1. Complex Formation Titrations
Metal ions react with electron pair donors to form
coordination complexes
Coordination number- number of covalent
bonds a cation tends to form
2,4,6
Example, Cu(NH3)4
2+
Donor species: Ligand
has at least one pair of unshared e-
Examples: NH3, H2O, Cl, CN

2. Complex Formation Titrations
Ligand that has single donor group:
monodentate
Ligand that has two donor groups:
bidentate
Example of bidentate,

4. Complex Formation Titrations
EDTA-ethylenediamine tetracetic acid
•Most widely used chelator in Analytical Chemistry
•Forms chelates with metal ions (1:1)
•Many are stable – serves as a basis for volumetric analysis
•polyprotic weak acid

7. H4Y, H3Y-
, H2Y-2
, HY-3
, Y4-
Dissociation Products
Fraction of EDTA in each of its
protonated forms is a function of pH
αY4- = [Y4-
] / [EDTA]
Low pH (3-6), H2Y-2
predominates
High pH (>10), Y4-
predominates

9. Use to find α at
different pHs
Chapter 11

10. Complex Formation Titrations
The formation constant (Kf) for a metal EDTA complex
Describes the reaction between Y-4
and metal ion
Mn+
+ Y4-
= MYn-4
Kf = [MYn-4
] / [Mn+
] [Y4-
]
Usually very largeFound in a table in Chapter

11. Use to find Kf

12. Complex Formation Titrations
Typically only a small percentage of EDTA is in the form
of Y4-
and it is pH dependent
Fix pH, define a conditional formation constant (Kf’)
[Y4-
] = α [EDTA]
Kf = [MYn-4
] / [Mn+
] [Y4-
]
Kf = [MYn-4
] / [Mn+
] α [EDTA]
Fix pH
Kf
’
= α Kf = [MYn-4
] / [Mn+
] [EDTA]

13. Clicker question
If the formation constant of Ag+
with EDTA
is 2.1 x 107
and the pH is 10 (alpha = 0.35),
what is the conditional formation constant.
a. 2.1 x 107
b. 4.8 x 10-8
c. 7.4 x 106
d. 1.36 x 10-7

14. Complex Formation Titrations
EDTA titrations – four main regions
Derive a titration curve for 50.0 mL of a 0.0150 M Fe2+
with 0.030 M EDTA at a pH of 7.0. Kf = 2.1 x 1014

15. Complex Formation Titrations
EDTA titrations – four main regions
Derive a titration curve for 50.0 mL of a 0.0150 M Fe2+
with 0.030 M EDTA at a pH of 7.0. Kf = 2.1 x 1014
A. Initial pM
50.0 mL of 0.0150 M Fe
pFe = - log [Fe] = -log [0.0150] = 1.82

16. Complex Formation Titrations
Derive a titration curve for 50.0 mL of a 0.0150 M Fe2+
with 0.030 M EDTA at a pH of 7.0. Kf = 2.1 x 1014
Before equiv. Point – add 10.0 mL
Fe2+
+ EDTA  FeY-2
Initial mmol
Reacts
After
0.300
0.300
0.750
0.300 0.300
0.450 0.00 0.300
[Fe] = 0.450/60.0 mL
pFe = 2.12

17. Complex Formation Titrations
• Derive a titration curve for 50.0 mL of a 0.0150 M
Fe2+
with 0.030 M EDTA at a pH of 7.0. Kf = 2.1 x
1014
.
• At equiv. Point – add 25.0 mL
Fe2+
+ EDTA  FeY-2
Initial mmol
Reacts
After
0.750
0.750
0.750
0.750 0.750
0.00 0.00 0.750
[FeY-2
] = 0.750/75.0 mL
= 0.010 M

18. FeY2-
= Fe2+
+ EDTA
I
C
E
0.010
x x-x
0 0
0.010-x x x
1/Kf’ = 1/(αKf) = [Fe2+
] [ EDTA] / [FeY-2
]
9.92 x 10-12
= x2
/ (0.010 – x)
pFe = 6.50
α = 0.00048 at pH 7; Kf = 2.1 x 1014 (From tables in chapter)

19. Complex Formation Titrations
• Derive a titration curve for 50.0 mL of a 0.0150 M
Fe2+
with 0.030 M EDTA at a pH of 7.0
At equiv. Point – add 30.0 mL
Fe2+
+ EDTA  FeY-2
Initial mmol
Reacts
After
0.900
0.750
0.750
0.750 0.750
0.00 0.150 0.750
[FeY2-
] = 0.750/80.0 = 0.00938 M
[EDTA] = 0.150/80.0 = 0.00188 M

20. Complex Formation Titrations
Derive a titration curve for 50.0 mL of a 0.0150 M Fe2+
with 0.030 M EDTA at a pH of 7.0
D. After equivalence point  add 30.0 mL
[FeY2-
] = 0.750/80.0 = 0.00938 M
[EDTA] = 0.150/80.0 = 0.00188 M
FeY2-
= Fe2+
+ EDTA
Initial 0.00938 0 0.00188
change -x x x
Equil 0.00938-x x 0.00188 + x

21. Complex Formation Titrations
1/Kf
’
= 1/(αKf) = [Fe2+
] [ EDTA] / [FeY-2
]
9.92 x 10-12
= (x) (0.00188 + x) / (0.00938 – x)
pFe = 10.30

23. Larger Kf,
larger change at equiv pt

25. Complex Formation Titrations
EDTA titrations:
Kf’ is sensitive to pH

26. Complex Formation Titrations
EDTA titrations:
Kf’ is sensitive to pH
Change pH – titrate one metal over
Titration (Direct or Back)
pH buffered; Ensure Kf’ is large
Auxiliary complexing agent may be used
Masking agent may be used

27. Metal ion indicators
MgIn + EDTA  MgEDTA + In

29. Complex Formation Titrations
Example 1:
A 100.0 mL sample of water contains Mg2+
and Ca2+
.
Sample 1 is titrated with 15.28 mL of a 0.01016 M
EDTA in pH 10 ammonium buffer. Another 100.0
mL sample is treated with NaOH to precipitate out
the magnesium hydroxide and titrated in pH 13
buffer with 10.43 mL of the same EDTA solution.
Calculate the ppm of CaCO3 and MgCO3.

31. Calculate the pAg of a solution prepared by mixing 25.0 mL
of 0.0100 M Ag+
with 15.0 mL of 0.0200 M EDTA.
The pH of the solution is 10, alpha is 0.35, and Kf = 2.1 x 107
.
Example 2