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Chemistry: Atoms First 2e

14.3 Relative Strengths of Acids and Bases

Chemistry: Atoms First 2e14.3 Relative Strengths of Acids and Bases
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  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
    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 Salts
    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 Coupled Equilibria
    5. Key Terms
    6. Key Equations
    7. Summary
    8. Exercises
  17. 16 Electrochemistry
    1. Introduction
    2. 16.1 Review of Redox Chemistry
    3. 16.2 Galvanic Cells
    4. 16.3 Electrode and Cell Potentials
    5. 16.4 Potential, Free Energy, and Equilibrium
    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
By the end of this section, you will be able to:
  • Assess the relative strengths of acids and bases according to their ionization constants
  • Rationalize trends in acid–base strength in relation to molecular structure
  • Carry out equilibrium calculations for weak acid–base systems

Acid and Base Ionization Constants

The relative strength of an acid or base is the extent to which it ionizes when dissolved in water. If the ionization reaction is essentially complete, the acid or base is termed strong; if relatively little ionization occurs, the acid or base is weak. As will be evident throughout the remainder of this chapter, there are many more weak acids and bases than strong ones. The most common strong acids and bases are listed in Figure 14.6.

This table has seven rows and two columns. The first row is a header row, and it labels each column, “6 Strong Acids,” and, “6 Strong Bases.” Under the “6 Strong Acids” column are the following: H C l O subscript 4 perchloric acid; H C l hydrochloric acid; H B r hydrobromic acid; H I hydroiodic acid; H N O subscript 3 nitric acid; H subscript 2 S O subscript 4 sulfuric acid. Under the “6 Strong Bases” column are the following: L i O H lithium hydroxide; N a O H sodium hydroxide; K O H potassium hydroxide; C a ( O H ) subscript 2 calcium hydroxide; S r ( O H ) subscript 2 strontium hydroxide; B a ( O H ) subscript 2 barium hydroxide.
Figure 14.6 Some of the common strong acids and bases are listed here.

The relative strengths of acids may be quantified by measuring their equilibrium constants in aqueous solutions. In solutions of the same concentration, stronger acids ionize to a greater extent, and so yield higher concentrations of hydronium ions than do weaker acids. The equilibrium constant for an acid is called the acid-ionization constant, Ka. For the reaction of an acid HA:

HA(aq)+H2O(l)H3O+(aq)+A(aq),HA(aq)+H2O(l)H3O+(aq)+A(aq),

the acid ionization constant is written

Ka=[H3O+][A][HA]Ka=[H3O+][A][HA]

where the concentrations are those at equilibrium. Although water is a reactant in the reaction, it is the solvent as well, so we do not include [H2O] in the equation. The larger the Ka of an acid, the larger the concentration of H3O+H3O+ and A relative to the concentration of the nonionized acid, HA, in an equilibrium mixture, and the stronger the acid. An acid is classified as “strong” when it undergoes complete ionization, in which case the concentration of HA is zero and the acid ionization constant is immeasurably large (Ka ≈ ∞). Acids that are partially ionized are called “weak,” and their acid ionization constants may be experimentally measured. A table of ionization constants for weak acids is provided in Appendix H.

To illustrate this idea, three acid ionization equations and Ka values are shown below. The ionization constants increase from first to last of the listed equations, indicating the relative acid strength increases in the order CH3CO2H < HNO2 < HSO4:HSO4:

CH3CO2H(aq)+H2O(l)H3O+(aq)+CH3CO2(aq)Ka=1.8×10−5CH3CO2H(aq)+H2O(l)H3O+(aq)+CH3CO2(aq)Ka=1.8×10−5
HNO2(aq)+H2O(l)H3O+(aq)+NO2(aq)Ka=4.6×10−4HNO2(aq)+H2O(l)H3O+(aq)+NO2(aq)Ka=4.6×10−4
HSO4(aq)+H2O(aq)H3O+(aq)+SO42−(aq)Ka=1.2×10−2HSO4(aq)+H2O(aq)H3O+(aq)+SO42−(aq)Ka=1.2×10−2

Another measure of the strength of an acid is its percent ionization. The percent ionization of a weak acid is defined in terms of the composition of an equilibrium mixture:

% ionization=[H3O+]eq[HA]0×100% ionization=[H3O+]eq[HA]0×100

where the numerator is equivalent to the concentration of the acid's conjugate base (per stoichiometry, [A] = [H3O+]). Unlike the Ka value, the percent ionization of a weak acid varies with the initial concentration of acid, typically decreasing as concentration increases. Equilibrium calculations of the sort described later in this chapter can be used to confirm this behavior.

Example 14.7

Calculation of Percent Ionization from pH Calculate the percent ionization of a 0.125-M solution of nitrous acid (a weak acid), with a pH of 2.09.

Solution The percent ionization for an acid is:

[H3O+]eq[HNO2]0×100[H3O+]eq[HNO2]0×100

Converting the provided pH to hydronium ion molarity yields

[H3O+]=102.09=0.0081M[H3O+]=102.09=0.0081M

Substituting this value and the provided initial acid concentration into the percent ionization equation gives

8.1×10−30.125×100=6.5%8.1×10−30.125×100=6.5%

(Recall the provided pH value of 2.09 is logarithmic, and so it contains just two significant digits, limiting the certainty of the computed percent ionization.)

Check Your Learning Calculate the percent ionization of a 0.10-M solution of acetic acid with a pH of 2.89.

Answer:

1.3% ionized

Just as for acids, the relative strength of a base is reflected in the magnitude of its base-ionization constant (Kb) in aqueous solutions. In solutions of the same concentration, stronger bases ionize to a greater extent, and so yield higher hydroxide ion concentrations than do weaker bases. A stronger base has a larger ionization constant than does a weaker base. For the reaction of a base, B:

B(aq)+H2O(l)HB+(aq)+OH(aq),B(aq)+H2O(l)HB+(aq)+OH(aq),

the ionization constant is written as

Kb=[HB+][OH][B]Kb=[HB+][OH][B]

Inspection of the data for three weak bases presented below shows the base strength increases in the order NO2<CH2CO2<NH3.NO2<CH2CO2<NH3.

NO2(aq)+H2O(l)HNO2(aq)+OH(aq)Kb=2.17×10−11 CH3CO2(aq)+H2O(l)CH3CO2H(aq)+OH(aq)Kb=5.6×10−10 NH3(aq)+H2O(l)NH4+(aq)+OH(aq)Kb=1.8×10−5NO2(aq)+H2O(l)HNO2(aq)+OH(aq)Kb=2.17×10−11 CH3CO2(aq)+H2O(l)CH3CO2H(aq)+OH(aq)Kb=5.6×10−10 NH3(aq)+H2O(l)NH4+(aq)+OH(aq)Kb=1.8×10−5

A table of ionization constants for weak bases appears in Appendix I. As for acids, the relative strength of a base is also reflected in its percent ionization, computed as

%ionization= [ OH ] eq / [B] 0 ×100% %ionization= [ OH ] eq / [B] 0 ×100%

but will vary depending on the base ionization constant and the initial concentration of the solution.

Relative Strengths of Conjugate Acid-Base Pairs

Brønsted-Lowry acid-base chemistry is the transfer of protons; thus, logic suggests a relation between the relative strengths of conjugate acid-base pairs. The strength of an acid or base is quantified in its ionization constant, Ka or Kb, which represents the extent of the acid or base ionization reaction. For the conjugate acid-base pair HA / A, ionization equilibrium equations and ionization constant expressions are

HA(aq)+H2O(l)H3O+(aq)+A(aq)Ka=[H3O+][A][HA]HA(aq)+H2O(l)H3O+(aq)+A(aq)Ka=[H3O+][A][HA]
A(aq)+H2O(l)OH(aq)+HA(aq)Kb=[HA][OH][A]A(aq)+H2O(l)OH(aq)+HA(aq)Kb=[HA][OH][A]

Adding these two chemical equations yields the equation for the autoionization for water:

HA(aq)+H2O(l)+A(aq)+H2O(l)H3O+(aq)+A(aq)+OH(aq)+HA(aq)HA(aq)+H2O(l)+A(aq)+H2O(l)H3O+(aq)+A(aq)+OH(aq)+HA(aq)
2H2O(l)H3O+(aq)+OH(aq)2H2O(l)H3O+(aq)+OH(aq)

As discussed in another chapter on equilibrium, the equilibrium constant for a summed reaction is equal to the mathematical product of the equilibrium constants for the added reactions, and so

Ka×Kb=[H3O+][A][HA]×[HA][OH][A]=[H3O+][OH]=KwKa×Kb=[H3O+][A][HA]×[HA][OH][A]=[H3O+][OH]=Kw

This equation states the relation between ionization constants for any conjugate acid-base pair, namely, their mathematical product is equal to the ion product of water, Kw. By rearranging this equation, a reciprocal relation between the strengths of a conjugate acid-base pair becomes evident:

K a = K w / K b or K b = K w / K a K a = K w / K b or K b = K w / K a

The inverse proportional relation between Ka and Kb means the stronger the acid or base, the weaker its conjugate partner. Figure 14.7 illustrates this relation for several conjugate acid-base pairs.

The diagram shows two horizontal bars. The first, labeled, “Relative acid strength,” at the top is red on the left and gradually changes to purple on the right. The red end at the left is labeled, “Stronger acids.” The purple end at the right is labeled, “Weaker acids.” Just outside the bar to the lower left is the label, “K subscript a.” The bar is marked off in increments with a specific acid listed above each increment. The first mark is at 1.0 with H subscript 3 O superscript positive sign. The second is ten raised to the negative two with H C l O subscript 2. The third is ten raised to the negative 4 with H F. The fourth is ten raised to the negative 6 with H subscript 2 C O subscript 3. The fifth is ten raised to a negative 8 with C H subscript 3 C O O H. The sixth is ten raised to the negative ten with N H subscript 4 superscript positive sign. The seventh is ten raised to a negative 12 with H P O subscript 4 superscript 2 negative sign. The eighth is ten raised to the negative 14 with H subscript 2 O. Similarly the second bar, which is labeled “Relative conjugate base strength,” is purple at the left end and gradually becomes blue at the right end. Outside the bar to the left is the label, “Weaker bases.” Outside the bar to the right is the label, “Stronger bases.” Below and to the left of the bar is the label, “K subscript b.” The bar is similarly marked at increments with bases listed above each increment. The first is at ten raised to the negative 14 with H subscript 2 O above it. The second is ten raised to the negative 12 C l O subscript 2 superscript negative sign. The third is ten raised to the negative ten with F superscript negative sign. The fourth is ten raised to a negative eight with H C O subscript 3 superscript negative sign. The fifth is ten raised to the negative 6 with C H subscript 3 C O O superscript negative sign. The sixth is ten raised to the negative 4 with N H subscript 3. The seventh is ten raised to the negative 2 with P O subscript 4 superscript three negative sign. The eighth is 1.0 with O H superscript negative sign.
Figure 14.7 Relative strengths of several conjugate acid-base pairs are shown.
This figure includes a table separated into a left half which is labeled “Acids” and a right half labeled “Bases.” A red arrow points up the left side, which is labeled “Increasing acid strength.” Similarly, a blue arrow points downward along the right side, which is labeled “Increasing base strength.” Names of acids and bases are listed next to each arrow toward the center of the table, followed by chemical formulas. Acids listed top to bottom are sulfuric acid, H subscript 2 S O subscript 4, hydrogen iodide, H I, hydrogen bromide, H B r, hydrogen chloride, H C l, nitric acid, H N O subscript 3, hydronium ion ( in pink text) H subscript 3 O superscript plus, hydrogen sulfate ion, H S O subscript 4 superscript negative, phosphoric acid, H subscript 3 P O subscript 4, hydrogen fluoride, H F, nitrous acid, H N O subscript 2, acetic acid, C H subscript 3 C O subscript 2 H, carbonic acid H subscript 2 C O subscript 3, hydrogen sulfide, H subscript 2 S, ammonium ion, N H subscript 4 superscript +, hydrogen cyanide, H C N, hydrogen carbonate ion, H C O subscript 3 superscript negative, water (shaded in beige) H subscript 2 O, hydrogen sulfide ion, H S superscript negative, ethanol, C subscript 2 H subscript 5 O H, ammonia, N H subscript 3, hydrogen, H subscript 2, methane, and C H subscript 4. The acids at the top of the listing from sulfuric acid through nitric acid are grouped with a bracket to the right labeled “Undergo complete acid ionization in water.” Similarly, the acids at the bottom from hydrogen sulfide ion through methane are grouped with a bracket and labeled, “Do not undergo acid ionization in water.” The right half of the figure lists bases and formulas. From top to bottom the bases listed are hydrogen sulfate ion, H S O subscript 4 superscript negative, iodide ion, I superscript negative, bromide ion, B r superscript negative, chloride ion, C l superscript negative, nitrate ion, N O subscript 3 superscript negative, water (shaded in beige), H subscript 2 O, sulfate ion, S O subscript 4 superscript 2 negative, dihydrogen phosphate ion, H subscript 2 P O subscript 4 superscript negative, fluoride ion, F superscript negative, nitrite ion, N O subscript 2 superscript negative, acetate ion, C H subscript 3 C O subscript 2 superscript negative, hydrogen carbonate ion, H C O subscript 3 superscript negative, hydrogen sulfide ion, H S superscript negative, ammonia, N H subscript 3, cyanide ion, C N superscript negative, carbonate ion, C O subscript 3 superscript 2 negative, hydroxide ion (in blue), O H superscript negative, sulfide ion, S superscript 2 negative, ethoxide ion, C subscript 2 H subscript 5 O superscript negative, amide ion N H subscript 2 superscript negative, hydride ion, H superscript negative, and methide ion C H subscript 3 superscript negative. The bases at the top, from perchlorate ion through nitrate ion are group with a bracket which is labeled “Do not undergo base ionization in water.” Similarly, the lower 5 in the listing, from sulfide ion through methide ion are grouped and labeled “Undergo complete base ionization in water.”
Figure 14.8 This figure shows strengths of conjugate acid-base pairs relative to the strength of water as the reference substance.

The listing of conjugate acid–base pairs shown in Figure 14.8 is arranged to show the relative strength of each species as compared with water, whose entries are highlighted in each of the table’s columns. In the acid column, those species listed below water are weaker acids than water. These species do not undergo acid ionization in water; they are not Bronsted-Lowry acids. All the species listed above water are stronger acids, transferring protons to water to some extent when dissolved in an aqueous solution to generate hydronium ions. Species above water but below hydronium ion are weak acids, undergoing partial acid ionization, wheres those above hydronium ion are strong acids that are completely ionized in aqueous solution.

If all these strong acids are completely ionized in water, why does the column indicate they vary in strength, with nitric acid being the weakest and perchloric acid the strongest? Notice that the sole acid species present in an aqueous solution of any strong acid is H3O+(aq), meaning that hydronium ion is the strongest acid that may exist in water; any stronger acid will react completely with water to generate hydronium ions. This limit on the acid strength of solutes in a solution is called a leveling effect. To measure the differences in acid strength for “strong” acids, the acids must be dissolved in a solvent that is less basic than water. In such solvents, the acids will be “weak,” and so any differences in the extent of their ionization can be determined. For example, the binary hydrogen halides HCl, HBr, and HI are strong acids in water but weak acids in ethanol (strength increasing HCl < HBr < HI).

The right column of Figure 14.8 lists a number of substances in order of increasing base strength from top to bottom. Following the same logic as for the left column, species listed above water are weaker bases and so they don’t undergo base ionization when dissolved in water. Species listed between water and its conjugate base, hydroxide ion, are weak bases that partially ionize. Species listed below hydroxide ion are strong bases that completely ionize in water to yield hydroxide ions (i.e., they are leveled to hydroxide). A comparison of the acid and base columns in this table supports the reciprocal relation between the strengths of conjugate acid-base pairs. For example, the conjugate bases of the strong acids (top of table) are all of negligible strength. A strong acid exhibits an immeasurably large Ka, and so its conjugate base will exhibit a Kb that is essentially zero:

strongacid: K a conjugatebase: K b = K w / K a = K w /0 strongacid: K a conjugatebase: K b = K w / K a = K w /0

A similar approach can be used to support the observation that conjugate acids of strong bases (Kb ≈ ∞) are of negligible strength (Ka ≈ 0).

Example 14.8

Calculating Ionization Constants for Conjugate Acid-Base Pairs Use the Kb for the nitrite ion, NO2,NO2, to calculate the Ka for its conjugate acid.

SolutionKb for NO2NO2 is given in this section as 2.17 ×× 10−11. The conjugate acid of NO2NO2 is HNO2; Ka for HNO2 can be calculated using the relationship:

Ka×Kb=1.0×10−14=KwKa×Kb=1.0×10−14=Kw

Solving for Ka yields

Ka=KwKb=1.0×10−142.17×10−11=4.6×10−4Ka=KwKb=1.0×10−142.17×10−11=4.6×10−4

This answer can be verified by finding the Ka for HNO2 in Appendix H.

Check Your LearningDetermine the relative acid strengths of NH4+NH4+ and HCN by comparing their ionization constants. The ionization constant of HCN is given in Appendix H as 4.9 ×× 10−10. The ionization constant of NH4+NH4+ is not listed, but the ionization constant of its conjugate base, NH3, is listed as 1.8 ×× 10−5.

Answer:

NH4+NH4+ is the slightly stronger acid (Ka for NH4+NH4+ = 5.6 ×× 10−10).

Acid-Base Equilibrium Calculations

The chapter on chemical equilibria introduced several types of equilibrium calculations and the various mathematical strategies that are helpful in performing them. These strategies are generally useful for equilibrium systems regardless of chemical reaction class, and so they may be effectively applied to acid-base equilibrium problems. This section presents several example exercises involving equilibrium calculations for acid-base systems.

This image shows two bottles containing clear colorless solutions. Each bottle contains a single p H indicator strip. The strip in the bottle on the left is red, and a similar red strip is placed on a filter paper circle in front of the bottle on surface on which the bottles are resting. Similarly, the second bottle on the right contains and orange strip and an orange strip is placed in front of it on a filter paper circle. Between the two bottles is a pack of p Hydrion papers with a p H color scale on its cover. This photo shows two glass containers filled with a transparent liquid. In between the containers is a p H strip indicator guide. There are p H strips placed in front of each glass container. The liquid in the container on the left appears to have a p H of 10 or 11. The liquid in the container on the right appears to have a p H of about 13 or 14.

Example 14.9

Determination of Ka from Equilibrium Concentrations Acetic acid is the principal ingredient in vinegar (Figure 14.9) that provides its sour taste. At equilibrium, a solution contains [CH3CO2H] = 0.0787 M and [H3O+]=[CH3CO2]=0.00118M.[H3O+]=[CH3CO2]=0.00118M. What is the value of Ka for acetic acid?

An image shows the label of a bottle of distilled white vinegar. The label states that the contents have been reduced with water to 5 percent acidity.
Figure 14.9 Vinegar contains acetic acid, a weak acid. (credit: modification of work by “HomeSpot HQ”/Flickr)

Solution The relevant equilibrium equation and its equilibrium constant expression are shown below. Substitution of the provided equilibrium concentrations permits a straightforward calculation of the Ka for acetic acid.

CH3CO2H(aq)+H2O(l)H3O+(aq)+CH3CO2(aq)CH3CO2H(aq)+H2O(l)H3O+(aq)+CH3CO2(aq)
Ka=[H3O+][CH3CO2][CH3CO2H]=(0.00118)(0.00118)0.0787=1.77×10−5Ka=[H3O+][CH3CO2][CH3CO2H]=(0.00118)(0.00118)0.0787=1.77×10−5

Check Your Learning The HSO4HSO4 ion, weak acid used in some household cleansers:

HSO4(aq)+H2O(l)H3O+(aq)+SO42−(aq)HSO4(aq)+H2O(l)H3O+(aq)+SO42−(aq)

What is the acid ionization constant for this weak acid if an equilibrium mixture has the following composition: [H3O+][H3O+] = 0.027 M; [HSO4]=0.29M;[HSO4]=0.29M; and [SO42−]=0.13M?[SO42−]=0.13M?

Answer:

Ka for HSO4HSO4 = 1.2 ×× 10−2

Example 14.10

Determination of Kb from Equilibrium Concentrations Caffeine, C8H10N4O2 is a weak base. What is the value of Kb for caffeine if a solution at equilibrium has [C8H10N4O2] = 0.050 M, [C8H10N4O2H+][C8H10N4O2H+] = 5.0 ×× 10−3 M, and [OH] = 2.5 ×× 10−3 M?

Solution The relevant equilibrium equation and its equilibrium constant expression are shown below. Substitution of the provided equilibrium concentrations permits a straightforward calculation of the Kb for caffeine.

C8H10N4O2(aq)+H2O(l)C8H10N4O2H+(aq)+OH(aq)C8H10N4O2(aq)+H2O(l)C8H10N4O2H+(aq)+OH(aq)
Kb=[C8H10N4O2H+][OH][C8H10N4O2]=(5.0×10−3)(2.5×10−3)0.050=2.5×10−4Kb=[C8H10N4O2H+][OH][C8H10N4O2]=(5.0×10−3)(2.5×10−3)0.050=2.5×10−4

Check Your Learning What is the equilibrium constant for the ionization of the HPO42−HPO42− ion, a weak base

HPO42−(aq)+H2O(l)H2PO4(aq)+OH(aq)HPO42−(aq)+H2O(l)H2PO4(aq)+OH(aq)

if the composition of an equilibrium mixture is as follows: [OH] = 1.3 ×× 10−6 M; [H2PO4]=0.042M;[H2PO4]=0.042M; and [HPO42−]=0.341M?[HPO42−]=0.341M?

Answer:

Kb for HPO42−=1.6×10−7HPO42−=1.6×10−7

Example 14.11

Determination of Ka or Kb from pH The pH of a 0.0516-M solution of nitrous acid, HNO2, is 2.34. What is its Ka?

HNO2(aq)+H2O(l)H3O+(aq)+NO2(aq)HNO2(aq)+H2O(l)H3O+(aq)+NO2(aq)

Solution The nitrous acid concentration provided is a formal concentration, one that does not account for any chemical equilibria that may be established in solution. Such concentrations are treated as “initial” values for equilibrium calculations using the ICE table approach. Notice the initial value of hydronium ion is listed as approximately zero because a small concentration of H3O+ is present (1 × 10−7 M) due to the autoprotolysis of water. In many cases, such as all the ones presented in this chapter, this concentration is much less than that generated by ionization of the acid (or base) in question and may be neglected.

The pH provided is a logarithmic measure of the hydronium ion concentration resulting from the acid ionization of the nitrous acid, and so it represents an “equilibrium” value for the ICE table:

[ H 3 O + ]= 10 2.34 =0.0046M [ H 3 O + ]= 10 2.34 =0.0046M

The ICE table for this system is then

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 N O subscript 2 plus sign H subscript 2 O equilibrium sign H subscript 3 O superscript positive sign plus sign N O subscript 2 superscript negative sign.” Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.0516, negative 0.0046, 0.0470. The second column is blank in all three rows. The third column has the following: approximately 0, positive 0.0046, 0.0046. The fourth column has the following: 0, positive 0.0046, 0.0046.

Finally, calculate the value of the equilibrium constant using the data in the table:

Ka=[H3O+][NO2][HNO2]=(0.0046)(0.0046)(0.0470)=4.6×10−4Ka=[H3O+][NO2][HNO2]=(0.0046)(0.0046)(0.0470)=4.6×10−4

Check Your Learning. The pH of a solution of household ammonia, a 0.950-M solution of NH3, is 11.612. What is Kb for NH3.

Answer:

Kb = 1.8 ×× 10−5

Example 14.12

Calculating Equilibrium Concentrations in a Weak Acid Solution Formic acid, HCO2H, is one irritant that causes the body’s reaction to some ant bites and stings (Figure 14.10).

A photograph is shown of a large black ant on the end of a human finger.
Figure 14.10 The pain of some ant bites and stings is caused by formic acid. (credit: John Tann)

What is the concentration of hydronium ion and the pH of a 0.534-M solution of formic acid?

HCO2H(aq)+H2O(l)H3O+(aq)+HCO2(aq)Ka=1.8×10−4HCO2H(aq)+H2O(l)H3O+(aq)+HCO2(aq)Ka=1.8×10−4

Solution

The ICE table for this system is

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 C O subscript 2 H plus sign H subscript 2 O equilibrium arrow H subscript 3 O superscript positive sign.” Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.534, blank, 0.534 minus x. The second column is blank in all three rows. The third column has the following: approximately 0, positive x, x. The fourth column has the following: 0, positive x, x.

Substituting the equilibrium concentration terms into the Ka expression gives

Ka=1.8×10−4=[H3O+][HCO2][HCO2H]Ka=1.8×10−4=[H3O+][HCO2][HCO2H]


=(x)(x)0.534x=1.8×10−4=(x)(x)0.534x=1.8×10−4


The relatively large initial concentration and small equilibrium constant permits the simplifying assumption that x will be much lesser than 0.534, and so the equation becomes

Ka=1.8×10−4=x20.534Ka=1.8×10−4=x20.534


Solving the equation for x yields

x2=0.534×(1.8×10−4)=9.6×10−5x2=0.534×(1.8×10−4)=9.6×10−5


x=9.6×10−5x=9.6×10−5


=9.8×10−3M=9.8×10−3M

To check the assumption that x is small compared to 0.534, its relative magnitude can be estimated:

x0.534=9.8×10−30.534=1.8×10−2(1.8%of 0.534)x0.534=9.8×10−30.534=1.8×10−2(1.8%of 0.534)

Because x is less than 5% of the initial concentration, the assumption is valid.

As defined in the ICE table, x is equal to the equilibrium concentration of hydronium ion:

x=[ H 3 O + ]=0.0098Mx=[ H 3 O + ]=0.0098M

Finally, the pH is calculated to be

pH=log[ H 3 O + ]=log(0.0098)=2.01 pH=log[ H 3 O + ]=log(0.0098)=2.01

Check Your Learning Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of a 0.100-M solution of acetic acid, CH3CO2H?

CH3CO2H(aq)+H2O(l)H3O+(aq)+CH3CO2(aq)Ka=1.8×10−5CH3CO2H(aq)+H2O(l)H3O+(aq)+CH3CO2(aq)Ka=1.8×10−5

Answer:

percent ionization = 1.3%

Example 14.13

Calculating Equilibrium Concentrations in a Weak Base Solution Find the concentration of hydroxide ion, the pOH, and the pH of a 0.25-M solution of trimethylamine, a weak base:

(CH3)3N(aq)+H2O(l)(CH3)3NH+(aq)+OH(aq)Kb=6.3×10−5(CH3)3N(aq)+H2O(l)(CH3)3NH+(aq)+OH(aq)Kb=6.3×10−5

Solution The ICE table for this system is

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 “( C H subscript 3 ) subscript 3 N plus sign H subscript 2 O equilibrium arrow ( C H subscript 3 ) subscript 3 N H superscript positive sign plus sign O H superscript positive sign.” Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.25, negative x, 0.25 plus sign negative x. The second column is blank in all three rows. The third column has the following: 0, x, 0 plus x. The fourth column has the following: approximately 0, x, and approximately 0 plus x.


Substituting the equilibrium concentration terms into the Kb expression gives

Kb=[(CH3)3NH+][OH][(CH3)3N]=(x)(x)0.25x=6.3×10−5Kb=[(CH3)3NH+][OH][(CH3)3N]=(x)(x)0.25x=6.3×10−5


Assuming x << 0.25 and solving for x yields

x=4.0×10−3Mx=4.0×10−3M


This value is less than 5% of the initial concentration (0.25), so the assumption is justified.
As defined in the ICE table, x is equal to the equilibrium concentration of hydroxide ion:

[OH]=~0+x=x=4.0×10−3M[OH]=~0+x=x=4.0×10−3M


=4.0×10−3M=4.0×10−3M


The pOH is calculated to be

pOH=−log(4.0×10−3)=2.40pOH=−log(4.0×10−3)=2.40


Using the relation introduced in the previous section of this chapter:

pH+pOH=pKw=14.00pH+pOH=pKw=14.00


permits the computation of pH:

pH=14.00pOH=14.002.40=11.60pH=14.00pOH=14.002.40=11.60

Check Your LearningCalculate the hydroxide ion concentration and the percent ionization of a 0.0325-M solution of ammonia, a weak base with a Kb of 1.76 ×× 10−5.

Answer:

7.56 ×× 10−4 M, 2.33%

In some cases, the strength of the weak acid or base and its formal (initial) concentration result in an appreciable ionization. Though the ICE strategy remains effective for these systems, the algebra is a bit more involved because the simplifying assumption that x is negligible can not be made. Calculations of this sort are demonstrated in Example 14.14 below.

Example 14.14

Calculating Equilibrium Concentrations without Simplifying Assumptions Sodium bisulfate, NaHSO4, is used in some household cleansers as a source of the HSO4HSO4 ion, a weak acid. What is the pH of a 0.50-M solution of HSO4?HSO4?

HSO4(aq)+H2O(l)H3O+(aq)+SO42−(aq)Ka=1.2×10−2HSO4(aq)+H2O(l)H3O+(aq)+SO42−(aq)Ka=1.2×10−2

Solution The ICE table for this system is

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 ( M ). The second column has the header of “H S O subscript 4 superscript negative sign plus sign H subscript 2 O equilibrium sign H subscript 3 O superscript positive sign plus sign S O subscript 4 superscript 2 superscript negative sign.” Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.50, negative x, 0.50 minus x. The second column is blank for all three rows. The third column has the following: approximately 0, positive x, x. The fourth column has the following: 0, positive x, x.


Substituting the equilibrium concentration terms into the Ka expression gives

Ka=1.2×10−2=[H3O+][SO42−][HSO4]=(x)(x)0.50xKa=1.2×10−2=[H3O+][SO42−][HSO4]=(x)(x)0.50x


If the assumption that x << 0.5 is made, simplifying and solving the above equation yields

x=0.077Mx=0.077M


This value of x is clearly not significantly less than 0.50 M; rather, it is approximately 15% of the initial concentration:
When we check the assumption, we calculate:

x[HSO4]ix[HSO4]i


x0.50=7.7×10−20.50=0.15(15%)x0.50=7.7×10−20.50=0.15(15%)


Because the simplifying assumption is not valid for this system, the equilibrium constant expression is solved as follows:

Ka=1.2×10−2=(x)(x)0.50xKa=1.2×10−2=(x)(x)0.50x


Rearranging this equation yields

6.0×10−31.2×10−2x=x26.0×10−31.2×10−2x=x2


Writing the equation in quadratic form gives

x2+1.2×10−2x6.0×10−3=0x2+1.2×10−2x6.0×10−3=0


Solving for the two roots of this quadratic equation results in a negative value that may be discarded as physically irrelevant and a positive value equal to x. As defined in the ICE table, x is equal to the hydronium concentration.

x=[ H 3 O + ]=0.072M pH=log[ H 3 O + ]=log(0.072)=1.14 x=[ H 3 O + ]=0.072M pH=log[ H 3 O + ]=log(0.072)=1.14

Check Your Learning Calculate the pH in a 0.010-M solution of caffeine, a weak base:

C8H10N4O2(aq)+H2O(l)C8H10N4O2H+(aq)+OH(aq)Kb=2.5×10−4C8H10N4O2(aq)+H2O(l)C8H10N4O2H+(aq)+OH(aq)Kb=2.5×10−4

Answer:

pH 11.16

Effect of Molecular Structure on Acid-Base Strength

Binary Acids and Bases

In the absence of any leveling effect, the acid strength of binary compounds of hydrogen with nonmetals (A) increases as the H-A bond strength decreases down a group in the periodic table. For group 17, the order of increasing acidity is HF < HCl < HBr < HI. Likewise, for group 16, the order of increasing acid strength is H2O < H2S < H2Se < H2Te.

Across a row in the periodic table, the acid strength of binary hydrogen compounds increases with increasing electronegativity of the nonmetal atom because the polarity of the H-A bond increases. Thus, the order of increasing acidity (for removal of one proton) across the second row is CH4 < NH3 < H2O < HF; across the third row, it is SiH4 < PH3 < H2S < HCl (see Figure 14.11).

This diagram has two rows and four columns. Red arrows point left across the bottom of the figure and down at the right side and are labeled “Increasing acid strength.” Blue arrows point left across the bottom and up at the right side of the figure and are labeled “Increasing base strength.” The first column is labeled 14 at the top and two white squares are beneath it. The first has the number 6 in the upper left corner and the formula C H subscript 4 in the center along with designation Neither acid nor base. The second square contains the number 14 in the upper left corner, the formula C H subscript 4 at the center and the designation Neither acid nor base. The second column is labeled 15 at the top and two blue squares are beneath it. The first has the number 7 in the upper left corner and the formula N H subscript 3 in the center along with the designation Weak base and K subscript b equals 1.8 times 10 superscript negative 5. The second square contains the number 15 in the upper left corner, the formula P H subscript 3 at the center and the designation Very weak base and K subscript b equals 4 times 10 superscript negative 28. The third column is labeled 16 at the top and two squares are beneath it. The first is shaded tan and has the number 8 in the upper left corner and the formula H subscript 2 O in the center along with the designation neutral. The second square is shaded pink, contains the number 16 in the upper left corner, the formula H subscript 2 S at the center and the designation Weak acid and K subscript a equals 9.5 times 10 superscript negative 8. The fourth column is labeled 17 at the top and two squares are beneath it. The first is shaded pink, has the number 9 in the upper left corner and the formula H F in the center along with the designation Weak acid and K subscript a equals 6.8 times 10 superscript negative 4. The second square is shaded a deeper pink, contains the number 17 in the upper left corner, the formula H C l at the center, and the designation Strong acid.
Figure 14.11 The figure shows trends in the strengths of binary acids and bases.

Ternary Acids and Bases

Ternary compounds composed of hydrogen, oxygen, and some third element (“E”) may be structured as depicted in the image below. In these compounds, the central E atom is bonded to one or more O atoms, and at least one of the O atoms is also bonded to an H atom, corresponding to the general molecular formula OmE(OH)n. These compounds may be acidic, basic, or amphoteric depending on the properties of the central E atom. Examples of such compounds include sulfuric acid, O2S(OH)2, sulfurous acid, OS(OH)2, nitric acid, O2NOH, perchloric acid, O3ClOH, aluminum hydroxide, Al(OH)3, calcium hydroxide, Ca(OH)2, and potassium hydroxide, KOH:

A diagram is shown that includes a central atom designated with the letter E. Single bonds extend above, below, left, and right of the E. An O atom is bonded to the right of the E, and an arrow points to the bond labeling it, “Bond a.” An H atom is single bonded to the right of the O atom. An arrow pointing to this bond connects it to the label, “Bond b.”

If the central atom, E, has a low electronegativity, its attraction for electrons is low. Little tendency exists for the central atom to form a strong covalent bond with the oxygen atom, and bond a between the element and oxygen is more readily broken than bond b between oxygen and hydrogen. Hence bond a is ionic, hydroxide ions are released to the solution, and the material behaves as a base—this is the case with Ca(OH)2 and KOH. Lower electronegativity is characteristic of the more metallic elements; hence, the metallic elements form ionic hydroxides that are by definition basic compounds.

If, on the other hand, the atom E has a relatively high electronegativity, it strongly attracts the electrons it shares with the oxygen atom, making bond a relatively strongly covalent. The oxygen-hydrogen bond, bond b, is thereby weakened because electrons are displaced toward E. Bond b is polar and readily releases hydrogen ions to the solution, so the material behaves as an acid. High electronegativities are characteristic of the more nonmetallic elements. Thus, nonmetallic elements form covalent compounds containing acidic −OH groups that are called oxyacids.

Increasing the oxidation number of the central atom E also increases the acidity of an oxyacid because this increases the attraction of E for the electrons it shares with oxygen and thereby weakens the O-H bond. Sulfuric acid, H2SO4, or O2S(OH)2 (with a sulfur oxidation number of +6), is more acidic than sulfurous acid, H2SO3, or OS(OH)2 (with a sulfur oxidation number of +4). Likewise nitric acid, HNO3, or O2NOH (N oxidation number = +5), is more acidic than nitrous acid, HNO2, or ONOH (N oxidation number = +3). In each of these pairs, the oxidation number of the central atom is larger for the stronger acid (Figure 14.12).

A diagram is shown that includes four structural formulas for acids. A red, right pointing arrow is placed beneath the structures which is labeled “Increasing acid strength.” At the top left, the structure of Nitrous acid is provided. It includes an H atom to which an O atom with two unshared electron pairs is connected with a single bond to the right. A single bond extends to the right and slightly below to a N atom with one unshared electron pair. A double bond extends up and to the right from this N atom to an O atom which has two unshared electron pairs. To the upper right is a structure for Nitric acid. This structure differs from the previous structure in that the N atom is directly to the right of the first O atom and a second O atom with three unshared electron pairs is connected with a single bond below and to the right of the N atom which has no unshared electron pairs. At the lower left, an O atom with two unshared electron pairs is double bonded to its right to an S atom with a single unshared electron pair. An O atom with two unshared electron pairs is bonded above and an H atom is single bonded to this O atom. To the right of the S atom is a single bond to another O atom with two unshared electron pairs to which an H atom is single bonded. This structure is labeled “Sulfurous acid.” A similar structure which is labeled “Sulfuric acid” is placed in the lower right region of the figure. This structure differs in that an H atom is single bonded to the left of the first O atom, leaving it with two unshared electron pairs and a fourth O atom with two unshared electron pairs is double bonded beneath the S atom, leaving it with no unshared electron pairs.
Figure 14.12 As the oxidation number of the central atom E increases, the acidity also increases.

Hydroxy compounds of elements with intermediate electronegativities and relatively high oxidation numbers (for example, elements near the diagonal line separating the metals from the nonmetals in the periodic table) are usually amphoteric. This means that the hydroxy compounds act as acids when they react with strong bases and as bases when they react with strong acids. The amphoterism of aluminum hydroxide, which commonly exists as the hydrate Al(H2O)3(OH)3, is reflected in its solubility in both strong acids and strong bases. In strong bases, the relatively insoluble hydrated aluminum hydroxide, Al(H2O)3(OH)3, is converted into the soluble ion, [Al(H2O)2(OH)4],[Al(H2O)2(OH)4], by reaction with hydroxide ion:

Al(H2O)3(OH)3(aq)+OH(aq)H2O(l)+[Al(H2O)2(OH)4](aq)Al(H2O)3(OH)3(aq)+OH(aq)H2O(l)+[Al(H2O)2(OH)4](aq)

In this reaction, a proton is transferred from one of the aluminum-bound H2O molecules to a hydroxide ion in solution. The Al(H2O)3(OH)3 compound thus acts as an acid under these conditions. On the other hand, when dissolved in strong acids, it is converted to the soluble ion [Al(H2O)6]3+[Al(H2O)6]3+ by reaction with hydronium ion:

3H3O+(aq)+Al(H2O)3(OH)3(aq)Al(H2O)63+(aq)+3H2O(l)3H3O+(aq)+Al(H2O)3(OH)3(aq)Al(H2O)63+(aq)+3H2O(l)

In this case, protons are transferred from hydronium ions in solution to Al(H2O)3(OH)3, and the compound functions as a base.

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