# Lewis Acids and Bases

### Complex Ions

The reaction of a Lewis acid (an electron-pair donor) with a Lewis base (an electron-pair acceptor) can form a complex ion.

Some chemical reactions occur when one reagent donates a pair of electrons to another to form new compounds. The electron-pair donor in a Lewis acid-base reaction is called a Lewis base, and the electron-pair acceptor in a Lewis acid-base reaction is called a Lewis acid. Because the electrons are donated and accepted in pairs in these reactions, any compounds formed by a Lewis acid-base reaction have at least one coordinate covalent bond, also called a dative bond, which is a covalent bond in which the shared electrons come from only one of the atoms. A product containing at least one coordinate covalent bond is called an adduct. Note that an adduct can in turn act as a Lewis acid or base in a new reaction.

Recall that AgCl is only slightly soluble in water, with $K_{\rm{sp}}=1.77\times10^{-10}$. However, when AgCl is added to an aqueous solution of NH3, the AgCl dissolves readily, according to the following equation:
${\rm{AgCl}}(s)+{\rm{2NH}_3}(aq)\rightleftharpoons {{\lbrack}\rm{Ag}(\rm{NH}_3)_2\rbrack^+}(aq)+{\rm{Cl}^-}(aq)$
The [Ag(NH3)2]+ is written in brackets to indicate that it is a complex ion, which is a molecule consisting of a central ion to which other groups of molecules or ions are bound. An ion or molecule that bonds to a central metal atom or ion as part of a complex is called a ligand.
Recall the reaction in which adding NH3 caused the dissolution of AgCl. The reason AgCl dissolves is more clearly understood if the reaction is thought of as two separate reactions happening at the same time:
\begin{aligned}{\rm{AgCl}}(s) &\rightleftharpoons{\rm{Ag}^+}(aq)+{\rm{Cl}^-}(aq)\\{\rm{Ag}}^+(aq)+{2\rm{NH}_3}(aq) &\rightleftharpoons {\lbrack\rm{Ag}(\rm{NH}_3)_2\rbrack}^ +(aq)\end{aligned}
In solution with Ag+, the two NH3 molecules act as Lewis bases, and each donates a pair of electrons to the Ag+ ion, which is the Lewis acid. Their product is an adduct because it contains two coordinate covalent bonds. This adduct is also a complex ion: the NH3 molecules are ligands, bound to the central Ag+ ion. [Ag(NH3)2]+ is a stable compound. It ties up the free Ag+ ion, which lowers the overall Ag+ concentration. This in turn shifts the equilibrium of the AgCl solution to the right, and the AgCl therefore dissolves in the presence of NH3. Just as the solubility product constant Ksp is the equilibrium constant for an aqueous solution, the equilibrium of complex ion formation is given its own constant, called the formation constant (Kf). For the silver nitrate complex ion, the formation constant is
${K_{\rm{f}}} = \frac{\left[{\left[ {{\rm{Ag}}( {{\rm{N}}{{\rm{H}}_3}} )_2} \right]^+}\right]}{{[ {{\rm{A}}{{\rm{g}}^ + }} ]{{[ {{\rm{N}}{{\rm{H}}_3}} ]}^2}}} = 1.6 \times {10^7}$
Note that for this constant, the concentrations of the reacting species on the left side of the equation, [Ag]+ and [NH3], are included because they are aqueous instead of solid as has previously been the case. Also note how large Kf is for [Ag(NH3)2]+. So far, Ksp values in the range of 10–10 to 10–17 have indicated that the equilibria of those solutions are shifted far to the left and favor the reactants, meaning low solubility. A Kf on the order of 106, on the other hand, indicates that the equilibrium for the [Ag(NH3)2]+ formation is shifted far to the right, showing that [Ag(NH3)2]+ is a stable complex.

### Equilibrium Calculations Using Kf

Equilibrium can be calculated by using the formation constant Kf.
It has already been qualitatively explained that the introduction of NH3 will cause solid AgCl to dissolve by shifting the equilibrium equation
${\rm{AgCl}}(s)\rightleftharpoons{\rm{Ag}^+}(aq)+{\rm{C}\rm{l}^-}(aq)$
to the right through the formation of the complex ion [Ag(NH3)2]+. The Kf value of [Ag(NH3)2]+ can be used to find what concentration of an ion is needed to prevent the CgCl from precipitating out of a solution containing 0.10 M Ag+ and 0.10 M Cl. The first step is to determine the lowest concentration of free Ag+ in solution that will cause solid AgCl to precipitate out. AgCl first precipitates when $Q_{\rm{sp}}={[\rm{Ag}^+]_{\rm{init}}}{[\rm{Cl}^-]_{\rm{init}}}=K_{\rm{sp}}$. The introduction of NH3 will not change the concentration of Cl from 0.10 M, so [Ag+] must therefore change:
\begin{aligned}\lbrack\rm{Ag}^+\rbrack(0.10)&\le{K_{\rm{sp}}}=1.77\times{10^{-10}}\\{\lbrack}\rm{Ag}^+\rbrack&\le 1.77\times10^{-9}\;\rm{M}\end{aligned}
Thus, the concentration of free Ag+ that will cause AgCl to just start to precipitate is $1.77 \times {10^{-9}}\;\rm{ M}$. Given that the starting concentration of [Ag+] is $1.0\times 10^{-1}\;\rm{M}$, essentially all of the [Ag+] must be used to form the complex ion in order to keep its concentration below $1.77 \times {10^{-9}}\;\rm{ M}$. It is therefore safe to assume that the concentration of [Ag(NH3)2]+ is equal to the initial concentration of [Ag+], 0.10 M. Next, use Kf to calculate the concentration of free NH3 required to keep $\rm{[Ag}^+\rm{]}\leq1.77\times10^{-9}\;\rm{ M}$:
\begin{aligned}K_{\rm{f}} &=\frac{\left[[\rm{Ag}(\rm{NH}_{3})_{2}]^+\right]}{[\rm{Ag}^{+}][\rm{NH}_{3}]^{2}}=\frac{0.10}{(1.77\times10^{-9})[\rm{NH}_{3}]^{2}}=1.6\times10^{7}\\\\\rm{[NH}_{3}]&=\sqrt{\frac{0.10}{(1.6\times10^{7})(1.77\times10^{-9})}} =1.9\,\rm{M}\end{aligned}
This is the concentration of free NH3 required, but two NH3 molecules are bound up in every mole of the complex ion. Since all of the [Ag+] is contained in the complex ion, the concentration of [Ag(NH3)2]+ must be equal to the initial concentration of the [Ag+], or 0.1 M. Thus, 0.2 M of NH3 is bound up in the complex, and the total concentration of NH3 required to keep AgCl from precipitating is
$[\rm{NH}_{3}]_{\rm{total}}=1.9\,\rm{M}+0.2\,\rm{M}=2.1\,\rm{M}$