Chapter 9. Chemical Equilibria
Faculty Resource and Organizational Guide (FROG)
Table of Contents
Materials for Chapter 9 Activities ................................................................................................6
Reagents for Chapter 9 Activities: ...............................................................................................7
Section 9.1. The Nature of Equilibrium ......................................................................................8
Learning Objectives for Section 9.1 ............................................................................................8
Investigate This 9.1. How can we model a system at equilibrium? ............................................8
Alternative procedure: .................................................................................................................9
Consider This 9.2. What is dynamic equilibrium? .....................................................................9
Investigate This 9.3. Do solutions of Fe(NO3)3 and KSCN react when mixed? ......................10
Consider This 9.4. How do solutions of Fe(NO3)3 and KSCN react when mixed? ..................11
Investigate This 9.8. What conditions affect CoCl2 in solution? ..............................................12
Consider This 9.9. How do solvents and temperature affect CoCl2 in solution?......................14
Consider This 9.10. What occurs when water is added to alcoholic Co2+ solutions?...............15
Section 9.2. Mathematical Expression for the Equilibrium Condition ..................................16
Learning Objectives for Section 9.2 ..........................................................................................16
Investigate This 9.12. What is the pH of an acetic acid solution? ............................................16
Consider This 9.13. How is an equilibrium system characterized quantitatively? ...................17
Consider This 9.16. What can equilibrium constant expressions tell us? .................................18
Section 9.3. Acid-Base Reactions and Equilibria .....................................................................19
Learning Objectives for Section 9.3 ..........................................................................................19
Investigate This 9.23. What is the pH of an acetate ion solution? ............................................20
Consider This 9.24. Does the pH of the acetate ion solution make sense?...............................21
Consider This 9.29. What is the relationship of pH and pOH? ................................................21
Section 9.4. Solutions of Conjugate Acid–Base Pairs: Buffer Solutions ................................22
Learning Objectives for Section 9.4 ..........................................................................................22
Investigate This 9.30. What are the pHs of conjugate acid-base pair solutions?......................22
Consider This 9.31. Do the pHs of conjugate acid-base pair solutions make sense? ...............24
Investigate This 9.35. How does a buffer respond to added H3O+(aq) or OH–(aq)? ................24
Consider This 9.36. Do the pHs of conjugate acid-base pair solutions make sense? ...............26
Consider This 9.41. How do you prepare a buffer solution of known concentration? .............27
Section 9.5. Acid-Base Properties of Proteins ..........................................................................28
Learning Objectives for Section 9.5 ..........................................................................................28
Investigate This 9.42. How is a protein solution affected by addition of H3O+(aq)? ...............28
Consider This 9.43. How are protein solubility and pH related?..............................................31
Consider This 9.44. How do side group charges on the model protein vary with pH? ............32
Consider This 9.45. What is the isoelectric pH for casein? ......................................................33
Consider This 9.47. What is the difference between HbA and HbS? .......................................33
Section 9.6. Solubility Equilibria for Ionic Salts .......................................................................34
Learning Objectives for Section 9.6 ..........................................................................................34
March 2005
ACS Chemistry FROG
1
Chemical Equilibria
Chapter 9
Investigate This 9.48. Is silver chromate insoluble? .................................................................34
Consider This 9.49. How insoluble is silver chromate? ...........................................................35
Consider This 9.57. Does a common ion always decrease solubility? .....................................36
Section 9.7. Thermodynamics and the Equilibrium Constant ................................................37
Learning Objectives for Section 9.7 ..........................................................................................37
Investigate This 9.61. How much urea will dissolve in water? ................................................37
Consider This 9.62. What is the solubility of urea in water? ....................................................38
Consider This 9.63. What are G, H, and S for dissolution of urea in water? ...............39
Section 9.8. Temperature Dependence of the Equilibrium Constant ....................................40
Learning Objectives for Section 9.8 ..........................................................................................40
Investigate This 9.64. How does temperature affect the solubility of PbI2(s)? ........................41
Consider This 9.65. Is the dissolution of PbI2(s) exothermic or endothermic? ........................42
Section 9.9. Thermodynamics in Living Systems .....................................................................43
Learning Objectives for Section 9.9 ..........................................................................................43
Consider This 9.73. What is the equilibrium cellular concentration of ATP? ..........................43
Consider This 9.74. How do you analyze a coupled reaction? .................................................44
Section 9.11. Extension -- Thermodynamic Basis for the Equilibrium Constant .................45
Learning Objectives for Section 9.7 ..........................................................................................45
Consider This 9.77. What are some properties of nitrogen oxides? .........................................45
Consider This 9.78. Which of the gases is colored: NO2(g) or N2O4(g)? .................................47
Consider This 9.80. What is the entropy of compressed carbon dioxide? ................................48
Consider This 9.81. What is Sreaction for 2NO2(g)  N2O4(g)? ..............................................48
Consider This 9.82. What is Greaction for 2NO2(g)  N2O4(g)? ..............................................49
Solutions for Chapter 9 Check This Activities ...........................................................................51
Check This 9.5. Concentrations in the Fe(NO3)3–KSCN mixture ............................................51
Check This 9.6. Dynamic equilibrium animation .....................................................................51
Check This 9.7. Adding SCN–(aq) to a Fe(NO3)3–KSCN solution ..........................................51
Check This 9.11. Temperature and Le Chatelier’s principle ....................................................52
Check This 9.15. Equilibrium constant expression and equilibrium constant ..........................52
Check This 9.18. Acid equilibrium constant for acetic acid .....................................................52
Check This 9.20. pH of an aqueous solution of benzoic acid ...................................................53
Check This 9.22. Extent of water autoionization in a lactic acid solution................................53
Check This 9.26. Base equilibrium constant for acetate ion.....................................................54
Check This 9.27. Ka·Kb for the acetic acid–acetate ion pair .....................................................54
Check This 9.28. Using the relationship of pKa to pKb.............................................................54
Check This 9.29. What is the relationship of pH and pOH?.....................................................54
Check This 9.33. Ka for acetic acid ...........................................................................................55
Check This 9.34. Conjugate base-to-acid ratios and pKa for acetic acid ..................................57
Check This 9.37. Stoichiometry in buffer solutions .................................................................57
Check This 9.38. pH and the conjugate acid–base ratio ...........................................................57
Check This 9.40. pH change when hydroxide ion is added to a buffer solution ......................58
Check This 9.46. Relating a Web Companion animation to the casein investigation ..............58
Check This 9.51. Solubility product, Ksp, for silver chromate, Ag2CrO4(s) .............................59
Check This 9.53. Solubility of silver phosphate, Ag3PO4 ........................................................60
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ACS Chemistry FROG
Chapter 9
Check This 9.55.
Check This 9.56.
Check This 9.58.
Check This 9.60.
Check This 9.66.
Check This 9.68.
Check This 9.70.
Check This 9.72.
Check This 9.75.
Check This 9.76.
Check This 9.79.
Chemical Equilibria
Solubility of Cu(IO3)2(s) in water and 0.25 M Cu2+(aq) solution ................60
Compare Figures 9.9 and 9.10 ......................................................................61
Agreement of units in equation (9.42) ..........................................................61
Reaction quotient and free energy for a gas phase reaction .........................61
Rearranging equation (9.58) .........................................................................62
H°reaction from solubility temperature dependence ......................................62
S°reaction and G°reaction from solubility temperature dependence ...............63
Graphical determination of H°reaction and S°reaction ....................................63
Transport of oxygen by hemoglobin ............................................................64
Relative sizes of oxygen-binding equilibrium constants ..............................65
H°reaction for 2NO2(g)  N2O4(g) ...............................................................65
Solutions for Chapter 9 End-of-Chapter Problems....................................................................66
Problem 9.1. ...............................................................................................................................66
Problem 9.2. ...............................................................................................................................66
Problem 9.3. ...............................................................................................................................67
Problem 9.4. ...............................................................................................................................67
Problem 9.5. ...............................................................................................................................67
Problem 9.6. ...............................................................................................................................68
Problem 9.7. ...............................................................................................................................68
Problem 9.8. ...............................................................................................................................68
Problem 9.9. ...............................................................................................................................69
Problem 9.10. .............................................................................................................................69
Problem 9.11. .............................................................................................................................69
Problem 9.12. .............................................................................................................................70
Problem 9.13. .............................................................................................................................70
Problem 9.14. .............................................................................................................................71
Problem 9.15. .............................................................................................................................71
Problem 9.16. .............................................................................................................................72
Problem 9.17. .............................................................................................................................73
Problem 9.18. .............................................................................................................................74
Problem 9.19. .............................................................................................................................74
Problem 9.20. .............................................................................................................................77
Problem 9.21. .............................................................................................................................77
Problem 9.22. .............................................................................................................................78
Problem 9.23. .............................................................................................................................78
Problem 9.24. .............................................................................................................................79
Problem 9.25. .............................................................................................................................80
Problem 9.26. .............................................................................................................................81
Problem 9.27. .............................................................................................................................82
Problem 9.28. .............................................................................................................................82
Problem 9.29. .............................................................................................................................83
Problem 9.30. .............................................................................................................................83
Problem 9.31. .............................................................................................................................83
Problem 9.32. .............................................................................................................................84
Problem 9.33. .............................................................................................................................84
Problem 9.34. .............................................................................................................................85
ACS Chemistry FROG
3
Chemical Equilibria
Chapter 9
Problem 9.35. .............................................................................................................................85
Problem 9.36. .............................................................................................................................86
`Problem 9.37. ............................................................................................................................87
Problem 9.38. .............................................................................................................................87
Problem 9.39. .............................................................................................................................88
Problem 9.40. .............................................................................................................................88
Problem 9.41. .............................................................................................................................89
Problem 9.42. .............................................................................................................................89
Problem 9.43. .............................................................................................................................90
Problem 9.44. .............................................................................................................................90
Problem 9.45. .............................................................................................................................90
Problem 9.46. .............................................................................................................................91
Problem 9.47. .............................................................................................................................91
Problem 9.48. .............................................................................................................................92
Problem 9.49. .............................................................................................................................92
Problem 9.50. .............................................................................................................................93
Problem 9.51. .............................................................................................................................93
Problem 9.52. .............................................................................................................................93
Problem 9.53. .............................................................................................................................94
Problem 9.54. .............................................................................................................................94
Problem 9.55. .............................................................................................................................95
Problem 9.56. .............................................................................................................................95
Problem 9.57. .............................................................................................................................96
Problem 9.58. .............................................................................................................................96
Problem 9.59. .............................................................................................................................97
Problem 9.60. .............................................................................................................................97
Problem 9.61. .............................................................................................................................97
Problem 9.62. .............................................................................................................................97
Problem 9.63. .............................................................................................................................99
Problem 9.64. ...........................................................................................................................100
Problem 9.65. ...........................................................................................................................100
Problem 9.66. ...........................................................................................................................101
Problem 9.67. ...........................................................................................................................102
Problem 9.68. ...........................................................................................................................102
Problem 9.69. ...........................................................................................................................102
Problem 9.70. ...........................................................................................................................103
Problem 9.71. ...........................................................................................................................103
Problem 9.72. ...........................................................................................................................104
Problem 9.73. ...........................................................................................................................104
Problem 9.74. ...........................................................................................................................105
Problem 9.75. ...........................................................................................................................106
Problem 9.76. ...........................................................................................................................107
Problem 9.77. ...........................................................................................................................107
Problem 9.78. ...........................................................................................................................108
Problem 9.79. ...........................................................................................................................109
Problem 9.80. ...........................................................................................................................110
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ACS Chemistry FROG
Chapter 9
Chemical Equilibria
Problem 9.81. ...........................................................................................................................111
Problem 9.82. ...........................................................................................................................112
Problem 9.83. ...........................................................................................................................112
Problem 9.84. ...........................................................................................................................112
Problem 9.85. ...........................................................................................................................113
Problem 9.86. ...........................................................................................................................113
Problem 9.87. ...........................................................................................................................114
Problem 9.88. ...........................................................................................................................115
Problem 9.89. ...........................................................................................................................115
Problem 9.90. ...........................................................................................................................116
Problem 9.91. ...........................................................................................................................117
Problem 9.92. ...........................................................................................................................118
Problem 9.93. ...........................................................................................................................118
Problem 9.94. ...........................................................................................................................118
Problem 9.95. ...........................................................................................................................120
ACS Chemistry FROG
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Chemical Equilibria
Chapter 9
Materials for Chapter 9 Activities
Activity
6
Material
Total Quantity
9.1
Clear plastic container at
least 10-cm deep
2
9.1
100-mL beaker
1
9.1; 9.64
250-mL beaker
1; 1
9.3
24-well plate
1
9.8
125-mL Erlenmeyer flask
(dry)
1
9.8
200-mm test tubes (dry)
4
9.8
Rack to support test tubes
1
9.8
Plastic transfer pipet
1
9.8
1000-mL beaker for ice
bath
1
9.8
1000-mL beaker on a hot
plate for 80-90 ºC water
bath
1
9.12; 9.23; 9.30; 9.35; 9.42
pH meter/combination pH
electrode
1
9.30; 9.35
Sample vials
3; 4
9.35; 9.48; 9.61
Thin-stem plastic pipets
2; 1; 1/group
9.42
Magnetic stirrer
1
9.42
Magnetic stir bar
1
9.42
400-mL beaker
1
9.42
25-mL dropper bottles or
small wash bottles
2
9.48
25-mL beaker
1
9.48
10-mL graduated cylinder
9.61
15-mL capped, graduated
plastic centrifuge tubes
2/group
9.64
Magnetic stirrer-hot plate
combination and stir bar
1
9.64
Thermometer, –10-110 C
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ACS Chemistry FROG
Chapter 9
Chemical Equilibria
Reagents for Chapter 9 Activities:
Activity
Reagents
Total Quantity
9.3
0.002 M potassium
thiocyanate, KSCN
0.0097 g in 50 mL of water.
9.3
0.2 M Iron (III) nitrate,
Fe(NO3)3·9H2O
4.04 g in 50 mL of water
9.8
CoCl2·6H2O
2g
9.8
91% isopropanol (isopropyl
alcohol) or reagent grade
50 mL
9.12; 9.30; 9.35
0.1 M acetic acid
Dilute from stock
9.12; 9.23; 9.30; 9.35; 9.42
Standardization buffers for
pH meter and electrode
1 set
9.23; 9.30; 9.35
0.1 M sodium acetate
0.82 g in 100 mL of water
9.35
1 M hydrochloric acid
Dilute from stock
9.35
1 M sodium hydroxide
4.0 g in 100 mL of water
9.42
3 M hydrochloric acid
25 mL concentrated HCl
diluted to 100 mL
9.42
3 M sodium hydroxide
12 g in 100 mL of water
9.42
0.25% casein(aq)
0.6 g of casein in 250 mL of
0.01 M NaOH
9.48
0.0015 M silver nitrate
25.5 mg in 100 mL of water
9.48
0.0015 M potassium
chromate
29.1 mg in 100 mL of water
9.61
Urea
4.0 g/group
9.64
1 M lead nitrate
1.7 g in 5 mL of water
9.64
0.05 M potassium iodide
1.7 g in 10 mL water
ACS Chemistry FROG
7
Chemical Equilibria
Chapter 9
Section 9.1. The Nature of Equilibrium
Learning Objectives for Section 9.1
 Use Le Chatelier’s principle to predict the direction of the response of an equilibrium system
to changes in the concentration(s) of reactants or products.
 Use Le Chatelier’s principle and the response of an equilibrium system to heating or cooling
to tell whether the reaction is endothermic or exothermic.
 Explain how Le Chatelier’s principle is a consequence of the dynamic nature of chemical
equilibrium.
Investigate This 9.1. How can we model a system at equilibrium?
Goal:
Model a system of dynamic equilibrium of water transfers between two reservoirs.
Set-up time:
 5-10 minutes.
Time for activity:
 From less than 10 minutes to 10-15 minutes, depending on the amount of discussion and the
variations (different starting conditions) you decide to try.
Materials:
 Two clear plastic basin or other containers at least 10-cm deep.
 One 100-mL beaker.
 One 250-mL beaker.
Reagents:
 Water.
Procedure:
 Conduct as a class investigation with two students and work in small groups to analyze the
results.
SAFETY NOTE
Wear your safety goggles.




Each student has a plastic basin about 10-cm deep.
One student has a 100-mL beaker and the other has a 250-mL beaker.
Add water to fill one of the basins about two-thirds full.
In unison, the students dip their beakers into their basins and transfer any water in their
beakers to the other investigator's basin.
 Continue the water transfers for a few minutes.
 Observe and record the water levels in the two basins.
NOTE: To make this activity work, each student should put her/his beaker into the basin with the
beaker lying on its side and then tilt the beaker upwards to retain whatever water it can for the
transfer. Ultimately, "equilibrium" (unchanging liquid levels in the two basins) will be reached
when the amount of water that the 250-mL beaker can retain from a shallow depth of water is
equal to the amount that the 100-mL beaker can retain from a deeper pool. The size of the basins
and initial amount of water should be chosen by experiment to produce the equilibrium state in
one or two minutes, so as not to make the activity messy or tedious.
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ACS Chemistry FROG
Chapter 9
Chemical Equilibria
Alternative procedure:
 One faculty member gives each student about a dozen pieces of hard candy or other small
identical items. Students pair up and transfer candy back and forth in prescribed fractions. For
example, one student transfers one-half the amount s/he has and the partner transfers onefourth the amount s/he has. After a few transfers, the number of pieces of candy each partner
has remains constant, although the pieces continue to be transferred back and forth. This
activity can be carried out in small groups, so many students have the opportunity to
participate and observe the results close up.
Follow-up discussion:
 Use Consider This 9.2 to initiate discussion of the outcomes of this activity.
Follow-up activities:
 Investigate This 9.3. Do solutions of Fe(NO3)3 and KSCN react when mixed?
 Consider This 9.4. How do solutions of Fe(NO3)3 and KSCN react when mixed?
 Check This 9.5. Concentrations in the Fe(NO3)3-KSCN mixture.
 Check This 9.6. Dynamic equilibrium animation.
 Check This 9.7. Adding SCN-(aq) to a Fe(NO3)3-KSCN solution?
 End of chapter problems 9.1 through 9.8.
Consider This 9.2. What is dynamic equilibrium?
Goal:
Define dynamic equilibrium and relate the definition to the observations in Investigate This 9.1.
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to complete this activity. Then, the
groups can share their analyses with the class, summarizing them on the chalkboard or an
overhead transparency.
 This activity could be conducted as an open class discussion.
Time for activity:
 5-10 minutes.
Instructor notes:
 Review Le Chatelier's principle, which was first introduced in Chapter 2, Section 2.14.
 Try to be sure the class agrees on the results from Investigate This 9.1.
Students should reason and conclude:
(a) After several transfers of water, the water level in each basin remains constant. It does not
make a difference which basin initially contains water. If the same amount of water is used in
each trial, the water levels in the basins will come to the same constant level (different in
each basin) after several transfers. We can predict that the result will be the same if the water
is initially distributed so some is in each basin. [If there is some controversy about this
prediction, it should be tested experimentally.]
(b) The water comes to constant levels in the basins even as the transfer of water continues.
If we imagine that the transfers of water represent the forward and reverse directions of a
chemical reaction, then we are led to think of chemical equilibrium as a dynamic situation in
which both the forward and reverse reactions continue to occur, but at rates that are equal so
that the amount of reactant and amount of product in the system remain constant.
Follow-up discussion:
 Use the wood-oxygen example (or another you prefer) to illustrate the difficulty deciding
whether systems are in equilibrium or just reacting so slowly as to appear to be unchanging.
ACS Chemistry FROG
9
Chemical Equilibria
Chapter 9
Follow-up activities:
 Investigate This 9.3. Do solutions of Fe(NO3)3 and KSCN react when mixed?
 Consider This 9.4. How do solutions of Fe(NO3)3 and KSCN react when mixed?
 Check This 9.5. Concentrations in the Fe(NO3)3-KSCN mixture.
 Check This 9.6. Dynamic equilibrium animation.
 Check This 9.7. Adding SCN-(aq) to a Fe(NO3)3-KSCN solution.
 End of chapter problems 9.1 through 9.8.
Investigate This 9.3. Do solutions of Fe(NO3)3 and KSCN react when mixed?
Goal:
Observe the results when solutions of Fe(NO3)3 and KSCN are mixed and the effect of addition
of more of each reagent.
Set-up time:
 15-20 minutes or somewhat longer if you prepare individual sets of reagents for small groups
to do the activity.
Time for activity:
 From 15 to 20 minutes, depending on how far the discussion is extended to include molecular
level interpretations.
Materials:
 24-well plate.
Reagents:
 0.002 M potassium thiocyanate, KSCN.
 Tiny crystal of potassium thiocyanate, KSCN.
 0.2 M iron (III) nitrate, Fe(NO3)3. It may be necessary to acidify this solution with nitric acid
to prevent formation of hydrous iron hydroxides that will give the solution a yellowish color.
Procedure:
 You can conduct this as a class investigation or in small groups. Use Consider This 9.4 and
questions that arise from the class to initiate discussion of the observations and their
interpretations as the activity is carried out.
SAFETY NOTE
Wear your safety goggles.

(a) To each of three adjacent wells in a 24-well plate (on an overhead projector for the class or
on a white sheet of paper in the groups), add about 1 mL (one-third of a well) of 0.002 M
potassium thiocyanate, KSCN, solution. To each of the three wells, add one drop of 0.2 M
iron(III) nitrate, Fe(NO3)3. Record your observations on the appearance of the two reagent
solutions and their mixtures.
 (b) To one of the mixtures add one more drop of the iron(III) nitrate solution. Record your
observations on the appearance of this mixture compared to the others.
 (c) To one of the remaining original mixtures, add a tiny crystal of solid KSCN and observe
what happens. Record your observations on the appearance of this mixture compared to the
others.
Anticipated results:
 (a) The starting solutions are essentially clear and colorless and all three mixed solutions
should be the same color orange.
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ACS Chemistry FROG
Chapter 9
Chemical Equilibria

(b) The mixed solution to which more iron(III) ion is added will be a deeper (darker) orange
than the two that have been left undisturbed.
 (c) The mixed solution to which more thiocyanate ion is added will be a deep orange to red,
darker than either the control or the solution with added iron(III). It is difficult to add such a
tiny amount of thiocyanate ion that the solution only darkens a little bit.
 This is a photograph of the results with the initial solution, the solution with added iron(III),
and the solution with added thiocyanate from left to right.
Disposal:
 Wash the solutions down the drain with copious amounts of water.
Follow-up discussion:
 Use Consider This 9.4 to guide discussion of this activity as it is being carried out.
Follow-up activities:
 Check This 9.5. Concentrations in the Fe(NO3)3-KSCN mixture.
 Check This 9.6. Dynamic equilibrium animation.
 Check This 9.7. Adding SCN-(aq) to a Fe(NO3)3-KSCN solution?
 End of chapter problems 9.1 through 9.8.
Consider This 9.4. How do solutions of Fe(NO3)3 and KSCN react when mixed?
Goal:
Conclude, based on Investigate This 9.3 observations, that there are unreacted reactants in the
initial mixture and more products can be formed if more of one or the other reactant is added.
Classroom options:
 You can conduct this activity as an open class discussion or have the students work in small
groups (the only option, if they are doing Investigate This 9.3 in groups) to formulate their
responses to the questions as the discussion proceeds while Investigate This 9.3 is carried out.
Time for activity:
 From 15 to 20 minutes, including Investigate This 9.3.
Instructor notes:
 This discussion should go on concurrently with Investigate This 9.3.
Students should reason and conclude:
(a) When clear, colorless Fe(NO3)3 and KSCN solutions are mixed the mixed solution is a
clear orange color. The color change is a good indication that a chemical reaction has
occurred.
(b) When more Fe(NO3)3 solution was added to one of the mixed solutions, a darker orange
color was observed indicating that more of the colored product of the reaction between
Fe(NO3)3 and KSCN had been formed. There must have been some unreacted thiocyanate
ACS Chemistry FROG
11
Chemical Equilibria
Chapter 9
ion in the mixed solution, so that added iron(III) could react to form more product. The
reaction is assumed to occur between the iron(III) ions and thiocyanate ions, because nitrate
anions and potassium cations are generally unreactive and usually do not form colored
compounds.
(c) When more KSCN solid was added to another of the mixed solutions, a darker orange to
red color was observed indicating that more of the colored product of the reaction between
Fe(NO3)3 and KSCN had been formed. There must have been some unreacted iron(III) ions
in the mixed solution so the added thiocyanate ion could react to form more product.
Follow-up discussion:
 Discuss the Fe-SCN complex ion formation and lead in to application of Le Chatelier's
principle to this reaction.
Follow-up activities:
 Check This 9.5. Concentrations in the Fe(NO3)3-KSCN mixture.
 Check This 9.6. Dynamic equilibrium animation.
 Check This 9.7. Adding SCN-(aq) to a Fe(NO3)3-KSCN solution?
 End of chapter problems 9.1 through 9.8.
Investigate This 9.8. What conditions affect CoCl2 in solution?
Goal:
Observe how solvent composition and temperature affect the color of CoCl2 in alcohol-water
solutions.
Set-up time:
 15 minutes (assuming reagents are available)
Time for activity:
 10-15 minutes (including discussion).
Materials:
 One dry 125-mL Erlenmeyer flask.
 Four dry 200-mm test tubes.
 Rack or other means to support the test tubes.
 Plastic transfer pipet.
 Ice water in a 1000-mL beaker.
 Water bath at 80-90 ºC. A 1000-mL beaker of water on a hot plate is sufficient.
Reagents:
 2 grams of solid CoCl2·6H2O.
 50 mL of 91% isopropanol (isopropyl alcohol is usually available in drugstores). The alcohol
has to be at least 91% or reagent grade. A lot of water in the solution ruins the color
formation. The solution of Co(II) in alcohol could be prepared in advance, but this ruins the
charm of seeing the deep reddish-purple solid form a beautiful blue solution.
Procedure and anticipated results:
 Conduct this as a class investigation, perhaps with student volunteers.
SAFETY NOTES
Wear your safety goggles.
Cobalt compounds are somewhat toxic; wear disposable
gloves when handling the solid and solutions.
Alcohols are flammable; extinguish all flames before
conducting this investigation.
12
ACS Chemistry FROG
Chapter 9
Chemical Equilibria

Place about 2 g of solid CoCl2·6H2O in a dry 125-mL Erlenmeyer flask. Add about 50 mL of
91% isopropanol. As this dissolution is carried out, have the class observe and record the
appearance, especially the colors, of all the solids, liquids, and solutions.
 Divide the solution equally among four dry 200-mm test tubes. The four aliquots of solution
in 91% isopropanol are shown in this photograph.
(a) Add water a few drops at a time to the Co2+ ion solution in one of the test tubes and mix
thoroughly after each addition. About 10 drops of water will be needed to turn the solution a
lavender color, if the original solvent is 91% isopropanol. The photograph in the text with this
activity shows the color of this solution. Repeat this procedure with another of the solutions.
Finally, add this same amount of water to a third test tube of the solution and then add about 2
ml more water. Have the class record the colors of the solutions in all four test tubes.
 (b) Place the test tube with one of the lavender solutions from part (a) in a beaker of ice water.
Place the test tube with the other lavender solution in a beaker of 80-90 ºC water. After one to
two minutes, have the class observe and record the colors of the solutions. The cold lavender
solution will turn pink and the hot one will turn blue as in these photographs.


Remove the test tubes from the water baths, allow them to return to room temperature, and
observe and have the class record the colors of the solutions. The room temperature solutions
will return to the lavender color.
Alternative procedure:
 One faculty member used a different equilibrium system, beginning with an aqueous solution
of cobalt(II) chloride. Use concentrated HCl to shift the pink aqueous cobalt to purple, then
divide the resulting solution into four aliquots. Add silver nitrate to one sample, water to a
second, and more chloride (HCl) to a third. The photograph below displays the results. The
reason for choosing the somewhat simpler (fewer reactants) system in the textbook is that it is
easier to isolate the temperature variable than to bring in concentrations, as here. A difficulty
for students is understanding that the only variable of importance in the aqueous system is the
chloride concentration and that the role of water (when added to the lavender solution) is to
ACS Chemistry FROG
13
Chemical Equilibria
Chapter 9
dilute the chloride and thus shift the equilibrium toward the aquo complex. Too often,
students believe that adding the water adds significantly to the concentration of water and that
that is the stress on the equilibrium.
Extension:
Each student can make a humidity indicator that they can observe over a period of several days
and then report back their observations with an explanation. The directions are:
 Wet a piece of filter paper with the original alcohol solution of CoCl2·6H2O.
 Allow the paper to dry in the air or heat it gently with a hair dryer to speed the drying.
 Tack the paper someplace in your room where you can observe any changes that occur.
 Keep a record of your observations and the weather over a couple of weeks.
Questions:
 What correlations do you find, if any, between your observations and the weather?
 How might you explain the correlations?
Disposal:
 Dispose of the solutions containing Co(II) in a properly labeled waste container according to
your campus/local guidelines/ordinances.
Follow-up discussion:
 Use Consider This 9.9 to guide discussion of this activity as it is being carried out.
Follow-up activities:
 Consider This 9.10. What occurs when water is added to alcoholic Co2+ solutions?
 Check This 9.11. Temperature and Le Chatelier's principle.
 End of chapter problems 9.1 through 9.8.
Consider This 9.9. How do solvents and temperature affect CoCl2 in solution?
Goal:
Try to interpret the effects of solvent composition and temperature on the color of CoCl2 in
alcohol-water solutions.
Classroom options:
 You can conduct this activity as an open class discussion or have the students work in small
groups to formulate their responses to the questions as the discussion proceeds while
Investigate This 9.8 is carried out.
Time for activity:
 Approximately 5-15 minutes, depending on the amount of molecular level interpretation the
discussion entails.
Instructor notes:
 This discussion should go on concurrently with Investigate This 9.8.
Students should reason and conclude:
14
ACS Chemistry FROG
Chapter 9
Chemical Equilibria
(a) The color of the solution is blue. [Some see a touch of lavender.] Since CoCl2·6H2O(s) is
a deep reddish purple solid, the blue color of the solution is surprising. We may recall, from
Investigate This 2.38, that the cobalt(II) cation in water is pink, but forms a deep blue
precipitate with the phosphate anion. Thus red/pink and blue colors seem to be common in
compounds and/or solutions containing cobalt(II).
(b) The color of solution that has the most water is pink. This is the color we might have
expected from the dissolution of the solid and is the color of the aqueous solutions in
Investigate This 2.38. Formation of a pink color on addition of water to the alcohol-water
solvent is consistent with these other observations. [The pink color is from the
Co(H2O)62+(aq) complex.]
(c) When the lavender solution is heated, it turns blue, and when it is cooled to ice-water
temperature, it turns pink. The blue color of the hot solution is essentially the same as in the
original solution before water was added. The pink color of the cold solution is the same as
the color when excess water was added to go from lavender to pink in part (a). It seems likely
that the lavender solution contains a mixture of a blue Co(II) species (probably a Lewis acidbase metal ion complex) and a pink Co(II) species (another complex). The pink complex
seems to depend on the presence of a good deal of water in the solvent (and/or low
temperature), so we might postulate that it is an aquo complex, Co(H2O)62+(aq), as we found
for several other complexes in Chapter 6, Section 6.6. The only other complexing ligand in
this solution is Cl–(aq), so we might suppose that a complex (or complexes) of Co(II) cation
with Cl– are responsible for the blue color. This complex seems to form when the proportion
of water in the solvent is low and/or at elevated temperatures. Thus, in a solution that
contains a mixture of the two complexes, high temperature causes the blue (chloride)
complex to dominate the color of the solution and low temperature favors the pink (aquo)
complex.
NOTE: Without further information we can’t know that the chloride complex can be represented
as CoCl42-(aq). It’s quite unlikely that much, if any, of the tetrachloro complex actual forms in
these solutions. There is some evidence that the coordination about the Co(II), and hence the
color, changes when two or more chloride ligands have bonded. The light absorptivity of the
chloro complex(es) is a great deal higher than for the aquo complex, so their blue color
dominates the solution color, even when they are present in much lower concentration than the
aquo complex. The color changes in this investigation, and others involving the chloro
complexes, rest on fairly small changes in the equilibrium concentrations of the complexes in
these solutions.
Follow-up activities:
 Consider This 9.10. What occurs when water is added to alcoholic Co2+ solutions?
 Check This 9.11. Temperature and Le Chatelier's principle.
 End of chapter problems 9.1 through 9.8.
Consider This 9.10. What occurs when water is added to alcoholic Co2+ solutions?
Goal:
Explain what occurs when water is added to an alcoholic solution containing Co2+ and Cl– ions.
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to formulate their response and then
have the groups share their work with the class.
 This activity could be conducted as an open class discussion.
ACS Chemistry FROG
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Chemical Equilibria
Chapter 9
Time for activity:
 5-10 minutes.
Instructor notes:
 Try to be sure that the class agrees on the results from Investigate This 9.8.
Students should reason and conclude:
Equation (9.3) explains the changes that we observed. If the system described by this
equation is in equilibrium, then addition of water, shown as a product in the equation,
disturbs the equilibrium by increasing the concentration of a product of the reaction. (There
is also a secondary effect of decreasing the concentrations of all species by increasing the
volume of the solution.) The equilibrium system responds, as Le Chatelier’s principle states,
by changing to minimize the effect of the disturbance. That is, it shifts toward the reactant
side and forms more of the pink aquo complex, thus using up some of the added water (and
also decreasing the amount of blue complex).
Follow-up discussion:
 Draw attention to and use the discussion to summarize points in the Reflection and Projection
section.
Follow-up activities:
 Check This 9.11. Temperature and Le Chatelier's principle.
 End of chapter problems 9.1 through 9.8.
Section 9.2. Mathematical Expression for the Equilibrium Condition
Learning Objectives for Section 9.2
 Write the equilibrium constant expressions, using appropriate “concentration” units, for
balanced chemical reactions.
Investigate This 9.12. What is the pH of an acetic acid solution?
Goal:
Measure the pH of 0.1 M aqueous solution of acetic acid and explain the result.
Set-up time:
 15 minutes (assumes that solutions, including buffers for standardization, are prepared
previously) to standardize the pH meter and electrode.
Time for activity:
 Less than 5-10 minutes (including discussion).
Materials:
 pH meter with combination pH electrode. If the pH can be projected for the class to see, that
enhances the presentation.
Reagents:
 0.1 M aqueous acetic acid.
 Standardization buffers for the pH meter and electrode.
Procedure:
 Conduct this as a class investigation, but have students work in small groups to discuss,
analyze, ad explain the results.
SAFETY NOTE
Wear your safety glasses
16
ACS Chemistry FROG
Chapter 9
Chemical Equilibria

Measure the pH (0.02 units) of a 0.1 M aqueous solution of acetic acid, using a calibrated pH
meter and electrode.
 Record the pH of the sample.
Anticipated results:
 The pH should be about 2.9. This value is sensitive to impurities in the sample and container,
but should be in this ballpark, well above what it would be for a strong acid, about pH 1.
Students should reason and conclude that:
Based on previous results, including Investigate This 6.30, we have concluded that acetic
(ethanoic) acid is a weak acid that transfers only some of its protons to water. If it had
transferred all of its protons to water, the resulting pH would have been approximately 1.
Disposal:
 Rinse the solution down the drain.
Follow-up discussion:
 Use the discussion to lead in to Consider This 9.13, which leads to the equilibrium constant
expression and equilibrium constant.
Follow-up activities:
 Consider This 9.13. How is an equilibrium system characterized quantitatively?
 Worked Example 9.14. Equilibrium constant expression and equilibrium constant.
 Check This 9.15. Equilibrium constant expression and equilibrium constant.
 Consider This 9.16. What can equilibrium constant expressions tell us?
 End of chapter problem 9.10.
Consider This 9.13. How is an equilibrium system characterized quantitatively?
Goal:
Analyze a series of experimental data for an acid-base equilibrium to find a mathematical
relationship among the concentrations.
Classroom options:
 Allow about 5 minutes for students, working in small groups, to try to find relationships and
then have the groups share their suggestions with the class on the chalkboard or an overhead
transparency.
 This activity could be conducted as an open class discussion.
 This activity could be assigned as homework, if your class is prepared for this challenge, and
discussed at the next class session.
Time for activity:
 From 5 to 15 minutes, depending on how far you want the class to go on its own without
being led too overtly.
Instructor notes:
 Show or refer students to the table of data from the series of experimental results.
Students should reason and conclude:
(a) The total dye concentration in the solutions is 5.2  10–5 M. This is the concentration of D
when the pH is high and essentially all the dye is unreacted and the concentration of HD+ at
low pH when essentially all the dye has reacted. At the intermediate pHs, the sum of [D] plus
[HD+] is 5.2  10–5 M, showing that it is all accounted for in the solution.
(b) As [H3O+(aq)] increases, we see that [D] decreases and [HD+] increases. The addition of
H3O+(aq) increases the concentration of a reactant and Le Chatelier’s principle predicts that
ACS Chemistry FROG
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Chapter 9
the system will respond by reacting to use up some of the added reactant, thus forming more
of the product, HD+, at the expense of the reactant, D.
(c) The ratio of the concentration of the acid form of the dye [HD+] to the concentration of
the base form of the dye [D], in the three solutions increases in the order [HD+]/[D] = 0.27,
1.1, 2.7 as [H3O+(aq)] increases from 1.0  10–5 M to 1.0  10–4 M.
(d) Consider the result in part (c) as you seek a mathematical relationship among [H3O+(aq)],
[D(aq)], and [HD+(aq) that has the same numeric value in the three intermediate-pH
solutions. Noting that the ratio [HD+]/[D] increases by a factor of 10 when [H3O+(aq)]
increases by 10, perhaps multiplying the ratio by 1/[H3O+(aq)] will give a constant. For the
three solutions we have:
[HD + ]
1
= 0.27/(1.0  10–5 M) = 2.7  104 M–1

+
[D] [H3O ]
= 1.1/(4.0  10–5 M) = 2.75  104 M–1
= 2.7/(1.0  10–4 M) = 2.7  104 M–1
The relationship shown in these equations gives a constant numeric value and solves the
problem. Note that the inverse relationship, [D][H3O+]/[HD+] also is constant, 3.7  10–5 M,
and we have no way to choose between these possibilities.
Follow-up discussion:
 Discuss and introduce the equilibrium constant expression and equilibrium constant, K.
Follow-up activities:
 Worked Example 9.14. Equilibrium constant expression and equilibrium constant.
 Check This 9.15. Equilibrium constant expression and equilibrium constant.
 Consider This 9.16. What can equilibrium constant expressions tell us?
 End of chapter problem 9.10.
Consider This 9.16. What can equilibrium constant expressions tell us?
Goal:
Analyze equilibrium constant expressions and show how the results are related to Le Chatelier’s
principle.
Classroom options:
 Allow about 5 minutes for students, working in small groups, to answer these questions and
then have the groups share their results with the class on the chalkboard or an overhead
transparency.
 This activity can be conducted as an open class discussion.
 This activity could also be assigned as a homework problem and then discussed at the next
class section.
Time for activity:
 Approximately 10-15 minutes.
Instructor notes:
 Define/discuss the equilibrium constant expression before conducting this activity.
Students should reason and conclude:
(a) The equilibrium constant expression for reaction (9.4), written in the ratio form shown
equation (9.6) is:
18
ACS Chemistry FROG
Chapter 9
Chemical Equilibria
 H 3O+ (aq)  OAc - (aq) 
K= 

 HOAc(aq) H 2O(l) eq
(b)/(c) If more acetic acid is added (without any other change occurring), the denominator of
the equilibrium constant expression in part (a) would become larger and the relationship
would no longer be equal to K. To restore the ratio to its equilibrium value, some of the
acetic acid would have to transfer protons to water forming more of the products, including
the hydronium ion. The resulting pH will be a bit lower than before the acetic acid is added.
Our answer is consistent with the predictions of Le Chatelier’s Principle. Addition of acetic
acid is a disturbance to the equilibrium system; more of one of the reactants has been added.
The system reacts to minimize the disturbance by reducing the concentration of reactants,
that is, by transferring some protons from the acetic acid and thus forming more product, just
as the analysis based on the equilibrium constant expression also told us.
(d) One relationship we found in Consider This 9.13(c) was (the state designation, (aq), is
omitted for all species):
[HD + ]
[D][H 3 O+ ]
The equilibrium constant expression for the reaction in Consider This 9.13 is:
[HD + ][H 2 O]
K=
[D][H 3O+ ]
These two expressions are the same, except for the appearance of the water concentration in
the equilibrium constant expression. We had no information about the water concentration in
Consider This 9.13, but we can guess that its concentration must not vary much in this
aqueous system, since the relationship we found is a constant without taking the water into
account.
Follow-up discussion:
 Part (d) is designed to be a lead-in to a discussion of standard states in solution, in order to
reconcile the empirical relationship from Consider This 9.13(c) with the equilibrium constant
expression modeled on equation (9.6).
 Apply the discussion to the acetic acid-water reaction and show how the equilibrium constant
expression in its conventional form without the concentration of water arises.
Follow-up activities:
 Worked Example 9.17. Acid equilibrium constant for acetic acid.
 Check This 9.18. Acid equilibrium constant for acetic acid.
 Worked Example 9.19. pH of an aqueous solution of lactic acid.
 Check This 9.20. pH of an aqueous solution of benzoic acid.
 End of chapter problems 9.9 through 9.14.
Section 9.3. Acid-Base Reactions and Equilibria
Learning Objectives for Section 9.3
 Use Le Chatelier’s principle to predict the direction of the response of an acid-base
equilibrium system to changes in the concentration(s) of reactants or products.
ACS Chemistry FROG
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Chapter 9

Find Ka and pKa for a weak acid (or Kb and pKb for a weak base) when you have the initial
concentrations in the solution and the pH or pOH at equilibrium.
 Find the pH and/or pOH of a solution of a weak acid or its conjugate base when you have the
initial concentrations in the solution and the Ka or pKa for the acid.
 Find the concentrations of a weak acid and its conjugate base in a solution when you have the
initial concentrations in the solutions and the Ka or pKa for the acid.
Investigate This 9.23. What is the pH of an acetate ion solution?
Goal:
Measure the pH of a 0.1 M aqueous solution of sodium acetate.
Set-up time:
 15 minutes (assumes that solutions, including buffers for standardization, are prepared
previously) to standardize the pH meter and electrode.
Time for activity:
 Approximately 10-15 minutes (including discussion).
Materials:
 pH meter with combination pH electrode. If the pH can be projected for the class to see, that
enhances the presentation.
Reagents:
 0.1 M aqueous sodium acetate solution.
 Standardization buffers for the pH meter and electrode.
Procedure:
 Conduct this as a class investigation, but have students work in small groups to discuss,
analyze, ad explain the results.
SAFETY NOTE
Wear your safety glasses

Measure the pH (0.02 units) of a 0.1 M aqueous solution of sodium acetate, using a
calibrated pH meter and electrode.
 Record the pH of the sample.
Anticipated results:
 The pH should be about 8.6-8.8. This value is sensitive to impurities in the sample and
container, but should be in this ballpark.
Disposal:
 Rinse the solution down the drain.
Follow-up discussion:
 Use Consider This 9.24 to initiate discussion of the results of this activity as it is being carried
out and use as a lead in to discussion of the equilibrium constant expression and equilibrium
constant for the acetate ion-water reaction.
NOTE: The reaction of Brønsted-Lowry conjugate bases with water is sometimes called
hydrolysis, but we have consciously avoided that term (which we reserve for reactions in which
water actually does “lyse” or break bonds, as in the hydrolysis of esters or amides) and simply
treat the reaction as another acid-base reaction, so as to keep the proton transfer at the center of
attention.
Follow-up activities:
 Worked Example 9.25. Base equilibrium constant for acetate ion.
20
ACS Chemistry FROG
Chapter 9





Chemical Equilibria
Check This 9.26. Base equilibrium constant for acetate ion.
Check This 9.27. Ka·Kb for the acetic acid-acetate ion pair.
Check This 9.28. Using the relationship of pKa to pKb
Consider This 9.29. What is the relationship of pH and pOH?
End of chapter problems 9.15 through 9.27.
Consider This 9.24. Does the pH of the acetate ion solution make sense?
Goal:
Conclude that the pH of a solution of sodium acetate (the conjugate base of a weak acid) should
be basic, pH > 7.
Classroom options:
 You can conduct this activity as an open class discussion or have the students work in small
groups to formulate their responses to the questions as the discussion proceeds (and is
extended) while Investigate This 9.23 is carried out.
Time for activity:
 From 5-10 minutes to 10-15 minutes, depending on the amount of discussion of conjugate
weak acid-weak base properties.
Instructor notes:
 Be sure the class agrees on the pH of the 0.1 M aqueous sodium acetate solution in
Investigate This 9.23.
Students should reason and conclude:
The acetate (ethanoate) ion is the conjugate base of a weak acid. We know that acetic acid
does not transfer most of its protons to water, so the acetate ion must have a good deal of
attraction for protons and water would be expected to transfer some protons to acetate ion,
thus forming acetic acid and hydroxide ion. The hydroxide ion should give the solution a
basic pH (pH > 7), as is observed in Investigate This 9.23. The observation is consistent with
our reasoning.
Follow-up discussion:
 Set up an equilibrium constant expression as shown in equation (9.19), pointing out that Kb
refers to a base reacting to accept a proton from water.
Follow-up activities:
 Worked Example 9.25. Base equilibrium constant for acetate ion.
 Check This 9.26. Base equilibrium constant for acetate ion.
 Check This 9.27. Ka·Kb for the acetic acid-acetate ion pair.
 Check This 9.28. Using the relationship of pKa to pKb
 Consider This 9.29. What is the relationship of pH and pOH?
 End of chapter problems 9.15 through 9.27.
Consider This 9.29. What is the relationship of pH and pOH?
Goal:
Derive the relationship between pH and pOH and use it to find [OH–(aq)] in a solution of known
pH.
Classroom options:
 This activity can be conducted as an open class discussion, perhaps following up a discussion
of student results for Check This 9.27 and 9.28.
Time for activity:
 About 5 minutes.
ACS Chemistry FROG
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Chapter 9
Instructor notes:
 Show or refer students to equation (9.16) as they discuss this activity.
Students should reason and conclude:
(a) Begin by taking the logarithm of both sides of equation (9.16):
log(Kw) = log{(H3O+(aq))(OH–(aq))} = log(H3O+(aq)) + log(OH–(aq))
Multiply through this equation by –1 and then apply the definition of p:
–log(Kw) = –log(H3O+(aq)) – log(OH–(aq))
pKw = pH + pOH
14.00 = pH + pOH
The sum of pH and pOH in an aqueous solution at 298 K (where pKw = 14.00) is always
14.00.
(b) If a solution has pH = 8.73, its pOH = 14.00 – 8.73 = 5.27. Using the definition of pOH
we have:
(OH–(aq)) = 10–pOH = 10–5.27 = 5.4  10–6
The concentration of hydroxide ion calculated in Worked Example 9.25 for a solution with
pH = 8.73 is the same as that calculated here (within round-off uncertainties). This is not
surprising, since the only difference is that this one is calculated using the logarithmic values
instead of from the corresponding concentrations and equilibrium constants.
Section 9.4. Solutions of Conjugate Acid–Base Pairs: Buffer Solutions
Learning Objectives for Section 9.4
 Use Le Chatelier's principle to predict the direction of the response of an equilibrium system
to changes in the concentration(s) of reactants or products.
 Find the pH of a buffer solution when you have the initial concentrations of the weak acid and
its conjugate base in the solution and the Ka or pKa for the acid.
 Describe how to prepare a buffer solution of a specified pH and specified total concentration
of an appropriately selected weak acid-base conjugate pair.
Investigate This 9.30. What are the pHs of conjugate acid-base pair solutions?
Goal:
Measure the pH of three acetic acid/sodium acetate (buffer) solutions.
Set-up time:
 15 minutes (assumes that solutions, including buffers for standardization, are prepared
previously) to standardize the pH meter and electrode.
Time for activity:
 From about 10 to 20 minutes including discussion based on Consider This 9.31, which should
be incorporated with this activity as it is being carried out.
Materials:
 pH meter with combination pH electrode. If the pH can be projected for the class to see, that
enhances the presentation.
 Test tubes or vials large enough to accommodate the pH electrode. You could place the
reagents in the containers before class, but there is some pedagogical advantage to showing
the mixture preparation to reinforce the ratio differences among the samples.
 A beaker or other container and water rinse bottle to clean the electrode between solutions.
22
ACS Chemistry FROG
Chapter 9
Chemical Equilibria
Reagents:
 0.1 M ethanoic (acetic) acid, CH3C(O)OH.
 0.1 M sodium ethanoate (acetate), CH3C(O)ONa.
 Water to rinse the electrode between solutions.
 Standardization buffers for the pH meter and electrode.
Procedure:
 Do this as a classroom activity with student volunteers to carry out the measurements.
SAFETY NOTE
Wear your safety goggles.

Use a calibrated pH meter to determine the pH (0.02 units) of these solutions:
Sample
#1
#2
#3
0.10 M acetic acid
9.0 mL
5.0 mL
1.0 mL
0.10 M sodium acetate
1.0 mL
5.0 mL
9.0 mL
NOTE: If time is short, you could just provide the pH values, but this is very unsatisfactory,
since the impact of the activity, which is central to the concept of buffer solutions, is lost.
Anticipated results:
Ideal results and a set of experimental results are given in this table:
Sample
Ideal pH
Expt’l pH
#1
3.81
3.66
#2
4.76
4.76
#3
5.71
5.87
NOTE: It appears that the experimental pHs for samples #1 and #3 were made on mixtures that
contained 10-12 mL of one solution and 1 mL of the other. This is probably why the pH value
for sample #1 is about 0.15 units too acidic and that for sample #3 is about 0.16 units too basic.
Disposal:
 Rinse reagents down the drain.
Follow-up discussion:
 Incorporate Consider This 9.31 with this activity to initiate and direct the discussion of the
results as they are obtained.
Follow-up activities:
 Worked Example 9.32. Ka for acetic acid.
 Check This 9.33. Ka for acetic acid.
 Check This 9.34. Conjugate base-to-acid ratios and pKa for acetic acid.
 Investigate This 9.35. How does a buffer respond to added H3O+(aq) or OH-(aq)?
 Consider This 9.36. How does buffer pH respond to added H3O+(aq) or OH-(aq)?
 End of Chapter problems 9.28 through 9.38.
ACS Chemistry FROG
23
Chemical Equilibria
Chapter 9
Consider This 9.31. Do the pHs of conjugate acid-base pair solutions make sense?
Goal:
Conclude that the pHs of conjugate acid-base pair solutions are lower when the proportion of
conjugate acid is higher and higher when the proportion of conjugate base is higher.
Classroom options:
 You can conduct this activity as an open class discussion or have the students work in small
groups to formulate their responses to the questions as the discussion proceeds while
Investigate This 9.30 is carried out.
Time for activity:
 Approximately 5-10 minutes, which are incorporated as a part of Investigate This 9.30.
Instructor notes:
 Conduct this activity concurrently with Investigate This 9.30.
Students should reason and conclude:
The pH of the solution that contains the most acid has the lowest pH value while the pH of
the solution that contains the most base has the highest pH value. These correlations make
sense in terms of the acid-base properties, since the pH of the conjugate acid alone is low and
the pH of the conjugate base alone is high, pH > 7.
Follow-up discussion:
 Point out that an order of magnitude change in the conjugate acid to conjugate base ratio
results in one unit of change in the pH. Smaller changes in the ratio will lead to
correspondingly smaller changes in pH.
 Use this discussion to introduce buffer solutions.
Follow-up activities:
 Worked Example 9.32. Ka for acetic acid.
 Check This 9.33. Ka for acetic acid.
 Check This 9.34. Conjugate base-to-acid ratios and pKa for acetic acid.
 Investigate This 9.35. How does a buffer respond to added H3O+(aq) or OH-(aq)?
 Consider This 9.36. How does buffer pH respond to added H3O+(aq) or OH-(aq)?
 End of Chapter problems 9.28 through 9.38.
Investigate This 9.35. How does a buffer respond to added H3O+(aq) or OH–(aq)?
Goal:
Observe the difference in pH change when H3O+(aq) or OH-(aq) is added to water compared to
an acetic acid/acetate buffer.
Set-up time:
 15 minutes (assumes that solutions, including buffers for standardization, are prepared
previously) to standardize the pH meter and electrode.
Time for activity:
 15-20 minutes (including discussion guided by Consider This 9.36).
Materials:
 pH meter with combination pH electrode. If the pH can be projected for the class to see, that
enhances the presentation. You could use pH paper or a universal indicator solution for these
measurements, since they are only semiquantitative and the differences are large.
 Small beakers/vials or well plate with wells large enough to accommodate the pH electrode.
 Two thin-stem plastic pipets for the HCl and NaOH solutions.
Reagents:
 0.1 M acetic acid, CH3C(O)OH.
24
ACS Chemistry FROG
Chapter 9
Chemical Equilibria




0.1 M sodium acetate, CH3C(O)ONa.
1.0 M hydrochloric acid, HCl.
1.0 M sodium hydroxide, NaOH.
Standardization buffers for the pH meter and electrode.
Procedure:
 Conduct this as a class activity with student volunteers. The rest of the class should work in
small groups to analyze the results (guided by Consider This 9.36) and share their analyses
with the class.
SAFETY NOTE
Wear your safety goggles.

Use four, clean, dry, labeled sample vials or test tubes large enough to accommodate a
combination pH electrode. Put 10 mL of water in two of the vials. In each of the other two
make a mixture of 5.0 mL 0.10 M aqueous acetic acid solution and 5.0 mL of 0.10 M aqueous
sodium acetate solution.
(a) Measure and record the pH of one of the water samples. Add one drop of 1.0 M aqueous
hydrochloric acid solution, mix the solution, and again measure and record the pH. Repeat
this procedure with one of the acetic acid-acetate mixtures in place of the water.
(b) Repeat the procedure with the other water sample and the other acetic acid-acetate
solution, but this time use one drop of 1.0 M aqueous sodium hydroxide solution instead of
hydrochloric acid.
Anticipated results:
 A set of experimental results are shown in this table:
Sample
pH
Distilled water
6.23
Distilled water + HCl
2.34
Distilled water + NaOH
12.00
Acetic acid/acetate mixture
4.77
Acetic acid/acetate mixture + HCl
4.71
Acetic acid/acetate mix + NaOH
4.80
Disposal:
 Rinse reagents down the drain with copious amounts of water.
Follow-up discussion:
 Use Consider This 9.36 to guide discussion of the results of this activity as it is being carried
out..
Follow-up activities:
 Check This 9.37. Stoichiometry in buffer solutions.
 Check This 9.38. pH and the conjugate acid-base ratio.
 Worked Example 9.39. pH change when hydronium ion is added to a buffer solution.
 Check This 9.40. pH change when hydroxide ion is added to a buffer solution.
 Consider This 9.41. How do you prepare a buffer solution of known concentration?
 End of chapter problems 9.28 through 9.38.
ACS Chemistry FROG
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Chemical Equilibria
Chapter 9
Consider This 9.36. Do the pHs of conjugate acid-base pair solutions make sense?
Goal:
Try to explain the difference in pH change when a strong acid (or base) is added to water
compared to a solution of a conjugate acid-base pair (buffer solution).
Classroom options:
 You can conduct this activity as an open class discussion or have the students work in small
groups to formulate their responses to the questions as the discussion proceeds while
Investigate This 9.35 is carried out.
Time for activity:
 Approximately 10 minutes incorporated with the time for Investigate This 9.35.
Instructor notes:
 Use this activity to guide discussion of the results from Investigate This 9.35 as they are
obtained.
Students should reason and conclude:
(a) When a drop of hydrochloric acid is added to water, the pH drops almost by about four
units, which is a 10,000-fold increase in the hydronium ion concentration in the solution.
When a drop of hydrochloric acid is added to the acetic acid/acetate mixture, the pH change
is only 0.06 units, which is an increase of only a factor of XX in the hydronium ion
concentration. When hydronium ion is added to water, its only possible reactions are to
transfer a proton to water (thus reproducing the reactants as products) or to transfer a proton
to the tiny amount of hydroxide ion produced by the reaction of water with itself (which uses
up a negligible amount of the added hydronium ion). Thus, the pH of this solution is
determined by the amount of hydronium ion added, assuming that it is all still present at
hydronium ion. We have carried out calculations on this sort of system in previous chapters
and found that the pH is fairly low, as observed here as well. When hydronium ion is added
to the acetic acid/acetate solution, it can transfer a proton to the acetate ion, a stronger base
than water (see Table 6.2 in Chapter 6, Section 6.4), which substantially reduces the amount
of hydronium in the solution and leads to a much smaller change in the pH, as observed here.
Based on this reasoning, the results make sense. Note that added hydronium ion does slightly
increase the acidity (lower the pH) of the acetic acid/acetate solution, as we would expect.
NOTE: This argument can also be made in terms of Le Chatelier’s principle applied to the acidbase equilibrium reaction for the conjugate acid and base in the solution:
HOAc(aq) + H2O(aq) Ó H3O+(aq) + OAc–(aq)
Addition of hydronium ion is a disturbance to this equilibrium that the system responds to by
decreasing the added hydronium ion by reaction with acetate ion, thus increasing the ratio of
acetic acid to acetate ion in the solution. In Consider This 9.31, we concluded (in the discussion)
that small changes in this ratio produce only small changes in the pH of the solution. Thus, the
small change observed here is reasonable and in the right direction, since the ratio of conjugate
acid to conjugate base is increased.
(b) When a drop of sodium hydroxide is added to water, the pH increases by about six units,
which is a 106-fold decrease in hydronium ion concentration in the solution (or, equivalently,
a 106-fold increase in the hydroxide ion concentration). When a drop of sodium hydroxide
solution is added to the acetic acid/acetate mixture, the pH change is only 0.03 units, which is
a decrease of only a factor of 7% in the hydronium ion concentration (or an increase of 7% in
the hydroxide ion concentration). Arguments similar to those in part (a) explain these results.
The acetic acid in the acetic acid/acetate solution is a stronger acid than water and donates
26
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Chapter 9
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protons to hydroxide ion, which substantially decreases its concentration in the solution and
leads to a much smaller change in the pH. Note that added hydroxide ion does slightly
decrease the acidity (raise the pH) of the acetic acid/acetate solution, as we would expect.
Reasoning based on Le Chatelier’s principle can also be used here.
Follow-up discussion:
 Use this activity as a lead-in to the discussion of acid-base buffer solutions (solutions that
decrease changes in pH, as any buffer is expected to decrease the effect of whatever stress it is
meant to deal with, as with the bumper on an automobile, which is supposed to buffer against
greater damage in a collision).
 Emphasize what is happening at the molecular level, as represented in Figure 9.4.
 Then go on to introduce the Henderson-Hasselbalch equation (useful to students in later
biochemistry courses), which is nothing more than a recasting of the acid-base equilibrium
constant expression for the conjugate acid and base in the buffer solution, and the summary of
all this discussion in Table 9.2.
Follow-up activities:
 Check This 9.37. Stoichiometry in buffer solutions.
 Check This 9.38. pH and the conjugate acid-base ratio.
 Worked Example 9.39. pH change when hydronium ion is added to a buffer solution.
 Check This 9.40. pH change when hydroxide ion is added to a buffer solution.
 Consider This 9.41. How do you prepare a buffer solution of known concentration?
 End of chapter problems 9.28 through 9.38.
Consider This 9.41. How do you prepare a buffer solution of known concentration?
Goal:
Calculate the quantities of a conjugate acid and its conjugate base required to prepare a buffer
solution of known pH and concentration.
Classroom options:
 Allow about 5 minutes for students, working in small groups, to answer these questions and
then have the groups share their results with the class on the chalkboard or an overhead
transparency.
 This activity could be conducted as an open class discussion.
Time for activity:
 About 10 minutes.
Instructor notes:
 Remind students that the solution to this problem is begun in the paragraph just preceding this
conjugate base 
activity where the necessary 
is calculated.
conjugate acid 
eq
Students should reason and conclude:
(a) For this buffer solution, we want [conjugate acid] + [conjugate base] = 0.050 M. Let us
set y = [conjugate base], so that [conjugate acid] = {(0.050 M) – y} and


conjugate base 
y
=


conjugate acid eq  (0.050 M) - y  eq
We know from the paragraph preceding this activity that:
ACS Chemistry FROG
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Chapter 9


conjugate base 
y
=
= 0.59


conjugate acid eq  (0.050 M) - y  eq
Solve for y to get y = [conjugate base] = 0.019 M and therefore [conjugate acid] = 0.031 M.
You can check this result by substituting these concentrations in the Henderson-Hasselbalch
equation to see you obtain the desired pH of the solution.
(b) For this solid acid and base, we need to find the mass of each solid that dissolved in one
liter of solution will give the concentrations calculated in part (a).
 0.031 mol   157.6 g 
0.031 M tris acid = 

 = 4.9 g

1 L   1 mol 
 0.019 mol   121.1 g 
0.019 M tris acid = 
 
 = 2.3 g

1L
1 mol 
Follow-up discussion:
 Use the discussion of this activity to reinforce the importance of buffer solutions and as a
lead-in to their roles in biological systems, which are introduced in the next section.
Follow-up activities:
 End of chapter problems 9.28 through 9.38.
Section 9.5. Acid-Base Properties of Proteins
Learning Objectives for Section 9.5
 Find the concentrations of a weak acid and its conjugate base in a solution when you have the
initial concentrations in the solution and the Ka or pKa for the acid.
 Predict the direction of change of the net charge on a protein and its consequent behavior in
electrophoresis if one amino acid is substituted for another in its structure.
Investigate This 9.42. How is a protein solution affected by addition of H3O+(aq)?
Goal:
Observe how an aqueous solution of a protein, casein, is affected by the pH of the solution.
Set-up time:
 20-25 minutes (assumes that the acid and base and buffers for standardization, are prepared
previously) to standardize the pH meter and electrode and prepare casein solution. For best
results, prepare the basic 0.25% casein the day before the activity is conducted. Since the pH
for this activity need not be accurately known (as long as it is within ±0.5 unit), you may not
need to standardize the pH meter and electrode.
Time for activity:
 10-15 minutes (including discussion).
Materials:
 pH meter with combination pH electrode. If the pH can be projected for the class to see, that
enhances the presentation.
 Magnetic stirrer.
 Magnetic stir bar.
 400-mL beaker.
Reagents:
 0.25% aqueous solution of casein. Dissolve 0.6 g casein in a mixture of 250 mL water and 1
mL of 3 M NaOH solution. Any grade casein will work, but the purer grades give colorless,
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almost clear solutions. You might need to add a drop or two more of 3 M NaOH, if the protein
does not dissolve relatively quickly.
 3 M HCl, hydrochloric acid. (Do not substitute other acids.)
 3 M NaOH, sodium hydroxide solution.
 Standardization buffers for the pH meter and electrode.
NOTE: Dispense the acid and base from dropper bottles or small wash bottles. You may need
25-50 mL of each, depending on how many times you wish to cycle back and forth to show the
reversibility of the precipitation process.
Procedure:
 Do this as a class activity with student volunteers to add the acid and base (and read the pH
meter, if necessary), while the rest of the class works in small groups to discuss and analyze
the results as they are obtained.
SAFETY NOTE
Wear your safety goggles.

Stir the protein solution in a 400-mL beaker at medium speed with a magnetic stirrer and a
magnetic stir bar.
 Monitor and record the pH of the solution as strong acid or base are slowly added.
 Add 3 M hydrochloric acid solution five drops at a time to the stirred casein solution. After
each addition, record your observations on the appearance of the solution and its pH. The
number of drops is not super critical. You are trying to make the changes rapidly enough not
to be boring, but without missing the interesting development of the precipitate and then its
disappearance. This may mean only a drop or two at a time as you pass through the pH of
maximum precipitation (the isoelectric pH)
 After the pH of solution has reached about 2, reverse the pH change by adding 3 M sodium
(or potassium) hydroxide solution five drops at a time to the stirred casein solution. After each
addition, record your observations on the appearance of the solution and its pH.
 Repeat this cycle to see if your observations are reproducible.
NOTE: When the casein precipitates, do not allow it to stay out of solution too long or it will
begin to clump up and not redissolve well as more acid (or base) is added. This means that you
need to pass through the isoelectric pH without long delays between additions of acid (or base in
the reverse direction).
Alternate procedure:
 One faculty member suggests the use of skim milk, a substance that is familiar to students,
instead of casein. If you choose to try this, you will have to experiment with the dilution of the
skim milk to get a solution that is not so opaque as to obscure the formation of precipitate
while still getting enough precipitate to give the activity appropriate drama.
Anticipated results:
 A representative series of results is shown in these tables. The first table shows the changes in
pH and appearance of the solution as HCl is added. The second table shows the results for
addition of NaOH solution taking the solution back through the isoelectric pH. In both cases,
the isoelectric point was passed too rapidly (too many drops at a time), so the increasing (and
then decreasing) amounts of precipitate are not captured. As the precipitate begins to form, the
rate of addition of acid or base should be decreased, while keeping in mind the warning in the
note about not going too slowly. Practice is useful so you can help direct your student
volunteers.
ACS Chemistry FROG
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Chapter 9
Total # of drops of HCl
pH
Observations
0
11.15
Slightly opalescent
5
11.09
Slightly opalescent
10
7.62
Slightly opalescent
15
4.76
A lot of white precipitate has formed.
20
2.74
Precipitate has dissolved and solution
returned to slightly opalescent.
Total # of drops of NaOH
pH
0
2.74
Slightly opalescent
5
2.86
Slightly opalescent
10
4.56
A lot of white precipitate has formed.
15
7.95
Precipitate has dissolved and solution
returned to slightly opalescent
20
10.88
Slightly opalescent
Observations
These photographs show the appearance of the solution at pHs far from and near the
isoelectric pH:
pH far from isoelectric pH
pH at approximately isoelectric pH
Disposal:
 Rinse the casein down the drain with copious amounts of water. Use any remaining acid and
base for other activities.
Follow-up discussion:
 Use Consider This 9.43 to initiate discussion of the results of this activity.
Follow-up activities:
 Consider This 9.44. How do side group charges on the model protein vary with pH?
 Consider This 9.45. What is the isoelectric pH for casein?
 Check This 9.46. Relating a Web Companion animation to the casein investigation.
 Consider This 9.47. What is the difference between HbA and HbS?
 End of chapter problems 9.39 and 9.40.
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Consider This 9.43. How are protein solubility and pH related?
Goal:
Correlate protein solubility and solution pH and consider how the acid-base properties of amino
acids might help explain the observed pattern.
Time for activity:
 10-15 minutes depending on how far the discussion is carried into acid-base properties of
amino acid side groups.
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to answer these questions. Then,
have groups share their ideas with the class.
 This activity could be conducted as an open class discussion.
Instructor notes:
 Try to be sure the class agrees on the results from Investigate This 9.42.
 Show or refer students to the data for amino acids in Table 9.2. You might remind students
that the amine and carboxylic acid groups directly bonded to the alpha carbon are used to
form amide bonds in protein, so are not relevant to the acid-base properties of proteins (except
at the protein chain termini, but these contribute little to the overall acid-base properties).
Students should reason and conclude:
(a) The observations in Investigate This 9.42 are probably somewhat surprising. Around pH
4 to 5, the protein precipitated out of solution and then redissolved into solution at higher or
lower pH values. The precipitation might not be so surprising for students who have
observed milk curdling when acid is added, but the re-dissolution will probably not be
familiar.
(b) See part (a) for the answers to the pH questions. There are two classes of amino acid side
groups in Table 9.2. In the first class, side groups in aspartic acid, glutamic acid, cysteine,
and tyrosine are electrically neutral in their acid forms (at lower pH), but are negatively
charged in their base forms (at higher pH). On the other side of the coin, the second class of
side groups in histidine, lysine, and arginine are positively charged in their acidic forms (at
lower pH), but electrically neutral in their base forms (at higher pH). If the protein is in a
solution at high pH, the side groups in the first class will be negatively charged and those in
the second class electrically neutral. Thus, the protein molecule will have a net negative
charge. In a solution at low pH, the side groups in the first class will be electrically neutral
and those in the second class positively charged. Thus, the protein molecule will have a net
positive charge. At some intermediate pH we can reason that the protein will have a net
charge of zero since there has to be a transition from net negative to net positive as the pH
changes from high (basic solution) to low (acidic solution). Perhaps it is at this intermediate
pH that the protein molecules come together to form a precipitate.
Follow-up discussion:
 Use the discussion as an introduction to acid-base side groups on proteins, their pKa values,
and the implications for the net charge on a protein molecule.
 Show or refer students to Figure 9.5, showing a protein with only two acid-base side groups,
to help make the discussion a bit more concrete.
Follow-up activities:
 Consider This 9.44. How do side group charges on the model protein vary with pH?
 Consider This 9.45. What is the isoelectric pH for casein?
 Check This 9.46. Relating a Web Companion animation to the casein investigation.
ACS Chemistry FROG
31
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

Chapter 9
Consider This 9.47. What is the difference between HbA and HbS?
End of chapter problems 9.39 and 9.40.
Consider This 9.44. How do side group charges on the model protein vary with pH?
Goal:
Determine how side group charges on the model protein vary with pH and hence the net charge
on the protein at various pHs.
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to answer the questions and
formulate their explanations. Then, have groups share their ideas with the class, using the
chalkboard or overhead transparency.
 This activity could be conducted as an open class discussion.
Time for activity:
 5-10 minutes.
Instructor notes:
 Show or refer students to Figure 9.5 as this activity is carried out.
Students should reason and conclude:
(a) This table shows the logarithm of the conjugate base to conjugate acid ratio at the three
pHs in Figure 9.5 for the carboxylic acid side group [from equation (9.33)] and the amine
side group [from equation (9.34)];
pH
12
6
3
 –COO–  
log 
 = pH – pKa1
 –COOH eq
8
2
–1
 –NH2 
log 
= pH – pKa2
+ 
  –NH3   eq
1
–5
–8
For the cases where the logarithm of the ratio is positive, the ratio is larger than unity and the
base form of the side group predominates. This is why the carboxylic acid group is shown in
its carboxylate ion form in Figures 9.5(a) and (b) and the amine side group in its neutral
amine form in Figure 9.5(a). When the logarithm is negative, the ratio is less than unity and
the acid form of the side group predominates. This is why the carboxylic acid group is shown
in its neutral acid form in Figure 9.5(c) and the amine side group in its protonated aminium
ion form in Figure 9.5(b) and (c).
(b) In Figure 9.5(b), the carboxylate ion has a 1– charge and the amine group has a charge of
1+. Adding these charges together gives the net charge of zero for the protein. In Figure
9.5(c), the carboxylic acid is in its acidic form and has zero charge and the amine group has a
1+ charge. Adding these charges together gives the net charge of 1+ for the protein.
Follow-up discussion:
 Introduce and discuss the isoelectric pH for proteins. Students should get the idea that the
same ideas apply to real proteins with multiple acid-base side groups as to this simple case.
Follow-up activities:
 Consider This 9.45. What is the isoelectric pH for casein?
 Check This 9.46. Relating a Web Companion animation to the casein investigation.
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Consider This 9.47. What is the difference between HbA and HbS?
End of chapter problems 9.39 and 9.40.
Consider This 9.45. What is the isoelectric pH for casein?
Goal:
Conclude that the behavior of casein in Investigate This 9.42 is consistent with the model for
protein behavior presented in the text and determine the isoelectric pH for casein.
Classroom options:
 This activity could be conducted as an open class discussion.
Time for activity:
 About 5 minutes.
Instructor notes:
 Try to be sure the class agrees on the results from Investigate This 9.42 and definition of
isoelectric pH.
Students should reason and conclude:
(a) The observations from Investigate This 9.42 are consistent with the model of protein
behavior presented in the text. The protein is soluble at low pH where, presumably, the
molecules have a net positive charge and at high pH where the molecules have a net negative
charge. At an intermediate pH around 4.7, casein precipitated out of solution. This
observation is consistent with the protein having no net charge at this pH.
(b) Based upon the results of Investigate This 9.42, the isoelectric pH of casein is
approximately 4.7  0.2. [Although it would have been better to get data at more closely
spaced pHs near the isoelectric pH, it is difficult to judge exactly when the most precipitate
has formed. Pinpointing the isoelectric pH to better than a few tenths of a pH unit is not
possible with this technique.]
Follow-up discussion:
 Introduce electrophoresis and its importance in analyzing biological molecules and use Figure
9.6 as a lead in to Consider This 9.47.
Follow-up activities:
 Check This 9.46. Relating a Web Companion animation to the casein investigation.
 Consider This 9.47. What is the difference between HbA and HbS?
 End of chapter problems 9.39 and 9.40.
Consider This 9.47. What is the difference between HbA and HbS?
Goal:
Determine the direction of the difference in charge between HbA and HbS at pH 8.6.
Classroom options:
 Allow about 3 minutes for students, working in small groups, to answer these questions. Then,
have the groups share their conclusions and explanations with the class.
 This activity could be conducted as an open class discussion.
 This activity could also be assigned as a homework problem and then discussed at the next
class session, either in small groups initially or open class discussion.
Time for activity:
 About 5 minutes.
Instructor notes:.
 Show or refer students to Figure 9.7 as they carry out this activity.
ACS Chemistry FROG
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Chapter 9
Students should reason and conclude:
More negatively charged proteins are attracted to the positive end of the gel. If HbA is closer
to the positive end of the gel, it must contain more side-group carboxylate anions than HbS.
Follow-up discussion:
 Discuss the molecular difference between HbA and HbS, with a stress on the difference in
biological properties such a small difference in chemical structure can make.
 Conclude with the important points in the "Reflection and Projection."
Follow-up activities:
 End of chapter problems 9.41 through 9.44.
Section 9.6. Solubility Equilibria for Ionic Salts
Learning Objectives for Section 9.6
 Find the solubility product, Ksp (or pKsp), for an ionic compound when you have data for the
solubility of the compound and vice versa.
 Find the solubility of an ionic compound of known Ksp (or pKsp) in a solution containing a
stoichiometric excess of one of the ions.
Investigate This 9.48. Is silver chromate insoluble?
Goal:
Discover that a precipitate of solid silver chromate forms only after the addition of several drops
of dilute chromate solution to a dilute solution of silver ion.
Set-up time:
 10-15 minutes (assuming reagents have been prepared previously).
Time for activity:
 5-10 minutes.
Materials:
 25 mL beaker.
 10-mL graduated cylinder.
 Thin stem pipet.
Reagents:
 0.0015 M silver nitrate, AgNO3(aq), solution.
 0.0015 M potassium chromate, K2CrO4(aq), solution
Procedure:
 Do this as a class investigation and have students work in small groups to discuss and analyze
the results. Due to the nature of the reagents, you should carry out this procedure, rather than
involving students directly with the manipulations.
SAFETY NOTES
Wear your safety goggles.
WARNING: Cr(VI) compounds are suspect carcinogens.
Solutions of silver ion will stain skin and clothing.
Wear disposable gloves when handling the solutions.

34
Place 10.0 mL of an aqueous 0.0015 M solution of silver nitrate, AgNO3, in a 25-mL beaker
on an overhead projector stage. If you are set up with a video camera to project small-scale
ACS Chemistry FROG
Chapter 9
Chemical Equilibria
apparatus, this would be a good activity to show this way, since the brick-red color of the
precipitate will not show up in the overhead projection.
 Add one drop of an aqueous 0.0015 M solution of potassium chromate, K2CrO4, to the silver
nitrate solution and swirl the mixture.
 Students should record their observations.
 Keep track of the number of drops of chromate solution added as you continue to add
potassium chromate solution, one drop at a time, until a permanent precipitate forms and
makes the solution cloudy.
Anticipated results:
 4-6 drops should produce a permanent brick-red silver chromate precipitate. It’s worth trying
the activity with your reagent solutions in advance to know about what to expect and how
vigorously you have to swirl the solution to redissolve any precipitate that forms in locally
higher concentrations of reagents.
Disposal:
 Dispose of all heavy metal solutions in a properly labeled waste container in accordance with
local regulations.
Follow-up discussion:
 Use Consider This 9.49 to initiate discussion of the results of this activity.
Follow-up activities:
 Worked Example 9.50. Solubility product, Ksp, for silver chromate, Ag2CrO4(s).
 Check This 9.51. Solubility product, Ksp, for silver chromate, Ag2CrO4(s).
 Worked Example 9.52. Solubility product, Ksp, for magnesium hydroxide, Mg(OH)2(s).
 Check This 9.53. Solubility of silver phosphate, Ag3PO4.
 Worked Example 9.54. Solubility of CaSO4(s) in 0.50 M SO42-(aq) solution.
 Check This 9.55. Solubility of Cu(IO3)2(s) in water and 0.25 M Cu2+(aq) solution.
 Check This 9.56. Compare Figures 9.9 and 9.10.
 End of chapter problems 9.45 through 9.57.
Consider This 9.49. How insoluble is silver chromate?
Goal:
Try to explain why several drops of dilute chromate solution have to be added before a
permanent precipitate of silver chromate is formed.
Classroom options:
 This activity can be conducted as an open class discussion.
Time for activity:
 About 5 minutes.
Instructor notes:
 Try to be sure the class agrees on the results from Investigate This 9.48: the number of drops
of the chromate solution required to produce a permanent precipitate of silver chromate.
Students should reason and conclude:
Four to six drops of chromate solution are needed to produce a permanent precipitate of the
brick-red silver chromate. This corresponds to about 0.25 mL [= (5 drops)(0.75 mL·drop–1)]
of 0.0015 M potassium chromate solution. Evidently, a precipitate that does not redissolve is
not formed until the added chromate has reached a high enough concentration in the solution.
Follow-up discussion:
 Use this activity as a lead in to define and discuss Ksp and the solubility product expression,
stressing that the form of the equilibrium constant expressions for solubility equilibria is the
ACS Chemistry FROG
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Chemical Equilibria
Chapter 9
same as for other equilibria. The special form taken by the solubility product expression is a
result of the fact that the concentration ratio for a pure solid is unity, so this ratio does not
appear in the solubility product.
NOTE: The activity in Investigate This 9.48 is not a good way to determine a solubility product
(as we do in Worked Example 9.50 and in Check This 9.51). There are too many sources of
uncertainty, some of which are noted in Worked Example 9.50. The primary purpose of the
investigation and this analysis is to show that a minimum concentration of the ions that form the
precipitate is required before the precipitate will form. Insoluble ionic compounds are not totally
insoluble; at least some of the ions are present in a solution in equilibrium with the solid and we
can characterize this equilibrium via the Ksp for the ionic compound.
Follow-up activities:
 Worked Example 9.50. Solubility product, Ksp, for silver chromate, Ag2CrO4(s).
 Check This 9.51. Solubility product, Ksp, for silver chromate, Ag2CrO4(s).
 Worked Example 9.52. Solubility product, Ksp, for magnesium hydroxide, Mg(OH)2(s).
 Check This 9.53. Solubility of silver phosphate, Ag3PO4.
 Worked Example 9.54. Solubility of CaSO4(s) in 0.50 M SO42-(aq) solution.
 Check This 9.55. Solubility of Cu(IO3)2(s) in water and 0.25 M Cu2+(aq) solution.
 Check This 9.56. Compare Figures 9.9 and 9.10.
 End of chapter problems 9.45 through 9.57.
Consider This 9.57. Does a common ion always decrease solubility?
Goal:
Conclude that a common ion does not always decrease solubility and try to think of a reason why
it might not.
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to analyze these data and try to
think of an explanation. Then have the groups share their ideas with the class.
 This activity could be conducted as an open class discussion.
 This activity could also be assigned as a homework problem (if your class is ready for this
challenge) and then discussed at the next class section, either in small groups initially or open
class discussion.
Instructor notes:
Students should reason and conclude:
(a) The solubility of HgI2(s) increases as the concentration of I–(aq) in the solution increases.
What we would expect to observe on the basis of the common ion effect is a continuous
decrease in solubility as the concentration of I–(aq) in the solution increases. The results are
just the opposite. We also note that the solubility is half the concentration of the I–(aq) in the
solution in all cases.
(b) The stoichiometry we just noted might lead us to suspect that some reaction between
HgI2 and I– is occurring in the solution. For example, complex ions such as HgI3–(aq) and
HgI42–(aq) might be formed in the solution and account for the increasing solubility of
HgI2(s) in the presence of I–(aq). We would need more experiments and data to prove such
an hypothesis, but it explains the observations and is consistent with the idea that metal ions
(Lewis acids) do form such complexes with Lewis bases.
Follow-up activities:
 End of chapter problems 9.45 through 9.57.
36
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Chapter 9
Chemical Equilibria
Section 9.7. Thermodynamics and the Equilibrium Constant
Learning Objectives for Section 9.7
 Find Greaction for a reaction when you have the equilibrium constant, K, for the reaction.
 Find the equilibrium constant, K, for a reaction when you have Gf for the reactants and
products.
NOTE: See the note at the beginning of Section 9.11.
Investigate This 9.61. How much urea will dissolve in water?
Goal:
Find the volume of water required to just dissolve a known mass of urea and the volume of the
resulting solution.
Set-up time:
 Time will depend on the number of sets of materials and reagents you need to prepare for the
groups in your class. Each set requires weighing out 4.0 g of urea into one of the graduated
tubes and adding about 10 mL of water to the other. This will probably take about 2 min/set.
Time for activity:
 About 20-25 minutes, incorporating Consider This 9.62 to begin the analysis of the results.
Materials:
NOTE: Amounts here are per group, so you need to multiply by the number of groups you will
have.
 Thin-stem plastic pipet.
 Two capped, graduated plastic (or glass with rubber stoppers) centrifuge tubes. Plastic is safer
because the tubes will not break if dropped and they have screw caps that are watertight.
Reagents:
 4.0 grams of urea, H2NCONH2(s).
 About 10 mL of water.
Procedure:
 This activity is much more effective if carried out in small groups, rather than as a class
activity (although you could do the latter with student volunteers, perhaps two or three teams
to check the reproducibility of the data).
SAFETY NOTE
Wear your safety glasses.

Each group should have a thin-stem plastic pipet and two capped, graduated conical plastic
tubes containing, respectively, 4.0 g of urea, H2NCONH2(s), and about 10 mL of water.
 Read and record the volume of water to 0.1 mL.
 Use the pipet to transfer 4.0 mL of water to the tube containing urea, cap the tube, and invert
several times to dissolve as much urea as possible.
NOTE: The 4.0 mL of water transferred to the urea-containing tube should be measured by
difference from the original amount in the tube containing water, not as the liquid level in the
urea solution. Mixing to dissolve the urea must be done vigorously with each addition of water.
The most common error in this activity is insufficiently vigorous mixing. Although urea
dissolves readily, the last few crystals dissolve more slowly in the saturated solution and it is
easy to add more water than necessary and overshoot the just-saturated solution that is the
ACS Chemistry FROG
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Chapter 9
desired goal. The 4.0 mL of water initially added is almost enough to dissolve all the urea. Only
about 0.2 mL (3-5 drops) more will be necessary, so adding a drop at a time with mixing
between drops is not too onerous or time consuming.
 Add water one drop at a time to the urea solution, cap the tube, and mix the contents
vigorously. Continue these additions until the urea is just dissolved.
NOTE: After 3-4 drops have been added, the liquid usually appears clear and colorless and is
still quite cold. If you hold the tube up to the light you can see glinting crystals of urea drifting
about and settling to the bottom. There will not be much solid still left, but what is left is easy to
see as it collects in the conical bottom of the tube. At this stage, another drop is all that is
required to produce a truly clear colorless solution with no crystals of urea remaining.
 Return all unused water from the pipet to the tube containing pure water.
 Measure and record the volume of liquid in each tube to 0.1 mL. Calculate the volume of
water used to make the solution [for part (a) of Consider This 9.62].
Anticipated results:
 The results should be quite consistent from one group to another. List each group’s results
(volume of solution and volume of water in the solution) on the chalkboard or overhead, so
the reproducibility is easy to see (and outliers – volumes that are too large – easy to spot). The
results are usually within about 0.1 mL of these:

Volume of just-saturated urea solution: 7.5 mL

Volume of water added to urea: 4.2 mL
Disposal:
 Dispose of the solutions down the drain with lots of water.
Follow-up discussion:
 Use Consider This 9.62 to initiate analysis and discussion of the results of this activity.
Follow-up activities:
 Consider This 9.63. What are G, H, and S for dissolution of urea in water?
 End of chapter problems 9.58 through 9.69.
Consider This 9.62. What is the solubility of urea in water?
Goal:
Determine the number of moles and the molarity of urea and of water in the just-saturated
solution of urea in Investigate This 9.61.
Classroom options:
 It is best for the groups to continue their analysis of their group results from Investigate This
9.61 and add them to the listing begun at the end of that activity, which incorporated part (a)
from this activity.
 This activity could be conducted as an open class discussion, perhaps using an average value
for the two critical liquid volumes.
 This activity could also be assigned as a homework problem and then discussed at the next
class section, either in small groups initially or open class discussion. The impact and
immediacy of the activity are lost this way, but timing might make it necessary.
Time for activity:
 10-15 minutes included in time suggested for Investigate This 9.61.
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Instructor notes:
 Each group should use its results from Investigate This 9.61 (unless it is clear that an error
was made, in which case you can assign values for them that are consistent with the rest) for
this activity.
 The molar mass of urea, H2NCONH2, is 60.0 g·mol-1.
Students should reason and conclude:
(a) From the results for Investigate This 9.61, the volume of the just-saturated urea solution
is 7.5 mL. The volume of water added to make this solution is 4.2 mL.
(b) Use the molar mass of urea to determine the number of moles of urea and then the
number of moles and volume of solution to get the molarity of urea, [H2NCONH2(aq)]:
 1 mol 
4.0 g urea = (4.0 g) 
= 6.7  10-2 mol
 60.0 g 
6.7  10-2 mol
= 8.9 M
0.0075 L
(c) Use the volume and density of water to find the mass of water in the solution and thence
the number of moles and molarity of water, [H2O(aq)], just as in part (b) for urea:
 1.00 g   1 mol 
4.2 mL water = (4.2 mL) 
= 0.23 mol
 1.00 mL   18.0 g 
0.23 mol
[H2O(aq)] =
= 31 M
0.0075 L
Follow-up discussion:
 Discuss the equilibrium constant expression (9.50) for dissolution, Kdiss. The reaction is
written (at saturation – a condition of equilibrium between the solid and aqueous solution) as a
reaction of solid urea with the water present in its saturation concentration to give the solvated
urea in the saturated solution.
Follow-up activities:
 Consider This 9.63. What are G, H, and S for dissolution of urea in water?
 End of chapter problems 9.58 through 9.69.
[H2NCONH2(aq)] =
Consider This 9.63. What are G, H, and S for dissolution of urea in water?
Goal:
Use the results from Consider This 9.62 and calorimetric data from Chapter 7 to calculate G,
H, and S for the dissolution of urea in water.
Classroom options:
 If there is time, the groups can continue the analysis of their group results from Investigate
This 9.61 and Consider This 9.62 and share their results with the class.
 This activity could be conducted as an open class discussion.
 This activity could also be assigned as a homework problem and then discussed at the next
class section, either in small groups initially or open class discussion.
Time for activity:
 Approximately 10-15 minutes.
Instructor notes:
 Students can use their own results from Consider This 9.62 (b) and (c) or you can provide the
class with an average figure for everyone to use.
ACS Chemistry FROG
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Chapter 9


Show or refer students to equations (9.50) and (9.51) as they begin the activity.
For part (c), remind students of the results from Check This 7.27 (or, if this activity is
assigned as homework, refer students to this activity for the necessary datum). We found
H = 15.7 kJ·mol–1 for dissolution of urea to produce a 1 M (standard state) solution.
Students should reason and conclude:
(a) The concentration ratio for water in the saturated solution, from equation (9.51), is:
 31 M 
(H2O(aq)) = 
= 0.56
 55.5 M 
Use equation (9.50) to find the equilibrium constant for urea dissolution at saturation:


H 2NCONH2 (aq)  = 8.9 = 16
Kdiss = 

H2 NCONH2 (s) H2 O(aq)
eq (1)(0.56)

(b) Use the relationship between the standard free energy change for the reaction and its
equilibrium constant to find the standard free energy change (assuming T = 298 K):
0.001 kJ
G = –RTlnKdiss = –(8.314 J·K-1)(298 K) 
[ln(16)]
1 J 
G = –6.8 kJ
The standard free energy change for dissolution is negative so the dissolution of urea is
favored, as we see from the fact that the saturated solution is almost 9 M.
(c) The dissolution reaction for urea in water is endothermic. Thermal energy is absorbed
from the surroundings to dissolve the urea. Indeed, in Investigate This 9.61, as urea dissolved
in water, we observed that the tube felt cold. In Chapter 7, Check This 7.27, we found
H = 15.7 kJ·mol–1 for dissolution of urea to produce a 1 M (standard state) solution.
(d) We use the relationship among G, H, S, and T, together with the values from parts
(b) and (c) to find S for the dissolution:
G = H – TS
–6.8 kJ = 15.7 kJ·mol–1 – (298 K)S
S= 76 J·K–1·mol–1
This relatively large positive value for S is consistent with the spontaneity of this
endothermic reaction. The entropy change of the system dominates the negative thermal
entropy change of the surroundings.
Follow-up activities:
 End of chapter problems 9.58 through 9.69.
Section 9.8. Temperature Dependence of the Equilibrium Constant
Learning Objectives for Section 9.8
 Find Hreaction, Sreaction, and Greaction for a reaction when you have values of K for the
reaction at two or more temperatures.
 Find Greaction and K for a reaction at a temperature T when you have Hreaction for the
reaction.
40
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Chapter 9
Chemical Equilibria
Investigate This 9.64. How does temperature affect the solubility of PbI2(s)?
Goal:
Observe that a precipitate of lead iodide, PbI2(s), produced in a cold solution dissolves when the
temperature of the solution is increased.
Set-up time:
 About 15 minutes, assuming that the solutions are prepared.
Time for activity:
 About 15-20 minutes, incorporating Consider This 9.65 to analyze the results.
Materials:
 250-mL beaker.
 Magnetic stirrer-hot plate combination and magnetic stir bar.
 Thermometer.
Reagents:
 5 mL of 1 M aqueous lead nitrate, Pb(NO3)2(aq), solution.
 2 mL of ice-cold 0.05 M aqueous potassium iodide, KI(aq), solution.
 100 mL ice-cold water. Have a container of ice and water available and take this aliquot for
use at the start of the activity. Also use the ice-water bath to cool the KI solution.
Procedure:
 Do this as a class investigation and have students work in small groups to discuss and analyze
the results. Due to the nature of the reagents, you should carry out this procedure, rather than
involving students directly with the manipulations.
SAFETY NOTES
Wear your safety goggles.
WARNING: Lead compounds are toxic.
Wear disposable gloves when handling the solutions.

Add 100 mL of ice-cold water and a stir bar to the beaker, set it on the stirrer, and begin
stirring at a moderate speed.
 Add 5 mL of 1 M Pb(NO3)2(aq) solution to the stirred cold water. (Occasionally, the solution
may appear a bit cloudy, if lead hydroxide begins to form. A drop or two of 1 M nitric acid
should bring the solution back to a clear, colorless state.)
 Add 2 mL of 0.05 M KI(aq) solution to the stirred cold water. Watch for the formation of a
precipitate of lead iodide, PbI2(s).
NOTE: The precipitate may not form immediately as the solution is close to saturation under
these conditions and formation of a noticeable precipitate could take a few minutes. Pb2+(aq)
also forms complexes with I–(aq) that increase solubility. You should practice this activity before
doing it in class so you have some idea how long you will have to wait. If the wait is too long,
you might add a bit more of the lead nitrate solution initially. Be careful, however, not to go too
far, because you want to be sure that the precipitate will dissolve in the next step.
 Turn on the hot plate and heat the stirred solution to about 70 C. Observe what happens to the
precipitate as the solution is heated.
 Turn off the hot plate and observe what happens as the solution cools.
Anticipated results
 Local or state ordinances may restrict your use of lead compounds in the classroom. If this is
the case, you might have to provide students with these results.
ACS Chemistry FROG
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Chapter 9

A yellow precipitate of PbI2(s) should form in the cold solution. You do not want copious
amounts of precipitate, because it would not redissolve.
 When the solution is heated, the precipitate should dissolve to form a clear, essentially
colorless solution.
 When the solution cools the precipitate should re-form, but this may take some time and not
occur before the end of the class session. This isn’t a serious problem, because it is the initial
formation cold and then dissolution when heated that is of most interest for the data analysis.
Disposal:
 Dispose of solutions containing lead(II) in the appropriate heavy metal waste container. You
can precipitate the lead as lead sulfide, PbS(s), which is quite insoluble, and then dispose of
the precipitate as solid heavy metal waste taking up much less volume.
Follow-up discussion:
 Use Consider This 9.65 to initiate analysis and discussion of the results of this activity.
Follow-up activities:
 Worked Example 9.67. Hreaction from solubility temperature dependence.
 Check This 9.68. H°reaction from solubility temperature dependence.
 Worked Example 9.69. S°reaction and G°reaction from solubility temperature dependence.
 Check This 9.70. S°reaction and G°reaction from solubility temperature dependence.
 Worked Example 9.71. Graphical determination of H°reaction and S°reaction.
 Check This 9.72. Graphical determination of H°reaction and S°reaction.
 End of chapter problems 9.70 through 9.84.
Consider This 9.65. Is the dissolution of PbI2(s) exothermic or endothermic?
Goal:
Reason, using Le Chatelier’s principle and the results from Investigate This 9.64, that the
dissolution of PbI2(s) is endothermic.
Classroom options:
 Allow about 3 minutes for students, working in small groups, to draw their conclusions and
formulate their explanation and then have the groups share with class.
 This activity could be conducted as an open class discussion.
 This activity could also be assigned as a homework problem and then discussed at the next
class session, either in small groups initially or open class discussion, but this loses the
immediacy of the investigation.
Time for activity:
 Approximately 5 minutes.
Instructor notes:
 Remind the class how they interpreted the temperature dependence of another reaction, the
change in complexation of cobalt(II) in Section 9.1, as a tie-in to the interpretation here.
Students should reason and conclude:
A yellow precipitate of PbI2(s) formed in cold solution and dissolved as the solution was
warmed. The reaction of interest is the dissolution of the solid:
PbI2(s)  Pb2+(aq) + 2I–(aq)
Addition of energy to this system (by heating the solution) is a disturbance to the solubility
equilibrium and, by Le Chatelier’s principle, the system responds in a way that minimizes the
disturbance, that is, it responds in a way that takes in energy (enthalpy). This means that
42
ACS Chemistry FROG
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Chemical Equilibria
dissolution, represented by this equation, takes in energy or is endothermic, since this is the
direction the reaction takes upon addition of energy.
Follow-up discussion:
 Use this discussion as a lead in to the more quantitative thermodynamic treatment of
temperature dependence, but emphasize that the qualitative (Le Chatelier) and the quantitative
approach always give results that are in the same direction.
Follow-up activities:
 Worked Example 9.67. H°reaction from solubility temperature dependence.
 Check This 9.68. H°reaction from solubility temperature dependence.
 Worked Example 9.69. S°reaction and G°reaction from solubility temperature dependence.
 Check This 9.70. S°reaction and G°reaction from solubility temperature dependence.
 Worked Example 9.71. Graphical determination of H°reaction and S°reaction.
 Check This 9.72. Graphical determination of H°reaction and S°reaction.
 End of chapter problems 9.70 through 9.84.
Section 9.9. Thermodynamics in Living Systems
Learning Objectives for Section 9.9
 Find the equilibrium constant, K, for a reaction when you have Gf for the reactants and
products.
 Find the free energy Greaction for a reaction that is not at equilibrium when you have
G°reaction or K for the reaction and the concentrations in the non-equilibrium system.
 Find G°reaction or Greaction for a coupled reaction when you have G°reaction or Greaction for
the individual reactions that are coupled.
Consider This 9.73. What is the equilibrium cellular concentration of ATP?
Goal:
Use the equilibrium constant expression and equilibrium constant for ATP hydrolysis, from
equation (9.65), to calculate the equilibrium cellular concentration of ATP.
Classroom options:
 Allow about 3 minutes for students, working in small groups, to do their calculations and
formulate their explanation and then have the groups share with class.
 This activity could be conducted as an open class discussion.
Time for activity:
 5-10 minutes, including discussion of the meaning of the notation in equation (9.65).
Instructor notes:
 Introduce students to how and why the equilibrium constant expression for hydrolysis of ATP
is formulated in terms of the overall concentrations of all forms of ATP, ADP, and Pi in the
solution.
Students should reason and conclude:
 ADPPi 
0.0002 0.001 ;
5
K´ = 
=
10
=
[ATP] = 2  10-12 M
 ATP   eq
ATP 
At equilibrium, the hydrolysis of ATP greatly favors the products, ADP and Pi, as indicated
by the large negative value of G°hydrolysis and the tiny concentration of ATP that would be
present if the cells were in equilibrium. [The ATP and ADP concentrations in cells are far
ACS Chemistry FROG
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Chapter 9
from equilibrium and the consequence of this fact is discussed further in the succeeding
paragraphs.]
Follow-up discussion:
 Discuss the free energy changes for cellular ATP reactions.
 Activities 9.73 through 9.76 should probably be completed in an open class discussion or a
recitation session, as they involve ideas that may be unfamiliar to some students.
Follow-up activities:
 Consider This 9.74. How do you analyze a coupled reaction?
 Check This 9.75. Transport of oxygen by hemoglobin.
 Check This 9.76. What are the relative sizes of oxygen-binding equilibrium constants?
 End of chapter problems 9.85 through 9.89.
Consider This 9.74. How do you analyze a coupled reaction?
Goal:
Show how a coupled reaction can change the equilibrium amount of a cellular component.
Classroom options:
 Activities 9.74 through 9.76 should probably be completed in an open class discussion or a
recitation session, as they involve ideas that may be unfamiliar to some students.
Time for activity:
 5-10 minutes, including any necessary review of the concept of summing thermodynamic
values for reactions that are summed algebraically.
Instructor notes:
 Try to be sure the class understands equations 9.69 through 9.71 before carrying out this
activity.
Students should reason and conclude:
(a) The sum of reaction (9.69) and the reverse of reaction (9.70) gives reaction (9.71). The
free energies also combine in the same way to give Greaction for reaction (9.71):
(9.69) ATP(aq) + H2O  ADP(aq) + PI(aq)
G = –29 kJ·mol-1
–(9.70) R-OH + PI(aq)  P-OPO32-(aq) + H2O
G = 9 kJ·mol-1
(9.71) ATP(aq) + R-OH  P-OPO32-(aq) + ADP(aq) Greaction = –20 kJ·mol-1
To find K , we use the relationship between the equilibrium constant and the standard
change in free energy, assuming T = 298 K for this system:
– –20,000 J mol-1 
–Greo action
lnK =
=
= 8.1
RT
8.314J  mol-1K -1298 K
K = 3  103
(b) Write the equilibrium constant expression and substitute the known values for the
equilibrium constant and the ATP to ADP ratio to find the ratio of glycerol phosphate to
glycerol at equilibrium under these conditions:
2
 ADP(aq) R-OPO3 2 (aq)  
 1    R-OPO 3  
3
K = 3  10 = 

 =   
10  R-OH 
 ATP (aq) R-OH(aq)  
eq
 R-OPO 
= 3  104
R-OH
2
3
44
ACS Chemistry FROG
Chapter 9
Chemical Equilibria
Since the problem statement indicates that glycerol phosphate is required by cells, this ratio
is quite advantageous to the cell because the glycerol phosphate concentration is more than
four orders of magnitude greater than its hydrolysis product, glycerol. This is one of the
many cases where it is important to the cell to maintain a higher concentration of ATP than
ADP, even though its hydrolysis equilibrium would reduce the ATP to very low levels, as we
found in Consider This 9.73.
Follow-up activities:
 Check This 9.75. Transport of oxygen by hemoglobin.
 Check This 9.76. What are the relative sizes of oxygen-binding equilibrium constants?
 End of chapter problems 9.85 through 9.89.
Section 9.11. Extension -- Thermodynamic Basis for the Equilibrium
Constant
Learning Objectives for Section 9.7
 Calculate the entropy of gases at nonstandard pressures.
 Show how the entropy of gases at nonstandard pressures is related to the free energy and
reaction quotient for a gas-phase reaction (and, by a leap of faith, to all reactions).
NOTE: The results from this Extension are introduced at the beginning of Section 9.7. This
allows you to avoid the more mathematical discussion here in order to get on with applications of
the free energy relationship to the reaction quotient and hence the equilibrium constant. If you
wish, you can introduce the material here before beginning Section 9.7, so that the free energy
relationship does not appear out of nowhere. Or, you can direct those students who are interested
in the derivation to this Extension and let them work through it as a group, perhaps with out-ofclass guidance from you or another instructor in the course.
Consider This 9.77. What are some properties of nitrogen oxides?
Goal:
Use Lewis structures to predict the reactivity of NO2. Use reasoning from concepts in Chapters
5, 7, and 8 to predict the direction of change of thermodynamic variables for the reaction.
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to answer these questions and then,
have the groups share with the class, summarizing answers on the chalkboard or an overhead
transparency.
 This activity could be conducted as an open class discussion.
 This activity could also be assigned as a homework problem and then discussed at the next
class session, either in small groups initially or open class discussion.
Time for activity:
 From 10 to 20 minutes, depending on how well students recall Lewis structures and bonding
and thermodynamic concepts.
Instructor notes:
 This activity provides a good review of Lewis structures.
Students should reason and conclude:
(a) The possible Lewis structures for NO2 are:
ACS Chemistry FROG
45
Chemical Equilibria
Chapter 9
N
O
or
O
N
N
O
O
O
O
or
N
O
O
There is no single Lewis structure that describes the electron distribution in the molecule.
Because the number of valence electrons in this molecule is odd (17), it is not possible for all
of them to be paired and it is not possible to supply each atom with an octet of electrons.
There are two structures with the odd electron on the N atom and two with the odd electron
on an O atom. The members of each pair have the same energy, but the two pairs of
structures probably have different energies. The fact that there are several possible structures,
some of which have the same energy, indicates that the pi electrons in the molecule are
delocalized over the whole structure.
(b) NO2 should be reactive, due to the fact that it has an unpaired electron (it’s a free radical)
that is available to interact with and pair up with another electron in another molecule.
(c) There are three possible ways for two molecules of NO2 [with the Lewis structures shown
in part (a)] to come together and pair up the unpaired electrons to form N2O4:
O
O
N
O
O
or
N
O
N
O
N
O
N
O
O
O
O
O
O
N
O
O
N
O
or
N
O
O
O
N
N
O
In each case, the N2O4 molecule is formed by bringing together the atoms with unpaired
electrons from two of the Lewis structures in part (a). Thus, we can get a molecule with an
N–N, an O–O, or an N–O bond. If we look back at bond enthalpies, Table 7.3, we find that
all three of these single bonds are weak, but the order is (from strongest to weakest): N–N >
N–O > O–O. As pictured here, the central structure has both the weakest bond and no
electron delocalization to stabilize the molecule, so we can probably eliminate it as a likely
structure for N2O4. The other two structures both have delocalized pi electrons. (There are, in
fact, several other Lewis structures that you can write for each one. Try it.) The N–N bond is
just a bit stronger than the N–O bond, so this factor favors the structure on the left and we
can tentatively conclude that this is the likely structure for N2O4. (This is the structure that
other experiments prove is correct.)
(d) The entropy change for reaction (9.76) should be negative. We found in Chapter 8 that
the entropy change for reactions with fewer gas phase products than gas phase reactants have
negative entropy changes. Another way to look at this is in terms of the number of
distinguishable arrangements of the atoms. In NO2, all N atoms are independent of one
another, but in N2O4, each N atom is paired with another, so only half of them are
independent of all the others. The number of possible arrangements is lower, so the entropy
of an N2O4 is lower than the entropy of two NO2. See Consider This 9.81 for quantitative
confirmation.
(e) We expect reaction (9.76) to be exothermic because a bond is formed and none is broken.
Bond formation is always exothermic. The product is a more stable arrangement of electrons
and atomic cores.
Follow-up discussion:
 Show or refer the class to Figure 9.16 to help make the transition from reactions that occur in
aqueous solutions to reactions that occur in the gas phase.
46
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Chapter 9
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Follow-up activities:
 Consider This 9.78. Which of the gases is colored: NO2(g) or N2O4(g)?
 Check This 9.79. Hreaction for 2NO2(g) and N2O4(g).
 Consider This 9.80. What is the entropy of a compressed gas?
 Consider This 9.81. What is Sreaction for 2NO2(g)  N2O4(g)?
 Consider This 9.82. What is Greaction for 2NO2(g)  N2O4(g)?
 End of chapter problems 9.90 and 9.91.
Consider This 9.78. Which of the gases is colored: NO2(g) or N2O4(g)?
Goal:
Use the data in Figure 9.16 and the predictions from Consider This 9.77 to identify which gas is
colored.
Classroom options:
 Allow about 5 minutes for students, working in small groups, to formulate their responses and
then share them with the class using the chalkboard or an overhead transparency.
 This activity could be conducted as an open class discussion.
Time for activity:
 Approximately 10 minutes.
Instructor notes:
 Show or refer students to Figure 9.16 as this activity is carried out.
Students should reason and conclude:
(a) Heating the tube containing the gas mixture adds energy to the equilibrium system. The
heated tube of gas has a deeper reddish-brown than the ones at room temperature, so it
contains more of the colored gas. Note that the gas mixture in the cooled tube is lighter in
color than those at room temperature, which is consistent with the colored gas being favored
when energy is added to the mixture.
(b) If the colored gas is NO2(g), then there is more NO2(g) in the equilibrium mixture when
energy is added. Addition of energy is a disturbance to the equilibrium and, by Le Chatelier’s
principle, the system will react to minimize the disturbance, that is, to use up some energy in
the present case. Thus, formation of the colored gas must require energy, since the system
reacts to form more of it when energy is to be used up. If NO2(g) is the colored gas, then
formation of NO2(g) is an endothermic process, requiring energy.
(c) If the colored gas is N2O4(g), then there is more N2O4(g) in the equilibrium mixture when
energy is added. By the same argument as in part (b) we reason that formation of the colored
gas must require energy, since the system reacts to form more of it when energy is to be used
up. If N2O4(g) is the colored gas, then formation of N2O4(g) is an endothermic process,
requiring energy.
(d) Part (c) agrees with our answer In Consider This 9.77(e), we argued that formation of a
molecule of N2O4(g) from two molecules of NO2(g) is exothermic. Our reasoning in part (c)
would make the formation of N2O4(g) endothermic, so the reasoning in part (b), which makes
the formation of NO2(g) endothermic (and hence formation of N2O4(g) exothermic) must be
correct. The colored gas is NO2(g). [It is NO2(g) that gives some polluted air a reddish-brown
color.]
Follow-up activities:
 Check This 9.79. Hreaction for 2NO2(g) and N2O4(g).
 Consider This 9.80. What is the entropy of compressed carbon dioxide?
ACS Chemistry FROG
47
Chemical Equilibria



Chapter 9
Consider This 9.81. What is Sreaction for 2NO2(g)  N2O4(g)?
Consider This 9.82. What is Greaction for 2NO2(g)  N2O4(g)?
End of chapter problems 9.90 and 9.91.
Consider This 9.80. What is the entropy of compressed carbon dioxide?
Goal:
Calculate the entropy of compressed carbon dioxide and predict the direction of entropy change
for expansion.
Classroom options:
 Allow about 5 minutes for students, working in small groups, to answer the questions and then
have groups share their answers with the class using the chalkboard or an overhead
transparency, as necessary.
 This activity could be conducted as an open class discussion.
Time for activity:
 10-15 minutes.
Instructor notes:
 You can use this activity to assess whether students have read and can use equation (9.83).
Students should reason and conclude:
(a) Use equation (9.83) to calculate the entropy of compressed carbon dioxide:
  2.8 bar  
Sgas = S°gas – Rln(P) = (213.7 J·K-1·mol-1) – (8.314 J·K-1·mol-1)  ln 

  1 bar  
Sgas = 205.1 J·K-1·mol-1
The direction of the change from standard condition makes sense. When the pressure
increases the gas is compressed and there is less volume available to the molecules. At
pressures above 1 bar we expect the entropy to decrease from its standard entropy value at 1
bar.
(b) When a gas is expanded to a pressure less than 1 bar, the volume available to the
molecules increases and, therefore, the entropy of the gas is higher than its standard entropy
at 1 bar. Use equation (9.83) to show that this is the case.
Follow-up activities:
 Consider This 9.81. What is Sreaction for 2NO2(g)  N2O4(g)?
 Consider This 9.82. What is Greaction for 2NO2(g)  N2O4(g)?
 End of chapter problems 9.90 and 9.91.
Consider This 9.81. What is Sreaction for 2NO2(g)  N2O4(g)?
Goal:
Use a table of standard entropies to calculate Sreaction for 2NO2(g)  N2O4(g) and then Sreaction
under nonstandard conditions
Classroom options:
 Allow about 5 minutes for students, working in small groups, to work on the activity and then
have groups share their results with the class using the chalkboard or an overhead
transparency, as necessary.
 This activity could be conducted as an open class discussion.
 This activity could also be assigned as a homework problem and then discussed at the next
class session, either in small groups initially or open class discussion.
48
ACS Chemistry FROG
Chapter 9
Chemical Equilibria
Time for activity:
 From 10 to 20 minutes, depending on difficulties students may have with equation (9.85) and
its derivation.
Instructor notes:
 You can use this activity to assess whether students have read the background for and can use
equation (9.85).
Students should reason and conclude:
(a) The combination of standard entropies to find Sreaction is shown implicitly in equation
(9.84):
Sreaction = (1 mol)(304.29 JK–1mol–1) – (2 mol)(240.06 JK–1mol–1) = –175.83 JK–1
In Consider This 9.77(d), we predicted that the entropy change for the dimerization of
NO2(g) to give N2O4(g) would be negative, and this is what we find.
(b) Although not shown explicitly, the coefficients for each of the two terms in equation
(9.84) have units of moles, so moles have cancelled out in the units for R in equation 9.85:

PN2 O4  

Sreaction = S°reaction – Rln 
2 

PNO2  





  0.437 bar

 
1
bar
 = –192 JK–1
Sreaction = –175.83 JK – [(8.314 JK ) ln
2


  0.243 bar 1 bar 

 
The pressures of both gases in this system are less than 1 bar, so the entropy of each gas is
higher than its standard entropy. The difference is larger for the NO2(g), because its pressure
is lower than that of N2O4(g). In addition, the difference for NO2(g) is multiplied by two,
because two molecules (moles) react. Both these factors increase the entropy of the reactants
more (compared to their standard entropy) than the entropy of the product, so the entropy
decreases more (is more negative) than it would be if the gases were in their standard states.
The mathematical consequence is that the logarithmic ratio is greater than unity, so the
logarithmic term is positive and the second term in equation (9.85) is subtracted from the
standard entropy change, making the entropy change for the nonstandard conditions more
negative, as we have reasoned.
Follow-up discussion:
 Introduce and discuss free energy change for a gas reaction.
Follow-up activities:
 End of chapter problems 9.90 and 9.91.
–1
–1
Consider This 9.82. What is Greaction for 2NO2(g)  N2O4(g)?
Goal:
Use the standard enthalpy and entropy changes to calculate Greaction for 2NO2(g)  N2O4(g)
and then Greaction under nonstandard conditions.
Classroom options:
 Allow about 5 minutes for students, working in small groups, to work on the activity and then
have groups share their results with the class using the chalkboard or an overhead
transparency, as necessary.
 This activity could be conducted as an open class discussion.
ACS Chemistry FROG
49
Chemical Equilibria
Chapter 9

This activity could also be assigned as a homework problem and then discussed at the next
class session, either in small groups initially or open class discussion.
Time for activity:
 From 10 to 20 minutes, depending on how far you wish to carry the generalization of the
analysis to all equilibrium systems.
Instructor notes:
 You can use this activity to assess whether students have read the background for and can use
equation (9.88).
Students should reason and conclude:
(a) Combining the standard enthalpy and entropy changes previously calculated gives:
G°reaction = H°reaction – TS°reaction
 1 kJ 
G°reaction = –57.20 kJ – (298 K)(–175.83 JK-1) 
= –4.80 kJ
 1000 J 
(The value calculated directly from the standard free energies of formation in Appendix B is
–4.73 kJ, which, within the likely uncertainty of the thermodynamic values, is the same as
the value calculated here.)
(b) Use equation (9.88) together with the result in part (a) to find Greaction for the system
described in Consider This 9.81(b) [remember that moles in the units for R have cancelled
out]:

PN2 O4  

Greaction = G°reaction + RTln 
2 

PNO2  





  0.437 bar

 
1
bar
  1 kJ  = 0.16 kJ
Greaction = –4.80 kJ + [(8.314 JK )(298 K) ln
2  
 0.243 bar
 1000 J 
 

1 bar 
 
The positive value for the free energy of reaction indicates that the reaction is not
spontaneous under the stated conditions. Even though the formation of the new bond is a
favorable process (final energy is lower), the change in thermal entropy of the surroundings
is not large enough to compensate for the negative positional entropy change in the reaction.
(Note that under standard conditions the reaction would be spontaneous, so, as is evident
from Figure 9.16, there are conditions that favor the reaction occurring.)
Follow-up discussion:
 Remind the class that they have previously been introduced to equation (9.42) and to the
reaction quotient that appears in this equation. In the event that you have chosen to introduce
the material in this Extension before beginning Section 9.7, simply extend the discussion to
the opening of that section.
Follow-up activities:
 End of chapter problems 9.90 and 9.91.
-1
50
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Chapter 9
Chemical Equilibria
Solutions for Chapter 9 Check This Activities
Check This 9.5. Concentrations in the Fe(NO3)3–KSCN mixture
The problem states that the Fe3+(aq) is diluted by a factor of 20 when one drop is added to 1 mL
of solution. This is almost, but not quite, correct, because it does not account for the actual final
volume of the solution, 1.05 mL. The dilution factor is (0.05 mL/(1.05 mL) = 1/21. Thus, we
have:
 0.05 mL 
[Fe3+(aq)] = (0.2 M) 
= 0.01 M
 1.05 mL 
Similarly, for [SCN–(aq)], we get:
 1 mL 
[SCN–(aq)] = (0.002 M) 
= 0.002 M
 1.05 mL 
Fe3+(aq) is present in higher concentration.
Check This 9.6. Dynamic equilibrium animation
(a) In the Web Companion, Chapter 9, Section 9.1, page 4, we choose the phrase, "on average
the number of complex ions in the solution remains constant" by observing that the overall
number of complexes remains the same during the movie. Some dissociate, but an equal number
form, so almost every frame shows the same number of complex ions. As the Companion states
“the rates of formation and dissociation of complex ions are equal.”
(b) Two metal ion complexes are shown in almost all frames of the movie. In a few frames one
of these dissociates, but in a few others a third is formed. The average is two. The movie
represents a dynamic equilibrium in which the number of complexes remains essentially constant
as some dissociate and some form: the rates of formation and dissociation are the same.
Check This 9.7. Adding SCN–(aq) to a Fe(NO3)3–KSCN solution
(a) The response of the system to the addition of more SCN–(aq) is for the solution to become
more deeply colored, which indicates that more of the Fe(SCN)2+(aq) complex ion has been
formed. If the solution is an equilibrium mixture, addition of SCN–(aq) is a disturbance to the
equilibrium. Le Chatelier’s principle tells us that a system in equilibrium reacts to minimize the
disturbance. In this case, that would be to use up some of the added SCN–(aq), which the system
can do by forming more of the Fe(SCN)2+(aq) complex. This would result in a more deeply
colored solution, which is the observed effect and is consistent with the assumption that reaction
(9.2) is in equilibrium.
(b) A diagram modeled after Figure 9.2(a) that represents the addition of more SCN–(aq) to the
equilibrium mixture is:
ACS Chemistry FROG
51
Chemical Equilibria
Chapter 9
[The equilibria represented in Figure 9.2(a) are analyzed numerically in Worked Example 9.14
and Check This 9.15. A similar analysis for the equilibrium represented on the right here is:
Fe(SCN) (aq) = 3c
K=
Fe (aq)SCN (aq) (17c)(9c)
2+
3+
–
=
1
51c
The equilibrium constant for all three analyses is the same, within the constraints of the small
numbers of ions depicted in these representations.]
Check This 9.11. Temperature and Le Chatelier’s principle
Le Chatelier’s principle, tells us that the system will respond to minimize the effect of the
disturbance. The disturbance in this case is an input of energy to the system (which raises its
temperature). The system reacts by using up some of the added energy. Since the reaction
forming the blue chloro complex from the pink aquo complex is endothermic (requires energy
input), the system can use up some energy by forming more of the blue complex, as is observed.
Similarly, removing energy from the system (lowering its temperature) is also a stress and results
in using up the blue chloro complex to form more of the pink aquo complex, since this reaction
is exothermic and will add some energy to the system. This why the solution returns to the
lavender color as it cools back to room temperature and why the lavender solution becomes pink
when it is cooled even further.
Check This 9.15. Equilibrium constant expression and equilibrium constant
(a) The equilibrium constant, using the expression developed in Worked Example 9.14 and the
data from the right-hand part of Figure 9.2, is:
Fe(SCN)2 + (aq)
3c
1

K=
=
=
3+
–
Fe (aq)SCN (aq) (33c)(5c) 55c
(b) This equilibrium constant is essentially the same as that obtained in Worked Example 9.14, K
= 1/54c. With small, countable numbers of “moles,” this is the level of agreement we can expect,
if we assume that the figure is a reasonable representation of actual equilibrium systems. That the
results agree is an indication that the representation does make sense.
(c) If we set K = 1.3  102 M–1 = 1/54.5c, then c = 1.4  10–4 M. This value for c gives an initial
thiocyanate ion concentration, [SCN–(aq)], of 8.4  10–4 M [= 6·(1.4  10–4 M)], in the solution
in Figure 9.2 and a concentration of iron(III) ion, [Fe3+(aq)] (before complexation reaction), of
2.8  10–3 M [= 20·(1.4  10–4 M)]. These values are lower than those used in the actual
experiment, Investigate This 9.3, but are comparable. The higher concentrations used in the
investigation make it easier to see the color of the complex.
Check This 9.18. Acid equilibrium constant for acetic acid
The solution here is based on a pH of 2.87 for a 0.10 M aqueous acetic acid solution.
(H3O+(aq)) = 10–pH = 10–2.87 = 1.3  10-3
From the stoichiometry of the reaction, we get:
(HOAc(aq)) = (H3O+(aq)) = 1.3  10-3
(HOAc(aq)) = 0.10 – 1.3  10-3 ≈ 0.10
52
ACS Chemistry FROG
Chapter 9
Chemical Equilibria
 H3O+ (aq)OAc - (aq)
1.3  10-3 

Ka = 
= 1.7  10-5
 =

HOAc(aq)

0.10

 eq
2
Within the precision of the data, this is the same value we calculated in Worked Example 9.14.
Check This 9.20. pH of an aqueous solution of benzoic acid
For simplicity, let’s represent benzoic acid as BzOH and the benzoate anion as BzO–, so the
reaction of interest and its equilibrium constant expression are:
BzOH(aq) + H2O(aq)  H3O+(aq) + BzO–(aq)
 H3O + (aq)BzO– (aq) 
Ka = 6.4  10 = 

BzOH(aq)


The change table for this system is:
–5
species
H3O+(aq)
BzO–(aq)
BzOH(aq))
initial mol
1.0  10–7
-0-
0.028
change in mol
x formed
x formed
x reacts
final mol
x
x
0.028 – x
Assume that x is small and can be neglected relative to 0.028, that is, 0.028 – x ≈ 0.028:
(x)2
 x x 
Ka = 6.4  10–4  
=
 0.028 
0.028
(x)2 =(0.028)(6.4  10–5) = 1.8 10–6
x = (H3O+(aq)) = 1.3  10–3
Check the assumption that the value of x is mall and can be neglected compared to 0.028:
0.0013
 100  4.6%
0.028
We suggested that a 5% error is acceptable for this sort of approximation, so we accept our result
for x = (H3O+(aq)) and use it to determine the pH of the solution:
pH = –log (H3O+(aq)) = –log (1.3  10–3) = 2.88
Check This 9.22. Extent of water autoionization in a lactic acid solution
(a) As in Worked Example 9.21, the only source of OH–(aq) in this solution is the autoionization
of water. We use (H3O+(aq)) = 2.6  10-3 (calculated in Worked Example 9.19) and equation
(9.16) to find (OH–(aq)), which equals the concentration of hydronium ion produced by water
autoionization.
Kw = 1.00  10-14 = {(2.6  10-3)(OH–(aq))}eq
(OH–(aq)) = 3.8  10-12 and [OH–(aq)] = 3.8  10-12 M
The [H3O+(aq)] contributed by the water autoionization is 3.8  10-12 M.
(b) The [H3O+(aq)] contributed by water autoionization in the acetic acid solution is about three
times larger than the [H3O+(aq)] contributed by water autoionization in this lactic acid solution.
ACS Chemistry FROG
53
Chemical Equilibria
Chapter 9
The [H3O+(aq)] contributed by the acetic acid in its solution is 9.6  10–4 M, while the
[H3O+(aq)] contributed by the lactic acid in its solution is 2.6  10–3 M. Since equation (9.16)
must be satisfied in any aqueous solution at 298 K, the larger concentration of H3O+(aq) in the
lactic acid solution suppresses the water autoionization more than in the acetic acid solution.
This accounts for the smaller contribution of water autoionization to the overall amount of
hydronium ion in the lactic acid solution.
Check This 9.26. Base equilibrium constant for acetate ion
The solution here is based on a pH of 8.65 for a 0.1 M aqueous sodium acetate solution.
(H3O+(aq)) = 10–pH = 10–8.65 = 2.2  10-9
1.00  10 –14
KW
(OH (aq)) =
=
= 4.5  10-6
H 3O(aq) 2.2  10 –9
–
From the stoichiometry of the reaction, we get:
(HOAc (aq) ) = (OH– (aq) ) = 4.5  10-6
(OAc– (aq) ) = 0.10 – 4.5  10-6 ≈ 0.10
 OH– (aq)HOAc(aq)
4.5  10-6 

Kb = 
= 2.0  10-10
 =
–
0.10
OAc (aq)  eq

2
The value for Kb in Worked Example 9.25 is about three times larger than the value here. A
likely source of the discrepancy could be small amounts of impurities in the solution that causes
the measured pH to be a bit off from what it should be. [Measuring the pHs of solutions of weak
acids or their conjugate bases is not a good way to determine Ka or Kb. In Section 9.4, we will see
a better way using buffer solutions.]
Check This 9.27. Ka·Kb for the acetic acid–acetate ion pair
Based upon our results, Ka·Kb = (1.7  10–5)(2.0  10–10) = 3.4  10-15. This result is about three
times smaller than predicted from equation (9.21). Given the uncertainty discussed at the end of
Check This (9.26), this result is probably consistent with equation (9.21). Your results, based on
your pH measurements, may be better.
Check This 9.28. Using the relationship of pKa to pKb
The equilibrium constant expression for the reaction of ammonia with water is:
 NH 4  (aq)OH – (aq)
Kb = 

NH3 (aq)

 eq
Table 9.2 gives the pKa = 9.24 for the ammonium ion, NH4+(aq), the conjugate acid of ammonia.
We can rearrange equation (9.23) to get pKb and Kb:
pKb = 14.00 – 9.24 = 4.76
Kb = 10 –pKb = 10-4.76 = 1. 7  10-5
Check This 9.29. What is the relationship of pH and pOH?
(a) Take the logarithm of both sides of equation (9.16):
logKw = log(H3O+(aq)) + log(OH–(aq))
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Multiply through this equation by –1:
–logKw = –log(H3O+(aq)) – log(OH–(aq))
Substitute the value for Kw from equation (9.17) as well as pH = –log(H3O+(aq)) and pOH = –
log(OH–(aq)):
–log(1.00  10–14) = 14.00 = pH + pOH
(b) For a solution with pH = 8.73, the pOH and (OH–(aq)) are:
pOH = 14.00 – pH = 14.00 – 8.73 = 5.27
(OH–(aq)) = 10–pOH = 10–5.27 = 5.4  10–6
This value for (OH–(aq)) is the same (within the limitations of round-off errors in the
calculations) as the one we found in Worked Example 9.25 for a solution with pH = 8.73. This
makes sense because the calculations here based on logarithms of concentrations and equilibrium
constants, pH, pOH, and pKw, are exactly the same as those in Worked Example 9.25 based on
the concentrations and equilibrium constants themselves.
Check This 9.33. Ka for acetic acid
NOTE: The solution provided here is based on the assumption that the experimental mixtures
whose pHs are given above in Investigate This 9.30 were made up with 10.0 mL of the more
abundant component instead of the 9.0 mL. For your experimental results, the volumes used are
not important, as long as you know what they are. With a calibrated pH meter and careful
attention to the volumes used, your results should be better than these.
(a) The strategy and implementation here are identical to those in Worked Example 9.32 with the
exception that the experimental values are different.
Sample #1:
(H3O+(aq)) = 10–3.66 = 2.2  10–4
0.10 M 0.0100 L
M
V
acetic acid: Mmixt = s tock s tock =
= 0.091 M
0.0110 L
Vmixt
acetate ion: Mmixt =
species
initial mol
change in mol
final mol
0.10 M 0.0010 L
Ms tock  Vs tock
=
= 0.009 M
0.0110 L
Vmixt
H3O+(aq)
HOAc(aq)
OAc–(aq)
1.0  10–7
0.091
0.009
2.2  10–4 formed
2.2  10–4 reacts
2.2  10–4 formed
2.2  10–4
0.091
0.009
The amounts of acetic acid that react and acetate ion that form are negligible compared to the
amounts present initially, so we neglect them in the final row of the table.
 H3O+ (aq) OAc – (aq)
(2.2  10 –4 ) (0.009)
Ka = 
=
= 2.2  10–5

0.091
HOAc(aq) 

 eq
pKa = –log(2.2  10–5) = 4.66
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Sample #2:
(H3O+(aq)) = 10–4.76 = 1.7  10–5
0.10 M 0.0050 L
M
V
acetic acid: Mmixt = s tock s tock =
= 0.050 M
0.0100 L
Vmixt
acetate ion: Mmixt =
H3O+(aq)
HOAc(aq)
OAc–(aq)
1.0  10–7
0.050
0.050
1.7  10–5 formed
1.7  10–5 reacts
1.7  10–5 formed
1.7  10–5
0.050
0.050
species
initial mol
change in mol
0.10 M 0.0050 L
Ms tock  Vs tock
=
= 0.050 M
0.0100 L
Vmixt
final mol
(1.7  10 –5 ) (0.050)
= 1.7  10–5
0.050
pKa = –log(1.7  10–5) = 4.76
Ka =
Sample #3:
(H3O+(aq)) = 10–5.87 = 1.3  10–6
0.10 M 0.0010 L
M
V
acetic acid: Mmixt = s tock s tock =
= 0.009 M
0.0110 L
Vmixt
acetate ion: Mmixt =
species
initial mol
change in mol
final mol
0.10 M 0.0100 L
Ms tock  Vs tock
=
= 0.091 M
0.0110 L
Vmixt
H3O+(aq)
HOAc(aq)
OAc–(aq)
1.0  10–7
0.009
0.091
1.3  10–6 formed
1.3  10–6 reacts
1.3  10–6 formed
1.3  10–6
0.009
0.091
(1.3  10 –6 ) (0.091)
= 1.3  10–5
0.009
pKa = –log(1.3  10–5) = 4.88
The Ka values differ by less than a factor of two and the pKa values agree to about ±0.1 unit. We
expect to calculate values that agree, because we are investigating the same chemical system in
each sample. [The agreement could be even better.]
(b) The values for Ka in part (a) compare favorably to the value calculated in Check This 9.15.
The values obtained here are probably more reliable. These buffer solutions are less prone to
problems caused by minor impurities because they resist their effects on the pH (see Consider
This 9.36).
Ka =
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Check This 9.34. Conjugate base-to-acid ratios and pKa for acetic acid
(a) To show how the base-to-acid ratio is equal to the volume ratio in Investigate This 9.30,
consider the base-to-acid concentration ratio for sample #3 calculated in Check This 9.33:
 0.10 M 0.0100 L 


 OAc – (aq)
0.0100 L 10.0 mL
0.0110 L
=
=
=
HOAc(aq)  0.10 M 0.0010 L  0.0010 L 1.0 mL


0.0110 L
The concentrations of the stock solutions and the total volume of the sample solution cancel out,
because they are the same for both species. (If the original stock solutions do not have the same
concentration, these concentrations must be included in the ratio.)
Sample #1:
 1.0 ml 
pKa = 3.66 – log 
= 3.66 – (–1.00) = 4.66
 10.0 ml 
Sample #2:
5.0 ml 
pKa = 4.76 – log 
= 4.76 – (0) = 4.76
5.0 ml 
Sample #3:
9.0 ml 
pKa = 5.87 – log 
= 5.87 – (1.00) = 4.87
1.0 ml 
(b) Within the round-off errors of the mathematics, the results in part (a) are the same as those in
Check This 9.33. This is what you should expect because the calculations use the same data. The
only difference is that they are treated logarithmically here while the data are used directly in
Check This 9.33.
Check This 9.37. Stoichiometry in buffer solutions
(a) The positive species represented in the solution before hydrochloric acid is added are sodium
cations (7) and hydronium ion (1). The negative species are acetate anions (8). The solution
contains eight 1+ ions and eight 1– ions, so it is electrically neutral.
(b) The positive species represented in the solution after hydrochloric acid is added are sodium
cations (7) and hydronium ions (2). The negative species are acetate anions (6) and chloride
anions (3). The solution contains nine 1+ ions and nine 1– ions, so it is electrically neutral.
Check This 9.38. pH and the conjugate acid–base ratio
 A- (aq)
(a) If [HA(aq)] > [A (aq)], then log 
 is negative and, from equation (9.31)
 HA(aq)  eq
[rearranged], we see that pH – pKa < 0, so pH < pKa. If [HA(aq)] < [A-(aq)], then
 A- (aq)
log 
 is positive and pH – pKa > 0, so pH > pKa.
 HA(aq)  eq
(b) When the slider in the Web Companion, Chapter 9, Section 9.4, page 6, is at 0% conjugate
base, no blue rectangle is shown in the graphic representation of the (conjugate base)/(conjugate
acid) ratio below the graph. As the slider is moved, the blue rectangle becomes visible and the
red rectangle begins to shrink. This change is accompanied on the graph by a curve showing
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Chapter 9
increasing pH as the (conjugate base)/(conjugate acid) ratio grows. When the ratio is one (equal
size rectangles of blue and red), the pH is equal to the pKa of the conjugate acid. As the ratio
continues to grow, the pH continues to climb. From the graph and the preceding pages of the
Companion, we find that, when the (conjugate base)/(conjugate acid) ratio is 1/10, the pH =
pKa – 1 and when the ratio is 10/1, the pH = pKa + 1. Thus, the pH changes by only 1 from the
pH = pKa value even as the (conjugate base)/(conjugate acid) ratio changes by a factor of 100. In
Investigate This 9.30, we measured the pH of solutions with (conjugate base)/(conjugate acid)
ratios of about 1/9, 1/1, and 9/1 and found that the pH changed by only about two units from one
extreme to the other, as the material in the Companion indicates it should. In Investigate This
9.35 we added some strong acid or some strong base to a solution containing equal
concentrations of acetate and acetic acid, a (conjugate base)/(conjugate acid) ratio equal to one
and found that the pH changed, but not very much compared to the change when the same
amount of strong acid or base was added to pure water. Addition of strong acid and strong base
are the experiments represented on pages 4 and 5 of the Companion and we saw there that the pH
did not change too much, just as we observed experimentally in Investigate This 9.35.
Check This 9.40. pH change when hydroxide ion is added to a buffer solution
The strategy and implementation for this problem are just like those in Worked Example 9.39,
with the exception that a bit of strong base is added to the buffer solution rather than a strong
acid.
initial mol HOAc(aq) = (0.050 M)·(0.010 L) = 5.0  10–4 mol
initial mol OAc–(aq) = (0.050 M)·(0.010 L) = 5.0  10–4 mol
mol OH–(aq) added = (1.0 M)·(0.00005 L) = 5.0  10–5 mol
H3O+(aq)
HOAc(aq)
OAc–(aq)
(pH = 4.76)
5.0  10–4
5.0  10–4
change in mol
?
5.0  10–5 reacted
5.0  10–5 formed
final mol
?
4.5  10–4
5.5  10–4
species
initial mol
Only a tiny volume of base is added, so the total volume of the solution is essentially constant
and we can write:
 conjugate base 
pH = pKa + log 
 conjugate acid 
eq
5.5  10 –4 mol

0.010L 
= 4.76 + log 
–4

4.5  10 mol 0.010L 

pH = = 4.76 + 0.09 = 4.85
Addition of strong base causes the pH of the solution to increase, as we would expect. The
change in pH is not very large, because the added hydroxide ion reacted with acetic acid. The
change in the (conjugate base)/(conjugate acid) ratio is, however, not very large, so the change in
pH is small as well. In Investigate This 9.35, the pH change was also small, about 0.03 pH units.
Check This 9.46. Relating a Web Companion animation to the casein investigation
(a) In the Web Companion, Chapter 9, Section 9.5, page 3, as the slider moves from left to right
the curve on the graph shows the pH of the solution increasing. The ratio of one form of the
peptide to another in the solution, represented by the symbols in the animation, changes as the
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ACS Chemistry FROG
Chapter 9
Chemical Equilibria
pH changes. On the left half of the curve (lower pH), the changing pH changes the peptide from
its net positive (blue) form at very low pH to its net neutral (green) form at a higher pH. On the
right half of the curve (higher pH), the changing pH changes the peptide from its net neutral form
to its net negative (red) form at high pH.
(b) In Investigate This 9.42, we started with a solution of casein at high pH, corresponding to the
right hand side of the plot in the Companion, where the peptide (or protein) has a net negative
charge. As hydrochloric acid was added to lower the pH, a precipitate began to form when the
pH was about 5 ± 1 unit. We attributed this to the formation of the electrically neutral form of the
protein, just as we see that the peptide exists mainly in its neutral form at an intermediate pH. As
yet more acid was added, the pH continued to go down and the precipitate redissolved. We
attributed this to the formation of the net positively charged protein, just as we see the peptide
change to its net positive form at lower pH.
(c) The left and right hand regions of the pH curve on this page of the Companion each look like
the pH curve for a buffer solution as the (conjugate base)/(conjugate acid) ratio changes.
Compare either half of the curve with the pH vs. (conjugate base)/(conjugate acid) ratio curve on
page 6 of the Companion, Chapter 9, Section 9.4 [see Check This 9.38(b)]. The graphic beneath
the plot on this latter page shows the (conjugate base)/(conjugate acid) ratio as different size
rectangles. The graphic on the present page of the Companion shows the relative amounts of two
forms of the peptide (symbols in a square array) as a function of the pH of the solution. The blue
and green symbols represent the conjugate acid-base pair associated with the carboxylic acid
group with a pKa = 3.1. Note that the intermediate point, where the two forms are present in
equal amount, occurs at pH = 3.1, just as it should for a (conjugate base)/(conjugate acid) ratio
equal to one. Similarly, the green and red symbols represent the conjugate acid-base pair
associated with the ammonium-like side group. The halfway point in the change from the green
to red form is at pH = 8.0, which is equal to the pKa for this side group. Thus, this graphic could
be used as another representation of the contents and action of buffer solutions.
Check This 9.51. Solubility product, Ksp, for silver chromate, Ag2CrO4(s)
To determine the solubility product, we need to find [Ag+(aq)] and [CrO42-(aq)] as the precipitate
just begins to form, that is, after 9.8 mL of the silver ion solution has been added. At this point,
the total volume of the solution is 109.8 mL. This value suggests that the volume of the solution
is known to four significant figures, but the way the data are presented, about the best we can
possibly do is two or three significant figures, so we will use 0.110 L as the volume of the
solution. The number of moles of each ion in the solution divided by this total volume gives their
concentrations:
–3
mol Ag + (aq) 0.0098 L 5.0  10 M 
+
[Ag (aq)] =
=
= 4.5  10–4 M
0.110 L
0.110 L
0.100 L 2.0  10 –4 M
mol CrO 2–
4 (aq)
=
=
= 1.8  10–4 M
0.110 L
0.110 L
Substitute the molar concentration ratios into the solubility product expression to find Ksp:
Ksp = (Ag+(aq))2(CrO42-(aq)) = (4.5  10-4)2(1.8  10-4) = 3.6  10-11
This value is about 30 times larger than the value given in Table 9.4 for the solubility product of
Ag2CrO4(s). [Because the data in this problem are made up, we can’t really make any judgments
about how well or poorly experiments like this imaginary one correspond to table values.]
[CrO42-(aq)]
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Check This 9.53. Solubility of silver phosphate, Ag3PO4
(a) The dissolution reaction and solubility constant expression for silver phosphate are:
Ag3PO4(s)  3Ag+(aq) + PO43–(aq)
Ksp = 2.8  10–18 = {(Ag+(aq))3(PO43≠–(aq))}eq
(b) Let the solubility of silver phosphate be s M. For every mole of Ag3PO4(s) that dissolves, we
get three moles of Ag+(aq) and one mole of PO43-(aq) in solution, so (Ag+(aq)) = 3s and (PO43–
(aq)) = s. Substituting these values into the solubility product expression gives:
Ksp = 2.8  10–18 = (3s)3(s) = 27s4
s = 1.8  10–5 M
(c) To compare the solubility in part (b) with the values from the handbooks, we either need our
solubility in g·L–1, or we have to convert the handbook solubilities in g·L–1 to molar solubilities.
We will do the former:
 419 g Ag 3PO4 
–1
s = (1.8  10–5 mol·L–1) 
 = 0.0075 g·L

1 mol
This solubility value is intermediate between the two handbook values, but closer to the lower of
the two. However, it is rare for the solubility product to give a higher solubility than that
measured experimentally, because most of the ionic interactions that occur in the solutions favor
solubility. Thus, the higher measured solubility, 0.065 g·L–1, may be correct, even though the
lower one, 0.0065 g·L–1, is closer to what we calculated from the solubility product.
Check This 9.55. Solubility of Cu(IO3)2(s) in water and 0.25 M Cu2+(aq) solution
(a) The solubility equilibrium reaction for copper(II) iodate is:
Cu(IO3)2(s)  Cu2+(aq) + IO3–(aq)
If s is the molar solubility of the solid, then (Cu2+(aq)) = s and (IO3–(aq)) = 2s. Write the
equilibrium constant expression, substitute these values and Ksp from Table 9.4, and solve for s:
Ksp = 1.4  10-7 = (Cu2+(aq))(IO3–(aq))2 = (s)(2s)2 = 4s3
s = 3.3  10-3 M
(b) Let s be the molar solubility of copper(II) iodate in the solution that initially contains
[Cu2+(aq)] = 0.25 M. When solubility equilibrium is attained in this solution, we will have
(Cu2+(aq)) = 0.25 + s and (IO3–(aq)) = 2s. We know that the solubility in pure water is low and
we expect the solubility to be even lower in a solution that already contains copper(II) cation. It
is likely that s is negligible compared to 0.25, so we will assume (Cu2+(aq)) ≈ 0.25. Now we can
use the equilibrium constant and equilibrium constant expression to find s:
Ksp = 1.4  10-7 = (Cu2+(aq))(IO3–(aq))2 = 0.25(2s)2 = s2
s = 3.7  10-4 M
As expected, this solubility is almost 10-fold lower than in pure water, because the solubility
product is satisfied with less solid dissolved when there is already Cu2+(aq) in solution.
(c) Because (IO3–(aq)) enters the equilibrium constant expression raised to the second power, we
expect that a solution containing the iodate anion as the common ion will have a larger effect on
the solubility than an equal concentration of copper(II) cation. Thus, we predict that the
solubility of copper(II) iodate in 0.25 M NaIO3 solution will be lower than the solubility
calculated in part (b). (Working out the solubility quantitatively gives s = 2.2  10-4 M, which
confirms our prediction..)
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Check This 9.56. Compare Figures 9.9 and 9.10
(a) Figure 9.9(b) and Figure 9.10(a) each contain six Ba2+ cations and six SO42+ anions. In each
figure, three of each ion are in solution and the other three are present in the ionic solid
BaSO4(s). This is supposed to represent the solubility equilibrium for the ionic solid. Figure
9.10(a) also contains 12 each of Na+ cations and Cl– anions and shows that the solubility
equilibrium in pure water represented in Figure 9.10(a) is not disturbed by the presence of these
other ions. (In reality, the other ions are likely to have some relatively minor effect on the
solubility of the ionic solid, because they change the electrical properties of the solution to some
extent.)
(b) You could make the solutions in Figure 9.10(a) identical to the one in Figure 9.9(b) by
dissolving NaCl(s) to give a solution that also contains equal numbers of Na+ cations and Cl–
anions. As we said in part (a), the figures suggest that the presence of these other ions, not
common to the ionic solid, does not affect the solubility equilibrium for BaSO4(s).
(c) Figures 9.9(c) and 9.10(b) each contain six Ba2+ cations and ten SO42+ anions. In each figure,
five of each ion are present in the ionic solid BaSO4(s). In each solution, the remaining Ba2+
cation and five SO42+ anions are in solution. In the final solution in Figure 10.10(b), electrical
neutrality is maintained by the presence of eight Na+ cations to give a total of ten positive and ten
negative charges in the solution. (The solids are electrically neutral as well, with five Ba2+
cations and five SO42+ anions.) The solution in Figure 9.9(c) is a bit more complicated, because it
also contains Cl– anions. There are 12 Cl– anions and 20 Na+ cations in this solution. The eight
extra Na+ cations balance the charge on four of the SO42+ anions, just as in Figure 9.10(b).
Overall, this solution has a total of 22 positive charges and 22 negative charges. The presence of
the extra common ion, SO42+ anion, in each of these solutions lowers the solubility of the ionic
solid, BaSO4(s). This is why more of it is present in these cases compared to the solutions
analyzed in part (a), which contain the same amount of Ba2+ cation.
(d) Again, you could make the solutions in Figure 9.10(b) identical to the one in Figure 9.9(c) by
dissolving NaCl(s) to give a solution that contains an additional 12 Na+ cations as well as 12 Cl–
anions. The implication of these figures is, once again, that the presence of these other ions, not
common to the ionic solid, does not affect the solubility equilibrium for BaSO4(s).
Check This 9.58. Agreement of units in equation (9.42)
The units of the term on the far right in equation (9.42) determine
the units of RT because the logarithmic term has no units. The units of RT are (J·mol–1·K–1)(K)
= J·mol–1, which are the units of free energy for a mole of reaction.
Check This 9.60. Reaction quotient and free energy for a gas phase reaction
Under these new conditions (see Worked Example 9.59 for the old conditions), the reaction
quotient is:
 (BrCl(g) )2  

(0.500 bar 1 bar)2
Q = 
=
 
 = 17.4
(Br2 (g))(Cl 2 (g) )  (0.120 bar 1 bar)(0.120 bar 1 bar) 
The free energy change for these conditions is:
Greaction = G°reaction + RTlnQ = –4.90 kJ + (8.314 J·mol–1·K–1)(298 K)ln(17.4)
Greaction = +2.18 kJ
The positive value for Greaction shows that the reaction is not spontaneous under these
conditions.
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Check This 9.66. Rearranging equation (9.58)
To obtain equation (9.59) from equation (9.58) we need to put the difference on the right over a
common denominator and change its sign:
H o

K 
 H oreaction   T1 – T2   H oreaction   T2 – T1 
1 
react ion  1


ln  2  = – 
=
–
=
–
 R T T 

 K1 
R   T2 T1  
R   T2 T1 

 2
1
Check This 9.68. H°reaction from solubility temperature dependence
(a) The solubility of Ca(OH)2(s) decreases as temperature increases, that is, as energy is added to
the system. The addition of energy is a disturbance to the system, which reacts by trying to
minimize the disturbance by using up some energy. Since the amount of precipitate increases as
energy is added, solid formation must be the process that requires energy, that is, is endothermic.
If precipitation is endothermic, then dissolution of Ca(OH)2(s) must be exothermic.
(b) To find Ksp we need to convert the solubilities in g·L–1 to molar solubilities, s, use the
stoichiometry of the dissolution reaction to find the solution concentrations, and substitute in the
equilibrium constant expression. The molar mass of Ca(OH)2 is 74.1 g:
1.85 g  L–1
s273 =
= 2.50  10–2 M
74.1 g  mol –1
0.77 g  L–1
–2
s373 =
M
–1 = 1.04  10
74.1 g  mol
The reaction equation and solubility product expression are:
Ca(OH)2(s)  Ca2+(aq) + 2OH–(aq)
Ksp = (Ca2+(aq))(OH–(aq))2 = (s)(2s)2 = 4s3
Ksp (at 273 K) = 4(2.50  10–2)3 = 6.25  10–5
Ksp (at 373 K) = 4(1.04  10–2)3 = 4.5  10–6
Substitute these temperatures (T1 = 373 K and T2 = 273 K) and equilibrium constants into
equation (9.59) and solve for H°reaction:
K 
 6.25  10 –5 
 H oreaction   T2 – T1   H oreaction   (273 K) – (373 K) 
ln  2  = ln 
=
=

 4.5  10 –6 
R   (273 K) (373 K) 
 K1 
R   T2 T1  
  6.25  10 –5    (273 K) (373 K) 
H°reaction = (8.314 J·mol ·K ) ln 
–6   

  4.5  10    (273 K) – (373 K) 
–1
–1
H°reaction = –22 kJ·mol–1
As predicted in part (a), using Le Chatelier’s principle, the reaction is exothermic.
(c) Use the H°f values in Appendix B to find H°reaction:
H°reaction = (1 mol)[H°f(Ca2+(aq))] + (2 mol)[H°f(OH–(aq))] – (1 mol)[H°f(Ca(OH)2(s))]
H°reaction = (1)[–542.83 kJ] + (2)[–229.99 kJ] – (1)[–986.09 kJ]
H°reaction = –16.72 kJ (per mole of reaction)
The value for H°reaction calculated from the temperature dependence of the solubility agrees well
with this value calculated from the thermodynamic tables. The agreement is not always so good
for comparisons like this, but it is usually in the right ballpark.
62
ACS Chemistry FROG
Chapter 9
Chemical Equilibria
Check This 9.70. S°reaction and G°reaction from solubility temperature dependence
(a) Following the strategy in Worked Example 9.69 gives:
(–22  10 3 J mol–1 )
Horeacti on
–1
–1
–5
S°reaction = RlnKsp +
= (8.314 J·mol ·K )·ln(6.25  10 ) +
273 K
T
2
–1
–1
S°reaction = –1.6  10 J·mol ·K
Combining this S°reaction with the H°reaction from Check This 9.68 gives:
G°reaction(298 K) = H°reaction – TS°reaction
= (–22  103 J·mol–1) – (298 K)·(–160 J·mol–1·K–1)
G°reaction(298 K) = 26 kJ·mol–1
lnKsp
3
–1
o


26  10 J mol
Greacti
on
=–
= –
 = –10.5
–1
RT
 8.314 J  mol  K  298 K 
Ksp (298 K) = 2.8  10–5
(b) We can combine standard entropies and standard free energies of formation from Appendix
B, exactly as standard enthalpies of formation were combined in Check This 9.68 to get:
S°reaction = –158.0 J·mol–1·K–1
G°reaction = 30.4 kJ·mol–1
The agreement of these thermodynamic values with those from part (a) is pretty good,
considering the quality of the data and the nonideality of the solution.
Ksp (from Table 9.4) = 6.5  10–6
The value for the free energy change calculated from the solubility data is less positive than the
value calculated from the thermodynamic data, which may be a factor that leads to a higher
solubility product than from the table. [Using the value for the thermodynamic free energy
change gives a solubility product of 4.7  10–6, which leads one to wonder about the source of
the table value, but this is not worth bringing up for students at this stage.]
Check This 9.72. Graphical determination of H°reaction and S°reaction
(a) The trans-1,2-dicholoroethene data from Table 9.5 and their conversion to the values we
need to plot are in this table:
v.p., torr
10.0
40.0
100.
400.
760.
v.p., bar
0.0133
0.0533
0.133
0.533
1.013
ln(v.p.)
–4.320
–2.932
–2.017
–0.629
0.0129
T, K
235.1
256.1
272.9
303.9
320.9
0.00425
0.00390
0.00366
0.00329
0.00312
1 , K–1
T
The plot of these data is:
ACS Chemistry FROG
63
G
are
Qraphics
uickTim
needed
decom
e™
toand
see
pressor
athis picture.
lnK
Chemical Equilibria
0.5
0
-0.5
-1
-1.5
-2
-2.5
-3
-3.5
-4
Chapter 9
trans-1,2-dichloroethene
lnK= -3,824(1/T) +11.96
0.003
0.0035
1/T
0.004
0.0045
We get the thermodynamic values using the-4.5
slope and intercept values noted on the plot:
H o

slope = –  reacti on  = (–3.82  103 K)
 R 
H°reaction = (8.314 J·mol–1·K–1)·(3.82  103 K) = 31.8 kJ·mol–1
Soreacti on
= 11.96
R
S°reaction = (8.314 J·mol–1·K–1)·(11.96) = 99.4 J·mol–1·K–1
We also use the slope and intercept values to obtain the vapor pressure at 298 K:
o
H reacti

1 
Soreacti on
on 1 
lnK = ln(v.p.) = – 
= (–3.82  103 K) 
+ 11.96
  +

 R  T
298 K 
R
intercept =
ln(v.p.) = –0.859
vapor pressure at 298 K = 0.424 bar = 318 torr
(b) It takes a bit more energy to vaporize the cis isomer, H°reaction = 33.5 kJ·mol–1 from Worked
Example 9.71, probably because it has a permanent dipole moment and there is more dipoledipole attraction in the liquid phase, which also accounts for its lower volatility, a vapor pressure
of 204 torr at 298 K. Similarly, the somewhat larger ordering in the liquid phase for the cis
isomer leads to a slightly higher entropy of vaporization, S°reaction = 99.4 J·mol–1·K–1.
Check This 9.75. Transport of oxygen by hemoglobin
(a) In the arterial blood, hemoglobin is essentially 100% oxygenated; all four binding sites are
occupied by oxygen. In the venous blood in tissues that are working, hemoglobin is less than
10% oxygenated; on average, only four in ten (40%) of the hemoglobin molecules have even one
binding site occupied by oxygen.
(b) When hemoglobin is in the arteries (close to the lungs), it is saturated with oxygen. When
these saturated molecules reach the tissues where the oxygen pressure is very low, they are no
longer in equilibrium with the surrounding oxygen pressure. The hemoglobins lose oxygen (give
it up to the cells that need it) to move toward their 10%-saturated equilibrium state under these
64
ACS Chemistry FROG
Chapter 9
Chemical Equilibria
conditions. This is the phenomenon that is represented on the right in the chapter opening
illustration where red blood cells saturated with oxygen (red) pass through the capillaries and
lose their oxygen to become deoxygenated (blue).
Check This 9.76. Relative sizes of oxygen-binding equilibrium constants
(a) Although no numerical scale is shown in Figure 9.14, it is always the case that a lower (more
negative) free energy change is associated with a larger equilibrium constant [shown in equation
(9.48)]. The free energy change is most positive for the binding of the first oxygen molecule and
most negative for binding of the fourth with the second and third at intermediate values and
about the same. The order of equilibrium constants is, therefore, K1 < K2 = K3 < K4.
(b) Myoglobin has a much higher affinity for O2(g) than hemoglobin. Even at a very low
pressure, myoglobin is 50% saturated, allowing it to store O2(g) more effectively than
hemoglobin. In a working muscle, myoglobin is about 80% saturated while hemoglobin is only
about 10% saturated. When hemoglobin is in the capillaries at low oxygen pressure, its O2(g)
binding decreases dramatically allowing O2(g) to readily leave and bind to myoglobin.
Check This 9.79. H°reaction for 2NO2(g)  N2O4(g)
In Appendix B, we find Hf(NO2(g)) = 33.18 kJ·mol-1 and Hf(N2O4(g)) = 9.16 kJ·mol-1.
Combining these values with the stoichiometry for reaction (9.76) gives:
Hreaction = [(1 mol)(9.16 kJ·mol-1)] – [(2 mol)(33.18 kJ·mol-1)]
Hreaction = –57.20 kJ
Our reasoning in Consider This 9.77(e) led to the prediction that this reaction is exothermic and
the negative Hreaction calculated here confirms this prediction.
ACS Chemistry FROG
65
Chemical Equilibria
Chapter 9
Solutions for Chapter 9 End-of-Chapter Problems
Problem 9.1.
(a) Balanced chemical equations for the formation of pale blue Cu(OH)2(s) and deep blue-violet
[Cu(NH3)4]2+(aq) when NH3(aq) is slowly added to a light blue solution of Cu2+(aq) are:
2NH3(aq) + 2H2O(l) + Cu2+(aq)  Cu(OH)2(s) + 2NH4+(aq)
light blue
pale blue
Cu(OH)2(s) + 4NH3(aq)  [Cu(NH3)4]2+(aq) + 2OH–(aq)
pale blue
deep blue-violet
(b) To understand how the formation of Cu(OH)2(s) from Cu2+(aq) will be affected by basic
conditions, we need to consider the reactions that make up the net reaction written in part (a):
2NH3(aq) + 2H2O(l)  2NH4+(aq) + 2OH–(aq)
Cu2+(aq) + 2OH–(aq)  Cu(OH)2(s)
Additional OH–(aq) is a disturbance to these equilibria (Le Chatelier’s principle) and the system
will respond by using up some of the additional OH–(aq). Thus, the first reaction will shift
toward reactants, because OH–(aq) is a product, and the second reaction will shift toward
products because OH–(aq) is a reactant. The net result will be to favor formation of Cu(OH)2(s)
from Cu2+(aq).
(c) The equation written in part (a) suggests that the formation of [Cu(NH3)4]2+(aq) from
Cu(OH)2(s) will be favored under mildly acidic conditions, because H3O+(aq) will react with
OH–(aq) ions to form water. This reaction is a disturbance to the equilibrium, because it removes
a product of the reaction, The system should respond by trying to form more OH–(aq), which
also would mean the formation of more [Cu(NH3)4]2+(aq). This analysis is flawed because it
neglects the basicity of NH3(aq), which will also react with added H3O+(aq):
NH3(aq) + H3O+(aq)  NH4+(aq)
This reaction removes a reactant, NH3(aq), necessary for formation of [Cu(NH3)4]2+(aq), so the
presence of H3O+(aq) in the solution will not favor formation of [Cu(NH3)4]2+(aq). Note that the
reaction of H3O+(aq) with OH–(aq) also removes a reactant in the second reaction in part (b), so
Cu(OH)2(s) will tend to dissolve or not form under acidic conditions.
(d) None of the reactions involves oxidation or reduction. Copper has an oxidation number of +2
in all cases.
Problem 9.2.
If Fe3+(aq) and SCN–(aq) formed a one-to-two metal ion complex, the complex in Figure 9.1
would have to show the new ratio of two SCN–(aq) ions to every Fe3+(aq) ion. Twice as many
SCN–(aq) ions would be used in forming the complex and there would be four more unreacted
Fe3+(aq) ions in the right-hand part of the figure, as shown in this alternative figure:
66
ACS Chemistry FROG
Chapter 9
Chemical Equilibria
Problem 9.3.
(a) Addition of CN–(aq) to the equilibrium reaction, Cd2+(aq) + 6CN–(aq)  Cd(CN)62–(aq),
will cause the reaction to shift toward formation of more of the product, Cd(CN)62–(aq). Addition
of more of a reactant is a disturbance to the equilibrium system (Le Chatelier’s principle) and the
system responds by minimizing the disturbance, using up some of the added CN–(aq) to form
more of the product.
(b) Identification of change in the CoCl2 aqueous equilibrium is possible because of color
differences between the reactants and the products. Both species in this equilibrium are colorless,
so there is no visual clue to the position of equilibrium.
Problem 9.4.
The statements in this problem refer to the equilibrium reaction: H2(g) + Cl2(g)  2HCl(g).
(a) When this reaction has reached a state of equilibrium, no further reaction occurs. This is not a
true statement. No further net reaction occurs, but there is a dynamic equilibrium in which
molecules of HCl(g) continuously form, even as other HCl(g) molecules are decomposing into
H2 and Cl2. While the macroscopic concentrations are not changing, there is constant change at
the microscopic level.
(b) When equilibrium is established, the number of moles of reactants equals the number of
moles of products for this reaction. This is not a true statement. The stoichiometric coefficients
that balance the equation do not give the relative number of moles present at equilibrium. It is
necessary to know the position of the equilibrium and to use the equilibrium constant at a given
temperature to find relative numbers of moles of reactants and products.
(c) The concentration of each substance in the system will be constant at equilibrium. This is a
true statement. The observable concentrations are constant if dynamic equilibrium has been
established, but there still is constant change at the microscopic level.
Problem 9.5.
The reaction of interest is: CO(g) + 2O2(g)  2CO2(g). In this reaction, a C–O triple bond (BH =
1071 kJ·mol–1) and two O–O double bonds (BH = 498.7 kJ·mol–1) are broken, and four C–O
double bonds in carbon dioxide (BH = 799 kJ·mol–1) are made. The overall change in enthalpy
for the reaction is –1127 kJ·mol–1, that is, the reaction is exothermic. If the reaction is at
equilibrium and is disturbed by adding energy to raise its temperature, the system will respond
(Le Chatelier’s principle) by “using up” some of the added energy which it can do by proceeding
in reverse (the endothermic direction) toward increase in the concentration of the reactants,
CO(g) and O2(g), at the expense of some of the reactant, CO2(g).
ACS Chemistry FROG
67
Chemical Equilibria
Chapter 9
Problem 9.6.
The reaction of interest is: CO(g) + 2H2(g)  CH3OH(g).
(a) If some CH3OH(g) is removed while the volume of the system is held constant, the
concentration of the product decreases. The system will respond to this disturbance
(Le Chatelier’s principle) by producing more product at the expense of a decrease in the
concentrations (amounts) of the reactants.
(b) Adding CO(g) to the system while the volume is held constant increases the concentration of
one of the reactants. The system will respond to this disturbance by reacting to reduce the
concentration of this reactant with the consequence that the concentration of the other reactant
will be reduced and the concentration of the product will increase.
(c) If the system is compressed to a smaller volume, the concentrations of all the species will be
increased. In response to this disturbance, the system will react to reduce the number of moles of
gas, so as to try to bring the concentrations as near as possible to where they had been. The
number of moles of gas will be reduced if reactants combine to form the product, so the result
will be an increase in the concentration of the product and reductions in the concentrations of the
reactants.
(d) N2(g) has no role in the reaction and adding it at constant volume does not affect the
concentrations of the reactants and products. There will be no change in the equilibrium system
when this addition is made.
Problem 9.7.
The reaction of interest is this Lewis acid-base equilibrium reaction between boric acid and
water:
B(OH)3(s) + 2H2O(aq)  [B(OH)4]–(aq) + H3O+(aq)
(a) Addition of B(OH)3(s) will have no effect on the equilibrium. Addition of a solid reactant
does not change its concentration.
(b) Addition of H3O+(aq) will cause this equilibrium reaction to shift toward the reactants.
Increasing the H3O+(aq) concentration is a disturbance to the equilibrium system (Le Chatelier’s
principle) and the system will respond by using up some of the added H3O+(aq). It can do this by
proceeding in reverse, using up some [B(OH)4]–(aq) and forming more B(OH)3(s).
(c) Addition of OH–(aq) will cause this equilibrium reaction to shift toward the products. Added
OH–(aq) will react with H3O+(aq), which will disturb the equilibrium system by decreasing the
concentration of a product. The system will respond by producing more H3O+(aq) to replace part
of what is lost, thus forming more [B(OH)4]–(aq) and using up some B(OH)3(s).
Problem 9.8.
The reaction of interest is: HA(aq) + H2O(aq)  H3O+(aq) + A–(aq).
(a) This reaction equation tells us that, in the forward direction, a molecule of HA, HA(aq),
(solvated by the water) interacts with a molecule of water, H2O, to transfer a proton to the water
to form a solvated hydronium ion, H3O+(aq), and a solvated anion, A–(aq), HA without its acidic
proton. In the reverse direction a hydronium ion interacts with a solvated anion to transfer a
proton to the anion to form a water molecule and a solvated molecule of the acid, HA(aq). The
way the equation is written, it appears that the same hydronium and anion that form in the
forward reaction react in the reverse reaction. This is an incorrect interpretation. The Web
Companion animation illustrates the correct interpretation.
68
ACS Chemistry FROG
Chapter 9
Chemical Equilibria
(b) The beginning (proton transfer from HA(aq) to H2O(aq)) and end (proton transfer from
H3O+(aq) to A–(aq)) of the reaction shown in the Web Companion, Chapter 9, Section 9.4,
page 1, animation are exactly consistent with the reaction written and described above. What we
see in the “middle” of the animation is that protons can be transferred between hydronium ions
and water molecules, that it is not the same proton transferred in each case, and that the anion to
which a proton is transferred from a hydronium is not the same one from which a proton
originally came. What is missing in the equation is a way to show: (1) that any one of the acid
molecules in the mixture might transfer its proton to a water molecule, (2) that the hydronium
ion formed is just one of many hydronium ions in the solution, (3) that the hydronium ions are
continuously transferring protons to water molecules to form different hydronium ions, and (4)
that any one of the acid anions and any one of the hydronium ions might be the ones that react to
transfer a proton from hydronium ion to anion. Although reaction equations are a very compact
way to represent the stoichiometry of reactions, they do not very well represent the details of
what is occurring at the molecular level. As you use such equations, always try to use what you
are learning about reactions to visualize what is going on at the molecular level as well.
Problem 9.9.
The equilibrium constant expression for ammonia synthesis is:
NH3 (g)
N 2(g)H2 (g) 3
2
Keq =
Since all species in this equilibrium constant expression are gases, the dimensionless
concentration ratios should be expressed in bar (standard state = 1 bar).
Problem 9.10.
It requires 50 drops of a 1.0 M weak acid, HA(aq), solution added to 10.0 mL of distilled water
to produce enough ions to light the bulb in a conductivity detector as brightly as 1 drop of 1 M
HCl(aq) added to 10.0 mL of distilled water. We can conclude that there are only about 2%
 1 drop

50 drops 100%  as many ions formed by the weak acid as by an equivalent number

of moles of HCl(aq), hydrochloric acid. We know that HCl(aq) reacts completely to form
H3O+(aq) and Cl–(aq) ions, so we estimate that only about 2% of the weak acid reacts to transfer
its proton to water. Thus, in a 1 M solution of HA(aq), the concentrations are [HA(aq)] = 1 M,
[H3O+(aq)] = 0.02 M, and [A–(aq)] = 0.02 M. The proton transfer reaction and equilibrium
constant (dissociation constant) for the weak acid are:
HA(aq) + H2O(aq)  H3O+(aq) + A–(aq)


 H3O+ (aq)A– (aq) 
 (0.02)(0.02) 
Keq = 
= 4  10–4
 = 


(1)(1)
 HA(aq) H2 O(aq)
eq
Problem 9.11.
(a) The reaction equation and equilibrium constant expression for the decomposition of calcium
carbonate are:
CaCO3(s)  CaO(s) + CO2(g)
ACS Chemistry FROG
69
Chemical Equilibria
Keq =
Chapter 9
CaO(s)CO2 (g)
CaCO 3(s)
For pure solids the dimensionless concentration ratio is 1, unity. For the gas we need to use bar
as the unit (standard state = 1 bar).
(b) If the equilibrium pressure of CO2(g) is 0.37 bar, then:
(0.37 bar)
1
(1 bar)
Keq =
= 0.37
1


Here, we have substituted the values for the concentration ratios to get Keq.
Problem 9.12.
(a) Addition of four more thiocyanate anions, SCN(aq)–, to the equilibrium solution depicted in
the center of Figure 9.2 leads to the formation of a total of three iron-thiocyanate complexes at
equilibrium in the solution. As in Worked Example 9.14, we count the number of each species
represented in our solutions and assume the numbers are proportional to their molarities in the
solution. After adding four more thiocyanate ions to the middle panel of Figure 9.2 and forming
three iron-cyanate complexes in the resulting mixture, there will be nine thiocyanate ions
unreacted (the original eight plus the four added minus the three in the complexes) and 17 iron
ions (the original 20 minus the three in the complexes). The equilibrium constant for this system
is then:
Fe(SCN)2 + (aq) = 3c = 1
K=
Fe3+ (aq)SCN– (aq) (17c)(9c) 51c
(b) The equilibrium constants in Worked Example 9.14 and Check This 9.15 are 1/54c and
1/55c, respectively. In part (a), we get 1/51c. Within the limitations of these simple pictures with
countable numbers of “molecules,” the agreement among these three values is quite good.
Problem 9.13.
(a) The pressure of each gas in the original mixture in the 1.00 L container at 500 K is given by
P = nRT/V, where the number of moles of each gas is given in the problem statement:
(2.21  10–3 mol)(8.314  10 –2 L  bar  K –1  mol –1 )(500K)
P(HI) =
= 9.19  10–2 bar
1.00 L
(1.46  10–3 mol)(8.314  10 –2 L  bar  K–1  mol –1 )(500K)
P(I2) =
= 6.07  10–2 bar
1.00 L
(2.09  10–5 mol)(8.314  10 –2 L  bar  K–1  mol –1 )(500K)
P(H2) =
= 8.69  10–4 bar
1.00 L
(b) The equilibrium constant expression for the reaction, H2(g) + I2(g)  2HI(g), is:
Keq =
HI(g) 2
H 2(g)I2 (g)
(c) Each of the dimensionless concentration terms in the equilibrium constant expression is a
ratio of the pressure (in bar) of the reactant to the standard pressure, 1 bar. Therefore, we can
70
ACS Chemistry FROG
Chapter 9
Chemical Equilibria
substitute the numeric values for the pressures in part (a) into the expression in part (b) to get the
numeric value of the equilibrium constant:
9.19  10 
Keq =
= 1.60  102
8.69

10
6.07

10



–2 2
–4
–2
Problem 9.14.
{NOTE: There is a typographical error in this problem. The concentrations should be [HIn(aq)]
= 6.3  10–5 M and [In–(aq)] = 8.7  10–5 M. Both are labeled [HIn(aq)].}
(a) For the reaction, HIn(aq) + H2O(aq)  H3O+(aq) + In–(aq), the equilibrium constant
expression is:
H O (aq) In
=
+
Keq
–
3
(aq)
HIn(aq)H 2O(aq)
(b) The molar concentrations of the acidic and basic forms of the indicator are given and we can
use the pH to get the molar concentration of the hydronium ion, [H3O+(aq)] = 10–7.41 M = 3.9 
10–8 M. The ratio of these molar concentrations to the standard concentration, 1 M, gives the
dimensionless concentration terms for these species. Since the solution is dilute, the water
concentration is essentially the same as in pure water, 55.5 M, so the dimensionless
concentration term for water is unity. The numeric value for the equilibrium constant is:
3.9 10 –8 8.7 10 –5 

Keq =
= 5.4  10–8
6.3 10 –5 1
Problem 9.15.
The information in this table is used in this problem and the next.
Acid name
acetic acid
chloroacetic acid
Formula
CH3C(O)OH
ClCH2C(O)OH
Ka
1.8  10
1.4  10–3
–5
(a) The acid equilibrium (proton transfer to water) equations for the acids are:
CH3C(O)OH(aq) + H2O(l)  H3O+(aq) + CH3C(O)O–(aq)
ClCH2C(O)OH(aq) + H2O(l)  H3O+(aq) + ClCH2C(O)O–(aq)
(b) The equilibrium constant expressions for Ka for the reactions in part (a) are:



 H3O+ (aq) CH3C(O)O – (aq) 

Ka(acetic acid) = 
 (CH 3C(O)OH(aq)) H2 O(aq)
eq



 H3O+ (aq) ClCH2 C(O)O– (aq) 

Ka(chloroacetic acid) = 
 (ClCH 2C(O)OH(aq)) H2 O(aq) 
eq
(c) The equilibrium constant for chloroacetic acid is nearly two orders of magnitude greater than
that for acetic acid, that is, chloroacetic acid is a stronger acid than acetic acid (but still does not
ACS Chemistry FROG
71
Chemical Equilibria
Chapter 9
transfer all its protons to water). For solutions of equal concentration, chloroacetic acid will
release more hydronium ions into solution, resulting in a lower pH than for acetic acid.
Problem 9.16.
(a) Use the equilibrium constant expressions and equilibrium constants from the previous
problem to find the pH of a 0.050 M solution of acetic acid:
Ka(acetic acid) = 1.8 10
species
initial mol
change in mol
final mol
–5



 H3O+ (aq) CH3C(O)O – (aq) 

=
 (CH 3C(O)OH(aq)) H2 O(aq)
eq
CH3C(O)O–(aq)
-0x formed
x
H3O+(aq)
1.0  10–7
x formed
x
CH3C(O)OH(aq)
0.050
x reacts
0.050 – x
Because Ka(acetic acid) is relatively small, start by assuming that the value of x is small and can
be neglected relative to 0.050:
x2
 x x 
 x x 
Ka = 1.8  10–5 = 

=


 0.050 – x 
0.050 
0.050
x2 = 0.050·(1.8  10–5) = 9.0  10–7
x = 9.5  10–4
Check the assumption that the value of x is small and can be neglected relative to 0.050:
 9.5  10 –4 
 0.050  ·100% = 1.9%
Our rule of thumb is that a 5% error is acceptable and this is well within the limit.
pH = –log (H3O+(aq)) = –log (9.5  10–4) = 3.02
(b) Use the equilibrium constant expressions and equilibrium constants from the previous
problem to find the pH of a 0.050 M solution of chloroacetic acid:
–3
Ka(chloroacetic acid) = 1.4 10
species
initial mol
change in mol
final mol
H3O+(aq)
1.0  10–7
x formed
x



 H3O+ (aq) ClCH2 C(O)O– (aq) 

=
 (ClCH 2C(O)OH(aq)) H2 O(aq) 
eq
ClCH2C(O)O–(aq)
-0x formed
x
ClCH2C(O)OH(aq)
0.050
x reacts
0.050 – x
Because Ka(chloroacetic acid) is relatively small, start by assuming that the value of x is small
and can be neglected relative to 0.050:
x2
 x x 
 x x 
–3
Ka = 1.4  10 = 

=
 0.050 – x 
 0.050 
0.050
2
–3
–5
x = 0.050·(1.4  10 ) = 7.0  10
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ACS Chemistry FROG
Chapter 9
Chemical Equilibria
x = 8.4  10–3
Check the assumption that the value of x is small and can be neglected relative to 0.050:
 8.4  10 –3 
 0.050  ·100% = 17%
Our rule of thumb is that a 5% error is acceptable and this value is far outside the limit, so we
will have to discard the assumption and solve the problem more exactly. One approach is to use
successive approximations. We start with the result just obtained for x, and use it to give us an
approximation for the concentration ratio of the unreacted acid:
0.050 – x = 0.050 – 0.0084 ≈ 0.042
Then we repeat the above calculation with this new value for the acid concentration:
x2
 x x 
–3
Ka = 1.4  10  
=
 0.042 
0.042
2
–3
x = 0.042·(1.4  10 ) = 5.9  10–5
x = 7.7  10–3
If we consider a third approximation we would use this new value of x to give us another
approximation for the concentration of the unreacted acid:
0.050 – x = 0.050 – 0.0077 ≈ 0.042
This value is identical (within the limits of uncertainty of the data given) to the one used in the
previous calculation, so it will give the same result, x = 7.7  10–3. Thus, we take this result as
the correct value for x = (H3O+(aq)). Another approach would have been to begin with the exact
equation, rearrange it to quadratic form, ax2 +bx + c = 0, and then solve using the quadratic
b  b 2  4ac
formula: x =
. The result is x = 7.7  10–3, so both methods give the same value.
2a
The pH of the 0.050 M chloroacetic acid solution is:
pH = –log (H3O+(aq)) = –log (7.7  10–3) = 2.11
(c) The results calculated in parts (a) and (b) confirm the prediction made in our solution to
Problem 9.15(c). For acetic acid and chloroacetic acid at the same concentrations, the stronger
acid, chloroacetic, transfers more protons to water and forms more hydronium ions. Thus, the pH
is lower for chloroacetic acid.
Problem 9.17.
(a) For the autoionization reaction, H2O(l) + H2O(l)  H3O+(aq) + OH–(aq), Kw =
(H3O+(aq))(OH–(aq)) and pKa = –log{(H3O+(aq))(OH–(aq))}. If the value of pKa decreases as
the temperature increases, this means that the value of Kw increases with increasing temperature:
at 25 C:
Kw = 10 – pKw = 10–14.0 = 1 
–14
–12
at 60 C:
Kw = 10 – pKw = 10–11.5 = 3 
The product concentrations increase at the higher temperature. Le Chatelier’s principle predicts
that equilibria will shift toward formation of more product if energy is added (the temperature is
increased) in endothermic reaction. The autoionization of water is an endothermic reaction.
–7
(b) An aqueous solution with H3O+(aq) concentration equal to 1.0 
M at 25 °C will have a
+
–7
pH of 7.0 [= –log(H3O (aq)) = –log(1.0 
)]. To find out whether the solution is acidic,
ACS Chemistry FROG
73
Chemical Equilibria
Chapter 9
basic, or neutral, we need to know how the concentrations of hydronium ion, [H3O+(aq)], and
hydroxide ion, [OH–(aq)], compare. Use Kw to find [OH–(aq)]:
–14
–7
Kw = 1 
= (H3O+(aq))(OH–(aq)) = (1.0 
)(OH–(aq))
–7
–7
(OH–(aq)) = 1 
[OH–(aq)] = 1 
M
+
–
Because [H3O (aq)] = [OH (aq)] (within the uncertainty of the data given), a solution in which
–7
[H3O+(aq)] = 1.0 
M at 25 °C is neutral.
–7
(c) An aqueous solution with H3O+(aq) concentration equal to 1.0 
M at 60 °C will have a
+
–7
pH of 7 [= –log(H3O (aq)) = –log(1.0 
)]. To find out whether the solution is acidic, basic,
or neutral, we need to know how the concentrations of hydronium ion, [H3O+(aq)], and
hydroxide ion, [OH–(aq)], compare. Use Kw to find [OH–(aq)]:
–12
–7
Kw = 3 
= (H3O+(aq))(OH–(aq)) = (1.0 
)(OH–(aq))
–5
–5
(OH–(aq)) = 3 
[OH–(aq)] = 3 
M
–7
Because [H3O+(aq)] < [OH–(aq)], a solution in which [H3O+(aq)] = 1.0 
M at 60 °C is
basic. Note that a pH of 7.0 at 60 °C means the solution is basic, but a pH of 7.0 at 25 °C
indicates a neutral solution. The relationship between [H3O+(aq)] and [OH–(aq)] determines
whether a solution is acidic, neutral, or basic. The “neutral pH” at a particular temperature
depends upon the equilibrium constant at that temperature. Neutral pH at 60 °C is about 5.8.
Problem 9.18.
We are asked to determine the order of acid strength for species acting as acids in these
reactions, given that the position of equilibrium lies to the right (products are favored over
reactants) in each reaction:
(i)
N2H5+(aq) + NH3(aq)  NH4+(aq) + N2H4(aq)
(ii)
NH3(aq) + HBr(aq)  NH4+(aq) + Br–(aq)
(iii)
N2H4(aq) + HBr(aq)  N2H5+(aq) + Br–(aq)
It is convenient to use the Brønsted-Lowry approach to identify the acids, proton donors, in these
reactions. The acids are: N2H5+(aq), NH4+(aq), and HBr(aq). Recall that the position of
equilibrium in an acid-base reaction favors the weaker acid (and base). Since these equilibria all
favor products, the weaker acid in each reaction is on the product side. Note that HBr(aq) is
never on the product side in these reactions, so we can conclude that it is probably the strongest
of the three acids. Conversely, NH4+(aq) is always found on product side, so we can conclude
that it is probably the weakest of the three acids. This leaves N2H5+(aq) as the intermediate
strength acid and you see in reaction (i) that it is stronger than NH4+(aq). In reaction (iii), you see
that N2H5+(aq) is weaker than HBr(aq). Thus, the relative acid strengths (strongest to weakest)
are: HBr(aq) > N2H5+(aq) > NH4+(aq).
Problem 9.19.
(a) The pH of a 0.1 M solution of lactic acid can be calculated from its equilibrium constant
expression and equilibrium constant (let LacH = lactic acid and Lac– = lactate ion):
LacH(aq) + H2O(aq)  Lac–(aq) + H3O+(aq)
 (H O + (aq))(Lac – (aq)) 
Ka(lactic acid) = 1.4  10–4 =  3

 (LacH(aq))(H 2 O(aq)) eq
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ACS Chemistry FROG
Chapter 9
species
initial mol
change in mol
final mol
Chemical Equilibria
H3O+(aq)
1.0  10–7
x formed
x
Lac–(aq)
-0x formed
x
LacH(aq)
0.1
x reacts
0.1 – x
Because Ka(lactic acid) is relatively small, start by assuming that the value of x is small and can
be neglected relative to 0.1:
x2
 x  x   x  x
Ka = 1.4  10–4 = 

=
 0.1 – x   0.1 
0.1
x2 = 0.1·(1.4  10–4) = 1.4  10–5
x = 3.7  10–3 (uncertainty is larger than the 4% implied here)
Check the assumption that the value of x is small and can be neglected relative to 0.1:
 3.7  10 –3 
·100% = 3.7%

0.1 
Our rule of thumb is that a 5% error is acceptable and this result is within the limit.
pH = –log (H3O+(aq)) = –log (3.7  10–3) = 2.4
(b) The pH of a 0.1 M solution of benzoic acid can be calculated from its equilibrium constant
expression and equilibrium constant (let BzOH = benzoic acid and BzO– = benzoate ion):
BzOH(aq) + H2O(aq)  BzO–(aq) + H3O+(aq)
 (H 3O + (aq))(BzO – (aq)) 
Ka(benzoic acid) = 6.5  10 = 
 (BzOH(aq))(H O(aq)) 
–5
2
species
initial mol
change in mol
final mol
H3O+(aq)
1.0  10–7
x formed
x
BzO–(aq)
-0x formed
x
eq
BzOH(aq)
0.1
x reacts
0.1 – x
Because Ka(benzoic acid) is relatively small, start by assuming that the value of x is small and
can be neglected relative to 0.1:
x2
 x  x   x  x
Ka = 6.5  10–5 = 

=
 0.1 – x   0.1 
0.1
x2 = 0.1·(6.5  10–5) = 6.5  10–6
x = 2.5  10–3 (uncertainty is larger than the 4% implied here)
Check the assumption that the value of x is small and can be neglected relative to 0.1:
 2.5  10 –3 
·100% = 2.5%

0.1 
Our rule of thumb is that a 5% error is acceptable and this result is within the limit.
pH = –log (H3O+(aq)) = –log (2.5  10–3) = 2.6
ACS Chemistry FROG
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Chemical Equilibria
Chapter 9
(c) The pH of a 0.1 M solution of aniline can be calculated from its equilibrium constant
expression and equilibrium constant (let An = aniline and AnH+ = protonated aniline):
An(aq) + H2O(aq)  AnH+(aq) + OH–(aq)
 (OH – (aq))(AnH + (aq)) 
Kb(aniline) = 4.3  10–10 = 

 (An(aq))(H 2O(aq)) eq
species
initial mol
change in mol
final mol
OH–(aq)
1.0  10–7
x formed
x
An(aq)
0.1
x reacts
0.1 – x
AnH+(aq)
-0x formed
x
Because Kb(aniline) is relatively small, start by assuming that the value of x is small and can be
neglected relative to 0.1:
x2
 x  x   x  x
Kb = 4.3  10-10 = 

=
 0.1 – x   0.1 
0.1
x2 = 0.1·(4.3  10-10) = 4.3  10-11
x = 6.6  10–6 (uncertainty is larger than the 2% implied here)
Check the assumption that the value of x is small and can be neglected relative to 0.1:
 6.6 10 –6 

 ·100% = 0.007%
 0.1 
Our rule of thumb is that a 5% error is acceptable and this result is well within the limit.
pOH = –log (OH-(aq)) = –log (6.6  10–6) = 5.2
pH = 14.00 – pOH = 14.0 – 5.2 = 8.8
(d) The pH of a 0.1 M solution of hydrazine can be calculated from its equilibrium constant
expression and equilibrium constant:
N2H4(aq) + H2O(aq)  N2H5+(aq) + OH–(aq)
 (OH – (aq))(N 2 H +5 (aq)) 
Kb(hydrazine) = 8.9  10–7 = 
 (N 2 H 4 (aq))(H 2O(aq))  eq
species
initial mol
change in mol
final mol
OH–(aq)
1.0  10–7
x formed
x
N2H4(aq)
0.1
x reacts
0.1 – x
N2H5+(aq)
-0x formed
x
Because Kb(hydrazine) is relatively small, start by assuming that the value of x is small and can
be neglected relative to 0.1:
x2
 x  x   x  x
Kb = 8.9  10–7 = 

=
 0.1 – x   0.1 
0.1
x2 = 0.1·(8.9  10–7) = 8.9  10–8
x = 3.0  10–4 (uncertainty is larger than the 3% implied here)
76
ACS Chemistry FROG
Chapter 9
Chemical Equilibria
Check the assumption that the value of x is small and can be neglected relative to 0.1:
 3.0 10 –4 

 ·100% = 0.3%
 0.1 
Our rule of thumb is that a 5% error is acceptable and this result is well within the limit.
pOH = –log (OH-(aq)) = –log (3.0  10–4) = 3.5
pH = 14.00 – pOH = 14.0 – 3.5 = 10.5
Problem 9.20.
From Table 9.2, pKa = 9.24 for NH4+(aq) + H2O  NH3(aq) + H3O+(aq). We are interested in
the basic reaction of ammonia, NH3(aq) + H2O(aq)  NH4+(aq) + OH–(aq), and its pKb, which
is 4.76 (= 14.00 – pKa). Using this value, we can calculate the pH of a 0.250 M aqueous
ammonia solution.
 (OH – (aq))(NH +4 (aq)) 
Kb(NH3) = 10 – pKb = 10–4.76 = 1.7  10–5 = 
 (NH 3 (aq))(H 2O(aq))  eq
species
initial mol
change in mol
final mol
OH–(aq)
1.0  10–7
x formed
x
NH3(aq)
0.250
x reacts
0.250 – x
NH4+(aq)
-0x formed
x
Because Kb(NH3) is relatively small, start by assuming that the value of x is small and can be
neglected relative to 0.250:
x2
 xx   xx 
Kb = 1.7  10–5 = 

=
 0.250 – x   0.250 
0.250
x2 = 0.250·(1.7  10–5) = 4.3  10–6
x = 2.1  10–3
Check the assumption that the value of x is small and can be neglected relative to 0.250:
 2.1  10 –3 
 0.250  ·100% = 0.84%
Our rule of thumb is that a 5% error is acceptable and this result is within the limit.
pOH = –log (OH-(aq)) = –log (2.1  10–3) = 2.68
pH = 14.00 – pOH = 14.00 – 2.68 = 11.32
Problem 9.21.
The [H3O+(aq)] in a 0.350 M solution of aqueous hydrogen fluoride, HF, solution can be
calculated from its equilibrium constant expression and equilibrium constant:
HF(aq) + H2O(aq)  F–(aq) + H3O+(aq)
 (H O + (aq))(F – (aq)) 
Ka(HF) = 6.46  10–4 =  3

 (HF(aq))(H 2O(aq)) eq
ACS Chemistry FROG
77
Chemical Equilibria
Chapter 9
species
initial mol
change in mol
final mol
H3O+(aq)
1.0  10–7
x formed
x
F–(aq)
-0x formed
x
HF(aq)
0.350
x reacts
0.350 – x
Because Ka(HF) is relatively small, start by assuming that the value of x is small and can be
neglected relative to 0.350:
 xx   xx 
x2
Ka = 6.46  10–4 = 
 
=
 0.350 – x   0.350  0.350
x2 = 0.350·(6.46  10–4) = 2.26  10–4
x = 1.50  10–2
Check the assumption that the value of x is small and can be neglected relative to 0.1:
 1.50  10 –2 

 ·100% = 4.30%
 0.350 
Our rule of thumb is that a 5% error is acceptable and this result is just within the limit. This
percentage also represents the percentage of the HF(aq) that transfers its protons to water.
Problem 9.22.
Assume that the only source of hydronium ions in pH 3.4 lemon juice is transfer of protons from
citric acid to water, H3Cit(aq) + H2O(aq)  H3O+(aq) + H2Cit–(aq), with Ka = 7.4  10-4. From
the reaction stoichiometry, (H3O+(aq)) = (H2Cit–(aq)) = 10–pH = 10–3.4 = 4  10–4. We can use the
equilibrium constant expression and solve to find (H3Cit(aq)) in the solution:
(H 3O+ (aq))(H2Cit – (aq))
(4 10 –4 )2
K = 7.4  10 =
=
(H 3Cit(aq))
(H 3Cit(aq))
–4
(4 10 –4 )2
= 2  10–4
[H3Cit(aq)] = 2  10–4 M
–4
7.4 10
This is the amount of citric acid left undissociated. The total concentration of citric acid present
in the solution is the concentration left unreacted, [H3Cit(aq)], plus the amount present as the
anion, [H2Cit–(aq)], that is, 2  10–4 M + 4  10–4 M = 6  10–4 M. Another way to do this
problem is to let the original concentration (before proton transfer) of citric acid be c M. Then
the equilibrium concentration is c – [H3O+(aq)] M, so we have:
(4  10 –4 )2
7.4  10–4 =
(c – (4  10–4 ))
(H3Cit(aq)) =
(7.4  10 –4 )(4  10 –4 )  (4  10 –4 )2
= 6  10–4 M
(7.4  10 –4 )
We see that both methods give the same result for the total concentration of citric acid in the
solution (unreacted and monoanionic base form).
c=
Problem 9.23.
(a) The aluminum cation, Al3+, is relatively small (it has three fewer electrons than an aluminum
atom) and has a high charge. The electric field produced by the ion is quite strong; water
78
ACS Chemistry FROG
Chapter 9
Chemical Equilibria
molecules can get fairly close to the ion and interact strongly with the field. When a water
molecule is part of the hydration layer of the cation, the positive charge on the cation attracts
electron density from the oxygen atoms toward the cation and makes the bonds between the
oxygen atom and hydrogen atoms in the water much more polar than in the water molecule
alone. Thus, it is easier for the hydrogen atoms on the waters in the hydration layer to be
transferred to adjacent water molecules, than for proton transfer between water molecules in pure
water. The result is that there is more proton transfer and higher hydronium ion concentration in
the cation solutions than in pure water. The sodium ion (with only one fewer electrons than its
atom) is a larger ion with lower charge than the aluminum ion. Water molecules are not nearly so
strongly held by this larger ion with lower charge, so there is negligible effect on the hydronium
ion concentration of the solution.
(b) For the reaction, Al(H2O)63+(aq) + H2O(aq)  H3O+(aq) + Al(OH)(H2O)52+(aq), the acid
equilibrium constant expression is:
H O (aq)Al(OH)(H O) (aq)
Ka =
Al(H O) (aq)H O(aq)
+
3
2+
5
2
2
3+
6
2
To find the numeric value for the equilibrium constant, we need to find the molar concentrations
of the reactants and products in this equation. The pH of the solution, 2.9, gives us:
[H3O+(aq)] = 10-pH =10–2.9 M = 1  10–3 M
The stoichiometry of the reaction tells us that:
[H3O+(aq)] = [Al(OH)(H2O)52+(aq)] = 1  10–3 M
The solution contains 0.1 M aluminum ion, so, assuming all of this ion gets hydrated, the
stoichiometry gives:
[Al(H2O)63+(aq)] = (0.1 M) – [Al(OH)(H2O)52+(aq)] = (0.1 M – (1  10–3 M) = 0.1 M
At these low solute concentrations, the water concentration is essentially the same as in pure
water, 55.5 M, so the dimensionless concentration term for water is unity. The numeric value for
the equilibrium constant is:
Ka
–3
–3
1  10 1  10 

=
= 1  10–5 ;
0.11
pKa = 5.0
(c) The smaller the value of pKa, the larger the value of Ka. Since the pKa for the Fe(III) ion, 2.5,
is much smaller than that for Al(III), 5.0 [from part (b)] the acid equilibrium reaction for Fe(III)
goes further toward products and produces more hydronium ions. Fe(III) solutions are more
acidic than Al(III) solutions.
Problem 9.24.
(a) For the reaction, HA(aq) + H2O(aq)  H3O+(aq) + A–(aq), the acid equilibrium constant
expression is:
H O (aq)A (aq) = H O (aq)A (aq)
Ka =
+
–
+
3
–
3
HA(aq)H 2O(aq)
HA(aq)
For a 0.215 M HA solution, with a pH of 4.66, we have:
(H3O+(aq)]) = 10–pH = 10–4.66 = 2.2  10–5
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and the stoichiometry of the reaction tells us that:
(H3O+(aq)) = (A–(aq)) = 2.2  10–5
It is easy to observe that much less than 5% of the original acid, [HA(aq)] = 0.215 M, has
reacted.
Ka =
2.2  10–5 2.2  10 –5  = 2.3  10–9
0.215
(b) For the reaction, A–(aq) + H2O(aq)  OH–(aq) + HA(aq), the base equilibrium constant
expression is:
OH (aq)HA(aq)
Kb =
A (aq)
–
–
We know that Ka·Kb = Kw = 1.00  10–14, so:
Kb = K w
Ka
–14
= (1.00 10 )
(2.3 10 –9 )
= 4.4  10–6
The stoichiometry of the base reaction shows us that [OH–(aq)] = [HA(aq)], so, assuming that a
negligible amount of A–(aq) reacts, we can rearrange the equilibrium constant expression to find
the hydroxide concentration and pH of the solution:
(OH–(aq))(HA(aq)) = (OH–(aq))2 = Kb·(A–(aq)) = (4.4  10–6)(0.175) = 7.7  10–7
(OH–(aq)) = 8.8  10–4;
pOH = 3.06
pH = 14.00 – pOH = 14.00 – 3.06 = 10.94
Check the assumption that the value of (OH–(aq))x is small and can be neglected relative to
0.175:
 8.8 10 –4 

 ·100% = 0.50%
 .175 
Our rule of thumb is that a 5% error is acceptable and this result is well within the limit.
Problem 9.25.
(a) For the reaction, HO(O)CC(O)O–(aq) + H2O(aq)  –O(O)CC(O)O–(aq) + H3O+(aq), pKa2 =
4.27 (“a2” because it is the second proton transferred from oxalic acid). Since pKa + pKb = 14.00
for a conjugate acid-base pair, the pKb for –O(O)CC(O)O–(aq) is 9.73.
(b) The molar mass of disodium oxalate is 134 g·mol–1, so a solution containing 35 g of this salt
in a liter of water (assumed to yield a liter of solution) gives [–O(O)CC(O)O–(aq)] = 0.26 M. The
basic reaction in this solution is:
–
O(O)CC(O)O–(aq) + H2O(aq)  HO(O)CC(O)O–(aq) + OH–(aq)
The stoichiometry gives [OH–(aq)] = [HO(O)CC(O)O–(aq)] and we can use the equilibrium
constant value from part (a), with the assumption that a negligible amount of the oxalate ion
reacts, to get the (OH–(aq)) and the pH of the solution:
–
OH (aq )HO(O)CC(O)O (aq OH (aq )

–9.73
–10
Kb = 10
= 1.9  10 =
=
0.26
 O(O)CC(O)O (aq
–
–
–
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–
2
Chapter 9
Chemical Equilibria
(OH–(aq)) =
(1.9  10 –10 )(0.26) = 7.0  10–6;
pOH = 5.15
pH = 14.00 – pOH = 14.00 – 5.15 = 8.85
Checking the assumption, we see that 7.0  10–6 M (concentration of hydrogen oxalate ion) is
much less than 5% of the original, 0.26 M, oxalate ion, so a negligible amount of the oxalate has
reacted.
Problem 9.26.
(a) For ease in writing, let HA = HC(O)OH (methanoic acid) and A– = HC(O)O– (methanoate
anion). The proton transfer reaction, equilibrium reaction (for a dilute solution), and equilibrium
constant are:
HA(aq) + H2O(aq)  H3O+(aq) + A–(aq)
H O (aq)A (aq) = H O (aq)A (aq) = 1.8  10-4
Ka =
–
+
+
3
–
3
HA(aq)H 2O(aq)
HA(aq)
From the reaction stoichiometry we see that (H3O+(aq)) = (A–(aq)), as in several of the previous
problems. Substitute this equality, rearrange the equilibrium constant equation, and solve for
(H3O+(aq)):
(H3O+(aq))2 = Ka·( HA(aq))
(H3O+(aq)) =
Ka  (HA(aq))
Assuming that a negligible amount of HA(aq) transfers its protons to water, we can begin with
the approximation that (HA(aq)) = c = initial concentration of HA(aq) in the solution, so:
(H3O+(aq)) =
Ka  c
In the present example, we have:
(H3O+(aq)) =
Ka  c = (1.8 10 –4 )(0.100) = 4.2  10–3
pH = –log(H3O+(aq)) = –log(4.2  10–3) = 2.38
We can check the assumption (approximation) and get the percent ionization of the acid in one
operation:
 4.2 10 –3 
% ionization = 
 ·100% = 4%
 0.100 
Although this result is within the limit we have set (5%) for accepting the approximation that a
negligible amount of acid has transferred its protons, we should check to see whether this much
ionization affects the result we get for (H3O+(aq)). One way to do this is to assume that 4% of the
acid reacts and use the lower concentration in the equation to see what (H3O+(aq)) is:
(H3O+(aq)) =
(1.8 10 –4 )(0.096) = 4.1  10–3
Within the precision of the data, this result is the same as the previous one, so the 4% ionization
result is acceptable.
(b) Le Chatelier’s principle predicts that disturbing an equilibrium system by decreasing the
concentration of a reactant results in the system responding to form more of the reactant, thereby
decreasing the concentration of the product(s). Le Chatelier’s principle is, however, powerless to
ACS Chemistry FROG
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Chemical Equilibria
Chapter 9
make a quantitative prediction of the result, so the best we can say is that the [H3O+(aq)] and [A–
(aq)] will be lower when [HA(aq)] is lower.
(c) Begin by using the simplest approach and then checking to see whether the results make
sense or whether further work is necessary. For a 0.050 M solution of HA(aq), we find:
(H3O+(aq)) = (1.8 10 –4 )(0.050) = 3.0  10–3
Our prediction of a lower concentration of H3O+(aq) [and A-(aq)] is borne out. The percent
ionization is:
 3.0 10 –3 
% ionization = 
 ·100% = 6%
 0.050 
This result, 6% ionization, is larger than for the more concentrated acid, a result that Le
Chatelier’s principle could not predict and is also larger than we usually are willing to accept
without trying a less approximate calculation. We should again try what we did in part (b),
substitute this tentative result back into the equation to see how it affects the outcome:
(H3O+(aq)) = (1.8 10 –4 )(0.047) = 2.9  10–3
Again we find that our result, within the precision of the data, is not changed, so we can have
reasonable confidence that our conclusion is valid: as the concentration of a weak acid is
decreased, the percent that ionizes increases.
Problem 9.27.
In essentially all the problems we have presented, the concentration of hydronium ion produced
by the transfer of protons from an acid is a good deal larger than 10–7 M, the maximum
concentration of hydronium ions that water autoionization can produce. In addition, the presence
of the hydronium ion from the acid decreases the amount of hydronium ion produced by water
autoionization because the added hydronium ion is a disturbance to the autoionization
equilibrium and the response of the system is to react to try to use up the added product which
reduces the amount of hydroxide ion and hydronium ion formed by autoionization. If the
concentration of hydronium ion from the acid we add to the system is very low, comparable to
10–7 M, then we have to account for the autoionization, because it will be adding a significant
amount of hydronium ion to the solution. The conditions under which we have to be account for
the autoionization of water can be determined by looking at the equation developed in the
solution for Problem 9.26(a): (H3O+(aq)) = Ka  c . When the added concentration of acid is
very low and/or when the acid dissociation constant for the added acid is very small, the amount
of hydronium ion from the added acid will be tiny and could be comparable to 10–7 M.
Problem 9.28.
A solution contains a buffer system, if significant concentrations of the protonated and
unprotonated forms of a weak acid-base conjugate pair are present. In case (i), a solution of
HClO4 and NaClO4, no significant concentration of the protonated form, HClO4(aq) exists in the
solution, because HClO4(aq) is a strong acid and transfers essentially all its protons to water. In
case (ii), a solution of Na2CO3 and NaHCO3, the HCO3–(aq) anion is a weak acid (also a weak
base that is not relevant to this discussion) that is in equilibrium with its conjugate base:
HOCO2–(aq) + H2O(aq)  H3O+(aq) + CO32–(aq)
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ACS Chemistry FROG
Chapter 9
Chemical Equilibria
Since the solution contains both the conjugate acid and conjugate base, this equilibrium reaction
provides buffering action.
Problem 9.29.
The reaction that is relevant to this problem is:
HOAc(aq) + H2O(aq)  H3O+(aq) + OAc–(aq)
If NaOAc(s) is added to an aqueous solution of acetic acid, the concentration of OAc–(aq) will
increase as the solid dissolves. Adding a product is a disturbance to the equilibrium system, so
the system will react by trying to decrease the effect of the disturbance, that is, by shifting
toward reactants to use up some of the added product. The result will also be to decrease the
concentration of H3O+(aq), because it is required to react with the added acetate ion. The result
of addition of the conjugate base to a weak acid-base equilibrium system is to lower the
(H3O+(aq)), that is, raise the pH of the solution.
Problem 9.30.
The reaction that is relevant to this problem is:
NH3(aq) + H2O(aq)  NH4+(aq) + OH–(aq)
If NH4Br(s) is added to an aqueous solution of ammonia, the concentration of NH4+(aq) will
increase as the solid dissolves. Adding a product is a disturbance to the equilibrium system, so
the system will react by trying to decrease the effect of the disturbance, that is, by shifting
toward reactants to use up some of the added product. The result will also be to decrease the
concentration of OH–(aq), because it is required to react with the added ammonium ion. The
result of addition of the conjugate acid to a weak acid-base equilibrium system is to lower the
(OH–(aq)), that is, lower the pH of the solution.
Problem 9.31.
The reaction that is relevant to this problem is:
HOAc(aq) + H2O(aq)  H3O+(aq) + OAc–(aq)
Because HOAc(aq) is a weak acid, we know that only a tiny amount of it will transfer its protons
to water in a 0.250 M solution, so (H3O+(aq)) = (OAc–(aq)) will be rather small in this solution.
Using the equation in the solution for Problem 9.26 we have:
Ka  c =
(H3O+(aq)) =
(1.7 10 –5 )(0.250) = 2.1  10–3
Addition of sufficient solid NaOAc to make the final sodium ion concentration, [Na+(aq)], 1.50
M will enormously increase (OAc–(aq)) and decrease the amount of HOAc(aq) that reacts by the
above equation. We can start with the approximation that (HOAc(aq)) and (OAc–(aq)) have the
values, 0.250 and 1.50, respectively, they would have, if no proton transfers took place. Thus, we
have:
H O (aq)OAc (aq) ≈  H O(aq)1.50
=
+
Ka = 1.7  10
-5
–
3
HOAc(aq)
2

0.250

(H3O+(aq)) = 2.8  10-6;
pH = 5.55
–
As we said, addition of the OAc (aq reduces the (H3O+(aq)) (by a factor of about 1000 in this
case), which is also the direction of the change predicted in the solution for Problem 9.29. The
hydronium ion that is no longer present reacted with the added OAc–(aq), the amount of OAc–
ACS Chemistry FROG
83
Chemical Equilibria
Chapter 9
–3
 


100% , so a negligible amount has been used
(aq) reacted is about 0.14%  (2.110 )

(1.50)
 

up and our assumptions about the concentrations of the conjugate acid and conjugate base are
valid.
Problem 9.32.
If x = (OAc–(aq)) in a 0.25 M acetic acid-acetate buffer solution, then (HOAc(aq)) = (0.25 – x) in
this solution. If the pH = 5.36, then (H3O+(aq)) = 4.4  10-6. Substitute these values into the
equilibrium constant expression and solve for x to find the concentrations of acetate and acetic
acid that will produce this buffer solution:
 x 4.4 10 -6 
-5

Ka = 1.7  10 = 

0.25

x


x = (OAc–(aq)) = 0.20;
(HOAc(aq)) = (0.25 – x) = 0.05
[OAc–(aq)] = 0.20 M
[HOAc(aq)] = 0.05 M
Problem 9.33.
(a) The equilibrium reaction, equilibrium constant expression, and equilibrium constant for
lactic acid transferring a proton to water are:
HC3H5O3(aq) + H2O(aq)  H3O+(aq) + C3H5O3–(aq)
H O (aq)C H O

Ka = 1.4 
–4
=
3
3
5
–
3
(aq)
HOC3H 5O2 (aq)H 2O(aq)
H O (aq)C H O
=

3
3
5
–
3
(aq)
HOC3H 5O2 (aq)
For a buffer solution composed of 0.15 M lactic acid, HOC3H5O2, and 0.10 M sodium lactate,
NaOC3H5O2, we begin by making the approximation that these concentrations are negligibly
affected by the proton transfers, and solve the equilibrium constant expression to find
(H3O+(aq)):
H O (aq)0.10

1.4 
–4
=
3
0.15
(H3O+(aq)) = 2.1  10–4;
pH = –log(2.1  10–4) = 3.68
Because (H3O+(aq)) is small compared to the initial concentrations of the conjugate acid-base
pair, the approximation that their concentrations are not appreciably affected is valid.
An alternative (equivalent) solution is to use the Henderson-Hasselbalch equation [with pKa =
–4
-logKa = –log(1.4 
) = 3.85]:
 conjugate base
 0.10 
 = 3.85 + log 
pH = pKa + log 
 = 3.85 + (–0.18) = 3.67
 0.15 
 conjugate acid  
Within the round-off uncertainty of the calculations, the pH calculated by each route is the same.
(b) Half way to the equivalence point in the titration between 0.15 M lactic acid and 0.15 M
NaOH, the concentration of the lactic acid and its conjugate base will be equal, and the pH will
equal the pKa, 3.85, as we see from the Henderson-Hasselbalch equation when (conjugate base)
= (conjugate acid):
84
ACS Chemistry FROG
Chapter 9
Chemical Equilibria
 conjugate base
 conjugate base
 = 3.85 + log 
 = 3.85 + log(1)
pH = pKa + log 
 conjugate acid  
 conjugate base
= 3.85 + 0 = 3.85
In part (a), where there was more conjugate acid than conjugate base, the pH of the buffer
solution was lower, more acidic. The solution had a higher concentration of hydronium ion than
in the solution here.
Problem 9.34.
(a) This is a buffer solution containing a weak acid, HA = phenol, pKa = 9.98, and its conjugate
base, A– = phenolate ion. The pH of the solution is:
 (A – (aq)) 
0.377 
pH = pKa + log 
= 9.81
 = 9.98 + log 
0.551

 (HA(aq)) 
Note that this solution with a higher concentration of the conjugate acid than the conjugate base
has a lower numeric value of pH (more acidic) than the numeric pKa value.
(b) Assume that all the hydronium ion produced by the added HCl (0.100 M in the 1 L of
solution) reacts with the phenolate anion to produce phenol. The new concentration of the
phenolate anion will be 0.227 M (= 0.377 M – 0.100 M) and the concentration of phenol will be
0.651 M (= 0.551 M + 0.100 M). The new pH will be:
0.277 
pH = 9.98 + log 
= 9.61

0.651

As we would expect, the addition of hydronium ion (from the HCl) lowers the pH, but by only
0.20 pH units in this buffered solution.
(c) Assume that all the added hydroxide ion (0.125 M in 1 L of solution) from the NaOH reacts
with phenol to produce phenolate anion. The new concentration of phenolate anion will be 0.502
M (= 0.377 M + 0.125 M) and the concentration of phenol will be 0.426 M (= 0.551 M – 0.125
M). The new pH will be:
0.502 
pH = 9.98 + log 
= 10.05
0.426 

As we would expect, the addition of hydroxide ion (from the NaOH) raises the pH, but by only
0.24 pH units in this buffered solution.
Problem 9.35.
(a) If the density of bleach and water are the same, then 1 L of bleach has a mass of 1.00 kg and
5% of this, about 50 g, is NaOCl. The molar mass of NaOCl is 74.5 g·mol–1, so the molarity of
NaOCl in the solution and, hence, the concentration of the aqueous anion is:
[OCl–(aq)] =
50 g  L–1
= 0.7 M
74.5 g  mol–1
Only one significant figure is justified by the data (5% solution).
(b) The reaction of the hypochlorite ion (a weak base) with water is:
OCl–(aq) + H2O(aq)  HOCl(aq) + OH–(aq)
The equilibrium constant expression for this reaction is:
ACS Chemistry FROG
85
Chemical Equilibria
Chapter 9
OH (aq)HOCl(aq)
Kb =
OCl (aq)H O(aq)
–
–
2
The symbol, Kb, we have used is the usual designation for the equilibrium constant that refers to
the reaction of a base with water to form hydroxide ion. The dimensionless concentration ratio
for water is included in the expression, but often (usually) it is left out, since, in dilute solutions,
it is unity. The numeric value for Kb is easily obtained from the pKa = 7.75 value for
hypochlorous acid, the conjugate acid of the hypochlorite anion:
pKb = 14.00 – pKa = 6.25; Kb = 10–6.25 = 5.6  10–7
(c) Assuming that a negligible amount of the hypochlorite anion reacts and that (HOCl(aq)) =
(OH–(aq)) in the resulting solution, we can write:
(OH–(aq))2 = Kb·(OCl–(aq)) = (5.6  10–7)(0.7) = 4  10–7
(OH–(aq)) = 6  10–4
(H3O+(aq)) =
1.00 10 –14
= 2  10–11;
–4
6 10
pH = –log(2  10–11) = 10.8
(d) To lower the pH of the bleach solution to 6.5, you would have to add hydrochloric acid to
react with the hypochlorite ion to convert much of it to hypochlorous acid. This target pH is a
lower numeric value than the pKa for hypochlorous acid, so the ratio of conjugate base to
conjugate acid (assuming this is a buffer solution) is less than unity.
(e) If we assume (as in part (d)) that the solution of hypochlorous acid and hypochlorite ion is a
buffer, we can use the equation for the pH of a buffer solution (the Henderson-Hasselbalch
equation) rearranged to find the ratio of the base to acid:
 (OCl– (aq)) 
pH – pKa = log 
 = 6.5 – 7.75 = –1.25
 (HOCl(aq)) 
 (OCl– (aq)) 
–1.25
= 0.06

 = 10
 (HOCl(aq)) 
Just as we reasoned in part (d), the ratio of conjugate base to conjugate acid is quite low; almost
all the hypochlorite ion has to be converted to its acid form to get the pH so low.
Problem 9.36.
We will rearrange the Henderson-Hasselbalch equation to find the (HO)2CO(aq) concentration in
human arterial blood, given that the pH of blood is 7.41, the HOCO2–(aq) concentration is
26  10-3 M, and Ka1 for (HO)2CO(aq) is 8.0  10-7 at 37 C [pKa1 = 6.10].
 HOCO – (aq) 
2

pH – pKa1 = log 

(HO)
CO(aq)


2


 26  10 –3


7.41– 6.10 = 1.31 = log 
 (HO)2 CO(aq)
 26  10 –3


 = 101.31 = 2.0  101
 (HO)2 CO(aq)
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ACS Chemistry FROG
Chapter 9
Chemical Equilibria
((HO)2CO(aq)) = 1.3  10-3;
[(HO)2CO(aq)] = 1.3  10-3 M
`Problem 9.37.
(a) The pKa for acetic (ethanoic) acid, HOAc(aq), is 4.76. The desired pH is not too different, so
we will be making a buffer solution by adding 6.0 M hydrochloric acid to 250. mL of 0.10 M
aqueous sodium acetate solution to give a solution with a pH of 4.30. We are starting with a
solution containing 0.10 M acetate anion, OAc–(aq), and adding hydronium ions (from the 6.0 M
hydrochloric acid solution) to convert some of the acetate anion to acetic acid. We can use the
buffer equation (Henderson-Hasselbalch equation) to determine the ratio of acetate anion to
acetic acid necessary to obtain the desired pH and then calculate how much hydrochloric acid is
needed to convert the requisite number of moles of acetate anion to acetic acid to give this ratio:
 (OAc – (aq)) 
pH – pKa = log 
 = 4.30 – 4.76 = –0.46
 (HOAc(aq)) 
 (OAc – (aq)) 
–0.46
= 0.35

 = 10
(HOAc(aq))


The values in this ratio can have any convenient units, so we will choose to use moles. The
original solution contains 0.0250 mol acetate [= (0.250 L)·(0.100 M)]. If we form x mol of acetic
acid, then we will have left (0.0250 – x) mol of acetate, and we use the ratio to find x:
 (OAc – (aq)) 
(0.0250 – x) mol
;
x = [HOAc(aq)] = 0.0185

 = 0.35 =
x mol
 (HOAc(aq)) 
mol
In order to form 0.0185 mol of acetic acid, we have to add 0.0185 mol of hydronium ion
(hydrochloric acid) to the original solution. The volume of 6.0 M hydrochloric acid required is:
0.0185 mol
volume 6.0 M HCl solution =
= 0.0031 L
6.0 mol  L–1
(b) The addition of the HCl solution brings the volume of the solution to 0.253 L, so we get:
0.0185 mol
[HOAc(aq)] =
= 0.073 M
0.253 L
(0.0250 – 0.0185) mol
[OAc–(aq)] =
= 0.026 M
0.253 L
Note that we started with a solution that was 0.100 M in acetate anion and end with one that has
been diluted by about one percent (250 to 253 mL) in which the sum of the molarities of acetate
anion and acetic acid is 0.099 M.
Problem 9.38.
Recall that the best buffers are those for which the pKa of the conjugate acid used is close to the
desired buffer pH. In Table 9.2 we see that pKa1, pKa2, and pKa3 for phosphoric acid are,
respectively, 2.15, 7.20, and 12.15. These values are close to the pHs of the buffers we are asked
to make, so it looks like we can use conjugate acid-base pairs of phosphoric acid and/or its salts
to prepare these buffers. The reactions to which each of these pKa values refers are these (with
the (aq) omitted for simplicity):
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(1) pKa1 = 2.15
(HO)3PO + H2O  (HO)2PO2– + H3O+
(2) pKa2 = 7.20
(HO)2PO2– +H2O  (HO)PO32– + H3O+
(3) pKa3 = 12.15
(HO)PO32– + H2O  PO43– + H3O+
From the reagents available, we could use these conjugate acid-base combinations for the
buffers:
(1) pH 2.00 (HO)3PO(aq) and Na(HO)2PO2(s)
(2) pH 7.00 Na(HO)2PO2(s) and K2(HO)PO3(s)
(3) pH 12.00 K2(HO)PO3(s) and Na3PO4(s)
Problem 9.39.
The isoelectric point (or isoelectric pH) is the pH of the solution in which a protein has a net
charge of zero. Within the protein, the net positive and net negative electric charges are equal, so
the overall net charge is zero. Generally, the protein will be least soluble at its isoelectric point
and will not move in an electrophoretic analysis. The charges on a protein are caused by loss or
gain on protons by acidic and basic side groups, respectively. The pH of the solution in which
the protein is dissolved controls the gain and loss of protons by the side groups through its effect
on the equilibria that characterize each of the acid-base side groups.
Problem 9.40.
(a) The equilibrium reactions and equilibrium constant expressions for glycine, leaving off the
(aq) designations and assuming that (H2O) = 1, are:
+
H3NCH2C(O)OH + H20 +  H3O+ + +H3NCH2C(O)O–
(  H3NCH 2C(O)O – )(H3O+ )
K1a =
( H3 NCH2C(O)OH)
H3NCH2C(O)O– + H20 +  H3O+ + H2NCH2C(O)O–
(H NCH 2C(O)O – )(H3O+ )
K2a = 2 
( H3NCH2C(O)O– )
+
(b) At the isoelectric pH, (+H3NCH2C(O)OH) = (H2NCH2C(O)O–). Multiplying the two
equilibrium constant expressions and canceling identical and equal terms gives:
(  H3NCH 2C(O)O – )(H3O+ ) (H2 NCH2C(O)O– )(H3O+ )
K1a·K2a =
·
= (H3O+)2


–
( H3 NCH2C(O)OH)
( H3NCH2C(O)O )
(H3O+) =
K1a  K2a  = K1a  K2a  2
1
isoelectric pH = –log(H3O+) = –log K1a  K2a  2 = 1 2 –logK1a  –logK2a 
1
= 1 2 pK1a  pK2a = 1 2 2.35  9.78 = 6.07
(c) The graphic in the Web Companion, Chapter 9, Section 9.5, page 3, is simplified somewhat to
indicate that all of the dipeptide goes to the zero net charge form at the isoelectric pH. In fact,
small, but equal, amounts of both the net positive and net negative forms also are present at this
point, as we assumed in part (b) for glycine. The dipeptide is like glycine in that it has an amino
and a carboxylic acid functional group. The mathematics to find the isoelectric pH is identical for
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the dipeptide and glycine and the result is the same: the isoelectric pH is “half the sum of the pKa
values of the two acid-base groups.”
Problem 9.41.
The large open circles in this sketch represent hemoglobin molecules with the positions of the
two amino acids that are changed from glutamate to valine in HbS denoted by the dark dots. The
red lozenge shapes represent water molecules.
1)
2)
3)
In normal hemoglobin, HbA, water molecules interact with the carboxylate anion of the
glutamate side groups by hydrogen bonding, a relatively strong interaction that helps keep the
protein in solution. When the glutamate is replaced by valine, which has a nonpolar hydrocarbon
side group water molecules are oriented around the nonpolar side group as they are around any
nonpolar hydrocarbon, but the intermolecular forces holding the water and side group together
are relatively weak dispersion interactions. If the nonpolar side groups on two HbS molecules
come together and “squeeze out” the water molecules between them, the entropy of the system is
increased by the release of the water molecules that are now free, as shown in line 2) of the
sketch, to move about anywhere in the solution. This is the same kind of interaction as is
represented in Figures 8.20 and 8.23 and in the Web Companion, Chapter 8, Section 8.13, to
show the formation of micelles and bilayer membranes. This process can continue, as shown in
line 3) of the sketch, to form long strands of HbS molecules that are responsible for the crescent
shape of the red blood cells in sickle-cell disease. Thus, the clumping of HbS molecules is
largely driven by an increase in entropy of the water.
Problem 9.42.
(a) Since protein B moves further toward the positive end of the gel, we can conclude that this
protein has the higher net negative charge. Table 9.2 shows that the pKa for the histidine side
group is 6.00, so, at pH 8.6 (the gel pH), it will be in its basic, uncharged form. The cysteine side
group has a pKa = 8.36, so most of it will also be in its basic, negatively-charged form. The
protein containing the cysteine will have a more negative charge at pH 8.6, so it is protein B that
contains the cysteine and A that contains the histidine.
(b) You can tell which protein is which from the electrophoresis data at pH 8.6. To help confirm
your identification and your assumption about the amino acids, you could run an electrophoresis
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analysis at a more acidic pH such as 5.6, where the histidine side group would be positively
charged and the cysteine side group would be uncharged. If the assumption and identification are
correct, protein A would be more attracted to the negative end of the gel than would protein B, so
protein A would be further toward the negative end of the gel. Note that this is the same as
saying protein B would be further toward the positive end. The relative positions of the proteins
are not changed by the change in pH, but the result is consistent with our assumption and
identification, so this is further evidence that they are correct.
Problem 9.43.
In sickle-cell disease, the substitution of an amino acid with a non-polar side for one with a polar
side group changes the solubility of the hemoglobin, by making it easier for the protein
molecules to stick to one another and precipitate, causing the sickling of the red blood cell.
Similarly, in Investigate This 9.42, as the net charge on the casein molecules changes with pH,
the protein molecules are better able to stick together and precipitate, as the hemoglobin
molecules do in sickle-cell disease.
Problem 9.44.
(a) A mixture of Fe(III) and Co(II) ions in an 8 M solution of hydrochloric acid was analyzed by
paper electrophoresis and, after several minutes of electrophoresis, a yellow spot on the paper
was closer to the positive end of the paper than a blue spot. Since the yellow spot, Fe(III)
chloride complex, is closer to the positive end of the paper than the blue spot, Co(II) chloride
complex, we can conclude that the net charge on the Fe(III) complex is more negative than the
net charge on the Co(II) complex. This means that the Fe(III) complex has to contain at least two
more negative chloride ions than the Co(II) complex. One of these extra chlorides compensates
for the difference in charge between the two cations (FeCl2+ and Co2+ have the same net charge)
and the second makes the Fe(III) complex more negative. If our explanation for the colors in
Investigate This 9.8 is correct, the Co(II) complex is CoCl42-(aq), so the Fe(III) complex must be
FeCl63–(aq).
(b) One of the most common complex-ion structures is the octahedral arrangement of the ligands
about the central cation. Thus, the structure of FeCl63–(aq) is likely to be like the structure shown
in Figure 6.5 with the water molecules replaced by chloride ions. For the CoCl42–(aq) complex,
two structures are possible, either a square planar arrangement like that shown for the Pt(II)
complexes in Figure 6.8 or a tetrahedral arrangement like that shown around Zn(II) in Figure 6.9.
Other data indicate that the Co(II) complex is tetrahedral, which we might have predicted from
its position near Zn in the periodic table.
Problem 9.45.
(a) The equilibrium reaction, equilibrium constant expression, and equilibrium constant for the
dissolution of barium sulfate are:
BaSO4(s)  Ba2+(aq) + SO42-(aq)



 Ba 2+ (aq) SO24 – (aq) 
2+
2-10
Keq = 
 = [(Ba (aq))(SO4 (aq))]eq = Ksp = 1.1  10
BaSO
(s)



 eq
4
From the stoichiometry of the reaction, we know that (Ba2+(aq)) = (SO42-(aq)) = s, where s is
numeric value of the molar solubility of the solid ionic compound. Substitution into the solubility
product equation gives:
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Ksp = 1.1  10–10 = s2
s = 1.0  10–5 = (Ba2+(aq)) = (SO42-(aq))
For the rest of the problem, we will use the solubility products and the stoichiometries to find the
concentrations of the ions in solution.
(b)
Ag2S(s)  2Ag+(aq) + S2-(aq)
(Ag+(aq)) = 2s;
(S2-(aq)) = s
Ksp = 8  10–51 = [(Ag+(aq))2(S2-(aq))]eq = (2s)2(s) = 4s3
s = 1.3  10–17 = (S2-(aq));
(Ag+(aq)) = 2.6  10–17
The uncertainty in the data (Ksp) justify only one significant figure in the results. Two are shown
to make clear that the concentration of silver ion is twice that of sulfide ion in the solution.
(c)
Mg(OH)2(s)  Mg2+(aq) + 2OH-(aq)
(Mg2+(aq)) = s;
(OH-(aq)) = 2s
Ksp = 7.1  10–12 = [(Mg2+(aq))(OH-(aq))2]eq = (s)(2s)2 = 4s3
s = 1.2  10–4 = (Mg2+(aq));
(OH-(aq)) = 2.4  10–4
(d)
PbBr2(s)  Pb2+(aq) + 2Br-(aq)
(Pb2+(aq)) = s;
(Br-(aq)) = 2s
Ksp = 2.1  10–6 = [(Pb2+(aq))(Br-(aq))2]eq = (s)(2s)2 = 4s3
s = 8.1  10–3 M = (Pb2+(aq));
(Br–(aq)) = 1.6  10–2 M
Problem 9.46.
To find the moles and mass of solid calcium carbonate, CaCO3(s), that will be obtained if 100.
mL of a solution saturated with calcium carbonate is evaporated to dryness, calculate the molar
solubility of the solid, s, in water, convert to moles in 100. mL, and then to mass of CaCO3(s).
CaCO3(s)  Ca2+(aq) + CO32-(aq)
(Ca2+(aq)) = (CO32-(aq)) = s
Ksp = 6.0  10–9 = [(Ca2+(aq))(CO32-(aq))]eq = (s)(s) = s2
s = 7.7  10–5
The molar solubility is 7.7  10–5 M, so the number of moles of CaCO3(s) dissolved in 100. mL
is:
mol in 100. mL = (7.7  10–5 mol·L-1)(0.100 L) = 7.7  10–6 mol
The molar mass of CaCO3 = 100 g·mol-1 , so the mass of CaCO3(s) dissolved in 100. mL is:
mass CaCO3(s) = (100 g·mol-1)(7.7  10–6 mol) = 7.7  10–4 g = 0.77 mg
Problem 9.47.
To calculate the mass of calcium fluoride, CaF2(s), that will dissolve in 100. mL of water
(assuming no volume change), calculate the molar solubility, s, and convert to mass (pKsp of
CaF2 = 10.57):
CaF2(s)  Ca2+(aq) + 2F–(aq)
(Ca2+(aq)); = s
(F–(aq)) = 2s
Ksp = 10–10.57 = 2.7  10-11 = [(Ca2+(aq))(F–(aq))2]eq = (s)(2s)2 = 4s3
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s = 1.9  10–4
The molar solubility is 1.9  10–4 M, so the number of moles of CaF2(s) dissolved in 100. mL is:
mol in 100. mL = (1.9  10–4 mol·L-1)(0.100 L) = 1.9  10–5 mol
The molar mass of CaF2 = 78 g·mol-1 , so the mass of CaF2(s) dissolved in 100. mL is:
mass CaF2(s) = (78 g·mol-1)(1.9  10–5 mol) = 1.5  10–3 g = 1.5 mg
Problem 9.48.
Assuming that the dissolution of 5.8  10–5 g of PbCrO4(s) in exactly 1 L of water does not
change the volume of liquid, we can calculate the molar solubility for PbCrO4(s). Assuming
complete dissociation of the ionic compound into ions, so that [Pb2+(aq)] = [CrO42–(s)] = molar
solubility of the solid, we can calculate the Ksp for the dissolution. The molar mass of PbCrO4 is
323.2 g·mol–1.
5.8  10 –5 g  L–1
molar solubility =
= 1.8  10–7 mol·L–1
–1
323.2 g  mol
2+
Ksp = (Pb (aq))(CrO42–(s)) = (1.8  10–7)2 = 3.2  10–14
Problem 9.49.
To do this problem, assume that the dissolution of Fe(OH)2(s) is a simple solubility represented
by this reaction equation:
Fe(OH)2(s)  Fe2+(aq) + 2OH–(aq)
Solutions containing transition metal ions are usually more complicated than this with complexes
(other than the aquo complex) often accounting for much of the metal ion species in the solution.
In the absence of any information about complexes in this system, we will ignore them and
assume that Fe(OH)2(s), Fe2+(aq) and OH–(aq) are the only species we need to account for in
these solutions.
(a) For dissolution in water, we use the stoichiometry of the reaction and Ksp = 7.0  10–16, to
find the solubility, (Fe2+(aq)), (OH–(aq)), and pH.
(Fe2+(aq)); = s
(OH–(aq)) = 2s
Ksp = 7.0  10–16 = [(Fe2+(aq))(OH–(aq))2]eq = (s)(2s)2 = 4s3
s = 5.6  10-6 = (Fe2+(aq));
(OH-(aq)) = 1.12  10-5 ;
pOH = 4.95
pH = 14.00 – pOH = 14.00 – 4.95 = 9.05
(b) For dissolution in 0.25 M NaOH solution, we use Ksp and the solubility product, with the
(OH–(aq)) provided by the NaOH, to find the solubility, (Fe2+(aq)), and pH.
(Fe2+(aq)); = s
(OH–(aq)) = 0.25
Ksp = 7.0  10–16 = [(Fe2+(aq))(OH–(aq))2]eq = (s)(0.25)2 = 0.0625s
s = 1.1  10-14 = (Fe2+(aq));
pOH = 0.60
pH = 14.00 – pOH = 14.00 – 0.60 = 13.40
As we would anticipate, the solubility of Fe(OH)2(s) in a solution that already contains
hydroxide ion is much lower than in pure water. Hydroxide from this dissolution is about 13
orders of magnitude less than that from the NaOH.
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(c) For dissolution in 0.25 M Fe(NO3)2 solution, we use Ksp and the solubility product, with the
(Fe2+(aq)) provided by the Fe(NO3)2, to find the solubility, (OH–(aq)), and pH.
(Fe2+(aq)); = 0.25
(OH–(aq)) = 2s
Ksp = 7.0  10–16 = [(Fe2+(aq))(OH–(aq))2]eq = 0.25(2s)2 = 1.00s2
s = 2.6  10-8;
(OH-(aq)) = 2s = 5.2  10-8 ; pOH = 7.28
pH = 14.00 – pOH = 14.00 – 7.28 = 6.72
Note that there is something peculiar about these results. The pH of the solution is less than the
pH of pure water, so there is less hydroxide ion in the solution than in pure water. The analysis
suggests that the Fe2+(aq) in a solution that is 0.25 M in Fe2+(aq) should react with hydroxide ion
from the water self ionization to precipitate a little Fe(OH)2(s). This result shows that this system
is more complex than our analysis indicates.
Problem 9.50.
A precipitate will form when 0.0025 mol of Ba2+(aq) (from Ba(NO3)2) and 0.00030 mol of CO32–
(aq) (from Na2CO3) are mixed in 250. mL of distilled water, if the product of the ion
concentrations in this solution exceeds the solubility product constant for dissolution of
BaCO3(s). Calculate (Ba2+(aq)) and (CO32–(aq)), find the ion product, and compare it with the
solubility product constant, Ksp = 6.0  10-9, for BaCO3(s) dissolution.
0.0025 mol Ba 2+ (aq)
= 1.0  10–2 M
0.250 L
0.0030 mol CO2–
3 (aq)
[CO32– (aq)] =
= 1.2  10–2 M
0.250 L
2+
ion product = (Ba (aq))( CO32– (aq)) = (1.0  10–2)( 1.2  10–2) = 1.2  10–4
The ion product far exceeds Ksp (by about 5 orders of magnitude) so a precipitate will form in
this case.
[Ba2+(aq)] =
Problem 9.51.
The minimum concentration of sulfate anion, [SO42–(aq)], required to begin the precipitation of
either 2.5  10-3 M Pb2+(aq) or 0.05 M Ca2+(aq) is the concentration that will make the ion
product just equal the solubility product constant for PbSO4(s) or CaSO4(s). Calculate the (SO42–
(aq)) required for each cation concentration and the one that requires the lower concentration
will be the one to precipitate first.
Ksp(PbSO4) = 6.3  10-7 = [(Pb2+(aq))(SO42-(aq))]eq = (2.5  10-3)( SO42-(aq))
( SO42-(aq)) = 2.5  10-4
Ksp(CaSO4) = 2.4  10-5 = [(Ca2+(aq))(SO42-(aq))]eq = (0.05)( SO42-(aq))
( SO42-(aq)) = 4.8  10-4
Therefore, PbSO4(s) will form once [SO42-(aq)] reaches about 2.5  10-4 M.
Problem 9.52.
(a) To determine how much Ba2+(aq) is in 200 mL of a saturated barium sulfate solution, we use
the solubility product, Ksp = 1.1  10–10, to find the molar solubility (or its equivalent) and then
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convert to mass of Ba2+(aq) [from the stoichiometry of the solubility equilibrium, (Ba2+(aq)) =
(SO42–(aq))]:
Ksp = 1.1  10–10 = (Ba2+(aq))(SO42–(aq)) = (Ba2+(aq))2
(Ba2+(aq)) = 1.0  10–5;
[Ba2+(aq)] = 1.0  10–5 M
mol Ba2+(aq) = (1.0  10–5 M)(0.200 L) = 2.1  10–6 mol
mass Ba2+(aq) = (2.1  10–6 mol)(137.3 g·mol–1) = 2.9  10–4 g = 0.29 mg (in 200 mL)
(b) The liquid the patient drinks for x-ray analysis contains a large amount of solid barium
sulfate suspended in water, but the amount of barium cation in the solution is, as part (a) shows,
quite small, so it is below the toxic limit for humans.
Problem 9.53.
Possible ways to decrease the [Ba2+(aq)], in order to minimize allergic discomfort when
ingesting the BaSO4 solution for x-ray analysis, are suggested and require a choice.
(i) Heating the saturated solution is not a good approach for decreasing the [Ba2+(aq)] in the
ingested solution, because the dissolution, BaSO4(s)  Ba2+(aq) + SO42-(aq), is endothermic.
Since energy is required to cause the dissolution, adding energy will increase the solubility (Le
Chatelier’s principle) and increase [Ba2+(aq)]. This is just the opposite of the effect we wish.
(ii) Adding sodium sulfate, , to the solution is a reasonable approach for decreasing the
[Ba2+(aq)] in the ingested solution. Because SO42-(aq) is a product of the dissolution, increasing
its concentration is a disturbance to the system, which will react to try to use up some of the
added SO42-(aq) and this will require reaction with Ba2+(aq) , thus reducing its concentration, as
we desire. The same result can be obtained by analyzing the system in terms of the solubility
product when the concentration of sulfate ion is increased.
(iii) Adding additional solid BaSO4, will have no effect on the [Ba2+(aq)] in the ingested
solution, because the solution is already saturated and there is a great deal of the solid ionic
compound in the mixture.
Problem 9.54.
Assume that the metal salt, M(OH)2 dissociates completely to it cations and hydroxide ions when
it dissolves in water. From the stoichiometry of the dissolution, two moles of hydroxide ion for
each mole of metal cation dissolved, we can calculate the concentration of the metal ion in terms
of the concentration of the hydroxide ion:
1 mol M 2+ 
2+
–
–
[M (aq)] = 
–  [OH (aq)] = 0.5[OH (aq)]
2 mol OH 
The pH of the solution is determined by the hydroxide from the dissolved salt. We can use the
pH to calculate the concentration of hydroxide in the solution and then the Ksp:
pOH = 14.00 – pH = 14.00 – 8.11 = 5.89;
(OH–(aq)) = 10–5.89 = 1.3  10–6
Ksp = (M2+(aq))(OH–(aq))2 = [0.5(OH–(aq))](OH–(aq))2 = 0.5(OH–(aq))3
= 0.5(1.3  10–6)3 = 1.1  10–18
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Problem 9.55.
This problem is like all others for which you have a g·L–1 solubility (0.000562 g Ag3AsO4(s)
dissolves in 1 liter of water) and wish to find the molar solubility and/or Ksp (pKsp). The only
difference in this case is the slightly greater complexity of the stoichiometry:
Ag3AsO4(s)  3Ag+(aq) + AsO43–(aq)
In the solution, therefore:
 3 mol Ag + 
3–
3–
[Ag+(aq)] = 
[AsO4 (aq)] = 3[AsO4 (aq)]
1 mol AsO 3–

4
The molar mass of Ag3AsO4 is 462.5 g·mol–1. Thus, the molar solubility of the salt and the molar
concentration of the arsenate ion are:
0.000562 g  L–1
molar solubility =
[AsO43–(aq)] = 1.22  10–6 M
–1
462.5 g  mol
Therefore,
Ksp = (Ag+(aq))3(AsO43–(aq)) = [3(AsO43–(aq))]3(AsO43–(aq)) = 27(AsO43–(aq))4
Ksp = 27(1.22  10–6)4 = 5.98  10–23;
pKsp = –log(5.98  10–23) = 22.223
Problem 9.56.
The solubility of silver chromate is somewhat greater than the solubility of silver chloride, so
essentially all the chloride in a solution will be precipitated by added silver ion before any red
Ag2CrO4(s) is formed to indicate that the titration of chloride ion is complete. Let us calculate
the ratio of (CrO42–(aq)) to (Cl–(aq)) at the point where the Ag2CrO4(s) just begins to form. At
this point, both these equilibria need to be satisfied:
AgCl(s)  Ag+(aq) + Cl–(aq)
Ksp(AgCl) = 1.8  10–10
Ag2CrO4(s)  2Ag+(aq) + CrO42–(aq)
Ksp(Ag2CrO4) = 1.2  10–12
Since all species are together in the same solution, we can solve both equilibrium constant
equations for (Ag+(aq)) and set them equal to find the desired ratio of anions in the solution:
(Ag+(aq)) =
Ksp (AgCl)
(Cl– (aq))
(CrO2–
4 (aq))
=
–
(Cl (aq))
=
K sp (Ag 2CrO 4 )
(CrO2–
4 (aq))
K sp (Ag 2 CrO 4 )
K sp (AgCl)
=
1.2  10 –12
= 6.1  104
–10
1.8  10
Assume that the few drops of the potassium chromate solution that is added to the chloride
solution to be titrated gives (CrO42–(aq))  0.01. The red color of Ag2CrO4(s) will then start to
appear when:
(CrO2–
0.01
4 (aq))
=
= 1.6  10–6
4
4
6.1  10
6.1  10
This is quite a low value for the concentration of Cl–(aq) remaining in the solution, so we can be
pretty confident that essentially all the original Cl–(aq) has precipitated as AgCl(s). Thus, the
titration is complete, that is, the equivalence point between the added silver ion and chloride ion
in solution has been reached.
(Cl–(aq)) =
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[Note that the assumption about the concentration of CrO42–(aq) in the solution is not critical.
Assuming a higher—but reasonable—concentration, say 0.1, still gives a low chloride ion
concentration when the red precipitate begins to form. Lower concentrations of CrO42–(aq) give
lower chloride concentrations. From a practical point of view, the concentration of CrO42–(aq)
has to be low enough to give the color change when essentially all the chloride has been titrated,
but high enough to form a visibly detectable reddish-pink color with only a tiny excess of the
silver ion titrant.]
Problem 9.57.
Solve the Ni(OH)2(s) (pKsp = 17.2) and Fe(OH)2(s) (pKsp = 13.8) solubility product expressions
for the hydroxide anion concentration that would be in equilibrium with the cations at their
specified concentrations: [Ni2+(aq)] = 0.001 M and [Fe2+(aq)] = 0.06 M. The cation that requires
the lower concentration of hydroxide anion to just be in equilibrium with the solid salt will be the
first to precipitate. We convert this lower hydroxide ion concentration to pH to answer the
question asked.
Ksp(Ni(OH)2) = 10–17.2 = 6.3  10–18 = (Ni2+(aq))(OH–(aq))2
(OH–(aq)) =
6.3  10 18
=
Ni2+ (aq)


6.3  10 18
= 7.9  10–8;
0.001
pOH = 7.10
Ksp(Fe(OH)2) = 10–13.8 = 1.6  10–14 = (Fe2+(aq))(OH–(aq))2
(OH–(aq)) =
1.6  10 14
=
Fe 2+ (aq)


1.6  10 14
= 5.1  10–7;
0.06
pOH = 6.29
Thus, the nickel cation will begin to precipitate first at a pOH = 7.10, a pH of 6.90. Note that
both of these hydroxide salts will begin to precipitate at about pH 7. In order to keep Ni(II) and
Fe(II) cations in solution, the solutions have to be somewhat acidic.
Problem 9.58.
The focus of this problem is this reaction (at equilibrium in a constant volume reaction vessel):
H2(g) + I2(g)  2HI(g)
(a) If the pressure of I2(g) is increased (by adding more I2 to the vessel), the equilibrium system
will act to decrease the disturbance by reacting to use up some of the added I2(g), which will also
use up some of the H2(g) and form more HI(g). Thus, the concentration of HI(g) would increase
while the concentration of H2(g) would decrease.
(b) If the pressure of HI(g) is reduced (by removing some HI from the vessel), the equilibrium
system will act to decrease the disturbance by reacting to produce more HI(g) to compensate for
what has been removed. In the system, both the forward and reverse reactions are always going
on. At equilibrium, the forward and reverse rates are the same, so there is no net change in the
concentrations. When the equilibrium is disturbed, however, the rates change and are no longer
the same. In order to produce more HI(g), the rate of the forward reaction has to be greater than
the rate of the reverse reaction. As the concentrations approach their new equilibrium values, the
rates of the forward and reverse reactions again approach one another, and, at equilibrium are
again the same. At equilibrium, the H2(g) and HI(g) concentrations will be lower than they were
before the disturbance. The equilibrium will have shifted toward the products. The concentration
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of HI(g) will have increased from what it was just after some was removed, but not back to its
original value before the disturbance.
Problem 9.59.
To determine which of these two reactions has the larger equilibrium constant, we need to
compare the G values for the two reactions:
(i) H2(g) + 1/2O2(g)  H2O(l)
(ii)
H2(g) + 1/2O2(g)  H2O(g)
These reactions are the reactions forming (i) liquid and (ii) gaseous water from the elements. The
standard free energies for these reactions are (i) G = Gf for liquid water and (ii) G = Gf
for gaseous water. The relationship between G for a reaction and the equilibrium constant is:
G = –RTln(Keq)
R is the gas law constant in units of J·K-1mol-1 and T is the Kelvin temperature. The
interpretation of this relationship is that the more negative the value of G, the larger are
ln(Keq) and Keq. Since Gf for liquid water is more negative than Gf for gaseous water, the
equilibrium constant for the formation of liquid water from its elements in their standard states,
reaction (i), will have a larger equilibrium constant.
Problem 9.60.
Use the data in Appendix B to determine G°rxn at 298 K for the formation of urea from carbon
dioxide and ammonia, CO2(g) + 2NH3(g)  (NH2)2CO(s) + H2O(l:
Grxn = (1 mol)Gf(urea(s)) + (1 mol)Gf(H2O(l))
– (1 mol)Gf(CO2(g)) – (2 mol)Gf(NH3(g))
Grxn = (1)(–197.15 kJ) + (1)(–237.13 kJ) – (1)(–394.36 kJ) – (2)(–16.45 kJ)
Grxn = –7.02 kJ
Under standard conditions at 298 K, Grxn = Grxn. Thus, Grxn = –7.02 kJ < 0 and the reaction
is spontaneous. Grxn < 0 is the criterion for spontaneity.
Problem 9.61.
We can use the equilibrium constant for H2(g) + I2(g)  2HI(g) at 500 K, from Problem 9.13,
Keq = 1.60  102, to calculate Grxn for the reaction forming hydrogen iodide gas from its
gaseous elements at 500 K:
Grxn = –RTlnKeq = –(8.314 J·K–1·mol–1)(500 K)ln(1.60  102)
Grxn = –21.1 kJ·mol–1
Note that the units for the standard free energy change are kJ·mol–1, and you should ask: “Per
mole of what?” This means “per mole of reaction,” that is, per Avogadro’s number of the
reaction events represented by the reaction equation.
Problem 9.62.
(a) For the oxidation of ammonia, 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(l), we have, from the
standard enthalpies of formation:
Hrxn = (4 mol)Hf(NO(g)) + (6 mol)Hf(H2O(l))
– (4 mol)Hf(NH3(g)) – (5 mol)Hf(O2(g))
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Hrxn = (4)(90.25 kJ) + (6)(–285.83 kJ) – (4)(–46.11 kJ) – (5)(0 kJ)
Hrxn = –1169.54 kJ
To calculate Hrxn from bond enthalpies we have to break 12 N–H bonds and 5 O=O bonds and
we get back the enthalpy of formation of 4 of the bonds in NO and 12 H–O bonds. Table 7.3 has
all these bond enthalpies except for NO, which is given as 630 kJ·mol–1 in the problem.
[NOTE: The bond enthalpy for NO is the enthalpy of reaction for NO(g)  N(g) + O(g), which
from Appendix B is: H = (472.70 kJ·mol–1) + (249.4 kJ·mol–1) – (90.25 kJ·mol–1) =
632 kJ·mol–1. This is the value we will use in this problem.]
Therefore, the enthalpy required to break the requisite bonds is:
H(breaking) = (12 mol)(BHN–H) + (5 mol)(BHO=O)
H(breaking) = (12)(393 kJ) + (5)(498.7 kJ)
H(breaking) = 7210 kJ
H(making) = (4 mol)(BHNO) + (12 mol)(BHH-O)
H(making) = (4)(632 kJ) + (12)(460 kJ)
H(making) = 8048 kJ
For the overall reaction:
Hrxn = H(breaking) – H(making) = –838 kJ
This result from bond enthalpies seems to predict that quite a bit less energy is released in the
ammonia oxidation reaction than we calculated from enthalpies of formation. However, recall
that bond enthalpies refer to molecules in the gas phase and the product water in this reaction is a
liquid. You can see in Appendix B that 44 kJ·mol–1 is released when water condenses from gas
to liquid and this energy is included in the calculation based on enthalpies of formation. If
(6 mol)(–44 kJ·mol–1) = –264 kJ is added to the result from the bond enthalpy calculation, we get
–1102 kJ for the overall enthalpy of the reaction and this value is within 10% of the value from
enthalpies of formation. The two calculations give comparable results when the states of the
reactants and products are accounted for.
(b) For the standard entropy change of the reaction, we get:
Srxn = (4 mol)S(NO(g)) + (6 mol)S(H2O(l))
– (4 mol)S(NH3(g)) – (5 mol)S(O2(g))
Srxn = (4)(210.76 J·K–1) + (6)(69.91 J·K–1) – (4)(192.45 J·K–1) – (5)(205.14 J·K–1)
Srxn = –533.64 J·K–1
This negative value of the standard entropy change for the reaction suggests that the reaction is
unfavorable as far as the entropy of the system is concerned. We see from the reaction equation
that 9 moles of gaseous reactants react to give 4 moles of a gaseous product and 6 moles of a
liquid product. We have seen that gases have higher entropies than liquids (larger volume per
molecule), so when the moles of gas decrease in the reaction, we expect a decrease in the
entropy, as we have calculated.
(c) The standard free energy change for the reaction, based on the results in parts (a) and (b), is:
Grxn = Hrxn – TSrxn = –1169.54 kJ – (298 K)(–533.64 J·K–1)
Grxn = –1169.54 kJ + 159.02 kJ = – 1010.52 kJ
Using the data in Appendix B, we get:
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Grxn = (4 mol)Gf(NO(g)) + (6 mol)Gf(H2O(l))
– (4 mol)Gf(NH3(g)) – (5 mol)Gf(O2(g))
Grxn = (4)(86.55 kJ) + (6)(–237.13 kJ) – (4)(–16.45 kJ) – (5)(0 kJ)
Grxn = –1010.78 kJ
Within the round-off errors of the data and calculations, the two results are identical.
(d) We can use the relationship between the equilibrium constant and the standard free energy of
reaction to calculate the equilibrium constant. In order to do this, we need to specify how many
“moles of reaction” we are considering for the standard free energy change. It makes sense to
think about one mole of ammonia being oxidized, so the thermodynamic quantities in parts (a) –
(c) have to be divided by four to get the values for one mole of ammonia in the balanced reaction
equation. With this understanding, the equilibrium constant for the ammonia oxidation reaction
is:
o
Grxn
–252.70  10 3 J  (mol NH 3 )–1
lnKeq = –
=–
RT
8.314 J  K –1  mol–1 298 K 


lnKeq = 102
Keq = 1.97  10 =
44
NO(g)H2O(l)3/2
NH 3 (g)O2 (g)5 / 4
The equilibrium constant is very large; products are enormously favored over reactants (as the
standard free energy change also suggests).
Problem 9.63.
(a) Hreaction for the ethyne-forming reaction, CaC2(s) + 2H2O(l)  Ca(OH)2(s) + C2H2(g), must
be negative (reaction is exothermic), since the problem statement says “the solution gets quite
hot.”
(b) Sreaction for the ethyne-forming reaction is almost certainly positive, since the reactants are
both condensed phases (liquid and solid) and one of the products is a gas (ethyne). The positional
entropy of the system increases.
(c) Greaction = Hreaction –TSreaction for the ethyne-forming reaction must be negative, since
Hreaction is negative and the positive Sreaction means that –TSreaction will be negative as well. The
reaction is spontaneous under standard conditions. Experimentally we note that the reaction is
spontaneous and the conditions are close to standard. The two solids and the liquid are pretty
much in their standard states and, if the gas is close to atmospheric pressure, it will also be
approximately in its standard state.
(d) For this reaction, the data from Appendix B give:
Hrxn = (1 mol)Hf(Ca(OH)2(s)) + (1 mol)Hf(C2H2(g))
– (1 mol)Hf(CaC2(s)) – (2 mol)Hf(H2O(l))
Hrxn = (1)(–986.09 kJ) + (1)(226.73 kJ) – (1)(–59.8 kJ) – (2)(–285.83 kJ)
Hrxn = –127.9 kJ
Srxn = (1 mol)S(Ca(OH)2(s)) + (1 mol)S(C2H2(g))
– (1 mol)S(CaC2(s)) – (2 mol)S(H2O(l))
Srxn = (1)(83.39 J·K–1) + (1)(200.94 J·K–1) – (1)(69.96 J·K–1) – (2)(69.91 J·K–1)
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Srxn = 74.55 J·K–1
Grxn = (1 mol)Gf(Ca(OH)2(s)) + (1 mol)Gf(C2H2(g))
– (1 mol)Gf(CaC2(s)) – (2 mol)Gf(H2O(l))
Grxn = (1l)(–898.49 kJ) + (1)(209.20 kJ) – (1)(–64.9 kJ) – (2)(–237.13 kJ)
Grxn = –150.1 kJ
Our reasoning in parts (a) – (c) is reinforced by these results: the signs we predicted were correct.
(e) We see that the standard enthalpy change accounts for more than 80% of the standard free
energy change, so it is the exothermicity of the reaction, which increases the thermal entropy of
the surroundings, that is the major factor contributing to the large negative free energy change
for this reaction.
Problem 9.64.
This is a subtle problem. The solution is not obvious and needs careful consideration of the
meaning of G. Starting from pure reactants or pure products, G always decreases going from
either pure reactants or pure products toward equilibrium. That is, reactions are spontaneous
proceeding toward equilibrium from either side. Since G decreases beginning from either
reactants or products, at some intermediate composition, G must reach a minimum. This is the
equilibrium composition for the given system. Since the reactants and products are in
equilibrium at this composition, very tiny changes one way or the other (toward the reactant side
or toward the product side) will result in no change in G. This is the criterion for equilibrium:
the change in G is zero for the change from reactants to products at this composition. In the
language of the calculus we say that the derivative of G with respect to the composition of the
reaction mixture (usually called the extent of reaction) is zero, which indicates that G has
reached a minimum. The dependence of G on reaction mixture composition looks like this
(where the position of equilibrium, the minimum in G, varies from one reaction to another):
Problem 9.65.
(a) The reaction equation and equilibrium constant expression for the calcite decomposition
reaction are:
CaCO3(s)  CaO(s) + CO2(g)
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Keq = 1.6  10–23 =
Chemical Equilibria
CaO(s) CO2 (g) 
CaCO3 (s)
= (CO2(g)) =
P(CO 2 (g))
1 bar
The dimensionless concentration ratios for the pure solids are unity, so the expression reduces to
the pressure of carbon dioxide, 1.6  10–23 bar, in equilibrium with these solids. The amount of
CO2(g) in the atmosphere varies somewhat but is in the neighborhood of about 10–2 bar.
Obviously, the CO2(g) contribution from chalk decomposition is negligible [and the
decomposition is inhibited by the presence of the high atmospheric content of CO2(g)].
(b) From Keq we can get Greaction for the calcite decomposition. Combine this value with the
Gf values for calcite and carbon dioxide and solve to get the Gf(CaO(s)).
Greaction = –RTlnKeq = –(8.314 J·K–1·mol–1)( 298 K) ln(1.6  10–23)
Greaction = 1.3  102 kJ·mol–1
The reaction, as written, involves one mole of the reactant, so we have:
Greaction = (1 mol)Gf(CaO(s)) + (1 mol)Gf(CO2(g)) – (1 mol)Gf(CaCO3(s))
1.3  102 kJ = (1 mol)Gf(CaO(s)) + (1)(–394.36 kJ) – (1)(–1128.8 kJ)
Gf(CaO(s)) = –604 kJ·mol–1
Problem 9.66.
Compounds A, B, and C were dissolved in water in a reaction
vessel and allowed to come to equilibrium. The graph shows
the variation of concentration with time for this reaction:
3A(aq)  B(aq) + 2C(aq)
(a) The reaction quotient, Q, and its value at time t1, the time
the compounds were mixed, are:
2
2
B(aq)C(aq)
0.010.02


Q=
=
= 0.008
3
3
A(aq)
0.08
(b) The graph shows that the concentration of A is
decreasing, while the concentrations of B and C are increasing to attain equilibrium. This means
that the reaction is shifting to the right, favoring the formation of additional product molecules.
(c) Macroscopic concentrations of A, B, and C do not appear to be changing after t2, so it is
reasonable to suppose that equilibrium has been reached in this system at t2.
(d) At t2, we assume that equilibrium has been reached, so the reaction quotient equals K for this
reaction:
2
 B(aq)C(aq)2 
0.02 0.04 

 =
Q = K = 
= 0.3
3
3

0.05
A(aq)





eq
Note that the stoichiometry of the reaction is included in the plot. As the concentration of A
decreases by 0.03 M, the sum of the concentrations of B and C increases by 0.03 M, so mass is
conserved.
(e) Use the relationship, Grxn = –RTlnK, and the K value from part (d) to find the standard free
energy change for this reaction at 298 K:
Grxn = –RTlnKeq = –(8.314 J·K–1·mol–1)( 298 K) ln(0.3) = 3 kJ·mol–1
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Problem 9.67.
(a) For the reaction, H2(g) + I2(g)  2HI(g), we use the equilibrium constant at 700 K to find
Greaction at 700 K:
Greaction = –RTlnK = –(8.314 J·K–1·mol–1)( 700 K) ln(54)
Greaction = –23 kJ·mol–1 (for reaction of one mole of each reactant)
(b) To get the reaction quotient, Q, in a mixture in which the pressures of H2(g), I2(g), and HI(g)
are, respectively, 0.040 bar, 0.35 bar, and 1.05 bar, we write the reaction quotient expression and
substitute these pressure values (as ratios to the standard pressure, 1 bar):
HI(g)
H 2 (g)I2 (g)
2
Q=
1.052
=
= 79
0.0400.35
(c) The free energy of a reaction is related to the standard free energy and the reaction quotient
under the conditions in the reaction system:
Greaction = Greaction + RTlnQ = –23 kJ·mol–1 + (8.314 J·K–1·mol–1)( 700 K) ln(79)
Greaction = 2.4 kJ·mol–1
(This result should really be reported only to the nearest 1 kJ·mol–1, because the value for
Greaction is only known to the nearest kJ·mol–1,) The positive sign of Greaction is consistent with
Q > K. When Q > K, the concentration of products compared to reactants is larger than the
equilibrium value. Going from reactants to products under this circumstance is not favored,
which is what the positive Greaction tells us.
(d) To attain equilibrium in the system described in part (b), some of the HI(g) has to be
converted to its elements, H2(g) and I2(g),so that the value of Q will be reduced to K. The
reaction proceeds to produce more reactants at the expense of the product.
Problem 9.68.
We calculate Greaction for methanol decomposition, CH3OH(l)  CO(g) + 2H2(g), and use its
sign to determine whether the reaction will occur spontaneously under standard conditions at 298
K:
Greaction = (1 mol)Gf(CO(g)) + (2 mol)Gf(H2(g)) – (1 mol)Gf(CH3OH(l))
Greaction = (1)(–137.17 kJ) + (2)(0 kJ) – (1)(–166.27 kJ)
Greaction = 29.10 kJ
Since Greaction > 0, methanol decomposition to carbon monoxide and hydrogen under standard
conditions is not spontaneous.
Problem 9.69.
Calculate Greaction for the reaction, CH2CH2(g) + H2O(l)  CH3CH2OH(l), and use its sign to
determine whether the reaction will occur spontaneously under standard conditions at 298 K:
Greaction = (1 mol)Gof(CH3CH2OH(l)) – (1 mol)Gf(C2H4(g)) – (1
mol)Gf(H2O(l))
Greaction = (1)(–174.78 kJ) – (1)(68.15 kJ) – (1)(–237.13 kJ)
Greaction = –5.80 kJ
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The reaction of water with ethene (ethylene) to form ethanol under standard conditions is favored
with an equilibrium constant of 10.4. Although the reaction is slow, its rate can be increased by
catalysis and a good deal of ethanol is manufactured by this reaction.
Problem 9.70.
To explain how temperature, pressure, and concentration affect a system at equilibrium, we can
use le Chatelier’s principle: a system at equilibrium responds to a disturbance by adjusting to
decrease the effect of the disturbance. The application to changes in temperature is through the
energetics of the equilibrium reaction. If the temperature of an equilibrium system is increased, it
gains energy, so the reaction shifts in a direction to use energy. An endothermic reaction requires
energy, so raising its temperature will shift the equilibrium toward more products, in order to use
energy. An exothermic reaction releases energy, so raising its temperature will shift the
equilibrium toward more reactants, in order to use energy (since the reverse reaction must be
endothermic). Pressure and concentration effects are related. If more reactant is added to a
constant volume system, the concentration of reactant is increased and the equilibrium shifts
toward formation of products to reduce the added reactant concentration. If a reactant is removed
from a constant volume system, the equilibrium shifts toward formation of more reactants to
compensate for the loss of the reactant that was removed. Similar arguments can be made about
adding or removing a product of the reaction. The effect of pressure changes is a special case of
concentration change for gaseous species. If the pressure of a reactant is increased by adding
more (or decreased by removing some) in a constant volume system, the effects are the same as
just noted for any concentration change. The more subtle effect is a change in the volume of a
system in which some or all of the reactants and products are gases. In this case, the
concentrations of all gases in the system are increased. To try to lower the concentration of
gases, the reaction will shift toward the side with the lower number of moles of gas. A third way
of changing the pressure in a system is by adding a gas that is not involved in the reaction. This
addition will have no affect on the equilibrium, because it does not change the concentration of
any reactant or product.
Problem 9.71.
[NOTE: The reaction equation in the problem statement is not balanced. The balanced equation
is used here.]
For this exothermic reaction, 2SO2(g) + O2(g)  2SO3(g), initially at equilibrium:
(a) Removing some SO3(g) shifts the position of equilibrium to the right, forming more product
SO3(g) to compensate for that which has been removed. The concentrations (pressures) of SO2(g)
and O2(g) will decrease (assuming that the volume remains constant) as they react to form more
SO3(g).
(b) Decreasing the pressure (presumably by increasing the volume of the system) shifts the
position of equilibrium to the left, forming more reactant, in order to form the largest number of
moles of gas possible to compensate for the decrease in pressure (concentration) of all the gases.
Thus more SO2(g) and O2(g) will form and some SO3(g) will be used up by the back reaction.
(c) Decreasing the temperature, by removing energy from the system, shifts the position of
equilibrium to the right, forming more product SO3(g), because the reaction, as written is
exothermic and releases energy. If energy has been removed, the system reacts to release more
energy to compensate for the loss. The final concentrations (pressures) in the system cannot be
predicted without knowing the actual temperature change that occurs (as well as the volume and
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initial pressures of the reactants and product), because the pressures of gases are temperature
dependent.
(d) Adding some O2(g) is added shifts the position of equilibrium to the right, forming more
product SO3(g), as the system compensates for the increase in concentration of a reactant by
using up some of it, which will also decrease the concentration of SO2(g). The new equilibrium
mixture will have more O2(g) and SO3(g) and less SO2(g), than the original equilibrium mixture.
Problem 9.72.
[NOTE: The problem statement gives the bond energy for NO as 630 kJ·mol–1, but a better value
is 632 kJ·mol–1 (See Problem 9.62), which we will use in this problem.]
(a) For the equilibrium reaction N2(g) + O2(g)  2NO(g), higher temperatures favor the
formation of NO(g). Since the reaction favors the products as temperature increases (energy is
added), we predict, using Le Chatelier’s principle, that the reaction is endothermic, that is
requires energy, Hrxn > 0. An endothermic reaction responds to added energy by forming more
product to use up some of the added energy and this is the direction of the observed effect.
(b) The bonds in N2 and O2 have to be broken for the reaction to proceed and this requires 941.4
kJ·mol–1 + 498.7 kJ·mol–1 = 1440.1 kJ·mol–1. Formation of the bonds in two moles of NO
releases 1264 kJ·mol–1, so the reaction overall is endothermic by 176 kJ·mol–1. This result is
consistent with our prediction in part (a), based on the Le Chatelier’s principle and the effect of
temperature on the equilibrium reaction.
(c) Since the formation of products is favored as temperature increases, the numerator of the
equilibrium constant expression increases at the expense of the denominator and, hence, the
equilibrium constant increases with temperature.
(d) If the equilibrium constant for this reaction increases as temperature increases, as we
reasoned in part (c), the standard free energy of reaction, Grxn = –RTlnK, must become more
negative (or less positive) as the temperature increases. We know Grxn = Hrxn – TSrxn, so,
since the standard enthalpy of reaction is positive, the only way that the standard free energy can
get more negative (or less positive) is for the standard entropy change for the reaction to be
positive. From Appendix B we have:
Srxn = (2 mol)S(NO) – (1 mol)S(N2) – (1 mol)S(O2)
Srxn = 2(210.76 J·K–1) – (191.61 J·K–1) – (205.14 J·K–1) = 24.77 J·K–1
Thus, the thermodynamic data confirm the deductions we make based solely on the experimental
information that the NO yield increases with increasing temperature.
Problem 9.73.
(a) To find the solubility product for PbI2(s) at 20. C (293 K), use the temperature dependence
of the solubility product constant, plus its value at one of the temperatures (273 K) and Hrxn
from Worked Example 9.67:
o
 K 293 
 K 293   H rxn   T2 – T1 
ln 
= ln 
=
 3.5  10 –9   R   T2 T1 
 K 273 
–1
 293 K – 273 K 
 K 293   56 kJ  mol
ln 
=

 = 1.68
 3.5  10 –9  
8.314 J  K– 1  mol – 1(293 K)(273 K)1 
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 K 293 
= 5.4;  K293 = 5.4(3.5  10–9) = 1.9  10–8

–9 

3.5  10
In Worked Example 9.67, the molar solubility is labeled s and the equilibrium constant is 4s3.
Therefore, at 20 C, the molar solubility (moles of PbI2 dissolved in 1 L of water at 20 C) is:
K 
s =  293 
 4 
1
3
 1.9  10 –8 
=

4

1
3
= 1.7  10–3
Use the molar solubility (mol·L–1) and the molar mass of PbI2 to find the solubility in g·L–1:
solubility in g·L–1 = (1.7  10–3 mol·L–1)(461 g·mol–1) = 0.77 g L–1
(b) The handbook value of 0.63 g L–1 for the solubility of PbI2 at 20 C is in the right ballpark
with our calculated value, although 20% lower. This is another example of the non-ideality of
aqueous ionic solutions. Our calculations are usually (but not always) of the correct order of
magnitude (compared to experiment), but rarely right on the money. This is why we have
stressed that these calculations are only approximations to the true state of affairs in the solutions
where many interactions of charged species are involved.
Problem 9.74.
(a) The reaction of interest is: CaSO4(s)  Ca2+(aq) + SO42–(aq). If the molar solubility of
CaSO4 is s, then Ksp = s2. The molar mass of CaSO4 is 135 g mol–1 so the molar solubilities and
solubility products are:
at 303 K: s =
2.09 g  L–1
= 1.55  10–2 mol L–1; Ksp = 2.40  10–4
135 g  mol–1
at 373 K: s =
1.62 g  L–1
= 1.20  10–2 mol L–1; Ksp = 1.44  10–4
–1
135 g  mol
(b) Use the temperature dependence of equilibrium to get Hrxn (assuming that the enthalpy
change is not temperature dependent over this range of temperature):
o
Ksp (373) 
1.44  10 –4  
  373 K – 303 K 
H rxn
ln 
–4  = 
 = ln 
–1
–1  
2.40  10   8.314 J  K  mol   (373 K)(303 K) 
Ksp (303) 
Hrxn = –6.86 kJ·mol–1
Note that the standard enthalpy change is negative, which we could have predicted from Le
Chatelier’s principle and the decrease in solubility as temperature increases. To get Srxn
(assuming that it also is not temperature dependent), we use Hrxn and the equilibrium constant
at one of the temperatures:
Grxn = –RTlnKsp = Hrxn – TSrxn
–(8.314 J·K–1·mol–1)( 303 K) ln(2.40  10–4) = –6.86  103 J·mol–1 – (303 K)Srxn
Srxn = –91.9 J·K–1·mol–1
(c) Use the relationship, Grxn = Hrxn – TSrxn, to find Grxn at 298 K and thence Ksp at
298 K:
Grxn = Hrxn – TSrxn = –6.86  103 J·mol–1 – (298 K)(–91.9 J·K–1·mol–1)
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Chapter 9
Grxn = 20.5  103 J·mol–1 (for one mole of CaSO4(s) dissolving)
–RTlnKsp = Grxn = –(8.314 J·K–1·mol–1)(298 K) ln(Ksp) = 20.5  103 J·mol–1
Ksp = 2.55  10–4
Note that the solubility product at 298 K is slightly greater than at 303 K, which is consistent
with the exothermicity of the reaction [part (b)]. We can also calculate Grxn from the data in
Appendix B:
Grxn = (1 mol)Gf(Ca2+(aq)) + (1 mol)Gf(SO42–(aq)) – (1 mol)Gf(CaSO4(s))
Grxn = (1)(–553.58 kJ) + (1)(–744.53 kJ) – (1)(–1321.79 kJ)
Grxn = 23.68 kJ
This value compares favorably (within about 10%) with that calculated from the solubility data.
The pKsp from our calculations is 3.60 compared with the value of 4.62 in Table 9.4. This is
about an order of magnitude difference in the solubility products with our value calculated from
the experimental solubilities being higher. The values given in tables like Table 9.4 are usually
obtained from data that are extrapolated from real solutions to high dilutions where interionic
interactions are lower and often the solubility products are lower than those calculated from
experimental solubilities.
Problem 9.75.
(a) For the coal gas reaction, C(s) + H2O(g)  CO(g) + H2(g), use the data in Appendix B to
calculate Hrxn, Srxn, and Grxn at 298 K (assuming the carbon is graphite):
Hrxn = (1 mol)Hf(CO(g)) + (1 mol)Hf(H2(g))
– (1 mol)Hf(C(s)) – (1 mol)Hf(H2O(g))
Hrxn = (1)(–110.53 kJ) + (1)(0 kJ) – (1)(0 kJ) – (1)(–241.82 kJ) = 131.29 kJ
Srxn = (1 mol)S(CO(g)) + (1 mol)S(H2(g))
– (1 mol)S(C(s)) – (1 mol)S(H2O(g))
Srxn = (1)(197.67 J·K–1) + (1)(130.68 J·K–1) – (1)(5.74 J·K–1) – (1)(188.83 J·K–1)
Srxn = 133.78 J·K–1
Grxn = (1 mol)Gf(CO(s)) + (1 mol)Gf(H2(g))
– (1 mol)Gf(C(s)) – (1 mol)Gf(H2O(g))
Grxn = (1)(–137.17 kJ) + (1)(0 kJ) – (1)(0 kJ) – (1)(–228.57 kJ) = 91.40 kJ
(b) Since both Hrxn and Srxn are positive, there should be a temperature at which TSrxn =
Hrxn, and Grxn = 0. Assuming that Hrxn and Srxn are not temperature dependent over the
temperature range of interest:
o
 H rxn

131.29 kJ
T= 
=
= 980 K
o 
0.13378 kJ  K –1
 Srxn 
We really ought to report this as 1000 K, since the assumptions of constant Hrxn and Srxn
are surely incorrect over such a large range.
(c) Since, from part (b), G°rxn = 0 at 980 K, the equilibrium constant at 980 K must be unity,
ln(K) = ln(1) = 0, in order to satisfy the equation, Grxn = –RTlnK.
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Problem 9.76.
(a) For a phase change, such as, liquid  gas, at equilibrium at one bar pressure, the
dimensionless concentration ratios for the reactant and product are both unity, since they are pure
substances. Thus, the reaction quotient (the ratio of the two concentrations) is also unity and lnQ
= 0. Therefore:
Grxn = Grxn + RTlnQ = Grxn
Since the reaction is in equilibrium, Grxn = 0 = Grxn.
(b) To estimate the temperature at which rhombic and monoclinic sulfur are in equilibrium,
S(s, rhombic)  S(s, monoclinic), at one bar, assume that H°rxn and S°rxn are not temperature
dependent over the temperature range of interest. Then, at the equilibrium temperature, where
Grxn = 0, we have TSrxn = Hrxn. We can use the values of Hrxn and Srxn to estimate the
equilibrium temperature. For the phase change of one mole of rhombic sulfur to monoclinic
sulfur, Appendix B gives Hrxn = 0.33 kJ [= (0.33kJ) – (0 kJ)) and Srxn = 0.8 J·K–1
[= (32.6 J·K–1) – (31.80 J·K–1)]. Thus:
o
 H rxn

0.33 kJ
T= 
=
 410 K (~140 C)
o 
0.0008 kJ  K –1
 Srxn 
This value is a bit high. Rhombic sulfur melts at 112 C and the triclinic form crystallizes out
from the melt. Our calculation does not account for the solid-liquid phase changes, but the
answer is not unreasonably out of line.
Problem 9.77.
(a) Given that the equilibrium constant for the reaction, H2(g) + I2(g)  2HI(g), is 794 at 298 K
and using other results from Problems 9.13 and 9.61, find Hrxn and Srxn. There are at least
two ways to solve this problem and the assumptions in each are the same, that Hrxn and Srxn
are not temperature dependent over the temperature range of interest. One way to do the problem
is to use the equilibrium constant values at 500 K (from Problem 9.13) and 298 K (from this
problem statement) with equation (9.59) to solve for Hrxn. Another way is to subtract the
values for Grxn at 500 K (from Problem 9.61) and 298 K (calculated from the equilibrium
constant in this problem), written in terms of Hrxn and Srxn, so that the Hrxn term cancels
out and then solve for Srxn. In either case, we then substitute our known values into the
expression for Grxn to get the unknown. Let’s use the second method.
Grxn(298) = –RTlnK = –(8.314 J·K–1·mol–1)(298 K) ln(794) = –16.5 kJ·mol–1
Grxn(500) = – 21.1 kJ·mol–1
Now we write the equations for G°rxn in terms of Hrxn and Srxn and subtract:
Grxn(500) = – 21.1 kJ·mol–1 = Hrxn – (500 K)Srxn
– {Grxn(298) = – 16.5 kJ·mol–1 = Hrxn – (298 K)Srxn}
(– 21.1 kJ·mol–1) – (– 16.5 kJ·mol–1) = – (500 K – 298 K)Srxn
Srxn = 22.8 J·K·mol–1
Now substitute this value for Srxn into either of the Grxn equations to get Hrxn:
Grxn(298) = – 16.5 kJ·mol–1 = Hrxn – (298 K)(0.0228 kJ·K·mol–1)
Hrxn = –9.71 kJ·mol–1
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Chapter 9
(b) If we had done part (a) by the first method (using equilibrium constants) we would not know
Grxn(298) at this point and would have to calculate it, either as we did in part (a) or by
combining the Hrxn and Srxn we got without knowing Grxn(298). The answers would all be
the same (within some calculational round-off uncertainty), no matter which method you choose.
For comparison, using the data from Appendix B, we get:
Grxn(298) = (2 mol)Gf(HI(g)) – (1 mol)Gf(H2(g)) – (1 mol)Gf(I2(g))
Grxn(298) = (2)(1.70 kJ) – (1)(0 k) – (1)(19.3 kJ) = –15.9 kJ·mol–1
Grxn(298) = –15.9 kJ·mol–1
This value for Grxn(298) is almost the same as the one we found above. It appears that our
assumptions are good (or that there are compensating effects that cancel out). Slight
discrepancies between values determined in different ways, free energies from thermodynamic
measurements and from equilibrium constants, for example, are usually due to small
uncertainties in each experimental value that combine to yield slightly different numeric results
for the same variable.
[For comparison, here are the values calculated by the other method.
o
 500 K – 298 K 
K(500)
Hrx
160  
n
ln 
 = ln   = 

–1
– 1
K(298) 
8.314 J  K  mol (500 K)(298 K) 
794
Hrxn = –9.82 kJ·mol–1
Grxn(500) = – 21.1 kJ·mol–1 = –9.82 kJ·mol–1 – (500 K)Srxn
Srxn = 22.6 J·K·mol–1
Grxn(298) = –9.82 kJ·mol–1 – (298 K)(0.0226 kJ·K·mol–1) = –16.6 kJ·mol–1
The small discrepancies are due to rounding off and retaining only three significant figures in the
intermediate values.]
Problem 9.78.
(a) Estimate Hrxn for ethanol vaporization. At its normal boiling point, a liquid is in
equilibrium, Grxn = 0, with its vapor at one bar pressure. Under these conditions, Hrxn =
TSrxn, where T is the boiling point. Since the total pressure is one bar, Hrxn and Srxn are equal
to Hrxn and Srxn, respectively. For ethanol at 351.6 K (its normal boiling point) we get:
Hrxn = TSrxn = (351.6 K)(122 J·K–1·mol–1) = 42.9 kJ·mol–1
(b) The equilibrium constant expression for ethanol vaporization is:
P
C2 H 5 OH
Keq =
(C2 H 5OH(g))
=
(C2 H 5OH(l))

1 bar 
1
The denominator of the equilibrium constant expression is unity because the component is a pure
liquid. The numerator is a ratio of the vapor pressure of ethanol (in bar) to the standard pressure,
1 bar. At the normal boiling temperature, 78.5 C, the vapor pressure of the ethanol is 1 bar, so
both top and bottom of the equilibrium constant expression are unity and Keq = 1. Thus, Grxn =
–RTlnKeq = –RTln(1) = 0. Also, since Grxn refers to the reaction at one bar total pressure and
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ACS Chemistry FROG
Chapter 9
Chemical Equilibria
the system is in equilibrium at the normal boiling point and one bar total pressure, we can make
the connection that Grxn = Grxn = 0.
(c) At 298 K (25 C), we can use Hrxn and Srxn from part (a), assuming that they are
independent of temperature, to find Grxn and Keq:
Grxn = Hrxn – TSrxn = 42.9 kJ·mol–1 – (298 K)(0.122 kJ·K–1·mol–1)
Grxn = 6.54 kJ·mol–1
Grxn = –RTlnKeq = 6.54 kJ·mol–1 = – (8.314 J·K–1·mol–1)(298 K)lnKeq
Keq = 0.0714
(d) From part (b), you see that the numeric value of Keq is the vapor pressure of ethanol (in bar).
Thus, the vapor pressure of ethanol at 298 K (25 C) is 0.0714 bar (= 7.14  103 Pa = 53.6 torr =
0.0705 atm).
Problem 9.79.
(a) To determine Hrxn and Srxn for the vaporization of 1-propanol (1-PrOH) and 2-propanol
(2-PrOH), use the vapor pressure vs. temperature data given, the equivalence between vapor
pressure and the equilibrium constant for vaporization, equation (9.61), and the relationship
among K, Hrxn, and Srxn, equation 9.55, as was done for similar data in Worked Example
9.71.
v. p., torr
v. p., bar
1.0
0.0013
ln K –6.65
1-PrOH, T, K 258.1
1/T, K–1 0.00387
2-PrOH, T, K 247.0
1/T, K–1 0.00405
10.0
0.0133
–4.320
287.8
0.00347
275.5
0.00363
40.0
0.0533
–2.932
309.5
0.00323
296.9
0.00337
100.
0.133
–2.017
325.9
0.00307
312.6
0.00320
400.
0.533
–0.629
355.1
0.00282
340.9
0.00293
760.
1.013
0.0129
370.9
0.00270
355.6
0.00281
The plots of these data (and the computer-generated equations of the lines through the data
points) are shown on this graph:
Values determined from the plot for 1-PrOH are:
ACS Chemistry FROG
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Chapter 9
Hrxn = –R·(slope) = – (8.314 J·K–1·mol–1)·(–5.70  103 K) = 47.4 kJ·mol–1
Srxn = R·(intercept) = (8.314 J·K–1·mol–1)·(15.46) = 129 J·K–1·mol–1
Values determined from the plot for 2-PrOH are:
Hrxn = – (8.314 J·K–1·mol–1)·(–5.36  103 K) = 44.6 kJ·mol–1
Srxn = (8.314 J·K–1·mol–1)·(15.10) = 126 J·K–1·mol–1
(b) The slightly higher standard enthalpy change for vaporization of 1-propanol compared to 2propanol probably reflects a bit greater bonding interactions (both hydrogen bonding and
dispersion forces) among the linear-chain 1-propanol molecules compared to the more compact
structure of 2-propanol.
(c) If we assume that these isomeric molecules in the gas phase have about the same entropy,
then the smaller change in entropy for vaporization of 2-propanol suggests that its liquid phase
entropy is a bit closer to the gas phase entropy, that is the liquid phase entropy of 2-propanol is a
bit higher than that of 1-propanol. If the molecules of the more compact isomer are freer to move
about in the liquid [as our reasoning in part (b) suggests], higher liquid phase entropy for 2propanol would be expected (higher positional entropy).
(d) The 2-propanol is more volatile. The temperature required to produce any given vapor
pressure is lower for 2-propanol. This is largely a result of the lower enthalpy of vaporization,
which gives a less positive free energy for the vaporization (at temperatures below the boiling
point). The positive entropies of vaporization also contribute to lowering the free energy, but
contribute less for 2-propanol, because the entropy change is a bit smaller.
Problem 9.80.
Assuming that the changes in standard enthalpy and entropy are not temperature dependent, we
can use the temperature dependence of the equilibrium constant for water autoionization, pKw
14.00 at 25 C and 12.97 at 60 C, to find Hrxn and then combine this with one of the
equilibrium constants to get Srxn, and thence Grxn:
Kw(298) = 1.00  10–14; Kw(333) = 1.07  10–13
1.07  10 –1 3  
 K(333) 
Horeacti on
 333 K – 298 K 
ln 
=
 = ln 


–1
–1 
–1 4 
K(298) 
1.00  10  8.314 J  K mol (333 K)(298 K) 
Hrxn = 55.9 kJ·mol–1
–RTlnKw(T) = Hrxn – TSrxn
–(8.314 J·K–1·mol–1)(298 K) ln(1.00  10–14) = (55.9  103 J·mol–1) – (298 K)Srxn
Srxn = –80.4 J·K–1·mol–1
Grxn can be determined either by combining Hrxn and Srxn or from one of the equilibrium
constants. There is no difference mathematically, since an equilibrium constant was used to get
Srxn in the first place.
Grxn = –RTlnKw(T) = 79.9 kJ·mol–1
The reaction of interest here is H2O(l)  H+(aq) + OH–(aq), for which we can calculate Hrxn,
Srxn, and Grxn from Appendix B to compare with the results above:
Hrxn = (1 mol)Hf(H+(aq)) + (1 mol)Hf(OH–(aq)) – (1 mol)Hf(H2O(l))
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Hrxn = (1)(0 kJ) + (1)(–229.29 kJ) – (1)(–285.83 kJ) = 55.84 kJ
Srxn = (1 mol)S(H+(aq)) + (1 mol)S(OH–(aq)) – (1 mol)S(H2O(l))
Srxn = (1)(0 J·K–1) + (1)(–10.75 J·K–1) – (1)(69.91 J·K–1) = 80.66 J·K–1
Grxn = (1 mol)Gf(H+(aq)) + (1 mol)Gf(OH–(aq)) – (1 mol)Gf(H2O(l))
Grxn = (1)(0 kJ) + (1)(–157.24 kJ) – (1)(–237.13 kJ) = 79.89 kJ
The thermodynamic values derived from the data in Appendix B are the same (for one mole of
reaction) as those determined from the temperature dependence of the equilibrium constant. This
makes sense, because it is likely to be equilibrium constant data like these that are used to
determine the thermodynamic values for the hydroxide ion. The values for the hydronium ion are
assigned and all other ions are given relative to the hydronium ion to make them consistent with
equilibrium data.
Problem 9.81.
[NOTE: The relationship, K = C + 273.1, should be added to this problem.]
(a) The numerical value of the equilibrium constant for a phase change from a condensed phase
to the gas phase, such as, PCl5(s)  PCl5(g), is equal to the equilibrium pressure of the gas (in
bar), equation (9.61). Assuming that the changes in standard enthalpy and entropy are not
temperature dependent, we can use the temperature dependence of the equilibrium constant to
find Hrxn and then combine this with one of the equilibrium constants to get Srxn. For
phosphorus pentachloride sublimation, the equilibrium constants are:
 40 torr 

–1 

750 torr  bar  = 0.053
K(375.6) = 
1 bar






 400 torr 

–1 

750 torr  bar  = 0.533
K(420.3) = 
1 bar






o
  420.3 K – 375.6 K 
Hrxn
K(420.3) 
0.533 
ln 
=
 = ln 
–1
–1


K(375.6) 
0.053  8.314 J  K  mol   (420.3 K)(375.6 K) 
Hrxn = 67.6 kJ·mol–1
–RTlnK(T) = Hrxn – TSrxn
– (8.314 J·K–1·mol–1)(420.3 K)ln(0.533) = 67.6  103 J·mol–1 – (420.3 K)Srxn
Srxn = 156 J·K–1·mol–1
(b) When the equilibrium sublimation pressure is 1 bar, K = 1 and lnK = 0, so
T=
o
67.6  10 3 J  mol –1
Hrxn
=
3
–1
–1 = 433 K = 160 C
Sorxn
156  10 J  K  mol
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Problem 9.82.
(a) A spontaneous endothermic reaction during which the entropy of the system increases must
have a negative free energy change, because all spontaneous reactions have G < 0.
(b) A spontaneous exothermic reaction during which the entropy of the system decreases must
have a negative free energy change, because all spontaneous reactions have G < 0.
(c) In part (a), H > 0 and S > 0. If the temperature is decreased, the –TS term in the free
energy expression contributes less to G. At low enough T, it is possible that H > |TS|, so that
G > 0 and the reaction would become non-spontaneous. In part (b), H < 0 and S < 0. If the
temperature is increased, the –TS term in the free energy expression contributes more to G. At
high enough T, it is possible that |H| < –TS, so that G > 0 and the reaction would become
non-spontaneous.
Problem 9.83.
To determine whether the reaction, CH3OH(g)  CO(g) + 2H2(g), might occur spontaneously at
1000 K and 1 bar, we can determine Hrxn and Srxn at 298 K and, assuming they are not
dependent on temperature, find Hrxn –TSrxn at 1000 K to see if Grxn < 0.
Hrxn = (1 mol)H°f(CO(g)) + (2 mol)H°f(H2(g)) – (1 mol)H°f(CH3OH(g))
Hrxn = (1)(–110.53 kJ) + (2)(0 kJ) – (1)(–200.66 kJ) = 90.13 kJ
Srxn = (1 mol)S(CO(g)) + (2 mol)S(H2(g)) – (1 mol)S(CH3OH(g))
Srxn = (1)(197.67 J·K–1) + (2)(130.68 J·K–1) – (1)(239.81 J·K–1) = 219.22 J·K–1
Grxn = Hrxn – TSrxn = (90.13  103 J) – (1000 K)(219.22 J·K–1)
Grxn = – 129 kJ
At 1 bar total pressure, the reaction is likely be spontaneous at 1000 K, (if the molecules do not
dissociate or undergo reactions in other ways) since, under these conditions, Grxn = Grxn has a
large negative value.
Problem 9.84.
(a) From Problem 9.79, we have Hrxn = 44.6 kJ·mol–1 and Srxn = 126 J·K–1·mol–1 for the
vaporization of 2-propanol. We can use these values to calculate K at 35 C and hence the vapor
pressure (in bar) for 2-propanol:
–RTlnK(T) = Hrxn – TSrxn
– (8.314 J·K–1·mol–1)·(308 K)lnK(308) = 44.6  103 J·mol–1 – (308 K)(126 J·K–1·mol–
1
)
lnK(308) = –2.262;
K(308) = 0.104
The vapor pressure of 2-propanol at skin temperature, 35 C, is 0.104 bar [= 78 torr), which
makes it evaporate (vaporize) relatively rapidly from the skin. The process requires the enthalpy
of vaporization, which is furnished by the skin and, hence, we feel a cooling sensation. See
Chapter 1, Section 1.12.
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(b) 12 g of 2-propanol, CH3CHOHCH3, is
12 g 
60 g  mol  = 0.20 mol. The amount of
–1
energy that is required to vaporize this sample is (0.20 mol)·(44.6 kJ·mol–1) = 8.9 kJ. Thus, the
thermal energy of your body is decreased by this amount when the alcohol evaporates.
Problem 9.85.
This data table for Problems 9.85 through
9.88 is repeated here for easy reference.
The free energies are for the hydrolysis at
pH 7 of the phosphorylated compounds to
give the non-phosphorylated compound
plus phosphate:
compound-P + H2O  compound + Pi
Free energy of hydrolysis at pH 7 and 25 C
Compound
phosphoenolpyruvate, PEP
creatine phosphate
ATP (gives ADP)
glucose-1-phosphate, Glu-1-P
glucose-6-phosphate, Glu-6-P
glycerol-3-phosphate, G3P
G, kJ·mol–1
–62
–43
–30
–21
–14
–9
(a) The hydrolysis reactions for glu-6-P
and glu-1-P and the combination to give the
isomerization reaction are:
glu-6-P + H2O  glu + Pi
––––––––––––––––––––––––––
[glu-1-P + H2O  glu + Pi]
glu-6-P  glu-1-P
(b) We can combine the standard free energy changes for the hydrolysis reactions, just as we
combined the reactions, to give Grxn for the isomerization reaction:
Grxn = (–14 kJ·mol–1) – (–21 kJ·mol–1) = 7 kJ·mol–1
(c) The equilibrium constant and equilibrium constant expression for the isomerization can be
obtained as:
o
 (glu-1-P) 
–Grxn

–7  103 J  mol –1

lnK = ln 
=
= 
 = –2.8
–1
–1


 (glu-6-P) 
(8.314 J  K mol )(298 K) 
RT
K=
(glu-1-P)
= 0.06
(glu-6-P)
Thus the equilibrium ratio of glu-6-P to glu-1-P is 1/0.06 or about 17 to 1. Since the free energy
change is positive for the isomerization, it makes sense that the reactant, glu-6-P should
predominate at equilibrium.
Problem 9.86.
(a) The hydrolysis reactions, their combination, ATP + Pyr  ADP + PEP (Pyr = pyruvate), the
Grxn for the combination, and the equilibrium constant are:
ATP + H2O  ADP + Pi
––––––––––––––––––––––––––
[PEP + H2O  Pyr + Pi]
ATP + Pyr  ADP + PEP
Grxn = (–30 kJ·mol–1) – (–62 kJ·mol–1) = 32 kJ·mol–1
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o
 (ADP)(PEP) 
–Grxn

–32  10 3 J  mol –1

lnK = ln 
=
=

 = –12.9
–1
–1

(8.314 J  K mol )(298 K) 
 (ATP)(Pyr) 
RT
(ADP)(PEP)
K=
= 2.5  10–6
(ATP)(Pyr)
(b) If the (ATP)/(ADP) ratio in the cell is 10/1, and the reaction in part (a) is at equilibrium, we
(Pyr)
can substitute this ratio in the equilibrium constant expression and solve for the
(PEP) ratio.
The result is:
(Pyr)
4
(PEP) = 4  10
The reaction in part (a) heavily favors the reactants, as the substantial positive free energy
suggests. Indeed, in the glycolysis pathway in cells, ATP is produced by reaction with PEP (the
favored reverse reaction).
Problem 9.87.
For each of these reactions, imagine starting with a mixture of all the reactants and products at
the same concentration in a pH 7 solution at 25 C. For each case, tell which direction the
reaction will go in order to approach equilibrium.
(i) ATP + creatine  ADP + creatine phosphate
(ii) ATP + Glu  ADP + Glu-6-P
For each case, we combine the hydrolysis reactions to give the desired reaction and then the free
energies in the same way to find out whether Grxn is positive or negative. If Grxn is positive,
the reaction favors reactants (is not spontaneous), so the mixture will go from equal
concentrations of reactants and products toward a state with higher concentrations of reactants
(that is, it will proceed in reverse of the direction it is written). If Grxn is negative, the reaction
favors products (is spontaneous), so the mixture will go from equal concentrations of reactants
and products toward a state with higher concentrations of products (that is, it will proceed in the
direction it is written).
(i)
ATP + H2O  ADP + Pi
––––––––––––––––––––––––––––––––––––––––
[creatine phosphate + H2O  creatine + Pi]
ATP + creatine  ADP + creatine phosphate
Grxn = (–30 kJ·mol–1) – (–43 kJ·mol–1) = 13 kJ·mol–1
This mixture will react to form reactants from products under the conditions stated.
(ii)
ATP + H2O  ADP + Pi
–––––––––––––––––––––––––––––
[Glu-6-P + H2O  Glu + Pi]
ATP + Glu  ADP + Glu-6-P
Grxn = (–30 kJ·mol–1) – (–14 kJ·mol–1) = –16 kJ·mol–1
This mixture will react to form products from reactants under the conditions stated.
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Problem 9.88.
[NOTE: The reactions written in this problem should be:
Glu +2ADP + 2Pi + 2NAD+  2Pyr +2ATP + 2NADH + 2H2O
2Pyr + 6ATP + 2NADH + 6H2O  Glu + 6ADP + 6Pi + 2NAD+ + 4H+
This stoichiometry is consistent with reaction (9.62) for the hydrolysis of ATP and with half
reaction (6.77) for reduction of NAD+. Using the charges on the species in equations (9.62) and
(6.77), you can show that the equations here are balanced in both atoms and charge.]
(a) The stoichiometry of the glycolysis pathway in organisms is:
Glu +2ADP + 2Pi + 2NAD+  2Pyr +2ATP + 2NADH + 2H2O
The Grxn, for glycolysis is about –84 kJ at 25 C. When a reaction is reversed, the signs of its
thermodynamic changes are reversed. Thus, if the glycolysis reaction is considered in reverse,
Grxn would be about 84 kJ. This is a very unfavorable change. Glycolysis is not reversible, so
gluconeogenesis must use a different pathway, for which Grxn = –38 kJ at 25 C:
2Pyr + 6ATP + 2NADH + 6H2O  Glu + 6ADP + 6Pi + 2NAD+ + 4H+
(b) If we sum the glycolysis and gluconeogenesis reactions, we get:
Glu +2ADP + 2Pi + 2NAD+  2Pyr +2ATP + 2NADH + 2H2O
2Pyr
+ 6ATP + 2NADH + 6H2O  Glu + 6ADP + 6Pi + 2NAD+ + 4H+
–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
4ATP + 4H2O  4ADP + 4Pi + 4H+
The standard free energy change for the net reaction (hydrolysis of 4 moles of ATP per mole of
glucose broken down and remade) is the sum of the standard free energy changes for the
individual reactions:
Gnet rxn = (–84 kJ) + (–38 kJ) = –122 kJ
The standard free energy change for hydrolysis of one mole of ATP is about –29 kJ, so the free
energy change for hydrolysis of four moles is –116 kJ, which is just about what we get for the
sum of the glycolysis and gluconeogenesis pathways. The difference between the two overall
reactions is consistent with their free energy changes.
(c) In Figure 7.17, the downward facing arrow on the far left and the upward facing arrow next to
it could represent glycolysis, a reaction pathway that has an overall negative free energy and
produces ATP. The upward facing arrow on the far right and the downward facing arrow next to
it could represent gluconeogenesis, a reaction that has an overall negative free energy and uses
(hydrolyzes) a good deal of ATP (note that the downward arrow is longer than the upward
arrow) to produce this overall negative free energy.
Problem 9.89.
(a) The reduction of pyruvate to lactate, Pyr + NADH + H+  Lac + NAD+, is a fermentation
reaction and is required to enable glycolysis to proceed in the absence of oxygen (which is
required form the complete glucose oxidation pathway). As you find in Chapter 6, Section 6.11,
and in Problem 9.88, glycolysis requires the presence of NAD+ for the oxidation of glucose to
pyruvate. There is only a small amount NAD+ in a cell, so it has to be recycled from NADH in
order to keep glycolysis going. Usually, NADH is oxidized back to NAD+ in reactions that
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involve oxygen as the ultimate oxidizing agent. If there is an insufficient supply of oxygen, as
considered in this problem, then some other pathway is required to oxidize the NADH and
fermentation is such a pathway.
(b) The combination of glycolysis and this pyruvate fermentation (taken twice, because each
glucose produces two pyruvate molecules to be reduced) is:
Glu +2ADP + 2Pi + 2NAD+  2Pyr +2ATP + 2NADH + 2H2O
2Pyr
+ 2NADH + 2H+  2Lac + 2NAD+
––––––––––––––––––––––––––––––––––––––––––––––––––––
Glu + 2ADP + 2Pi + 2H+  2Lac + 2ATP + 2H2O
(c) We had to combine two moles of the fermentation reaction to give the net reaction of one
mole of glucose to lactate. Thus, with Grxn = –84 kJ for glycolysis (from Problem 9.88) and
Grxn = –25 kJ for this fermentation, we find:
Gnet rxn = (–84 kJ) + 2(–25 kJ) = –134 kJ (for one mole of glucose reacting)
(d) The equilibrium constant expression for the net reaction in part (b), with the standard state
for H+(aq) taken as pH 7, and also assuming (H2O) = 1, is:
 (Lac) 2 (ATP ) 2 
K = 

(Glu)(ADP) 2 (Pi )2 eq
(e) We get the free energy change for the reaction from the relationship:
Gnet rxn = Gnet rxn + RTlnQ
Q is the reaction quotient (at pH 7), which has the same form as the equilibrium constant
expression in part (d), but with the actual system dimensionless concentration ratios, not
necessarily equilibrium concentrations.
Gnet rxn = (–134  103 J·mol–1


(5  10 –5 )2 (2  10 –3 )2
–1
–1
+ (8.314 J·K ·mol )·(298 K)ln 
–3
–4 2
–3 2 
(5  10 )(2  10 ) (1  10 ) 
Gnet rxn = –124 kJ·mol–1
Since Gnet rxn < 0, the reaction converting reactants at the specified concentrations to products
at the specified concentrations is spontaneous. Glycolysis followed by fermentation is a favored
process under cellular conditions.
Problem 9.90.
(a) We can write the entropy for a gas at pressures P1 and P2, equation (9.83), and take the
difference to get S for the change of pressure.
S2 = So – RlnP2
–––––––––––––––––––––––––––––––––––
[S1 = So – RlnP1]
S2 –S1 = S = R(lnP1 – lnP2) = Rln(P1/P2)
(b) When a gas is compressed, the final pressure, P2, is higher than the initial pressure, P1. Thus,
P1/P2 < 1 and ln(P1/P2) < 0, that is, the entropy change [part (a)] is negative. When a gas is
compressed at constant temperature, the final volume of the gas is less than the initial volume.
We found in Chapter 8 that positional entropy decreases as the volume available to molecules
decreases, so the entropy will decrease upon compression, just as the equation here also shows.
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Problem 9.91.
(a) The equilibrium constant expression for the reaction, 2NO2(g)  N2O4(g), and the
equilibrium constant when the system is at equilibrium with NO2(g) and N2O4(g) pressures of
0.225 bar and 0.438 bar, respectively, are:
K=
P
P
N2 O 4
NO2
 = 0.438
0.225
1 bar 
1 bar
2
2
= 8.65
(b) If the volume is doubled, the pressure is halved (at constant temperature). Thus, P(NO2) =
0.113 bar and P(N2O4) = 0.219 bar, and the reaction quotient is:
0.219
Q=
= 17.2
0.1132
(c) In order to move the reaction quotient toward the equilibrium value, the quotient has to
decrease. This means that the numerator, P(N2O4), will decrease and the denominator, P(NO2),
will increase. In other words, the reaction will go in reverse to produce more reactant. The stress
on this system is lowering the pressure of the gases. The response of the system, according to Le
Chatelier’s principle, will be to try to produce more moles of gas to compensate for the lowering
of the concentrations (pressures). That means going in reverse to form two molecules of NO2(g)
from one molecule of N2O4(g). The predictions from thermodynamics (reaction quotients) and
Le Chatelier are the same.
(d) If N2O4(g) reacts to give NO2(g), P(N2O4) will be reduced and P(NO2) will increase. After
this occurs, we can write the pressures of the gases as P(N2O4) = 0.219 – x and P(NO2) = 0.113 +
2x, since the stoichiometry produces two NO2(g) for each N2O4(g) that reacts. Substitute these
values into the equilibrium constant expression from part (a) and solve for x and hence the new
equilibrium pressures.
0.219 – x
0.219 – x
8.65 =
=
2
0.0127  0.452x  4x 2
0.113  2x
This equation can be rearranged to a quadratic form and solved:
34.6x2 + 4.91x – 0.109 = 0
4.91  (4.91) 2 – 4  34.6 (–0.109)
x=
= 0.0195 (the positive root)
2  34.6
Thus, P(N2O4) = 0.219 – 0.0195 = 0.1995 and P(NO2) = 0.113 + 2(0.0195) = 0.152. This
calculation carries too many significant figures, but the result is quite sensitive to round-off
uncertainties, so they are retained. Check to see if these are correct by substituting into the
equilibrium expression:
0.1995
K=
= 8.63
0.1522
Within the round-off uncertainties of the calculation, this value for K is the same as in part (a), so
we can have some confidence that the pressures we calculated are correct.
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Problem 9.92.
From Problem 9.13, K = 1.60  102 for the reaction, H2(g) + I2(g)  2HI(g), at 500 K. If 4.00 
10–3 mol of HI are placed in a 450. mL reaction vessel heated to 500 K and no reaction occurred,
the pressure of HI(g) in the reaction vessel would be:
(4.00  10 3 mol)(8.314  10 -2 L  bar  K-1mol -1)(500 K)
nRT
P(HI) =
=
0.450 L
V
P(HI) = 0.3695 bar
After reaction occurs and equilibrium is established, let the pressures of H2(g) and I2(g) be x bar.
From the stoichiometry of the reaction, we know that P(HI) = (0.3695 – 2x) bar at equilibrium.
Therefore, we have:
0.3695  2x 2 0.1365 –1.478x + 4x 2
1.60  102 =
=
x x
x2
156x2 + 1.478x – 0.1365 = 0
1.478  (1.478)2 – 4 156 (–0.1365)
= 0.0252
2 156
You can check this result by substituting the pressures back into the equilibrium constant
expression and finding that the numeric value is 160. The numeric value of x is the pressure of
I2(g) (and H2(g)) in bar. Use the ideal gas equation once again to find what we are asked for, the
number of moles of I2(g):
(0.0252 bar)(0.450 L)
PV
n=
=
= 2.73  10–4 mol
-2
-1
-1
RT
(8.314  10 L bar  K mol )(500 K)
x=
Problem 9.93.
The equilibrium proton transfer reaction, HOAc(aq) + H2O(aq)  OAc–(aq) + H3O+(aq), for
acetic acid will be upset by the addition of hydrochloric acid, HCl(aq), to the system. The
hydrochloric acid solution contains H3O+(aq) [and Cl–(aq)] that will increase the concentration
of a product of the acetic acid system. The system will respond by reacting to use up some of the
added H3O+(aq) by going in reverse. This will, of necessity, use up some of the OAc–(aq), so the
equilibrium concentration of OAc–(aq) will decrease. In the new equilibrium state, the
concentration of H3O+(aq) will be higher than in the initial equilibrium. This change can be
characterized as common ion effect, where the common ion is H3O+(aq), which is present in both
the acetic acid and hydrochloric acid solutions that are mixed.
Problem 9.94.
[NOTE: The data in this problem are from Science, 2002, 297, 1665 and also reported in C&EN,
September 9, 2002, page 35.]
(a) The reaction and its equilibrium constant expression (with BzH and BzF used to symbolize
benzene and fluorobenzene, respectively) are:
BzH(g) + CuF2(s)  BzF(g) + Cu(s) + HF(g)
P P
(BzF(g))(Cu(s))(HF(g)) (BzF(g))(HF(g))
K=
=
= BzF HF
PBzH
(BzH(g))(CuF2 (s))
(BzH(g))
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Chapter 9
Chemical Equilibria
The dimensionless concentration ratios for the solids are unity (Table 9.1) and the numeric
values of the concentration ratios for the gases are equal to the pressures of each gas in bar. From
the stoichiometry of the reaction we know that PBzF = PHF in the reaction mixture. We also know,
from the problem statement, that the total pressure of gases in the reactor is about 0.1 bar, that is,
PBzH + PBzF + PHF = PBzH + 2PBzF = 0.1 bar.
At 350 C (623 K) there is a 5% conversion of BzH to BzF, so
PBzF/PBzH = 5%/95% = 1/19;
PBzF = (1/19)PBzH
0.1 bar = PBzH + 2(1/19)PBzH = (21/19)PBzH
PBzH = (19/21)(0.1 bar) = .09 bar
PBzF = PHF = (1/19)PBzH = 0.005 bar
(0.005)(0.005)
P P
K(623) = BzF HF =
= 0.0003
PBzH
(0.09)
At 450 C (723 K) there is a 30% conversion of BzH to BzF, so
PBzF/PBzH = 30%/70% = 3/7;
PBzF = (3/7)PBzH
0.1 bar = PBzH + 2(3/7)PBzH = (13/7)PBzH
PBzH = (7/13)(0.1 bar) = .054 bar
PBzF = PHF = (3/7)PBzH = 0.023 bar
(0.023)(0.023)
P P
K(723) = BzF HF =
= 0.01
PBzH
(0.054)
(b) We get the standard enthalpy change, Hrxn, for this process from the temperature
dependence of the equilibrium constant [see equation (9.59)], assuming that Hrxn is
independent of temperature over this temperature range:
o
  723 K – 623 K 
K(723) 
H rxn
 0.01  
ln 
=
 = ln 



–1
–1
K(623) 
0.0003  8.314 J  K  mol   (723 K)(623 K) 
Hrxn = 1.3  102 kJ·mol–1
(c) We get the standard entropy change, Srxn, for this process from Hrxn and one of the
equilibrium constant values [see equation (9.54)], assuming that Srxn is independent of
temperature over this temperature range:
–RTlnK(T) = Hrxn – TSrxn
– (8.314 J·K–1·mol–1)·(723 K)ln(0.01) = 1.3  105 J·mol–1 – (723 K)Srxn
Srxn = 1.4  102 J·K–1·mol–1
(d) For 50% conversion, we have:
PBzF/PBzH = 50%/50% = 1/1;
PBzF = PBzH
0.1 bar = PBzH + 2PBzH = 3PBzH
PBzH = (1/3)(0.1 bar) = .033 bar
PBzF = PHF = PBzH = 0.033 bar
P P
(0.033)(0.033)
K(T) = BzF HF =
= 0.03
PBzH
(0.033)
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Chemical Equilibria
Chapter 9
We need to find the temperature, T, at which the equilibrium constant is 0.03. We can do this by
rearranging equation (9.54):
–RTlnK = Hrxn – TSrxn
TSrxn – RTlnK = Hrxn
o
1.3  10 5 J  mol – 1
H rxn
T=
=
o
(1.4  10 2 J  K – 1  mol – 1) – (8.314 J  K– 1  mol – 1)ln(0.03)
Srxn
– Rln K
T = 770 K (500 C)
Problem 9.95.
\The equilibrium constant expression for our generalized reaction, A + B  C + D, is
(C) c (D)d 
K =  a b  [equation (9.47)]
(A) (B) eq
Le Chatelier’s principle says that if we disturb the system at equilibrium, that is, when the
concentrations give the correct K, by increasing the concentration of a reactant, the system will
respond by reacting to use up some of this reactant (the disturbance), in order to restore
equilibrium. In the equation, you see that increasing the concentration of a reactant, say A,
without changing the concentrations of any other species, results in a reaction quotient, Q, with a
numeric value less than K. In order to restore the system to a state where Q equals K, the
numerator must increase and the denominator decrease. That is, some of the added A (as well as
some B) will react to form more C and D. You can make similar arguments for any change in the
concentrations, so Le Chatelier’s principle and the equilibrium constant expression give the same
directionality for the change that occurs to restore equilibrium. The power of the equilibrium
constant is that it allows you also to quantify the change.
When a system at equilibrium is disturbed by changing its temperature, Le Chatelier’s principle
says that the reaction will proceed to form more products if the reaction is endothermic,
Hrxn > 0, and will proceed in reverse to form more reactants if the reaction is exothermic,
Hrxn < 0. The thermodynamic relationship between the equilibrium constant and Hrxn is:
o 
o
 H rxn
1
Srxn
lnK = – 
[equation (9.55)]
 +
R
 R T
Here we see that lnK is a linear function of 1/T, if Hrxn and Srxn are constants, independent of
temperature. As T increases, 1/T decreases and we can ask what happens to lnK (or K). The
direction of change of lnK, depends on the sign of the coefficient of 1/T, that is, –Hrxn/R. It’s
easiest to see the direction of the effect on plots of lnK vs. 1/T, as shown here.
120
ACS Chemistry FROG
Chapter 9
Chemical Equilibria
If Hrxn > 0, the coefficient is negative and the slope of the line is negative; lnK increases as T
increases (as 1/T decreases). Conversely, if Hrxn < 0, the coefficient is positive and the slope of
the line is positive; lnK increases as T decreases (as 1/T increases). Here we see that increasing
the temperature of an endothermic reaction system at equilibrium, increases its equilibrium
constant, that is, increases the concentrations of products relative to the concentrations of
reactants. This is the direction of change predicted by Le Chatelier’s principle. Increasing the
temperature of an exothermic reaction system at equilibrium, decreases its equilibrium constant,
that is, increases the concentrations of reactants relative to the concentrations of products, just as
Le Chatelier’s principle also predicts.
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