Complex Ion Equilibria
In an aqueous solution, a complex ion exists in equilibrium with it's components, the central metallic ion and the ligands.
 
Complex ions are formed when Lewis bases (ligands) attach themselves to metal ions in solution. An example is the blue Cu(NH3)42+ ion, which is formed by attaching four ammonia molecules to a Cu2+ ion. Complexes of this type can be quite complicated because as each ligand attaches it forms it's own unique equilibrium. When the equilibrium concentration is large we can simplify the problem by realizing that the equilibrium is just:
 
Cu2+(aq) + 4 NH3(aq) <======>  Cu(NH3)42+(aq)
The K value is given a new name, the Kform or the K of Formation for the complex.
 

 
The Kform is often called a stability constant. The larger it's magnitude the more stable is the complex. Most of the metal ions that form complexes are those of the transition elements. In fact, the tendency to form complex ions is often listed as one of the general properties of the transition metals.
 
Instability Constants
In many chemical references the stabilities of complex ions are indicated as inverses. The inverses of formation constants are called instability constants, Kinst.
 
Kinst =                                Kinst is called the instability constant because
            Kform                                the larger it's value the more unstable the complex is.
 
Effect of Complex Ion Formation on Solubility
The Ksp of AgBr is 5.0 x 10-13. Suppose we have a saturated solution of this salt, with some AgBr(s) left undissolved at the bottom of the beaker. We next add some aqueous ammonia to the system. Its molecules are strong ligands for silver ions, so they begin to form Ag(NH3)2+ ions from the trace amount of Ag+(aq) initially present in solution.
 
Ag+(aq) + 2 NH3(aq) <======> Ag(NH3)2+(aq)
The ammonia represents upsets the equilibrium present in the saturated solution of
 
AgBr(s) <======> Ag+(aq) + Br-(aq)
By pulling Ag+ ions out of this equilibrium, the equilibrium must shift to replace them as best it can. i.e. more AgBr(s) must go into solution. The solubility of a slightly soluble salt increases when one of its ions can be changed to a soluble complex ion.
 
Adding the ammonia introduces another equilibrium and we can sum the two equilibrium present.
 
Original Equilibrium
AgBr(s) <=====>  Ag+(aq) + Br-(aq)      Ksp = [Ag+][Br-]
 
New Equilibrium
Ag+(aq) + 2 NH3(aq) <======>  Ag(NH3)2+(aq) 
__________________________________________

Sum of the equilibriums

AgBr(s) + 2 NH3(aq) <======> Ag(NH3)2+(aq) + Br-(aq)
The equilibrium constant for this new overall equilibrium is:
 

 
The value of Kc can be found by multiplying Kform * Ksp
 
i.e. Kc = Kform * Ksp
 
The [Ag+] cancel out. If we know the values of Kform and Ksp we can calculate the new value of Kc.
 
Calculating the Solubility of a Slightly Soluble Salt in the Presence of a Ligand
How many moles of AgBr can dissolve in 1.0 L of 1.0 M NH3?
 
AgBr(s) + 2 NH3(aq) <=====>  Ag(NH3)2+(aq) + Br-(aq)

 
Ksp = 5.0 x 10-13 Kform = 1.6 x 107 Kc = 5.0 x 10-13 * 1.6 x 107

      = 8.0 x 10-6
 

                    AgBr(s) + 2 NH3(aq) <======> Ag(NH3)2+(aq) + Br-(aq)
initial [ ]'s                         1.0                                  0.0                    0.0

Changes in
[ ] caused
by NH3                         -2x                                   +x                     +x

Becomes                    1.0-2x                                 +x                     +x 1
 

Therefore 
 
8.0 x 10-6 = (x)(x)
                   (1.0-2x)2
 

 
0.0028 = __x__
                1.0-2x

x = 0.0028 - 0.0056x

1.0056x = 0.0028

x = 2.78 x 10-3
 

So 2.8 x 10-3 mol of AgBr dissolves in 1.0 L of 1.0 M NH3. this is not very much but it is approximately 4000 times more than when AgBr dissolves in pure water.
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