15.4 Lewis Acids & Bases
If you recall from the Acid and Base chapter, Lewis acids are the compounds/ions that can accept a pair of electrons, and Lewis bases are the compounds/ions that can donate a pair of electrons in solution. Due to this ability to accept or donate a pair of electrons, these compounds/ions can make complex ions in solution. Let’s have a look at this process of complex ion formation, which is responsible for the higher solubility of some compounds in solution.
We saw the example of the dissolution of AgCl earlier in this chapter. It has very low solubility in water. However, in the presence of Ammonia (NH3), the solubility of AgCl increases dramatically. Why will the solubility of AgCl increase in the presence of NH3?
To answer this question, we need to understand the process of complex ion formation. We will begin this discussion with the Lewis structure of NH3.
In the Lewis structure of ammonia, the central nitrogen atom has a lone pair of electrons. This lone pair of electrons is donated to the empty orbitals of metal ion; and a coordinate covalent bond is formed between metal ion and ammonia to produce a complex ion. Since nitrogen in NH3 is donating a pair of electrons, you can say NH3 is acting as a Lewis base and, the silver metal ion, accepting a pair of electrons, is serving as a Lewis acid in this reaction.
To show the overall reaction for the formation of the complex ion, we can start with the dissolution of AgCl in solution.
AgCl (s) 
Ag+ (aq) + Cl–(aq)
In the next step, these Ag+ ions will accept a pair of electrons from NH3 and make a complex ion.
Ag+ (aq) + NH3 (aq) [Ag(NH3)+] (aq) Kf = 2.1 x 103
The equilibrium constant for complex ion formation reactions is known as formation constant Kf. The value of Kf for the reaction above is 2.1 x 103.
Molecules or ions attached around the metal ions are called ligands. NH3 is a ligand in this reaction. Ligands are added to metal ions one at a time, and each step has its unique equilibrium constant value or formation constant.
Ag(NH3)+ (aq) + NH3 (aq) Ag(NH3)2+ (aq) Kf = 8.2 x 103
The total number of ligands present around the metal ion is called the coordination number. Most seen coordination numbers are 2, 4, and 6; however other numbers are also observed.
The high value of formation constant Kf in these reactions indicates that the reaction is highly favored in the forward direction.
Now, to answer the original problem, why the solubility of AgCl will increase in the presence of NH3, we can apply Le Chatelier’s principle to the dissolution reaction of AgCl.
AgCl (s) Ag+ (aq) + Cl–(aq)
Since NH3 is combining with these Ag+ ions to produce complex ion, the presence of ammonia will result in the removal of silver ions from the solution; hence equilibrium will shift to the right, or more AgCl will dissolve in water.
Watch this youtube video to see more complex ions with different Lewis acid and Bases.
We can use the value of Kf to calculate the concentration of these ions present in the solution.
Example 7: a) Find the solubility of AgCl in pure water, given Ksp for AgCl = 1.7 x 10-10
b) find the solubility of AgCl in 0.15 M NH3 solution, given Kf for Ag(NH3)2 + is 1.6 x 107
To solve for the solubility of AgCl in pure water, we will begin with the dissolution of AgCl in pure water.
AgCl (s) Ag+ (aq) + Cl–(aq)
- Ksp = [Ag+] [Cl–]
If the solubility of AgCl in pure water = s mol/L,
based on mol ratio, we can say the concentration of [Ag+ ] = [Cl– ] = s mol/L.
1.7 x 10-10 = (s) (s)
x = 
s = 1.3 x 10-5 M
Therefore, the solubility of AgCl in pure water will be 1.3 x 10-5 M.
- In the presence of NH3, a complex ion will be produced, as shown in the reaction below.
Ag+ (aq) + 2NH3 (aq) [Ag(NH3)2+] (aq) Kf = 1.6 x 107
So if we assume the solubility of AgCl in the presence of NH3 is S mol/L,
Then the concentration of [Ag+ ] = [ Cl–] = S mol /L
Since the value of Kf is very high, we can assume the reaction will go to completion, and all the Ag+ ions are present in the solution will convert to [Ag(NH3)2+].
Therefore, the concentration of complex ion [Ag(NH3)2+] = S mol /L
The concentration of chloride ions will still be S mol/L.
Based on Ksp value of AgCl from the reference table and the concentration of chloride ions, we can solve for the remaining concentration of Ag+ ions in the solution.
Ksp = 1.7 x 10-10 = [Ag+] [Cl–]
1.7 x 10-10 = [Ag+] [S]
[Ag+] = 
The concentration of [NH3] = 0.15 M
Plugging in these values in the equation for Kf we can solve for S
Kf =
= 1.6 x 107
S = 0.0078 M
Comparing the results of part a and b, you can see a big increase in the solubility of AgCl in the NH3. In part a, we found solubility of AgCl in pure water was 1.3 x 10-5 M. However, in the presence of 0.15 M NH3 it came out to be 0.0078 M.
Watch this video below to further understand the calculations involving complex ions.
https://www.youtube.com/watch?v=LPG_MR5YXYA
We can have similar Lewis acid and base reactions with ions acting as a ligand also.
In the example below, Cu+ ions can combine with CN– ions, acting as ligands. Here CN– ions being an electron-pair donor, will act as a Lewis base, while Cu+ ions, being electron acceptor, will serve as a Lewis acid.
Cu + (aq) + 2CN– (aq) [Cu(CN)2–](aq)
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Learning Check
- What will be the effect of adding in ammonia on the solubility of CuS.
Answer: Solubility will increase because Cu+2 ions will make a complex ion with ammonia; hence equilibrium for the dissolution of CuS will shift to the right.
CuS (s) Cu+2 (aq) + S-2 (aq)
Cu +2 (aq) + 4NH3 (aq) [Cu(NH3)4+2 ] (aq)