This document discusses complexometric reactions and titrations. It defines complexes as compounds formed from the combination of metal ions and ligands. Ligands donate electron pairs to form coordination bonds with electron deficient metal ions. The stability of metal-ligand complexes depends on factors like the nature of the metal ion and ligand, ligand basicity, and size of the chelate ring formed. Example calculations are provided to determine metal ion concentrations in solutions containing ligands using known formation constants.

1. 1
Complexometric Reactions
and Titrations

2. 2
Complexes are compounds formed from
combination of metal ions with ligands
(complexing agents). A metal ionis an
electron deficient species while a
ligand is an electron rich, and thus,
electron donating species. A metal ion
will thus accept electrons from a ligand
where coordination bonds are formed.
Electrons forming coordination bonds
come solely from ligands.

3. 3
A ligand is called a monodentate if it donates a
single pair of electrons (like :NH3) while a
bidentate ligand (like ethylenediamine,
:NH2CH2CH2H2N:) donates two pairs of
electrons. Ethylenediaminetetraacetic acid
(EDTA) is a hexadentate ligand. The ligand
can be as simple as ammonia which forms a
complex with Cu2+, for example, giving the
complex Cu(NH3)4
2+. When the ligand is a
large organic molecule having two or more of
the complexing groups, like EDTA, the ligand
is called a chelating agent and the formed
complex, in this case, is called a chelate.

4. 4

5. 5

6. 6
The tendency of complex formation is
controlled by the formation constant of the
reaction between the metal ion (Lewis acid)
and the ligand (Lewis base). As the formation
constant increases, the stability of the
complex increases.
Let us look at the complexation reaction of Ag+
with NH3:
Ag+ + NH3 D Ag(NH3)+ kf1 = [Ag(NH3)+]/[Ag+][NH3]
Ag(NH3)+ + NH3 D Ag(NH3)2
+ kf2 = [Ag(NH3)2
+]/[Ag(NH3)+][NH3]

7. 7
Kf1 x kf2 = [Ag(NH3)2
+]/[Ag+][NH3]2
Now look at the overall reaction:
Ag+ + 2 NH3 D Ag(NH3)2
+
kf = [Ag(NH3)2
+]/[Ag+][NH3]2
It is clear from inspection of the values of the kf
that:
Kf = kf1 x kf2
For a multistep complexation reaction we will
always have the formation constant of the
overall reaction equals the product of all step
wise formation constants

8. 8
Stability of Metal-Ligand Complexes
The stability of complexes is influenced by a
number of factors related to the ligand and
metal ions.
1. Nature of the metal ion: Small ions with high
charges lead to stronger complexes.
2. Nature of the ligand: The ligands forming
chelates impart extra stability (chelon
effect). For example the complex of nickel
with a multidentate ligand is more stable
than the one formed with ammonia.
3. Basicity of the ligand: Greater basicity of the
ligand results in greater stability of the
complex.

9. 9
4. Size of chelate ring: The formation of
five- or six-membered rings provides
the maximum stability.
5. Number of metal chelate rings: The
stability of the complex is directly
related to the number of chelate rings
formed between the ligand and metal
ion. Greater the number of such rings,
greater is the stability.
7. Steric effects: These also play an
important role in the stability of the
complexes.

10. 10
Example
A divalent metal ion reacts with a ligand to form a
1:1 complex. Find the concentration of the metal
ion in a solution prepared by mixing equal
volumes of 0.20 M M2+ and 0.20 M ligand (L). kf =
1.0x108.
The formation constant is very large and
essentially the metal ions will almost
quantitatively react with the ligand.
The concentration of metal ions and ligand will be
half that given as mixing of equal volumes of the
ligand and metal ion will make their
concentrations half the original concentrations
since the volume was doubled.

11. 11
[M2+] = 0.10 M, [L] = 0.10 M
M2+ + L D ML2+
Kf = ( 0.10 –x )/x2
Assume 0.10>>x since kf is very large
1.0x108 = 0.10/x2, x = 3.2x10-5
Relative error = (3.2x10-5/0.10) x 100 = 3.2x10-2 %
The assumption is valid.
[M2+] = 3.2x10-5 M

12. 12
Silver ion forms a stable 1:1 complex with trien.
Calculate the silver ion concentration at equilibrium
when 25 mL of 0.010 M silver nitrate is added to 50 mL
of 0.015 M trien. Kf = 5.0x107
Ag+ + trien D Ag(trien)+
mmol Ag+ added = 25x0.01 = 0.25
mmol trien added = 50x0.015 = 0.75
mmol trien excess = 0.75 – 0.25 =
0.50
[Trien] = 0.5/75 M
[Ag(trien)+] = 0.25/75 M

13. 13
Kf = ( 0.25/75 – x )/(x * 0.50/75 + x)
Assume 0.25/75>>x since kf is very large
5.0x107 = (0.25/75)/(x * 0.50/75)
x = 1.0x10-8
Relative error = (1.0x10-8/(0.25/75)) x 100 = 3.0x10-4 %
The assumption is valid.
[Ag+] = 1.0x10-8 M