Chapter 16: Equilibria of Other Reaction Classes

16.2 Lewis Acids and Bases

Learning Outcomes

  • 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 [latex]\ce{H+}[/latex]. A few examples involving other Lewis acids and bases are described below.

The boron atom in boron trifluoride, [latex]\ce{BF3}[/latex], has only six electrons in its valence shell. Being short of the preferred octet, [latex]\ce{BF3}[/latex] 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.”

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 [latex]\ce{H2O}[/latex] or [latex]\ce{NH3}[/latex], or ions such as [latex]\ce{CN-}[[/latex] or [latex]\ce{OH-}[/latex]. 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 (Kf) (sometimes called a stability constant). For example, the complex ion [latex]\ce{Cu(CN)2-}[/latex]

A Cu atom is bonded to two C atoms. Each of these C atoms is triple bonded to an N atom. Each N atom has two dots on the side of it.
is produced by the reaction

[latex]\ce{Cu+}(aq)+\ce{2CN-}(aq)\rightleftharpoons \ce{Cu(CN)2-}(aq)[/latex]

The formation constant for this reaction is

[latex]{K}_{\text{f}}=\dfrac{\ce{[Cu(CN)2-]}} { \ce{[Cu+]} \ce{[CN-]^2}}[/latex]

Alternatively, the reverse reaction (decomposition of the complex ion) can be considered, in which case the equilibrium constant is a dissociation constant (Kd). Per the relation between equilibrium constants for reciprocal reactions described, the dissociation constant is the mathematical inverse of the formation constant, Kd = Kf–1. A tabulation of formation constants is provided in Formation Constants for Complex Ions.

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 [latex]\ce{Ag+}[/latex] ([latex]\ce{[Ag+]}[/latex] = 1.3 [latex]\times[/latex] 10–5 M):

[latex]\ce{AgCl}(s)\rightleftharpoons \ce{Ag+}(aq)+\ce{Cl-}(aq)[/latex]

However, if [latex]\ce{NH3}[/latex] is present in the water, the complex ion, [latex]\ce{Ag(NH3)2+}[/latex], can form according to the equation:

[latex]\ce{Ag+}(aq)+\ce{2NH3}(aq)\rightleftharpoons \ce{Ag(NH3)2+}(aq)[/latex]

with

[latex]{K}_{\text{f}}=\dfrac{\left[\ce{Ag}{\left({\ce{NH}}_{3}\right)}_{2}{}^{+}\right]}{\left[{\ce{Ag}}^{+}\right]{\left[{\ce{NH}}_{3}\right]}^{2}}=1.7\times {10}^{7}[/latex]

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

[latex]Q = \ce{[Ag+][Cl-]} < K \text{sp}[/latex]

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

Example 16.2.1: Dissociation of a Complex Ion

Calculate the concentration of the silver ion in a solution that initially is 0.10 M with respect to [latex]\ce{Ag(NH3)2+}[/latex].

Show 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.”

Step 1. Determine the direction of change. The complex ion [latex]\ce{Ag(NH3)2+}[/latex] is in equilibrium with its components, as represented by the equation:[latex]\ce{Ag+}(aq)+\ce{2NH3}(aq)\rightleftharpoons \ce{Ag(NH3)2+}(aq)[/latex]We write the equilibrium as a formation reaction because Formation Constants for Complex Ions lists formation constants for complex ions. Before equilibrium, the reaction quotient is larger than the equilibrium constant [Kf = 1.6 [latex]\times[/latex] 107, and [latex]Q=\frac{0.10}{0\times 0}[/latex], 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 [latex]\ce{Ag+}[/latex] be x. Dissociation of 1 mol of [latex]\ce{Ag(NH3)2+}[/latex] gives 1 mol of [latex]\ce{Ag+}[/latex] and 2 mol of [latex]\ce{NH3}[/latex], so the change in [latex]\ce{[NH3]}[/latex] is 2x and that of [latex]\ce{Ag(NH3)2+}[/latex] is –x. In summary:

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.

Step 3. Solve for x and the equilibrium concentrations. Substituting these equilibrium concentration terms into the Kf expression gives

[latex]{K}_{\text{f}}=\dfrac{\left[\ce{Ag}{\left({\ce{NH}}_{3}\right)}_{2}{}^{+}\right]}{\left[{\ce{Ag}}^{+}\right]{\left[{\ce{NH}}_{3}\right]}^{2}}[/latex]

[latex]1.7\times {10}^{7}\text{=}\dfrac{0.10-x}{\left(x\right){\left(2x\right)}^{2}}[/latex]

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:

[latex]1.7\times {10}^{7}=\dfrac{0.10-x}{\left(x\right){\left(2x\right)}^{2}}[/latex]

[latex]{x}^{3}=\dfrac{0.10}{4\left(1.7\times {10}^{7}\right)}=1.5\times {10}^{-9}[/latex]

[latex]x=\sqrt[3]{1.5\times {10}^{-19}}=1.1\times {10}^{-3}[/latex]

Because only 1.1% of the [latex]\ce{Ag(NH3)2+}[/latex] dissociates into [latex]\ce{Ag+}[/latex] and [latex]\ce{NH3}[/latex], 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:

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

[latex]\ce{[NH3]}=0+2x=2.2\times {10}^{-3}M[/latex]

[latex]\ce{[Ag(NH3)2+]}=0.10-x=0.10 - 0.0011=0.099[/latex]

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 Kf within the error associated with the significant figures in the calculation.

Check Your Learning

Key Concepts and Summary

A Lewis acid is a species that can accept an electron pair, whereas a Lewis base has an electron pair available for donation to a Lewis acid. Complex ions are examples of Lewis acid-base adducts and comprise central metal atoms or ions acting as Lewis acids bonded to molecules or ions called ligands that act as Lewis bases. The equilibrium constant for the reaction between a metal ion and ligands produces a complex ion called a formation constant; for the reverse reaction, it is called a dissociation constant.

Try It

  1. Under what circumstances, if any, does a sample of solid [latex]\ce{AgCl}[/latex] completely dissolve in pure water?
  2. Sometimes equilibria for complex ions are described in terms of dissociation constants, Kd. For the complex ion [latex]{\ce{AlF}}_{6}{}^{\ce{3-}}[/latex] the dissociation reaction is:[latex]\ce{AlF6^3-}\rightleftharpoons \ce{Al^3+}+\ce{6F-}[/latex] and [latex]{K}_{\text{d}}=\frac{\left[{\ce{Al}}^{\ce{3+}}\right]{\left[{\ce{F}}^{-}\right]}^{6}}{\left[{\ce{AlF}}_{6}^{3-}\right]}=2\times {10}^{-24}[/latex]Calculate the value of the formation constant, Kf, for [latex]\ce{AlF6^3-}[/latex].
  3. Using the dissociation constant, Kd = 7.8 [latex]\times[/latex] 10–18, calculate the equilibrium concentrations of [latex]\ce{Cd^2+}[/latex] and [latex]\ce{CN-}[/latex] in a 0.250-M solution of [latex]\ce{Cd(CN)4^2-}[/latex].
  4. Calculate the mass of potassium cyanide ion that must be added to 100 mL of solution to dissolve 2.0 [latex]\times[/latex] 10–2 mol of silver cyanide, [latex]\ce{AgCN}[/latex].
  5. Write the Lewis structures of the reactants and product of each of the following equations, and identify the Lewis acid and the Lewis base in each:
    1. [latex]\ce{CO2}+\ce{OH-}\longrightarrow \ce{HCO3-}[/latex]
    2. [latex]\ce{B(OH)3}+\ce{OH-}\longrightarrow \ce{B(OH)4-}[/latex]
    3. [latex]\ce{I-}+\ce{I2}\longrightarrow \ce{I3-}[/latex]
    4. [latex]\ce{AlCl3}+\ce{Cl-}\longrightarrow \ce{AlCl4-}[/latex] (use Al-Cl single bonds)
    5. [latex]\ce{O^2-}+\ce{SO3}\longrightarrow \ce{SO4^2-}[/latex]
Show Selected Solutions

1. When the amount of solid is so small that a saturated solution is not produced.

2. For the formation reaction:

[latex]\begin{array}{l}{\ce{Al}}^{\ce{3+}}\left(aq\right)+6{\ce{F}}^{-}\left(aq\right)\rightleftharpoons {\ce{AlF}}_{6}{}^{\ce{3-}}\left(aq\right)\\{K}_{\ce{f}}=\frac{\left[{\ce{AlF}}_{6}{}^{\ce{3-}}\right]}{\left[{\ce{Al}}^{\ce{3+}}\right]{\left[{\ce{F}}^{-}\right]}^{6}}=\frac{1}{{K}_{\text{d}}}=\frac{1}{2\times {10}^{-24}}=5\times {10}^{23}\end{array}[/latex]

3.

[Cd(CN)42−] [CN−] [Cd2+]
Initial concentration (M) 0.250 0 0
Equilibrium (M) 0.250 − x 4x x

[latex]{K}_{\text{d}}=\frac{\left[{\ce{Cd}}^{\ce{2+}}\right]\left[{\ce{CN}}^{-}\right]}{\left[\ce{Cd}{\left(\ce{CN}\right)}_{4}{}^{2-}\right]}=7.8\times {10}^{-18}=\frac{x{\left(4x\right)}^{4}}{0.250-x}[/latex]

Assume that x is small when compared with 0.250 M.

256x5 = 0.250 [latex]\times[/latex] 7.8 [latex]\times[/latex] 10–18

x5 = 7.617 [latex]\times[/latex] 10–21

x = [latex]\ce{[Cd^2+]}[/latex] = 9.5 [latex]\times[/latex] 10–5M

4x =[latex]\ce{[CN-]}[/latex] = 3.8 [latex]\times[/latex] 10–4M

 

4. Because Ksp is small and Kf is large, most of the [latex]\ce{Ag+}[/latex] is used to form [latex]\ce{Ag(CN)2-}[/latex]; that is:

[latex]\begin{array}{rll}\left[\ce{Ag}^{+}\right]&\lt&\left[\ce{Ag}\left(\ce{CN}\right)_{2}^{-}\right]\\\left[\ce{Ag}\left(\ce{CN}\right)_{2}^{-}\right]&\approx&2.0\times{10}^{-1}M\end{array}[/latex]

The [latex]\ce{CN-}[/latex] from the dissolution and the added [latex]\ce{CN-}[/latex] exist as [latex]\ce{CN-}[/latex] and [latex]\ce{Ag(CN)2-}[/latex]. Let x be the change in concentration upon addition of [latex]\ce{CN-}[/latex]. Its initial concentration is approximately 0.

[latex]\ce{[CN-]}[/latex] + 2 [latex]\left[\text{Ag}{\left(\ce{CN}\right)}_{2}{}^{-}\right][/latex] = 2 [latex]\times[/latex] 10–1 + x

Because Ksp is small and Kf is large, most of the [latex]\ce{CN-}[/latex] is used to form [latex]\ce{[Ag(CN)2-]}[/latex]; that is:

[latex]\begin{array}{rll}\left[{\ce{CN}}^{-}\right]&<&2\left[\ce{Ag}{\left(\ce{CN}\right)}_{2}{}^{-}\right]\\2\left[\ce{Ag}{\left(\ce{CN}\right)}_{2}{}^{-}\right]&\approx&2.0\times {10}^{-1}+x\end{array}[/latex]

2(2.0 [latex]\times[/latex] 10–1) – 2.0 [latex]\times[/latex] 10–1 = x

2.0 [latex]\times[/latex] 10–1M [latex]\times[/latex] L = mol [latex]\ce{CN-}[/latex] added

The solution has a volume of 100 mL.

2 [latex]\times[/latex] 10–1 mol/L [latex]\times[/latex] 0.100 L = 2 [latex]\times[/latex] 10–2 mol

mass [latex]\ce{KCN}[/latex] = 2.0 [latex]\times[/latex] 10–2 mol [latex]\ce{KCN}[/latex] [latex]\times[/latex] 65.120 g/mol = 1.3 g

5.

(a)

This figure shows a chemical reaction modeled with structural formulas. On the left side is a structure with a central C atom. O atoms, each with two unshared electron pairs, are double bonded to the left and right sides of the C atom. Following a plus sign is another structure in brackets which has an O atom with three unshared electron dot pairs single bonded to an H atom on the right. Outside the brackets is superscript negative sign. Following a right pointing arrow is a structure in brackets that has a central C atom to which 3 O atoms are bonded. Above and slightly to the right, one of the O atoms is connected with a double bond. This O atom has two unshared electron pairs. The second O atom is single bonded below and slightly to the right. This O atom has three unshared electron pairs. The third O atom is bonded to the left of the C atom. This O atom has two unshared electron pairs and an H atom single bonded to its left. Outside the brackets to the right is a superscript negative symbol.

(b)

This figure shows a chemical reaction modeled with structural formulas. On the left side is a structure that has a central B atom to which 3 O atoms are bonded. The O atoms above and below slightly right of the B atom each have an H atom single bonded to the right. The third O atom is single bonded to the left side of the B atom. This O atom has an H atom single bonded to its left side. All O atoms in this structure have two unshared electron pairs. Following a plus sign is another structure which has an O atom single bonded to an H atom on its right. The O atom has three unshared electron pairs. The structure appears in brackets with a superscript negative sign. Following a right pointing arrow is a structure in brackets has a central B atom to which 4 O atoms are bonded. The O atoms above, below, and right of the B atom each hav an H atom single bonded to the right. The third O atom is single bonded to the left side of the B atom. This O atom has an H atom single bonded to its left side. All O atoms in this structure have two unshared electron pairs. Outside the brackets to the right is a superscript negative symbol.

(c)

This figure illustrates a chemical reaction using structural formulas. On the left, two I atoms, each with 3 unshared electron pairs, are joined with a single bond. Following a plus sign is another structure which has an I atom with four pairs of electron dots and a superscript negative sign. Following a right pointing arrow is a structure in brackets that has three I atoms connected in a line with single bonds. The two end I atoms have three unshared electron dot pairs and the I atom at the center has two unshared electron pairs. Outside the brackets is a superscript negative sign.

(d)

This figure illustrates a chemical reaction using structural formulas. On the left, an A l atom is positioned at the center of a structure and three Cl atoms are single bonded above, leftt, and below. Each C l atom has three pairs of electron dots. Following a plus sign is another structure which has an F atom is surrounded by four electron dot pairs and a superscript negative symbol. Following a right pointing arrow is a structure in brackets that has a central A l atom to which 4 C l atoms are connected with single bonds above, below, to the left, and to the right. Each C l atom in this structure has three pairs of electron dots. Outside the brackets is a superscript negative symbol.

(e)

This figure illustrates a chemical reaction using structural formulas. On the left is a 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. Following a plus sign is an O atom which is surrounded by four electron dot pairs and has a superscript 2 negative. 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.

Glossary

complex ion: ion consisting of a transition metal central atom and surrounding molecules or ions called ligands

dissociation constant: (Kd) equilibrium constant for the decomposition of a complex ion into its components in solution

formation constant: (Kf) (also, stability constant) equilibrium constant for the formation of a complex ion from its components in solution

Lewis acid: any species that can accept a pair of electrons and form a coordinate covalent bond

Lewis acid-base adduct: compound or ion that contains a coordinate covalent bond between a Lewis acid and a Lewis base

Lewis base: any species that can donate a pair of electrons and form a coordinate covalent bond

ligand: molecule or ion that surrounds a transition metal and forms a complex ion; ligands act as Lewis bases

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