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Chemistry: Atoms First

15.3 Multiple Equilibria

Chemistry: Atoms First15.3 Multiple Equilibria
  1. Preface
  2. 1 Essential Ideas
    1. Introduction
    2. 1.1 Chemistry in Context
    3. 1.2 Phases and Classification of Matter
    4. 1.3 Physical and Chemical Properties
    5. 1.4 Measurements
    6. 1.5 Measurement Uncertainty, Accuracy, and Precision
    7. 1.6 Mathematical Treatment of Measurement Results
    8. Key Terms
    9. Key Equations
    10. Summary
    11. Exercises
  3. 2 Atoms, Molecules, and Ions
    1. Introduction
    2. 2.1 Early Ideas in Atomic Theory
    3. 2.2 Evolution of Atomic Theory
    4. 2.3 Atomic Structure and Symbolism
    5. 2.4 Chemical Formulas
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  4. 3 Electronic Structure and Periodic Properties of Elements
    1. Introduction
    2. 3.1 Electromagnetic Energy
    3. 3.2 The Bohr Model
    4. 3.3 Development of Quantum Theory
    5. 3.4 Electronic Structure of Atoms (Electron Configurations)
    6. 3.5 Periodic Variations in Element Properties
    7. 3.6 The Periodic Table
    8. 3.7 Molecular and Ionic Compounds
    9. Key Terms
    10. Key Equations
    11. Summary
    12. Exercises
  5. 4 Chemical Bonding and Molecular Geometry
    1. Introduction
    2. 4.1 Ionic Bonding
    3. 4.2 Covalent Bonding
    4. 4.3 Chemical Nomenclature
    5. 4.4 Lewis Symbols and Structures
    6. 4.5 Formal Charges and Resonance
    7. 4.6 Molecular Structure and Polarity
    8. Key Terms
    9. Key Equations
    10. Summary
    11. Exercises
  6. 5 Advanced Theories of Bonding
    1. Introduction
    2. 5.1 Valence Bond Theory
    3. 5.2 Hybrid Atomic Orbitals
    4. 5.3 Multiple Bonds
    5. 5.4 Molecular Orbital Theory
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  7. 6 Composition of Substances and Solutions
    1. Introduction
    2. 6.1 Formula Mass and the Mole Concept
    3. 6.2 Determining Empirical and Molecular Formulas
    4. 6.3 Molarity
    5. 6.4 Other Units for Solution Concentrations
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  8. 7 Stoichiometry of Chemical Reactions
    1. Introduction
    2. 7.1 Writing and Balancing Chemical Equations
    3. 7.2 Classifying Chemical Reactions
    4. 7.3 Reaction Stoichiometry
    5. 7.4 Reaction Yields
    6. 7.5 Quantitative Chemical Analysis
    7. Key Terms
    8. Key Equations
    9. Summary
    10. Exercises
  9. 8 Gases
    1. Introduction
    2. 8.1 Gas Pressure
    3. 8.2 Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law
    4. 8.3 Stoichiometry of Gaseous Substances, Mixtures, and Reactions
    5. 8.4 Effusion and Diffusion of Gases
    6. 8.5 The Kinetic-Molecular Theory
    7. 8.6 Non-Ideal Gas Behavior
    8. Key Terms
    9. Key Equations
    10. Summary
    11. Exercises
  10. 9 Thermochemistry
    1. Introduction
    2. 9.1 Energy Basics
    3. 9.2 Calorimetry
    4. 9.3 Enthalpy
    5. 9.4 Strengths of Ionic and Covalent Bonds
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  11. 10 Liquids and Solids
    1. Introduction
    2. 10.1 Intermolecular Forces
    3. 10.2 Properties of Liquids
    4. 10.3 Phase Transitions
    5. 10.4 Phase Diagrams
    6. 10.5 The Solid State of Matter
    7. 10.6 Lattice Structures in Crystalline Solids
    8. Key Terms
    9. Key Equations
    10. Summary
    11. Exercises
  12. 11 Solutions and Colloids
    1. Introduction
    2. 11.1 The Dissolution Process
    3. 11.2 Electrolytes
    4. 11.3 Solubility
    5. 11.4 Colligative Properties
    6. 11.5 Colloids
    7. Key Terms
    8. Key Equations
    9. Summary
    10. Exercises
  13. 12 Thermodynamics
    1. Introduction
    2. 12.1 Spontaneity
    3. 12.2 Entropy
    4. 12.3 The Second and Third Laws of Thermodynamics
    5. 12.4 Free Energy
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  14. 13 Fundamental Equilibrium Concepts
    1. Introduction
    2. 13.1 Chemical Equilibria
    3. 13.2 Equilibrium Constants
    4. 13.3 Shifting Equilibria: Le Châtelier’s Principle
    5. 13.4 Equilibrium Calculations
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  15. 14 Acid-Base Equilibria
    1. Introduction
    2. 14.1 Brønsted-Lowry Acids and Bases
    3. 14.2 pH and pOH
    4. 14.3 Relative Strengths of Acids and Bases
    5. 14.4 Hydrolysis of Salt Solutions
    6. 14.5 Polyprotic Acids
    7. 14.6 Buffers
    8. 14.7 Acid-Base Titrations
    9. Key Terms
    10. Key Equations
    11. Summary
    12. Exercises
  16. 15 Equilibria of Other Reaction Classes
    1. Introduction
    2. 15.1 Precipitation and Dissolution
    3. 15.2 Lewis Acids and Bases
    4. 15.3 Multiple Equilibria
    5. Key Terms
    6. Key Equations
    7. Summary
    8. Exercises
  17. 16 Electrochemistry
    1. Introduction
    2. 16.1 Balancing Oxidation-Reduction Reactions
    3. 16.2 Galvanic Cells
    4. 16.3 Standard Reduction Potentials
    5. 16.4 The Nernst Equation
    6. 16.5 Batteries and Fuel Cells
    7. 16.6 Corrosion
    8. 16.7 Electrolysis
    9. Key Terms
    10. Key Equations
    11. Summary
    12. Exercises
  18. 17 Kinetics
    1. Introduction
    2. 17.1 Chemical Reaction Rates
    3. 17.2 Factors Affecting Reaction Rates
    4. 17.3 Rate Laws
    5. 17.4 Integrated Rate Laws
    6. 17.5 Collision Theory
    7. 17.6 Reaction Mechanisms
    8. 17.7 Catalysis
    9. Key Terms
    10. Key Equations
    11. Summary
    12. Exercises
  19. 18 Representative Metals, Metalloids, and Nonmetals
    1. Introduction
    2. 18.1 Periodicity
    3. 18.2 Occurrence and Preparation of the Representative Metals
    4. 18.3 Structure and General Properties of the Metalloids
    5. 18.4 Structure and General Properties of the Nonmetals
    6. 18.5 Occurrence, Preparation, and Compounds of Hydrogen
    7. 18.6 Occurrence, Preparation, and Properties of Carbonates
    8. 18.7 Occurrence, Preparation, and Properties of Nitrogen
    9. 18.8 Occurrence, Preparation, and Properties of Phosphorus
    10. 18.9 Occurrence, Preparation, and Compounds of Oxygen
    11. 18.10 Occurrence, Preparation, and Properties of Sulfur
    12. 18.11 Occurrence, Preparation, and Properties of Halogens
    13. 18.12 Occurrence, Preparation, and Properties of the Noble Gases
    14. Key Terms
    15. Summary
    16. Exercises
  20. 19 Transition Metals and Coordination Chemistry
    1. Introduction
    2. 19.1 Occurrence, Preparation, and Properties of Transition Metals and Their Compounds
    3. 19.2 Coordination Chemistry of Transition Metals
    4. 19.3 Spectroscopic and Magnetic Properties of Coordination Compounds
    5. Key Terms
    6. Summary
    7. Exercises
  21. 20 Nuclear Chemistry
    1. Introduction
    2. 20.1 Nuclear Structure and Stability
    3. 20.2 Nuclear Equations
    4. 20.3 Radioactive Decay
    5. 20.4 Transmutation and Nuclear Energy
    6. 20.5 Uses of Radioisotopes
    7. 20.6 Biological Effects of Radiation
    8. Key Terms
    9. Key Equations
    10. Summary
    11. Exercises
  22. 21 Organic Chemistry
    1. Introduction
    2. 21.1 Hydrocarbons
    3. 21.2 Alcohols and Ethers
    4. 21.3 Aldehydes, Ketones, Carboxylic Acids, and Esters
    5. 21.4 Amines and Amides
    6. Key Terms
    7. Summary
    8. Exercises
  23. A | The Periodic Table
  24. B | Essential Mathematics
  25. C | Units and Conversion Factors
  26. D | Fundamental Physical Constants
  27. E | Water Properties
  28. F | Composition of Commercial Acids and Bases
  29. G | Standard Thermodynamic Properties for Selected Substances
  30. H | Ionization Constants of Weak Acids
  31. I | Ionization Constants of Weak Bases
  32. J | Solubility Products
  33. K | Formation Constants for Complex Ions
  34. L | Standard Electrode (Half-Cell) Potentials
  35. M | Half-Lives for Several Radioactive Isotopes
  36. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
    11. Chapter 11
    12. Chapter 12
    13. Chapter 13
    14. Chapter 14
    15. Chapter 15
    16. Chapter 16
    17. Chapter 17
    18. Chapter 18
    19. Chapter 19
    20. Chapter 20
    21. Chapter 21
  37. Index

Learning Objectives

By the end of this section, you will be able to:
  • Describe examples of systems involving two (or more) simultaneous chemical equilibria
  • Calculate reactant and product concentrations for multiple equilibrium systems
  • Compare dissolution and weak electrolyte formation

There are times when one equilibrium reaction does not adequately describe the system being studied. Sometimes we have more than one type of equilibrium occurring at once (for example, an acid-base reaction and a precipitation reaction).

The ocean is a unique example of a system with multiple equilibria, or multiple states of solubility equilibria working simultaneously. Carbon dioxide in the air dissolves in sea water, forming carbonic acid (H2CO3). The carbonic acid then ionizes to form hydrogen ions and bicarbonate ions (HCO3),(HCO3), which can further ionize into more hydrogen ions and carbonate ions (CO32−):(CO32−):

CO2(g)CO2(aq)CO2(g)CO2(aq)
CO2(aq)+H2OH2CO3(aq)CO2(aq)+H2OH2CO3(aq)
H2CO3(aq)H+(aq)+HCO3(aq)H2CO3(aq)H+(aq)+HCO3(aq)
HCO3(aq)H+(aq)+CO32−(aq)HCO3(aq)H+(aq)+CO32−(aq)

The excess H+ ions make seawater more acidic. Increased ocean acidification can then have negative impacts on reef-building coral, as they cannot absorb the calcium carbonate they need to grow and maintain their skeletons (Figure 15.7). This in turn disrupts the local biosystem that depends upon the health of the reefs for its survival. If enough local reefs are similarly affected, the disruptions to sea life can be felt globally. The world’s oceans are presently in the midst of a period of intense acidification, believed to have begun in the mid-nineteenth century, and which is now accelerating at a rate faster than any change to oceanic pH in the last 20 million years.

This figure contains two photographs of coral reefs. In a, a colorful reef that includes hues of purple and pink corals is shown in blue green water with fish swimming in the background. In b, grey-green mossy looking coral is shown in a blue aquatic environment. This photo does not have the colorful appearance or fish that were shown in figure a.
Figure 15.7 Healthy coral reefs (a) support a dense and diverse array of sea life across the ocean food chain. But when coral are unable to adequately build and maintain their calcium carbonite skeletons because of excess ocean acidification, the unhealthy reef (b) is only capable of hosting a small fraction of the species as before, and the local food chain starts to collapse. (credit a: modification of work by NOAA Photo Library; credit b: modification of work by “prilfish”/Flickr)

Slightly soluble solids derived from weak acids generally dissolve in strong acids, unless their solubility products are extremely small. For example, we can dissolve CuCO3, FeS, and Ca3(PO4)2 in HCl because their basic anions react to form weak acids (H2CO3, H2S, and H2PO4).H2PO4). The resulting decrease in the concentration of the anion results in a shift of the equilibrium concentrations to the right in accordance with Le Châtelier’s principle.

Of particular relevance to us is the dissolution of hydroxylapatite, Ca5(PO4)3OH, in acid. Apatites are a class of calcium phosphate minerals (Figure 15.8); a biological form of hydroxylapatite is found as the principal mineral in the enamel of our teeth. A mixture of hydroxylapatite and water (or saliva) contains an equilibrium mixture of solid Ca5(PO4)3OH and dissolved Ca2+, PO43−,PO43−, and OH ions:

Ca5(PO4)3OH(s)5Ca2+(aq)+3PO43−(aq)+OH(aq)Ca5(PO4)3OH(s)5Ca2+(aq)+3PO43−(aq)+OH(aq)
This figure includes an image of two large light blue apatite crystals in a mineral conglomerate that includes white, grey, and tan crystals. The blue apatite crystals have a dull, dusty, or powdered appearance.
Figure 15.8 Crystal of the mineral hydroxylapatite, Ca5(PO4)3OH, is shown here. Pure apatite is white, but like many other minerals, this sample is colored because of the presence of impurities.

When exposed to acid, phosphate ions react with hydronium ions to form hydrogen phosphate ions and ultimately, phosphoric acid:

PO43−(aq)+H3O+(aq)H2PO42−(aq)+H2O(l)PO43−(aq)+H3O+(aq)H2PO42−(aq)+H2O(l)
H2PO42−(aq)+H3O+(aq)H2PO4(aq)+H2O(l)H2PO42−(aq)+H3O+(aq)H2PO4(aq)+H2O(l)
H2PO4(aq)+H3O+(aq)H3PO4(aq)+H2O(l)H2PO4(aq)+H3O+(aq)H3PO4(aq)+H2O(l)

Hydroxide ion reacts to form water:

OH(aq)+H3O+2H2OOH(aq)+H3O+2H2O

These reactions decrease the phosphate and hydroxide ion concentrations, and additional hydroxylapatite dissolves in an acidic solution in accord with Le Châtelier’s principle. Our teeth develop cavities when acid waste produced by bacteria growing on them causes the hydroxylapatite of the enamel to dissolve. Fluoride toothpastes contain sodium fluoride, NaF, or stannous fluoride [more properly named tin(II) fluoride], SnF2. They function by replacing the OH ion in hydroxylapatite with F ion, producing fluorapatite, Ca5(PO4)3F:

NaF+Ca5(PO4)3OHCa5(PO4)3F+Na++OHNaF+Ca5(PO4)3OHCa5(PO4)3F+Na++OH

The resulting Ca5(PO4)3F is slightly less soluble than Ca5(PO4)3OH, and F is a weaker base than OH. Both of these factors make the fluorapatite more resistant to attack by acids than hydroxylapatite. See the Chemistry in Everyday Life feature on the role of fluoride in preventing tooth decay for more information.

Chemistry in Everyday Life

Role of Fluoride in Preventing Tooth Decay

As we saw previously, fluoride ions help protect our teeth by reacting with hydroxylapatite to form fluorapatite, Ca5(PO4)3F. Since it lacks a hydroxide ion, fluorapatite is more resistant to attacks by acids in our mouths and is thus less soluble, protecting our teeth. Scientists discovered that naturally fluorinated water could be beneficial to your teeth, and so it became common practice to add fluoride to drinking water. Toothpastes and mouthwashes also contain amounts of fluoride (Figure 15.9).

Replace with updated art; figure in “99” art folder is current and correct
Figure 15.9 Fluoride, found in many toothpastes, helps prevent tooth decay (credit: Kerry Ceszyk).

Unfortunately, excess fluoride can negate its advantages. Natural sources of drinking water in various parts of the world have varying concentrations of fluoride, and places where that concentration is high are prone to certain health risks when there is no other source of drinking water. The most serious side effect of excess fluoride is the bone disease, skeletal fluorosis. When excess fluoride is in the body, it can cause the joints to stiffen and the bones to thicken. It can severely impact mobility and can negatively affect the thyroid gland. Skeletal fluorosis is a condition that over 2.7 million people suffer from across the world. So while fluoride can protect our teeth from decay, the US Environmental Protection Agency sets a maximum level of 4 ppm (4 mg/L) of fluoride in drinking water in the US. Fluoride levels in water are not regulated in all countries, so fluorosis is a problem in areas with high levels of fluoride in the groundwater.

When acid rain attacks limestone or marble, which are calcium carbonates, a reaction occurs that is similar to the acid attack on hydroxylapatite. The hydronium ion from the acid rain combines with the carbonate ion from calcium carbonates and forms the hydrogen carbonate ion, a weak acid:

H3O+(aq)+CO32−(aq)HCO3(aq)+H2O(l)H3O+(aq)+CO32−(aq)HCO3(aq)+H2O(l)

Calcium hydrogen carbonate, Ca(HCO3)2, is soluble, so limestone and marble objects slowly dissolve in acid rain.

If we add calcium carbonate to a concentrated acid, hydronium ion reacts with the carbonate ion according to the equation:

2H3O+(aq)+CO32−(aq)H2CO3(aq)+2H2O(l)2H3O+(aq)+CO32−(aq)H2CO3(aq)+2H2O(l)

(Acid rain is usually not sufficiently acidic to cause this reaction; however, laboratory acids are.) The solution may become saturated with the weak electrolyte carbonic acid, which is unstable, and carbon dioxide gas can be evolved:

H2CO3(aq)CO2(g)+H2O(l)H2CO3(aq)CO2(g)+H2O(l)

These reactions decrease the carbonate ion concentration, and additional calcium carbonate dissolves. If enough acid is present, the concentration of carbonate ion is reduced to such a low level that the reaction quotient for the dissolution of calcium carbonate remains less than the solubility product of calcium carbonate, even after all of the calcium carbonate has dissolved.

Example 15.14

Prevention of Precipitation of Mg(OH)2

Calculate the concentration of ammonium ion that is required to prevent the precipitation of Mg(OH)2 in a solution with [Mg2+] = 0.10 M and [NH3] = 0.10 M.

Solution

Two equilibria are involved in this system:

Reaction (1): Mg(OH)2(s)Mg2+(aq)+2OH(aq);Ksp=8.9×10−12Mg(OH)2(s)Mg2+(aq)+2OH(aq);Ksp=8.9×10−12

Reaction (2): NH3(aq)+H2O(l)NH4+(aq)+OH(aq)Kb=1.8×10−5NH3(aq)+H2O(l)NH4+(aq)+OH(aq)Kb=1.8×10−5

To prevent the formation of solid Mg(OH)2, we must adjust the concentration of OH so that the reaction quotient for Equation (1), Q = [Mg2+][OH]2, is less than Ksp for Mg(OH)2. (To simplify the calculation, we determine the concentration of OH when Q = Ksp.) [OH] can be reduced by the addition of NH4+,NH4+, which shifts Reaction (2) to the left and reduces [OH].

  1. Step 1.

    We determine the [OH] at which Q = Ksp when [Mg2+] = 0.10 M:

    Q=[Mg2+][OH]2=(0.10)[OH]2=8.9×10−12Q=[Mg2+][OH]2=(0.10)[OH]2=8.9×10−12
    [OH]=9.4×10−6M[OH]=9.4×10−6M

    Solid Mg(OH)2 will not form in this solution when [OH] is less than 9.4 ×× 10–6 M.

  2. Step 2.

    We calculate the [NH4+][NH4+] needed to decrease [OH] to 9.4 ×× 10–6 M when [NH3] = 0.10.

    Kb=[NH4+][OH][NH3]=[NH4+](9.4×10−6)0.10=1.8×10−5Kb=[NH4+][OH][NH3]=[NH4+](9.4×10−6)0.10=1.8×10−5
    [NH4+]=0.19M[NH4+]=0.19M

When [NH4+][NH4+] equals 0.19 M, [OH] will be 9.4 ×× 10–6 M. Any [NH4+][NH4+] greater than 0.19 M will reduce [OH] below 9.4 ×× 10–6 M and prevent the formation of Mg(OH)2.

Check Your Learning

Consider the two equilibria:
ZnS(s)Zn2+(aq)+S2−(aq)Ksp=1×10−27ZnS(s)Zn2+(aq)+S2−(aq)Ksp=1×10−27
2H2O(l)+H2S(aq)2H3O+(aq)+S2−(aq)K=1.0×10−262H2O(l)+H2S(aq)2H3O+(aq)+S2−(aq)K=1.0×10−26

and calculate the concentration of hydronium ion required to prevent the precipitation of ZnS in a solution that is 0.050 M in Zn2+ and saturated with H2S (0.10 M H2S).

Answer:

[H3O+]>0.2M[H3O+]>0.2M

([S2–] is less than 2 ×× 10–26 M and precipitation of ZnS does not occur.)

Therefore, precise calculations of the solubility of solids from the solubility product are limited to cases in which the only significant reaction occurring when the solid dissolves is the formation of its ions.

Example 15.15

Multiple Equilibria

Unexposed silver halides are removed from photographic film when they react with sodium thiosulfate (Na2S2O3, called hypo) to form the complex ion Ag(S2O3)23−Ag(S2O3)23− (Kf = 4.7 ×× 1013). The reaction with silver bromide is: A chemical reaction is shown using structural formulas. On the left, A g superscript plus is followed by a plus sign, the number 2, and a structure in brackets. The structure is composed of a central S atom which has O atoms single bonded above, right, and below. A second S atom is single bonded to the left. Each of these bonded atoms has 6 dots around it. Outside the brackets is a superscript 2 negative. Following a bidirectional arrow is a structure in brackets with a central A g atom. To the left and right, S atoms are single bonded to the A g atom. Each of these S atoms has four dots around it, and an S atom connected with a single bond moving out from the central A g atom, forming the ends of the structure. Each of these atoms has three O atoms attached with single bonds above, below, and at the end of the structure. Each O atom has six dots around it. Outside the brackets is a superscript 3 negative.

What mass of Na2S2O3 is required to prepare 1.00 L of a solution that will dissolve 1.00 g of AgBr by the formation of Ag(S2O3)23−?Ag(S2O3)23−?

Solution

Two equilibria are involved when AgBr dissolves in a solution containing the S2O32−S2O32− ion:

Reaction (1): AgBr(s)Ag+(aq)+Br(aq)Ksp=5.0×10−13AgBr(s)Ag+(aq)+Br(aq)Ksp=5.0×10−13

Reaction (2): Ag+(aq)+S2O32−(aq)Ag(S2O3)23−(aq)Kf=4.7×1013Ag+(aq)+S2O32−(aq)Ag(S2O3)23−(aq)Kf=4.7×1013

In order for 1.00 g of AgBr to dissolve, the [Ag+] in the solution that results must be low enough for Q for Reaction (1) to be smaller than Ksp for this reaction. We reduce [Ag+] by adding S2O32−S2O32− and thus cause Reaction (2) to shift to the right. We need the following steps to determine what mass of Na2S2O3 is needed to provide the necessary S2O32−.S2O32−.

  1. Step 1.

    We calculate the [Br] produced by the complete dissolution of 1.00 g of AgBr (5.33 ×× 10–3 mol AgBr) in 1.00 L of solution:

    [Br]=5.33×10−3M[Br]=5.33×10−3M
  2. Step 2.

    We use [Br] and Ksp to determine the maximum possible concentration of Ag+ that can be present without causing reprecipitation of AgBr:

    [Ag+]=9.4×10−11M[Ag+]=9.4×10−11M
  3. Step 3.

    We determine the [S2O32−][S2O32−] required to make [Ag+] = 9.4 ×× 10–11 M after the remaining Ag+ ion has reacted with S2O32−S2O32− according to the equation:

    Ag++2S2O32−Ag(S2O3)23−Kf=4.7×1013Ag++2S2O32−Ag(S2O3)23−Kf=4.7×1013

    Because 5.33 ×× 10–3 mol of AgBr dissolves:

    (5.33×10−3)(9.4×10−11)=5.33×10−3molAg(S2O3)23−(5.33×10−3)(9.4×10−11)=5.33×10−3molAg(S2O3)23−

    Thus, at equilibrium: [Ag(S2O3)23−][Ag(S2O3)23−] = 5.33 ×× 10–3 M, [Ag+] = 9.4 ×× 10–11 M, and Q = Kf = 4.7 ×× 1013:

    Kf=[Ag(S2O3)23−][Ag+][S2O32−]2=4.7×1013Kf=[Ag(S2O3)23−][Ag+][S2O32−]2=4.7×1013
    [S2O32−]=1.1×10−3M[S2O32−]=1.1×10−3M

    When [S2O32−][S2O32−] is 1.1 ×× 10–3 M, [Ag+] is 9.4 ×× 10–11 M and all AgBr remains dissolved.

  4. Step 4.

    We determine the total number of moles of S2O32−S2O32− that must be added to the solution. This equals the amount that reacts with Ag+ to form Ag(S2O3)23−Ag(S2O3)23− plus the amount of free S2O32−S2O32− in solution at equilibrium. To form 5.33 ×× 10–3 mol of Ag(S2O3)23−Ag(S2O3)23− requires 2 ×× (5.33 ×× 10–3) mol of S2O32−.S2O32−. In addition, 1.1 ×× 10–3 mol of unreacted S2O32−S2O32− is present (Step 3). Thus, the total amount of S2O32−S2O32− that must be added is:


    2×(5.33×10−3molS2O32−)+1.1×10−3molS2O32−=1.18×10−2molS2O32−2×(5.33×10−3molS2O32−)+1.1×10−3molS2O32−=1.18×10−2molS2O32−
  5. Step 5.

    We determine the mass of Na2S2O3 required to give 1.18 ×× 10–2 mol S2O32−S2O32− using the molar mass of Na2S2O3:

    1.18×10−2molS2O32−×158.1gNa2S2O31molNa2S2O3=1.9gNa2S2O31.18×10−2molS2O32−×158.1gNa2S2O31molNa2S2O3=1.9gNa2S2O3

    Thus, 1.00 L of a solution prepared from 1.9 g Na2S2O3 dissolves 1.0 g of AgBr.

Check Your Learning

AgCl(s), silver chloride, is well known to have a very low solubility: AgCl(s)Ag+(aq)+Cl(aq),AgCl(s)Ag+(aq)+Cl(aq), Ksp = 1.6 ×× 10–10. Adding ammonia significantly increases the solubility of AgCl because a complex ion is formed: Ag+(aq)+2NH3(aq)Ag(NH3)2+(aq),Ag+(aq)+2NH3(aq)Ag(NH3)2+(aq), Kf = 1.7 ×× 107. What mass of NH3 is required to prepare 1.00 L of solution that will dissolve 2.00 g of AgCl by formation of Ag(NH3)2+?Ag(NH3)2+?

Answer:

1.00 L of a solution prepared with 4.81 g NH3 dissolves 2.0 g of AgCl.

Dissolution versus Weak Electrolyte Formation

We can determine how to shift the concentration of ions in the equilibrium between a slightly soluble solid and a solution of its ions by applying Le Châtelier’s principle. For example, one way to control the concentration of manganese(II) ion, Mn2+, in a solution is to adjust the pH of the solution and, consequently, to manipulate the equilibrium between the slightly soluble solid manganese(II) hydroxide, manganese(II) ion, and hydroxide ion:

Mn(OH)2(s)Mn2+(aq)+2OH(aq)Ksp=[Mn2+][OH]2Mn(OH)2(s)Mn2+(aq)+2OH(aq)Ksp=[Mn2+][OH]2

This could be important to a laundromat because clothing washed in water that has a manganese concentration exceeding 0.1 mg per liter may be stained by the manganese. We can reduce the concentration of manganese by increasing the concentration of hydroxide ion. We could add, for example, a small amount of NaOH or some other base such as the silicates found in many laundry detergents. As the concentration of OH ion increases, the equilibrium responds by shifting to the left and reducing the concentration of Mn2+ ion while increasing the amount of solid Mn(OH)2 in the equilibrium mixture, as predicted by Le Châtelier’s principle.

Example 15.16

Solubility Equilibrium of a Slightly Soluble Solid

What is the effect on the amount of solid Mg(OH)2 that dissolves and the concentrations of Mg2+ and OH when each of the following are added to a mixture of solid Mg(OH)2 in water at equilibrium?

(a) MgCl2

(b) KOH

(c) an acid

(d) NaNO3

(e) Mg(OH)2

Solution

The equilibrium among solid Mg(OH)2 and a solution of Mg2+ and OH is:
Mg(OH)2(s)Mg2+(aq)+2OH(aq)Mg(OH)2(s)Mg2+(aq)+2OH(aq)

(a) The reaction shifts to the left to relieve the stress produced by the additional Mg2+ ion, in accordance with Le Châtelier’s principle. In quantitative terms, the added Mg2+ causes the reaction quotient to be larger than the solubility product (Q > Ksp), and Mg(OH)2 forms until the reaction quotient again equals Ksp. At the new equilibrium, [OH] is less and [Mg2+] is greater than in the solution of Mg(OH)2 in pure water. More solid Mg(OH)2 is present.

(b) The reaction shifts to the left to relieve the stress of the additional OH ion. Mg(OH)2 forms until the reaction quotient again equals Ksp. At the new equilibrium, [OH] is greater and [Mg2+] is less than in the solution of Mg(OH)2 in pure water. More solid Mg(OH)2 is present.

(c) The concentration of OH is reduced as the OH reacts with the acid. The reaction shifts to the right to relieve the stress of less OH ion. In quantitative terms, the decrease in the OH concentration causes the reaction quotient to be smaller than the solubility product (Q < Ksp), and additional Mg(OH)2 dissolves until the reaction quotient again equals Ksp. At the new equilibrium, [OH] is less and [Mg2+] is greater than in the solution of Mg(OH)2 in pure water. More Mg(OH)2 is dissolved.

(d) NaNO3 contains none of the species involved in the equilibrium, so we should expect that it has no appreciable effect on the concentrations of Mg2+ and OH. (As we have seen previously, dissolved salts change the activities of the ions of an electrolyte. However, the salt effect is generally small, and we shall neglect the slight errors that may result from it.)

(e) The addition of solid Mg(OH)2 has no effect on the solubility of Mg(OH)2 or on the concentration of Mg2+ and OH. The concentration of Mg(OH)2 does not appear in the equation for the reaction quotient:

Q=[Mg2+][OH]2Q=[Mg2+][OH]2

Thus, changing the amount of solid magnesium hydroxide in the mixture has no effect on the value of Q, and no shift is required to restore Q to the value of the equilibrium constant.

Check Your Learning

What is the effect on the amount of solid NiCO3 that dissolves and the concentrations of Ni2+ and CO32−CO32− when each of the following are added to a mixture of the slightly soluble solid NiCO3 and water at equilibrium?

(a) Ni(NO3)2

(b) KClO4

(c) NiCO3

(d) K2CO3

(e) HNO3 (reacts with carbonate giving HCO3HCO3 or H2O and CO2)

Answer:

(a) mass of NiCO3(s) increases, [Ni2+] increases, [CO32−][CO32−] decreases; (b) no appreciable effect; (c) no effect except to increase the amount of solid NiCO3; (d) mass of NiCO3(s) increases, [Ni2+] decreases, [CO32−][CO32−] increases; (e) mass of NiCO3(s) decreases, [Ni2+] increases, [CO32−][CO32−] decreases

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