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8/2/2019 anas atttia

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Complexometric Reactions

and TitrationsBy

Anas atttia

2nd pharmacy (2050)Alazhar


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Complexes are compounds formed from

combination of metal ions with ligands(complexing agents). A metal is anelectron deficient species while a ligand is

an electron rich, and thus, electrondonating species. A metal will thus acceptelectrons from a ligand where coordinationbonds are formed. Electrons formingcoordination bonds come solely fromligands.


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A ligand is called a monodentate if it donates asingle pair of electrons (like :NH3) while abidentate ligand (like ethylenediamine,:NH2CH2CH2H2N:) donates two pairs ofelectrons. Ethylenediaminetetraacetic acid(EDTA) is a hexadentate ligand. The ligand can

be as simple as ammonia which forms acomplex with Cu2+, for example, giving thecomplex Cu(NH3)4

2+. When the ligand is a largeorganic molecule having two or more of the

complexing groups, like EDTA, the ligand iscalled a chelating agent and the formedcomplex, in this case, is called a chelate.


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The tendency of complex formation is controlled bythe formation constant of the reaction betweenthe metal ion (Lewis acid) and the ligand (Lewisbase). As the formation constant increases, thestability of the complex increases.

Let us look at the complexation reaction of Ag+with NH3:

Ag+ + NH3D Ag(NH3)+ kf1 = [Ag(NH3)+]/[Ag+][NH3]Ag(NH3)

+

+ NH3D Ag(NH3)2+ kf2 = [Ag(NH3)2+]/[Ag(NH3)+][NH3]


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Kf1 x kf2 = [Ag(NH3)2+]/[Ag+][NH3]

2

Now look at the overall reaction:

Ag+ + 2 NH3D Ag(NH3)2+kf = [Ag(NH3)2

+]/[Ag+][NH3]2

It is clear fro inspection of the values of the kf that:

Kf = kf1 x kf2

For a multistep complexation reaction we willalways have the formation constant of theoverall reaction equals the product of all step

wise formation constants


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The formation constant is also called thestability constant and if the equilibrium is

written as a dissociation the equilibriumconstant in this case is called theinstability constant.

Ag(NH3)2+

D Ag+ + 2 NH3kinst = [Ag

+][NH3]2/[Ag(NH3)2

+]

Therefore, we have:Kinst = 1/kf


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Example

A divalent metal ion reacts with a ligand to form a 1:1complex. Find the concentration of the metal ion in asolution prepared by mixing equal volumes of 0.20

M M2+ and 0.20 M ligand (L). kf = 1.0x108.

Solution

The formation constant is very high and essentially themetal ions will almost quantitatively react with theligand.

The concentration of metal ions and ligand will be half

that given as mixing of equal volumes of the ligandand metal ion will make their concentrations half theoriginal concentrations since the volume wasdoubled.


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[M2+] = 0.10 M, [L] = 0.10 M

M2+ + L D ML2+

Kf = ( 0.10x )/x2

Assume 0.10>>x since kf is very large

1.0x108 = 0.10/x2, x = 3.2x10-5

Relative error = (3.2x10-5/0.10) x 100 = 3.2x10-2 %

The assumption is valid.

[M2+] = 3.2x10-5 M


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Example

Silver ion forms a stable 1:1 complex with trien. Calculatethe silver ion concentration at equilibrium when 25 mL of

0.010 M silver nitrate is added to 50 mL of 0.015 M trien.Kf = 5.0x10

7

Solution

Ag+ + trien D Ag(trien)+mmol Ag+ added = 25x0.01 = 0.25

mmol trien added = 50x0.015 = 0.75

The reaction occurs in a 1:1 ratiommol trien excess = 0.75 0.25 = 0.50

[Trien] = 0.5/75 M

[Ag(trien)+] = 0.25/75 M


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Kf = ( 0.25/75 x )/(x * 0.50/75 + x)Assume 0.25/75>>x since kf is very large

5.0x107 = (0.25/75)/(x* 0.50/75)

x = 1.0x10-8

Relative error = (1.0x10-8/(0.25/75)) x 100 = 3.0x10-4 %

The assumption is valid.

[Ag+] = 1.0x10-8 M


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The Chelon Effect

We have seen earlier that large multidentateligands can form complexes with metalions. These complexes are called

chelates. The question is which is morestable a chelate formed from a chelatingagent with four chelating groups or a

complex formed from the same metal withfour moles of ligand having the samedonating group?


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This can be simply answered by looking at thethermodynamics of the process. We know from

simple thermodynamics that spontaneousprocesses are favored if an increase in entropyresults. Now look at the dissociation of the chelateand the complex mentioned above, dissociation ofthe chelate will give two molecules whiledissociation of the complex will give five molecules.Therefore, dissociation of the complex results inmore disorder and thus more entropy. Thedissociation of the complex is thus more favored

and therefore the chelate is more stable as itsdissociation is not favored.


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EDTA Titrations

Ethylenediaminetetraacetic acid disodium salt(EDTA) is the most frequently used chelate incomplexometric titrations. Usually, the disodiumsalt is used due to its good solubility. EDTA is

used for titrations of divalent and polyvalentmetal ions. The stoichiometry of EDTA reactionswith metal ions is usually 1:1. Therefore,calculations involved are simple andstraightforward. Since EDTA is a polydentateligand, it is a good chelating agent and itschelates with metal ions have good stability.


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EDTA Equilibria

EDTA can be regarded as H4Y where in solution

we will have, in addition to H4Y, the followingspecies: H3Y-, H2Y

2-, HY3-, and Y4-. The amountof each species depends on the pH of thesolution where:

a4 = [Y4-]/CT where:CT = [H4Y] + [H3Y

-] + [H2Y2-] + [HY3-] + [Y4-]

a4 = ka1ka2ka3ka4/([H+]4 + ka1 [H

+]3 + ka1ka2[H+]2 +

ka1ka2ka3[H+] + ka1ka2ka3ka4)

The species Y4- is the ligand species in EDTAtitrations and thus should be looked at carefully.


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The Formation Constant

Reaction of EDTA with a metal ion to form a chelate isa simple reaction. For example, EDTA reacts withCa2+ ions to form a Ca-EDTA chelate forming the

basis for estimation of water hardness. The reactioncan be represented by the following equation:

Ca

2+

+ Y

4-

= CaY

2-

kf = 5.0x10

10

Kf = [CaY

2-]/[Ca2+][Y4-]

The formation constant is very high and the reactionbetween Ca2+ and Y4- can be considered

quantitative. Therefore, if equivalent amounts ofCa2+ and Y4- were mixed together, an equivalentamount of CaY2- will be formed.


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The question now is how to calculate the amount of Ca2+ atequilibrium?

CaY2-D Ca2+ + Y4-However, [Ca2+] # [Y4-] at this point since the amount of Y4-

is pH dependent and Y4- will disproportionate to form allthe following species, depending on the pH

CT = [H4Y] + [H3Y-] + [H2Y

2-] + [HY3-] + [Y4-]

Where, CT is the sum of all species derived from Y4- which

is equal to [Ca2+].

Therefore, the [Y4-] at equilibrium will be less than the[Ca2+] and in fact it will only be a fraction of CT where:

a4 = [Y4-]/CT

a4 = ka1ka2ka3ka4/([H+]4 + ka1 [H+]3 + ka1ka2[H+]2 + ka1ka2ka3[H+] + ka1ka2ka3ka4)


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Formation Constants for EDTA Complexes

Cation KMY Cation KMY

Ag+ 2.1 x 107 Cu2+ 6.3 x 1018Mg2+ 4.9 x 108 Zn2+ 3.2 x 1016

Ca2+ 5.0 x1010 Cd2+ 2.9 x 1016

Sr2+ 4.3 x 108 Hg2+ 6.3 x 1021Ba2+ 5.8 x 107 Pb2+ 1.1 x 1018

Mn2+ 6.2 x1013 Al3+ 1.3 x 1016

Fe2+

2.1 x1014

Fe3+

1.3 x 1025

Co2+ 2.0 x1016 V3+ 7.9 x 1025

Ni2+ 4.2 x1018 Th4+ 1.6 x 1023


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The Conditional Formation Constant

We have seen that for the reaction

Ca2+ + Y4-D CaY2- kf = 5.0x1010We can write the formation constant expression

Kf = [CaY2-]/[Ca2+][Y4-]

However, we do not know the amount of Y4- at equilibrium

but we can say that sincea

4 = [Y

4-

]/CT,

then we have:[Y4-] = a4CT

Substitution in the formation constant expression we get:

Kf = [CaY2-]/[Ca2+]a4CT or at a given pH we can write

Kf' = [CaY2-]/[Ca2+]CTWhere Kf

' is called the conditional formation constant. It isconditional since it is now dependent on pH.


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Titration Curves

In most cases, a titration is performed by addition of thetitrant (EDTA) to the metal ion solution adjusted to

appropriate pH and in presence of a suitable indicator.The break in the titration curve is dependent on:

1. The value of the formation constant.

2. The concentrations of EDTA and metal ion.

3. The pH of the solution

As for acid-base titrations, the break in the titration curveincreases as kf increases and as the concentration ofreactants is increased. The pH effect on the break of thetitration curve is such that sharper breaks are obtained athigher pH values.


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Minimum pH for effective titrations

of various metal ions with EDTA.

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