Multiple-Choice Questions
Answer the following questions in 45 minutes. You may not use a calculator. You may use
the periodic table and the equation sheet at the back of this book.
1. A 0.1 molar solution of acetic acid, CH3COOH, has a pH of about:
(A)
(B)
(C)
(D)
1
3
7
10
2.
Using the given information, choose the best answer for preparing a buffer with a pH
of 8.
(A) K2HPO4 + KH2PO4
(B) H3PO4
(C) K2HPO4 + K3PO4
(D) K3PO4
Use the following information for questions 3–4.
3. Which of the following is a solution with an initial KCOOH concentration of 1 M and
an initial K2HPO2 concentration of 1 M?
(A)
(B)
(C)
(D)
a solution with a pH > 7, which is a buffer
a solution with a pH < 7, which is not a buffer
a solution with a pH < 7, which is a buffer
a solution with a pH > 7, which is not a buffer
4. Which of the following is a solution with an initial H3PO2 concentration of 1 M and an
initial KH2PO2 concentration of 1 M?
(A)
(B)
(C)
(D)
a solution with a pH > 7, which is a buffer
a solution with a pH < 7, which is not a buffer
a solution with a pH < 7, which is a buffer
a solution with a pH > 7, which is not a buffer
5. A solution of a weak base is titrated with a solution of a standard strong acid. The
progress of the titration is followed with a pH meter. Which of the following
observations would occur?
(A) The pH of the solution gradually decreases throughout the experiment.
(B) Initially the pH of the solution drops slowly, and then it drops much more rapidly.
(C) At the equivalence point, the pH is 7.
(D) After the equivalence point, the pH becomes constant because this is the buffer
region.
6. What is the ionization constant, Ka, for a weak monoprotic acid with a 0.30 molar
solution having a pH of 4.0?
(A) 3.3 × 10−8
(B) 4.7 × 10−2
(C) 1.7 × 10−6
(D) 3.0 × 10−4
7. Phenol, C6H5OH, has Ka = 1.0 × 10−10. What is the pH of a 0.010 M solution of
phenol?
(A) between 3 and 7
(B) 10
(C) 2
(D) between 7 and 10
8. You are given equimolar solutions of each of the following. Which has the lowest pH?
(A) NH4Cl
(B) NaCl
(C) K3PO4
(D) Na2CO3
9. When sodium nitrite dissolves in water:
(A) The solution is acidic because of the hydrolysis of the sodium ion.
(B) The solution is basic because of the hydrolysis of the NO2− ion
(C) The solution is basic because of the hydrolysis of the sodium ion.
(D) The solution is acidic because of the hydrolysis of the NO2− ion.
10. Which of the following solutions has a pH nearest 7?
(A) 1 M H2C2O4 (oxalic acid) and 1 M KHC2O4 (potassium hydrogen oxalate)
(B) 1 M KNO3 (potassium nitrate) and 1 M HNO3 (nitric acid)
(C) 1 M NH3 (ammonium nitrate) and 1 M NH4NO3 (ammonium nitrate)
(D) 1 M CH3NH2 (methylamine) and 1 M HC2H3O2 (acetic acid)
11. Determine the OH−(aq) concentration in a 1.0 M aniline (C6H5NH2) solution.
The Kbfor aniline is 4.0 × 10−10.
(A) 2.0 × 10−5 M
(B) 4.0 × 10−10 M
(C) 3.0 × 10−6 M
(D) 5.0 × 10−7 M
12. A student wishes to reduce the zinc ion concentration in a saturated zinc iodate
solution to 1 × 10−6 M. How many moles of solid KIO3 must be added to 1.00 L of
solution? [Ksp Zn(IO3)2 = 4 × 10−6 at 25°C.]
(A)
(B)
(C)
(D)
1 mole
0.5 mole
2 moles
4 moles
13. At constant temperature, a change in volume will NOT affect the moles of substances
present in which of the following?
(A) H2(g) + I2(g)
2 HI(g)
(B) CO(g) + Cl2(g)
(C) PCl5(g)
COCl2(g)
PCl3(g) + Cl2(g)
(D) N2(g) + 3 H2(g)
2 NH3(g)
14. The equilibrium constant for the hydrolysis of C2O42− is best represented by which of
the following?
K = [OH−] [C2O42−]/[HC2O4−]
(B) K = [H3O+] [C2O42−]/[HC2O4−]
(C) K = [HC2O4−] [OH−]/[C2O42−]
(D) K = [C2O42−]/[HC2O4−] [OH−]
(A)
15. ZnS(s) + 2 H+(aq)
Zn2+(aq) + H2S(aq)
What is the equilibrium constant for the above reaction? The successive acid
dissociation constants for H2S are 9.5 × 10−8 (Ka1) and 1 × 10−19 (Ka2). The Ksp, the
solubility product constant, for ZnS equals 1.6 × 10−24.
(A) 1.6 × 10−24/9.5 × 10−8
(B) 1 × 10−79/1.6 × 10−24
(C) 1.6 × 10−24/9.5 × 10−27
(D) 9.5 × 10−8/1.6 × 10−24
16. C(s) + H2O(g)
CO(g) + H2(g) endothermic
An equilibrium mixture of the reactants is placed in a sealed container at 150°C. The
amount of the products may be increased by which of the following changes?
(A) decreasing the volume of the container
(B) raising the temperature of the container and increasing the volume of the
container
(C) lowering the temperature of the container
(D) adding 1 mole of C(s) to the container
17. CH4(g) + CO2(g)
2 CO(g) + 2 H2(g)
A 1.00 L flask is filled with 0.30 mole of CH4 and 0.40 mole of CO2 and allowed to come
to equilibrium. At equilibrium, there are 0.20 mole of CO in the flask. What is the value
of Kc, the equilibrium constant, for the reaction?
(A)
(B)
(C)
(D)
1.2
0.027
0.30
0.060
18. 2 NO2(g)
2 NO(g) + O2(g)
The above materials were sealed in a flask and allowed to come to equilibrium at a
certain temperature. A small quantity of O2(g) was added to the flask, and the mixture
was allowed to return to equilibrium at the same temperature. Which of the following
has increased over its original equilibrium value?
(A) the quantity of NO2(g) present
(B) the quantity of NO(g) present
(C) the equilibrium constant, K, increases
(D) the rate of the reaction
19. 2 CH4(g) + O2(g)
2 CO(g) + 4 H2(g) ΔH < 0
To increase the value of the equilibrium constant, K, which of the following changes
must be made to the above equilibrium?
(A)
(B)
(C)
(D)
increase the temperature
increase the volume
decrease the temperature
add CO(g)
20. The addition of nitric acid increases the solubility of which of the following
compounds?
(A) KCl(s)
(B) Pb(CN)2(s)
(C) Cu(NO3)2(s)
(D) NH4NO3(s)
21. At constant temperature, a change in the volume of the system will NOT affect the
moles of the substances present in which of the following?
(A) C(s) + H2O(g)
H2(g) + CO(g)
(B) 3 O2(g)
2 O3(g)
(C) Xe(g) + 2 F2(g)
XeF4(g)
(D) 6 CO2(g) + 6 H2O(l)
6 O2(g) + C6H12O6(aq)
22. Which of the following is the strongest Brønsted–Lowry acid?
(A) HBrO
(B) HBrO3
(C) HBrO2
(D) HBrO4
23. Ionization Constants
How would a solution with an initial NH4Cl (NH4+ + Cl−) concentration of 1 M and an
initial CH3NH3Cl (CH3NH3+ + Cl−) concentration of 1 M be classified?
(A)
(B)
(C)
(D)
a solution with a pH > 7, which is a buffer
a solution with a pH < 7, which is not a buffer
a solution with a pH < 7, which is a buffer
a solution with a pH > 7, which is not a buffer
24. Which of the following species CANNOT serve as both a Brønsted base and a
Brønsted acid?
(A) H2PO42−
(B) CO32−
(C) HSeO4−
(D) HCO3−
25. A chemistry student adds some dilute ammonia solution to some insoluble silver
chloride. The solid dissolves. Which of the following is the correct net ionic equation for
the reaction?
(A) AgCl + 2 NH3 → [Ag(NH3)2]+ + Cl−
(B) AgCl + 2 NH4+→ [Ag(NH4)2]3+ + Cl−
(C) AgCl + NH4+→ Ag++ NH4Cl
(D) AgCl + NH3 → Ag++ NH3Cl
26. Which of the following contains only equilibria that are normally always
homogeneous?
(A) Ka, Kb, and Ksp
Ka, Ksp, and Kf
(C) Ka, Kb, and Kf
(D) Ksp, Kb, and Kf
(B)
27. For which of the following equilibriums will Kc = Kp?
(A) H2(g) + I2(g)
(B) CO(g) + Cl2(g)
2 HI(g)
COCl2(g)
(C) PCl5(g)
PCl3(g) + Cl2(g)
(D) N2(g) + 3 H2(g)
2 NH3(g)
28. CaCO3(s)
CaO(s) + CO2(g)
Which of the following is the correct Kp expression for the above reaction?
(A) Kp =
(B) Kp = [CO2]
(C) Kp =
(D) Kp =
Answers and Explanations
1. B—Any acid will have a pH below 7; thus, C and D can be eliminated. A 0.1 molar
solution of a strong acid would have a pH of 1. Acetic acid is not a strong acid, which
eliminates A.
2. A—The K nearest 10−8 will give a pH near 8. The answer must involve the H2PO4−ion.
3. D—The two substances are not a conjugate acid–base pair, so this is not a buffer.
Both compounds are salts of a strong base and a weak acid; such salts are basic (pH >
7).
4. C—The two substances constitute a conjugate acid–base pair, so this is a buffer. The
pH should be near –log Ka1. This is about 2 (acid).
5. B—Any time an acid is added, the pH will drop. The reaction of the weak base with
the acid produces the conjugate acid of the weak base. The combination of the weak
base and its conjugate is a buffer, so the pH will not change very much until all the
base is consumed. After all the base has reacted, the pH will drop much more rapidly.
The equivalence point of a weak base–strong acid titration is always below 7 (only
strong base–strong acid titrations will give a pH of 7 at the equivalence point). The
value of pOH is equal to pKb halfway to the equivalence point.
6. A—If pH = 4.0, then [H+] = 1 × 10−4 = [A−], and [HA] = 0.30 – 1 × 10−4 ≈ 0.30. The
generic Ka is
, and when the values are entered into this equation,
= 3.3 × 10−8. Since you can estimate the answer, no actual calculations
are necessary.
7. A—This is an acid-dissociation constant; thus, the solution must be acidic (pH < 7).
The pH of a 0.010 M strong acid would be 2.0. This is not a strong acid, so the pH
must be above 2.
8. A—A is the salt of a strong acid and a weak base; therefore, this salt is acidic. B is a
salt of a strong acid and a strong base; such salts are neutral. C and D are salts of a
weak acid and a strong base; such salts are basic. Since you can estimate the answer,
no actual calculations are necessary.
9. B—Sodium nitrite is a salt of a weak acid and a strong base. Ions from strong bases
(Na+ in this case) do not undergo hydrolysis and do not affect the pH. Ions from weak
acids (NO2− in this case) undergo hydrolysis to produce basic solutions.
10. D—The weak acid and the weak base partially cancel each other to give a nearly
neutral solution.
11. A—The equilibrium constant expression is Kb = 4.0 × 10−10 =
This expression becomes
= 4.0 × 10−10, which simplifies to
.
= 4.0 ×
10−10. Taking the square root of each side gives x = 2.0 × 10−5 = [OH−]. Since you
can estimate the answer, no actual calculations are necessary.
12. C—The solubility-product constant expression is Ksp = [Zn2+ ] [IO3−]2 = 4 × 10−6.
This may be rearranged to [IO3−]2 =
concentration gives [IO3−]2 =
. Inserting the desired zinc ion
= 4. Taking the square of each side leaves a
desired IO3− concentration of 2 M. Two moles of KIO3 must be added to 1.00 L of
solution to produce this concentration. Since you can estimate the answer, no actual
calculations are necessary.
13. A—When dealing with gaseous equilibriums, volume changes are important when
there is a difference in the total number of moles of gas on opposite sides of the
equilibrium arrow. All the answers, except A, have differing numbers of moles of gas
on opposite sides of the equilibrium arrow.
14. C—The hydrolysis of any ion begins with the interaction of that ion with water. Thus,
both the ion and water must be on the left side of the equilibrium arrow and hence in
the denominator of the equilibrium-constant expression (water, like all solvents, will
be left out of the expression). The oxalate ion is the conjugate base of a weak acid. As
a base, it will produce OH− in solution along with the conjugate acid, HC2O4−, of the
base. The equilibrium reaction is HC2O4−(aq) + H2O(l)
OH−(aq) + H2 HC2O4(aq).
15. C—The equilibrium given is actually the sum of the following three equilibriums:
Summing these equations means you need to multiply the equilibrium constants:
16. B—The addition or removal of some solid, as long as some remains present, will not
change the equilibrium. An increase in volume will cause the equilibrium to shift
toward the side with more moles of gas (right). Raising the temperature of an
endothermic process will shift the equilibrium to the right. Any shift to the right will
increase the amounts of the products.
17. B—Using the following table:
The presence of 0.20 mole of CO (0.20 M) at equilibrium means that 2x = 0.20 and
that x = 0.10. Using this value for x, the bottom line of the table becomes
The equilibrium expression is K =
the equilibrium expression gives K =
. Entering the equilibrium values into
.
18. A—The addition of a product will cause the equilibrium to shift to the left. The
amounts of all the reactants will increase, and the amounts of all the products will
decrease (the O2 will not go below its earlier equilibrium value since excess was
added). The value of K is constant, unless the temperature is changed. The rates of the
forward and reverse reactions are equal at equilibrium.
19. C—The only way to change the value of K is to change the temperature. For an
exothermic process (ΔH < 0), K increases with a decrease in temperature.
20. B—Nitric acid, being an acid, will react with a base. In addition to obvious bases
containing OH−, the salts of weak acids are also bases. All the anions, except CN−, are
from strong acids and, as such, are not bases.
21. D—If there are equal numbers of moles of gas on each side of the equilibrium arrow,
then volume or pressure changes will not affect the equilibrium. The presence of
solids, liquids, or aqueous phases does not make any difference as long as some of
the phase is present.
22. D—Perbromic acid, HBrO4, is expected to be the strongest acid in this group because
the greater number of oxygen atoms pulls more electron density from the hydrogen
atom, making it easier to be lost as a hydrogen ion.
23. B—The two substances (NH4Cl and CH3NH3Cl) do not contain a conjugate acid–base
pair, so this is not a buffer, which eliminates A and C. Both compounds are salts of a
weak base (NH3 or CH3NH2) and a strong acid (HCl); such salts are acidic (pH < 7).
The Kb values shown in the table indicate that all three compounds are weak bases.
24. B—All the ions can serve as Brønsted bases (accept a hydrogen ion to form H3PO4,
HCO3−, H2SeO4, and H2CO3). All but B can behave as Brønsted acids (donate a
hydrogen ion to form HPO42−, SeO42−, and CO32).
25. A—Aqueous ammonia contains primarily NH3, which eliminates B and C. NH3Cl does
not exist, which eliminates D. The reaction produces the silver–ammonia complex,
[Ag(NH3)2]+.
26. A—A Ksp is never homogeneous; all the others are normally always homogenous.
27. A—Kc = Kp occurs whenever there are equal moles of gaseous molecules on each
side of the equilibrium arrow.
28. D—This is a heterogeneous equilibrium; therefore, the solids (CaCO3 and CaO) will
not appear in the equilibrium expression. B is the Kc expression, not the Kp expression.