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Chapter 15: Acid-Based Equilibria

15.1 Brønsted-Lowry Acids and Bases

Learning Outcomes

  • Identify acids, bases, and conjugate acid-base pairs according to the Brønsted-Lowry definition
  • Write equations for acid and base ionization reactions
  • Use the ion-product constant for water to calculate hydronium and hydroxide ion concentrations
  • Describe the acid-base behavior of amphiprotic substances

The acid-base reaction class has been studied for quite some time. In 1680, Robert Boyle reported traits of acid solutions that included their ability to dissolve many substances, to change the colors of certain natural dyes, and to lose these traits after coming in contact with alkali (base) solutions. In the eighteenth century, it was recognized that acids have a sour taste, react with limestone to liberate a gaseous substance (now known to be COA2), and interact with alkalis to form neutral substances. In 1815, Humphry Davy contributed greatly to the development of the modern acid-base concept by demonstrating that hydrogen is the essential constituent of acids. Around that same time, Joseph Louis Gay-Lussac concluded that acids are substances that can neutralize bases and that these two classes of substances can be defined only in terms of each other. The significance of hydrogen was reemphasized in 1884 when Svante Arrhenius defined an acid as a compound that dissolves in water to yield hydrogen cations (now recognized to be hydronium ions) and a base as a compound that dissolves in water to yield hydroxide anions.

Johannes Brønsted and Thomas Lowry proposed a more general description in 1923 in which acids and bases were defined in terms of the transfer of hydrogen ions, H+. (Note that these hydrogen ions are often referred to simply as protons, since that subatomic particle is the only component of cations derived from the most abundant hydrogen isotope, A1A221H.) A compound that donates a proton to another compound is called a Brønsted-Lowry acid, and a compound that accepts a proton is called a Brønsted-Lowry base. An acid-base reaction is, thus, the transfer of a proton from a donor (acid) to an acceptor (base).

The concept of conjugate pairs is useful in describing Brønsted-Lowry acid-base reactions (and other reversible reactions, as well). When an acid donates HA+, the species that remains is called the conjugate base of the acid because it reacts as a proton acceptor in the reverse reaction. Likewise, when a base accepts HA+, it is converted to its conjugate acid. The reaction between water and ammonia illustrates this idea. In the forward direction, water acts as an acid by donating a proton to ammonia and subsequently becoming a hydroxide ion, OHA−, the conjugate base of water. The ammonia acts as a base in accepting this proton, becoming an ammonium ion, NHA4A+,NHA4A+,the conjugate acid of ammonia. In the reverse direction, a hydroxide ion acts as a base in accepting a proton from ammonium ion, which acts as an acid.

This figure has three parts in two rows. In the first row, two diagrams of acid-base pairs are shown. On the left, a space filling model of H subscript 2 O is shown with a red O atom at the center and two smaller white H atoms attached in a bent shape. Above this model is the label “H subscript 2 O (acid)” in purple. An arrow points right, which is labeled “Remove H superscript plus.” To the right is another space filling model with a single red O atom to which a single smaller white H atom is attached. The label in purple above this model reads, “O H superscript negative (conjugate base).” Above both of these red and white models is an upward pointing bracket that is labeled “Conjugate acid-base pair.” To the right is a space filling model with a central blue N atom to which three smaller white H atoms are attached in a triangular pyramid arrangement. A label in green above reads “N H subscript 3 (base).” An arrow labeled “Add H superscript plus” points right. To the right of the arrow is another space filling model with a blue central N atom and four smaller white H atoms in a tetrahedral arrangement. The green label above reads “N H subscript 3 superscript plus (conjugate acid).” Above both of these blue and white models is an upward pointing bracket that is labeled “Conjugate acid-base pair.” The second row of the figure shows the chemical reaction, H subscript 2 O ( l ) is shown in purple, and is labeled below in purple as “acid,” plus N H subscript 3 (a q) in green, labeled below in green as “base,” followed by a double sided arrow arrow and O H superscript negative (a q) in purple, labeled in purple as “conjugate base,” plus N H subscript 4 superscript plus (a q)” in green, which is labeled in green as “conjugate acid.” The acid on the left side of the equation is connected to the conjugate base on the right with a purple line. Similarly, the base on the left is connected to the conjugate acid on the right side.

The reaction between a Brønsted-Lowry acid and water is called acid ionization. For example, when hydrogen fluoride dissolves in water and ionizes, protons are transferred from hydrogen fluoride molecules to water molecules, yielding hydronium ions and fluoride ions:

This figure has two rows. In both rows, a chemical reaction is shown. In the first, structural formulas are provided. In this model, in red, is an O atom which has H atoms singly bonded above and to the right. The O atom has lone pairs of electron dots on its left and lower sides. This is followed by a plus sign. The plus sign is followed, in blue, by an N atom with one lone pair of electron dots. The N atom forms a double bond with a C atom, which forms a single bond with a C atom. The second C atom forms a double bond with another C atom, which forms a single bond with another C atom. The fourth C atom forms a double bond with a fifth C atom, which forms a single bond with the N atom. This creates a ring structure. Each C atom is also bonded to an H atom. An equilibrium arrow follows this structure. To the right, in brackets is a structure where an N atom bonded to an H atom, which is red, appears. The N atom forms a double bond with a C atom, which forms a single bond with a C atom. The second C atom forms a double bond with another C atom, which forms a single bond with another C atom. The fourth C atom forms a double bond with a fifth C atom, which forms a single bond with the N atom. This creates a ring structure. Each C atom is also bonded to an H atom. Outside the brackets, to the right, is a superscript positive sign. This structure is followed by a plus sign. Another structure that appears in brackets also appears. An O atom with three lone pairs of electron dots is bonded to an H atom. There is a superscript negative sign outside the brackets. Under the initial equation, is this equation: H subscript 2 plus C subscript 5 N H subscript 5 equilibrium arrow C subscript 5 N H subscript 6 superscript positive sign plus O H superscript negative sign. H subscript 2 O is labeled, “acid,” in red. C subscript 5 N H subscript 5 is labeled, “base,” in blue. C subscript 5 N H subscript 6 superscript positive sign is labeled, “acid” in blue. O H superscript negative sign is labeled, “base,” in red.

Base ionization of a species occurs when it accepts protons from water molecules. In the example below, pyridine molecules, CA5NHA5, undergo base ionization when dissolved in water, yielding hydroxide and pyridinium ions:

This figure has two rows. In both rows, a chemical reaction is shown. In the first, structural formulas are provided. In this model, in red, is an O atom which has H atoms singly bonded above and to the right. The O atom has lone pairs of electron dots on its left and lower sides. This is followed by a plus sign. The plus sign is followed, in blue, by an N atom with one lone pair of electron dots. The N atom forms a double bond with a C atom, which forms a single bond with a C atom. The second C atom forms a double bond with another C atom, which forms a single bond with another C atom. The fourth C atom forms a double bond with a fifth C atom, which forms a single bond with the N atom. This creates a ring structure. Each C atom is also bonded to an H atom. An equilibrium arrow follows this structure. To the right, in brackets is a structure where an N atom bonded to an H atom, which is red, appears. The N atom forms a double bond with a C atom, which forms a single bond with a C atom. The second C atom forms a double bond with another C atom, which forms a single bond with another C atom. The fourth C atom forms a double bond with a fifth C atom, which forms a single bond with the N atom. This creates a ring structure. Each C atom is also bonded to an H atom. Outside the brackets, to the right, is a superscript positive sign. This structure is followed by a plus sign. Another structure that appears in brackets also appears. An O atom with three lone pairs of electron dots is bonded to an H atom. There is a superscript negative sign outside the brackets. To the right, is the equation: k equals [ C subscript 5 N H subscript 6 superscript positive sign ] [ O H superscript negative sign] all divided by [ C subscript 5 N H subscript 5 ]. Under the initial equation, is this equation: H subscript 2 plus C subscript 5 N H subscript 5 equilibrium arrow C subscript 5 N H subscript 6 superscript positive sign plus O H superscript negative sign. H subscript 2 O is labeled, “acid,” in red. C subscript 5 N H subscript 5 is labeled, “base,” in blue. C subscript 5 N H subscript 6 superscript positive sign is labeled, “acid” in blue. O H superscript negative sign is labeled, “base,” in red.

The preceding ionization reactions suggest that water may function as both a base (as in its reaction with hydrogen fluoride) and an acid (as in its reaction with ammonia). Species capable of either donating or accepting protons are called amphiprotric, or more generally, amphoteric, a term that may be used for acids and bases per definitions other than the Brønsted-Lowry one. The equations below show the two possible acid-base reactions for two amphiprotic species, bicarbonate ion and water:

HCOA3A−(aq)+HA2O(l)COA3A2−(aq)+HA3OA+(aq)

HCOA3A−(aq)+HA2O(l)HA2COA3(aq)+OHA−(aq)

The first equation represents the reaction of bicarbonate as an acid with water as a base, whereas the second represents reaction of bicarbonate as a base with water as an acid. When bicarbonate is added to water, both these equilibria are established simultaneously and the composition of the resulting solution may be determined through appropriate equilibrium calculations, as described later in this chapter.

In the liquid state, molecules of an amphiprotic substance can react with one another as illustrated for water in the equations below:

This figure has two rows. In both rows, a chemical reaction is shown. In the first, structural formulas are provided. In this model, in purple, O atom which has H atoms singly bonded above and to the right. The O atom has pairs of electron dots on its left and lower sides. This is followed by a plus sign, which is followed in green by an O atom which has H atoms singly bonded above and to the right. The O atom has pairs of electron dots on its left and lower sides. A double arrow follows. To the right, in brackets is a structure with a central O atom in green, with green H atoms singly bonded above and to the right. A pair of green electron dots is on the lower side of the O atom. To the left of the green O atom, a purple H atom is singly bonded. Outside the brackets to the right is a superscript plus. This is followed by a plus sign and an O atom in purple with pairs of electron dots above, left, and below. An H atom is singly bonded to the right. This atom has a superscript negative sign. The reaction is written in symbolic form below. H subscript 2 O is labeled in purple below as “Acid subscript 1.” This is followed by plus H subscript 2 O, which is labeled in green below as “Base subscript 2.” A double sided arrow follows. To the right is H subscript 3 O superscript plus, which is labeled in green as below in as “Acid subscript 2.” This is followed by plus and O with pairs of dots above, below, and to the left with a singly bonded H on the right with a superscript negative. The label below in purple reads, “ Base subscript 1.”

The process in which like molecules react to yield ions is called autoionization. Liquid water undergoes autoionization to a very slight extent; at 25 °C, approximately two out of every billion water molecules are ionized. The extent of the water autoionization process is reflected in the value of its equilibrium constant, the ion-product constant for water, Kw:

HA2O(l)+HA2O(l)⇌HA3OA+(aq)+OHA−(aq)Kw=[HA3OA+][OHA−]

The slight ionization of pure water is reflected in the small value of the equilibrium constant; at 25 °C, Kw has a value of 1.0 × 10−14. The process is endothermic, and so the extent of ionization and the resulting concentrations of hydronium ion and hydroxide ion increase with temperature. For example, at 100 °C, the value for Kw is about 5.1 × 10−13, roughly 100-times larger than the value at 25 °C.

Example 15.1.1: Ion Concentrations in Pure Water

What are the hydronium ion concentration and the hydroxide ion concentration in pure water at 25 °C?

Show Solution

The autoionization of water yields the same number of hydronium and hydroxide ions. Therefore, in pure water, [H3O+]=[OH−]. At 25 °C: Kw=[H3O+][OH−]=[H3O+]2+=[OH−]2+=1.0×10−14

So:

[H3O+]=[OH−]=1.0×10−14=1.0×10−7M

The hydronium ion concentration and the hydroxide ion concentration are the same, and we find that both equal 1.0 × 10−7M.

Check Your Learning

Example 15.1.2: The Inverse Proportionality of [H3O+] and [OH−]

A solution of carbon dioxide in water has a hydronium ion concentration of 2.0 × 10−6M. What is the concentration of hydroxide ion at 25 °C?

Show Solution

We know the value of the ion-product constant for water at 25 °C:

2HA2O(l)⇌HA3OA+(aq)+OHA−(aq)Kw=[HA3OA+][OHA−]=1.0×10−14

Thus, we can calculate the missing equilibrium concentration.

Rearrangement of the Kw expression yields that OHA− is directly proportional to the inverse of [HA3OA+]:

[OHA−]=Kw[HA3OA+]=1.0×10−142.0×10−6=5.0×10−9

The hydroxide ion concentration in water is reduced to 5.0 × 10−9M as the hydrogen ion concentration increases to 2.0 × 10−6M. This is expected from Le Châtelier’s principle; the autoionization reaction shifts to the left to reduce the stress of the increased hydronium ion concentration and the OHA− is reduced relative to that in pure water.

A check of these concentrations confirms that our arithmetic is correct:

Kw=[HA3OA+][OHA−]=(2.0×10−6)(5.0×10−9)=1.0×10−14

Check Your Learning

Example 15.1.3: Representing the Acid-Base Behavior of an Amphoteric Substance

Write separate equations representing the reaction of HSOA3A−

  1. as an acid with OHA−
  2. as a base with HI
Show Solution
  1. HSOA3A−(aq)+OHA−(aq)⇌SOA3A2−(aq)+HA2O(l)
  2. HSOA3A−(aq)+HI(aq)⇌HA2SOA3(aq)+IA−(aq)

Check Your Learning

Key Concepts and Summary

A compound that can donate a proton (a hydrogen ion) to another compound is called a Brønsted-Lowry acid. The compound that accepts the proton is called a Brønsted-Lowry base. The species remaining after a Brønsted-Lowry acid has lost a proton is the conjugate base of the acid. The species formed when a Brønsted-Lowry base gains a proton is the conjugate acid of the base. Thus, an acid-base reaction occurs when a proton is transferred from an acid to a base, with formation of the conjugate base of the reactant acid and formation of the conjugate acid of the reactant base. Amphiprotic species can act as both proton donors and proton acceptors. Water is the most important amphiprotic species. It can form both the hydronium ion, HA3OA+, and the hydroxide ion, OHA− when it undergoes autoionization:

2HA2O(l)⇌HA3OA+(aq)+OHA−(aq)

The ion product of water, Kw is the equilibrium constant for the autoionization reaction:

Kw=[HA2OA+][OHA−]=1.0×10−14 at 25∘C

Key Equations

  • Kw=[HA2OA+][OHA−]=1.0×10−14 at 25∘C

Try It

  1. Write equations that show NHA3 as both a conjugate acid and a conjugate base.
  2. Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry acid:
    1. HA3OA+
    2. HCl
    3. NHA3
    4. CHA3COA2H
    5. NHA4A+
    6. HSOA4A−
  3. Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry base:
    1. HA2O
    2. OHA−
    3. NHA3
    4. CNA−
    5. SA2−
    6. HA2POA4A−
  4. Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:
    1. HNOA3+HA2O⟶HA3OA++NOA3A−
    2. CNA−+HA2O⟶HCN+OHA−
    3. HA2SOA4+ClA−⟶HCl+HSOA4A−
    4. HSOA4A−+OHA−⟶SOA4A2−+HA2O
    5. OA2−+HA2O⟶2OH−
    6. [Cu(H2O)3(OH)]++[Al(H2O)6]3+⟶[Cu(H2O)4]2++[Al(H2O)5(OH)]2+
    7. H2S+NH2−⟶HS−+NH3
Show Selected Solutions

1. One example for NHA3 as a conjugate acid: NHA2A−+HA+⟶NHA3; as a conjugate base: NHA4A+(aq)+OHA−(aq)⟶NHA3(aq)+HA2O(l)

2. The net ionic equations are as follows:

  1. HA3OA+(aq)⟶HA+(aq)+HA2O(l)
  2. HCl(l)⟶HA+(aq)+ClA−(aq)
  3. NHA3(aq)⟶HA+(aq)+NHA2A−(aq)
  4. CHA3COA2H(aq)⟶HA+(aq)+CHA3COA2A−(aq)
  5. NHA4A+(aq)⟶HA+(aq)+NHA3(aq)
  6. HSOA4A−(aq)⟶HA+(aq)+SOA4A2−(aq)

3. The net ionic equations are as follows:

  1. HA2O(l)+HA+(aq)⟶HA3OA+(aq)
  2. OHA−(aq)+HA+(aq)⟶HA2O(l)
  3. NHA3(aq)+HA+(aq)⟶NHA4A+(aq)
  4. CNA−(aq)+HA+(aq)⟶HCN(aq)
  5. (e) SA2−(aq)+HA+(aq)⟶HSA−(aq)
  6. (f) HA2POA4A−(aq)+HA+(aq)⟶HA3POA4(aq)

4. The labels are Brønsted-Lowry acid = BA; its conjugate base = CB; Brønsted-Lowry base = BB; its conjugate acid = CA.

  1. HNOA3(BA), HA2O(BB), HA3OA+(CA), NO3−(CB)
  2. CN−(BB), HA2O(BA), HCN(CA), OH−(CB)
  3. HA2SOA4(BA), Cl−(BB), HCl(CA), HSO4−(CB)
  4. HSO4−(BA), OH−(BB), SO42- (CB), H2O(CA)
  5. OA2−(BB), HA2O(BA) OHA−(CB and CA)
  6. [Cu(H2O)3(OH)]+(BB), [Al(H2O)6]3+(BA), [Cu(H2O)4]2+(CA), [Al(H2O)5(OH)]2+(CB)
  7. H2S(BA), NH2−(BB), HS−(CB), NH3(CA)

Glossary

acid ionization: reaction involving the transfer of a proton from an acid to water, yielding hydronium ions and the conjugate base of the acid

amphiprotic: species that may either gain or lose a proton in a reaction

amphoteric: species that can act as either an acid or a base

autoionization: reaction between identical species yielding ionic products; for water, this reaction involves transfer of protons to yield hydronium and hydroxide ions

base ionization: reaction involving the transfer of a proton from water to a base, yielding hydroxide ions and the conjugate acid of the base

Brønsted-Lowry acid: proton donor

Brønsted-Lowry base: proton acceptor

conjugate acid: substance formed when a base gains a proton

conjugate base: substance formed when an acid loses a proton

ion-product constant for water (Kw): equilibrium constant for the autoionization of water

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