Skip to main content

Introductory Chemistry - 1st Canadian Edition: Neutralization Reactions

Introductory Chemistry - 1st Canadian Edition
Neutralization Reactions
    • Notifications
    • Privacy
  • Project HomeNatural Sciences Collection: Anatomy, Biology, and Chemistry
  • Projects
  • Learn more about Manifold

Notes

Show the following:

  • Annotations
  • Resources
Search within:

Adjust appearance:

  • font
    Font style
  • color scheme
  • Margins
table of contents
  1. Cover
  2. Title Page
  3. Copyright
  4. Table Of Contents
  5. Acknowledgments
  6. Dedication
  7. About BCcampus Open Education
  8. Chapter 1. What is Chemistry
    1. Some Basic Definitions
    2. Chemistry as a Science
  9. Chapter 2. Measurements
    1. Expressing Numbers
    2. Significant Figures
    3. Converting Units
    4. Other Units: Temperature and Density
    5. Expressing Units
    6. End-of-Chapter Material
  10. Chapter 3. Atoms, Molecules, and Ions
    1. Acids
    2. Ions and Ionic Compounds
    3. Masses of Atoms and Molecules
    4. Molecules and Chemical Nomenclature
    5. Atomic Theory
    6. End-of-Chapter Material
  11. Chapter 4. Chemical Reactions and Equations
    1. The Chemical Equation
    2. Types of Chemical Reactions: Single- and Double-Displacement Reactions
    3. Ionic Equations: A Closer Look
    4. Composition, Decomposition, and Combustion Reactions
    5. Oxidation-Reduction Reactions
    6. Neutralization Reactions
    7. End-of-Chapter Material
  12. Chapter 5. Stoichiometry and the Mole
    1. Stoichiometry
    2. The Mole
    3. Mole-Mass and Mass-Mass Calculations
    4. Limiting Reagents
    5. The Mole in Chemical Reactions
    6. Yields
    7. End-of-Chapter Material
  13. Chapter 6. Gases
    1. Pressure
    2. Gas Laws
    3. Other Gas Laws
    4. The Ideal Gas Law and Some Applications
    5. Gas Mixtures
    6. Kinetic Molecular Theory of Gases
    7. Molecular Effusion and Diffusion
    8. Real Gases
    9. End-of-Chapter Material
  14. Chapter 7. Energy and Chemistry
    1. Formation Reactions
    2. Energy
    3. Stoichiometry Calculations Using Enthalpy
    4. Enthalpy and Chemical Reactions
    5. Work and Heat
    6. Hess’s Law
    7. End-of-Chapter Material
  15. Chapter 8. Electronic Structure
    1. Light
    2. Quantum Numbers for Electrons
    3. Organization of Electrons in Atoms
    4. Electronic Structure and the Periodic Table
    5. Periodic Trends
    6. End-of-Chapter Material
  16. Chapter 9. Chemical Bonds
    1. Lewis Electron Dot Diagrams
    2. Electron Transfer: Ionic Bonds
    3. Covalent Bonds
    4. Other Aspects of Covalent Bonds
    5. Violations of the Octet Rule
    6. Molecular Shapes and Polarity
    7. Valence Bond Theory and Hybrid Orbitals
    8. Molecular Orbitals
    9. End-of-Chapter Material
  17. Chapter 10. Solids and Liquids
    1. Properties of Liquids
    2. Solids
    3. Phase Transitions: Melting, Boiling, and Subliming
    4. Intermolecular Forces
    5. End-of-Chapter Material
  18. Chapter 11. Solutions
    1. Colligative Properties of Solutions
    2. Concentrations as Conversion Factors
    3. Quantitative Units of Concentration
    4. Colligative Properties of Ionic Solutes
    5. Some Definitions
    6. Dilutions and Concentrations
    7. End-of-Chapter Material
  19. Chapter 12. Acids and Bases
    1. Acid-Base Titrations
    2. Strong and Weak Acids and Bases and Their Salts
    3. Brønsted-Lowry Acids and Bases
    4. Arrhenius Acids and Bases
    5. Autoionization of Water
    6. Buffers
    7. The pH Scale
    8. End-of-Chapter Material
  20. Chapter 13. Chemical Equilibrium
    1. Chemical Equilibrium
    2. The Equilibrium Constant
    3. Shifting Equilibria: Le Chatelier’s Principle
    4. Calculating Equilibrium Constant Values
    5. Some Special Types of Equilibria
    6. End-of-Chapter Material
  21. Chapter 14. Oxidation and Reduction
    1. Oxidation-Reduction Reactions
    2. Balancing Redox Reactions
    3. Applications of Redox Reactions: Voltaic Cells
    4. Electrolysis
    5. End-of-Chapter Material
  22. Chapter 15. Nuclear Chemistry
    1. Units of Radioactivity
    2. Uses of Radioactive Isotopes
    3. Half-Life
    4. Radioactivity
    5. Nuclear Energy
    6. End-of-Chapter Material
  23. Chapter 16. Organic Chemistry
    1. Hydrocarbons
    2. Branched Hydrocarbons
    3. Alkyl Halides and Alcohols
    4. Other Oxygen-Containing Functional Groups
    5. Other Functional Groups
    6. Polymers
    7. End-of-Chapter Material
  24. Chapter 17. Kinetics
    1. Factors that Affect the Rate of Reactions
    2. Reaction Rates
    3. Rate Laws
    4. Concentration–Time Relationships: Integrated Rate Laws
    5. Activation Energy and the Arrhenius Equation
    6. Reaction Mechanisms
    7. Catalysis
    8. End-of-Chapter Material
  25. Chapter 18. Chemical Thermodynamics
    1. Spontaneous Change
    2. Entropy and the Second Law of Thermodynamics
    3. Measuring Entropy and Entropy Changes
    4. Gibbs Free Energy
    5. Spontaneity: Free Energy and Temperature
    6. Free Energy under Nonstandard Conditions
    7. End-of-Chapter Material
  26. Appendix A: Periodic Table of the Elements
  27. Appendix B: Selected Acid Dissociation Constants at 25°C
  28. Appendix C: Solubility Constants for Compounds at 25°C
  29. Appendix D: Standard Thermodynamic Quantities for Chemical Substances at 25°C
  30. Appendix E: Standard Reduction Potentials by Value
  31. Glossary
  32. About the Authors
  33. Versioning History

Neutralization Reactions

Learning Objectives

  1. Identify an acid and a base.
  2. Identify a neutralization reaction and predict its products.

In the section called “Acids”, we defined an acid as an ionic compound that contains H+ as the cation. This is slightly incorrect, but until additional concepts were developed, a better definition needed to wait. Now we can redefine an acid: an acid is any compound that increases the amount of hydrogen ion (H+) in an aqueous solution. The chemical opposite of an acid is a base. The equivalent definition of a base is that a base is a compound that increases the amount of hydroxide ion (OH−) in an aqueous solution. These original definitions were proposed by Arrhenius (the same person who proposed ion dissociation) in 1884, so they are referred to as the Arrhenius definitions of an acid and a base, respectively.

You may recognize that, based on the description of a hydrogen atom, an H+ ion is a hydrogen atom that has lost its lone electron; that is, H+ is simply a proton. Do we really have bare protons moving about in aqueous solution? No. What is more likely is that the H+ ion has attached itself to one (or more) water molecule(s). To represent this chemically, we define the hydronium ion H3O+(aq), a water molecule with an extra hydrogen ion attached to it, as H3O+, which represents an additional proton attached to a water molecule. We use the hydronium ion as the more logical way a hydrogen ion appears in an aqueous solution, although in many chemical reactions H+ and H3O+ are treated equivalently.

The reaction of an acid and a base is called a neutralization reaction. Although acids and bases have their own unique chemistries, the acid and base cancel each other’s chemistry to produce a rather innocuous substance—water. In fact, the general reaction between an acid and a base is

acid + base → water + salt

where the term salt is generally used to define any ionic compound (soluble or insoluble) that is formed from a reaction between an acid and a base. (In chemistry, the word salt refers to more than just table salt.) For example, the balanced chemical equation for the reaction between HCl(aq) and KOH(aq) is

HCl(aq) + KOH(aq) → H2O(ℓ) + KCl(aq)

where the salt is KCl. By counting the number of atoms of each element, we find that only one water molecule is formed as a product. However, in the reaction between HCl(aq) and Mg(OH)2(aq), additional molecules of HCl and H2O are required to balance the chemical equation:

2HCl(aq) + Mg(OH)2(aq) → 2H2O(ℓ) + MgCl2(aq)

Here, the salt is MgCl2. (This is one of several reactions that take place when a type of antacid—a base—is used to treat stomach acid.)

Example 4.13

Problems

Write the neutralization reactions between each acid and base.

  1. HNO3(aq) and Ba(OH)2(aq)
  2. H3PO4(aq) and Ca(OH)2(aq)

Solutions

First, we will write the chemical equation with the formulas of the reactants and the expected products; then we will balance the equation.

  1. The expected products are water and barium nitrate, so the initial chemical reaction is HNO3(aq) + Ba(OH)2(aq) → H2O(ℓ) + Ba(NO3)2(aq). To balance the equation, we need to realize that there will be two H2O molecules, so two HNO3 molecules are required:

    2HNO3(aq) + Ba(OH)2(aq) → 2H2O(ℓ) + Ba(NO3)2(aq)

    This chemical equation is now balanced.

  2. The expected products are water and calcium phosphate, so the initial chemical equation is H3PO4(aq) + Ca(OH)2(aq) → H2O(ℓ) + Ca3(PO4)2(s). According to the solubility rules, Ca3(PO4)2 is insoluble, so it has an (s) phase label. To balance this equation, we need two phosphate ions and three calcium ions; we end up with six water molecules to balance the equation:

    2H3PO4(aq) + 3Ca(OH)2(aq) → 6H2O(ℓ) + Ca3(PO4)2(s)

    This chemical equation is now balanced.

Test Yourself

Write the neutralization reaction between H2SO4(aq) and Sr(OH)2(aq).

Answer

H2SO4(aq) + Sr(OH)2(aq) → 2H2O(ℓ) + SrSO4(aq)

Neutralization reactions are one type of chemical reaction that proceeds even if one reactant is not in the aqueous phase. For example, the chemical reaction between HCl(aq) and Fe(OH)3(s) still proceeds according to the equation:

3HCl(aq) + Fe(OH)3(s) → 3H2O(ℓ) + FeCl3(aq)

even though Fe(OH)3 is not soluble. When one realizes that Fe(OH)3(s) is a component of rust, this explains why some cleaning solutions for rust stains contain acids—the neutralization reaction produces products that are soluble and wash away. (Washing with acids like HCl is one way to remove rust and rust stains, but HCl must be used with caution!)

Complete and net ionic reactions for neutralization reactions will depend on whether the reactants and products are soluble, even if the acid and base react. For example, in the reaction of HCl(aq) and NaOH(aq):

HCl(aq) + NaOH(aq) → H2O(ℓ) + NaCl(aq)

The complete ionic reaction is:

H+(aq) + Cl−(aq) + Na+(aq) + OH−(aq) → H2O(ℓ) + Na+(aq) + Cl−(aq)

The Na+(aq) and Cl−(aq) ions are spectator ions, so we can remove them to have:

H+(aq) + OH−(aq) → H2O(ℓ)

as the net ionic equation. If we wanted to write this in terms of the hydronium ion, H3O+(aq), we would write it as:

H3O+(aq) + OH−(aq) → 2H2O(ℓ)

With the exception of the introduction of an extra water molecule, these two net ionic equations are equivalent.

However, for the reaction between HCl(aq) and Cr(OH)2(s), because chromium(II) hydroxide is insoluble, we cannot separate it into ions for the complete ionic equation:

2H+(aq) + 2Cl−(aq) + Cr(OH)2(s) → 2H2O(ℓ) + Cr2+(aq) + 2Cl−(aq)

The chloride ions are the only spectator ions here, so the net ionic equation is:

2H+(aq) + Cr(OH)2(s) → 2H2O(ℓ) + Cr2+(aq)

Example 4.14

Problem

Oxalic acid, H2C2O4(s), and Ca(OH)2(s) react very slowly. What is the net ionic equation between these two substances if the salt formed is insoluble? (The anion in oxalic acid is the oxalate ion, C2O42−.)

Solution

The products of the neutralization reaction will be water and calcium oxalate:

H2C2O4(s) + Ca(OH)2(s) → 2H2O(ℓ) + CaC2O4(s)

Because nothing is dissolved, there are no substances to separate into ions, so the net ionic equation is the equation of the three solids and one liquid.

Test Yourself

What is the net ionic equation between HNO3(aq) and Ti(OH)4(s)?

Answer

4H+(aq) + Ti(OH)4(s) → 4H2O(ℓ) + Ti4+(aq)

Key Takeaways

  • The Arrhenius definition of an acid is a substance that increases the amount of H+ in an aqueous solution.
  • The Arrhenius definition of a base is a substance that increases the amount of OH− in an aqueous solution.
  • Neutralization is the reaction of an acid and a base, which forms water and a salt.
  • Net ionic equations for neutralization reactions may include solid acids, solid bases, solid salts, and water.

Exercises

Questions

  1. What is the Arrhenius definition of an acid?
  2. What is the Arrhenius definition of a base?
  3. Predict the products of each acid-base combination listed. Assume that a neutralization reaction occurs.
    1. HCl and KOH
    2. H2SO4 and KOH
    3. H3PO4 and Ni(OH)2
  4. Predict the products of each acid-base combination listed. Assume that a neutralization reaction occurs.
    1. HBr and Fe(OH)3
    2. HNO2 and Al(OH)3
    3. HClO3 and Mg(OH)2
  5. Write a balanced chemical equation for each neutralization reaction in Exercise 3.
  6. Write a balanced chemical equation for each neutralization reaction in Exercise 4.
  7. Write a balanced chemical equation for the neutralization reaction between each given acid and base. Include the proper phase labels.
    1. HI(aq) + KOH(aq) → ?
    2. H2SO4(aq) + Ba(OH)2(aq) → ?
  8. Write a balanced chemical equation for the neutralization reaction between each given acid and base. Include the proper phase labels.
    1. HNO3(aq) + Fe(OH)3(s) → ?
    2. H3PO4(aq) + CsOH(aq) → ?
  9. Write the net ionic equation for each neutralization reaction in Exercise 7.
  10. Write the net ionic equation for each neutralization reaction in Exercise 8.
  11. Write the complete and net ionic equations for the neutralization reaction between HClO3(aq) and Zn(OH)2(s). Assume the salt is soluble.
  12. Write the complete and net ionic equations for the neutralization reaction between H2C2O4(s) and Sr(OH)2(aq). Assume the salt is insoluble.
  13. Explain why the net ionic equation for the neutralization reaction between HCl(aq) and KOH(aq) is the same as the net ionic equation for the neutralization reaction between HNO3(aq) and RbOH.
  14. Explain why the net ionic equation for the neutralization reaction between HCl(aq) and KOH(aq) is different from the net ionic equation for the neutralization reaction between HCl(aq) and AgOH.
  15. Write the complete and net ionic equations for the neutralization reaction between HCl(aq) and KOH(aq) using the hydronium ion in place of H+. What difference does it make when using the hydronium ion?
  16. Write the complete and net ionic equations for the neutralization reaction between HClO3(aq) and Zn(OH)2(s) using the hydronium ion in place of H+. Assume the salt is soluble. What difference does it make when using the hydronium ion?

Answers

  1. An Arrhenius acid increases the amount of H+ ions in an aqueous solution.
    1. KCl and H2O
    2. K2SO4 and H2O
    3. Ni3(PO4)2 and H2O
    1. HCl + KOH → KCl + H2O
    2. H2SO4 + 2KOH → K2SO4 + 2H2O
    3. 2H3PO4 + 3Ni(OH)2 → Ni3(PO4)2 + 6H2O
    1. HI(aq) + KOH(aq) → KCl(aq) + H2O(ℓ)
    2. H2SO4(aq) + Ba(OH)2(aq) → BaSO4(s) + 2H2O(ℓ)
    1. H+(aq) + OH−(aq) → H2O(ℓ)
    2. 2H+(aq) + SO42−(aq) + Ba2+(aq) + 2OH−(aq) → BaSO4(s) + 2H2O(ℓ)
  1. Complete ionic equation:

    2H+(aq) + 2ClO3−(aq) + Zn2+(aq) + 2OH−(aq) → Zn2+(aq) + 2ClO3−(aq) + 2H2O(ℓ)

    Net ionic equation:

    2H+(aq) + 2OH−(aq) → 2H2O(ℓ)

  1. Because the salts are soluble in both cases, the net ionic reaction is just H+(aq) + OH−(aq) → H2O(ℓ).
  1. Complete ionic equation:

    H3O+(aq) + Cl−(aq) + K+(aq) + OH−(aq) → 2H2O(ℓ) + K+(aq) + Cl−(aq)

    Net ionic equation:

    H3O+(aq) + OH−(aq) → 2H2O(ℓ)

    The difference is simply the presence of an extra water molecule as a product.

Annotate

Next Chapter
End-of-Chapter Material
PreviousNext
Chemistry

Copyright © 2014

                                by Jessie A. Key

            Introductory Chemistry - 1st Canadian Edition by Jessie A. Key is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.
Powered by Manifold Scholarship. Learn more at
Opens in new tab or windowmanifoldapp.org