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