Lewis Acids and Bases

Learning Objectives

By the end of this module, you will be able to:

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

In 1923, G. N. Lewis proposed a generalized definition of acid-base behavior in which acids and bases are identified by their ability to accept or to donate a pair of electrons and form a coordinate covalent bond.

A coordinate covalent bond (or dative bond) occurs when one of the atoms in the bond provides both bonding electrons. For example, a coordinate covalent bond occurs when a water molecule combines with a hydrogen ion to form a hydronium ion. A coordinate covalent bond also results when an ammonia molecule combines with a hydrogen ion to form an ammonium ion. Both of these equations are shown here.

This figure shows two reactions represented with Lewis structures. The first shows an O atom bonded to two H atoms. The O atom has two lone pairs of electrons. There is a plus sign and then an H atom with a superscript positive sign followed by a right-facing arrow. The next Lewis structure is in brackets and shows an O atom bonded to three H atoms. There is one lone pair of electrons on the O atom. Outside of the brackets is a superscript positive sign. The second reaction shows an N atom bonded to three H atoms. The N atom has one lone pair of electrons. There is a plus sign and then an H superscript positive sign. After the H superscript positive sign is a right-facing arrow. The next Lewis structure is in brackets. It shows an N atom bonded to four H atoms. There is a superscript positive sign outside the brackets.

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 acid-base reaction occurs when a base donates a pair of electrons to an acid. A Lewis acid-base 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, 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:

This figure illustrates a chemical reaction using structural formulas. On the left, an F atom is surrounded by four electron dot pairs and has a superscript negative symbol. This structure is labeled below as “Lewis base.” Following a plus sign is another structure which has a B atom at the center and three F atoms single bonded above, right, and below. Each F atom has three pairs of electron dots. This structure is labeled below as “Lewis acid.” Following a right pointing arrow is a structure in brackets that has a central B atom to which 4 F atoms are connected with single bonds above, below, to the left, and to the right. Each F atom in this structure has three pairs of electron dots. Outside the brackets is a superscript negative symbol. This structure is labeled below as “Acid-base adduct.”

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

This figure illustrates a chemical reaction using structural formulas. On the left side, a 2 preceeds an N atom which has H atoms single bonded above, to the left, and below. A single electron dot pair is on the right side of the N atom. This structure is labeled below as “Lewis base.” Following a plus sign is an A g atom which has a superscript plus symbol. Following a right pointing arrow is a structure in brackets that has a central A g atom to which N atoms are connected with single bonds to the left and to the right. Each of these N atoms has H atoms bonded above, below, and to the outside of the structure. Outside the brackets is a superscript plus symbol. This structure is labeled below as “Acid-base adduct.”

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

This figure illustrates a chemical reaction using structural formulas. On the left, an O atom is surrounded by four electron dot pairs and has a superscript 2 negative. This structure is labeled below as “Lewis base.” Following a plus sign is another structure which has an S atom at the center. O atoms are single bonded above and below. These O atoms have three electron dot pairs each. To the right of the S atom is a double bonded O atom which has two pairs of electron dots. This structure is labeled below as “Lewis acid.” Following a right pointing arrow is a structure in brackets that has a central S atom to which 4 O atoms are connected with single bonds above, below, to the left, and to the right. Each of the O atoms has three pairs of electron dots. Outside the brackets is a superscript 2 negative. This structure is labeled below as “Acid-base adduct.”

Many Lewis acid-base reactions are displacement reactions in which one Lewis base displaces another Lewis base from an acid-base adduct, or in which one Lewis acid displaces another Lewis acid:

This figure shows three chemical reactions in three rows using structural formulas. In the first row, to the left, in brackets is a structure that has a central A g atom to which N atoms are connected with single bonds to the left and to the right. Each of these N atoms has H atoms bonded above, below, and to the outside of the structure. Outside the brackets is a superscript plus symbol. This structure is labeled below as “Acid-base adduct.” Following a plus sign is a 2 and another structure in brackets that shows a C atom triple bonded to an N atom. The C atom has an unshared electron pair on its left side and the N atom has an unshared pair on its right side. Outside the brackets to the right is a superscript negative symbol. This structure is labeled below as “Base.” Following a right pointing arrow is a structure in brackets that has a central A g atom to which 4 FC atoms are connected with single bonds to the left and to the right. At each of the two ends, N atoms are triple bonded to the C atoms. The N atoms each have an unshared electron pair at the end of the structure. Outside the brackets is a superscript negative symbol. This structure is labeled below as “New adduct.” Following a plus sign is an N atom which has H atoms single bonded above, to the left, and below. A single electron dot pair is on the left side of the N atom. This structure is labeled below as “New base.” In the second row, on the left side in brackets is a structure with a central C atom. O atoms, each with three unshared electron pairs, are single bonded above and below and a third O atom, with two unshared electron pairs, is double bonded to the right. Outside the brackets is a superscript 2 negative. This structure is labeled below as “Acid-base adduct.” Following a plus sign is another structure which has an S atom at the center. O atoms are single bonded above and below. These O atoms have three electron dot pairs each. To the right of the S atom is a double bonded O atom which has two pairs of electron dots. This structure is labeled below as “Acid.” Following a right pointing arrow is a structure in brackets that has a central S atom to which 4 O atoms are connected with single bonds above, below, to the left, and to the right. Each of the O atoms has three pairs of electron dots. Outside the brackets is a superscript 2 negative. This structure is labeled below as “New adduct.”

The last displacement reaction shows how the reaction of a Brønsted-Lowry acid with a base fits into the Lewis concept. A Brønsted-Lowry acid such as HCl is an acid-base adduct according to the Lewis concept, and proton transfer occurs because a more stable acid-base 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 acid-base 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:

A structure is shown in brackets. The structure has a central A g atom to which N atoms are single bonded to the left and right. Each of these atoms N atom has H atoms single bonded above, below, and to the outer end of the structure. Outside the brackets is a superscripted plus.

The equations for the dissolution of AgCl in a solution of NH₃ are:

[latex]\begin{array}{rrll}{}&\text{AgCl}\left(s\right)&\longrightarrow&{\text{Ag}}^{\text{+}}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)\\{}&{\text{Ag}}^{\text{+}}\left(aq\right)+2{\text{NH}}_{3}\left(aq\right)&\longrightarrow&\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{\text{+}}\left(aq\right)\\\text{Net: }&\text{AgCl}\left(s\right)+2{\text{NH}}_{3}\left(aq\right)&\longrightarrow &\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{\text{+}}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)\end{array}[/latex]

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}{}^{-}[/latex]. The Lewis structure of the [latex]\text{Al}{\left(\text{OH}\right)}_{4}{}^{-}[/latex] ion is:

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 an A l 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.

The equations for the dissolution are:

[latex]\begin{array}{rrll}{}&\text{Al}{\left(\text{OH}\right)}_{3}\left(s\right)&\longrightarrow& {\text{Al}}^{\text{3+}}\left(aq\right)+3{\text{OH}}^{-}\left(aq\right)\\{}&{\text{Al}}^{\text{3+}}\left(aq\right)+4{\text{OH}}^{-}\left(aq\right)&\longrightarrow&\text{Al}{\left(\text{OH}\right)}_{4}{}^{-}\left(aq\right)\\\text{Net:}&\text{Al}{\left(\text{OH}\right)}_{3}\left(s\right)+{\text{OH}}^{-}\left(aq\right)&\longrightarrow&\text{Al}{\left(\text{OH}\right)}_{4}{}^{-}\left(aq\right)\end{array}[/latex]

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

[latex]\begin{array}{rrll}{}&\text{HgS}\left(s\right)&\longrightarrow&{\text{Hg}}^{\text{2+}}\left(aq\right)+{\text{S}}^{2-}\left(aq\right)\\{}&{\text{Hg}}^{\text{2+}}\left(aq\right)+2{\text{S}}^{2-}\left(aq\right)&\longrightarrow&{\text{HgS}}_{2}{}^{2-}\left(aq\right)\\\text{Net:}&\text{HgS}\left(s\right)+{\text{S}}^{2-}\left(aq\right)&\longrightarrow &{\text{HgS}}_{2}{}^{2-}\left(aq\right)\end{array}[/latex]

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 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. 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}{}^{-}[/latex] is shown here:

It forms 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]

At equilibrium:

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

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]}^{2-}[/latex]	1.3 [latex]\times [/latex] 10⁷
[latex]\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}[/latex]	1.7 [latex]\times [/latex] 10⁷
[latex]{\left[{\text{AlF}}_{6}\right]}^{\text{3-}}[/latex]	7 [latex]\times [/latex] 10¹⁹

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}}=\frac{\left[\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}\right]}{\left[{\text{Ag}}^{+}\right]{\left[{\text{NH}}_{3}\right]}^{2}}=1.6\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 Answer

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.”

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.
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:
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{0.10-x}{\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{0.10-x}{\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 NH₃, 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]=0.10-x=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 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 calculate the [Ag⁺] produced by dissociation of the complex.)

Show Answer

Key Takeaways

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 acid-base 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.

Exercises

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 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 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)

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

(b)

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

(c)

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

(d)

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

(e)

This figure illustrates a chemical reaction using structural formulas. On the left is a structure which has an S atom at the center. O atoms are single bonded above and below. These O atoms have three electron dot pairs each. To the right of the S atom is a double bonded O atom which has two pairs of electron dots. Following a plus sign is an O atom which is surrounded by four electron dot pairs and has a superscript 2 negative. Following a right pointing arrow is a structure in brackets that has a central S atom to which 4 O atoms are connected with single bonds above, below, to the left, and to the right. Each of the O atoms has three pairs of electron dots. Outside the brackets is a superscript 2 negative.

17.

(a)

This figure represents a chemical reaction in two rows. The top row shows the reaction using chemical formulas. The second row uses structural formulas to represent the reaction. The first row contains the equation H C l ( g ) plus P H subscript 3 ( g ) right pointing arrow left bracket P H subscript 4 right bracket superscript plus plus left bracket C l with 4 pairs of electron dots right bracket superscript negative sign. The second row begins on the left with H left bracket C l with four unshared electron pairs right bracket plus a structure in brackets with a central P atom with H atoms single bonded at the left, above, and to the right. A single unshared electron pair is on the central P atom. Outside the brackets to the right is a superscript plus sign. Following a right pointing arrow is a structure in brackets with a central P atom with H atoms single bonded at the left, above, below, and to the right. Outside the brackets is a superscript plus sign. This structure is followed by a plus and a C l atom in brackets with four unshared electron pairs and a superscript negative sign.

(b) [latex]{\text{H}}_{3}{\text{O}}^{+}+{\text{CH}}_{3}{}^{-}\longrightarrow {\text{CH}}_{4}+{\text{H}}_{2}\text{O}[/latex]

This figure represents a chemical reaction using structural formulas. A structure is shown in brackets on the left which is composed of a central O atom with one unshared electron pair and three single bonded H atoms to the left, right, and above the atom. Outside the brackets to the right is a superscript plus sign. Following a plus sign, is another structure in brackets composed of a central C atom with one unshared electron pair and three single bonded H atoms to the left, right, and above the atom. Outside the brackets to the right is a superscript negative sign. Following a right pointing arrow is a structure with a central C atom with H atoms single bonded above, below, left and right. Following a plus sign is a structure with a central O atom with two unshared electron pairs and two H atoms connected with single bonds.

This figure represents a chemical reaction using structural formulas. On the left, C a superscript 2 plus is just left of bracket O with four unshared electron pairs right bracket superscript 2 negative plus a structure with a central S atom to which two O atoms are single bonded at the left and right, and a single O atom is double bonded above. The two single bonded O atoms each have three unshared electron pairs and the double bonded O atom has two unshared electron pairs. Following a right pointing arrow is C a superscript 2 plus just left of a structure in brackets with a central S atom which has 4 O atoms single bonded at the left, above, below, and to the right. Each of the O atoms has three unshared electron pairs. Outside the brackets to the right is a superscript two negative.

(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

Unit 17: Solubility Equilibria