Complex Complex FormationFormation

CE 541CE 541

Complex ions consist of one Complex ions consist of one or more central ions (usually or more central ions (usually metals) associate with one or metals) associate with one or more ions or molecules more ions or molecules (ligands). Ligands act to (ligands). Ligands act to stabilize the central ion stabilize the central ion (metal) and keep it in (metal) and keep it in solution.solution.

Mononuclear ComplexesMononuclear ComplexesConsist of one single central ion to Consist of one single central ion to which is bound a number of which is bound a number of neutral or anionic ligandsneutral or anionic ligands

The number of ligands attached to The number of ligands attached to the central ion is called the central ion is called coordination numbercoordination number

If we take CuIf we take Cu2+2+ as a central ion and as a central ion and NHNH33 as ligand as ligand

103.9.....)()(

104.5.....)()(

102.2.....)(

108.9......

42433

233

23

2333

223

32

2233

23

31

233

2

KNHCuNHNHCu

KNHCuNHNHCu

KNHCuNHCuNH

KCuNHNHCu

Table 4 shows more Table 4 shows more formation constants formation constants for various ligands for various ligands and complexes.and complexes.

When formation When formation constants are constants are available, the available, the concentration of the concentration of the complex ion can be complex ion can be performed. The performed. The significance of significance of complexes can be complexes can be illustrated graphically illustrated graphically using Logarithmic using Logarithmic Concentration Concentration Diagrams.Diagrams.

)4.......(]][[

][

)3.......(]][[

][

)2.(..........]][[

][

)1......(..........]][[

][

4

32

33

2

43

3

32

23

2

33

23

23

2

23

13

2

23

KNHNHCu

NHCu

KNHNHCu

NHCu

KNHCuNH

NHCu

KNHCu

CuNH

pNHpNH33 (-log NH (-log NH33))

if concentration of Cuif concentration of Cu2+2+ is fixed, is fixed, equations (1) to (4) become equations (1) to (4) become straight lines on the Logarithmic straight lines on the Logarithmic Concentration Diagram.Concentration Diagram.

Equation (1) becomesEquation (1) becomes

from Table 4from Table 4

log Klog K11 = 3.99 = 3.99

3122

3 loglog]log[]log[ NHKCuCuNH

323

2

72

33

01.3]log[

7]log[

10][

]log[

pNHCuNH

then

Cu

MCu

if

pNHNH

Similarly,Similarly,

3

2

43

32

33

32

23

403.5]log[

306.3]log[

233.0]log[

pNHNHCu

pNHNHCu

pNHNHCu

Mixed Ligand ComplexesMixed Ligand ComplexesNatural waters contain several Natural waters contain several ligands (Clligands (Cl--, NH, NH33, S, S2-2-, OH, OH--) competing ) competing for complex formation with metal ionsfor complex formation with metal ions

How to solve the problem:How to solve the problem: All stepwise equilibrium for each ligand All stepwise equilibrium for each ligand

and the metal must be consideredand the metal must be considered Assume the unassociated Assume the unassociated

(uncomplexed) ion concentration(uncomplexed) ion concentration Determine concentration of each Determine concentration of each

complex for each ligandcomplex for each ligand

Problem 4.66Problem 4.66

Draw a logarithmic concentration Draw a logarithmic concentration diagram illustrating the effect of diagram illustrating the effect of chloride concentration on the chloride concentration on the relative concentrations of various relative concentrations of various chloride complexes of mercury, chloride complexes of mercury, assuming [Hgassuming [Hg2+2+] = 10] = 10-7-7 mol/l. Which mol/l. Which complex predominates when the complex predominates when the chloride concentration equals 0.1 chloride concentration equals 0.1 mg/l, 1 mg/l, 10 mg/l, and 100 mg/l.mg/l, 1 mg/l, 10 mg/l, and 100 mg/l.

Solubility of SaltsSolubility of Salts““All salts are soluble to some All salts are soluble to some degree”degree”

AgCl AgCl Ag Ag++ + Cl + Cl--

AgCl is known to be insoluble. AgCl is known to be insoluble. With insoluble salts the saturation With insoluble salts the saturation value is reached very quickly.value is reached very quickly.

The equilibrium that exists between The equilibrium that exists between the solids state of a compound and the solids state of a compound and its ion can be represented by the its ion can be represented by the following equilibrium:following equilibrium: K

AgCl

ClAg

][

]][[

Since AgCl is in solid state, it can Since AgCl is in solid state, it can be considered as constant (say Kbe considered as constant (say Kss))

KKspsp = solubility product constant = solubility product constant

sps

s

KKKClAg

or

KK

ClAg

]][[

]][[

For more complex substances:For more complex substances:

spKPOCa

POCaPOCa

]][[

2334

2

34

2243

Logarithmic Concentration Logarithmic Concentration Diagrams for Simple Solubility Diagrams for Simple Solubility

DeterminationDeterminationThe diagram was The diagram was draw based on:draw based on:Solubility Solubility product product constants (Table constants (Table 2.5)2.5)The following The following solubility solubility product product relationships:relationships:[M[M2+2+][CO][CO33

2-2-]=K]=Kspsp

log[Mlog[M2+2+] = pCO] = pCO33 + log K+ log Kspsp

The result is straight line with slope = 1The result is straight line with slope = 1[CO[CO33

2-2-] is represented by a line with slope = -1 and ] is represented by a line with slope = -1 and as a function of pCOas a function of pCO33

PbPb2+2+ is the least soluble and Mg is the least soluble and Mg2+2+ is the most is the most solublesolubleIntersection of COIntersection of CO33

2-2- line with cation line (say Pb) line with cation line (say Pb) indicates the solubility condition for that cation indicates the solubility condition for that cation (Pb)(Pb)[M[M2+2+] = [CO] = [CO33

2-2-] = solubility] = solubility

ExampleExampleDetermine the solubility of CaDetermine the solubility of Ca2+2+ in water in water containing 10containing 10-3-3 M carbonate @ 25 M carbonate @ 25 C. C.

SolutionSolutionpCOpCO33 = -log 10 = -log 10-3-3 = 3 = 3from the figurefrom the figureat pCOat pCO33 = 3, log[Ca = 3, log[Ca2+2+] = -5.3] = -5.3so, solubility = S = 10so, solubility = S = 10-5.3-5.3 = 5 = 5 10 10-6-6 M M

Complex Solubility Complex Solubility RelationshipsRelationships

““Solubility relationships are generally Solubility relationships are generally complex”complex”

ionic strength affects the ion activity (using ionic strength affects the ion activity (using molar concentration will not give exact molar concentration will not give exact solution)solution)

Cations and anions may form complexes with Cations and anions may form complexes with other materials in solution (reduce the other materials in solution (reduce the effective concentration of cations and effective concentration of cations and anions)anions)

Other ions may form salts with less solubility Other ions may form salts with less solubility than the one under considerationthan the one under consideration

Other equilibria beside solubility product Other equilibria beside solubility product affect the concentration of the ions presentaffect the concentration of the ions present

(1)(1) Effect of pH on Solubility of Effect of pH on Solubility of Hydroxide saltsHydroxide salts

wspz

ww

spz

sp

pKpHZKM

then

OHpKpOHpKpH

pHoffunctionaisOH

OHZKM

KOHM

log]log[

]log[

]log[

]log[log]log[

]][[ 2

The relationship can be drawn as shown The relationship can be drawn as shown in the figure. Slope of the line = -Zin the figure. Slope of the line = -Z

(2)(2) Effect of Weak Acids and BasesEffect of Weak Acids and Bases

pHpH Affects solubility of metal Affects solubility of metal

hydroxides, andhydroxides, and Affects other equilibria which affect Affects other equilibria which affect

solubility (such as that of weak solubility (such as that of weak acids)acids)

Take CaCOTake CaCO33 as an example as an example

)4(][][][

:)(lub

)3(107.4][

]][[

)2(103.4][

]][[

](

)1(105]][[

lub

CO CaCaCO

233

*32

112

3

23

71*

32

3

233

3*32

*32

23

923

2

-23

23

COHCOCOHC

isCcarbonlesoofionconcentrattotal

KHCO

COH

KCOH

HCOH

bothofproductionization

COHHCO

HCOHCOH

COHacidweakofanionanisCO

KCOCa

productilityso

T

T

A

A

sp

If NaOH was added to adjust the pH and if CaClIf NaOH was added to adjust the pH and if CaCl22 was added to the solutionwas added to the solution

At what [CaAt what [Ca2+2+] will the solution be saturated with ] will the solution be saturated with respect to CaCOrespect to CaCO33

How the saturation value will be affected by pH and CTHow the saturation value will be affected by pH and CT

How can we solve the problem?How can we solve the problem?Use equations (1) to (4) in addition to:Use equations (1) to (4) in addition to:

Equations (1) to (6) can be solved by trial and Equations (1) to (6) can be solved by trial and error. The relationship between CT, pH, and error. The relationship between CT, pH, and saturation value of [Casaturation value of [Ca2+2+] can be represented ] can be represented graphically.graphically.

)6]......([][][2][][][][2

arg

)5......(10]][[

233

2

14

ClOHCOHCOHNaCa

balanceechand

KOHH w

Study Examples 21 to 23Study Examples 21 to 23

(3)(3) Effect of ComplexesEffect of Complexes

““Complex formation affects the Complex formation affects the solubility of salts”solubility of salts”

Take Zn as an exampleTake Zn as an example

Base is added to increase pH to form Base is added to increase pH to form insoluble Zn(OH)insoluble Zn(OH)22

If excess base was added, Zn forms If excess base was added, Zn forms complexes with OHcomplexes with OH- - (soluble)(soluble)

This behavior can be expressed by the This behavior can be expressed by the following:following:

(a) Solubility product(a) Solubility product OHZnOHZn

c2)( 2

)(2

(b) Complex formation(b) Complex formation

(c) Water ionization(c) Water ionization

243

32

2

2

)()(

)()(

)(

OHZnOHOHZn

OHZnOHOHZn

OHZnOHZnOH

ZnOHOHZn

OHHOH 2

Equilibrium Equations for the AboveEquilibrium Equations for the Above

)7(])([])([])([][][

)6(101]][[

)5(108.1]][)(][)([

)4(103.1]][)(][)([

)3(101]][][)([

)2(104.1]][][[

)1(108]][[

2432

2

14

143

24

4323

622

41

2

182

OHZnOHZnOHZnZnOHZnC

additionIn

KOHH

KOHOHZnOHZn

KOHOHZnOHZn

KOHZnOHOHZn

KOHZnZnOH

KOHZn

T

w

sp

How to construct the diagram How to construct the diagram (solubility of Zn(OH)(solubility of Zn(OH)22 as a function of as a function of pH)pH)

For a given pH, find OHFor a given pH, find OH-- from equation from equation (6)(6)

Use it in Equation (1) to find ZnUse it in Equation (1) to find Zn2+2+

Use ZnUse Zn2+2+ in equation (2) to find [ZnOH in equation (2) to find [ZnOH++]] Continue the approach to find all Continue the approach to find all

complexescomplexes Use equation (7) to find CUse equation (7) to find CTT

logC-pH equations for each Zn logC-pH equations for each Zn species can be developedspecies can be developed

take log of equation (1)take log of equation (1)

pHZn

equationstwothesolving

HpHOHH

ce

OHKZn sp

9.10]log[

)]log[(14]log[]log[

sin

]log[2log]log[

2

2

Similarly, taking the log of equation Similarly, taking the log of equation (2)(2)

following the same approach, similar following the same approach, similar equations for the remaining Zn-equations for the remaining Zn-hydroxide complexes can be developedhydroxide complexes can be developed

pHZnOH

OHandZnforsubstitute

OHZnKZnOH

05.1]log[

]log[]log[

]log[]log[log]log[2

21

See Table 4-5 for other metalsSee Table 4-5 for other metalsStudy Examples 24 to 25Study Examples 24 to 25

The solubility of metals can be The solubility of metals can be complicated by the presence of complicated by the presence of ligands other than hydroxides ligands other than hydroxides (ammonia as an example)(ammonia as an example)

Now, how ammonia will affect the Now, how ammonia will affect the solubility of Zn(OH)solubility of Zn(OH)22. There are . There are four (4) Zn–NHfour (4) Zn–NH33 complexes as complexes as shown in Table 4-4. shown in Table 4-4.

243

233

223

2334,

3

243

233

223

23

2432

2,

43

233

243

33

223

233

23

23

223

13

2

23

)(4)(3)()(

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

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

2.91]][)([

])([

2.204]][)([

])([

8.177]][)([

])([

4.151]][[

])([

NHZnNHZnNHZnNHZnNHNHC

NHforbalancemass

NHZnNHZnNHZn

NHZnOHZnOHZnOHZnZnOHZnC

Znforbalancemass

KNHNHZn

NHZn

KNHNHZn

NHZn

KNHNHZn

NHZn

KNHZn

NHZn

NT

ZnT

Assuming that the concentration of Assuming that the concentration of Zn-NHZn-NH33 complexes is small compared complexes is small compared to [NHto [NH44

++] and [NH] and [NH33], then], then

Use these equations together with Use these equations together with those of Zn-OH to construct the those of Zn-OH to construct the diagram of Zn(OH)diagram of Zn(OH)22 solubility in the solubility in the presence of ammonia (NHpresence of ammonia (NH33))

34, NHNHC NT

Oxidation – Reduction Oxidation – Reduction ReactionsReactions

““Oxidation - reduction reactions tend Oxidation - reduction reactions tend towards a state of equilibrium”towards a state of equilibrium”

Oxidation-reduction reactions are Oxidation-reduction reactions are complexcomplex Understanding the equilibrium of such Understanding the equilibrium of such reactions will help to indicate whether reactions will help to indicate whether certain reactions can occur under certain certain reactions can occur under certain conditionsconditions

ExampleExampleUse half reactions for the reduction of Use half reactions for the reduction of NONO33

-- to NH to NH44 (Table 2-4) (Table 2-4)

)42(25@9.16

3.23.2

]log[

log

3.23.2][][log

log3.2,

sin

)96485.0(882.0)96485.0(1

08.85

)tan(ln

08.85)5.110(8

1)2.237(

8

3)50.79(

8

1

tan

)2.237(8

3)5.79(

8

100)5.110(

8

1

)13(

8

3

8

1

4

5

8

1

45

81

3

81

4

45

81

3

81

4

298

298

243

tableseeCEpE

RT

zFE

RT

GpE

epEwhere

HNO

NHpEpE

formfollowingthetoconvertedbecanequationthe

RT

zFE

RT

G

eHNO

NH

thusRT

zFE

RT

GKlinKthen

linKRTGce

factorconversionaisVE

potentialelectrodardstheisEKzF

RT

zF

GE

from

KJG

reactiontheforenergyfreedardstheisG

TableenergyfreetheisG

OHNHeHNO

The Redox potential The Redox potential can be indicated by can be indicated by pEpE

If pE is large, then If pE is large, then system oxidizing. system oxidizing. So, half reactions So, half reactions are driven to the are driven to the leftleft

If pE is small, then If pE is small, then system is reducing. system is reducing. So, half reactions So, half reactions are driven to the are driven to the rightright

Logarithmic Concentration Logarithmic Concentration DiagramDiagram

It can be used to illustrate the It can be used to illustrate the equilibrium concentrations for NHequilibrium concentrations for NH44

++ and NOand NO33

-- and a function of pE or E. and a function of pE or E. The diagram was constructed @ 25The diagram was constructed @ 25 C using:C using:

pEpE value can be found in Table 2-4 value can be found in Table 2-4

4

58

1

3

81

4log

HNO

NHpEpE

At pH = 7At pH = 7[H[H++] = 10] = 10-7-7

ThusThus

]log[2

177.137

][][

1log77.20

2

1)(

4

1

]log[2

177

][log0.0

)(2

1

Re

10][][

]log[8

1]log[

8

116.6

)10log(4

5]log[

8

1]log[

8

191.14

2

41

2

22

21

2

2

22

334

34

734

OpEpHat

HOpE

and

OHeHgO

Oxidation

HpEpHat

H

HpE

and

gHeH

duction

OandHlineswaterfor

assumedwasNONHC

NONHpE

NONHpE

T

pE-pH DiagramspE-pH Diagrams

As shownAs shown

pE is a function of [HpE is a function of [H++] which means it is a function ] which means it is a function of pHof pH

The general form of equation to be used in The general form of equation to be used in constructing the pE-pH diagrams is:constructing the pE-pH diagrams is:

(reduced) represents species on the reduced side(reduced) represents species on the reduced side(oxidized) represents species on the oxidized side(oxidized) represents species on the oxidized side

4

58

1

3

81

4log

HNO

NHpEpE

oxidized

reduced

zpEpE log

1

ThenThen

[H[H++] was ] was separated from separated from the log termthe log term

In order to draw In order to draw pE-pH diagram for pE-pH diagram for nitrogen system, nitrogen system, the following the following relationships will relationships will be needed:be needed:

3

4log8

1

4

591.14

NO

NHpHpE

34

33

32

24

34

23

/

/

/

/

/

/

NHNH

NONH

NONO

NONH

NONH

NONH