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Metal Complexes

Skills to develop

Metal Complexes

Metal are Lewis acids because of their positive charge. When dissolved in water, they react with water to form hydrated compound such as Na(H2O)6+ and Cu(H2O)62+. These are called metal complexes, or coordination compounds. These coordination reactions are Lewis acid-base reactions. The Neutral molecules such as H2O and NH3, and anions such as CN-, CH3COO- are called ligands. The coordination reaction can be represented by
CuSO4 + 6 H2O =  Cu(H2O)62+ + SO42-

Usually, copper sulphate solids are CuSO4*(H2O)5, and its color is light blue. When heated, it loses water of crystallization, and becomes CuSO4. It is colorless.

Most people know that when ammonia is added to a Cu(H2O)62+ solution, it turns deep blue. This is due to the formation of complexes:
Cu(H2O)62+ + 4 NH3Cu(H2O)2(NH3)42+ + 4 H2O

Actually, the above reaction takes place in steps. The H2O molecules are displaced one at a time as the ammonia concentration, [NH3], increases. As more NH3 is bonded to Cu2+, the blue color deepens.

Ammonia forms complex with many metals. It forms a very strong complex with Ag+ such that AgCl solid will dissolve in ammonia solution.

AgCl (s) + 2 NH3 = Ag(NH3)2+ + 2 Cl- The silver ammonia complex is colorless, however.

Other commonly encountered ligands are CN-, SCN-, Cl-, ethylene diamine (NH2CH3CH3NH2), and acetate (CH3COO-). For example,

Fe(H2O)63+ + SCN- = Fe(H2O)5SCN2+ + H2O
brown blood red

 

 

Formation Constant of Complexes

Complex Kf
Ag(NH3)2 1.6e7
Ag(S2O3)23- 1.7e13
Al(OH)4- 7.7e33
AlF63- 6.7e19
Zn(EDTA)2- 3.8e16
Formation of complexes is also a thermal dynamic phenomenon. For the equilibrium, Ag+ + 2 NH3 = Ag(NH3)2+, the formation constant is very large,         [Ag(NH3)2+]
Kf = ---------------- = 1.6e7 M-2
        [Ag+] [NH3]2
because [Ag+] is very small in such a equilibrium.

The formation constants of some other complexes are given in a table form on the right. Although only three metal ions are involved, the complex are formed by five ligands, and the overall formation constants range from 1.6e7 to 7.7e33. The EDTA is a complicated organic molecule, (-OOCCH2)2N-CH2-CH2-N(CH2COO-)2 with six sites (4 O and 2 N) embracing the zinc ion in the zinc complex.

 

 

 

Example 1

Calculate [Ag+] in a solution containing 0.10 M AgNO3 and 1.0 M NH3.

Solution
For simplicity of formulation, let M = [Ag+]. The equilibrium equation and concentration are:

Ag+ +   2 NH3     =   Ag(NH3)2+     Kf = 1.6e7 M-2
x     1.00-0.20+x       0.10-x

  0.10-x
------------ = 1.6e7 M-2
x (0.80+x)2

x = 9.7e-9 M

Discussion
If we assume x M = [Ag(NH3)2+], the calculation will be very difficult. Try it and find out why.

Suppose we now introduce 0.10 M Cl- into the equilibrium, will a precipitate form? For AgCl, Ksp = 1.8e-10. Ans. Since 9.7e-9*0.10 = 9.7e-10 > Ksp, a precipitate will form.

 

Example 2

What is the solubility of AgCl in a solution which contains 1.0 M NH3. For Ag(NH3)2+, Kf = 1.6e7, and for AgCl, Ksp = 1.8e-10

Solution
First consider the equilibria:

AgCl = Ag+ + Cl-     Ksp = 1.8e-10 M2
Ag+ +   2 NH3 = Ag(NH3)2+     Kf = 1.6e7 M-2
Add the two equations together results in the equilibrium equation below. AgCl + 2 NH3 = Ag(NH3)2+ + Cl-     K = Ksp Kf = 2.9e-3
            1.0-x           x               x <= equilibrium concentrations
where x is the molar solubility of AgCl.

      x2
--------- = 2.9e-3
(1.0-x)2

x / (1.0-x) = 0.054

x = 0.051 M

Discussion
Answer the following questions and review the discussion in Example 1.

The solubility product of AgBr is 5.0e-13 M2. Estimate the molar solubility of AgBr in a 1.0 M NH3 solution. Ans. 2.8e-3 M

 

Stepwise Formation Constants and Overall Constants

As indicated earlier, the formation of a complex takes place in steps. The formation constants in these steps are called stepwise formation constants.

For the reaction,

Ag+ + NH3 = Ag(NH3)+

        [Ag(NH3)+]
Kf1 = --------------- = 2.2e3 M
        [Ag+] [NH3]

And for the reaction, Ag(NH3)+ + NH3 = Ag(NH3)2+

          [Ag(NH3)2+]
Kf2 = -------------------- = 7.2e3 M
        [Ag(NH3)+] [NH3]

And obviously, for the overall reaction, Ag+ + 2 NH3 = Ag(NH3)2+

          [Ag(NH3)2+]
Kf = ------------------- = Kf1 * Kf2 = 1.6e7 M2
        [Ag+] [NH3]+
A generalized formula is Kf = Kf1 * Kf2 * Kf3 * ...

 

Dissociation Constants and the Formation Constants

The reverse reaction of the complex formation is called dissociation, and the equilibrium constant is called dissociation constant Kd.

Ag(NH3)2+ = Ag+ + 2 NH3,     Kd.

Obviously, we have

Kd = 1 / Kf. Similar to stepwise formation constant, we can also apply the concept to give a stepwise dissociation constant.

 

Example 3

Calculate [Ag+] when equal volume of 0.10 M AgNO3 and 0.10 M Na2S2O3 solutions are mixed, giving Kf = 1.7e13 M-2 for Ag(S2O3)23-.

Solution
When the solutions are mixed, [Ag(S2O3)23-] = 0.05 M. Let x M = [Ag+] at equilibrium. The dissociation equilibrium equation rather than the formation equilibrium equation is more convenient in this case.

Ag(S2O3)23- = Ag+ + 2 S2O33-
0.05-x             x         2 x

x (2 x)2
--------- = 1 / 1.7e13 = 5.9e-14 M2
0.05-x

Since x is very small, 0.05-x ~ 0.050, and
[Ag+] = x = (5.9e-14 / 4)1/3 = 2.5e-5 M
The approximation is justified.

Discussion
If you use the formation equilibrium equation and let [Ag(S2O3)23-] = x, the equation is very difficult to solve.

 

Confidence Building Questions

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