Ch 13 - EDTA Titrations
EthyleneDiamineTetraacetic Acid
Formation Constants, Kf
M  L  ML
Cu2  NH3  Cu(NH3 )2
Kf 
[ML]
[M][L]
[Cu(NH3 )2 ]
4
K1 

1.9x10
[Cu2 ][NH3 ]
Cu(NH3 )  NH3  Cu(NH )
[Cu(NH3 )22 ]
3
K2 

3.6x10
[Cu(NH3 )2 )][NH3 ]
Cu(NH3 )22  NH3  Cu(NH3 )32
[Cu(NH3 )32 ]
2
K3 

7.9x10
[Cu(NH3 )22 ][NH3 ]
Cu(NH3 )32  NH3  Cu(NH3 )24
[Cu(NH3 )24 ]
2
K4 

1.5x10
[Cu(NH3 )32 ][NH3 ]
2
2
3 2
EDTA Formation Constants
EDTA is a hexaprotic weak acid that
complexes 1:1 with metal cations -
Notice that the first 4 protons are much more
acidic than the last two, so the dominant form
of EDTA in solution will be H2Y2- ....
Mn+ + H2Y2- = MYn-4 + 2H+
low pH
high pH
By Le Chatelier's Principle, the complex will
dissociate at low pH's, and it will be more
stable at high pH's.
EDTA titrations are therefore pH dependent
and analyte solutions must be buffered to the
optimum pH.
Minimum pH for Titration
from "Fundamentals of Analytical Chemistry", Skoog and West, 4th ed.
The pH-Dependent Metal-EDTA Equilibrium
(Sec. 12-5 and Sec 13-5)
Fractional Composition Equations - the
fraction (percentage) of each species of an acid
or base existing at a given pH
HA = H+ + A-
A-
CT = [HA] + [A-]
= fraction of HA dissociated to A-
[A ]
[A  ]



[HA]  [A ]
CT
HA
= fraction of HA still existing as HA
[HA]
[HA]



[HA]  [A ]
CT


[H
][A
]


f or HA  H  A , K a 
[HA]
the mass balance equation is -
CHA  [HA]  [A - ]  CT
rearranging and substituting into Ka -
[A  ]  C T  [HA]
[H ](CT - [HA])
Ka 
[HA]
[HA] K a  [H ]CT - [H ][HA]
solve for [HA] [HA] K a  [H ]CT - [H ][HA] so [HA] K a  [H ][HA]  [H ]CT
[HA]
[H ]

CT
K a  [H ]
so αHA
[H ]

K a  [H ]
to solve for A- substitute [HA] = CT - [A-] into Ka
[H ][A ] [H ][A ]
Ka 

[HA]
CT - [A- ]
so (C T - [A - ])K a  [H ][A ]
CTK a - [A - ]K a  [H ][A ]
and
C TK a  [A  ]([H ]  K a )
Ka
so f inally α A- 
K a  [H ]
so
C TK a  [H ][A ]  [A - ]K a
Ka
[A ]


[H ]  K a
CT
Fractional Composition Diagram for Monoprotic Acids
αHA
[H ]

K a  [H ]
Ka
α A- 
K a  [H ]
cross where
pH = pKa
Fractional Composition Diagrams for
Polyprotic Acids: General Forms
diprotic
triprotic
Fractional Composition Diagram H 2CO3
1.00
0.90
0.80
alpha
0.70
Ka1 = 4.46 x 10-7
0.60
0.50
Ka2 = 4.69 x 10-11
0.40
0.30
0.20
0.10
0.00
0
2
4
6
8
10
12
14
pH
Fractional Composition Diagram H 3PO4
1.00
0.90
0.80
Ka1 = 7.11 x 10-3
alpha
0.70
0.60
Ka2 = 6.34 x 10-8
0.50
0.40
Ka3 = 4.50 x 10-13
0.30
0.20
0.10
0.00
0
2
4
6
8
pH
10
12
14
Fractional Composition Diagrams for EDTA
Conditional (Effective) Formation Constant, K'f
For EDTA titration curves, it's convenient to base
calcuations on the Y4- form of EDTA and derive a
new, pH-dependent formation constant K'f
Mn  Y 4  MY n4
[MY n4 ]
K f  n 4 
[M ][Y ]
CT = all forms of EDTA (Y4-, HY3- etc)
αY4 
[Y4 ]

CT
and theref ore αY4 CT  [Y4 ]
[MY n4 ]
[MY n4 ]
substituting into K f  n 4  n
[M ][Y ] [M ]α Y4 CT
[MY n4 ]
[MY n4 ]
and f inally K f  n
or K f α Y4   n
[M ]αY4 CT
[M ]CT
the conditiona l f ormation constant K 'f  K f α Y4 
[MY n4 ]
 n
[M ]CT
All EDTA equilibrium calculations will use K'f at
the pH of the titration. The value of Y4- at this
pH is taken from Table 13-3.
Example, p. 300
The formation constant for FeY- is 1.3 x 1025
(Fe3+). Calculate the concentration of free Fe3+
in a solution of 0.10 M FeY- at pH = 4.00 and pH
= 1.00.
EDTA Titration Curves, Sec 13-6
A complex formation titration curve plots pM (analogous to
pH) vs. volume of titrant (see next slide). To save time, we
will only calculate the pM = -log[Mn+] at the equivalence
point in order to select the correct indicator.
p.302 - Calculate the titration curve for the reaction of 50.0 mL
of 0.0500 M Mg2+, buffered to a pH of 10.0, with 0.0500 M
EDTA. The equivalence pt. volume is 50.0 mL.
e.g. Mg 2  Y 4  MgY 2K 'f  K f α Y4  
[MY n4 ]
K f  n 4  6.2 x 10 8
[M ][Y ]
At the equivalence pt. virtually all of the Mg is in the form
MgY2-.
[MgY2-] at the eq. pt. =
The ICE table for the reaction is:
Mg 2 
Initial:
Change:
Equilibrium:
Y4 
MgY 2-
Kf CaY2- > Kf MgY2- so the endpoint is sharper for Ca2+
The equiv. pt. becomes sharper
as the pH of the titration
approaches the optimal value
for the analyte, e.g. for Ca2+…..
The equiv. pt. becomes sharper
as the Kf of the EDTA-metal
complexes becomes larger…..
H2In- + H2O = HIn2- + H3O+
red
blue
HIn2- + H2O = In3- + H3O+
blue
pKa2 = 8.1
pKa3 = 12.4
orange
end point reaction with metal cation…..
MIn- + Y4- = MY2- + In3red
blue
Kf < Kf analyte
In the titration of Ca, Mg is added to the titrant in
order to sharpen the end point.