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Metal Complexes
Skills to develop
 Explain what are complex ions or metal complexes.
 Apply the concept of equilibrium in complex formation.
 Calculate concentrations by applying the formation constant.
 Derive the overall formation constant from stepwise formation constants.
 Derive the dissociation constant from formation constant.
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(H_{2}O)_{6}^{+} and
Cu(H_{2}O)_{6}^{2+}. These are called
metal complexes, or coordination compounds.
These coordination reactions are
Lewis acidbase reactions.
The Neutral molecules such as H_{2}O and NH_{3}, and anions
such as CN^{}, CH_{3}COO^{} are called
ligands.
The coordination reaction can be represented by
CuSO_{4} + 6 H_{2}O =
 Cu(H_{2}O)_{6}^{2+} + SO_{4}^{2}


Usually, copper sulphate solids are
CuSO_{4}*(H_{2}O)_{5},
and its color is light blue. When heated, it loses water of crystallization,
and becomes CuSO_{4}. It is colorless.
Most people know that when ammonia is added to a
Cu(H_{2}O)_{6}^{2+} solution,
it turns deep blue. This is due to the formation of complexes:
Cu(H_{2}O)_{6}^{2+} + 4 NH_{3} =
 Cu(H_{2}O)_{2}(NH_{3})_{4}^{2+} + 4 H_{2}O


Actually, the above reaction takes place in steps. The H_{2}O molecules
are displaced one at a time as the ammonia concentration, [NH_{3}],
increases. As more NH_{3} is bonded to Cu^{2+}, 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 NH_{3} = Ag(NH_{3})_{2}^{+} + 2 Cl^{}
The silver ammonia complex is colorless, however.
Other commonly encountered ligands are CN^{}, SCN^{},
Cl^{}, ethylene diamine (NH_{2}CH_{3}CH_{3}NH_{2}),
and acetate (CH_{3}COO^{}).
For example,
Fe(H_{2}O)_{6}^{3+} + SCN^{}
 =
 Fe(H_{2}O)_{5}SCN^{2+} + H_{2}O


brown

 blood red

Formation Constant of Complexes
Complex  K_{f}


Ag(NH_{3})_{2}  1.6e7

Ag(S_{2}O_{3})_{2}^{3}
 1.7e13

Al(OH)_{4}^{}  7.7e33

AlF_{6}^{3}  6.7e19

Zn(EDTA)^{2}  3.8e16

Formation of complexes is also a thermal dynamic phenomenon.
For the equilibrium,
Ag^{+} + 2 NH_{3} = Ag(NH_{3})_{2}^{+},
the formation constant is very large,
[Ag(NH_{3})_{2}^{+}]
K_{f} =  = 1.6e7 M^{2}
[Ag^{+}] [NH_{3}]^{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,
(^{}OOCCH_{2})_{2}NCH_{2}CH_{2}N(CH_{2}COO^{})_{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 AgNO_{3}
and 1.0 M NH_{3}.
Solution
For simplicity of formulation, let M = [Ag^{+}]. The equilibrium
equation and concentration are:
Ag^{+} + 2 NH_{3} = Ag(NH_{3})_{2}^{+}
K_{f} = 1.6e7 M^{2}
x 1.000.20+x 0.10x
0.10x
 = 1.6e7 M^{2}
x (0.80+x)^{2}
x = 9.7e9 M
Discussion
If we assume x M = [Ag(NH_{3})_{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, K_{sp} = 1.8e10.
Ans. Since 9.7e9*0.10 = 9.7e10 > K_{sp},
a precipitate will form.
Example 2
What is the solubility of AgCl in a solution which contains 1.0 M
NH_{3}. For Ag(NH_{3})_{2}^{+},
K_{f} = 1.6e7, and for AgCl, K_{sp}
= 1.8e10
Solution
First consider the equilibria:
AgCl = Ag^{+} + Cl^{}
K_{sp} = 1.8e10 M^{2}
Ag^{+} + 2 NH_{3} = Ag(NH_{3})_{2}^{+}
K_{f} = 1.6e7 M^{2}
Add the two equations together results in the equilibrium equation below.
AgCl + 2 NH_{3} = Ag(NH_{3})_{2}^{+} + Cl^{}
K = K_{sp} K_{f} = 2.9e3
1.0x
x
x
<= equilibrium concentrations
where x is the molar solubility of AgCl.
x^{2}
 = 2.9e3
(1.0x)^{2}
x / (1.0x) = 0.054
x = 0.051 M
Discussion
Answer the following questions and review the discussion in Example 1.
The solubility product of AgBr is 5.0e13 M^{2}. Estimate the molar
solubility of AgBr in a 1.0 M NH_{3} solution.
Ans. 2.8e3 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^{+} + NH_{3} = Ag(NH_{3})^{+}
[Ag(NH_{3})^{+}]
K_{f1} =  = 2.2e3 M
[Ag^{+}] [NH_{3}]
And for the reaction,
Ag(NH_{3})^{+} + NH_{3} = Ag(NH_{3})_{2}^{+}
[Ag(NH_{3})_{2}^{+}]
K_{f2} =  = 7.2e3 M
[Ag(NH_{3})^{+}] [NH_{3}]
And obviously, for the overall reaction,
Ag^{+} + 2 NH_{3} = Ag(NH_{3})_{2}^{+}
[Ag(NH_{3})_{2}^{+}]
K_{f} =  = K_{f1} * K_{f2} = 1.6e7 M^{2}
[Ag^{+}] [NH_{3}]^{+}
A generalized formula is
K_{f} = K_{f1} * K_{f2} * K_{f3} * ...
Dissociation Constants and the Formation Constants
The reverse reaction of the complex formation is called dissociation,
and the equilibrium constant is called dissociation constant
K_{d}.
Ag(NH_{3})_{2}^{+} = Ag^{+} + 2 NH_{3},
K_{d}.
Obviously, we have
K_{d} = 1 / K_{f}.
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 AgNO_{3}
and 0.10 M Na_{2}S_{2}O_{3} solutions are mixed,
giving K_{f} = 1.7e13 M^{2} for
Ag(S_{2}O_{3})_{2}^{3}.
Solution
When the solutions are mixed, [Ag(S_{2}O_{3})_{2}^{3}]
= 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(S_{2}O_{3})_{2}^{3}
= Ag^{+} + 2 S_{2}O_{3}^{3}
0.05x x
2 x
x (2 x)^{2}
 = 1 / 1.7e13 = 5.9e14 M^{2}
0.05x
Since x is very small, 0.05x ~ 0.050, and
[Ag^{+}] = x = (5.9e14 / 4)^{1/3} = 2.5e5 M
The approximation is justified.
Discussion
If you use the formation equilibrium equation and let
[Ag(S_{2}O_{3})_{2}^{3}] = x,
the equation is very difficult to solve.
Confidence Building Questions

Which of the following is a complex ion?
CH_{4},
H_{2}O,
NH_{3},
Al(OH)_{4}^{},
CCl_{4},
CO_{3}^{2},
NH_{4}^{+}
Answer Al(OH)_{4}^{}
Consider...
Explain what are complex ions or metal complexes

To a sample of 9.0 mL solution with [NH_{3}] = 1.1 M,
1.0 mL of 0.1 M AgNO_{3} solution is added.
Calculate [Ag^{+}]. Assume the final volume to be 10.0 mL.
The K_{f} for Ag(NH_{3})_{2} = 1.6e7
Answer [Ag^{+}] = 6.25e10
Consider...
Assume [Ag^{+}] = x, then
Ag^{+} + 2 NH_{3} = Ag(NH_{3})_{2}^{+},
K_{f} = 1.6e7
x 1.0 0.01 < equilibrium concentration
0.01
 = 1.6e7; x = ?
x 1.0^{2}
Apply the concept of equilibrium in complex formation.

Calculate the dissociation constant K_{d} for
Ag(NH_{3})^{+}, if K_{f} for
Ag(NH_{3})_{2} = 1.6e7.
Answer = K_{d} = 6.25e8
Consider...
Derive the dissociation constant from formation constant, and show that
K_{d} = 1 / K_{f}.

The solubility product of AgBr is 5.0e13 M^{2},
and K_{f} = 1.6e7 M^{2} for Ag(NH_{3})_{2}^{+}.
Estimate the molar solubility of AgBr in a 1.0 M NH_{3} solution.
Answer Molar solubility = 2.8e3 M
Hint...
Review Example 2

The solubility product of AgI is 8.3e17 M^{2}.
and K_{f} = 1.6e7 M^{2}for Ag(NH_{3})_{2}^{+}.
Estimate the molar solubility of AgI in a 1.0 M NH_{3} solution.
Answer Molar solubility = 3.6e5 M
Discussion...
The solubilities for AgCl, AgBr, and AgI in 1.0 M NH_{3}
solution are 0.051, 0.0028, and 0.000036 M respectively.

The solubility product of AgBr is 5.0e13 M^{2},
and K_{f} = 1.7e13 M^{2} for Ag(S_{2}O_{3})_{2}^{3}.
Estimate the molar solubility of AgBr in a 1.0 M
Na_{2}S_{2}O_{3} solution.
Answer The molar solubility is 0.49 M
Discussion...
Use the method of Example 2, but no approximation can be made.
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