L L M

L

L

Xn+

m+ L - ligand

M - central species

X - counter ion

Another factor is the coordination number of the central species - bonds formed.

Common numbers 2 - linear 4 - tetrahedral / square planer 6 - octahedral

Both the coordination and dentate numbers must be known to know the number of ligands/central species.

Possess only one accessible donor group.

H2O is a good example since all metal ions exist as aqua complexes in water.

Although two e- pair are available, only one is accessible.

The other will always point the wrong way

In aqueous systems, complexes form by the stepwise displacement of water.

M(H2O)nm+ + L M(H2O)n-1(L)m+ + H2O

M(H2O)n-1m+ + L M(H2O)n-2(L)2

m+ + H2O . . .

M(H2O)(L)n-1m+ + L M(L)m+ + H2O

K1

Kn-1

K2

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.

Zn2+ + 2 en Zn C N

C N

C N

C N

O-

N Zn2+ + 2

O

N Zn

2

C N

O

H

H3C

CH3C

N

O

Ni

N

O

C

CN

OH

CH3

CH3

Ni (dmg)2

N

N

Fe(II)

3

EDTA is typically used as the disodium salt to increase solubility.

HOOC

N

Na+ -OOC

C C N

COO- Na+

COOH

H2Y2-

The molecule contains 6 donor groups.

Regardless of the coordination number of the central species, the molecule will adapt to the number needed. Mg2+ + H2Y2- MgY2- + 2H+ Fe3+ + H2Y2- FeY- + 2H+

HOOC

N

Na+ -OOC

C C N

COO- Na+

COOH

H6Y2+ H5Y+ H4Y H3Y- H2Y2- HY3- Y4- K4 K3 K2 K1 K K’

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)

0 2 4 6 8 10 12 14

1

.5

0

!

H6Y2+

H5Y+

H4Y

H3Y-

H2Y2- HY3-

Y4-

Under normal conditions, the H6Y2+ and H5Y+ forms are not present at significant levels.

The effect of the hydrogen ion can then be calculated using !Y.

!Y = [Y4-] [H4Y] + [H3Y-] + [H2Y2-] + [HY3-] + [Y4-]

= [ Y ] / [ Y’ ]

[H+] [ H2Y2-] [H3Y-]

[H+] [ H3Y-] [H4Y] Similar expressions

can be written for the other two

equilibria

We can proceed through a series of substitutions leading to an equation in terms of [H+].

1 !Y

[H+] K4

[H+]2 K3K4

[H+]3 K2K3K4

[H+]4 K1K2K3K4

= + + + + 1

We can then calculate !Y and plotted as a function of pH.

0 7 14

0

-18

-9

pH

log

!Y

pH ~11.5 All in Y4+ form

While the actual formation constant for a metal-EDTA complex is:

KMY =

A conditional constant can be calculated at any known, constant pH as:

[MY4-] [M] [Y4-]

[MY4-] [M] [Y’] KMY !Y = KMY’=

In order to get sharp endpoints, solutions are typically buffered at basic conditions.

This also insures that: Y4- form is available. EDTA will be in a soluble form.

In addition, only one !Y must be used in calculating equivalence and over titration conditions.

As with our other types of titrations, we have four regions to deal with.

0% titration >0% and < 100% titration The equivalence point Over-titration region

We’ll outline the basic steps for each region 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 100 ml + 50 ml KMY’ = KMY !Y = [NiY2-]

[Ni2+] [Y’]

KMY = 3.98 x 1018

We’ll need to calculate !Y at pH 10.2.

1 !Y

[H+] K4

[H+]2 K3K4

[H+]3 K2K3K4

[H+]4 K1K2K3K4

= + + + + 1

K1 = 1.02x10-2 K2 = 2.14x10-3 K3 = 6.92x10-7 K4 = 5.50x10-11 [H+] = 6.31x10-11

If you’re willing to trust me, then !Y = 0.47

KMY !Y = [NiY2-]

[Ni2+] [Y’] = 3.98x1018 x 0.47

= 1.87x1018

At the equivalence point, we also know that virtually all of our nickel exists as NiY2-.

So, [NiY2-] = 0.0050 M due to dilution.

In addition [Ni2+] = [Y’] so:

1.87x1018 =

[Ni2+] = (0.0050 M / 1.87x1018 )1/2

= 5.17 x10-11 M

pNi = 10.3

0.0050 M [Ni2+]2

KMY’ = [NiY2-]

[Ni2+] [Y’] = 1.87x1018

[Ni2+] = 5.35x10-19 pNi = 18.27

pNi

12

8

2 0 100 200

The 200% titration mark is useful for estimating the shape of a titration curve because:

pM = log KMY’

Since you also know the 0% value (-log[species]) and the general shape of a titration curve, a reasonable estimate is possible.

p[

]

12

8

2

Ni2+

Ca2+

To be a viable indicator, we need a species that:

Competes for our metal ion.

Is a weaker complexer than EDTA.

Exhibits a measurable change between the complexed and uncomplexed form.

M-Ind + Y’ MY + Ind (color 1) (color 2)

In the presence of an indicator, our reaction proceeds in two steps.

M + Y’ MY M-Ind + Y’ MY + Ind

Since it is easier to EDTA to react with the uncomplexed metal, that reaction occurs first.

M-Ind is harder for EDTA to react with so we must insure that only a small amount of indicator is used.

Effect of !Y

Effect of !M

Effects of pH and indicator

The red plot shows a titration curve if our metal is not complexed by an indicator and EDTA is not complexed by H+.

- N = N -

OH HO

SO3-

pH 8.1 pH 12.4

(orange) (blue) (red)

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 block the indicator.

- N = N -

OH OH

O2N

-O3S

Complexes are red Uncomplexed form is blue

- N = N -

AsO3H2 OH

SO3-

OH

-O3S

- N = N -

-O3S

OH N

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.