Make your own free website on Tripod.com
Chapter 13, 14, & 14 Review

1. (3 pts)  In which of the following systems would the number of moles of the substances present at equilibrium NOT be shifted by a change in the volume of the system at constant temperature?

 A) CO(g) + NO(g) -->  CO2(g) + ½ N2(g)
 B) N2(g) + 3 H2(g) -->  2 NH3(g)
 C) N2(g) + 2 O2(g) -->  2 NO2(g)
 D) N2O4(g)  --> 2 NO2(g)
 E) NO(g) + O3(g) -->  NO2(g) + O2(g)

7. (5 pts)  Consider the following reaction:

  PCl5(g) -->   PCl3(g) + Cl2(g)  dH = 92.5 kJ

 Predict the direction of the shift in equilibrium when a) the temperature is raised, b) more chlorine gas is added to the reaction mixture, c) some PCl3 is removed from the mixture, d) the pressure on the gases is increased, and e) a catalyst is added to the reaction mixture.

8. At elevated temperatures, SbCl5 gas decomposes into SbCl3 gas and Cl2 gas, as shown by the following equation.
     SbCl5(g) -->   SbCl3(g) + Cl2(g)

 A) An 89.7 g sample of SbCl5 (MW = 299.0) is placed in an evacuated 15.0 L container at 182 C.

  1) (2 pts)  What is the concentration in moles per liter of SbCl5 in the container before any decomposition occurs?
  2) (3 pts)  What is the pressure in atmospheres of SbCl5 in the container before any decomposition occurs?

 B) (5 pts)  If the SbCl5 is 29.2 percent decomposed (i.e. 29.2% of the original mass is gone) when equilibrium is established at 182 C, calculate the value for either equilibrium constant, Kp or Kc, for this decomposition reaction.  Indicate whether you are calculating Kp or Kc.

 C) (5 pts)  In another experiment, 1.00 mole of each of the reactants and products above is placed in a 1.00 liter container.  At the temperature the experiment is run, Kc for the reaction equals 0.117.  What will the concentration of SbCl5 be at equilibrium?

 D) (6 pts)  In order to produce some SbCl5, a 1.00-mole sample of SbCl3 is first placed in an empty 2.00-liter container maintained at the same temperature in C above (i.e. Kc = 0.117).  How many moles of Cl2 must be added to this container to reduce the number of moles of SbCl3 to 0.700 mole at equilibrium?


Chapter 14 & 15 Test

Questions 1-3
 A) A solution with a pH less than 7 that is not a buffer solution
 B) A buffer solution with a pH between 4 and 7
 C) A buffer solution with a pH between 7 and 10
 D) A solution with a pH greater than 7 that is not a buffer solution
 E) A solution with a pH of 7

Ionization constants: CH3COOH  =  1.8 x 10-5;  H2CO3    K1  = 4 x 10-7,   K2  = 4 x 10-11;  NH3  =  1.8 x 10-5

1. (3 pts)  A solution prepared to be initially 1 M in NaCl and 1 M in HCl.
2. (3 pts)  A solution prepared to be initially 1 M in Na2CO3 and 1 M in CH3COONa.
3. (3 pts)  A solution prepared to be initially 0.5 M in CH3COOH and 1 M in CH3COONa.

4. (3 pts)  In the titration of a weak acid of unknown concentration with a standard solution of a strong base, a pH meter was used to follow the progress of the titration.  Which of the following is true for this experiment?
 A) The pH is 7 at the equivalence point.
 B) The pH at the equivalence point depends on the indicator used.
 C) The graph of pH versus volume of base added rises gradually at first and then much more rapidly.
 D) The graph of pH versus volume of base added shows no sharp rise.
 E) The [H+] at the equivalence point equals the ionization constant of the acid.

5. (3 pts)  All of the following species can function as Bronsted-Lowry bases in solution EXCEPT

 A) H2O  B) NH3  C) S2-  D) NH4+ E) HCO3-

6. (3 pts)  When phenolphthalein is used as the indicator in a titration of an HCl solution with a solution of NaOH, the indicator undergoes a color change from clear to red at the end point of the titration.  This color change occurs abruptly because
 A) phenolphthalein is a very strong acid that is capable of rapid dissociation
 B) the solution being titrated undergoes a large pH change near the end point of the titration
 C) phenolphthalein undergoes an irreversible reaction in basic solution
 D) OH- acts as a catalyst for the decomposition of phenolphthalein
 E) phenolphthalein is involved in the rate determining step of the reaction between H3O+ and OH-

7. (3 pts)  As the number of oxygen atoms increases in any series of oxygen acids, such as HXO, HXO2, HXO3...., which of the following is generally true?
 A) The acid strength varies unpredictably.
 B) The acid strength decreases only if X is a nonmetal.
 C) The acid strength decreases only if X is a metal.
 D) The acid strength decreases whether X is a nonmetal or a metal.
 E) The acid strength increases.

8. (5 pts)  A 0.20-molar solution of a weak monoprotic acid, HA, has a pH of 3.00.  The ionization constant of this acid is

 A) 5.0 x 10-7 B) 2.0 x 10-7 C) 5.0 x 10-6 D) 5.0 x 10-3 E) 2.0 x 10-3

9. (3 pts)   H2PO4- + HBO32-  -->   HPO42- + H2BO3-

 The equilibrium constant for the reaction represented by the equation above is greater than 1.0.  Which of the following gives the correct relative strengths of the acids and bases in the reaction?
 

   Acids    Bases
 A)  H2PO4- > H2BO3- and HBO32- > HPO42-
 B)  H2BO3- > H2PO4- and HBO32- > HPO42-
 C)  H2PO4- > H2BO3- and HPO42- > HBO32-
 D)  H2BO3- > H2PO4- and  HPO42- > HBO32-
 E)  H2PO4- = H2BO3- and HPO42- = HBO32-

10. (5 pts)  The solubility of CuI is 2 x 10-6 molar.  What is the solubility product constant, Ksp, for CuI?

 A) 1.4 x 10-3  C) 4 x 10-12  E) 8 x 10-18
 B) 2 x 10-6  D) 2 x 10-12

11. (5 pts)   MnS(s) + 2 H+  -->   Mn2+ + H2S(g)

 At 25 C the solubility product constant, Ksp, for MnS is 5 x 10-15 and the acid dissociation constants K1 and K2 for H2S are 1 x 10-7 and 1 x 10-13, respectively.  What is the equilibrium constant for the reaction represented by the equation above at 25 C?

 Had several answers here, but I was too lazy to include them, because they were all graphics...  However, see below for the answer.

12. (3 pts)  Equal volumes of 0.10-molar H3PO4 and 0.20-molar KOH are mixed.  After equilibrium is established, the type of ion is solution in largest concentration, other than the K+ ion, is

 A) H2PO4- B) HPO42- C) PO43- D) OH-  E) H3O+

Free Response

1. In an experiment to determine the molecular weight and the ionization constant for ascorbic acid (vitamin C), a student dissolved 1.3717 grams of the acid in water to make 50.00 milliliters of solution.  The entire solution was titrated with a 0.2211-molar NaOH solution.  The pH was monitored throughout the titration.  The equivalence point was reached when 35.23 milliliters of the base had been added.  Under the conditions of this experiment, ascorbic acid acts as a monoprotic acid that can be represented as HA.

 A) (5 pts)  From the information above, calculate the molecular weight of ascorbic acid.
 B) (5 pts)  When 20.00 milliliters of NaOH had been added during the titration, the pH of the solution was 4.23.  Calculate the acid ionization constant for ascorbic acid.
 C) (4 pts)  Calculate the equilibrium constant for the reaction of the ascorbate ion, A-, with water.
 D) (5 pts)  Calculate the pH of the solution at the equivalence point of the titration.

2. The solubility of iron(II) hydroxide, Fe(OH)2, is 1.43 x 10-3 gram per liter at 25 C.

 A) Write a balanced equation for the solubility equilibrium.
 B) Write the expression for the solubility product constant, Ksp, and calculate its value.
 C) Calculate the pH of a saturated solution of Fe(OH)2 at 25 C.
 D) A 50.0-milliliter sample of 3.00 x 10-3-molar FeSO4 solution is added to 50.0 milliliters of 4.00 x10-6-molar NaOH solution.  Does a precipitate of Fe(OH)2 form?  Explain and show calculations to support your answer.


Answers

Chapter 13 Test

1.    E
7.    a.    right
       b.    left
       c.    right
       d.    left
       e.    no change
8.
    a.
        1.  .0200 M
        2.  .748 M
    b.    Kp = .0901, Kc = .00242
    c.    1.571 M
    d.    .400 mol

Chapter 14&15 Test

1.    A
2.    D
3.    B
4.    C
5.    D
6.    B
7.    E
8.    C
9.    A
10.    C
11.    D
12.    B

Free Response

1.
    A.    176.1 g/mol
    B.    7.75 x 10-5
    C.    1.29 x 10-10
    D.    pH = 8.54

2.
    A.    Fe(OH)2(s) <-->  Fe2+ + 2 OH-
    B.    Ksp = 1.61 x 10-14
    C.    pH = 9.50
    D.    Q = 6 x 10-15 which is < Ksp  therefore, there will be no precipitate