Chapter 16: Equilibria of Other Reaction Classes
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.
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:
In the following reaction, each of two ammonia molecules, Lewis bases, donates a pair of electrons to a silver ion, the Lewis acid:
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 one Lewis base displaces another Lewis base from an acid-base adduct, or in which one Lewis acid displaces another Lewis acid:
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]
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].
[reveal-answer q=”576838″]Show Solution[/reveal-answer]
[hidden-answer a=”576838″]
We use the familiar path to solve this problem:
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:
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.
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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
- Under what circumstances, if any, does a sample of solid [latex]\ce{AgCl}[/latex] completely dissolve in pure water?
- 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].
- 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].
- 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].
- 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:
- [latex]\ce{CO2}+\ce{OH-}\longrightarrow \ce{HCO3-}[/latex]
- [latex]\ce{B(OH)3}+\ce{OH-}\longrightarrow \ce{B(OH)4-}[/latex]
- [latex]\ce{I-}+\ce{I2}\longrightarrow \ce{I3-}[/latex]
- [latex]\ce{AlCl3}+\ce{Cl-}\longrightarrow \ce{AlCl4-}[/latex] (use Al-Cl single bonds)
- [latex]\ce{O^2-}+\ce{SO3}\longrightarrow \ce{SO4^2-}[/latex]
[reveal-answer q=”246861″]Show Selected Solutions[/reveal-answer]
[hidden-answer a=”246861″]
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)
(b)
(c)
(d)
(e)
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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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any species that can donate a pair of electrons and form a coordinate covalent bond
any species that can accept a pair of electrons and form a coordinate covalent bond
compound or ion that contains a coordinate covalent bond between a Lewis acid and a Lewis base
ion consisting of a transition metal central atom and surrounding molecules or ions called ligands
molecule or ion that surrounds a transition metal and forms a complex ion; ligands act as Lewis bases
(Kf) (also, stability constant) equilibrium constant for the formation of a complex ion from its components in solution
(Kd) equilibrium constant for the decomposition of a complex ion into its components in solution