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(NH_{3})_{4}^{2+}
ion, which is formed by attaching four ammonia molecules to a Cu^{2+}
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: 



The K value is given a new name, the K_{form} or the K of
Formation for the complex. 



The K_{form} 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,
K_{inst}. 

K_{inst} = 1
K_{inst }is called the instability constant because K_{form }the larger it's value the more unstable the complex is. 

Effect of Complex Ion Formation on Solubility  
The K_{sp }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(NH_{3})_{2}^{+}
ions from the trace amount of Ag^{+}(aq) initially present in solution.




The ammonia represents upsets the equilibrium present in the saturated
solution of 



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)
K_{sp }= [Ag^{+}][Br^{}] 

New Equilibrium  


__________________________________________
Sum of the equilibriums 

AgBr(s) + 2 NH_{3}(aq) <======> Ag(NH_{3})_{2}^{+}(aq) + Br^{}(aq)  
The equilibrium constant for this new overall equilibrium is:




The value of K_{c} can be found by multiplying K_{form
}* K_{sp} 

i.e. K_{c }= K_{form} * K_{sp} =


The [Ag^{+}] cancel out. If we know the values of K_{form}
and K_{sp }we can calculate the new value of K_{c}. 

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 NH_{3}?






K_{sp }= 5.0 x 10^{13 }K_{form} = 1.6 x 10^{7}
K_{c }= 5.0 x 10^{13 }* 1.6 x 10^{7}
= 8.0 x 10^{6} 

AgBr(s) + 2 NH_{3}(aq) <======> Ag(NH_{3})_{2}^{+}(aq)
+ Br^{}(aq) initial [ ]'s 1.0 0.0 0.0 Changes in Becomes
1.02x
+x
+x ^{1} 

Therefore


8.0 x 10^{6 }= (x)(x) (1.02x)^{2} 



0.0028 = __x__ 1.02x 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 NH_{3}. this is not very much but it is approximately 4000 times more than when AgBr dissolves in pure water. 