South Pasadena • AP Chemistry
Name
Semester 1 Final Exam
Period
Date
PRACTICE
TEST
Part 1 – Multiple Choice
You should allocate 45 minutes to finish this portion of the test. No calculator should be used. A periodic table
and data table will be provided. Select the answer that best responds to each question.
1. A 12.0 g sample of a hydrate of cobalt(II)
chloride, CoCl2·nH2O, was heated, driving off all
the water. The mass of the remaining solid CoCl2
(molar mass 130) was 6.5 g. What is the formula
of the hydrate?
(A) CoCl2·3H2O
(B) CoCl2·6H2O
(C) CoCl2·9H2O
(D) CoCl2·12H2O
2. A 50.0 mL sample of a Cu(NO3)2 solution, which
is a blue solution, and a NaOH solution, which is
colorless, are combined forming a blue Cu(OH)2
(molar mass 81.5) precipitate. The 0.815 g
precipitate is collected through filtration, and the
colorless filtrate is set apart. What is the
concentration of the original Cu(NO3)2 solution?
(A) 0.10 M
(B) 0.20 M
(C) 0.40 M
(D) The concentration of Cu(NO3)2 solution
cannot be determined because it was the
excess reactant.
3. A student prepared various solutions of CuBr2 and
measured their absorbance at 635 nm.
Absorbance
Concentration
0.0030
2 × 10−3 M
0.0075
5 × 10−3 M
0.0150
1 × 10−2 M
What is the concentration of a CuBr2 solution with
an absorbance of 0.0135?
(A) 4 × 10−4 M
(B) 3 × 10−3 M
(C) 9 × 10−3 M
(D) 2 × 10−2 M
4. A student placed 0.0243 g Mg in 50.0 mL of a
0.200 M HCl solution, producing H2 gas according
to the following reaction:
Mg (s) + 2 HCl (aq) → H2 (g) + MgCl2 (aq)
What is the volume of H2 (g) produced at 300 K
and 0.800 atm?
(A) 10.0 mL
(B) 30.0 mL
(C) 100. mL
(D) 300. mL
5. A student placed a sample of water under a bell
jar, and used a vacuum pump to remove the air
under the jar, lowering the pressure. The student
claims that the water is boiling as bubbles are seen
in the sample of water. Which of the following
describes the student’s claim?
(A) This is a chemical process as the hydrogen
bonds in the water sample are being broken to
form the gas.
(B) This is a chemical process as water is breaking
covalent bonds to become hydrogen and
oxygen gases.
(C) This is a physical process as inter-particle
attractions between water molecules are
weakening to become a gas.
(D) This is a physical process as the water
molecules move faster to become a gas.
6. A sample of solid calcium carbonate was added to
vinegar. What is the net ionic equation for this
reaction?
(A) CaCO3 (s) + 2 H+ (aq) →
H2CO3 (aq) + Ca2+ (aq)
(B) CaCO3 (s) + 2 HC2H3O2 (aq) →
H2CO3 (aq) + Ca2+ (aq) + 2 C2H3O2− (aq)
(C) CaCO3 (s) + 2 HC2H3O2 (aq) → H2O (ℓ) +
CO2 (g) + Ca2+ (aq) + 2 C2H3O2− (aq)
(D) CO32− (aq) + 2 H+ (aq) → H2O (ℓ) + CO2 (g)
9. The following kinetics data was collected for the
reaction 2 A → 3 B.
Time
[A]
0s
0.100 M
0.50 s
0.070 M
What is the rate of formation of B during this
time?
(A) 0.02 M/s
(B) 0.06 M/s
(C) 0.09 M/s
(D) 0.14 M/s
10. Consider the following rate data for the reaction
2A + 2 B → C.
Trial
[A] (M)
[B] (M)
1
0.20
0.20
2
0.40
0.40
3
0.60
0.20
What is the rate law of the reaction?
(A) Rate = k[A]1[B]1
(B) Rate = k[A]2[B]0
(C) Rate = k[A]2[B]1
(D) Rate = k[A]3[B]2
Rate (M/s)
R
4R
9R
Time
[A]
ln[A]
1/[A]
0.00 s
1.0 M
0.0
1.0 M−1
0.20 s
0.4 M
−1.0
2.5 M−1
What is the magnitude of the rate constant, k?
(A) 2.5
(B) 3.0
(C) 5.0
(D) 12.5
12. The mechanism for the reaction X → Y is given:
(1) X → A
(2) A → B
(3) B → Y
The potential energy graph for this process is
shown below.
Potential energy
8. What is the sum of the coefficients when the
following redox reaction is balanced with lowest
whole number coefficients using the half-reaction
method?
MnO4− (aq) + H2O2 (aq) → Mn2+ (aq) + O2 (g)
(A) 7
(B) 14
(C) 21
(D) 28
11. Data for the first order reaction A → B is shown
below.
Reaction progress
List the substances in increasing potential energy.
(A) A < X < B < Y
(B) X < A < B < Y
(C) Y < B < A < X
(D) Y < B < X < A
13. Below, the 1/[A] vs. time graph for the second
order reaction 2A → B.
Expt 1
1/[A]
7. In which of the following process does nitrogen
have the greatest increase in its oxidation state?
(A) N2 → NO2
(B) NH3 → NO3−
(C) NO → N2O5
(D) N2H4 → N2O4
Expt 2
Time
Which of the following changes could have been
made between experiments 1 and 2?
(A) Experiment 1 had a catalyst in the reaction
mixture, while Experiment 2 did not.
(B) Experiment 1 had a higher initial
concentration of A than Experiment 2.
(C) Experiment 1 had a lower initial concentration
of B than Experiment 2.
(D) Experiment 1 was performed at a lower
temperature than Experiment 2.
14. Consider the following proposed mechanism:
(1) 2 NO2  N2O4
Fast Equil.
(2) N2O4 + H2O → HNO2 + HNO3
Slow
What is the rate law for this mechanism?
(A) Rate = k[H2O]
(B) Rate = k[NO2][H2O]
(C) Rate = k[NO2]2
(D) Rate = k[NO2]2[H2O]
15. The value of Ka for HF is 7 × 10−4, and the Ksp for
PbF2 is 4 × 10−8. What is the approximate value of
Kc for the following reaction?
2 HF (aq) + Pb2+ (aq)  PbF2 (s) + 2 H+ (aq)
(A) 2.8 × 10−11
(B) 1.8 ×10−4
(C) 8.0 ×10−2
(D) 1.2 × 101
16. Consider the following reaction:
180 kJ + CaCO3 (s) CaO (s) + CO2 (g)
Which of the following changes would result in
the greatest amount of CaO (s)?
(A) Adding CaCO3 (s) to the reaction mixture.
(B) Decrease the volume of the container.
(C) Raising the temperature.
(D) None of the changes would increase the
amount of CaO (s).
17. Consider the following reversible reaction:
2 SO2 (g) + O2 (g)  2 SO3 (g)
Samples of SO2 (g), O2 (g), and SO3 (g) are all
placed in a container so each has a partial pressure
of 1.0 atm. After equilibrium has been reached,
the partial pressure of SO3 (g) is 1.33 atm. What is
the value of Kp for this process?
(A) 3.0
(B) 4.8
(C) 6.0
(D) 12
18. A saturated solution of which of the following has
the highest pH?
(A) Hg(OH)2
Ksp = 3 × 10−26
(B) Sn(OH)2
Ksp = 3 × 10−27
(C) Cr(OH)3
Ksp = 7 × 10−31
(D) Al(OH)3
Ksp = 2 × 10−32
19. BaSO4 is expected to be more soluble in which of
the following solutions than in pure water?
(A) BaCl2
(B) HCl
(C) KCl
(D) KHSO4
20. What is the maximum amount of KCl can be
added to a 50.0 mL sample of a 0.200 M Pb(NO3)2
solution before a precipitate forms? The Ksp of
PbCl2 is 1.3 ×10−5.
(A) 0.00040 mol
(B) 0.0020 mol
(C) 0.0080 mol
(D) 0.010 mol
21. Which one of the following has a pH > 7?
(A) 0.10 M CH3NH2
(B) 0.10 M CH3OH
(C) 0.10 M NaClO3
(D) 0.10 M NH4Cl
22. What is the pH of a solution when 900 mL of
water is added to 100 mL of a 0.010 M NaOH
solution?
(A) 2.0
(B) 3.0
(C) 11.0
(D) 12.0
23. A 0.20 M solution of a weak acid has a pH = 3.0.
What is the value of Ka for this solution?
(A) 2.0 × 10−7
(B) 5.0 × 10−6
(C) 2.0 × 10−4
(D) 5.0 × 10−3
24. 0.100 M of solutions of which of the following
substances has the highest pH? The Ka of HClO2
= 1.2 × 10−2, and the Kb of NH3 is 1.8 × 10−5.
(A) HClO2
(B) KClO2
(C) NH4Cl
(D) NH4ClO2
25. Consider a 0.100 M solution of KHSO4. Which of
the following has the greatest concentration in this
solution?
(A) H2SO4
(B) HSO4−
(C) OH−
(D) SO42−
28. What volume of a 0.15 M KOH solution is needed
to titrate a 40.0 mL sample of a 0.75 M HClO2 to
its half equivalence (halfway) point?
(A) 50 mL
(B) 100 mL
(C) 150 mL
(D) 200 mL
26. Which of the following lists acids in decreasing
acid strength?
(A) HBrO2 > HBrO > HIO > HI
(B) HI > HBrO2 > HBrO > HIO
(C) HI > HIO > HBrO > HBrO2
(D) HIO > HBrO > HBrO2 > HI
29. A 0.100 M HNO2 solution is prepared, and NaOH
is added until the pH of the mixture is 5.00. How
do the concentrations of the species compare? The
pKa of HNO2 is 3.40.
(A) H+ = NO2− > HNO2
(B) HNO2 = NO2− > H+
(C) HNO2 > H+ = NO2−
(D) NO2− > HNO2 > H+
27. 200 mL of which of the following solutions would
require the greatest volume of 0.050 M NaOH to
reach a pH 7?
(A) 0.100 mol HNO2
(B) 0.100 mol HNO2 + 0.100 mol NaNO2
(C) 0.100 mol HNO3
(D) 0.100 mol NaNO3
30. Which of the following mixtures would produce a
solution with the greatest buffering capacity?
(A) 0.100 mol HNO2 + 0.050 mol NaNO2
(B) 0.100 mol HNO2 + 0.050 mol NaOH
(C) 0.100 mol HNO2 + 0.100 mol NaNO2
(D) 0.100 mol HNO2 + 0.100 mol NaOH
Part 2 – Free Response
You should allocate 50 minutes to finish this portion of the test. You may use a scientific calculator. A periodic
table and data table will be provided. Respond to each part of the questions completely. Be sure to show your
work clearly for questions that involve calculators.
31. A student prepares a 100.0 mL solution of potassium permanganate, KMnO4, using 10.0 mg KMnO4 (molar
mass 158.04 g/mol).
(a) Find the concentration of MnO4− in mol/L.
10.0 mg KMnO4  1 g KMnO4   1 mol KMnO4   1 mol MnO4−  1000 mL
[MnO4−] =
100.0 mL
1000 mg KMnO4 158.04 g KMnO4 1 mol KMnO4  1 L 
−4
= 6.33 × 10 M
(b) The absorbance spectrum over visible light for the solution is shown below.
2.0
Absorbance
1.5
1.0
Color
Wavelengths
Reds
Oranges
Yellows
Greens
Blues
Violets
625-740 nm
590-625 nm
565-590 nm
520-565 nm
440-520 nm
380-440 nm
0.5
0.0
400
i.
500
600
Wavelength (nm)
700
What color is this solution? Justify your answer using the spectrum.
The solution is purple/violet because the green/yellow/orange wavelengths are most absorbed.
ii. The colorimeter that the student is using measures absorbance at 430 nm, 470 nm, 565 nm, and 635
nm, and the student decides to use 565 nm. Is this the best wavelength at which the colorimeter be set
for this experiment? Justify your answer.
Yes, because among those four wavelengths, the absorbance is greatest at 565 nm.
iii. At 565 nm, the absorbance of the KMnO4 solution is found to be 0.800. The student then transfers
5.00 mL of the KMnO4 solution to a test tube, and adds 5.00 mL of distilled water to it. What is the
absorbance of this new solution in the test tube?
The concentration of the solution is cut in half, so the absorbance is also cut in half. A = 0.400.
A student performs an experiment by adding a solution of tin(II) chloride, SnCl2, to the solution of potassium
permanganate, KMnO4, undergoing the following reaction:
… MnO4− (aq) + … Sn2+ (aq) → … Mn2+ (aq) + … Sn4+ (aq)
(c) For the reaction that occurs,
i. Write the balanced net ionic equation for this reduction-oxidation process. Use the method of halfreactions.
Oxidation:
5 Sn2+ → 5 Sn4+ + 10 e−
Reduction:
10 e− + 16 H+ + 2 MnO4− → 2 Mn2+ + 8 H2O
Overall:
16 H+ + 2 MnO4− + 5 Sn2+ → 2 Mn2+ + 5 Sn4+ + 8 H2O
ii. How many electrons are transferred in the equation shown?
10 electrons
(d) The student performs an experiment by observing how the absorbance changes immediately after the
solutions are combined. The data for the experiment and their graphs are given below.
i.
Time (s)
Absorbance
ln(Absorbance)
1/Absorbance
0
1
2
3
4
5
6
7
8
0.8000
0.4852
0.2943
0.1785
0.1083
0.0657
0.0398
0.0242
0.0147
−0.22
−0.72
−1.22
−1.72
−2.22
−2.72
−3.22
−3.72
−4.22
1.25
2.06
3.40
5.60
9.24
15.23
25.11
41.39
68.25
The rate law depends only on [MnO4−] (in other words, it is zero order with respect to the reactants).
Write the rate law for this reaction. Justify your answer.
The reaction is first order with respect to MnO4− because the ln(A) vs. time graph is linear.
Rate = k[MnO4−]1
ii. Find the value of the rate constant, k.
ln (A) – ln (A)0 = −k·t
Use t = 1 s
(−0.72) – (−0.22) = −k(1 s)
k = 0.50 s−1
iii. What is the absorbance of the solution at 10.0 seconds?
ln (A)10 – ln (A)0 = −k·t
ln (A)10 – (−0.22) = −(0.50 s−1)(10 s)
A = e−5.22 = 0.0054
32. Ascorbic acid, HC6H7O6, a weak acid, has an acid-dissociation constant, Ka, 7.9 × 10−5 at 25°C.
a. Write the expression for the reaction for the ascorbate ion, C6H7O6−, in water.
C6H7O6− (aq) + H2O (ℓ)  HC6H7O6 (aq) + OH− (aq)
b. Write the expression of the base dissociation constant, Kb, for C6H7O6− and find its value.
[HC6H7O6][OH−] Kw 1.0 × 10−14
Kb =
=
=
= 1.3 × 10−10
[C6H7O6−]
Ka 7.9 × 10−5
c. Two 100.0 mL solutions of 0.500 M NaC6H7O6 are prepared, and labeled beakers A and B.
i. What is the pH of the solutions?
(x)(x)
= 1.3 × 10−10
0.500 − x
x = [OH−] = 8.0 × 10−6 M
pOH = 5.10
pH = 8.90
ii. 0.010 mol NaOH is added to beaker A. What is the pH of this solution?
0.010 mol
[OH−] =
= 0.100 M
pOH = 1.00
pH = 13.00
0.100 L
iii. 0.010 mol HCl is added to beaker B. What is the pH of this solution?
mol C6H7O6− = (0.100 L)(0.500 M) = 0.050 mol
C6H7O6− + H+ → HC6H7O6
Start
0.050
0.010
0
Change−0.010
−0.010
+0.010
End
0.040
0
0.010
Buffer
−
[C6H7O6 ]
0.040
pH = pKa + log
= −log(7.9 × 10−5) + log
= 4.70
[HC6H7O6]
0.010
d. Vitamin C is a nutritional supplement consisting of ascorbic acid, the ascorbate ion, or both. A particular
product claims to be “buffered,” with the following information from the product label for one tablet:
Vitamin C (as ascorbic acid and sodium ascorbate) 500 mg (HC6H7O6 and C6H7O6−)
Sodium (as sodium ascorbate)
23 mg (Na+)
i. Calculate the moles of ascorbic acid and sodium ascorbate in this tablet.
1 g Na+   1 mol Na+ 
Moles of NaC6H7O6:
23 mg Na+ 
1000 mg Na+ 22.99 g Na+ = 0.00100 mol NaC6H7O6
175.116 g C6H7O6−
−
Mass of C6H7O6−:
0.00100 mol C6H7O6− 
 1 mol C6H7O6−  = 0.175 g C6H7O6
Mass of HC6H7O6:
0.500 g total – 0.175 g C6H7O6− = 0.325 g HC6H7O6
1 mol HC6H7O6 
Moles of HC6H7O6:
0.325 g HC6H7O6 
176.124 g HC6H7O6 = 0.00185 mol HC6H7O6
ii. Find the pH when this tablet is fully dissolved in 250. mL distilled water. Assume that no other
ingredient in the tablet affects the pH of the system, and that volume change due to the tablet is
negligible.
[C6H7O6−]
0.00100
pH = pKa + log
= −log(7.9 × 10−5) + log
[HC6H7O6]
0.00185 = 3.84
iii. Does the pH increase, decrease, or remain the same if two tablets are dissolved? Explain.
[C6H7O6−]
The pH will remain the same because the
does not change.
[HC6H7O6]
e. A second H+ can be removed from ascorbic acid, and has the following dissociation equation:
C6H7O6− (aq) + H2O (ℓ)  C6H6O62− (aq) + H3O+ (aq)
Ka = 1.6 × 10−12
Provide a quantitative explanation for why this is reaction is ignored when determining the pH of a
NaC6H7O6 solution, as in part c.i.
The Kb of C6H7O6− (1.3 × 10−10) is larger than the Ka of C6H7O6− (1.6 × 10−12), so the C6H7O6− is a
stronger base than it is an acid.
33. Consider the salt, Ag2CO3, which has a solubility product, Ksp, of 8.1 × 10−12 at 25°C.
a. Saturated solutions of Ag2CO3 are prepared by adding the salt to pure water.
i. Write the dissociation equation for Ag2CO3 in pure water.
Ag2CO3 (s)  2 Ag+ (aq) + CO32− (aq)
ii. Calculate the molar solubility of Ag2CO3 in pure water.
Ksp = [Ag+]2[CO32−] = 8.1 × 10−12
(2s)2(s) = 8.1 × 10−12
s = 1.3 × 10−4 M
iii. Predict whether the [Ag+] will increase, decrease, or remain the same when the following are added.
 AgClO3
Increase
 HNO3
Increase
 Na2CO3
Decrease
b. The Ksp of AgCl is 1.6 × 10−10 at 25°C.
i. A 1.0 L solution contains 0.100 mol Cl− and 0.100 mol CO32−. As AgNO3 is added to the solution,
will Ag2CO3 or AgCl precipitate first? Explain using calculations.
Ksp
8.1 × 10−12
= 9.00 × 10−6 M
2− =
[CO3 ]
0.100
Ksp 1.6 × 10−10
AgCl:
[Ag+] =
=
= 1.60 ×10−9 M
[Cl−]
0.100
Because AgCl needs smaller [Ag+] to precipitate, it will precipitate first.
Ag2CO3:
[Ag+] =
ii. What is the value of equilibrium constant for the following reaction:
2 AgCl (s) + CO32− (aq)  Ag2CO3 (s) + 2 Cl− (aq)
AgCl (s)  Ag+ (aq) + Cl− (aq)
K1 = [Ag+][Cl−] = 1.6 × 10−10
+
2−
Ag2CO3 (s)  2 Ag (aq) + CO3 (aq)
K2 = [Ag+]2[CO32−] = 8.1 × 10−12
Kc =
(K1)2 (1.6 × 10−10)2
=
= 3.2 × 10−9
(K2) (8.1 × 10−12)
South Pasadena • AP Chemistry
Name
Semester 1 Final Exam
Period
Date
TEST BLUEPRINT
Part 1: Multiple Choice
 Format: 30 questions, four answer choices: (A)-(D)
 Expected time: 45 minutes
 Allowed resources: Periodic Table, Equations and Constants. No calculators.
 Topics:
o Unit 1 – 8 questions
o Unit 2 – 6 questions
o Unit 3 – 6 questions
o Unit 4 – 6 questions
o Unit 5 – 4 questions
Part 2: Free Response
 Format:
o 2 long questions (5-8 parts)
o 1 short question (2-4 parts)
 Expected time: 50 minutes
 Allowed resources: Periodic Table, Equations and Constants, and scientific calculators.
 Topics:
o Unit 1 – 12 points
o Unit 2 – 10 points
o Unit 3 – 10 points
o Unit 4 – 10 points
o Unit 5 – 6 points
Question 32.c with ICE Boxes
i.
WEAK BASE SOLUTION (Strategy: Use Kb and ICE Box)
Molarity
C6H7O6− (aq) + H2O (ℓ)  HC6H7O6 (aq) + OH− (aq)
Initial
0.500
0
0
Change
−x
+x
+x
Equilibrium
0.500 – x
x
x
−
[HC6H7O6][OH ]
(x)(x)
Kb =
= 0.500 − x = 1.3 × 10−10
[C6H7O6−]
x = [OH−] = 8.0 × 10−6 M
pOH = −log(8.0 × 10−6) = 5.10
pH = 14.00 – 5.10 = 8.90
ii. WEAK BASE + STRONG BASE SOLUTION (Strategy: Strong base wins!)
0.010 mol
[OH−] = 0.100 L = 0.100 M
Molarity
C6H7O6− (aq) + H2O (ℓ)  HC6H7O6 (aq) + OH− (aq)
Initial
0.500
0
0.100
Change
−x
+x
+x
Equilibrium 0.500 – x
x
0.100 + x
[HC6H7O6][OH−] (x)(0.100 + x)
Kb =
= 0.500 − x = 1.3 × 10−10
x = 6.5 × 10−10 M
[C6H7O6−]
[OH−] = 0.100 M
pOH = −log(0.100) = 1.00
pH = 14.00 – 1.00 = 13.00
iii. WEAK BASE + STRONG ACID SOLUTION (Strategy: Stoichiometry first.)
mol C6H7O6− = (0.100 L)(0.500 M) = 0.050 mol
Moles
C6H7O6− (aq) + H+ (aq)  HC6H7O6 (aq)
Initial
0.050
0.010
0
Change
−0.010
−0.010
+0.010
End
0.040
0
0.010
This is a buffer solution because solution contains HC6H7O6 and C6H7O6−.
Option 1: ICE Box
0.010 mol
0.040 mol
[HC6H7O6] = 0.100 L = 0.100 M
[C6H7O6−] = 0.100 L = 0.400 M
Molarity HC6H7O6 (aq)  H+ (aq) + C6H7O6− (aq)
Initial
0.100
0
0.400
Change
−x
+x
+x
Equilibrium 0.100 – x
x
0.400 + x
[H+][C6H7O6−] (x)(0.400 + x)
Ka = [HC H O ] = (0.100 − x) = 7.9 × 10−5
6 7 6
x = [H+] = 1.98 × 10−5
pH = −log(1.98 × 10−5) = 4.70
Option 2: Henderson Hasselbach Equation
[C6H7O6−]
0.040
pH = pKa + log[HC H O ] = −log(7.9 × 10−5) + log0.010 = 4.70


6 7 6