Lecture 578 Complexation

7/28/2019 Lecture 578 Complexation

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COORDINATIONCHEMISTRY

(COMPLEXATION)


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WHY IS CHEMICAL SPECIATION SO

IMPORTANT? The biological availability (bioavailability) of

metals and their physiological and toxicological

effects depend on the actual species present. Example: CuCO3

0, Cu(en)20, and Cu2+ all affect the

growth of algae differently

Example: Methylmercury (CH3Hg+) is readily formed

in biological processes, kinetically inert, and readilypasses through cell walls. It is far more toxic thaninorganic forms.

Solubility and mobility depend on speciation.


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Figure 6-20 from Stumm & Morgan: Effect of free Cu2+ ongrowth of algae in seawater.


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DEFINITIONS - I

Coordination (complex formation) - any combination of

cations with molecules or anions containing free pairsof electrons. Bonding may be electrostatic, covalent or amix.

Central atom (nucleus) - the metal cation.

Ligand- anion or molecule with which a cation formscomplexes.

Multidentate ligand- a ligand with more than one possible

binding site.Chelation - complex formation with multidentate ligands.

Multi- or poly-nuclear complexes - complexes with more

than one central atom or nucleus.


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N

N

HO

HO

MULTIDENTATE LIGANDS

Oxalate (bidentate)

Ethylendiamine (bidentate) Ethylendiaminetetraacetic acidor EDTA (hexadentate)


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Chelation

Polynuclear complexes

Sb2S42- Hg3(OH)4

2+


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DEFINITIONS - II

Species - refers to the actual form in which a molecule orion is present in solution.

Coordination number- total number of ligands

surrounding a metal ion.Ligation number- number of a specific type of ligand

surrounding a metal ion.

Colloid- suspension of particles composed of severalunits, whereas in true solution we have hydration of asingle molecule, atom or ion.


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FORMS OF OCCURRENCE OFMETAL SPECIES

Free metal

ion

Inorganic

ion pairsand

complexes

Organic

complexes,chelates

Metals

bound tohigh mol.

wt.species

Highly

dispersedcolloids

Metals

sorbed oncolloids

Cu2+ Cu2(OH)22+ Me-SR Me-lipids FeOOH Mex(OH)y

Fe3+

PbCO30

Me-OOCR Me-humicacid

Fe(OH)3 MeCO3,MeS,etc.on clays

Pb2+

CuCO30

Mn(IV)

oxides

FeOOH or

Mn(IV) onoxidesNa

+AgSH

0Ag2S

Al3+

CdCl+

Zn2+

CoOH+

1000 100 10


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Coordination numbers

CN = 2 (linear)

CN = 4 (tetrahedral) CN = 6 (octahedral)

CN = 4 (square planar)

Coordination numbers 2, 4, 6, 8, 9 and 12 are most commonfor cations.


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STABILITY CONSTANTS MEASURE THESTRENGTH OF COMPLEXATION

Stepwise constants

MLn-1 + L MLn

Cumulative constants

M + nL MLn

]][[

][

1 LML

MLK

n

nn

n

nn

LM

ML

]][[

][

n = K1K2K3Kn


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For a protonated ligand we have:

Stepwise complexation

MLn-1 + HL MLn + H+

Cumulative complexation

M + nHL MLn + nH+

]][[]][[

1

*

HLMLHMLK

n

nn

n

n

n

n HLM

HML

]][[

]][[*

The larger the value of the stability constant, the more

stable the complex, and the greater the proportion ofthe complex formed relative to the simple ion.


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STABILITY CONSTANTS FOR

POLYNUCLEAR COMPLEXESmM + nL MmLn

mM + nHL MmLn + nH+

nm

nmnm

LM

LM

][][

][

nm

n

nmnm

HLM

HLM

][][

]][[*

If m = 1, the second subscript on nm is omitted andthe expression simplifies to the previous expressionsfor mononuclear complexes.


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METAL-ION TITRATIONS

Metal ions can be titrated by ligands in thesame way that acids and bases can be

titrated. According to the Lewis definition, metal

ions are acids because they accept

electrons; ligands are bases because theydonate electrons.


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Figure 6-3 from Stumm & Morgan: Titration of H+ and Cu2+ withammonia and tetramine (trien)


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HYDROLYSISThe waters surrounding a cation may function as acids. The

acidity is expected to increase with decreasing ionicradius and increasing ionic charge. For example:

Zn(H2O)62+ Zn(H2O)5(OH)

+ + H+

Hydrolysis products may range from cationic to anionic.For example:

Zn2+ ZnOH+ Zn(OH)20 (ZnO0)

Zn(OH)3

- (HZnO2

-) Zn(OH)4

2- (ZnO2

2-)

May also get polynuclear species.

Kinetics of formation of mononuclear hydrolysis products

is rather fast, polynuclear formation may be slow.


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GENERAL RULES OF HYDROLYSIS

The tendency for a metal ion to hydrolyze willincrease with dilution and increasing pH(decreasing [H+])

The fraction of polynuclear products will decrease

on dilution Compare

Cu2+ + H2O CuOH+ + H+ log *K1 = -8.0

Mg2+ + H2O MgOH+ + H+ log *K1 = -11.4

][

]][[21

*

M

HMOHK


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At infinite dilution, pH 7 so

CuOH+ = (1 + 10-7/10-8)-1 = 1/11 = 0.091

MgOH+ = (1 + 10-7/10-11.4)-1 = 1/25119 = 4x10-5

Only salts with p*K1 < (1/2)pKw or p*n < (n/2)pKw

will undergo significant hydrolysis upon dilution.Progressive hydrolysis is the reason some salts

precipitate upon dilution. This is why it is necessary

to add acid when diluting standards.

1

*

2 ]][[][

][

][][

][

K

HMOHMOH

MOH

MMOH

MOHMOH

1

*

][1

1

K

HMOH


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POLYNUCLEAR SPECIES DECREASE INIMPORTANCE WITH DILUTION

Consider the dimerization of CuOH+:

2CuOH+ Cu2(OH)22+ log *K22 = 1.5

Assuming we have a system where:

CuT = [Cu2+] + [Cu(OH)+] + 2[Cu2(OH)2

2+]

we can write:

22*22

22

2

2

222

2

22

]))([2][(

])([

][

])([KOHCuCuCu

OHCu

CuOH

OHCu

T

So [Cu2(OH)22+] is clearly dependent on total Cu

concentration!


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HYDROLYSIS OF IRON(III)

Example 1: Compute the equilibrium composition of ahomogeneous solution to which 10-4 (10-2) M ofiron(III) has been added and the pH adjusted in therange 1 to 4.5 with acid or base.

The following equilibrium constants are available at I =3 M (NaClO4) and 25C:

Fe3+ + H2O FeOH2+ + H+ log *K1 = -3.05

Fe3+

+ 2H2O Fe(OH)2+

+ 2H+

log*

2 = -6.312Fe3+ + 2H2O Fe2(OH)2

4+ + 2H+ log *22 = -2.91

FeT = [Fe3+] + [FeOH2+] + [Fe(OH)2

+] + 2[Fe2(OH)24+]


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Now we define: 0 = [Fe3+]/FeT; 1= [FeOH

2+]/FeT;2= [Fe(OH)2

+]/FeT; and 22= 2[Fe2(OH)24+]/FeT.

2

22

*3

2

2

*

1

*3

][

][2

][][

1][

H

Fe

HH

KFeFeT

1

2

22

*

0

2

2

*

1

*

0][

2

][][1

H

Fe

HH

K T

01][][

1][

22

2

*

1

*

02

22

*2

0

HH

KHFeT


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These equations can then be employed to calculatethe speciation diagrams on the next slide.

2

22

*2

022

][

2

H

FeT

][

1

*

01

H

K

2

22

*

02

][

H

This last equation can be solved for0 at given valuesof FeT and pH. The remaining values are obtained

from the following equations:


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FeT

= 10-4

M

%Fe

0

20

40

60

80

100

FeT

= 10-2

M

pH

1 2 3 4

%Fe

0

20

40

60

80

100

Fe3+

FeOH2+

Fe(OH)2

+

Fe2(OH)

2

4+

Fe3+

FeOH2+

Fe(OH)2

+


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Example 2: Compute the composition of a Fe(III)solution in equilibrium with amorphous ferrichydroxide given the additional equilibriumconstants:

Fe(OH)3(s) + 3H+ Fe3+ + 3H2O log

*Ks0 = 3.96

Fe(OH)3(s) + H2O Fe(OH)4- + H+ log *Ks4 = -18.7

Fe3+

log [Fe3+] = log *Ks0 - 3pH

Fe(OH)4-

log [Fe(OH)4-] = log *Ks4 + pH


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FeOH+


*Ks0 = 3.96

Fe3+

+ H2O FeOH2+

+ H+

log*

K1 = -3.05Fe(OH)3(s) + 2H+ FeOH2+ + 2H2O log

*Ks1 = 0.91

log [FeOH2+] = log *Ks1 - 2pH

Fe(OH)2+


*Ks0 = 3.96

Fe3+ + 2H2O Fe(OH)2+ + 2H+ log *2 = -6.31Fe(OH)3(s) + H

+ Fe(OH)2+ + H2O log

*Ks2 = -2.35

log [Fe(OH)2

+] = log *Ks2

- pH


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Fe2(OH)24+

2Fe(OH)3

(s) + 6H+ 2Fe3+ + 6H2

O 2log *Ks0

= 7.92

2Fe3+ + 2H2O Fe2(OH)24+ + 2H+ log *22 = -2.91

2Fe(OH)3(s) + 4H+ Fe2(OH)2

4+ + 4H2O

log *Ks22 = 5.01

log [Fe2(OH)24+] = log *Ks22 - 4pH

These equations can be used to obtain the concentration ofeach of the Fe(III) species as a function of pH. They canall be summed to give the total solubility.


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pH

0 2 4 6 8 10 12 14

log

conce

ntration

-15

-10

-5

0

5

Fe(OH)3(s)

Fe(OH)4

-

Fe(OH)2

+

FeOH2+

Fe2(OH)

2

4+

Fe3+

Figure 6 4a from Stumm and Morgan: Predominant pH range


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Figure 6-4a from Stumm and Morgan: Predominant pH rangefor the occurrence of various species for various oxidation states


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Figure 6-4b from Stumm & Morgan: The linear dependence of thefirst hydrolysis constant on the ratio of the charge to the M-Odistance (z/d) for four groups of cations at 25C.

Figure 6 6 from


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Figure 6-6 fromStumm & Morgan:Correlationbetween solubilityproduct of solidoxide/hydroxideand the firsthydrolysis constant.


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PEARSON HARD-SOFT ACID-BASE

(HSAB) THEORY Hard ions (class A)

small

highly charged d0electron

configuration

electron clouds not

easily deformedprefer to form ionic

bonds

Soft ions (class B) large

low charge d10 electron

configuration

electron clouds easily

deformedprefer to form covalent

bonds


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Pearsons Principle -In a competitive

situation, hard acids tend to form complexeswith hard bases, and soft acids tend to form

complexes with soft bases.

In other words - metals that tend to bond covalentlypreferentially form complexes with ligands thattend to bond covalently, and similarly, metals thattend to bond electrostatically preferentially form

complexes with ligands that tend to bondelectrostatically.

Classification of metals and ligands in terms of Pearsons (1963) HSAB principle.


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Hard Borderline Soft

Acids

H+

Li+

> Na+

> K+

> Rb+

> Cs+

Be

2+

> Mg

2+

> Ca

2+

> Sr

2+

> Ba

2+

Al3+

> Ga3+

Sc3+

> Y3+

; REE3+

(Lu3+

> La3+

);

Ce4+; Sn4+

Ti4+ > Ti3+, Zr4+ Hf4+

Cr6+

> Cr3+

; Mo6+

> Mo5+

>

Mo4+

; W6+

> W4+

; Nb5+

, Ta5+

;

Re

7+

> Re

6+

> Re

4+

; V

6+

> V

5+

>V4+

; Mn4+

; Fe3+

; Co3+

; As5+

; Sb5+

Th4+; U6+ > U4+

PGE6+

> PGE4+

, etc. (Ru, Ir, Os)

Acids

Fe2+, Mn2+, Co2+, Ni2+,

Cu2+

, Zn2+

, Pb2+

, Sn2+

,

As

3+

, Sb

3+

, Bi

3+

Acids

Au+ > Ag+ > Cu+

Hg2+

> Cd2+

Pt

2+

> Pd

2+

other PGE2+

Tl3+

> Tl+

Bases

F-; H2O, OH

-, O

2-; NH3; NO3

-;

CO32-

> HCO3-

; SO42-

> HSO4-

;PO43-

> HPO42-

> H2PO4-;

carboxylates (i.e., acetate,

oxalate, etc.);

MoO42-

; WO42-

Bases

Cl-

Bases

I-> Br

-; CN

-; CO;

S

2-

> HS

-

> H2S;organic phosphines (R3P);

organic thiols (RP);

polysulfide (SnS2-

),

thiosulfate (S2O32-

),

sulfite (SO32-

);

HSe-, Se

2-, HTe

-, Te

2-;

AsS2-

; SbS2-


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ION PAIRS VS. COORDINATION

COMPLEXES ION PAIRS

formed solely byelectrostatic attraction

ions often separated bycoordinated waters

short-lived association

no definite geometry also called outer-

sphere complexes

COORDINATIONCOMPLEXES

large covalentcomponent to bonding

ligand and metal joineddirectly

longer-lived species definite geometry

also called inner-sphere complexes


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STABILITY CONSTANTS OF ION PAIRSCAN BE ESTIMATED FROM

ELECTROSTATIC MODELSFor 1:1 pairs (e.g., NaCl0, LiF0, etc.)

log K 0 - 1 (I = 0)

For 2:2 pairs (e.g., CaSO40, MgCO30, etc.)log K 1.5 - 2.4 (I = 0)

For 3:3 pairs (e.g., LaPO40, AlPO4

0, etc.)

log K 2.8 - 4.0 (I = 0)Stability constants for covalently bound coordination

complexes cannot be estimated as easily.


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COMPLEX FORMATION AND

SOLUBILITY Total solubility of a system is given by:

[Me]T = [Me]free + [MemHkLn(OH)i]

Solubilities of relatively insoluble phases such as:Ag2S (pKs0 = 50); HgS (pKs0 = 52); FeOOH (pKs0 =38); CuO (pKs0 = 20); Al2O3 (pKs0 = 34) are probablynot determined by simple ions and solubility products

alone, but by complexes such as: AgHS0, HgS22- orHgS2H

-, Fe(OH)+, CuCO30 and Al(OH)4

-.


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Calculate the concentration of Ag+ in a solution inequilibrium with Ag2S with pH = 13 and ST = 0.1

M (20C, 1 atm., I = 0.1 M NaClO4).Ks0 = 10

-49.7 = [Ag+]2[S2-]

At pH = 13, [H2S0]


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[S2-

] = 9.1x10-3

M[Ag+]2 = 10-49.7/10-2.04 = 10-47.66

[Ag+] = 10-23.85 = 1.41x10-24 M

Obviously, in the absence of complexation, the solubilityof Ag2Sis exceedingly low under these conditions.

The concentration obtained corresponds to ~1 Ag ion perliter. What happens if we take 100 mL of such a

solution? Do we then have 1/10 of an Ag ion? No, thephysical interpretation of concentration does not makesense here. However, an Ag+ ion-selective electrodewould read [Ag+] = 10-23.85 nevertheless.

][11][10

][101.0 22

14

213

SSS


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Estimate the concentration of all species in a solution ofST = 0.02 M and saturated with respect to Ag2S as afunction of pH (in other words, calculate a solubility

diagram).[Ag]T = [Ag

+] + [AgHS0] + [Ag(HS)2-] + 2[Ag2S3H2

2-]

Ks0 = [Ag+]2[S2-], but [S2-] = 2ST so

Ks0 = [Ag+]22ST

2

0][T

s

S

KAg

Ag+ + HS-AgHS0 log K1 = 13.3

AgHS0 + HS-Ag(HS)2- log K2 = 3.87

Ag2S(s) + 2HS-Ag2S3H2

2- log Ks3 = -4.82

0A S


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]][[

][ 0

1

HSAg

AgHSK

TSAgAgHSK

1

0

1][

][

TSHS 1][

][][ 110

AgSKAgHS T

2

0

11

0

][ T

s

T S

K

SKAgHS

]][[

])([0

22

HSAgHS

HSAgK ]][[])([

0

22

HSAgHSKHSAg

2

02

1

2

122 ])([

T

sT

S

KSKKHSAg


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2

2

2323

][

][

HS

HSAgKs

2

1

2

3

2

232 ][ Ts SKHSAg

212321212112

0 21][

TsTT

T

sT SKSKKSK

S

KAg


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pH

0 2 4 6 8 10 12 14

log

concentration

-24

-22

-20

-18

-16

-14

-12

-10

-8

-6

Ag+

AgHS0

Ag(HS)2

-

Ag2S

3H

2

2-

pH = pK1(H

2S)


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pH

0 2 4 6 8 10 12 14

log

conc

entration

-10

-9

-8

-7

AgHS0

Ag(HS)2

-

Ag2S

3H

2

2-

pH = pK1(H

2S)

1

2

3

45


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Region 1: AgHS0 and H2S0 are predominant

Ag2S(s) + H2S0 2AgHS0

log [AgHS0] = 1/2log [H2S0] + 1/2log K

0]log[

2

SH

T

pH

Ag

Region 2: Ag(HS)2- and H2S

0 are predominant

Ag2S(s) + 3H2S0 2Ag(HS)2

- + 2H+

log [Ag(HS)2-] = 3/2log [H2S0] + 1/2log K + pH

1]log[

2

SH

T

pH

Ag


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Region 3: Ag(HS)2- and HS- are predominant

Ag2S(s) + 3HS- + H+ 2Ag(HS)2

-

log [Ag(HS)2-] = 3/2log [HS-] + 1/2log K - 1/2pH

2/1]log[

HS

T

pH

Ag

Region 4: Ag2S3H22- and HS- are predominant

Ag2S(s) + 2HS-Ag2S3H2

2-

log [Ag2S3H22-] = 2log [HS-] + log K

0]log[

HS

T

pH

Ag


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Region 5: Ag2S3H22- and S2- are predominant

Ag2S(s) + 2S2- + 2H+Ag2S3H22-log [Ag2S3H2

2-] = 2log [S2-] + log K - 2pH

2]log[

2

S

T

pH

Ag


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WHAT IS THE CAUSE OF THECHELATE EFFECT?

Gro = Hr

o - TSr0

For many ligands, Hro is about the same in multi-

and mono-dentate complexes, but there is a larger

entropy increase upon chelation!

Cu(H2O)42+ + 4NH3

0 Cu(NH3)42+ + 4H2O

Cu(H2O)42+ + N4 Cu(N4)2+ + 4H2O

The second reaction results in a greater increase inS

r

0.

Figure 6-11 from Stumm and Morgan. Effect of dilution on degree of complexation.


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g g g p

Figure 6 12a from


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Figure 6-12a fromStumm & Morgan.Complexing of Fe(III).The degree ofcomplexation isexpressed as pFe forvarious ligands at aconcentration of 10-2

M. The complexingeffect is highly pH-dependent because ofthe competing effects of

H+

and OH-

at low andhigh pH, respectively.

Figure 6-12b from Stumm & Morgan. Chelation of Zn(II).


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g g ( )


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METAL-ION BUFFERSAnalogous to pH buffers. Consider:

Me + L MeL

][

][

][ L

MeLK

Me

If we add MeL and L in approximately equalquantities, [Me] will be maintained approximately

constant unless a large amount of additional metalor ligand is added.

If [MeL] = [L], then pMe = pK!


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Example: Calculate [Ca2+] of a solution with thecomposition - EDTA = YT = 1.95x10

-2 M, CaT =

9.82x10-3

M, pH = 5.13 and I = 0.1 M (20C).For EDTA, pK1 = 2.0; pK2 = 2.67; pK3 = 6.16; and

pK4 = 10.26.

6.10

42

2

10]][[

][

CaYKYCa

CaY 5.332 10]][[

][

CaHYKHYCa

CaHY

1241

4

42

22

][]][[][1][

][][][)(

Ca

CaHYCaY

T

CaYHKKYKCa

CaHYCaYCaCai

T

CaCa

Ca ][ 2

C HYC YYHYYHYHYHYii 24320 ][][][][][][][)(


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CaHYCaY

T

KKYHCaKYCaY

CaHYCaYYHYYHYHYHYii

1

4

42421*

4

4

2432

23

0

4

]][][[]][[)]([

][][][][][][][)(

1

1234

4

234

3

34

2

4

4

0

4

4*

4

][][][][1

][

][

KKKKH

KKK

H

KK

H

K

H

YH

Y

i

i

i

Equations (i) and (ii) must be solved by trial and error.We know pH so we can calculate 4

* directly. We canthen assume that [HiY

4-i] YT - CaT. This permits usto calculate [Y4-] and then solve (i) for [Ca2+]. This

approach leads to: [CaY2-] = 9.66x10-3 M; [CaHY-] =1.09x10-4 M; [Ca2+] = 4.12x10-5 M; [Y4-] = 6.05x10-9M; [H3Y

-] = 3.07x10-5 M; [H2Y2-] = 8.8x10-3 M; [HY3-]

= 8.21x10-4 M; [H4

Y0] = 2.26x10-8 M.


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MIXED COMPLEXES

Examples: Zn(OH)2Cl22-, Hg(OH)(HS)0, PdCl3Br2-, etc.

Generalized complexation reaction:

M + mA + nB MAmBn

Snm

n

nm

mnmnmnm MBMABMA

loglogloglog

Log S is a statistical factor. For example, the probability

of forming MAB relative to MA2 and MB2 is S = 2because there are two distinct ways of forming MAB,i.e., A-M-B and B-M-A. The probability of formingMA2B relative to MA3 and MB3 is S = 3.


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In general, mixed complexes usually onlypredominate under a very restricted set of

conditions.

!!

)!(

nm

nmS

In simple cases we can use the followingformula:

Figure 6-15 from


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gStumm andMorgan.Predominance of

Hg(II) species as afunction of pCland pH. Inseawater, HgCl4

2-

predominates.


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COMPETITION FOR LIGANDS

The ratio of inorganic to organic substances in most

natural waters are usually very high. Does a large excess of, say, Ca2+ or Mg2+, decrease the

potential of organic ligands to complex trace metals?

Example: Fe3+, Ca2+ and EDTA

Fe3+ + Y4- FeY- log KFeY = 25.1

Ca2+

+ Y4-

CaY2-

log KCaY = 10.7

These data suggest that Fe3+ should be complexed byEDTA.


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But, let us combine the two above expressions to get:

CaY2- + Fe3+ FeY- + Ca2+ log Kexchange = 14.4

][

][4.14

][

][ 2

3

2

FeY

CaY

Fe

Ca

Thus, the relative importance of the two EDTAcomplexes depends also on the ratio of calcium toiron in solution.

For an exact solution to this problem, we also needto consider the species FeYOH and FeY(OH)2.

Figure 6.17a from Stumm & Morgan. Competitive effect of Ca2+ on


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complexation of Fe(III) with EDTA. Fe(OH)3(s) precipitates at pH > 8.6.

Figure 6 17b from Stumm & Morgan Competitive effect of Ca2+ on


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Figure 6.17b from Stumm & Morgan. Competitive effect of Ca2+ oncomplexation of Fe(III) with EDTA.


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Figure 6.17c from Stumm & Morgan. Competitive effect of Ca2+ oncomplexation of Fe(III) with citrate.

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Lecture 578 Complexation

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