1. m+
L L - ligand
M - central species
L M L Xn+
X - counter ion
L

2. Another factor is the coordination number of the Possess only one accessible donor group.
central species - bonds formed.
H2O is a good example since all metal ions exist
Common numbers as aqua complexes in water.
2 - linear
4 - tetrahedral / square planer Although two e- pair are
6 - octahedral available, only one is
accessible.
Both the coordination and dentate numbers
must be known to know the number of The other will always point
ligands/central species. the wrong way
In aqueous systems, complexes form by the
stepwise displacement of water.
K1
M(H2O)nm+ + L M(H2O)n-1(L)m+ + H2O
K2
M(H2O)n-1m+ + L M(H2O)n-2(L)2m+ + H2O
.
.
. Kn-1
M(H2O)(L)n-1m+ + L M(L)m+ + H2O
Form two binds / central species.
A good example is ethylene diamine.
NH2CH2CH2NH2 - (en)
The amino groups are far enough apart to
permit both to interact.
C N N C
Zn2+ + 2 en Zn
C N N C

3. H
O O
H 3C CH3
N C N N C
N
Zn Ni
Zn2+ + 2 C N N C
O- O H 3C
CH3
O O
2
H
N
Ni (dmg)2
Fe(II)
N
3

4. EDTA is typically used as the disodium salt to The molecule contains 6 donor groups.
increase solubility.
HOOC COO - Na+
HOOC COO - Na+ N C C N
Na + -
OOC COOH
N C C N
Regardless of the coordination number of the
+ - COOH central species, the molecule will adapt to the
Na OOC
number needed.
H2Y2- Mg2+ + H2Y2- MgY2- + 2H+
Fe3+ + H2Y2- FeY- + 2H+
K’ K K1 K2 K3 K4
H6Y2+ H5Y+ H 4Y H3Y- H2Y2- HY3- Y4-
The process is further complicated by the
facts that:
The amine groups can be protonated
The equilibria are not well behaved
(more than two species may exist at
significant levels at any given pH)

5. H5Y+ H 3 Y-
Under normal conditions, the H6Y2+ and H5Y+
1
H4Y Y4- forms are not present at significant levels.
H6Y2+
H2Y2- HY3-
The effect of the hydrogen ion can then be
calculated using !Y.
! .5
[Y4-]
!Y = [H Y] + [H Y-] + [H Y2-] + [HY3-] + [Y4-]
4 3 2
0 = [ Y ] / [ Y’ ]
0 2 4 6 8 10 12 14
We can proceed through a series of
substitutions leading to an equation in terms
of [H+].
1 [H+]4 [H+]3 [H+]2 [H+]
[H+] [ H3Y-] = + + + +1
[H4Y] Similar expressions !Y K1K2K3K4 K2K3K4 K3K4 K4
can be written
[H+] [ H2Y2-]
for the other two
[H3Y-] We can then calculate !Y and plotted as
equilibria
a function of pH.
0 While the actual formation constant for a metal-
EDTA complex is:
[MY4-]
KMY = [M] [Y4-]
log !Y
pH ~11.5
-9 All in Y4+ form
A conditional constant can be calculated at any
known, constant pH as:
-18 [MY4-]
0 7 14 KMY !Y = KMY’=
pH [M] [Y’]

6. In order to get sharp endpoints, solutions are As with our other types of titrations, we have
typically buffered at basic conditions. four regions to deal with.
This also insures that: 0% titration
Y4- form is available. >0% and < 100% titration
EDTA will be in a soluble form. The equivalence point
Over-titration region
In addition, only one !Y must be used in
calculating equivalence and over titration We’ll outline the basic steps for each region
conditions. using an example.
Initial [Ni2+] x VNi2+ - [EDTA] x VEDTA added
VNi2+ + VEDTA added
0.0100M x 100.0 ml - 0.0100M x 50.0ml
[NiY2-]
100 ml + 50 ml KMY’ = KMY !Y =
[Ni2+] [Y’]
KMY = 3.98 x 1018

7. We’ll need to calculate !Y at pH 10.2. If you’re willing to trust me, then
!Y = 0.47
1 [H+]4 [H+]3 [H+]2 [H+]
= + + + +1 [NiY2-]
!Y K1K2K3K4 K2K3K4 K3K4 K4 KMY !Y = = 3.98x1018 x 0.47
[Ni2+] [Y’]
= 1.87x1018
K1 = 1.02x10-2 K2 = 2.14x10-3
K3 = 6.92x10-7 K4 = 5.50x10-11 At the equivalence point, we also know that
[H+] = 6.31x10-11 virtually all of our nickel exists as NiY2-.
So, [NiY2-] = 0.0050 M due to dilution.
In addition [Ni2+] = [Y’] so:
0.0050 M
1.87x1018 =
[Ni2+]2
[Ni2+] = (0.0050 M / 1.87x1018 )1/2 [NiY2-]
KMY’ = = 1.87x1018
[Ni2+] [Y’]
= 5.17 x10-11 M
pNi = 10.3 [Ni2+] = 5.35x10-19 pNi = 18.27
12 The 200% titration mark is useful for estimating
the shape of a titration curve because:
pM = log KMY’
pNi
8
Since you also know the 0% value (-log
[species]) and the general shape of a titration
curve, a reasonable estimate is possible.
2
0 100 200

8. 12 Ni2+
Ca2+
p[ ]
8
2
To be a viable indicator, we need a species In the presence of an indicator, our reaction
that: proceeds in two steps.
Competes for our metal ion. M + Y’ MY
Is a weaker complexer than EDTA. M-Ind + Y’ MY + Ind
Exhibits a measurable change Since it is easier to EDTA to react with the
between the complexed and uncomplexed metal, that reaction occurs first.
uncomplexed form. M-Ind is harder for EDTA to react with so we
M-Ind + Y’ MY + Ind must insure that only a small amount of
(color 1) (color 2) indicator is used.
Effects of pH and indicator
The red plot
shows
Effect of a titration curve
!Y if our metal is OH HO
Effect of not complexed by
an indicator and -N=N-
!M
EDTA is not
complexed by H+.
SO3-

9. pH 8.1 pH 12.4
(red) (blue) (orange) OH OH
-O
3S -N=N-
Since the metal complexes are red, the indicator
is only useful in the range of pH 8.1 - 12.4.
Presence of Cu(II), Ni(II), Fe(III) or Al(III) can O2N Complexes are red
block the indicator. Uncomplexed form is blue
AsO3H2 OH OH
OH
N
-N=N- -N=N-
-O
3S SO3- -O
3S
SO3-
As with other titrations, ion selective electrodes
can be used to monitor a titration.
Once combined with EDTA, a metal ion’s
presence is masked from the electrode.
One simply needs to monitor E and plot verses
the ml titrant.
Many electrodes exist for this purpose, including
ones to detect hardness.