Edward J. Neth; Paul Flowers; William R. Robinson, PhD; Klaus Theopold; Richard Langley

Learning Objectives

By the end of this section, you will be able to:

Explain the Lewis model of acid-base chemistry
Write equations for the formation of adducts and complex ions
Perform equilibrium calculations involving formation constants

In 1923, G. N. Lewis proposed a generalized definition of acid-base behavior in which acids and bases are identified by their ability to accept or to donate a pair of electrons and form a coordinate covalent bond.

A coordinate covalent bond (or dative bond) occurs when one of the atoms in the bond provides both bonding electrons. For example, a coordinate covalent bond occurs when a water molecule combines with a hydrogen ion to form a hydronium ion. A coordinate covalent bond also results when an ammonia molecule combines with a hydrogen ion to form an ammonium ion. Both of these equations are shown here.

This figure shows two reactions represented with Lewis structures. The first shows an O atom bonded to two H atoms. The O atom has two lone pairs of electrons. There is a plus sign and then an H atom with a superscript positive sign followed by a right-facing arrow. The next Lewis structure is in brackets and shows an O atom bonded to three H atoms. There is one lone pair of electrons on the O atom. Outside of the brackets is a superscript positive sign. The second reaction shows an N atom bonded to three H atoms. The N atom has one lone pair of electrons. There is a plus sign and then an H superscript positive sign. After the H superscript positive sign is a right-facing arrow. The next Lewis structure is in brackets. It shows an N atom bonded to four H atoms. There is a superscript positive sign outside the brackets.

A Lewis acid is any species (molecule or ion) that can accept a pair of electrons, and a Lewis base is any species (molecule or ion) that can donate a pair of electrons.

A Lewis acid-base reaction occurs when a base donates a pair of electrons to an acid. A Lewis acid-base adduct, a compound that contains a coordinate covalent bond between the Lewis acid and the Lewis base, is formed. The following equations illustrate the general application of the Lewis concept.

The boron atom in boron trifluoride, BF₃, has only six electrons in its valence shell. Being short of the preferred octet, BF₃ is a very good Lewis acid and reacts with many Lewis bases; a fluoride ion is the Lewis base in this reaction, donating one of its lone pairs:

This figure illustrates a chemical reaction using structural formulas. On the left, an F atom is surrounded by four electron dot pairs and has a superscript negative symbol. This structure is labeled below as “Lewis base.” Following a plus sign is another structure which has a B atom at the center and three F atoms single bonded above, right, and below. Each F atom has three pairs of electron dots. This structure is labeled below as “Lewis acid.” Following a right pointing arrow is a structure in brackets that has a central B atom to which 4 F atoms are connected with single bonds above, below, to the left, and to the right. Each F atom in this structure has three pairs of electron dots. Outside the brackets is a superscript negative symbol. This structure is labeled below as “Acid-base adduct.”

In the following reaction, each of two ammonia molecules, Lewis bases, donates a pair of electrons to a silver ion, the Lewis acid:

This figure illustrates a chemical reaction using structural formulas. On the left side, a 2 preceeds an N atom which has H atoms single bonded above, to the left, and below. A single electron dot pair is on the right side of the N atom. This structure is labeled below as “Lewis base.” Following a plus sign is an A g atom which has a superscript plus symbol. Following a right pointing arrow is a structure in brackets that has a central A g atom to which N atoms are connected with single bonds to the left and to the right. Each of these N atoms has H atoms bonded above, below, and to the outside of the structure. Outside the brackets is a superscript plus symbol. This structure is labeled below as “Acid-base adduct.”

Nonmetal oxides act as Lewis acids and react with oxide ions, Lewis bases, to form oxyanions:

This figure illustrates a chemical reaction using structural formulas. On the left, an O atom is surrounded by four electron dot pairs and has a superscript 2 negative. This structure is labeled below as “Lewis base.” Following a plus sign is another structure which has an S atom at the center. O atoms are single bonded above and below. These O atoms have three electron dot pairs each. To the right of the S atom is a double bonded O atom which has two pairs of electron dots. This structure is labeled below as “Lewis acid.” Following a right pointing arrow is a structure in brackets that has a central S atom to which 4 O atoms are connected with single bonds above, below, to the left, and to the right. Each of the O atoms has three pairs of electron dots. Outside the brackets is a superscript 2 negative. This structure is labeled below as “Acid-base adduct.”

Many Lewis acid-base reactions are displacement reactions in which one Lewis base displaces another Lewis base from an acid-base adduct, or in which one Lewis acid displaces another Lewis acid:

This figure shows three chemical reactions in three rows using structural formulas. In the first row, to the left, in brackets is a structure that has a central A g atom to which N atoms are connected with single bonds to the left and to the right. Each of these N atoms has H atoms bonded above, below, and to the outside of the structure. Outside the brackets is a superscript plus symbol. This structure is labeled below as “Acid-base adduct.” Following a plus sign is a 2 and another structure in brackets that shows a C atom triple bonded to an N atom. The C atom has an unshared electron pair on its left side and the N atom has an unshared pair on its right side. Outside the brackets to the right is a superscript negative symbol. This structure is labeled below as “Base.” Following a right pointing arrow is a structure in brackets that has a central A g atom to which 4 FC atoms are connected with single bonds to the left and to the right. At each of the two ends, N atoms are triple bonded to the C atoms. The N atoms each have an unshared electron pair at the end of the structure. Outside the brackets is a superscript negative symbol. This structure is labeled below as “New adduct.” Following a plus sign is an N atom which has H atoms single bonded above, to the left, and below. A single electron dot pair is on the left side of the N atom. This structure is labeled below as “New base.” In the second row, on the left side in brackets is a structure with a central C atom. O atoms, each with three unshared electron pairs, are single bonded above and below and a third O atom, with two unshared electron pairs, is double bonded to the right. Outside the brackets is a superscript 2 negative. This structure is labeled below as “Acid-base adduct.” Following a plus sign is another structure which has an S atom at the center. O atoms are single bonded above and below. These O atoms have three electron dot pairs each. To the right of the S atom is a double bonded O atom which has two pairs of electron dots. This structure is labeled below as “Acid.” Following a right pointing arrow is a structure in brackets that has a central S atom to which 4 O atoms are connected with single bonds above, below, to the left, and to the right. Each of the O atoms has three pairs of electron dots. Outside the brackets is a superscript 2 negative. This structure is labeled below as “New adduct.”

The last displacement reaction shows how the reaction of a Brønsted-Lowry acid with a base fits into the Lewis concept. A Brønsted-Lowry acid such as HCl is an acid-base adduct according to the Lewis concept, and proton transfer occurs because a more stable acid-base adduct is formed. Thus, although the definitions of acids and bases in the two theories are quite different, the theories overlap considerably.

Many slightly soluble ionic solids dissolve when the concentration of the metal ion in solution is decreased through the formation of complex (polyatomic) ions in a Lewis acid-base reaction. For example, silver chloride dissolves in a solution of ammonia because the silver ion reacts with ammonia to form the complex ion $Ag {({NH}_{3})}_{2}^{+} .$ The Lewis structure of the $Ag {({NH}_{3})}_{2}^{+}$ ion is:

A structure is shown in brackets. The structure has a central A g atom to which N atoms are single bonded to the left and right. Each of these atoms N atom has H atoms single bonded above, below, and to the outer end of the structure. Outside the brackets is a superscripted plus.

The equations for the dissolution of AgCl in a solution of NH₃ are:

AgCl (s) ⇌ {Ag}^{+} (a q) + {Cl}^{−} (a q)

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{Ag}^{+} (a q) + 2 {NH}_{3} (a q) ⇌ Ag {({NH}_{3})}_{2}^{+} (a q)

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Net: AgCl (s) + 2 {NH}_{3} (a q) ⇌ Ag {({NH}_{3})}_{2}^{+} (a q) + {Cl}^{−} (a q)

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Aluminum hydroxide dissolves in a solution of sodium hydroxide or another strong base because of the formation of the complex ion $Al {(OH)}_{4}^{−} .$ The Lewis structure of the $Al {(OH)}_{4}^{−}$ ion is:

An H atom is bonded to an O atom. The O atom has 2 dots above it and 2 dots below it. The O atom is bonded to an A l atom, which has three additional O atoms bonded to it as well. Each of these additional O atoms has 4 dots arranged around it, and is bonded to an H atom. This entire molecule is contained in brackets, to the right of which is a superscripted negative sign.

The equations for the dissolution are:

Al {(OH)}_{3} (s) ⇌ {Al}^{3+} (a q) + 3 {OH}^{−} (a q)

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{Al}^{3+} (a q) + 4 {OH}^{−} (a q) ⇌ Al {(OH)}_{4}^{−} (a q)

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Net: Al {(OH)}_{3} (s) + {OH}^{−} (a q) ⇌ Al {(OH)}_{4}^{−} (a q)

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Mercury(II) sulfide dissolves in a solution of sodium sulfide because HgS reacts with the S^2– ion:

HgS (s) ⇌ {Hg}^{2+} (a q) + S^{2−} (a q)

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{Hg}^{2+} (a q) + 2 S^{2−} (a q) ⇌ {HgS}_{2}^{2−} (a q)

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Net: HgS (s) + S^{2−} (a q) ⇌ {HgS}_{2}^{2−} (a q)

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A complex ion consists of a central atom, typically a transition metal cation, surrounded by ions, or molecules called ligands. These ligands can be neutral molecules like H₂O or NH₃, or ions such as CN^– or OH^–. Often, the ligands act as Lewis bases, donating a pair of electrons to the central atom. The ligands form bonds with the central atom or ion, creating a new ion with a charge equal to the sum of the charges of the ligands and the central atom or ion. This more complex arrangement is why the resulting ion is called a complex ion. The complex ion formed in these reactions cannot be predicted; it must be determined experimentally. The types of bonds formed in complex ions are called coordinate covalent bonds, as electrons from the ligands are being shared with the central atom. Because of this, complex ions are sometimes referred to as coordination complexes. This will be studied further in upcoming chapters.

The equilibrium constant for the reaction of the components of a complex ion to form the complex ion in solution is called a formation constant (K_f) (sometimes called a stability constant). For example, the complex ion $Cu {(CN)}_{2}^{−}$ is shown here:

It forms by the reaction:

{Cu}^{+} (a q) + 2 {CN}^{−} (a q) ⇌ Cu {(CN)}_{2}^{−} (a q)

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At equilibrium:

K_{f} = Q = \frac{[Cu {(CN)}_{2}^{−}]}{[{Cu}^{+}] {[{CN}^{−}]}^{2}}

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The inverse of the formation constant is the dissociation constant (K_d), the equilibrium constant for the decomposition of a complex ion into its components in solution. We will work with dissociation constants further in the exercises for this section. Appendix K and Table 15.2 are tables of formation constants. In general, the larger the formation constant, the more stable the complex.

Common Complex Ions by Decreasing Formation Constants
Substance	K_f at 25 °C
${AlF}_{6}^{3−}$	7 $\times$ 10¹⁹
$Ag {({NH}_{3})}_{2}^{+}$	1.7 $\times$ 10⁷
${Cd {(CN)}_{4}}^{2−}$	3 $\times$ 10¹⁸

Table 15.2

As an example of dissolution by complex ion formation, let us consider what happens when we add aqueous ammonia to a mixture of silver chloride and water. Silver chloride dissolves slightly in water, giving a small concentration of Ag⁺ ([Ag⁺] = 1.3 $\times$ 10^–5 M):

AgCl (s) ⇌ {Ag}^{+} (a q) + {Cl}^{−} (a q)

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However, if NH₃ is present in the water, the complex ion, $Ag {({NH}_{3})}_{2}^{+},$ can form according to the equation:

{Ag}^{+} (a q) + 2 {NH}_{3} (a q) ⇌ Ag {({NH}_{3})}_{2}^{+} (a q)

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with

K_{f} = \frac{[Ag {({NH}_{3})}_{2}^{+}]}{[{Ag}^{+}] {[{NH}_{3}]}^{2}} = 1.7 \times 10^{7}

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The large size of this formation constant indicates that most of the free silver ions produced by the dissolution of AgCl combine with NH₃ to form $Ag {({NH}_{3})}_{2}^{+} .$ As a consequence, the concentration of silver ions, [Ag⁺], is reduced, and the reaction quotient for the dissolution of silver chloride, [Ag⁺][Cl^–], falls below the solubility product of AgCl:

Q = [{Ag}^{+}] [{Cl}^{−}] < K_{sp}

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More silver chloride then dissolves. If the concentration of ammonia is great enough, all of the silver chloride dissolves.

Example 15.13

Dissociation of a Complex Ion

Calculate the concentration of the silver ion in a solution that initially is 0.10 M with respect to

Ag {({NH}_{3})}_{2}^{+} .

Solution

We use the familiar path to solve this problem: Four boxes are shown side by side, with three right facing arrows connecting them. The first box contains the text “Determine the direction of change.” The second box contains the text “Determine x and the equilibrium concentrations.” The third box contains the text “Solve for x and the equilibrium concentrations.” The fourth box contains the text “Check the math.”

Four boxes are shown side by side, with three right facing arrows connecting them. The first box contains the text “Determine the direction of change.” The second box contains the text “Determine x and the equilibrium concentrations.” The third box contains the text “Solve for x and the equilibrium concentrations.” The fourth box contains the text “Check the math.”

Step 1.
Determine the direction of change. The complex ion $Ag {({NH}_{3})}_{2}^{+}$ is in equilibrium with its components, as represented by the equation:

${Ag}^{+} (a q) + 2 {NH}_{3} (a q) ⇌ Ag {({NH}_{3})}_{2}^{+} (a q)$
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We write the equilibrium as a formation reaction because Appendix K lists formation constants for complex ions. Before equilibrium, the reaction quotient is larger than the equilibrium constant [K_f = 1.7 $\times$ 10⁷, and $Q = \frac{0.10}{0 \times 0},$ it is infinitely large], so the reaction shifts to the left to reach equilibrium.
Step 2.
Determine x and equilibrium concentrations. We let the change in concentration of Ag⁺ be x. Dissociation of 1 mol of $Ag {({NH}_{3})}_{2}^{+}$ gives 1 mol of Ag⁺ and 2 mol of NH₃, so the change in [NH₃] is 2x and that of $Ag {({NH}_{3})}_{2}^{+}$ is –x. In summary:
Step 3.
Solve for x and the equilibrium concentrations. At equilibrium:

$K_{f} = \frac{[Ag {({NH}_{3})}_{2}^{+}]}{[{Ag}^{+}] {[{NH}_{3}]}^{2}}$
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$1.7 \times 10^{7} = \frac{0.10 - x}{(x) {(2 x)}^{2}}$
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Both Q and K_f are much larger than 1, so let us assume that the changes in concentrations needed to reach equilibrium are small. Thus 0.10 – x is approximated as 0.10:

$1.7 \times 10^{7} = \frac{0.10}{(x) {(2 x)}^{2}}$
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$x^{3} = \frac{0.10}{4 (1.7 \times 10^{7})} = 1.5 \times 10^{- 9}$
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$x = \sqrt[3]{1.5 \times 10^{- 9}} = 1.1 \times 10^{- 3}$
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Because only 1.1% of the $Ag {({NH}_{3})}_{2}^{+}$ dissociates into Ag⁺ and NH₃, the assumption that x is small is justified.
Now we determine the equilibrium concentrations:

$[{Ag}^{+}] = 0 + x = 1.1 \times 10^{- 3} M$
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$[{NH}_{3}] = 0 + 2 x = 2.2 \times 10^{- 3} M$
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$[Ag {({NH}_{3})}_{2}^{+}] = 0.10 - x = 0.10 - 0.0011 = 0.099$
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The concentration of free silver ion in the solution is 0.0011 M.
Step 4.
Check the work. The value of Q calculated using the equilibrium concentrations is equal to K_f within the error associated with the significant figures in the calculation.

Check Your Learning

Calculate the silver ion concentration, [Ag⁺], of a solution prepared by dissolving 1.00 g of AgNO₃ and 10.0 g of KCN in sufficient water to make 1.00 L of solution. (Hint: Because Q < K_f, assume the reaction goes to completion then calculate the [Ag⁺] produced by dissociation of the complex.)

Answer:

2.5 $\times$ 10^–22 M

15.2 Lewis Acids and Bases

Dissociation of a Complex Ion

Solution

Check Your Learning