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.

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:

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 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 [latex]\text{Cu}{\left(\text{CN}\right)}_{2}{}^{-}[/latex]

is produced by the reaction

[latex]{\text{Cu}}^{\text{+}}\left(aq\right)+2{\text{CN}}^{-}\left(aq\right)\rightleftharpoons \text{Cu}{\left(\text{CN}\right)}_{2}{}^{-}\left(aq\right)[/latex]

The formation constant for this reaction is

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

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

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

However, if NH₃ is present in the water, the complex ion, [latex]\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}[/latex], can form according to the equation:

[latex]{\text{Ag}}^{\text{+}}\left(aq\right)+2{\text{NH}}_{3}\left(aq\right)\rightleftharpoons \text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{\text{+}}\left(aq\right)[/latex]

with

[latex]{K}_{\text{f}}=\dfrac{\left[\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}\right]}{\left[{\text{Ag}}^{+}\right]{\left[{\text{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 AgCl combine with NH₃ to form [latex]\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}[/latex]. 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:

[latex]Q=\left[{\text{Ag}}^{+}\right]\left[{\text{Cl}}^{-}\right]<{K}_{\text{sp}}[/latex]

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

Example 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]\text{Ag}{\left({\text{NH}}_{3}\right)}_{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]\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}[/latex] is in equilibrium with its components, as represented by the equation:[latex]{\text{Ag}}^{\text{+}}\left(aq\right)+2{\text{NH}}_{3}\left(aq\right)\rightleftharpoons \text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{\text{+}}\left(aq\right)[/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 [K_f = 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 Ag⁺ be x. Dissociation of 1 mol of [latex]\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}[/latex] gives 1 mol of Ag⁺ and 2 mol of NH₃, so the change in [NH₃] is 2x and that of [latex]\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}[/latex] is –x. In summary:

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

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

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

[latex]\left[{\text{NH}}_{3}\right]=0+2x=2.2\times {10}^{-3}M[/latex]

[latex]\left[\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}\right]=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 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 K_f is very large, assume the reaction goes to completion then calculate the [Ag⁺] produced by dissociation of the complex.)

Show Solution

Key Takeaways

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 AgCl completely dissolve in pure water?
Explain why the addition of NH₃ or HNO₃ to a saturated solution of Ag₂CO₃ in contact with solid Ag₂CO₃ increases the solubility of the solid.
Calculate the cadmium ion concentration, [Cd²⁺], in a solution prepared by mixing 0.100 L of 0.0100 M Cd(NO₃)₂ with 1.150 L of 0.100 NH₃(aq).
Explain why addition of NH₃ or HNO₃ to a saturated solution of Cu(OH)₂ in contact with solid Cu(OH)₂ increases the solubility of the solid.
Sometimes equilibria for complex ions are described in terms of dissociation constants, K_d. For the complex ion [latex]{\text{AlF}}_{6}{}^{\text{3-}}[/latex] the dissociation reaction is:[latex]{\text{AlF}}_{6}^{3-}\rightleftharpoons {\text{Al}}^{\text{3+}}+6{\text{F}}^{-}[/latex] and [latex]{K}_{\text{d}}=\frac{\left[{\text{Al}}^{\text{3+}}\right]{\left[{\text{F}}^{-}\right]}^{6}}{\left[{\text{AlF}}_{6}^{3-}\right]}=2\times {10}^{-24}[/latex]Calculate the value of the formation constant, K_f, for [latex]{\text{AlF}}_{\text{6}}^{\text{3}-}[/latex].
Using the value of the formation constant for the complex ion [latex]\text{Co}{\left({\text{NH}}_{3}\right)}_{6}{}^{\text{2+}}[/latex], calculate the dissociation constant.
Using the dissociation constant, K_d = 7.8 [latex]\times[/latex] 10^–18, calculate the equilibrium concentrations of Cd²⁺ and CN^– in a 0.250-M solution of [latex]\text{Cd}{\left(\text{CN}\right)}_{4}^{2-}[/latex].
Using the dissociation constant, K_d = 3.4 [latex]\times[/latex] 10^–15, calculate the equilibrium concentrations of Zn²⁺ and OH^– in a 0.0465-M solution of [latex]\text{Zn}{\left(\text{OH}\right)}_{4}^{2-}[/latex].
Using the dissociation constant, K_d = 2.2 [latex]\times[/latex] 10^–34, calculate the equilibrium concentrations of Co³⁺ and NH₃ in a 0.500-M solution of [latex]\text{Co}{\left({\text{NH}}_{3}\right)}_{6}^{\text{3+}}[/latex].
Using the dissociation constant, K_d = 1 [latex]\times[/latex] 10^–44, calculate the equilibrium concentrations of Fe³⁺ and CN^– in a 0.333 M solution of [latex]\text{Fe}{\left(\text{CN}\right)}_{6}^{3-}[/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, AgCN.
Calculate the minimum concentration of ammonia needed in 1.0 L of solution to dissolve 3.0 [latex]\times[/latex] 10^–3 mol of silver bromide.
A roll of 35-mm black and white photographic film contains about 0.27 g of unexposed AgBr before developing. What mass of Na₂S₂O₃.5H₂O (sodium thiosulfate pentahydrate or hypo) in 1.0 L of developer is required to dissolve the AgBr as [latex]\text{Ag}{\left({\text{S}}_{2}{\text{O}}_{3}\right)}_{2}{}^{\text{3-}}[/latex] (K_f = 4.7 [latex]\times[/latex] 10¹³)?
We have seen an introductory definition of an acid: 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 compound that donates a proton (a hydrogen ion, H⁺) to another compound is called a Brønsted-Lowry acid, and a Lewis acid is any species that 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.
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]{\text{CO}}_{2}+{\text{OH}}^{-}\longrightarrow {\text{HCO}}_{3}{}^{-}[/latex]
2. [latex]\text{B}{\left(\text{OH}\right)}_{3}+{\text{OH}}^{-}\longrightarrow \text{B}{\left(\text{OH}\right)}_{4}{}^{-}[/latex]
3. [latex]{\text{I}}^{-}+{\text{I}}_{2}\longrightarrow {\text{I}}_{3}{}^{-}[/latex]
4. [latex]{\text{AlCl}}_{3}+{\text{Cl}}^{-}\longrightarrow {\text{AlCl}}_{4}{}^{-}[/latex] (use Al-Cl single bonds)
5. [latex]{\text{O}}^{2-}+{\text{SO}}_{3}\longrightarrow {\text{SO}}_{4}{}^{2-}[/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:
1. [latex]{\text{CS}}_{2}+{\text{SH}}^{-}\longrightarrow {\text{HCS}}_{3}{}^{-}[/latex]
2. [latex]{\text{BF}}_{3}+{\text{F}}^{-}\longrightarrow {\text{BF}}_{4}{}^{-}[/latex]
3. [latex]{\text{I}}^{-}+{\text{SnI}}_{2}\longrightarrow {\text{SnI}}_{3}{}^{-}[/latex]
4. [latex]\text{Al}{\left(\text{OH}\right)}_{3}+{\text{OH}}^{-}\longrightarrow \text{Al}{\left(\text{OH}\right)}_{4}{}^{-}[/latex]
5. [latex]{\text{F}}^{-}+{\text{SO}}_{3}\longrightarrow {\text{SFO}}_{3}{}^{-}[/latex]
Using Lewis structures, write balanced equations for the following reactions:
1. [latex]\text{HCl}\left(g\right)+{\text{PH}}_{3}\left(g\right)\longrightarrow[/latex]
2. [latex]{\text{H}}_{3}{\text{O}}^{+}+{\text{CH}}_{3}{}^{-}\longrightarrow[/latex]
3. [latex]\text{CaO}+{\text{SO}}_{3}\longrightarrow[/latex]
4. [latex]{\text{NH}}_{4}{}^{+}+{\text{C}}_{2}{\text{H}}_{5}{\text{O}}^{-}\longrightarrow[/latex]
Calculate [latex]\left[{\text{HgCl}}_{4}{}^{2-}\right][/latex] in a solution prepared by adding 0.0200 mol of NaCl to 0.250 L of a 0.100-M HgCl₂ solution.
In a titration of cyanide ion, 28.72 mL of 0.0100 M AgNO₃ is added before precipitation begins. [The reaction of Ag⁺ with CN^– goes to completion, producing the [latex]\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}[/latex] complex.] Precipitation of solid AgCN takes place when excess Ag⁺ is added to the solution, above the amount needed to complete the formation of [latex]\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}[/latex]. How many grams of NaCN were in the original sample?
What are the concentrations of Ag⁺, CN^–, and [latex]\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}[/latex] in a saturated solution of AgCN?
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, HNO₃ acts like a base and accepts protons. The acidity of liquid HF can be increased by adding one of several inorganic fluorides that are Lewis acids and accept F^– ion (for example, BF₃ or SbF₅). Write balanced chemical equations for the reaction of pure HNO₃ with pure HF and of pure HF with BF₃. Write the Lewis structures of the reactants and products, and identify the conjugate acid-base pairs.
The simplest amino acid is glycine, H₂NCH₂CO₂H. The common feature of amino acids is that they contain the functional groups: an amine group, –NH₂, and a carboxylic acid group, –CO₂H. An amino acid can function as either an acid or a base. For glycine, the acid strength of the carboxyl group is about the same as that of acetic acid, CH₃CO₂H, and the base strength of the amino group is slightly greater than that of ammonia, NH₃.
1. Write the Lewis structures of the ions that form when glycine is dissolved in 1 M HCl and in 1 M KOH.
2. Write the Lewis structure of glycine when this amino acid is dissolved in water. (Hint: Consider the relative base strengths of the –NH₂ and [latex]-{\text{CO}}_{2}{}^{-}[/latex] groups.)
Boric acid, H₃BO₃, is not a Brønsted-Lowry acid but a Lewis acid.
1. Write an equation for its reaction with water.
2. Predict the shape of the anion thus formed.
3. What is the hybridization on the boron consistent with the shape you have predicted?

Show Selected Solutions

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

3. Cadmium ions associate with ammonia molecules in solution to form the complex ion [latex]{\left[\text{Cd}{\left({\text{NH}}_{3}\right)}_{4}\right]}^{\text{2+}}[/latex], which is defined by the following equilibrium:

[latex]{\text{Cd}}^{\text{2+}}\left(aq\right)+4{\text{NH}}_{3}\left(aq\right)\longrightarrow {\left[\text{Cd}{\left({\text{NH}}_{3}\right)}_{4}\right]}^{\text{2+}}\left(aq\right){K}_{\text{f}}=4.0\times {10}^{6}[/latex]

The formation of the complex ion requires 4 mol of NH₃ for each mol of Cd²⁺. First, calculate the initial amounts of Cd²⁺ and of NH₃ available for association:

[latex]\left[{\text{Cd}}^{\text{2+}}\right]=\frac{\left(0.100\text{L}\right)\left(0.0100\text{mol}{\text{L}}^{-1}\right)}{0.250\text{L}}=4.00\times {10}^{-3}M[/latex]

[latex]\left[{\text{NH}}_{3}\right]=\frac{\left(0.150\text{L}\right)\left(0.100\text{mol}{\text{L}}^{-1}\right)}{0.250\text{L}}=6.00\times {10}^{-2}M[/latex]

For the reaction, 4.00 [latex]\times[/latex] 10^–3 mol/L of Cd²⁺ would require 4(4.00 [latex]\times[/latex] 10^–3 mol/L) of NH₃ or a 1.6 [latex]\times[/latex] 10^–2–M solution. Due to the large value of K_f and the substantial excess of NH₃, it can be assumed that the reaction goes to completion with only a small amount of the complex dissociating to form the ions. After reaction, concentrations of the species in the solution are

[NH₃] = 6.00 [latex]\times[/latex] 10^–2 mol/L – 1.6 [latex]\times[/latex] 10^–2 mol L^–1 = 4.4 [latex]\times[/latex] 10^–2M

Let x be the change in concentration of [Cd²⁺]:

	[Cd(NH₃)₄²⁺]	[Cd²⁺]	[NH₃]
Initial concentration (M)	4.00 × 10⁻³	0	4.4 × 10⁻²
Equilibrium (M)	4.00 × 10⁻³ − x	x	4.4 × 10⁻² + 4x

[latex]{K}_{\text{f}}=4.0\times {10}^{6}=\frac{\left[\text{Cd}{\left({\text{NH}}_{3}\right)}_{4}{}^{\text{2+}}\right]}{\left[{\text{Cd}}^{\text{2+}}\right]{\left[{\text{NH}}_{3}\right]}^{4}}[/latex]

[latex]4.00\times {10}^{6}=\frac{\left(4.00\times {10}^{-3}-x\right)}{\left(x\right){\left(4.4\times {10}^{-2}+4x\right)}^{4}}[/latex]

As x is expected to be about the same size as the number from which it is subtracted, the entire expression must be expanded and solved, in this case, by successive approximations where substitution of values for x into the equation continues until the remainder is judged small enough. This is a slightly different method than used in most problems. We have:

4.0 [latex]\times[/latex] 10⁶x (4.4 [latex]\times[/latex] 10^–2 + 4x)⁴ = 4.00 [latex]\times[/latex] 10^–3 – x

4.0 [latex]\times[/latex] 10⁶x (3.75 [latex]\times[/latex] 10^–6 + 1.36 [latex]\times[/latex] 10^–3x + 0.186x² + 11.264x³ +256x⁴) = 4.00 [latex]\times[/latex] 10^–3

16x + 5440x² + 7.44 [latex]\times[/latex] 10⁵x³ + 4.51 [latex]\times[/latex] 10⁷x⁴ + 1.024 [latex]\times[/latex] 10⁹x ⁵ = 4.00 [latex]\times[/latex] 10^–3

Substitution of different values x will give a number to be compared with 4.00 [latex]\times[/latex] 10^–3. Using 2.50 [latex]\times[/latex] 10^–4 for x gives 4.35 [latex]\times[/latex] 10^–3 compared with 4.00 [latex]\times[/latex] 10^–3. Using 2.40 [latex]\times[/latex] 10^–4 gives 4.16 [latex]\times[/latex] 10^–3 compared with 4.00 [latex]\times[/latex] 10^–3. Using 2.30 [latex]\times[/latex] 10^–4 gives 3.98 [latex]\times[/latex] 10^–3 compared with 4.00 [latex]\times[/latex] 10^–3. Thus 2.30 [latex]\times[/latex] 10^–4 is close enough to the true value of x to make the difference equal to zero. If the approximation to drop 4x is compared with 4.4 [latex]\times[/latex] 10^–2, the value of x obtained is 2.35 [latex]\times[/latex] 10^–4M.

5. For the formation reaction:

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

	[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[{\text{Cd}}^{\text{2+}}\right]\left[{\text{CN}}^{-}\right]}{\left[\text{Cd}{\left(\text{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 = [Cd²⁺] = 9.5 [latex]\times[/latex] 10^–5M

4x = [CN^–] = 3.8 [latex]\times[/latex] 10^–4M

	[Co(NH₃)₆³⁺]	[Co³⁺]	[NH₃]
Initial concentration (M)	0.500	0	0
Equilibrium (M)	0.500 − x	x	6x

[latex]{K}_{\text{d}}=\frac{\left[{\text{Co}}^{\text{2+}}\right]{\left[{\text{NH}}_{3}\right]}^{6}}{\left[\text{Co}{\left({\text{NH}}_{3}\right)}_{6}{}^{\text{3+}}\right]}=\frac{x{\left(6x\right)}^{6}}{0.500-x}=2.2\times {10}^{-34}[/latex]

Assume that x is small when compared with 0.500 M.

4.67 [latex]\times[/latex] 104x⁷ = 0.500 [latex]\times[/latex] 2.2 [latex]\times[/latex] 10^–34

x⁷ = 2.358 [latex]\times[/latex] 10^–39

x = [Co³⁺] = 3.0 [latex]\times[/latex] 10^–6M

6x = [NH₃] = 1.8 [latex]\times[/latex] 10^–5M

11. Because K_sp is small and K_f is large, most of the Ag⁺ is used to form [latex]\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}[/latex]; that is:

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

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

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

Because K_sp is small and K_f is large, most of the CN^– is used to form [latex]\left[\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}\right][/latex]; that is:

[latex]\begin{array}{rll}\left[{\text{CN}}^{-}\right]&<&2\left[\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}\right]\\2\left[\text{Ag}{\left(\text{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 CN^– 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 KCN = 2.0 [latex]\times[/latex] 10^–2 mol KCN [latex]\times[/latex] 65.120 g/mol = 1.3 g

13. The reaction is governed by two equilibria, both of which must be satisfied:

[latex]\begin{array}{l}\text{AgBr}\left(s\right)\rightleftharpoons\text{Ag}^{+}\left(aq\right)+\text{Br}^{-}\left(aq\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,{K}_{\text{sp}}=3.3\times{10}^{-13}\\\text{Ag}^{+}\left(aq\right)+2{\text{S}}_2\text{O}_{3}^{2-}\left(aq\right)\rightleftharpoons\text{Ag}\left(\text{S}_{2}\text{O}_{3}\right)_{2}^{3-}\left(aq\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,{K}_{\text{f}}=4.7\times{10}^{13}\end{array}[/latex]

The overall equilibrium is obtained by adding the two equations and multiplying their Ks:

[latex]\frac{\left[\text{Ag}{\left({\text{S}}_{2}{\text{O}}_{3}\right)}^{\text{3-}}\right]\left[{\text{Br}}^{-}\right]}{{\left[{\text{S}}_{2}{\text{O}}_{3}{}^{2-}\right]}^{2}}=15.51[/latex]

If all Ag is to be dissolved, the concentration of the complex is the molar concentration of AgBr.

formula mass (AgBr) = 187.772 g/mol

[latex]\text{moles present}=\frac{0.27\text{g}\text{AgBr}}{187.772\text{g}{\text{mol}}^{-1}}=1.438\times {10}^{-3}\text{mol}[/latex]

Let x be the change in concentration of [latex]{\text{S}}_{2}{\text{O}}_{3}{}^{2-}:[/latex]

	[Ag⁺]	[S₂O₃²⁻]
Initial concentration (M)	0	0
Equilibrium (M)	[latex]\frac{1}{2}x[/latex]	x

[latex]\frac{\left(1.438\times {10}^{-3}\right)\left(1.438\times {10}^{-3}\right)}{{x}^{2}}=15.51[/latex]

x² = 1.333 [latex]\times[/latex] 10^–7

[latex]x=3.65\times {10}^{-4}M=\left[{\text{S}}_{2}{\text{O}}_{3}{}^{2-}\right][/latex]

The formula mass of Na₂S₂O₃•5H₂O is 248.13 g/mol. The total [latex]\left[{\text{S}}_{2}{\text{O}}_{3}{}^{2-}\right][/latex] needed is:

2(1.438 [latex]\times[/latex] 10^–3) + 3.65 [latex]\times[/latex] 10^–4 = 3.241 [latex]\times[/latex] 10^–3 mol

g(hypo) = 3.241 [latex]\times[/latex] 10^–3 mol [latex]\times[/latex] 248.13 g/mol = 0.80 g

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]

(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. The equilibrium is:[latex]{\text{Ag}}^{\text{+}}\left(aq\right)+2{\text{CN}}^{-}\left(aq\right)\rightleftharpoons \text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}\left(aq\right){K}_{\text{f}}=1\times {10}^{20}[/latex]

The number of moles of AgNO₃ added is:

0.02872 L [latex]\times[/latex] 0.0100 mol/L = 2.87 [latex]\times[/latex] 10^–4 mol

This compound reacts with CN^– to form [latex]\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}[/latex], so there are 2.87 [latex]\times[/latex] 10^–4 mol [latex]\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}[/latex]. This amount requires 2 [latex]\times[/latex] 2.87 [latex]\times[/latex] 10^–4 mol, or 5.74 [latex]\times[/latex] 10^–4 mol, of CN^–. The titration is stopped just as precipitation of AgCN begins:

[latex]{\text{AgCN}}_{2}{}^{-}\left(aq\right)+{\text{Ag}}^{\text{+}}\left(aq\right)\rightleftharpoons 2\text{AgCN}\left(s\right)[/latex]

so only the first equilibrium is applicable. The value of K_f is very large.

mol CN^– < [latex]\left[\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}\right][/latex]

mol NaCN = 2 mol [ [latex]\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}[/latex] ] = 5.74 [latex]\times[/latex] 10^–4 mol

[latex]\text{mass}\left(\text{NaCN}\right)=5.74\times {10}^{-4}\text{mol}\times \frac{49.007\text{g}}{1\text{mol}}=0.0281\text{g}[/latex]

21. [latex]{\text{HNO}}_{3}\left(l\right)+\text{HF}\left(l\right)\longrightarrow {\text{H}}_{2}{\text{NO}}_{3}{}^{+}+{\text{F}}^{-}[/latex]; [latex]\text{HF}\left(l\right)+{\text{BF}}_{3}\left(g\right)\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) First, form a symmetrical structure with the unique atom, B, as the central atom. Then include the 32e^– to form the Lewis structure:

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 a B 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.

Because there are four bonds and no lone pair (unshared pair) on B, the electronic and molecular shapes are the same—both tetrahedral.

Glossary

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

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

formation constant: (K_f) (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

Module 15: Equilibria of Other Reaction Classes