Which of the Following Species Can Be Both Lewis Acid and Lewis Base?

Chapter 15. Equilibria of Other Reaction Classes

15.2 Lewis Acids and Bases

Learning Objectives

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

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

In 1923, 1000. 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 grade 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 instance, a coordinate covalent bail 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.

A Lewis acrid is whatever species (molecule or ion) that can have 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 of operations reaction occurs when a base of operations donates a pair of electrons to an acid. A Lewis acid-base adduct, a chemical compound that contains a coordinate covalent bond betwixt the Lewis acrid and the Lewis base, is formed. The following equations illustrate the general application of the Lewis concept.

The boron cantlet in boron trifluoride, BF₃, has only vi electrons in its valence crush. 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 ane of its lone pairs:

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

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

Many Lewis acid-base reactions are displacement reactions in which ane Lewis base of operations displaces another Lewis base of operations from an acrid-base adduct, or in which ane Lewis acid displaces another Lewis acrid:

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

Many slightly soluble ionic solids dissolve when the concentration of the metallic ion in solution is decreased through the formation of complex (polyatomic) ions in a Lewis acid-base reaction. For example, argent chloride dissolves in a solution of ammonia because the silver ion reacts with ammonia to class the circuitous ion [latex]\text{Ag(NH}_3)_2^{\;\;+}[/latex]. The Lewis structure of the [latex]\text{Ag(NH}_3)_2^{\;\;+}[/latex] ion is:

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

[latex]\text{AgCl}(s)\;{\longrightarrow}\;\text{Ag}^{+}(aq)\;+\;\text{Cl}^{-}(aq)[/latex]

[latex]\text{Ag}^{+}(aq)\;+\;2\text{NH}_3(aq)\;{\longrightarrow}\;\text{Ag(NH}_3)_2^{\;\;+}(aq)[/latex]

[latex]\text{Net:\;AgCl}(s)\;+\;2\text{NH}_3(aq)\;{\longrightarrow}\;\text{Ag(NH}_3)_2^{\;\;+}(aq)\;+\;\text{Cl}^{-}(aq)[/latex]

Aluminum hydroxide dissolves in a solution of sodium hydroxide or another stiff base because of the formation of the complex ion [latex]\text{Al(OH)}_4^{\;\;-}[/latex]. The Lewis structure of the [latex]\text{Al(OH)}_4^{\;\;-}[/latex] ion is:

The equations for the dissolution are:

[latex]\text{Al(OH)}_3(south)\;{\longrightarrow}\;\text{Al}^{three+}(aq)\;+\;iii\text{OH}^{-}(aq)[/latex]

[latex]\text{Al}^{three+}(aq)\;+\;4\text{OH}^{-}(aq)\;{\longrightarrow}\;\text{Al(OH)}_4^{\;\;-}(aq)[/latex]

[latex]\text{Net:\;Al(OH)}_3(due south)\;+\;\text{OH}^{-}(aq)\;{\longrightarrow}\;\text{Al(OH)}_4^{\;\;-}(aq)[/latex]

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

[latex]\text{HgS}(south)\;{\longrightarrow}\;\text{Hg}^{2+}(aq)\;+\;\text{S}^{2-}(aq)[/latex]

[latex]\text{Hg}^{2+}(aq)\;+\;2\text{S}^{ii-}(aq)\;{\longrightarrow}\;\text{HgS}_2^{\;\;2-}(aq)[/latex]

[latex]\text{Net:\;HgS}(southward)\;+\;\text{Due south}^{2-}(aq)\;{\longrightarrow}\;\text{HgS}_2^{\;\;2-}(aq)[/latex]

A complex ion consists of a central atom, typically a transition metallic 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 human action every bit Lewis bases, donating a pair of electrons to the primal cantlet. The ligands aggregate themselves around the central atom, creating a new ion with a accuse equal to the sum of the charges and, most often, a transitional metallic ion. This more complex arrangement is why the resulting ion is called a circuitous ion. The complex ion formed in these reactions cannot exist predicted; it must be determined experimentally. The types of bonds formed in complex ions are called coordinate covalent bonds, every bit electrons from the ligands are beingness shared with the cardinal atom. Because of this, circuitous ions are sometimes referred to as coordination complexes. This will be studied farther in upcoming capacity.

The equilibrium constant for the reaction of the components of a complex ion to form the circuitous ion in solution is chosen a germination abiding (K _f) (sometimes called a stability constant). For example, the circuitous ion [latex]\text{Cu(CN)}_2^{\;\;-}[/latex] is shown here:

It forms past the reaction:

[latex]\text{Cu}^{+}(aq)\;+\;two\text{CN}^{-}(aq)\;{\rightleftharpoons}\;\text{Cu(CN)}_2^{\;\;-}(aq)[/latex]

At equilibrium:

[latex]K_{\text{f}} = Q = \frac{[\text{Cu(CN)}_2^{\;\;-}]}{[\text{Cu}^{+}][\text{CN}^{-}]^two}[/latex]

The inverse of the formation constant is the dissociation constant (Thou _d), the equilibrium constant for the decomposition of a complex ion into its components in solution. We will work with dissociation constants farther in the exercises for this section. Appendix K and Table 2 are tables of formation constants. In general, the larger the formation constant, the more stable the complex; however, as in the example of G _sp values, the stoichiometry of the compound must exist considered.

Substance	Chiliad f at 25 °C
[latex][\text{Cd(CN)}_4]^{2-}[/latex]	iii × tenxviii
[latex]\text{Ag(NH}_3)_2^{\;\;+}[/latex]	ane.seven × 10seven
[latex][\text{AlF}_6]^{3-}[/latex]	7 × 1019
Table 2. Common Complex Ions by Decreasing Formulation Constants

As an example of dissolution by complex ion formation, let the states consider what happens when we add aqueous ammonia to a mixture of silver chloride and water. Silver chloride dissolves slightly in water, giving a modest concentration of Ag⁺ ([Ag⁺] = 1.three × 10^–5 G):

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

Yet, if NH_iii is present in the water, the complex ion, [latex]\text{Ag(NH}_3)_2^{\;\;+}[/latex], can form co-ordinate to the equation:

[latex]\text{Ag}^{+}(aq)\;+\;2\text{NH}_3(aq)\;{\rightleftharpoons}\;\text{Ag(NH}_3)_2^{\;\;+}(aq)[/latex]

with

[latex]K_{\text{f}} = \frac{[\text{Ag(NH}_3)_2^{\;\;+}]}{[\text{Ag}^{+}][\text{NH}_3]^two} = 1.7\;\times\;10^7[/latex]

The large size of this formation constant indicates that most of the free argent ions produced by the dissolution of AgCl combine with NH₃ to form [latex]\text{Ag(NH}_3)_2^{\;\;+}[/latex]. Every bit a consequence, the concentration of silver ions, [Ag⁺], is reduced, and the reaction quotient for the dissolution of silverish chloride, [Ag⁺][Cl^–], falls below the solubility product of AgCl:

[latex]Q = [\text{Ag}^{+}][\text{Cl}^{-}]\;{\textless}\;K_{\text{sp}}[/latex]

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

Example 1

Dissociation of a Complex Ion
Calculate the concentration of the silver ion in a solution that initially is 0.x One thousand with respect to [latex]\text{Ag(NH}_3)_2^{\;\;+}[/latex].

Solution
We apply the familiar path to solve this trouble:

1. Determine the direction of modify. The complex ion [latex]\text{Ag(NH}_3)_2^{\;\;+}[/latex] is in equilibrium with its components, equally represented past the equation:

   [latex]\text{Ag}^{+}(aq)\;+\;2\text{NH}_3(aq)\;{\rightleftharpoons}\;\text{Ag(NH}_3)_2^{\;\;+}(aq)[/latex]

   We write the equilibrium as a germination reaction because Appendix K lists formation constants for complex ions. Before equilibrium, the reaction quotient is larger than the equilibrium abiding [G _f = ane.7 × 10⁷, and [latex]Q = \frac{0.ten}{0\;\times\;0}[/latex], it is infinitely big], then the reaction shifts to the left to attain equilibrium.

2. Determine x and equilibrium concentrations. We let the change in concentration of Ag⁺ exist 10. Dissociation of one mol of [latex]\text{Ag(NH}_3)_2^{\;\;+}[/latex] gives one mol of Ag⁺ and 2 mol of NH₃, so the modify in [NH₃] is 210 and that of [latex]\text{Ag(NH}_3)_2^{\;\;+}[/latex] is –ten. In summary:
   
3. Solve for x and the equilibrium concentrations. At equilibrium:

   [latex]K_{\text{f}} = \frac{[\text{Ag(NH}_3)_2^{\;\;+}]}{[\text{Ag}^{+}][\text{NH}_3]^two}[/latex]

   [latex]ane.7\;\times\;10^7 = \frac{0.ten\;-\;x}{(10)(2x)^2}[/latex]

   Both Q and Grand _f are much larger than i, so let us presume that the changes in concentrations needed to reach equilibrium are pocket-size. Thus 0.10 – x is approximated as 0.10:

   [latex]1.7\;\times\;x^seven = \frac{0.10\;-\;ten}{(x)(2x)^2}[/latex]

   [latex]10^3 = \frac{0.ten}{four(ane.7\;\times\;10^7)} = 1.5\;\times\;10^{-9}[/latex]

   [latex]x = \sqrt[3]{1.five\;\times\;10^{-9}} = 1.1\;\times\;x^{-iii}[/latex]

   Considering only 1.1% of the [latex]\text{Ag(NH}_3)_2^{\;\;+}[/latex] dissociates into Ag⁺ and NH₃, the assumption that x is small is justified.

   Now we decide the equilibrium concentrations:

   [latex][\text{Ag}^{+}] = 0\;+\;ten = ane.1\;\times\;10^{-three}\;M[/latex]

   [latex][\text{NH}_3] = 0\;+\;2x = 2.2\;\times\;10^{-3}\;Thou[/latex]

   [latex][\text{Ag(NH}_3)_2^{\;\;+}] = 0.10\;-\;ten = 0.10\;-\;0.0011 = 0.099[/latex]

   The concentration of free silverish ion in the solution is 0.0011 K.

4. Bank check the work. The value of Q calculated using the equilibrium concentrations is equal to K _f inside the fault associated with the meaning figures in the adding.

Cheque Your Learning
Summate the silverish 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 summate the [Ag⁺] produced by dissociation of the circuitous.)

Key Concepts and Summary

G.N. Lewis proposed a definition for acids and bases that relies on an atom's or molecule's power to accept or donate electron pairs. 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. In a circuitous ion, nosotros accept a central cantlet, often consisting of a transition metal cation, which acts every bit a Lewis acid, and several neutral molecules or ions surrounding them called ligands that act as Lewis bases. Complex ions course past sharing electron pairs to class coordinate covalent bonds. The equilibrium reaction that occurs when forming a complex ion has an equilibrium constant associated with it called a germination constant, One thousand _f. This is often referred to as a stability constant, every bit it represents the stability of the circuitous ion. Formation of complex ions in solution can have a profound upshot on the solubility of a transition element chemical compound.

Chemistry Terminate of Affiliate Exercises

1. Under what circumstances, if whatever, does a sample of solid AgCl completely dissolve in pure water?
2. Explain why the addition of NH_iii or HNO₃ to a saturated solution of Ag₂CO₃ in contact with solid Ag_twoCO_three increases the solubility of the solid.
3. Calculate the cadmium ion concentration, [Cd²⁺], in a solution prepared by mixing 0.100 L of 0.0100 M Cd(NO₃)_two with ane.150 Fifty of 0.100 NH_iii(aq).
4. Explain why addition of NH_iii or HNO₃ to a saturated solution of Cu(OH)_two in contact with solid Cu(OH)₂ increases the solubility of the solid.
5. Sometimes equilibria for complex ions are described in terms of dissociation constants, K _d. For the circuitous ion [latex]\text{AlF}_6^{\;\;3-}[/latex] the dissociation reaction is:

   [latex]\text{AlF}_6^{\;\;3-}\;{\rightleftharpoons}\;\text{Al}^{3+}\;+\;6\text{F}^{-}[/latex] and [latex]K_{\text{d}} = \frac{[\text{Al}^{3+}][\text{F}^{-}]^six}{[\text{AlF}_6^{\;\;iii-}]} = 2\;\times\;10^{-24}[/latex]

   Calculate the value of the formation abiding, Yard _f, for [latex]\text{AlF}_6^{\;\;3-}[/latex].

6. Using the value of the formation constant for the complex ion [latex]\text{Co(NH}_3)_6^{\;\;two+}[/latex], calculate the dissociation constant.
7. Using the dissociation constant, 1000 _d = 7.viii × 10^–18, calculate the equilibrium concentrations of Cd²⁺ and CN^– in a 0.250-M solution of [latex]\text{Cd(CN)}_4^{\;\;2-}[/latex].
8. Using the dissociation constant, K _d = three.four × ten^–15, calculate the equilibrium concentrations of Znⁱⁱ⁺ and OH^– in a 0.0465-One thousand solution of [latex]\text{Zn(OH)}_4^{\;\;2-}[/latex].
9. Using the dissociation constant, K _d = two.2 × 10^–34, calculate the equilibrium concentrations of Coⁱⁱⁱ⁺ and NH₃ in a 0.500-M solution of [latex]\text{Co(NH}_3)_6^{\;\;3+}[/latex].
10. Using the dissociation constant, G _d = 1 × ten^–44, summate the equilibrium concentrations of Fe^three+ and CN^– in a 0.333 K solution of [latex]\text{Atomic number 26(CN)}_6^{\;\;3-}[/latex].
11. Calculate the mass of potassium cyanide ion that must be added to 100 mL of solution to dissolve 2.0 × 10^–2 mol of silver cyanide, AgCN.
12. Calculate the minimum concentration of ammonia needed in 1.0 L of solution to dissolve iii.0 × 10^–iii mol of silver bromide.
13. A scroll of 35-mm black and white photographic film contains near 0.27 chiliad of unexposed AgBr before developing. What mass of Na₂S_iiO₃·5H_twoO (sodium thiosulfate pentahydrate or hypo) in 1.0 L of developer is required to dissolve the AgBr equally [latex]\text{Ag(S}_2\text{O}_3)_2^{\;\;3-}[/latex] (K _f = 4.7 × 10¹³)?
14. We have seen an introductory definition of an acrid: An acid is a compound that reacts with water and increases the amount of hydronium ion present. In the chapter on acids and bases, we saw two more definitions of acids: a chemical compound that donates a proton (a hydrogen ion, H⁺) to another chemical compound is called a Brønsted-Lowry acid, and a Lewis acrid is any species that tin can accept a pair of electrons. Explain why the introductory definition is a macroscopic definition, while the Brønsted-Lowry definition and the Lewis definition are microscopic definitions.
15. 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:

    (a) [latex]\text{CO}_2\;+\;\text{OH}^{-}\;{\longrightarrow}\;\text{HCO}_3^{\;\;-}[/latex]

    (b) [latex]\text{B(OH)}_3\;+\;\text{OH}^{-}\;{\longrightarrow}\;\text{B(OH)}_4^{\;\;-}[/latex]

    (c) [latex]\text{I}^{-}\;+\;\text{I}_2\;{\longrightarrow}\;\text{I}_3^{\;\;-}[/latex]

    (d) [latex]\text{AlCl}_3\;+\;\text{Cl}^{-}\;{\longrightarrow}\;\text{AlCl}_4^{\;\;-}[/latex] (use Al-Cl single bonds)

    (e) [latex]\text{O}^{2-}\;+\;\text{SO}_3\;{\longrightarrow}\;\text{SO}_4^{\;\;2-}[/latex]

16. Write the Lewis structures of the reactants and product of each of the post-obit equations, and identify the Lewis acid and the Lewis base of operations in each:

    (a) [latex]\text{CS}_2\;+\;\text{SH}^{-}\;{\longrightarrow}\;\text{HCS}_3^{\;\;-}[/latex]

    (b) [latex]\text{BF}_3\;+\;\text{F}^{-}\;{\longrightarrow}\;\text{BF}_4^{\;\;-}[/latex]

    (c) [latex]\text{I}^{-}\;+\;\text{SnI}_2\;{\longrightarrow}\;\text{SnI}_3^{\;\;-}[/latex]

    (d) [latex]\text{Al(OH)}_3\;+\;\text{OH}^{-}\;{\longrightarrow}\;\text{Al(OH)}_4^{\;\;-}[/latex]

    (due east) [latex]\text{F}^{-}\;+\;\text{SO}_3\;{\longrightarrow}\;\text{SFO}_3^{\;\;-}[/latex]

17. Using Lewis structures, write counterbalanced equations for the following reactions:

    (a) [latex]\text{HCl}(g)\;+\;\text{PH}_3(g)\;{\longrightarrow}[/latex]

    (b) [latex]\text{H}_3\text{O}^{+}\;+\;\text{CH}_3^{\;\;-}\;{\longrightarrow}[/latex]

    (c) [latex]\text{CaO}\;+\;\text{SO}_3\;{\longrightarrow}[/latex]

    (d) [latex]\text{NH}_4^{\;\;+}\;+\;\text{C}_2\text{H}_5\text{O}^{-}\;{\longrightarrow}[/latex]

18. Summate [latex][\text{HgCl}_4^{\;\;2-}][/latex] in a solution prepared by adding 0.0200 mol of NaCl to 0.250 L of a 0.100-M HgCl₂ solution.
19. In a titration of cyanide ion, 28.72 mL of 0.0100 M AgNO₃ is added before atmospheric precipitation begins. [The reaction of Ag⁺ with CN^– goes to completion, producing the [latex]\text{Ag(CN)}_2^{\;\;-}[/latex] complex.] Precipitation of solid AgCN takes place when excess Ag⁺ is added to the solution, above the corporeality needed to complete the formation of [latex]\text{Ag(CN)}_2^{\;\;-}[/latex]. How many grams of NaCN were in the original sample?
20. What are the concentrations of Ag⁺, CN^–, and [latex]\text{Ag(CN)}_2^{\;\;-}[/latex] in a saturated solution of AgCN?
21. In dilute aqueous solution HF acts as a weak acid. However, pure liquid HF (boiling point = 19.5 °C) is a strong acid. In liquid HF, HNO_iii acts like a base and accepts protons. The acidity of liquid HF can be increased past calculation one of several inorganic fluorides that are Lewis acids and take F^– ion (for example, BF_iii or SbF_v). Write balanced chemical equations for the reaction of pure HNO₃ with pure HF and of pure HF with BF₃.
22. The simplest amino acid is glycine, H₂NCH_iiCO_iiH. The common feature of amino acids is that they contain the functional groups: an amine group, –NH_ii, and a carboxylic acid group, –CO₂H. An amino acrid can office as either an acrid or a base. For glycine, the acrid strength of the carboxyl group is almost the same as that of acetic acid, CH₃CO₂H, and the base of operations forcefulness of the amino grouping is slightly greater than that of ammonia, NH_iii.

    (a) Write the Lewis structures of the ions that course when glycine is dissolved in 1 M HCl and in 1 K KOH.

    (b) Write the Lewis structure of glycine when this amino acid is dissolved in h2o. (Hint: Consider the relative base strengths of the –NH₂ and [latex]-\text{CO}_2^{\;\;-}[/latex] groups.)

23. Boric acid, H₃BO₃, is not a Brønsted-Lowry acid but a Lewis acid.

    (a) Write an equation for its reaction with water.

    (b) Predict the shape of the anion thus formed.

    (c) What is the hybridization on the boron consistent with the shape you have predicted?

Glossary

complex ion

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

coordinate covalent bond

(as well, dative bond) bond formed when one atom provides both electrons in a shared pair

dissociation constant

( K _d ) equilibrium constant for the decomposition of a complex ion into its components in solution

germination abiding

( Yard _f ) (also, stability abiding) 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 bail between a Lewis acid and a Lewis base

Lewis base

any species that tin donate a pair of electrons and form a coordinate covalent bond

ligand

molecule or ion that surrounds a transition element and forms a complex ion; ligands human action as Lewis bases

Solutions

Answers to Chemistry Cease of Chapter Exercises

1. When the corporeality of solid is so pocket-sized that a saturated solution is not produced

3. eight × ten^–five K

5. 5 × 10²³

7.

[Cd²⁺] = 9.5 × 10^–5 G; [CN^–] = 3.8 × x^–four One thousand

ix. [Co^three+] = iii.0 × x^–6 M; [NH_iii] = one.viii × 10^–5 Chiliad

11. 1.iii k

thirteen. 0.79 k

15. (a)

(b)

(c)

(d)

(e)

17. (a)

(b) [latex]\text{H}_3\text{O}^{+}\;+\;\text{CH}_3^{\;\;-}\;{\longrightarrow}\;\text{CH}_4\;+\;\text{H}_2\text{O}[/latex]

(c) [latex]\text{CaO}\;+\;\text{And so}_3\;{\longrightarrow}\;\text{CaSO}_4[/latex]

(d) [latex]\text{NH}_4^{\;\;+}\;+\;\text{C}_2\text{H}_5\text{O}^{-}\;{\longrightarrow}\;\text{C}_2\text{H}_5\text{OH}\;+\;\text{NH}_3[/latex]

19. 0.0281 g

21. [latex]\text{HNO}_3(l)\;+\;\text{HF}(fifty)\;{\longrightarrow}\;\text{H}_2\text{NO}_3^{\;\;+}\;+\;\text{F}^{-}[/latex]; [latex]\text{HF}(fifty)\;+\;\text{BF}_3(g)\;{\longrightarrow}\;\text{H}^{+}\;+\;\text{BF}_4[/latex]

23. (a) [latex]\text{H}_3\text{BO}_3\;+\;\text{H}_2\text{O}\;{\longrightarrow}\;\text{H}_4\text{BO}_4^{\;\;-}\;+\;\text{H}^{+}[/latex]; (b) The electronic and molecular shapes are the same—both tetrahedral. (c) The tetrahedral construction is consistent with sp ³ hybridization.

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