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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)
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