1. EDTA Titration EDTA = Ethylenediaminetetraacetic acid
Lewis acid : electron pair acceptor eg metal ions
Lewis base : electron pair donor eg ligands
A ligand that attaches to a metal ion through more than one ligand atom is called a chelating ligand or multidentate ligand

2. A titration based on complex formation is called complexometic titration
The equilibrium constant for the reaction of a metal with a ligand is called the formation constant, Kf , or the stability constant
Consider :
Ag+ + NH3 Ag(NH3)+
Ag(NH3)+ + NH3 Ag(NH3)2+
Kf1 =
Kf2 =
[Ag(NH3)+]
[Ag+][NH3]
[Ag(NH3)2+]
[Ag(NH3)+][NH3]

3. and the overall formation constant : Kf =Kf1 Kf2 =
The overall reaction :
Ag+ + 2NH3 = Ag(NH3)2+
and the overall formation constant :
Kf =Kf1 Kf2 =
[Ag(NH3)2+]
[Ag+][NH3]2
Example :
A divalent metal M2+ reacts with a ligand L to form a 1:1 complex :
M L ML2+
Calculate the concentration of M2+ in a solution prepared by mixing equal volumes of 0.20 M M2+ and 0.20 M L. Given Kf = 1.0 x 108
Given Kf = 1.0 x complex is sufficiently strong such that the reaction is virtually complete
Since equal volumes were added initial concentration is halved

4. Let x = residual concentration of M2+ M2+ + L ML2+ x x 0.1 – x
Kf = = 1.0 x 108
x = 3.2 x 10-5 M x2
0.1 - x
Similarly, if L is a multidentate ligand and
M + nL MLn
then
Kf =
[MLn]
[M][L]n

5. EDTA Complexes It can be represented as having four Ka values :
EDTA is a hexaprotic system – H6Y2+
Neutral acid is tetraprotic – H4Y
It can be represented as having four Ka values :
H4Y H+ + H3Y- Ka1 = 1.0 x 10-2
H3Y- H+ + H2Y2- Ka2 = 2.2 x 10-3
H2Y H+ + HY3- Ka3 = 6.9 x 10-7
HY H+ + Y Ka4 = 5.5 x 10-11

7. CH Y = [Y4-] + [HY3-] + [H2Y2-] + [H3Y-] + [H4Y] 4
Hence in complexing with -
+1 cation :
Ag Y = AgY3- Kf =
+2 cation :
Hg Y = HgY2- Kf =
+3 cation :
Fe Y = FeY- Kf =
+n cation :
Mn+ + Y = MYn-4 Kf =
[MYn-4]
[Mn+] [Y4-]
[HgY2-]
[Hg2+][Y4-]
[FeY-]
[Fe3+][Y4-]
[AgY3-]
[Ag+][Y4-]
Fraction of the total EDTA species that exists as Y4- = a4 = [Y4-]/CH Y
where
CH Y = [Y4-] + [HY3-] + [H2Y2-] + [H3Y-] + [H4Y] 4

8. Substituting [Y4- ] = a4CH Y Kf = 4
a4Kf = = Kf’ 4
[MYn-4]
[Mn+] a4CH Y
[Mn+]CH Y
Conditional formation constant
– holds only for a particular pH
– describes the formation of MYn-4 at any particular pH
– allows us to look at EDTA complex formation as if the uncomplexed EDTA were all in one form
Example :
Consider the reaction: Fe3+ + EDTA FeY- (Kf = 1.3 x 1025; aY (at pH1) = 1.9 x and aY (at pH4) = 3.8 x 10-9). Calculate the concentration of free Fe3+ in a solution of 0.10 M FeY- at pH4 and at pH1. 4-

9. Let x = [Fe3+] = [EDTA], thus = = 4.9 x 1016 (at pH4)
Using Kf’ = a4Kf ,
at pH = 4 : Kf’ = (3.8 x 10-9)(1.3 x 1025)
= 4.9 x 1016
at pH = 1 : Kf’ = (1.9 x 10-18)(1.3 x 1025)
= 2.5 x 107
Let x = [Fe3+] = [EDTA], thus
= = 4.9 x 1016 (at pH4)
x = 1.4 x 10-9 M
Similarly, at pH1: x = 6.4 x 10-5 M
pH affects the stability of the complex.
The Kf’ values show that the metal-EDTA complex becomes less stable at lower pH
[FeY-]
[Fe3+][EDTA]
0.1 - x x2

10. For a titration reaction to be effective, the equilibrium constant must be large the analyte and titrant should be completely reacted at the equivalence point
Titration of Ca2+ with EDTA as a function of pH :

11. EDTA Titration Curves
Titration is carried out by adding the chelating agent to the sample :
Mn+ + EDTA MYn-4 Kf’ = a4Kf
The titration curve is a graph of pM (= -log[M]) versus the volume of added EDTA
Titration curve consists of 3 regions-
Before the equivalence point :
There is excess Mn+ in the solution after the EDTA has been consumed.
Concentration of free metal ion = concentration of excess unreacted Mn+

12. At the equivalence point :
[Mn+] = [EDTA]
Free Mn+ is from the dissociation of MYn-4 :
MYn-4 Mn+ + EDTA
After the equivalence point :
There is excess EDTA and virtually all the metal ion is in the MYn-4 form
Example :
Calculate the shape of the titration curve for the reaction of 50.0 ml of M Mg2+ (buffered to pH10.0) with M EDTA (Kf = 6.2 x 108 ; at pH10.0 : aY = 0.36)
Mg EDTA MgY2-
Using Kf’ = a4Kf = (0.36)(6.2 x 108)
= 2.2 x 108

13. Since Kf’ is large the reaction goes to completion with each addition of titrant
Before equivalence point :
Consider the addition of 5.0 ml EDTA to the solution
Moles of EDTA added = (0.005 l)( M)
= 2.5 x 10-4
Moles of Mg2+ present initially= (0.050 l)(0.050M)
= 2.5 x 10-3
Moles of Mg2+ present after the addition of EDTA =
(2.5 x 10-3 ) – (2.5 x 10-4) =
[Mg2+] = = M
pMg2+ = -log [Mg2+] =1.39
At equivalence point : [Mn+] = [EDTA]
Volume of EDTA added = (2.5 x 10-3)/(0.0500M)
= 50.0 ml
0.055

14. Since there is negligible dissociation, [MgY2-] = = 0.025 M From:
mol
0.100 l
[MgY2-]
[Mg2+][EDTA] x x2
Since there is negligible dissociation,
[MgY2-] = = M
From:
Mg EDTA MgY2-
Let x =[Mg2+] = [EDTA], then
Kf’ = =
= 2.2 x 108
x = 1.07 x 10-5 M
pMg2+ = -log [Mg2+] =4.97
After equivalence point :
If 51.0 ml of EDTA is added there will be 1.0 ml excess EDTA in the solution

15. [EDTA] ={(0.001)(0.05M)}/(0.101 l) =4.95 x 10-4 M
[MgY2-] =(2.5 x 10-3)/(0.101 l)
=2.48 x 10-2 M
Using
Kf’ = =
= 2.2 x 108
[Mg2+] = 2.3 x 10-7 M
p[Mg2+] = -log [Mg2+]
= 6.64
[MgY2-]
[Mg2+][EDTA]
2.48 x 10-2
[Mg2+](4.95 x 10-4)

16. Metal Ion Indicators
Methods to detect the end point in EDTA titrations are :
metal ion indicators
A metal ion indicator is a compound whose color changes when it binds to a metal ion (eg Eriochrome black T)
This compound must bind metal less strongly than EDTA
MgIn + EDTA MgEDTA + In
(red) (colorless) (colorless) (blue)
mercury electrode : measurement to potential
glass pH electrode
ion-selective electrode

17. EDTA Titration Techniques
Direct Titration
analyte is titrated with standard EDTA
analyte is buffered to an appropriate pH at which the conditional formation constant for the metal-EDTA complex is large and the color of the free indicator is distinctly different from that of the metal-indicator complex
Back titration
a known excess of EDTA is added to the analyte
the excess EDTA is titrated with a standard solution of a second metal
this method is useful if
- analyte precipitates in the absence of EDTA
- the analyte reacts too slowly with EDTA under titration conditions
- analyte blocks the indicator
- metal used in back titration must not displace the analyte metal ion from its EDTA complex

18. Displacement titration
Method is useful is the metal ions do not have a satisfactory indicator
analyte is treated with excess Mg(EDTA)2- to displace Mg2+
Mn+ + MgY2- MYn-4 + Mg2+
- the displaced Mg2+ is titrated with standard EDTA
Example :
2Ag+ + Ni(CN) Ag(CN)2- + Ni2+
(iv) Indirect titration
Anions (such as SO42-, CrO42-, CO32- and S2-) that precipitate with certain metal ions can be analyzed with EDTA through indirect titration
For example, SO42 can be precipitated with excess Ba2+. The BaSO4(s) is washed and boiled with excess EDTA at pH10 to bring Ba2+ back into solution as Ba(EDTA)2-. The excess EDTA is back titrated with Mg2+