EDTA TitrationEDTA Titration

EDTA = Ethylenediaminetetraacetic acid

Lewis acid : electron pair acceptor eg metal ions

Lewis base : electron pair donor eg ligandsA ligand that attaches to a metal ion through more than one ligand atom is called a chelating ligand or multidentate ligand

Consider :

Ag+ + NH3 Ag(NH3)+

Ag(NH3)+ + NH3 Ag(NH3)2+

Kf1 =

Kf2 =

[Ag(NH3)+][Ag+][NH3]

[Ag(NH3)2+]

[Ag(NH3)+][NH3]

A titration based on complex formation is called complexometic titration

The equilibrium constant for the reaction of a metal with a ligand is called the formation constant, Kf , or the stability constant

The overall reaction :

Ag+ + 2NH3 = Ag(NH3)2+

and the overall formation constant :

Kf =Kf1 Kf2 =[Ag(NH3)2

+][Ag+][NH3]2

Example :

A divalent metal M2+ reacts with a ligand L to form a 1:1 complex :

M2+ + L ML2+

Calculate the concentration of M2+ in a solution prepared by mixing equal volumes of 0.20 M M2+ and 0.20 M L. Given Kf = 1.0 x 108

Given Kf = 1.0 x 108 complex is

sufficiently strong such that the reaction is virtually complete

Since equal volumes were added initial concentration is halved

Let x = residual concentration of M2+

M2+ + L ML2+

x x 0.1 – x

Kf = = 1.0 x 108

x = 3.2 x 10-5 M

x2

0.1 - x

Similarly, if L is a multidentate ligand and

M + nL MLn

then

Kf = [MLn]

[M][L]n

EDTA ComplexesEDTA Complexes

EDTA is a hexaprotic system – H6Y2+

Neutral acid is tetraprotic – H4Y

It can be represented as having four Ka values :

H4Y H+ + H3Y- Ka1 = 1.0 x 10-2

H3Y- H+ + H2Y2- Ka2 = 2.2 x 10-3

H2Y2- H+ + HY3- Ka3 = 6.9 x 10-7

HY2- H+ + Y4- Ka4 = 5.5 x 10-11

Fraction of the total EDTA species that exists as Y4- = 4 = [Y4-]/CH Y

where

CH Y = [Y4-] + [HY3-] + [H2Y2-] + [H3Y-] +

[H4Y]

4

4

Hence in complexing with -

+1 cation :

Ag+ + Y4- = AgY3- Kf =

+2 cation :

Hg2+ + Y4- = HgY2- Kf =

+3 cation :

Fe3+ + Y4- = FeY- Kf =

+n cation :

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

[Mn+] [Y4-]

[HgY2-]

[Hg2+][Y4-]

[FeY-][Fe3+][Y4-]

[AgY3-][Ag+][Y4-]

Substituting [Y4- ] = 4CH Y

Kf =

4Kf = = Kf’

4

[MYn-4][Mn+] 4CH Y

4

[MYn-4][Mn+]CH Y

4

Conditional formation constant

– holds only for a particular pH

– describes the formation of MYn-4 at any particular pH

– allows us to look at EDTA complex formation as if the uncomplexed EDTA were all in one form

Example :

Consider the reaction: Fe3+ + EDTA FeY- (Kf = 1.3 x 1025; Y (at pH1) =

1.9 x 10-18 and Y (at pH4) = 3.8 x 10-9).

Calculate the concentration of free Fe3+ in a solution of 0.10 M FeY- at pH4 and at pH1.

4-

4-

Using Kf’ = 4Kf ,

at pH = 4 : Kf’ = (3.8 x 10-9)(1.3 x 1025)

= 4.9 x 1016

at pH = 1 : Kf’ = (1.9 x 10-18)(1.3 x 1025)

= 2.5 x 107

Let x = [Fe3+] = [EDTA], thus

= = 4.9 x 1016 (at pH4)

x = 1.4 x 10-9 M

Similarly, at pH1: x = 6.4 x 10-5 M

pH affects the stability of the complex.

The Kf’ values show that the metal-EDTA

complex becomes less stable at lower pH

[FeY-][Fe3+][EDTA]

0.1 - xx2

For a titration reaction to be effective, the equilibrium constant must be large the analyte and titrant should be completely reacted at the equivalence point

Titration of Ca2+ with EDTA as a function of pH :

EDTA Titration EDTA Titration CurvesCurves

Titration is carried out by adding the chelating agent to the sample :

Mn+ + EDTAMYn-4 Kf’ = 4Kf

The titration curve is a graph of pM (= -log[M]) versus the volume of added EDTA

Titration curve consists of 3 regions-

Before the equivalence point :

There is excess Mn+

in the solution after the EDTA has been consumed.

Concentration of free metal ion = concentration of excess unreacted Mn+

At the equivalence point :

[Mn+] = [EDTA]

Free Mn+ is from the dissociation of MYn-4 :

MYn-4 Mn+ + EDTA

After the equivalence point :

There is excess EDTA and virtually all the metal ion is in the MYn-4 form

Example :

Calculate the shape of the titration curve for the reaction of 50.0 ml of 0.0500 M Mg2+ (buffered to pH10.0) with 0.0500 M EDTA (Kf = 6.2 x 108 ; at pH10.0 : Y =

0.36)Mg2+ + EDTA MgY2-

Using Kf’ = 4Kf = (0.36)(6.2 x 108)

= 2.2 x 108

Since Kf’ is large the reaction goes to completion with each addition of titrant

Before equivalence point :

Consider the addition of 5.0 ml EDTA to the solution

Moles of EDTA added = (0.005 l)(0.0500 M)

= 2.5 x 10-4

Moles of Mg2+ present initially= (0.050 l)(0.050M)

= 2.5 x 10-3

Moles of Mg2+ present after the addition of EDTA =

(2.5 x 10-3 ) – (2.5 x 10-4) = 0.00225

[Mg2+] = = 0.0409 M

pMg2+ = -log [Mg2+] =1.39

At equivalence point : [Mn+] = [EDTA]

Volume of EDTA added = (2.5 x 10-3)/(0.0500M)

= 50.0 ml

0.002250.055

0.0025 mol0.100 l

[MgY2-][Mg2+][EDTA]

0.025 - xx2

Since there is negligible dissociation,

[MgY2-] = = 0.025 M

From:

Mg2+ + EDTA MgY2-

Let x =[Mg2+] = [EDTA], then

Kf’ = =

= 2.2 x 108

x = 1.07 x 10-5 M

pMg2+ = -log [Mg2+] =4.97

After equivalence point :

If 51.0 ml of EDTA is added there will be 1.0 ml excess EDTA in the solution

[EDTA] ={(0.001)(0.05M)}/(0.101 l)

=4.95 x 10-4 M

[MgY2-] =(2.5 x 10-3)/(0.101 l)

=2.48 x 10-2 M

Using

Kf’ = =

= 2.2 x 108

[Mg2+] = 2.3 x 10-7 M

p[Mg2+] = -log [Mg2+]

= 6.64

[MgY2-]

[Mg2+][EDTA]2.48 x 10-2

[Mg2+](4.95 x 10-4)

Metal Ion IndicatorsMetal Ion Indicators

Methods to detect the end point in EDTA titrations are :

- metal ion indicators

A metal ion indicator is a compound whose color changes when it binds to a metal ion (eg Eriochrome black T)

This compound must bind metal less strongly than EDTA

MgIn + EDTA MgEDTA + In

(red) (colorless) (colorless) (blue)

- mercury electrode : measurement to potential

- glass pH electrode

- ion-selective electrode

EDTA Titration EDTA Titration TechniquesTechniques

(i) Direct Titration

- analyte is titrated with standard EDTA

- analyte is buffered to an appropriate pH at which the conditional formation constant for the metal-EDTA complex is large and the color of the free indicator is distinctly different from that of the metal-indicator complex

(ii) Back titration

- a known excess of EDTA is added to the analyte

- the excess EDTA is titrated with a standard solution of a second metal

- this method is useful if

- analyte precipitates in the absence of EDTA

- the analyte reacts too slowly with EDTA under titration conditions

- analyte blocks the indicator

- metal used in back titration must not displace the analyte metal ion from its EDTA complex

(iii) Displacement titration

Method is useful is the metal ions do not have a satisfactory indicator

- analyte is treated with excess Mg(EDTA)2- to displace Mg2+

Mn+ + MgY2- MYn-4 + Mg2+

- the displaced Mg2+ is titrated with standard EDTA

Example :

2Ag+ + Ni(CN)42- 2Ag(CN)2

- + Ni2+

(iv) Indirect titration

Anions (such as SO42-, CrO4

2-, CO32- and S2-)

that precipitate with certain metal ions can be analyzed with EDTA through indirect titration

For example, SO42 can be precipitated

with excess Ba2+. The BaSO4(s) is washed and boiled with excess EDTA at pH10 to bring Ba2+ back into solution as Ba(EDTA)2-. The excess EDTA is back titrated with Mg2+