Skip to Content
OpenStax Logo
Chemistry

15.1 Precipitation and Dissolution

Chemistry15.1 Precipitation and Dissolution
  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. 2.5 The Periodic Table
    7. 2.6 Molecular and Ionic Compounds
    8. 2.7 Chemical Nomenclature
    9. Key Terms
    10. Key Equations
    11. Summary
    12. Exercises
  4. 3 Composition of Substances and Solutions
    1. Introduction
    2. 3.1 Formula Mass and the Mole Concept
    3. 3.2 Determining Empirical and Molecular Formulas
    4. 3.3 Molarity
    5. 3.4 Other Units for Solution Concentrations
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  5. 4 Stoichiometry of Chemical Reactions
    1. Introduction
    2. 4.1 Writing and Balancing Chemical Equations
    3. 4.2 Classifying Chemical Reactions
    4. 4.3 Reaction Stoichiometry
    5. 4.4 Reaction Yields
    6. 4.5 Quantitative Chemical Analysis
    7. Key Terms
    8. Key Equations
    9. Summary
    10. Exercises
  6. 5 Thermochemistry
    1. Introduction
    2. 5.1 Energy Basics
    3. 5.2 Calorimetry
    4. 5.3 Enthalpy
    5. Key Terms
    6. Key Equations
    7. Summary
    8. Exercises
  7. 6 Electronic Structure and Periodic Properties of Elements
    1. Introduction
    2. 6.1 Electromagnetic Energy
    3. 6.2 The Bohr Model
    4. 6.3 Development of Quantum Theory
    5. 6.4 Electronic Structure of Atoms (Electron Configurations)
    6. 6.5 Periodic Variations in Element Properties
    7. Key Terms
    8. Key Equations
    9. Summary
    10. Exercises
  8. 7 Chemical Bonding and Molecular Geometry
    1. Introduction
    2. 7.1 Ionic Bonding
    3. 7.2 Covalent Bonding
    4. 7.3 Lewis Symbols and Structures
    5. 7.4 Formal Charges and Resonance
    6. 7.5 Strengths of Ionic and Covalent Bonds
    7. 7.6 Molecular Structure and Polarity
    8. Key Terms
    9. Key Equations
    10. Summary
    11. Exercises
  9. 8 Advanced Theories of Covalent Bonding
    1. Introduction
    2. 8.1 Valence Bond Theory
    3. 8.2 Hybrid Atomic Orbitals
    4. 8.3 Multiple Bonds
    5. 8.4 Molecular Orbital Theory
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  10. 9 Gases
    1. Introduction
    2. 9.1 Gas Pressure
    3. 9.2 Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law
    4. 9.3 Stoichiometry of Gaseous Substances, Mixtures, and Reactions
    5. 9.4 Effusion and Diffusion of Gases
    6. 9.5 The Kinetic-Molecular Theory
    7. 9.6 Non-Ideal Gas Behavior
    8. Key Terms
    9. Key Equations
    10. Summary
    11. 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 Kinetics
    1. Introduction
    2. 12.1 Chemical Reaction Rates
    3. 12.2 Factors Affecting Reaction Rates
    4. 12.3 Rate Laws
    5. 12.4 Integrated Rate Laws
    6. 12.5 Collision Theory
    7. 12.6 Reaction Mechanisms
    8. 12.7 Catalysis
    9. Key Terms
    10. Key Equations
    11. Summary
    12. 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 Thermodynamics
    1. Introduction
    2. 16.1 Spontaneity
    3. 16.2 Entropy
    4. 16.3 The Second and Third Laws of Thermodynamics
    5. 16.4 Free Energy
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  18. 17 Electrochemistry
    1. Introduction
    2. 17.1 Balancing Oxidation-Reduction Reactions
    3. 17.2 Galvanic Cells
    4. 17.3 Standard Reduction Potentials
    5. 17.4 The Nernst Equation
    6. 17.5 Batteries and Fuel Cells
    7. 17.6 Corrosion
    8. 17.7 Electrolysis
    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 Organic Chemistry
    1. Introduction
    2. 20.1 Hydrocarbons
    3. 20.2 Alcohols and Ethers
    4. 20.3 Aldehydes, Ketones, Carboxylic Acids, and Esters
    5. 20.4 Amines and Amides
    6. Key Terms
    7. Summary
    8. Exercises
  22. 21 Nuclear Chemistry
    1. Introduction
    2. 21.1 Nuclear Structure and Stability
    3. 21.2 Nuclear Equations
    4. 21.3 Radioactive Decay
    5. 21.4 Transmutation and Nuclear Energy
    6. 21.5 Uses of Radioisotopes
    7. 21.6 Biological Effects of Radiation
    8. Key Terms
    9. Key Equations
    10. Summary
    11. 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:
  • Write chemical equations and equilibrium expressions representing solubility equilibria
  • Carry out equilibrium computations involving solubility, equilibrium expressions, and solute concentrations

The preservation of medical laboratory blood samples, mining of sea water for magnesium, formulation of over-the-counter medicines such as Milk of Magnesia and antacids, and treating the presence of hard water in your home’s water supply are just a few of the many tasks that involve controlling the equilibrium between a slightly soluble ionic solid and an aqueous solution of its ions.

In some cases, we want to prevent dissolution from occurring. Tooth decay, for example, occurs when the calcium hydroxylapatite, which has the formula Ca5(PO4)3(OH), in our teeth dissolves. The dissolution process is aided when bacteria in our mouths feast on the sugars in our diets to produce lactic acid, which reacts with the hydroxide ions in the calcium hydroxylapatite. Preventing the dissolution prevents the decay. On the other hand, sometimes we want a substance to dissolve. We want the calcium carbonate in a chewable antacid to dissolve because the CO32−CO32− ions produced in this process help soothe an upset stomach.

In this section, we will find out how we can control the dissolution of a slightly soluble ionic solid by the application of Le Châtelier’s principle. We will also learn how to use the equilibrium constant of the reaction to determine the concentration of ions present in a saturated solution.

The Solubility Product

Silver chloride is what’s known as a sparingly soluble ionic solid (Figure 15.2). Recall from the solubility rules in an earlier chapter that halides of Ag+ are not normally soluble. However, when we add an excess of solid AgCl to water, it dissolves to a small extent and produces a mixture consisting of a very dilute solution of Ag+ and Cl ions in equilibrium with undissolved silver chloride:

AgCl(s)precipitationdissolutionAg+(aq)+Cl(aq)AgCl(s)precipitationdissolutionAg+(aq)+Cl(aq)

This equilibrium, like other equilibria, is dynamic; some of the solid AgCl continues to dissolve, but at the same time, Ag+ and Cl ions in the solution combine to produce an equal amount of the solid. At equilibrium, the opposing processes have equal rates.

Two beakers are shown with a bidirectional arrow between them. Both beakers are just over half filled with a clear, colorless liquid. The beaker on the left shows a cubic structure composed of alternating green and slightly larger grey spheres. Evenly distributed in the region outside, 11 space filling models are shown. These are each composed of a central red sphere with two smaller white spheres attached in a bent arrangement. In the beaker on the right, the green and grey spheres are no longer connected in a cubic structure. Nine green spheres, 10 grey spheres, and 11 red and white molecules are evenly mixed and distributed throughout the liquid in the beaker.
Figure 15.2 Silver chloride is a sparingly soluble ionic solid. When it is added to water, it dissolves slightly and produces a mixture consisting of a very dilute solution of Ag+ and Cl ions in equilibrium with undissolved silver chloride.

The equilibrium constant for the equilibrium between a slightly soluble ionic solid and a solution of its ions is called the solubility product (Ksp) of the solid. Recall from the chapter on solutions and colloids that we use an ion’s concentration as an approximation of its activity in a dilute solution. For silver chloride, at equilibrium:

AgCl(s)Ag+(aq)+Cl(aq)Ksp=[Ag+(aq)][Cl(aq)]AgCl(s)Ag+(aq)+Cl(aq)Ksp=[Ag+(aq)][Cl(aq)]

Note that the Ksp expression does not contain a term in the denominator for the concentration of the reactant, AgCl. According to the guidelines for deriving mass-action expressions described in an earlier chapter on equilibrium, only gases and solutes are represented. Solids and liquids are assigned concentration values of one and thus do not appear in equilibrium constant expressions; therefore, [AgCl] does not appear in the expression for Ksp.

Some common solubility products are listed in Table 15.1 according to their Ksp values, whereas a more extensive compilation of solubility products appears in Appendix J. Each of these equilibrium constants is much smaller than 1 because the compounds listed are only slightly soluble. A small Ksp represents a system in which the equilibrium lies to the left, so that relatively few hydrated ions would be present in a saturated solution.

Common Solubility Products by Decreasing Equilibrium Constants
Substance Ksp at 25 °C
CuCl 1.2 ×× 10–6
CuBr 6.27 ×× 10–9
AgI 1.5 ×× 10–16
PbS 7 ×× 10–29
Al(OH)3 2 ×× 10–32
Fe(OH)3 4 ×× 10–38
Table 15.1

Example 15.1

Writing Equations and Solubility Products

Write the ionic equation for the dissolution and the solubility product expression for each of the following slightly soluble ionic compounds:

(a) AgI, silver iodide, a solid with antiseptic properties

(b) CaCO3, calcium carbonate, the active ingredient in many over-the-counter chewable antacids

(c) Mg(OH)2, magnesium hydroxide, the active ingredient in Milk of Magnesia

(d) Mg(NH4)PO4, magnesium ammonium phosphate, an essentially insoluble substance used in tests for magnesium

(e) Ca5(PO4)3OH, the mineral apatite, a source of phosphate for fertilizers

(Hint: When determining how to break (d) and (e) up into ions, refer to the list of polyatomic ions in the section on chemical nomenclature.)

Solution

(a) AgI(s)Ag+(aq)+I(aq)Ksp=[Ag+][I]AgI(s)Ag+(aq)+I(aq)Ksp=[Ag+][I]

(b) CaCO3(s)Ca2+(aq)+CO32−(aq)Ksp=[Ca2+][CO32−]CaCO3(s)Ca2+(aq)+CO32−(aq)Ksp=[Ca2+][CO32−]

(c) Mg(OH)2(s)Mg2+(aq)+2OH(aq)Ksp=[Mg2+][OH]2Mg(OH)2(s)Mg2+(aq)+2OH(aq)Ksp=[Mg2+][OH]2

(d) Mg(NH4)PO4(s)Mg2+(aq)+NH4+(aq)+PO43−(aq)Ksp=[Mg2+][NH4+][PO43−]Mg(NH4)PO4(s)Mg2+(aq)+NH4+(aq)+PO43−(aq)Ksp=[Mg2+][NH4+][PO43−]

(e) Ca5(PO4)3OH(s)5Ca2+(aq)+3PO43−(aq)+OH(aq)Ksp=[Ca2+]5[PO43−]3[OH]Ca5(PO4)3OH(s)5Ca2+(aq)+3PO43−(aq)+OH(aq)Ksp=[Ca2+]5[PO43−]3[OH]

Check Your Learning

Write the ionic equation for the dissolution and the solubility product for each of the following slightly soluble compounds:

(a) BaSO4

(b) Ag2SO4

(c) Al(OH)3

(d) Pb(OH)Cl

Answer:

(a) BaSO4(s)Ba2+(aq)+SO42−(aq)Ksp=[Ba2+][SO42];BaSO4(s)Ba2+(aq)+SO42−(aq)Ksp=[Ba2+][SO42]; (b) Ag2SO4(s)2Ag+(aq)+SO42−(aq)Ksp=[Ag+]2[SO42−];Ag2SO4(s)2Ag+(aq)+SO42−(aq)Ksp=[Ag+]2[SO42−]; (c) Al(OH)3(s)Al3+(aq)+3OH(aq)Ksp=[Al3+][OH]3;Al(OH)3(s)Al3+(aq)+3OH(aq)Ksp=[Al3+][OH]3; (d) Pb(OH)Cl(s)Pb2+(aq)+OH(aq)+Cl(aq)Ksp=[Pb2+][OH][Cl]Pb(OH)Cl(s)Pb2+(aq)+OH(aq)+Cl(aq)Ksp=[Pb2+][OH][Cl]

Now we will extend the discussion of Ksp and show how the solubility product is determined from the solubility of its ions, as well as how Ksp can be used to determine the molar solubility of a substance.

Ksp and Solubility

The Ksp of a slightly soluble ionic compound may be simply related to its measured solubility provided the dissolution process involves only dissociation and solvation, for example:

MpXq(s)pMm+(aq)+qXn−(aq)MpXq(s)pMm+(aq)+qXn−(aq)

For cases such as these, one may derive Ksp values from provided solubilities, or vice-versa. Calculations of this sort are most conveniently performed using a compound’s molar solubility, measured as moles of dissolved solute per liter of saturated solution.

Example 15.2

Calculation of Ksp from Equilibrium Concentrations

We began the chapter with an informal discussion of how the mineral fluorite (Figure 15.1) is formed. Fluorite, CaF2, is a slightly soluble solid that dissolves according to the equation:
CaF2(s)Ca2+(aq)+2F(aq)CaF2(s)Ca2+(aq)+2F(aq)

The concentration of Ca2+ in a saturated solution of CaF2 is 2.15 ×× 10–4 M; therefore, that of F is 4.30 ×× 10–4 M, that is, twice the concentration of Ca2+. What is the solubility product of fluorite?

Solution

First, write out the Ksp expression, then substitute in concentrations and solve for Ksp:
CaF2(s)Ca2+(aq)+2F(aq)CaF2(s)Ca2+(aq)+2F(aq)

A saturated solution is a solution at equilibrium with the solid. Thus:

Ksp=[Ca2+][F]2=(2.15×104)(4.30×104)2=3.98×1011Ksp=[Ca2+][F]2=(2.15×104)(4.30×104)2=3.98×1011

As with other equilibrium constants, we do not include units with Ksp.

Check Your Learning

In a saturated solution that is in contact with solid Mg(OH)2, the concentration of Mg2+ is 1.31 ×× 10–4 M. What is the solubility product for Mg(OH)2?
Mg(OH)2(s)Mg2+(aq)+2OH(aq)Mg(OH)2(s)Mg2+(aq)+2OH(aq)

Answer:

8.99 ×× 10–12

Example 15.3

Determination of Molar Solubility from Ksp

The Ksp of copper(I) bromide, CuBr, is 6.3 ×× 10–9. Calculate the molar solubility of copper bromide.

Solution

The solubility product of copper(I) bromide is 6.3 ×× 10–9.

The reaction is:

CuBr(s)Cu+(aq)+Br(aq)CuBr(s)Cu+(aq)+Br(aq)

First, write out the solubility product expression:

Ksp=[Cu+][Br]Ksp=[Cu+][Br]

Create an ICE table (as introduced in the chapter on fundamental equilibrium concepts), leaving the CuBr column empty as it is a solid and does not contribute to the Ksp:

This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header of, “C u B r equilibrium arrow C u superscript positive sign plus B r superscript negative sign.” Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0, x, 0 plus x equals x. The third column has the following 0, x, 0 plus x equals x.

At equilibrium:

Ksp=[Cu+][Br]Ksp=[Cu+][Br]
6.3×109=(x)(x)=x26.3×109=(x)(x)=x2
x=(6.3×109)=7.9×105x=(6.3×109)=7.9×105

Therefore, the molar solubility of CuBr is 7.9 ×× 10–5 M.

Check Your Learning

The Ksp of AgI is 1.5 ×× 10–16. Calculate the molar solubility of silver iodide.

Answer:

1.2 ×× 10–8 M

Example 15.4

Determination of Molar Solubility from Ksp, Part II

The Ksp of calcium hydroxide, Ca(OH)2, is 1.3 ×× 10–6. Calculate the molar solubility of calcium hydroxide.

Solution

The solubility product of calcium hydroxide is 1.3 ×× 10–6.

The reaction is:

Ca(OH)2(s)Ca2+(aq)+2OH(aq)Ca(OH)2(s)Ca2+(aq)+2OH(aq)

First, write out the solubility product expression:

Ksp=[Ca2+][OH]2Ksp=[Ca2+][OH]2

Create an ICE table, leaving the Ca(OH)2 column empty as it is a solid and does not contribute to the Ksp:

This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header of, “C a ( O H ) subscript 2 equilibrium arrow C a superscript 2 positive sign plus 2 O H superscript negative sign.” Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0, x, and 0 plus x equals x. The third column has the following 0, 2 x, and 0 plus 2 x equals 2 x.

At equilibrium:

Ksp=[Ca2+][OH]2Ksp=[Ca2+][OH]2
1.3×106=(x)(2x)2=(x)(4x2)=4x31.3×106=(x)(2x)2=(x)(4x2)=4x3
x=1.3×10643=6.9×103x=1.3×10643=6.9×103

Therefore, the molar solubility of Ca(OH)2 is 6.9 ×× 10–3 M.

Check Your Learning

The Ksp of PbI2 is 1.4 ×× 10–8. Calculate the molar solubility of lead(II) iodide.

Answer:

1.5 ×× 10–3 M

Note that solubility is not always given as a molar value. When the solubility of a compound is given in some unit other than moles per liter, we must convert the solubility into moles per liter (i.e., molarity) in order to use it in the solubility product expression. Example 15.5 shows how to perform those unit conversions before determining the solubility product equilibrium.

Example 15.5

Determination of Ksp from Gram Solubility

Many of the pigments used by artists in oil-based paints (Figure 15.3) are sparingly soluble in water. For example, the solubility of the artist’s pigment chrome yellow, PbCrO4, is 4.6 ×× 10–6 g/L. Determine the solubility product for PbCrO4.
A photograph is shown of a portion of an oil painting which reveals colors of orange, brown, yellow, green, blue, and purple colors in its strokes. A few water droplets rest on the surface.
Figure 15.3 Oil paints contain pigments that are very slightly soluble in water. In addition to chrome yellow (PbCrO4), examples include Prussian blue (Fe7(CN)18), the reddish-orange color vermilion (HgS), and green color veridian (Cr2O3). (credit: Sonny Abesamis)

Solution

We are given the solubility of PbCrO4 in grams per liter. If we convert this solubility into moles per liter, we can find the equilibrium concentrations of Pb2+ and CrO42−,CrO42−, then Ksp: This figure shows four horizontally oriented rectangles. The first three from the left are shaded green and the last one at the right is shaded white. Right pointing arrows between the rectangles are labeled “1,” “2,” and “3” moving left to right across the diagram. The first rectangle is labeled “Solubility of P b C r O subscript 4, in g divdided by L.” The second rectangle is labeled “[ P b C r O subscript 4 ], in m o l divided by L.” The third is labeled “[ P b superscript 2 plus] and [ C r O subscript 4 superscript 2 negative ].” The fourth rectangle is labeled “K subscript s p.”


  1. Step 1.

    Use the molar mass of PbCrO4 (323.2g1mol)(323.2g1mol) to convert the solubility of PbCrO4 in grams per liter into moles per liter:

    [ PbCrO4 ]=4.6×106g PbCrO41L×1mol PbCrO4323.2g PbCrO4=1.4×108mol PbCrO41L=1.4×108M[ PbCrO4 ]=4.6×106g PbCrO41L×1mol PbCrO4323.2g PbCrO4=1.4×108mol PbCrO41L=1.4×108M
  2. Step 2.

    The chemical equation for the dissolution indicates that 1 mol of PbCrO4 gives 1 mol of Pb2+(aq) and 1 mol of CrO42−(aq):CrO42−(aq):

    PbCrO4(s)Pb2+(aq)+CrO42−(aq)PbCrO4(s)Pb2+(aq)+CrO42−(aq)

    Thus, both [Pb2+] and [CrO42−][CrO42−] are equal to the molar solubility of PbCrO4:

    [Pb2+]=[CrO42−]=1.4×108M[Pb2+]=[CrO42−]=1.4×108M
  3. Step 3.

    Solve. Ksp = [Pb2+][CrO42−][CrO42−] = (1.4 ×× 10–8)(1.4 ×× 10–8) = 2.0 ×× 10–16

Check Your Learning

The solubility of TlCl [thallium(I) chloride], an intermediate formed when thallium is being isolated from ores, is 3.46 grams per liter at 20 °C. What is its solubility product?

Answer:

2.08 ×× 10–4

Example 15.6

Calculating the Solubility of Hg2Cl2

Calomel, Hg2Cl2, is a compound composed of the diatomic ion of mercury(I), Hg22+,Hg22+, and chloride ions, Cl. Although most mercury compounds are now known to be poisonous, eighteenth-century physicians used calomel as a medication. Their patients rarely suffered any mercury poisoning from the treatments because calomel is quite insoluble:
Hg2Cl2(s)Hg22+(aq)+2Cl(aq)Ksp=1.1×1018Hg2Cl2(s)Hg22+(aq)+2Cl(aq)Ksp=1.1×1018

Calculate the molar solubility of Hg2Cl2.

Solution

The molar solubility of Hg2Cl2 is equal to the concentration of Hg22+Hg22+ ions because for each 1 mol of Hg2Cl2 that dissolves, 1 mol of Hg22+Hg22+ forms: This figure shows four horizontally oriented light green rectangles. Right pointing arrows are placed between them. The first rectangle is labeled “Determine the direction of change.” The second rectangle is labeled “Determine x and the equilibrium concentrations.” The third is labeled “Solve for x and the equilibrium concentrations.” The fourth rectangle is labeled “Check the math.”
  1. Step 1.

    Determine the direction of change. Before any Hg2Cl2 dissolves, Q is zero, and the reaction will shift to the right to reach equilibrium.

  2. Step 2.

    Determine x and equilibrium concentrations. Concentrations and changes are given in the following ICE table:

    This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of, “H g subscript 2 C l subscript 2 equilibrium arrow H g subscript 2 superscript 2 positive sign plus 2 C l superscript negative sign.” Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0, x, 0 plus x equals x. The third column has the following: 0, 2 x, 0 plus 2 x equals 2 x.

    Note that the change in the concentration of Cl (2x) is twice as large as the change in the concentration of Hg22+Hg22+ (x) because 2 mol of Cl forms for each 1 mol of Hg22+Hg22+ that forms. Hg2Cl2 is a pure solid, so it does not appear in the calculation.

  3. Step 3.

    Solve for x and the equilibrium concentrations. We substitute the equilibrium concentrations into the expression for Ksp and calculate the value of x:

    Ksp=[Hg22+][Cl]2Ksp=[Hg22+][Cl]2
    1.1×1018=(x)(2x)21.1×1018=(x)(2x)2
    4x3=1.1×10184x3=1.1×1018
    x=(1.1×10184)3=6.5×107Mx=(1.1×10184)3=6.5×107M
    [Hg22+]=6.5×107M=6.5×107M[Hg22+]=6.5×107M=6.5×107M
    [Cl]=2x=2(6.5×107)=1.3×106M[Cl]=2x=2(6.5×107)=1.3×106M

    The molar solubility of Hg2Cl2 is equal to [Hg22+],[Hg22+], or 6.5 ×× 10–7 M.

  4. Step 4.

    Check the work. At equilibrium, Q = Ksp:

    Q=[Hg22+][Cl]2=(6.5×107)(1.3×106)2=1.1×1018Q=[Hg22+][Cl]2=(6.5×107)(1.3×106)2=1.1×1018

    The calculations check.

Check Your Learning

Determine the molar solubility of MgF2 from its solubility product: Ksp = 6.4 ×× 10–9.

Answer:

1.2 ×× 10–3 M

Tabulated Ksp values can also be compared to reaction quotients calculated from experimental data to tell whether a solid will precipitate in a reaction under specific conditions: Q equals Ksp at equilibrium; if Q is less than Ksp, the solid will dissolve until Q equals Ksp; if Q is greater than Ksp, precipitation will occur at a given temperature until Q equals Ksp.

How Sciences Interconnect

Using Barium Sulfate for Medical Imaging

Various types of medical imaging techniques are used to aid diagnoses of illnesses in a noninvasive manner. One such technique utilizes the ingestion of a barium compound before taking an X-ray image. A suspension of barium sulfate, a chalky powder, is ingested by the patient. Since the Ksp of barium sulfate is 1.1 ×× 10–10, very little of it dissolves as it coats the lining of the patient’s intestinal tract. Barium-coated areas of the digestive tract then appear on an X-ray as white, allowing for greater visual detail than a traditional X-ray (Figure 15.4).

This figure contains one image. A black and white abdominal x-ray image is shown in which the intestinal tract of a person is clearly visible in white.
Figure 15.4 The suspension of barium sulfate coats the intestinal tract, which allows for greater visual detail than a traditional X-ray. (credit modification of work by “glitzy queen00”/Wikimedia Commons)

Further diagnostic testing can be done using barium sulfate and fluoroscopy. In fluoroscopy, a continuous X-ray is passed through the body so the doctor can monitor, on a TV or computer screen, the barium sulfate’s movement as it passes through the digestive tract. Medical imaging using barium sulfate can be used to diagnose acid reflux disease, Crohn’s disease, and ulcers in addition to other conditions.

Visit this website for more information on how barium is used in medical diagnoses and which conditions it is used to diagnose.

Predicting Precipitation

The equation that describes the equilibrium between solid calcium carbonate and its solvated ions is:

CaCO3(s)Ca2+(aq)+CO32−(aq)CaCO3(s)Ca2+(aq)+CO32−(aq)

We can establish this equilibrium either by adding solid calcium carbonate to water or by mixing a solution that contains calcium ions with a solution that contains carbonate ions. If we add calcium carbonate to water, the solid will dissolve until the concentrations are such that the value of the reaction quotient (Q=[Ca2+][CO32−])(Q=[Ca2+][CO32−]) is equal to the solubility product (Ksp = 8.7 ×× 10–9). If we mix a solution of calcium nitrate, which contains Ca2+ ions, with a solution of sodium carbonate, which contains CO32−CO32− ions, the slightly soluble ionic solid CaCO3 will precipitate, provided that the concentrations of Ca2+ and CO32−CO32− ions are such that Q is greater than Ksp for the mixture. The reaction shifts to the left and the concentrations of the ions are reduced by formation of the solid until the value of Q equals Ksp. A saturated solution in equilibrium with the undissolved solid will result. If the concentrations are such that Q is less than Ksp, then the solution is not saturated and no precipitate will form.

We can compare numerical values of Q with Ksp to predict whether precipitation will occur, as Example 15.7 shows. (Note: Since all forms of equilibrium constants are temperature dependent, we will assume a room temperature environment going forward in this chapter unless a different temperature value is explicitly specified.)

Example 15.7

Precipitation of Mg(OH)2

The first step in the preparation of magnesium metal is the precipitation of Mg(OH)2 from sea water by the addition of lime, Ca(OH)2, a readily available inexpensive source of OH ion:
Mg(OH)2(s)Mg2+(aq)+2OH(aq)Ksp=8.9×1012Mg(OH)2(s)Mg2+(aq)+2OH(aq)Ksp=8.9×1012

The concentration of Mg2+(aq) in sea water is 0.0537 M. Will Mg(OH)2 precipitate when enough Ca(OH)2 is added to give a [OH] of 0.0010 M?

Solution

This problem asks whether the reaction:
Mg(OH)2(s)Mg2+(aq)+2OH(aq)Mg(OH)2(s)Mg2+(aq)+2OH(aq)

shifts to the left and forms solid Mg(OH)2 when [Mg2+] = 0.0537 M and [OH] = 0.0010 M. The reaction shifts to the left if Q is greater than Ksp. Calculation of the reaction quotient under these conditions is shown here:

Q=[Mg2+][OH]2=(0.0537)(0.0010)2=5.4×108Q=[Mg2+][OH]2=(0.0537)(0.0010)2=5.4×108

Because Q is greater than Ksp (Q = 5.4 ×× 10–8 is larger than Ksp = 8.9 ×× 10–12), we can expect the reaction to shift to the left and form solid magnesium hydroxide. Mg(OH)2(s) forms until the concentrations of magnesium ion and hydroxide ion are reduced sufficiently so that the value of Q is equal to Ksp.

Check Your Learning

Use the solubility product in Appendix J to determine whether CaHPO4 will precipitate from a solution with [Ca2+] = 0.0001 M and [HPO42−][HPO42−] = 0.001 M.

Answer:

No precipitation of CaHPO4; Q = 1 ×× 10–7, which is less than Ksp

Example 15.8

Precipitation of AgCl upon Mixing Solutions

Does silver chloride precipitate when equal volumes of a 2.0 ×× 10–4-M solution of AgNO3 and a 2.0 ×× 10–4-M solution of NaCl are mixed?

(Note: The solution also contains Na+ and NO3NO3 ions, but when referring to solubility rules, one can see that sodium nitrate is very soluble and cannot form a precipitate.)

Solution

The equation for the equilibrium between solid silver chloride, silver ion, and chloride ion is:
AgCl(s)Ag+(aq)+Cl(aq)AgCl(s)Ag+(aq)+Cl(aq)

The solubility product is 1.6 ×× 10–10 (see Appendix J).

AgCl will precipitate if the reaction quotient calculated from the concentrations in the mixture of AgNO3 and NaCl is greater than Ksp. The volume doubles when we mix equal volumes of AgNO3 and NaCl solutions, so each concentration is reduced to half its initial value. Consequently, immediately upon mixing, [Ag+] and [Cl] are both equal to:

12(2.0×104)M=1.0×104M12(2.0×104)M=1.0×104M

The reaction quotient, Q, is momentarily greater than Ksp for AgCl, so a supersaturated solution is formed:

Q=[Ag+][Cl]=(1.0×104)(1.0×104)=1.0×108>KspQ=[Ag+][Cl]=(1.0×104)(1.0×104)=1.0×108>Ksp

Since supersaturated solutions are unstable, AgCl will precipitate from the mixture until the solution returns to equilibrium, with Q equal to Ksp.

Check Your Learning

Will KClO4 precipitate when 20 mL of a 0.050-M solution of K+ is added to 80 mL of a 0.50-M solution of ClO4?ClO4? (Remember to calculate the new concentration of each ion after mixing the solutions before plugging into the reaction quotient expression.)

Answer:

No, Q = 4.0 ×× 10–3, which is less than Ksp = 1.05 ×× 10–2

In the previous two examples, we have seen that Mg(OH)2 or AgCl precipitate when Q is greater than Ksp. In general, when a solution of a soluble salt of the Mm+ ion is mixed with a solution of a soluble salt of the Xn– ion, the solid, MpXq precipitates if the value of Q for the mixture of Mm+ and Xn– is greater than Ksp for MpXq. Thus, if we know the concentration of one of the ions of a slightly soluble ionic solid and the value for the solubility product of the solid, then we can calculate the concentration that the other ion must exceed for precipitation to begin. To simplify the calculation, we will assume that precipitation begins when the reaction quotient becomes equal to the solubility product.

Example 15.9

Precipitation of Calcium Oxalate

Blood will not clot if calcium ions are removed from its plasma. Some blood collection tubes contain salts of the oxalate ion, C2O42−,C2O42−, for this purpose (Figure 15.5). At sufficiently high concentrations, the calcium and oxalate ions form solid, CaC2O4·H2O (which also contains water bound in the solid). The concentration of Ca2+ in a sample of blood serum is 2.2 ×× 10–3 M. What concentration of C2O42−C2O42− ion must be established before CaC2O4·H2O begins to precipitate?
A photograph is shown of 6 vials of blood resting on and near a black and white document. Two of the vials have purple caps, three have tan caps, and one has a red cap. Each has a label and the vials with tan caps have a small amount of an off-white material present in a layer at the base of the vial.
Figure 15.5 Anticoagulants can be added to blood that will combine with the Ca2+ ions in blood serum and prevent the blood from clotting. (credit: modification of work by Neeta Lind)

Solution

The equilibrium expression is:
CaC2O4(s)Ca2+(aq)+C2O42−(aq)CaC2O4(s)Ca2+(aq)+C2O42−(aq)

For this reaction:

Ksp=[Ca2+][C2O42−]=1.96×108Ksp=[Ca2+][C2O42−]=1.96×108

(see Appendix J)

CaC2O4 does not appear in this expression because it is a solid. Water does not appear because it is the solvent.

Solid CaC2O4 does not begin to form until Q equals Ksp. Because we know Ksp and [Ca2+], we can solve for the concentration of C2O42−C2O42− that is necessary to produce the first trace of solid:

Q=Ksp=[Ca2+][C2O42−]=1.96×108Q=Ksp=[Ca2+][C2O42−]=1.96×108
(2.2×103)[C2O42−]=1.96×108(2.2×103)[C2O42−]=1.96×108
[C2O42−]=1.96×1082.2×103=8.9×106[C2O42−]=1.96×1082.2×103=8.9×106

A concentration of [C2O42−][C2O42−] = 8.9 ×× 10–6 M is necessary to initiate the precipitation of CaC2O4 under these conditions.

Check Your Learning

If a solution contains 0.0020 mol of CrO42−CrO42− per liter, what concentration of Ag+ ion must be reached by adding solid AgNO3 before Ag2CrO4 begins to precipitate? Neglect any increase in volume upon adding the solid silver nitrate.

Answer:

4.5 ×× 10–9 M

It is sometimes useful to know the concentration of an ion that remains in solution after precipitation. We can use the solubility product for this calculation too: If we know the value of Ksp and the concentration of one ion in solution, we can calculate the concentration of the second ion remaining in solution. The calculation is of the same type as that in Example 15.9—calculation of the concentration of a species in an equilibrium mixture from the concentrations of the other species and the equilibrium constant. However, the concentrations are different; we are calculating concentrations after precipitation is complete, rather than at the start of precipitation.

Example 15.10

Concentrations Following Precipitation

Clothing washed in water that has a manganese [Mn2+(aq)] concentration exceeding 0.1 mg/L (1.8 ×× 10–6 M) may be stained by the manganese upon oxidation, but the amount of Mn2+ in the water can be reduced by adding a base. If a person doing laundry wishes to add a buffer to keep the pH high enough to precipitate the manganese as the hydroxide, Mn(OH)2, what pH is required to keep [Mn2+] equal to 1.8 ×× 10–6 M?

Solution

The dissolution of Mn(OH)2 is described by the equation:
Mn(OH)2(s)Mn2+(aq)+2OH(aq)Ksp=2×1013Mn(OH)2(s)Mn2+(aq)+2OH(aq)Ksp=2×1013

We need to calculate the concentration of OH when the concentration of Mn2+ is 1.8 ×× 10–6 M. From that, we calculate the pH. At equilibrium:

Ksp=[Mn2+][OH]2Ksp=[Mn2+][OH]2

or

(1.8×106)[OH]2=2×1013(1.8×106)[OH]2=2×1013

so

[OH]=3.3×104M[OH]=3.3×104M

Now we calculate the pH from the pOH:

pOH=−log[OH]=−log(3.3×104)=3.48pH=14.00pOH=14.003.80=10.52pOH=−log[OH]=−log(3.3×104)=3.48pH=14.00pOH=14.003.80=10.52

If the person doing laundry adds a base, such as the sodium silicate (Na4SiO4) in some detergents, to the wash water until the pH is raised to 10.52, the manganese ion will be reduced to a concentration of 1.8 ×× 10–6 M; at that concentration or less, the ion will not stain clothing.

Check Your Learning

The first step in the preparation of magnesium metal is the precipitation of Mg(OH)2 from sea water by the addition of Ca(OH)2. The concentration of Mg2+(aq) in sea water is 5.37 ×× 10–2 M. Calculate the pH at which [Mg2+] is diminished to 1.0 ×× 10–5 M by the addition of Ca(OH)2.

Answer:

10.97

Due to their light sensitivity, mixtures of silver halides are used in fiber optics for medical lasers, in photochromic eyeglass lenses (glass lenses that automatically darken when exposed to sunlight), and—before the advent of digital photography—in photographic film. Even though AgCl (Ksp = 1.6 ×× 10–10), AgBr (Ksp = 5.0 ×× 10–13), and AgI (Ksp = 1.5 ×× 10–16) are each quite insoluble, we cannot prepare a homogeneous solid mixture of them by adding Ag+ to a solution of Cl, Br, and I; essentially all of the AgI will precipitate before any of the other solid halides form because of its smaller value for Ksp. However, we can prepare a homogeneous mixture of the solids by slowly adding a solution of Cl, Br, and I to a solution of Ag+.

When two anions form slightly soluble compounds with the same cation, or when two cations form slightly soluble compounds with the same anion, the less soluble compound (usually, the compound with the smaller Ksp) generally precipitates first when we add a precipitating agent to a solution containing both anions (or both cations). When the Ksp values of the two compounds differ by two orders of magnitude or more (e.g., 10–2 vs. 10–4), almost all of the less soluble compound precipitates before any of the more soluble one does. This is an example of selective precipitation, where a reagent is added to a solution of dissolved ions causing one of the ions to precipitate out before the rest.

Chemistry in Everyday Life

The Role of Precipitation in Wastewater Treatment

Solubility equilibria are useful tools in the treatment of wastewater carried out in facilities that may treat the municipal water in your city or town (Figure 15.6). Specifically, selective precipitation is used to remove contaminants from wastewater before it is released back into natural bodies of water. For example, phosphate ions (PO43−)(PO43−) are often present in the water discharged from manufacturing facilities. An abundance of phosphate causes excess algae to grow, which impacts the amount of oxygen available for marine life as well as making water unsuitable for human consumption.

A color photograph is shown of a high volume wastewater treatment facility. Nineteen large circular pools of water undergoing treatment are visible across the center of the photograph. A building and parking lot are visible in the foreground.
Figure 15.6 Wastewater treatment facilities, such as this one, remove contaminants from wastewater before the water is released back into the natural environment. (credit: “eutrophication&hypoxia”/Wikimedia Commons)

One common way to remove phosphates from water is by the addition of calcium hydroxide, known as lime, Ca(OH)2. As the water is made more basic, the calcium ions react with phosphate ions to produce hydroxylapatite, Ca5(PO4)3OH, which then precipitates out of the solution:

5Ca2++3PO43−+OHCa5(PO4)3·OH(s)5Ca2++3PO43−+OHCa5(PO4)3·OH(s)

The precipitate is then removed by filtration and the water is brought back to a neutral pH by the addition of CO2 in a recarbonation process. Other chemicals can also be used for the removal of phosphates by precipitation, including iron(III) chloride and aluminum sulfate.

View this site for more information on how phosphorus is removed from wastewater.

Selective precipitation can also be used in qualitative analysis. In this method, reagents are added to an unknown chemical mixture in order to induce precipitation. Certain reagents cause specific ions to precipitate out; therefore, the addition of the reagent can be used to determine whether the ion is present in the solution.

Example 15.11

Precipitation of Silver Halides

A solution contains 0.0010 mol of KI and 0.10 mol of KCl per liter. AgNO3 is gradually added to this solution. Which forms first, solid AgI or solid AgCl?

Solution

The two equilibria involved are:
AgCl(s)Ag+(aq)+Cl(aq)Ksp=1.6×1010AgCl(s)Ag+(aq)+Cl(aq)Ksp=1.6×1010
AgI(s)Ag+(aq)+I(aq)Ksp=1.5×1016AgI(s)Ag+(aq)+I(aq)Ksp=1.5×1016

If the solution contained about equal concentrations of Cl and I, then the silver salt with the smallest Ksp (AgI) would precipitate first. The concentrations are not equal, however, so we should find the [Ag+] at which AgCl begins to precipitate and the [Ag+] at which AgI begins to precipitate. The salt that forms at the lower [Ag+] precipitates first.

For AgI: AgI precipitates when Q equals Ksp for AgI (1.5 ×× 10–16). When [I] = 0.0010 M:

Q=[Ag+][I]=[Ag+](0.0010)=1.5×1016Q=[Ag+][I]=[Ag+](0.0010)=1.5×1016
[Ag+]=1.5×10160.10=1.6×109[Ag+]=1.5×10160.10=1.6×109

AgI begins to precipitate when [Ag+] is 1.6 ×× 10–9 M.

For AgCl: AgCl precipitates when Q equals Ksp for AgCl (1.6 ×× 10–10). When [Cl] = 0.10 M:

Qsp=[Ag+][Cl]=[Ag+](0.10)=1.6×1010Qsp=[Ag+][Cl]=[Ag+](0.10)=1.6×1010
[Ag+]=1.6×10100.10=1.6×109M[Ag+]=1.6×10100.10=1.6×109M

AgCl begins to precipitate when [Ag+] is 1.6 ×× 10–9 M.

AgI begins to precipitate at a lower [Ag+] than AgCl, so AgI begins to precipitate first.

Check Your Learning

If silver nitrate solution is added to a solution which is 0.050 M in both Cl and Br ions, at what [Ag+] would precipitation begin, and what would be the formula of the precipitate?

Answer:

[Ag+] = 1.0 ×× 10–11 M; AgBr precipitates first

Common Ion Effect

As we saw when we discussed buffer solutions, the hydronium ion concentration of an aqueous solution of acetic acid decreases when the strong electrolyte sodium acetate, NaCH3CO2, is added. We can explain this effect using Le Châtelier’s principle. The addition of acetate ions causes the equilibrium to shift to the left, decreasing the concentration of H3O+H3O+ to compensate for the increased acetate ion concentration. This increases the concentration of CH3CO2H:

CH3CO2H+H2OH3O++CH3CO2CH3CO2H+H2OH3O++CH3CO2

Because sodium acetate and acetic acid have the acetate ion in common, the influence on the equilibrium is called the common ion effect.

The common ion effect can also have a direct effect on solubility equilibria. Suppose we are looking at the reaction where silver iodide is dissolved:

AgI(s)Ag+(aq)+I(aq)AgI(s)Ag+(aq)+I(aq)

If we were to add potassium iodide (KI) to this solution, we would be adding a substance that shares a common ion with silver iodide. Le Châtelier’s principle tells us that when a change is made to a system at equilibrium, the reaction will shift to counteract that change. In this example, there would be an excess of iodide ions, so the reaction would shift toward the left, causing more silver iodide to precipitate out of solution.

Example 15.12

Common Ion Effect

Calculate the molar solubility of cadmium sulfide (CdS) in a 0.010-M solution of cadmium bromide (CdBr2). The Ksp of CdS is 1.0 ×× 10–28.

Solution

The first thing you should notice is that the cadmium sulfide is dissolved in a solution that contains cadmium ions. We need to use an ICE table to set up this problem and include the CdBr2 concentration as a contributor of cadmium ions:
CdS(s)Cd2+(aq)+S2−(aq)CdS(s)Cd2+(aq)+S2−(aq)
This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header, “C d S equilibrium arrow C d to the second power plus S to the second power superscript negative sign.” Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0.010, x, 0.010 plus x. The third column has the following: 0, x, 0 plus x equals x.
Ksp=[Cd2+][S2−]=1.0×1028Ksp=[Cd2+][S2−]=1.0×1028
(0.010+x)(x)=1.0×1028(0.010+x)(x)=1.0×1028
x2+0.010x1.0×1028=0x2+0.010x1.0×1028=0

We can solve this equation using the quadratic formula, but we can also make an assumption to make this calculation much simpler. Since the Ksp value is so small compared with the cadmium concentration, we can assume that the change between the initial concentration and the equilibrium concentration is negligible, so that 0.010 + x ~ 0.010. Going back to our Ksp expression, we would now get:

Ksp=[Cd2+][S2−]=1.0×1028Ksp=[Cd2+][S2−]=1.0×1028
(0.010)(x)=1.0×1028(0.010)(x)=1.0×1028
x=1.0×1026x=1.0×1026

Therefore, the molar solubility of CdS in this solution is 1.0 ×× 10–26 M.

Check Your Learning

Calculate the molar solubility of aluminum hydroxide, Al(OH)3, in a 0.015-M solution of aluminum nitrate, Al(NO3)3. The Ksp of Al(OH)3 is 2 ×× 10–32.

Answer:

4 ×× 10–11 M

Citation/Attribution

Want to cite, share, or modify this book? This book is Creative Commons Attribution License 4.0 and you must attribute OpenStax.

Attribution information
  • If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:
    Access for free at https://openstax.org/books/chemistry/pages/1-introduction
  • If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:
    Access for free at https://openstax.org/books/chemistry/pages/1-introduction
Citation information

© Sep 15, 2020 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License 4.0 license. The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.