LEARNING OBJECTIVES
By the end of this module, you will be able to:
 Explain the Lewis model of acidbase 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 acidbase 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.
A Lewis acid is any species (molecule or ion) that can accept a pair of electrons, and a Lewis base is any species (molecule or ion) that can donate a pair of electrons.
A Lewis acidbase reaction occurs when a base donates a pair of electrons to an acid. A Lewis acidbase adduct, a compound that contains a coordinate covalent bond between the Lewis acid and the Lewis base, is formed. The following equations illustrate the general application of the Lewis concept.
The boron atom in boron trifluoride, BF_{3}, has only six electrons in its valence shell. Being short of the preferred octet, BF_{3} 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 acidbase reactions are displacement reactions in which one Lewis base displaces another Lewis base from an acidbase adduct, or in which one Lewis acid displaces another Lewis acid:
The last displacement reaction shows how the reaction of a BrønstedLowry acid with a base fits into the Lewis concept. A BrønstedLowry acid such as HCl is an acidbase adduct according to the Lewis concept, and proton transfer occurs because a more stable acidbase adduct is formed. Thus, although the definitions of acids and bases in the two theories are quite different, the theories overlap considerably.
Many slightly soluble ionic solids dissolve when the concentration of the metal ion in solution is decreased through the formation of complex (polyatomic) ions in a Lewis acidbase reaction. For example, silver chloride dissolves in a solution of ammonia because the silver ion reacts with ammonia to form the complex ion [latex]\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}[/latex]. The Lewis structure of the [latex]\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}[/latex] ion is:
The equations for the dissolution of AgCl in a solution of NH_{3} are:
Aluminum hydroxide dissolves in a solution of sodium hydroxide or another strong base because of the formation of the complex ion [latex]\text{Al}{\left(\text{OH}\right)}_{4}{}^{\text{}}[/latex]. The Lewis structure of the [latex]\text{Al}{\left(\text{OH}\right)}_{4}{}^{\text{}}[/latex] ion is:
The equations for the dissolution are:
Mercury(II) sulfide dissolves in a solution of sodium sulfide because HgS reacts with the S^{2–} ion:
A complex ion consists of a central atom, typically a transition metal cation, surrounded by ions, or molecules called ligands. These ligands can be neutral molecules like H_{2}O or NH_{3}, or ions such as CN^{–} or OH^{–}. Often, the ligands act as Lewis bases, donating a pair of electrons to the central atom. The ligands aggregate themselves around the central atom, creating a new ion with a charge equal to the sum of the charges and, most often, a transitional metal ion. This more complex arrangement is why the resulting ion is called a complex ion. The complex ion formed in these reactions cannot be predicted; it must be determined experimentally. The types of bonds formed in complex ions are called coordinate covalent bonds, as electrons from the ligands are being shared with the central atom. Because of this, complex ions are sometimes referred to as coordination complexes. This will be studied further in upcoming chapters.
The equilibrium constant for the reaction of the components of a complex ion to form the complex ion in solution 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}{}^{\text{}}[/latex] is shown here:
It forms by the reaction:
At equilibrium:
The inverse of the formation constant is the dissociation constant (K_{d}), the equilibrium constant for the decomposition of a complex ion into its components in solution. We will work with dissociation constants further in the exercises for this section. Formation Constants for Complex Ions and Table 1 are tables of formation constants. In general, the larger the formation constant, the more stable the complex; however, as in the case of K_{sp} values, the stoichiometry of the compound must be considered.
Table 1. Common Complex Ions by Decreasing Formulation Constants  

Substance  K_{f} at 25 °C 
[latex]{\left[\text{Cd}{\left(\text{CN}\right)}_{4}\right]}^{\text{2}}[/latex]  1.3 [latex]\times [/latex] 10^{7} 
[latex]\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}[/latex]  1.7 [latex]\times [/latex] 10^{7} 
[latex]{\left[{\text{AlF}}_{6}\right]}^{\text{3}}[/latex]  7 [latex]\times [/latex] 10^{19} 
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^{–5 }M):
However, if NH_{3} is present in the water, the complex ion, [latex]\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}[/latex], can form according to the equation:
with
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 [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:
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].
Solution
We use the familiar path to solve this problem:

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] 10^{7}, and [latex]Q=\frac{0.10}{0\times 0}[/latex], it is infinitely large], so the reaction shifts to the left to reach equilibrium.

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 NH_{3}, so the change in [NH_{3}] is 2x and that of [latex]\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}[/latex] is –x. In summary:

Solve for x and the equilibrium concentrations. At equilibrium:
[latex]{K}_{\text{f}}=\frac{\left[\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}\right]}{\left[{\text{Ag}}^{+}\right]{\left[{\text{NH}}_{3}\right]}^{2}}[/latex][latex]1.6\times {10}^{7}\text{=}\frac{0.10x}{\left(x\right){\left(2x\right)}^{2}}[/latex]Both Q and K_{f} are much larger than 1, so let us assume that the changes in concentrations needed to reach equilibrium are small. Thus 0.10 – x is approximated as 0.10:
[latex]1.6\times {10}^{7}=\frac{0.10x}{\left(x\right){\left(2x\right)}^{2}}[/latex][latex]{x}^{3}=\frac{0.10}{4\left(1.6\times {10}^{7}\right)}=1.6\times {10}^{9}[/latex][latex]x=\sqrt[3]{1.6\times {10}^{19}}=1.2\times {10}^{3}[/latex]Because only 1.2% of the [latex]\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}[/latex] dissociates into Ag^{+} and NH_{3}, the assumption that x is small is justified.
Now we determine the equilibrium concentrations:
[latex]\left[{\text{Ag}}^{+}\right]=0+x=1.2\times {10}^{3}M[/latex][latex]\left[{\text{NH}}_{3}\right]=0+2x=2.4\times {10}^{3}M[/latex][latex]\left[\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}\right]=0.10x=0.10  0.0012=0.099[/latex]The concentration of free silver ion in the solution is 0.0012 M.

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 AgNO_{3} 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 calculate the [Ag^{+}] produced by dissociation of the complex.)
Key Concepts and Summary
G.N. Lewis proposed a definition for acids and bases that relies on an atom’s or molecule’s ability 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 acidbase adducts. In a complex ion, we have a central atom, often consisting of a transition metal cation, which acts as a Lewis acid, and several neutral molecules or ions surrounding them called ligands that act as Lewis bases. Complex ions form by sharing electron pairs to form coordinate covalent bonds. The equilibrium reaction that occurs when forming a complex ion has an equilibrium constant associated with it called a formation constant, K_{f}. This is often referred to as a stability constant, as it represents the stability of the complex ion. Formation of complex ions in solution can have a profound effect on the solubility of a transition metal compound.
Chemistry End of Chapter Exercises
 Under what circumstances, if any, does a sample of solid AgCl completely dissolve in pure water?
 Explain why the addition of NH_{3} or HNO_{3} to a saturated solution of Ag_{2}CO_{3} in contact with solid Ag_{2}CO_{3} increases the solubility of the solid.
 Calculate the cadmium ion concentration, [Cd^{2+}], in a solution prepared by mixing 0.100 L of 0.0100 M Cd(NO_{3})_{2} with 1.150 L of 0.100 NH_{3}(aq).
 Explain why addition of NH_{3} or HNO_{3} 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}{}^{\text{3}}\rightleftharpoons {\text{Al}}^{\text{3+}}+6{\text{F}}^{\text{}}[/latex] and [latex]{K}_{\text{d}}=\frac{\left[{\text{Al}}^{\text{3+}}\right]{\left[{\text{F}}^{\text{}}\right]}^{6}}{\left[{\text{AlF}}_{6}{}^{\text{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 Cd^{2+} and CN^{–} in a 0.250M solution of [latex]\text{Cd}{\left(\text{CN}\right)}_{4}{}^{\text{2}}[/latex].
 Using the dissociation constant, K_{d} = 3.4 [latex]\times [/latex] 10^{–15}, calculate the equilibrium concentrations of Zn^{2+} and OH^{–} in a 0.0465M solution of [latex]\text{Zn}{\left(\text{OH}\right)}_{4}{}^{\text{2}}[/latex].
 Using the dissociation constant, K_{d} = 2.2 [latex]\times [/latex] 10^{–34}, calculate the equilibrium concentrations of Co^{3+} and NH_{3} in a 0.500M 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 Fe^{3+} and CN^{–} in a 0.333 M solution of [latex]\text{Fe}{\left(\text{CN}\right)}_{6}{}^{\text{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 35mm black and white photographic film contains about 0.27 g of unexposed AgBr before developing. What mass of Na_{2}S_{2}O_{3}.5H_{2}O (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] 10^{13})?
 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 BrønstedLowry 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 BrønstedLowry 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:
 [latex]{\text{CO}}_{2}+{\text{OH}}^{\text{}}\longrightarrow {\text{HCO}}_{3}{}^{\text{}}[/latex]
 [latex]\text{B}{\left(\text{OH}\right)}_{3}+{\text{OH}}^{\text{}}\longrightarrow \text{B}{\left(\text{OH}\right)}_{4}{}^{\text{}}[/latex]
 [latex]{\text{I}}^{\text{}}+{\text{I}}_{2}\longrightarrow {\text{I}}_{3}{}^{\text{}}[/latex]
 [latex]{\text{AlCl}}_{3}+{\text{Cl}}^{\text{}}\longrightarrow {\text{AlCl}}_{4}{}^{\text{}}[/latex] (use AlCl single bonds)
 [latex]{\text{O}}^{\text{2}}+{\text{SO}}_{3}\longrightarrow {\text{SO}}_{4}{}^{\text{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:
 [latex]{\text{CS}}_{2}+{\text{SH}}^{\text{}}\longrightarrow {\text{HCS}}_{3}{}^{\text{}}[/latex]
 [latex]{\text{BF}}_{3}+{\text{F}}^{\text{}}\longrightarrow {\text{BF}}_{4}{}^{\text{}}[/latex]
 [latex]{\text{I}}^{\text{}}+{\text{SnI}}_{2}\longrightarrow {\text{SnI}}_{3}{}^{\text{}}[/latex]
 [latex]\text{Al}{\left(\text{OH}\right)}_{3}+{\text{OH}}^{\text{}}\longrightarrow \text{Al}{\left(\text{OH}\right)}_{4}{}^{\text{}}[/latex]
 [latex]{\text{F}}^{\text{}}+{\text{SO}}_{3}\longrightarrow {\text{SFO}}_{3}{}^{\text{}}[/latex]
 Using Lewis structures, write balanced equations for the following reactions:
 [latex]\text{HCl}\left(g\right)+{\text{PH}}_{3}\left(g\right)\longrightarrow [/latex]
 [latex]{\text{H}}_{3}{\text{O}}^{+}+{\text{CH}}_{3}{}^{\text{}}\longrightarrow [/latex]
 [latex]\text{CaO}+{\text{SO}}_{3}\longrightarrow [/latex]
 [latex]{\text{NH}}_{4}{}^{+}+{\text{C}}_{2}{\text{H}}_{5}{\text{O}}^{\text{}}\longrightarrow [/latex]
 Calculate [latex]\left[{\text{HgCl}}_{4}{}^{\text{2}}\right][/latex] in a solution prepared by adding 0.0200 mol of NaCl to 0.250 L of a 0.100M HgCl_{2} solution.
 In a titration of cyanide ion, 28.72 mL of 0.0100 M AgNO_{3} is added before precipitation begins. [The reaction of Ag^{+} with CN^{–} goes to completion, producing the [latex]\text{Ag}{\left(\text{CN}\right)}_{2}{}^{\text{}}[/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}{}^{\text{}}[/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}{}^{\text{}}[/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, HNO_{3} 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, BF_{3} or SbF_{5}). Write balanced chemical equations for the reaction of pure HNO_{3} with pure HF and of pure HF with BF_{3}. Write the Lewis structures of the reactants and products, and identify the conjugate acidbase pairs.
 The simplest amino acid is glycine, H_{2}NCH_{2}CO_{2}H. The common feature of amino acids is that they contain the functional groups: an amine group, –NH_{2}, and a carboxylic acid group, –CO_{2}H. 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, CH_{3}CO_{2}H, and the base strength of the amino group is slightly greater than that of ammonia, NH_{3}.
 Write the Lewis structures of the ions that form when glycine is dissolved in 1 M HCl and in 1 M KOH.
 Write the Lewis structure of glycine when this amino acid is dissolved in water. (Hint: Consider the relative base strengths of the –NH_{2} and [latex]{\text{CO}}_{2}{}^{\text{}}[/latex] groups.)
 Boric acid, H_{3}BO_{3}, is not a BrønstedLowry acid but a Lewis acid.
 Write an equation for its reaction with water.
 Predict the shape of the anion thus formed.
 What is the hybridization on the boron consistent with the shape you have predicted?
Selected Answers
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 NH_{3} for each mol of Cd^{2+}. First, calculate the initial amounts of Cd^{2+} and of NH_{3} 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 Cd^{2+} would require 4(4.00 [latex]\times [/latex] 10^{–3} mol/L) of NH_{3} or a 1.6 [latex]\times [/latex] 10^{–2}–M solution. Due to the large value of K_{f} and the substantial excess of NH_{3}, 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
[NH_{3}] = 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^{–2}M
Let x be the change in concentration of [Cd^{2+}]:
[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] 10^{6}x (4.4 [latex]\times [/latex] 10^{–2} + 4x)^{4} = 4.00 [latex]\times [/latex] 10^{–3} – x
4.0 [latex]\times [/latex] 10^{6}x (3.75 [latex]\times [/latex] 10^{–6} + 1.36 [latex]\times [/latex] 10^{–3}x + 0.186x^{2} + 11.264x^{3} +256x^{4}) = 4.00 [latex]\times [/latex] 10^{–3}
16x + 5440x^{2} + 7.44 [latex]\times [/latex] 10^{5}x^{3} + 4.51 [latex]\times [/latex] 10^{7}x^{4} + 1.024 [latex]\times [/latex] 10^{9}x ^{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^{–4}M.
5. For the formation reaction:
[latex]\begin{array}{l}{\text{Al}}^{\text{3+}}\left(aq\right)+6{\text{F}}^{\text{}}\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}}^{\text{}}\right]}^{6}}=\frac{1}{{K}_{\text{d}}}=\frac{1}{2\times {10}^{24}}=5\times {10}^{23}\end{array}[/latex]
7.
[latex]{K}_{\text{d}}=\frac{\left[{\text{Cd}}^{\text{2+}}\right]\left[{\text{CN}}^{\text{}}\right]}{\left[\text{Cd}{\left(\text{CN}\right)}_{4}{}^{\text{2}}\right]}=7.8\times {10}^{18}=\frac{x{\left(4x\right)}^{4}}{0.250x}[/latex]
Assume that x is small when compared with 0.250 M.
256x^{5} = 0.250 [latex]\times [/latex] 7.8 [latex]\times [/latex] 10^{–18}
x^{5} = 7.617 [latex]\times [/latex] 10^{–21}
x = [Cd^{2+}] = 9.5 [latex]\times [/latex] 10^{–5}M
4x = [CN^{–}] = 3.8 [latex]\times [/latex] 10^{–4}M
9.
[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}}{0.500x}=2.2\times {10}^{34}[/latex]
Assume that x is small when compared with 0.500 M.
4.67 [latex]\times [/latex] 104x^{7} = 0.500 [latex]\times [/latex] 2.2 [latex]\times [/latex] 10^{–34}
x^{7} = 2.358 [latex]\times [/latex] 10^{–39}
x = [Co^{3+}] = 3.0 [latex]\times [/latex] 10^{–6}M
6x = [NH_{3}] = 1.8 [latex]\times [/latex] 10^{–5}M
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}{}^{\text{}}[/latex]; that is:
[latex]\begin{array}{l}\left[{\text{Ag}}^{+}\right]<\left[\text{Ag}\left(\text{CN}{\right)}_{2}{}^{\text{}}\right]\\ \left[\text{Ag}\left(\text{CN}{\right)}_{2}{}^{\text{}}\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}{}^{\text{}}[/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}{}^{\text{}}\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}{}^{\text{}}\right][/latex]; that is:
[latex]\left[{\text{CN}}^{\text{}}\right]<2\left[\text{Ag}{\left(\text{CN}\right)}_{2}{}^{\text{}}\right][/latex].
[latex]2\left[\text{Ag}{\left(\text{CN}\right)}_{2}{}^{\text{}}\right]\approx 2.0\times {10}^{1}+x[/latex]
2(2.0 [latex]\times [/latex] 10^{–1}) – 2.0 [latex]\times [/latex] 10^{–1} = x
2.0 [latex]\times [/latex] 10^{–1}M [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}}^{\text{+}}\left(aq\right)+{\text{Br}}^{\text{}}\left(aq\right){K}_{\text{sp}}=3.3\times {10}^{13}\\ {\text{Ag}}^{\text{+}}\left(aq\right)+2{\text{S}}_{2}{\text{O}}_{3}{}^{\text{2}}\left(aq\right)\rightleftharpoons \text{Ag}\left({\text{S}}_{2}{\text{O}}_{3}{\right)}_{2}{}^{\text{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}}^{\text{}}\right]}{{\left[{\text{S}}_{2}{\text{O}}_{3}{}^{\text{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}\text{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}{}^{\text{2}}:[/latex]
[latex]\frac{\left(1.438\times {10}^{3}\right)\left(1.438\times {10}^{3}\right)}{{x}^{2}}=15.51[/latex]
x^{2} = 1.333 [latex]\times [/latex] 10^{–7}
[latex]x=3.65\times {10}^{4}M=\left[{\text{S}}_{2}{\text{O}}_{3}{}^{\text{2}}\right][/latex]
The formula mass of Na_{2}S_{2}O_{3}•5H_{2}O is 248.13 g/mol. The total [latex]\left[{\text{S}}_{2}{\text{O}}_{3}{}^{\text{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}{}^{\text{}}\longrightarrow {\text{CH}}_{4}+{\text{H}}_{2}\text{O}[/latex]
(c) [latex]\text{CaO}+{\text{SO}}_{3}\longrightarrow \text{CaSO}4[/latex]
(d) [latex]{\text{NH}}_{4}{}^{+}+{\text{C}}_{2}{\text{H}}_{5}{\text{O}}^{\text{}}\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}}^{\text{}}\left(aq\right)\rightleftharpoons \text{Ag}{\left(\text{CN}\right)}_{2}{}^{\text{}}\left(aq\right){K}_{\text{f}}=1\times {10}^{20}[/latex]
The number of moles of AgNO_{3} 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}{}^{\text{}}[/latex], so there are 2.87 [latex]\times [/latex] 10^{–4} mol [latex]\text{Ag}{\left(\text{CN}\right)}_{2}{}^{\text{}}[/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}{}^{\text{}}\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}{}^{\text{}}\right][/latex]
mol NaCN = 2 mol [ [latex]\text{Ag}{\left(\text{CN}\right)}_{2}{}^{\text{}}[/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}}^{\text{}}[/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{}}+{\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:
Because there are four bonds and no lone pair (unshared pair) on B, the electronic and molecular shapes are the same—both tetrahedral.
(c) The tetrahedral structure is consistent with sp^{3} hybridization.
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 acidbase 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