Chapter 12 EDTA Titrations - California State University ...

Chapter 12 EDTA Titrations

Acids and Bases

? Definitions: Lewis ? Electrons (acid: electron pair acceptor); Br?nsted-Lowry (acid: proton donor)

Lewis acid-base concept in Metal-Chelate Complexes

Metal ions (electron pair acceptor) Lewis acid Ligand (electron pair donor) Lewis base

Coordination Number

? The atom of the ligand that supplies the nonbonding electrons for the metal-ligand bond is the donor atom.

? The number of these atoms is the coordination number.

Complex Formation

Formation of coordinate bonds between Lewis Acids/Bases

Ag + (aq ) + NH 3 (aq) K1 Ag ( NH 3 ) + (aq ) Ag ( NH 3 ) + ( aq ) + NH 3 (aq) K2 Ag ( NH 3 ) 2 + ( aq )

K = [ Ag (NH3 ) + ] 1 [ Ag + ][NH ]

3

K 2

=

[ Ag(NH ) + ] 32

[ Ag(NH ) + ][NH

]

3

3

Formation constants (Kf) are the equilibrium constants for complex ion

formation. The overall, or cumulative, formation constants are

denoted i

Ag+ (aq) + 2NH (aq) Kf Ag(NH ) + (aq)

3

32

K

=

[

Ag ( NH 3

)+ 2

]

=

=K K

f [ Ag + ][NH ]2

2

1

2

3

Geometries

There are two common geometries for metals with a coordination number of four:

Tetrahedral & Square planar

By far the mostencountered geometry, when the coordination number is six, is octahedral.

Ligands

? Monodentate ligand: binds to a metal ion through only one atom, e.g., CN-

? Multidentate ligand or chelating ligand: has more than one ligand donor atoms.

? In ethylenediamine, NH2CH2CH2NH2 (i.e., en), each N is a donor atom. en is bidentate.

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Chelating Effect

A multidentate ligand to form more stable metal complexes than those formed by similar monodentate ligand

Chelating Agents

? Porphyrins (tetradentate ligands, in heme and chlorophyll)

? Adenosine triphosphate (ATP)

EDTA

Ethylenediaminetetraacetic acid (H4EDTA or H4Y) Ethylenediaminetetraacetate anion (EDTA-4 or Y-4)

Ethylenediaminetetraacetate, mercifully abbreviated EDTA, has six donor atoms. EDTA is a primary standard material.

Some Metals Form 7 or 8 Coordinate Complexes

EDTA Complexes

? EDTA forms 1:1 complexes with most metals (Not with Group 1A metals)

? EDTA complexes are usually stable water soluble complexes with high formation constants

? Formation constant, Kf, (or stability constant):

Mn+ + Y4- MYn-4

[ [ ][ ] ] Kf

=

MY n - 4 Mn+ Y4-

? Kf could have been defined for any form of EDTA

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pH affects EDTA titration

Acid-Base Properties of EDTA:

HO2CH2C

CH2CO2H

HNCH2CH2NH

HO2CH2C

CH2CO2H

pK1 = 0.0 pK5 = 6.16

H6Y2+

pK2 = 1.5 pK3 = 2.0 pK6 = 10.24

EDTA is a hexaprotic system (H6Y2+) with 4 carboxylic acids and 2 ammoniums

pK4 = 2.66

Mn+ + Y4- MYn-4 pH:

Hn Yn-4

Fraction of EDTA in the form Y4-

[ ] [ ] [[ ] ] [ ] [ ] [ ] [ ] Y4-

=

H 6Y 2+

+

H 5Y +

+

H 4Y

+

Y 4- H 3Y -

+

H 2Y 2-

+

HY 3-

+ Y 4-

Y 4-

= [Y 4- ] [EDTA]

=

[Y 4- ] CEDTA

{[ ] [ ] [ ] [ ] [ ] [ ] } == Y44-Y-

K1K2K3K4K5K6 H+ 6 + H+ 5 K1 + H+ 4 K1K2 + H+ 3 K1K2K3 + H+ 2 K1K2K3K4 + H+ K1K2K3K4K5 + K1K2K3K4K5K6

pH Dependence of Y4-

Fractional Composition Diagram for EDTA

Conditional Formation Constant

M n+ + Y 4- MY n-4

Kf

=

[MY n-4 ] [M n+ ][Y 4-

]

[ ] [ ] [ ] = Y4-

Y 4EDTA

[ ] Y4- = Y4- EDTA

[ [ ][ ] ] [ ][ ][ ] Kf =

MYn-4 Mn+ Y4-

=

MY n-4

M n+ Y 4-

EDTA

Conditional formation constant:

[ [ ] ] [ ] K

' f

= Y4- Kf

=

MY n-4 Mn+ EDTA

Fixing the pH by buffering, then Y 4- is a constant.

At any fixed pH, find Y4- and evaluate Kf'

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Example: Calculate the concentration of free Ca2+ in a

solution of 0.10 M CaY2- at pH 10 and pH 6. Kf for CaY2- is 1010.65 (Table 12-2)

Ca 2+ + EDTA CaY2-

K

' f

= Y4- Kf

at

pH

= 10.00, K ' f

= K Y4-

f

= (0.30)(1010.65 ) = 1.3?1010

at

pH

=

6.00,

K

' f

= K Y4-

f

= (1.8 ?10-5 )(1010.65 ) = 8.0 ?105

Ca 2+ + EDTA CaY2-

Conci Concf

0

0

x

x

0.1 0.1- x

[ [ ] ] K

' f

=

CaY 2-

Ca2+ [EDTA]

=

0.1 - x2

x

[ ] x = Ca2+ = 2.7 ?10-6 M at pH = 10

= 3.5?10-4 M at pH = 6

EDTA Titration Curve

1. Excess Mn+ left after each addition of EDTA. Conc. of free metal equal to conc. of unreacted Mn+.

2. Equivalence point: [Mn+] = [EDTA] Some free Mn+ generated by MYn-4 Mn+ + EDTA

3. Excess EDTA. Virtually all metal in MYn-4 form.

EDTA Titration Curve

1. Titration reaction: Mn+ + EDTA MYn-4 K'f = Y4- Kf

2. Reaction completes at each point in the titration if Kf' is large.

3. Plot pM (= -log[Mn+]) vs. volume of EDTA added

EDTA Titration Curve

EXAMPLE: Derive a titration curve for the titration of 50.0 mL of 0.040 M Ca2+ (buffered at pH=10) with 5.00, 25.00, and

26.00 mL of 0.080 M EDTA.

Ca2+ + EDTA CaY 2-

K

' f

= Y 4- Kf

= 0.30 *1010.65

= 1.3?1010

At equivalence point, Vol. of EDTA = 25.0 mL

5.00 ml, 25.00 ml, 26.00 ml,

Before the equivalence point At the equivalence point After the equivalence point

EDTA Titration Curve

EXAMPLE: Derive a titration curve for the titration of 50.0 mL of 0.040 M Ca2+ (buffered at pH=10) with 5.00, 25.00, and 26.00 mL of 0.080 M EDTA.

At equivalence point, Vol. of EDTA = 25.0 mL 5mL before the equivalence point

[ ] Ca2+ = 25.0 - 5.0 (0.040) 50.0 = 0.0291M

25.0

5.00 + 50.0

Fraction Remaining

pCa2+ = - log(0.0291) = 1.54

EDTA Titration Curve

EXAMPLE: Derive a titration curve for the titration of 50.0 mL of 0.040 M Ca2+ (buffered at pH=10) with 5.00, 25.00, and

26.00 mL of 0.080 M EDTA.

At equivalence point, Vol. of EDTA = 25.0 mL almost all the metal is in the form, CaY2-

Conci Concf

[ ] CaY2- = (0.040) 50.0

= 0.0267 M

25.00 + 50.0

Ca2+ + EDTA CaY2-

0

0

0.0267

x

x

0.0267 - x

[ [ ] ] K

' f

=

Ca

CaY2-

2+ [EDTA]

=

0.0267 - x x2

= 1.8?1010

x = 1.2?10-6 M

pCa 2+ = - log(1.2?10-6 ) = 5.91

4

EDTA Titration Curve

EXAMPLE: Derive a titration curve for the titration of 50.0 mL of 0.040 M Ca2+ (buffered at pH=10) with 5.00, 25.00, and

26.00 mL of 0.080 M EDTA.

At equivalence point, Vol. of EDTA = 25.0 mL 26.00 mL 1.0 mL excess EDTA, after the equivalence point

[EDTA] = (0.080)

1.00

= 1.05?10-3 M

50.0 + 26.00

[ ] CaY2- = (0.040)

50.0

= 2.63?10-2 M

50.0 + 26.00

[ [ ] ] [ ] [ ] K

' f

=

CaY2- Ca 2+ EDTA

=

2.63?10-2 Ca 2+ (1.05?10-3 )

= 1.8?1010

Ca 2+ = 1.4 ?10-9 M pCa2+ = 8.86

pH affects the titration of Ca2+ with EDTA

Auxiliary Complexing Agents

? A ligand that binds strongly enough to the metal to prevent hydroxide precipitation, but weak enough to be displaced by EDTA (e.g., ammonia, tartrate, citrate, or trithanolamine)

Ammonia is a common auxiliary complex for transition metals

like zinc (p. 239)

K = K ''

f

Zn 2+ Y 4- f

Kf'' is the effective formation constant at a fixed concentration of auxiliary complexing agent.

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Metal Ion Indicators

? To detect the end point of EDTA titrations, we usually use a metal ion indicator or an ion-selective electrode (Ch. 15)

? Metal ion indicators change color when the metal ion is

bound to EDTA:

MgEbT + EDTA MgEDTA + EbT

(Red)

(Colorless) (Blue)

? Eriochrome black T is an organic ion

? The indicator must bind less strongly than EDTA

Metal Ion Indicators

EDTA Titration Techniques

? Direct titration: analyte is titrated with standard EDTA with solution buffered at a pH where Kf' is large

? Back titration: known excess of EDTA is added to analyte. Excess EDTA is titrated with 2nd metal ion.

Example: 25.0 mL of an unknown Ni2+ solution was treated with 25.00 mL of 0.05283 M Na2EDTA. The pH of the solution was buffered to 5.5 and than back-titrated with 17.61 mL of 0.02299 M Zn2+. What was the unknown Ni2+ in M?

mol EDTA = (25.00 mL)(0.05283 M) = 1.32 mmol EDTA

Zn 2+ + Y4- ZnY2mol Zn2+ = (17.61mL)(0.02299 M) = 0.4049 mmol Zn2+

Ni2+ + Y4- NiY2mol Ni2+ = 1.321mmol EDTA - 0.4049 mmol Zn2+ = 0.916 mmol

M Ni2+ = (0.916 mmol)/(25.00 mL) = 0.0366 M

EDTA Titration Techniques

? Direct titration: analyte is titrated with standard EDTA with solution buffered at a pH where Kf' is large

? Back titration: known excess of EDTA is added to analyte. Excess EDTA is titrated with 2nd metal ion.

? Displacement titration: For metals without a good indicator ion, the analyte can be treated with excess Mg(EDTA)2-. The analyte displaces Mg, and than Mg can be titrated with standard EDTA

? Indirect titration: Anions can be analyzed by precipitation with excess metal ion and then titration of the metal in the dissolved precipitate with EDTA.

? Masking agent: protects some componet of the analyte from reaction with EDTA (render metal ions inactive without actually removing them from solution). Demasking: releasing metal ion from a masking agent.

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