low temperature and high pressure. (200 °C and 100 kPa)
D.
Higher temperature → faster reaction rate even though reduced yield.
C
D
June 1995 Scholarship
WR
B.
Lower pressure → safer and easier to attain.
12.
January - August 2002
A
13.
January - August 2002
A
14.
June 1994 Provincial
B
15.
January - August 2000
D
16.
January - August 2002
A
— Chemistry 12 Provincial Exam Review — Page 46
Chemistry 12
17.
January - August 2002
Answers to Questions
WR
H2 + I2
January - August 2002
2HI
0.100 M
0.100 M
0.00 M
–x
–x
+2x
0.1 – x
0.1 – x
2x
0.0200 M
0.0200 M
0.160
Keq =
18.
→
←
[HI]2
(0.160)2
=
= 64.0
[H2][I2] (0.0200)2
WR
H2 + Br2
→
←
2HBr
0.020 M
0.020 M
0.00 M
–x
–x
+2x
0.020 – x
0.020 – x
2x
Keq =
[HBr]2
(2x)2
⇒
= 12.0
[H2][Br2]
(0.020–x)2
x = 0.0127 M
[HBr] = 2x = 0.0254 M
19.
January - August 2002
WR
Q=
[NO2]2 (0.010)2
=
= 0.0033
[N2O4] (0.030)
Q (0.0033) < Keq (0.0095) Equilibrium shifts right, forward direction.
— Chemistry 12 Provincial Exam Review — Page 47
Chemistry 12
Answers to Questions
UNIT III SOLUBILITY EQUILIBRIA
Q
SOURCE
ANS
1.
January - August 2002
C
2.
April - June 2001
D
3.
April - June 2001
D
4.
January - August 2002
B
5.
January - August 2002
C
6.
January - August 2002
D
7.
January - August 2002
B
8.
April - June 2001
B
9.
January - August 2002
D
10.
January - August 2002
B
11.
January - August 2002
A
12.
January - August 2002
WR
13.
January - August 2002
D
14.
January - August 2002
B
15.
Aug 1998 – Aug 1999
D
16.
January - August 2002
C
17.
January - August 2002
B
18.
January - August 2002
B
19.
January - August 2002
A
20.
January - August 2002
C
21.
Aug 1998 - Aug 1999
C
22.
January - August 2002
C
23.
January - August 2002
C
24.
January - August 2002
WR
COMMENTS
A.
Pb(NO3)2(aq) + 2KCl(aq) → 2KNO3(aq) + PbCl2(s)
B.
Pb2+ + 2NO3– + 2K+ + 2Cl– → 2K+ + 2NO3– + PbCl2(s)
C.
Pb2+ + 2Cl– → PbCl2(s)
mol AgNO3 = 0.03200 L x 0.125 M = 0.00400 mol Ag+
1 Cl–
mol of Cl– = 0.00400 mol Ag+ x
= 0.00400 mol Cl–
1 Ag+
[Cl–] =
0.00400 mol
= 0.200 M
0.02000 L
— Chemistry 12 Provincial Exam Review — Page 48
Chemistry 12
Answers to Questions
UNIT IV ACIDS AND BASES
Q
SOURCE
ANS
1.
January 2001
C
2.
January - August 1999
A
3.
January - August 2002
B
4.
January 2001
B
5.
January 1995 Provincial
D
6.
April 1994
B
7.
January - August 2002
C
8.
January - August 2000
C
9.
April 2001 - August 2001
D
10.
January - August 2000
WR
COMMENTS
A.
B.
11.
April 2001 - August 2001
B
12.
January - August 2002
C
13.
April 2001 - August 2001
A
14.
January - August 2002
B
15.
April 2001 - August 2001
B
16.
January 2001
A
17.
January - August 1999
D
18.
January - August 2002
A
19.
January - August 1999
WR
20.
January - August 2002
A
21.
January 1995 Provincial
B
22.
Bronsted–Lowry conjugate acid–base pairs are two substances that
differ from each other by one proton (H+).
H3PO4 and H4PO4–
H3O+ is the strongest acid that can exist in water because any acid
stronger will completely dissociate to form H3O+ ions.
A
23.
April 2001 – Aug 2001
C
24.
January - August 2002
D
25.
January - August 1999
C
26.
January - August 2002
B
27.
January - August 2002
C
28.
January - August 2002
D
29.
January - August 1999
D
30.
August 1996 Provincial
A
31.
January - August 2002
C
32.
January - August 2002
D
— Chemistry 12 Provincial Exam Review — Page 49
Chemistry 12
Answers to Questions
33.
January - August 2000
D
34.
January - August 2002
D
35.
January - August 2002
WR
H2CO3 + H2O
→
←
HCO3– + H3O+
0.35 M
—
0.00 M
0.00 M
–x
—
+x
+x
0.35 – x
≈ 0.35
—
x
x
Ka =
[HCO3–][H3O+]
x2
=
= 4.3 x 10–7
[H2CO3]
0.35
x = 3.9 x 10–4 M
pH = –log (3.9 x 10–4) = 3.41
36.
January - August 2000
WR
CH3NH2 + H2O
→
←
CH3NH3+ + OH–
0.0200 M
—
0.00 M
0.00 M
–x
—
+x
+x
0.0200 – x
—
x
x
0.0175
—
0.00251 M
0.00251 M
pOH = 14.00 – 11.40 = 2.60
[OH–] = antilog (–2.60) = 0.00251 M
[CH3NH3+][ OH–] (0.00251)2
Kb =
=
= 3.6 x 10–4
[ CH3NH2]
0.0175
37.
January - August 2002
B
38.
January - August 2002
WR
H2S + H2O
→
←
HS– + H3O+
1.0 M
—
0.00 M
0.00 M
–x
—
+x
+x
1.0 – x
—
x
x
≈ 1.0
—
1.79 x 10–4
1.79 x 10–4
[H3O+] = antilog (–3.75) = 1.79 x 10–4 M
[HS–][ H3O+] (1.79 x 10–4)2
Ka =
=
= 3.2 x 10–8
1.0
[ H2S]
39.
January - August 2002
A
40.
January - August 2002
B
41.
January - August 2002
B
42.
January - August 2002
A
— Chemistry 12 Provincial Exam Review — Page 50
Chemistry 12
Answers to Questions
43.
January - August 2002
B
44.
January - August 2000
A
45.
January - August 2002
B
46.
January - August 2002
B
47.
January - August 2002
D
48.
January - August 2002
C
49.
January - August 2002
A
50.
January - August 2002
D
51.
January - August 2000
WR
52.
April - August 2001
WR
2H+ + SO42– + Sr2+ + 2OH– → 2H2O(l) + SrSO4(s)
[H3O+] =
(15.0 mL)(0.50 M)
= 0.15 M
(50.0 mL)
[OH–] =
(35.0 mL)(1.0 M)
= 0.70 M
(50.0 mL)
[OH–] = 0.70 – 0.15 = 0.55 M
pOH = – log (0.55) = 0.26
pH = 14.00 – 0.26 = 13.74
53.
January - August 2002
WR
A.
B.
54.
January 1993
C
55.
January - August 2002
D
56.
January - August 2002
B
57.
D
58.
WR
A.
B.
59.
August 1996 Provincial
C
60.
April 2001 - Aug 2001
A
61.
April 2001 - Aug 2001
A
62.
January - August 2002
B
63.
January - August 2002
D
64.
WR
When a strong acid is titrated with a strong base, the salt which is
formed in a neutral salt so the pH of the solution is 7.00. When a weak
acid is titrated with a strong base, the salt which is formed in a basic
salt so the pH of the solution is > 7.00.
Phenolphthalein
∅
NH3 + H2O ♦ NH4+ + OH–
When a strong base is added to the buffer, the conjugate acid, NH4+,
reacts with the excess OH– and the equilibrium shifts left to compensate
and returns the [OH–] to a level close to what it was initially.
SO2 dissolves into the rain clouds to produce acid rain which then has
environmental impacts such as destroying plant and animal habitats by
upsetting the pH balance.
SO2 + H2O → H2SO3
— Chemistry 12 Provincial Exam Review — Page 51
Chemistry 12
Answers to Questions
UNIT V ELECTROCHEMISTRY
Q
SOURCE
ANS
1.
January - April 2002
B
2.
January - August 2000
D
3.
January - August 1999
B
4.
June - August 2002
B
5.
January - August 2000
C
6.
January - April 2002
A
7.
June - August 2002
B
8.
June - August 2002
D
9.
January - April 2002
D
10.
January - August 2000
C
11.
June - August 2002
WR
COMMENTS
5+
2+
2.5+
6H+ + ClO3– + 6S2O32–
12.
January - August 2001
WR
4+
+
2H + 3SO3
13.
January - August 2000
D
14.
January - August 2001
A
15.
June - August 2002
D
16.
January - April 2002
WR
7+
2–
+ 2MnO4
January - August 2001
B
18.
January - August 2001
D
19.
January - August 2000
WR
6+
–
4+
→ 3SO4
2–
+ 2MnO2 + H2O
0.02252 L x 0.015 M = 3.38 x 10–4 mol
5Fe2+
3.38 x 10–4 mol MnO4– x
= 1.69 x 10–3 mol Fe2+
MnO4–
[Fe2+] =
17.
1–
→ 3S4O62– + Cl– + 3H2O
1.69 x 10–3 mol
= 0.068 M
0.0250 L
Cu → Cu2+ + 2e–
–0.34
Au3+ + 3e– → Au
+1.50
1 M NaNO3
Au
Cu
1 M Au(NO3)3(aq)
20.
June - August 2002
C
21.
June - August 2002
D
— Chemistry 12 Provincial Exam Review — Page 52
1 M Cu(NO3)2(aq)
Chemistry 12
Answers to Questions
22.
June - August 2002
D
23.
August 1996
A
24.
January - August 1999
D
25.
January - August 1999
B
26.
April 1996
B
27.
January - August 1999
D
28.
June - August 2002
A
29.
January - April 2002
A
30.
January - August 1999
C
31.
June - August 2002
C
32.
June - August 2002
WR
33.
January - April 2002
C
34.
January - August 2000
D
35.
January - August 2000
A
36.
June - August 2002
B
37.
June - August 2002
B
38.
January - August 2000
D
39.
January - August 2000
WR
1.
Paint the iron → prevents H2O and O2 from coming in contact with Fe.
2.
Cathodic protection → attaching a stronger reducing agent such as Zn.
Stronger reducing agent will oxidize first leaving Fe intact.
–
D.C
.
+
Ag(s)
Inert Electrode
1 M AgNO3(aq)
40.
June - August 2002
WR
–
D.C
.
+
Impure Lead
Pure Lead
1 M Pb(NO3)2(aq)
— Chemistry 12 Provincial Exam Review — Page 53
Chemistry 12
41.
Answers to Questions
WR
A.
–
D.C
.
+
Inert Electrode
(Pt)
Inert Electrode
NaCl(l)
B.
Na+ + e– → Na(s)
C.
2Cl– → Cl2(g) + 2e– (anode reaction)
Cl2(g) is produced at the anode.
— Chemistry 12 Provincial Exam Review — Page 54