Complex-Ion Equilibria and Ligand Formation

A complex ion forms when a metal cation (a Lewis acid) binds to ligands (Lewis bases) through coordinate bonds, and the formation constant Kf measures how favorable this is: for Ag+ + 2NH3 ⇌ [Ag(NH3)2]+, Kf = [Ag(NH3)2+] / ([Ag+][NH3]2) ≈ 107. In brief: because of this complex formation, insoluble AgCl dissolves in 1.0 M ammonia to about 0.047 M.

Complex ions explain why a precipitate may dissolve when an excess of reagent is added, why gold dissolves in a mixture containing cyanide and why photographic film is “fixed” with thiosulfate. Complex-ion equilibria combine with solubility equilibria to determine the actual solubility of many salts. This article defines the terms, describes formation constants, shows how to calculate the solubility of a salt in a solution of a ligand and gives examples of applications.

What is a complex ion?

A complex ion consists of a central metal ion surrounded by molecules or ions called ligands. Each ligand donates a lone pair of electrons to the metal ion to form a coordinate (dative) bond. The number of donor atoms attached to the metal is the coordination number, typically 2, 4 or 6. Common ligands are H2O, NH3, Cl−, CN−, OH− and S2O32−. In water, metal ions exist as aqua complexes, for example [Cu(H2O)6]2+, and the formation of another complex involves the replacement of water molecules by other ligands: [Cu(H2O)6]2+ (pale blue) + 4NH3 ⇌ [Cu(NH3)4(H2O)2]2+ (deep blue) + 4H2O. The water molecules are usually left out of the equation, which is written for the central ion and the ligand.

What is the formation constant?

The equilibrium constant for the formation of the complex from the free metal ion and the ligands is called the formation constant (or stability constant) Kf:

Ag+ + 2NH3 ⇌ [Ag(NH3)2]+, Kf = [Ag(NH3)2+] / ([Ag+][NH3]2)

A large Kf means a stable complex, with very little free metal ion in solution. In reality, complexes form in steps (Ag+ + NH3 ⇌ Ag(NH3)+, then a second ammonia), but when the second step is much more favorable than the first, or when the ligand is in excess, the overall constant is used. The reverse reaction has the dissociation constant Kd = 1/Kf.

Formation constants of some complex ions

Approximate formation constants at 25 °C
Complex ion Formation reaction Kf
[Ag(NH3)2]+Ag+ + 2NH31.5 × 107
[Zn(NH3)4]2+Zn2+ + 4NH32.9 × 109
[Cu(NH3)4]2+Cu2+ + 4NH35.0 × 1013
[Ag(CN)2]−Ag+ + 2CN−1.0 × 1021

Values from different sources differ, sometimes by a factor of 2 or more, depending on the conditions. They show that cyanide forms an extremely strong complex with silver.

How do complexes increase solubility?

When a ligand is present, it removes the metal ion from the dissolution equilibrium by forming a complex, and so more solid dissolves (Le Châtelier’s principle). The overall reaction is the sum of the dissolving equilibrium and the complex formation:

AgCl(s) ⇌ Ag+ + Cl− (Ksp)
Ag+ + 2NH3 ⇌ [Ag(NH3)2]+ (Kf)
AgCl(s) + 2NH3 ⇌ [Ag(NH3)2]+ + Cl− (K = Ksp × Kf)

For the AgCl–ammonia system, K = (1.8 × 10−10)(1.5 × 107) = 2.7 × 10−3.

AgCl(s) ⇌ Ag+ + Cl− Ag+ + 2NH3 ⇌ [Ag(NH3)2]+ AgCl(s) + 2NH3 ⇌ [Ag(NH3)2]+ + Cl− K = Ksp Kf = (1.8 × 10−10)(1.5 × 107) = 2.7 × 10−3

Worked examples

Example 1: solubility of AgCl in ammonia

What is the solubility of AgCl in 1.0 M NH3?

  1. Let x be the amount of AgCl that dissolves: [[Ag(NH3)2]+] = x, [Cl−] = x and [NH3] = 1.0 − 2x.
  2. K = x2 / (1.0 − 2x)2 = 2.7 × 10−3.
  3. Take the square root: x / (1.0 − 2x) = 0.0520.
  4. x = 0.0520 − 0.104x, so 1.104x = 0.0520 and x = 0.047 M.
K = x2 / (1.0 − 2x)2 = 2.7 × 10−3 x / (1.0 − 2x) = 0.0520 1.104x = 0.0520 , x = 0.047 M

The solubility is 0.047 M, about 3500 times higher than in pure water (1.3 × 10−5 M).

Example 2: the free metal ion concentration

A solution contains 0.10 M [Ag(NH3)2]+ and 1.0 M free NH3. What is [Ag+]? From Kf: [Ag+] = [Ag(NH3)2+] / (Kf[NH3]2) = 0.10 / (1.5 × 107 × 1.0) = 6.7 × 10−9 M. Almost all of the silver is in the form of the complex.

[Ag+] = 0.10 / (1.5 × 107 × 1.0) = 6.7 × 10−9 M

Example 3: will AgCl precipitate in the presence of ammonia?

With [Ag+] = 6.7 × 10−9 M, Qsp for AgCl in the presence of 0.010 M Cl− is 6.7 × 10−9 × 0.010 = 6.7 × 10−11, which is less than Ksp (1.8 × 10−10), so no precipitate forms. The ammonia keeps the silver ion concentration low.

Qsp = 6.7 × 10−9 × 0.010 = 6.7 × 10−11 6.7 × 10−11 < 1.8 × 10−10 = Ksp

Example 4: photographic fixing

Silver bromide that was not exposed to light is removed from film with thiosulfate: AgBr(s) + 2S2O32− ⇌ [Ag(S2O3)2]3− + Br−. AgBr is too insoluble to dissolve in ammonia to a useful degree (Ksp = 5.4 × 10−13), but the stable thiosulfate complex dissolves it.

What are amphoteric hydroxides?

Some metal hydroxides, for example Zn(OH)2, Al(OH)3 and Pb(OH)2, dissolve in an excess of hydroxide because they form hydroxo complexes: Al(OH)3(s) + OH− ⇌ [Al(OH)4]−. They are also soluble in acid, and are said to be amphoteric. In qualitative analysis, adding NaOH to a solution of Zn2+ gives a white precipitate that dissolves in excess base, whereas Mg2+ gives a precipitate that does not.

Where are complex ions important?

Who explained coordination compounds?

The structure of coordination compounds was a puzzle in the 19th century, since compounds such as CoCl3·6NH3 and CoCl3·5NH3 had different properties. Alfred Werner proposed in 1893 that the metal has a primary valence (the oxidation state) and a secondary valence (the coordination number), and that the ligands are arranged around the metal in definite geometries such as the octahedron. He confirmed his theory by preparing isomers, and received the Nobel Prize in Chemistry in 1913.

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