15.3: Lewis Acids and Bases
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Learning Objectives
By the end of this section, you will be able to:
- Explain the Lewis model of acid-base chemistry
- Write equations for the formation of adducts and complex ions
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, 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 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 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. 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
is produced by the reaction
\[\ce{Cu^{+}(aq) + 2 CN^{-}(aq) <=> Cu(CN)2^{-}(aq)} \nonumber \]
The formation constant for this reaction is
\[K_{ f }=\frac{\left[ \ce{Cu(CN)2^{-}} \right]}{\left[ \ce{Cu^{+}} \right]\left[ \ce{CN^{-}}\right]^2} \nonumber \]
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 Appendix K.
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 + (\([\ce{Ag^{+}}] = 1.3 \times 10^{–5} ~\text{M}\)):
\[\ce{AgCl(s) <=> Ag^{+}(aq) + Cl^{-}(aq)} \nonumber \]
However, if NH 3 is present in the water, the complex ion, \(\ce{Ag(NH3)2^{+}}\) can form according to the equation:
\[\ce{Ag^{+}(aq) + 2 NH3(aq) <=> Ag(NH3)2^{+}(aq)} \nonumber \]
with
\[K_{ f }=\dfrac{[ \ce{Ag(NH3)2^{+}} ]}{[ \ce{Ag^{+}} ][\ce{NH3}]^2}=1.7 \times 10^7 \nonumber \]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 \(\ce{Ag(NH3)2^{+}}\). As a consequence, the concentration of silver ions, \(\ce{[Ag^{+}]}\), is reduced, and the reaction quotient for the dissolution of silver chloride, [Ag + ][Cl – ], falls below the solubility product of AgCl:
\[Q=\left[ \ce{Ag^{+}} \right]\left[ \ce{Cl^{-}} \right]<K_{ sp } \nonumber \]More silver chloride then dissolves. If the concentration of ammonia is great enough, all of the silver chloride dissolves.
Example \(\PageIndex{1}\): Dissociation of a Complex Ion
Calculate the concentration of the silver ion in a solution that initially is 0.10 M with respect to \(\ce{Ag(NH3)2^{+}}\)).
Solution
Applying the standard ICE approach to this reaction yields the following:
Substituting these equilibrium concentration terms into the K f expression gives
\[\begin{align*}
K_{ f } &=\dfrac{[ \ce{Ag(NH3)2^{+}} ]}{[ \ce{Ag^{+}} ][ \ce{NH3} ]^2} \\[4pt]
1.7 \times 10^7 &=\frac{0.10-x}{(x)(2 x)^2}
\end{align*} \nonumber \]
The very large equilibrium constant means the amount of the complex ion that will dissociate, x, will be very small. Assuming \(x \ll 0.1\) permits simplifying the above equation:
\[\begin{align*}
1.7 \times 10^7 &=\frac{0.10}{(x)(2 x)^2} \\[4pt]
x^3&=\frac{0.10}{4\left(1.7 \times 10^7\right)}=1.5 \times 10^{-9} \\[4pt]
x&=\sqrt[3]{1.5 \times 10^{-9}}=1.1 \times 10^{-3}
\end{align*} \nonumber \]
Because only 1.1% of the \(\ce{Ag(NH3)2^{+}}\)) dissociates into \(\ce{Ag^{+}}\) and \(\ce{NH3}\), 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:
\[\begin{align*}
[ \ce{Ag^{+}} ] &=0+x=1.1 \times 10^{-3}~\text{M} \\[4pt]
[ \ce{NH3}] &=0+2 x=2.2 \times 10^{-3} ~\text{M} \\[4pt]
[ \ce{Ag(NH3)2^{+}}] &=0.10-x=0.10-0.0011 \\[4pt] &=0.099 ~\text{M}
\end{align*} \nonumber \]
The concentration of free silver ion in the solution is 0.0011 M .
Exercise \(\PageIndex{1}\)
Calculate the silver ion concentration, [Ag + ], of a solution prepared by dissolving 1.00 g of \(\ce{AgNO3}\) and 10.0 g of \(\ce{KCN}\) 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.)
- Answer
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\(2.9 \times 10^{–22} ~\text{M}\)