Equilibrium Multiple Choice Review Questions for AP Chemistry (page 2)
Review the following concepts if necessary:
- Le Chatelier's Principle for AP Chemistry
- Acid–Base Equilibrium for AP Chemistry
- Ka, Kw, Kb - The Acid, Water, and Base Dissociation Constant for AP Chemistry
- Buffers for AP Chemistry
- Titration Equilibria for AP Chemistry
- Solubility Equilibria for AP Chemistry
Answer the following questions in 35 minutes. You may not use a calculator. You may use the periodic table at the back of the book.
- A 0.1-molar solution of acetic acid (CH3COOH) has a pH of about
- K2HPO4 + KH2PO4
- K2HPO4 + K3PO4
- K2HPO4 + H3PO4
- Ka, the acid dissociation constant, for an acid is 9 × 10–4 at room temperature. At this temperature, what is the approximate percent dissociation of the acid in a 1.0 M solution?
- a solution with pH = 7.
- 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
- a solution with a pH > 7, which is a buffer
Using the above information, choose the best answer for preparing a pH = 8 buffer.
Use the following information for questions 4–6.
- The pH of the solution gradually decreases throughout the experiment.
- Initially the pH of the solution drops slowly, and then it drops much more rapidly.
- At the equivalence point the pH is 7.
- After the equivalence point, the pH becomes constant because this is the buffer region.
- The pOH at the equivalence point equals the pKb of the base.
- 9.7 × 10–10
- 4.7 × 10–2
- 1.7 × 10–6
- 3.0 × 10–4
- 3.3 × 10–8
- between 3 and 7
- between 7 and 10
- the solution is acidic because of hydrolysis of the sodium ion
- the solution is neutral
- the solution is basic because of hydrolysis of the sodium ion
- the solution is acidic because of hydrolysis of the NO2– ion
- the solution is basic because of hydrolysis of the NO2– ion
Questions 12–15 refer to the following aqueous solutions. All concentrations are 1 M.
- HNO2C2O4 (oxalic acid) and KHC2O4 (potassium hydrogen oxalate)
- KNO3 (potassium nitrate) and HNO3 (nitric acid)
- NH3 (ammonia) and NH4NO3 (ammonium nitrate)
- C2H5NH2 (ethylamine) and KOH (potassium hydroxide)
- CH3NH2 (methylamine) and HC2H3O2 (acetic acid)
- 2.0 × 10–5 M
- 4.0 × 10–10 M
- 3.0 × 10–6 M
- 5.0 × 10–7 M
- 1.0 × 100 M
- 1 mol
- 0.5 mol
- 0.0001 mol
- 4 mol
- 2 mol
The above equation has an equilibrium constant that is less than 1. What are the relative strengths of the acids and bases?
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). Ksp, the solubility product constant, for ZnS equals 1.6 × 10–24.
- 1.6 × 10–24/9.5 × 10–8
- 1 × 10–79/1.6 × 10–24
- 9.5 × 10–27/1.6 × 10–24
- 9.5 × 10–8/1.6 × 10–24
- 1.6 × 10–24/9.5 × 10–27
Which species, in the above equilibrium, behave as bases?
- I only
- I and II
- II and III
- I and III
- III only
As shown above, malonic acid is a diprotic acid. The successive equilibrium constants are 1.5 × 10–3 (Ka1) and 2.0 × 10–6 (Ka2). What is the equilibrium constant for the above reaction?
- 1.0 × 10–14
- 2.0 × 10–6
- 4.0 × 10–12
- 3.0 × 10–9
- 1.5 × 10–3
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?
- raising the temperature of the container
- increasing the volume of the container
- adding 1 mol of C(s) to the container
- II only
- I and II
- I only
- II and III
- III only
An equal number of moles of each of the reactants are sealed in a container and allowed to come to the equilibrium shown above. At equilibrium which of the following must be true?
- [CO2] must equal [H2O]
- [O2] must be less than [C2H4]
- [CO2] must be greater than [C2H4]
- II and III
- I only
- III only
- II only
- I and II
A 1.00-L flask is filled with 0.30 mol of CH4 and 0.40 mol of CO2, and allowed to come to equilibrium. At equilibrium, there are 0.20 mol of CO in the flask. What is the value of Kc, the equilibrium constant, for the reaction?
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 allowed to return to equilibrium at the same temperature. Which of the following has increased over its original equilibrium value?
- the quantity of NO2(g) present
- the quantity of NO(g) present
- the equilibrium constant, K
- the rate of the reaction
- the partial pressure of NO(g)
In order to increase the value of the equilibrium constant, K, which of the following changes must be made to the above equilibrium?
- increase the temperature
- increase the volume
- decrease the temperature
- add CO(g)
- add a catalyst
The equilibrium constant, K, for the above equilibrium is 7.2 × 10–2. This value implies which of the following?
- A solution with equimolar amounts of HC3H5O2(aq) and HCOO–(aq) is neutral.
- C3H5O2–(aq) is a stronger base than HCOO–(aq).
- HC3H5O2(aq) is a stronger acid than HCOOH(aq).
- HCOO–(aq) is a stronger base than C3H5O2–(aq).
- The value of the equilibrium does not depend on the temperature.
Answers and Explanations
- B—An acid, any acid, will give a pH below 7; thus, answers C–E are eliminated. A 0.1-molar solution of a strong acid would have a pH of 1. Acetic acid is not a strong acid, so answer A is eliminated.
- A—The K nearest 10–8 will give a pH near 8. The answer must involve the H2PO4– ion.
- C—The generic Ka is:
- 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).
- 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).
- E—The two substances constitute a conjugate acid–base pair, so this is a buffer. The pOH should be near –log Kb. This is about 4. The pH would be about 14 – 4 = 10.
- 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 used. 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 half-way to the equivalence point.
- E—If pH = 4.0, then [H+] = 1 × 10–4 = [A–], and [HA] = 0.30 – 1 × 10–4. The generic Ka is [H+][A–]/[HA], and when the values are entered into this equation: (1 × 10–4)2/0.30 = 3.3 × 10–8. Since you can estimate the answer, no actual calculations need be done.
- A—This is an acid-dissociation constant, thus the solution must he 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.
- A—A is the salt of a strong acid and a weak base; it is acidic. B and E are salts of a strong acid and a strong base; they are neutral. C and D are salts of a weak acid and a strong base; they are basic. The lowest pH would be the acidic choice.
- E—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.
- B—The presence of a strong acid, HNO3, would make this the most acidic (lowest pH).
- E—The weak acid and the weak base partially cancel each other to give a nearly neutral solution.
- C—Both A and C are buffers, because they have conjugate acid–base pairs of either a weak acid (A) or a weak base (C). The weak acid buffer would have a pH below 7, and the weak base buffer would have a pH above 7.
- A—See the answer to question 14.
- A—The equilibrium constant expression is: Kb = 4.0 × 10–10 = [OH–][C6H5NH3+ ]/[C6H5NH2]. This expression becomes: (x)(x)/(1.0 – x) = 4.0 × 10–10, which simplifies to: x2/1.0 = 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 need be done.
- E—The solubility-product constant expression is: Ksp = [Zn2+][IO3–]2 = 4 × 10–6. This may be rearranged to: [IO3–]2 = 4 × 10–6/[Zn2+]. Inserting the desired zinc ion concentration gives: [IO3–]2 = 4 × 10–6/(1 × 10–6) = 4. Taking the square root of each side leaves a desired IO3– concentration of 2 M. 2 mol of KIO3 must be added to 1.00 L of solution to produce this concentration. Since you can estimate the answer, no actual calculations need be done.
- A—When dealing with gaseous equilibria, 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.
- C—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, as with 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: C2O42–(aq) + H2O(l) OH–(aq) + HC2O4–(aq).
- D—The low value for the equilibrium constant means that the equilibrium lies to the left. For this to be true, the weaker acid and the weaker base must be on the left side.
- E—The equilibrium given is actually the sum of the following three equilibria:
Ka = [H+][A–]/[HA] = 9 × 10–4 = x2/1.0 – x
x = 3 × 10–2 M and the percent dissociation is (3 × 10–2/1.0) × 100% = 3%
You will need to be able to do calculations at this level without a calculator.
Summing these equations means you need to multiply the equilibrium constants:
Ksum = KspKK' = Ksp/Ka2Ka1
- = 1.6 × 10–24/[(1 × 10–19) (9.5 × 10–8)]
The presence of 0.20 mol 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 = [CO]2[H2]2/[CH4][CO2]. Entering the equilibrium values into the equilibrium expression gives: K = (0.20)2(0.20)2/(0.20)(0.30)
Ksp = [Mn2+][OH–]2 = 1.6 × 10–13
If s is used to indicate the molar solubility, the equilibrium expression becomes:
Ksp = (s)(2s)2 = 4s3 = 1.6 × 10–13
This rearranges to:
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