Paul Flowers; Edward J. Neth; 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.

Reactions involving the formation of coordinate covalent bonds are classified as Lewis acid-base chemistry. The species donating the electron pair that compose the bond is a Lewis base, the species accepting the electron pair is a Lewis acid, and the product of the reaction is a Lewis acid-base adduct. As the two examples above illustrate, Brønsted-Lowry acid-base reactions represent a subcategory of Lewis acid reactions, specifically, those in which the acid species is H⁺. A few examples involving other Lewis acids and bases are described below.

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:

Two chemical reactions in two rows using structural formulas. First row, to the left, in brackets is a structure with a central A g atom to which N atoms are connected with single bonds to the left and right. Each N atom has H atoms bonded above, below, and to the outside. Outside the brackets is a superscript plus symbol. This structure is labeled “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 “Base.” Following a right pointing arrow is a structure in brackets with a central A g atom to which 4 FC atoms are connected with single bonds to the left and 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 “New adduct.” Following a plus sign is an N atom with 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 “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 “Acid-base adduct.” Following a plus sign is another structure with 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 “Acid.” Following a right pointing arrow is a structure in brackets with a central S atom to which 4 O atoms are connected with single bonds above, below, left, and right. Each of the O atoms has three pairs of electron dots. Outside the brackets is a superscript 2 negative. This structure is labeled “New adduct.” Following a plus sign is a structure with a central C atom that has two O atoms, each with two unshared electron pairs, double bonded to the left and right.

Another type of Lewis acid-base chemistry involves the formation of a complex ion (or a coordination complex) comprising 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. These types of Lewis acid-base reactions are examples of a broad subdiscipline called coordination chemistry—the topic of another chapter in this text.

The equilibrium constant for the reaction of a metal ion with one or more ligands to form a coordination complex is called a formation constant (K_f) (sometimes called a stability constant). For example, the complex ion $Cu {(CN)}_{2}^{−}$

is produced by the reaction

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

The formation constant for this reaction is

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

Alternatively, the reverse reaction (decomposition of the complex ion) can be considered, in which case the equilibrium constant is a dissociation constant (K_d). Per the relation between equilibrium constants for reciprocal reactions described, the dissociation constant is the mathematical inverse of the formation constant, K_d = K_f^–1. A tabulation of formation constants is provided in Appendix K.

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)

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)

with

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

The large size of this formation constant indicates that most of the free silver ions produced by the dissolution of AgCl combine with NH ₃ 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}

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

Example 15.14

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

Applying the standard ICE approach to this reaction yields the following:

This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header, “A g superscript positive sign plus 2 N H subscript 3 equilibrium sign A g ( N H subscript 3 ) subscript 2 superscript positive sign.” Under the second column is a subgroup of three rows and three columns. The first column contains: 0, positive x, x. The second column contains: 0, positive 2 x, 2 x. The third column contains 0.10, negative x, and 0.10 minus x.

Substituting these equilibrium concentration terms into the K_f expression gives

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

1.7 \times 10^{7} = \frac{0.10 - x}{(x) {(2 x)}^{2}}

The very large equilibrium constant means the amount of the complex ion that will dissociate, x, will be very small. Assuming x << 0.1 permits simplifying the above equation:

1.7 \times 10^{7} = \frac{0.10}{(x) {(2 x)}^{2}}

x^{3} = \frac{0.10}{4 (1.7 \times 10^{7})} = 1.5 \times 10^{- 9}

x = \sqrt[3]{1.5 \times 10^{- 9}} = 1.1 \times 10^{- 3}

Because only 1.1% of the $Ag {({NH}_{3})}_{2}^{+}$ dissociates into Ag⁺ and NH₃, the assumption that x is small is justified.

Using this value of x and the relations in the above ICE table allows calculation of all species’ equilibrium concentrations:

[{Ag}^{+}] = 0 + x = 1.1 \times 10^{- 3} M

[{NH}_{3}] = 0 + 2 x = 2.2 \times 10^{- 3} M

[Ag {({NH}_{3})}_{2}^{+}] = 0.10 - x = 0.10 - 0.0011 = 0.099

The concentration of free silver ion in the solution is 0.0011 M.

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 K_f is very large, the silver/cyanide complex ion (see Appendix K), assume the reaction goes to completion and then calculate the [Ag⁺] produced by dissociation of the complex.)

Answer:

2.9 $\times$ 10^–22 M

15.2 Lewis Acids and Bases

Dissociation of a Complex Ion

Solution

Check Your Learning