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


Ligand that has single donor group: monodentate

Ligand that has two donor groups: bidentate

Example of bidentate,


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


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


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


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


EDTA titrations – four main regions


A. Initial pM

50.0 mL of 0.0150 M Fe

pFe = - log [Fe] = -log [0.0150] = 1.82



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


• 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


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)


• 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


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


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


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


EDTA titrations:

Kf’ is sensitive to pH


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


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

edta

Technology

edta kf

edta fix ph kf

edta fey

edta initial

metal edta complex

change x x x equil

edta solution

edta mgedta