edta

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)42+

Donor species: Ligandhas at least one pair of unshared e-

Examples: NH3, H2O, Cl, CN

Complex Formation Titrations

Ligand that has single donor group: monodentate

Ligand that has two donor groups: bidentate

Example of bidentate,

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

H4Y, H3Y-, H2Y-2, HY-3, Y4-

Dissociation Products

Fraction of EDTA in each of its protonated forms is a function of pH

Y4- = [Y4-] / [EDTA]

Low pH (3-6), H2Y-2 predominatesHigh pH (>10), Y4- predominates

Use to find at different pHs

Chapter 11

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

Use to find Kf

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-] = [EDTA]

Kf = [MYn-4] / [Mn+] [Y4-]

Kf = [MYn-4] / [Mn+ ] [EDTA]

Fix pH

Kf’ = Kf = [MYn-4] / [Mn+] [EDTA]

Clicker question

If the formation constant of Ag+ with EDTAis 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

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

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

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

0.3000.3000.750

0.300 0.3000.450 0.00 0.300

[Fe] = 0.450/60.0 mLpFe = 2.12

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 mLFe2+ + EDTA FeY-2

Initial mmol Reacts After

0.7500.7500.750

0.750 0.7500.00 0.00 0.750

[FeY-2] = 0.750/75.0 mL= 0.010 M

FeY2- = Fe2+ + EDTA

I

C

E

0.010

x x-x

0 0

0.010-x x x

1/Kf’ = 1/(Kf) = [Fe2+] [ EDTA] / [FeY-2]

9.92 x 10-12 = x2 / (0.010 – x)

pFe = 6.50

= 0.00048 at pH 7; Kf = 2.1 x 1014 (From tables in chapter)

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

0.9000.7500.750

0.750 0.7500.00 0.150 0.750

[FeY2-] = 0.750/80.0 = 0.00938 M

[EDTA] = 0.150/80.0 = 0.00188 M

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

Complex Formation Titrations

1/Kf’ = 1/(Kf) = [Fe2+] [ EDTA] / [FeY-2]

9.92 x 10-12 = (x) (0.00188 + x) / (0.00938 – x)

pFe = 10.30

Larger Kf, larger change at equiv pt

Complex Formation Titrations

EDTA titrations:

Kf’ is sensitive to pH

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

Metal ion indicators

MgIn + EDTA MgEDTA + In

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.

Calculate the pAg of a solution prepared by mixing 25.0 mLof 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