The Effect of Complex Ion Formation on Solubility

When a metal cation combines with a Lewis base, a complex ion forms.  We can define a complex ion as an ion containing a central metal cation bonded (covalently) to one or more molecules or ions called ligands.  Transition metal ions have a particular tendency to form complex ions because they have more than one oxidation state.  This property allows them to act effectively as Lewis acids in reactions with many molecules or ions that serve as electron donors or as Lewis bases.  

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)

     [Cu(NH3)42+]
Kf = ------------
     [Cu2+][NH3]4

The equilibrium constant for this reaction, Kf, is know as the K of formation and is sometimes called the stability constant, the K for the reverse reaction is known as the instability constant.  The formation of a complex ion tends to increase the solubility of a salt because the metal cations are removed from the solubility equilibria to form the complex ion.

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 an upset to 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 the right 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)

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:

     [Ag(NH3)2+][Br-]
Kc = ----------------
         [NH3]2

The value of Kc can be found by multiplying   Kf * Ksp

 Example 

Calculate the molar solubility of AgBr in 2.6 M NH3. The Ksp of AgBr is 5.0 x 10-13, and the Kf for Ag(NH3)2+ is 1.7 x 107.

First set up an equilibrium table and let x = change concentration of Ca2+.

Concentrations(M) AgBr(s)  + 2 NH3(s)  <==> Ag(NH3)2+(aq)
+
 Br-(aq)
Starting    2.6  0 0
Change   -2x + x  + x
Equilibrium   2.6 - 2x  x  x

 Substitute these values into the equilibrium constant expression and solve for x.  Kc = Kf x Ksp = (1.7 x 107) x (5.0 x 10-13) = 8.5 x 10-6 . 

     [Ag(NH3)2+][Br-]       (x)(x)   
Kc = ---------------- =  ----------- = 8.5 x 10-6 
         [NH3]2          (2.6 - 2x)2

 Take the square root of each side and solve for x, the molar solubility of AgBr in 2.6 M NH3. 

  x   
--------- = 2.9 x 10-3 
 2.6 - 2x
x = 0.0075 M

 

Back to Solubility Equilibria Index Page