What happens to the absorbed energy?

Energy

so

s1

t1

EDTA Titrations

Outline What is EDTA?What is EDTA? Metal-Chelate ComplexesMetal-Chelate Complexes

ATP4- with Mg2+

Fe(NTA)23-

Fe(DTPA)2-

Chelate EffectChelate Effect EDTAEDTA

Acid Base Properties Y nomenclature Conditional Formation Constants

EDTA TitrationEDTA Titration

Metal-Chelate Complexes

Lewis Acid/Base ChemistryMonodentate

Multidentate and Chelates

Review:

What is a Lewis Acid? Examples?

And a Lewis Base? Examples?

Transition Metal with ligand

Central Metal ion is a Lewis Acid

Ligand – All ligands are Lewis Bases

Multidentate

Multidentate or chelating ligand attaches to a metal ion through

more than one atom is said to be multidentate, or a chelating

ligand.Examples?

ATP4- can also form complexes with metals

Complex of Iron and NTA

Fe3+ +

Fe(NTA)23-

2

Medical ApplicationsMedical Applications

The Thalassemia Story

The Chelate EffectThe Chelate Effect

Question: Describe in your own words, the “chelate

effect”.

The Chelate Effect!

Cd(H2O)62+ + 2

Cd(H2O)62+ + 4CH3NH2

H2N NH2

H2N

NH2

Cd

H2N

NH2

OH2

OH2

2+

+ 4 H2O

K = B2 = 8 x 109

H3CH2N

H3CH2N

Cd

NH2CH3

NH2CH3

OH2

OH2

2+

+ 4 H2O

K = B2 = 4 x 106

13-2 EDTA

“EDTA is by far, the most widely used chelator in analytical chemistry. By direct titration or through indirect series of reactions, virtually every element of the periodic table can be

measured with EDTA.” - Daniel Harris

Acid/Base Properties

H

H

(H6Y2+)

Acid/Base Properties

H

(H5Y+)

pKa = 0.0

Acid/Base Properties

(H4Y)

pKa = 0.0

pKa = 1.5

Acid/Base Properties

(H3Y-)

pKa = 0.0

pKa = 1.5

-pKa = 2.0

Acid/Base Properties

(H2Y-2)

pKa = 0.0

pKa = 1.5

-pKa = 2.0

- pKa = 2.7

Acid/Base Properties

(HY-3)

pKa = 0.0

pKa = 1.5

-pKa = 2.0

- pKa = 2.7

pKa = 6.16

Acid/Base Properties

(Y-4)

pKa = 0.0

pKa = 1.5

-pKa = 2.0

- pKa = 2.7

pKa = 6.16pKa = 10.24

Fraction as Y4-

The fraction of EDTA in form Y4- is given as 4-

][][ 44 EDTAY

Y (13-3)

Concentration in the form Y4-

Total Concentration of EDTA

Fraction of EDTA ion the form YFraction of EDTA ion the form Y4-4-

Fraction as Y4-

Equation 13-4 in text

654321543211

43212

3213

214

156

654321

][][][][][][4

KKKKKKKKKKKHKKKKHKKKHKKHKHH

KKKKKKY

Example

You make a solution of 0.10 M EDTA and you buffer the pH to (a) 10.0. What is Y4-

? (b) What is Y4- if the pH of the

solution is buffered to 11.0?

][][ 44 EDTAY

Y

)10.0(36.0][ 4 MY

MY pH 036.0][ 0.104

)10.0(85.0][ 4 MY

MY pH 085.0][ 0.114

EDTA reactions with Metals

Silver – Ag+

Mercury - Hg2+

Iron (III) – Fe3+

EDTA

ethylenediaminetetraacetate anion => EDTA-4 => Y-4

+1 cationAg+ + Y-4 AgY-3

EDTA

ethylenediaminetetraacetate anion => EDTA-4 => Y-4

+2 cation Hg+2 + Y-4 HgY-2

EDTA

ethylenediaminetetraacetate anion => EDTA-4 => Y-4

+3 cationFe+3 + Y-4 FeY-1

EDTA

ethylenediaminetetraacetate anion => EDTA-4 => Y-4

+n ionM+n + Y-4 MY(n-4)+

EDTA [MY(n-4)+]

KMY = -------------- [M][Y-4]

[MY(n-4)+]KMY = -------------------

[M+n] * 4 * [EDTA]

[MY(n-4)+]K'MY = KMY x 4 = -------------------

[M+n] [EDTA]

][][ 44 EDTAY

Y

Conditional formation constant!Conditional formation constant!

Example

Calculate the concentration of Ni2+ in a solution that was prepared by mixing 50.0 mL of 0.0300 M Ni2+ with 50.0 mL of 0.0500 M EDTA. The solution was buffered to pH of 3.00. Two Parts

1. Reaction2. Then equilibrium is established

EDTA Titrations

Figure 13-10 Theoretical titration curves

EXAMPLE:

Calculate the conditional constant:

Derive a curve (pCa as a function of volume of EDTA) for the titration of 50.0 mL of 0.0500 M Ca+2 with 0.1000 M EDTA in a solution buffered to a constant pH of 10.0.

pCa at Equivalence

Equivalence Volume

pCa at Pre-Equivalence Point

pCa at Post-Equivalence Point

pCa at Initial Point

Example

Derive a curve (pCa as a function of volume of EDTA) for the titration of 50.0 mL of 0.0500 M Ca+2 with 0.1000 M EDTA in a solution buffered to a constant pH of 10.0.

[CaY-2]K'CaY = KCaY * 4 = ----------------

[Ca+2] * [EDTA]

where Y4- = 0.36 at pH = 10.0

K'CaY = KCaY * 4 = 4.9 x 1010 * 0.36 = 1.8 x 1010

KCaY = 4.9 x 1010

EXAMPLE:

Derive a curve (pCa as a function of volume of EDTA) for the titration of 50.0 mL of 0.0500 M Ca+2 with 0.1000 M EDTA in a solution buffered to a constant pH of 10.0.Equivalence Volume 1 Mole of EDTA = 1 Mole of MetalM1V1 = M2V2 (Careful of Stoichiometry)

50.0 mL (0.0500 M) = 0.1000 M (V2)V2 = 25.0 mL

EXAMPLE:

K'CaY = 1.8 x 1010

0.00 mL EDTA added

pCa = - log[Ca+2]

Initial Point

= - log(0.00500 M) = 2.301

Derive a curve (pCa as a function of volume of EDTA) for the titration of 50.0 mL of 0.0500 M Ca+2 with 0.1000 M EDTA in a solution buffered to a constant pH of 10.0.

EXAMPLE:

Derive a curve (pCa as a function of volume of EDTA) for the titration of 50.0 mL of 0.0500 M Ca+2 with 0.1000 M EDTA in a solution buffered to a constant pH of 10.0.At 25.0 mL (Equivalence Point)At 25.0 mL (Equivalence Point)

Ca2+ + Y4- -> CaY2-

Before 0.0025 moles 0.0025 moles

-

After - - 0.0025 moles

What can contribute to Ca2+ “after” reaction?

EXAMPLE:

Derive a curve (pCa as a function of volume of EDTA) for the titration of 50.0 mL of 0.0500 M Ca+2 with 0.1000 M EDTA in a solution buffered to a constant pH of 10.0.

Ca2+ + Y4- CaY2-

I - - 0.0025 moles/V

C +x +x -x

E +x + x 0.0333 –x

]][[

][2

2'

CaEDTA

CaYK CaY

2' 0333.0

x

xK CaY

X = [Ca2+] = 1.4 x10-6

pX = p[Ca2+] = 5.866

0.0025moles/0.075 L0.0025moles/0.075 L

Pre-Equivalence Point

Let’s try 15 mL

EXAMPLE:

Derive a curve (pCa as a function of volume of EDTA) for the titration of 50.0 mL of 0.0500 M Ca+2 with 0.1000 M EDTA in a solution buffered to a constant pH of 10.0.At 15.0 mL Ca2+ + Y4- -> CaY2-

Before 0.0025 moles 0.0015 moles

-

After 0.0010 moles - 0.0015 moles

What can contribute to Ca2+ after reaction?

K’CaY = 1.8 x 1010

negligiblenegligible

EXAMPLE:

Derive a curve (pCa as a function of volume of EDTA) for the titration of 50.0 mL of 0.0500 M Ca+2 with 0.1000 M EDTA in a solution buffered to a constant pH of 10.0.At 15.0 mL

[Ca2+] = 0.0010 moles/0.065 L[Ca2+] = 0.015384 Mp [Ca2+] = 1.812

Post Equivalence Point

Let’s Try 28 ml

EXAMPLE:

Derive a curve (pCa as a function of volume of EDTA) for the titration of 50.0 mL of 0.0500 M Ca+2 with 0.1000 M EDTA in a solution buffered to a constant pH of 10.0.At 28.0 mL Ca2+ + Y4- -> CaY2-

Before 0.0025 moles 0.0028 moles

-

After - 0.0003 moles

0.0025 moles

What can contribute to Ca2+ after titration?

EXAMPLE:

Derive a curve (pCa as a function of volume of EDTA) for the titration of 50.0 mL of 0.0500 M Ca+2 with 0.1000 M EDTA in a solution buffered to a constant pH of 10.0.

Ca2+ + Y4- CaY2-

I - 0.0003 moles/V 0.0025 moles/V

C +x +x -x

E +x 0.003846 + x 0.03205 –x

]][[

][2

2'

CaEDTA

CaYK CaY

))(003846.0(

03205.0'

xx

xK CaY

X = [Ca2+] = 4.6 x10-10

pX = p[Ca2+] = 9.334

0.078 L

Experimental Considerations

EDTA Titration Techniques

Erichrome Black T

MgIn + EDTA MgEDTA + In

(red) (colorless) (blue)

Figure 13-13 Guide to EDTA titrations, light color, pH range for quantitative analysis, dark area where ammonia must be present