COORDINATION CHEMISTRY COMPLEXATION

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COORDINATION CHEMISTRY (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 CuCO30, Cu(en)20, and Cu2 all affect
  the growth of algae differently
• Example Methylmercury (CH3Hg) is readily formed
  in biological processes, kinetically inert, and
  readily passes through cell walls. It is far more
  toxic than inorganic forms.
• Solubility and mobility depend on speciation.

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Figure 6-20 from Stumm Morgan Effect of free
Cu2 on growth of algae in seawater.
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DEFINITIONS - I

• Coordination (complex formation) - any
  combination of cations with molecules or anions
  containing free pairs of electrons. Bonding may
  be electrostatic, covalent or a mix.
• Central atom (nucleus) - the metal cation.
• Ligand - anion or molecule with which a cation
  forms complexes.
• 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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MULTIDENTATE LIGANDS
Oxalate (bidentate)
Ethylendiaminetetraacetic acid or EDTA
(hexadentate)
Ethylendiamine (bidentate)
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Chelation
Polynuclear complexes
Sb2S42-
Hg3(OH)42
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DEFINITIONS - II

• Species - refers to the actual form in which a
  molecule or ion 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
  several units, whereas in true solution we have
  hydration of a single molecule, atom or ion.

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FORMS OF OCCURRENCE OF METAL SPECIES
1000 Å
100 Å
10 Å
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Coordination numbers
CN 2 (linear)
CN 4 (square planar)
CN 4 (tetrahedral)
CN 6 (octahedral)
Coordination numbers 2, 4, 6, 8, 9 and 12 are
most common for cations.
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STABILITY CONSTANTS MEASURE THE STRENGTH OF
COMPLEXATION

• Stepwise constants
• MLn-1 L ? MLn
• Cumulative constants
• M nL ? MLn

?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

The larger the value of the stability constant,
the more stable the complex, and the greater the
proportion of the complex formed relative to the
simple ion.
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STABILITY CONSTANTS FOR POLYNUCLEAR COMPLEXES

• mM nL ? MmLn
• mM nHL ? MmLn nH

If m 1, the second subscript on ?nm is omitted
and the expression simplifies to the previous
expressions for mononuclear complexes.
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METAL-ION TITRATIONS

• Metal ions can be titrated by ligands in the same
  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 they donate electrons.

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Figure 6-3 from Stumm Morgan Titration of H
and Cu2 with ammonia and tetramine (trien)
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HYDROLYSIS

• The waters surrounding a cation may function as
  acids. The acidity is expected to increase with
  decreasing ionic radius 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)42- (ZnO22-)
• 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 will
  increase 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

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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 pK1 lt (1/2)pKw or p?n lt
  (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.

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POLYNUCLEAR SPECIES DECREASE IN IMPORTANCE 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) 2Cu2(OH)22
• we can write

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
  a homogeneous solution to which 10-4 (10-2) M of
  iron(III) has been added and the pH adjusted in
  the range 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 -6.31
• 2Fe3 2H2O ? Fe2(OH)24 2H log ?22 -2.91
• FeT Fe3 FeOH2 Fe(OH)2
  2Fe2(OH)24

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• Now we define ?0 Fe3/FeT ?1 FeOH2/FeT
  ?2 Fe(OH)2/FeT and ?22 2Fe2(OH)24/FeT.

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This last equation can be solved for ?0 at given
values of FeT and pH. The remaining ? values are
obtained from the following equations

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

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• Example 2 Compute the composition of a Fe(III)
  solution in equilibrium with amorphous ferric
  hydroxide given the additional equilibrium
  constants
• 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 -3.05
• Fe(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 -6.31
• Fe(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 6H2O 2log Ks0 7.92
• 2Fe3 2H2O ? Fe2(OH)24 2H log ?22 -2.91
• 2Fe(OH)3(s) 4H ? Fe2(OH)24 4H2O
• log Ks22 5.01
• log Fe2(OH)24 log Ks22 - 4pH
• These equations can be used to obtain the
  concentration of each of the Fe(III) species as a
  function of pH. They can all be summed to give
  the total solubility.

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Figure 6-4a from Stumm and Morgan Predominant pH
range for the occurrence of various species for
various oxidation states
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Figure 6-4b from Stumm Morgan The linear
dependence of the first hydrolysis constant on
the ratio of the charge to the M-O distance (z/d)
for four groups of cations at 25C.
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Figure 6-6 from Stumm Morgan Correlation
between solubility product of solid
oxide/hydroxide and the first hydrolysis constant.
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PEARSON HARD-SOFT ACID-BASE (HSAB) THEORY

• Hard ions (class A)
• small
• highly charged
• d0 electron configuration
• electron clouds not easily deformed
• prefer to form ionic bonds

• Soft ions (class B)
• large
• low charge
• d10 electron configuration
• electron clouds easily deformed
• prefer to form covalent bonds

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Pearsons Principle - In a competitive situation,
hard acids tend to form complexes with hard
bases, and soft acids tend to form complexes with
soft bases. In other words - metals that tend to
bond covalently preferentially form complexes
with ligands that tend to bond covalently, and
similarly, metals that tend to bond
electrostatically preferentially form complexes
with ligands that tend to bond electrostatically.
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ION PAIRS VS. COORDINATION COMPLEXES

• ION PAIRS
• formed solely by electrostatic attraction
• ions often separated by coordinated waters
• short-lived association
• no definite geometry
• also called outer-sphere complexes

• COORDINATION COMPLEXES
• large covalent component to bonding
• ligand and metal joined directly
• longer-lived species
• definite geometry
• also called inner-sphere complexes

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STABILITY CONSTANTS OF ION PAIRS CAN BE ESTIMATED
FROM ELECTROSTATIC MODELS

• For 11 pairs (e.g., NaCl0, LiF0, etc.)
• log K ? 0 - 1 (I 0)
• For 22 pairs (e.g., CaSO40, MgCO30, etc.)
• log K ? 1.5 - 2.4 (I 0)
• For 33 pairs (e.g., LaPO40, AlPO40, 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
• MeT Mefree ?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 probably not determined by simple ions and
  solubility products alone, but by complexes such
  as AgHS0, HgS22- or HgS2H-, Fe(OH), CuCO30 and
  Al(OH)4-.

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• Calculate the concentration of Ag in a solution
  in equilibrium with Ag2S with pH 13 and ST
  0.1 M (20C, 1 atm., I 0.1 M NaClO4).
• Ks0 10-49.7 Ag2S2-
• At pH 13, H2S0 ltlt HS- because pK1 6.68
  and pK2 14.0 so
• ST HS- S2- 0.1 M

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• S2- 9.1x10-3 M
• Ag2 10-49.7/10-2.04 10-47.66
• Ag 10-23.85 1.41x10-24 M
• Obviously, in the absence of complexation, the
  solubility of Ag2S is exceedingly low under these
  conditions.
• The concentration obtained corresponds to 1 Ag
  ion per liter. What happens if we take 100 mL of
  such a solution? Do we then have 1/10 of an Ag
  ion? No, the physical interpretation of
  concentration does not make sense here. However,
  an Ag ion-selective electrode would read Ag
  10-23.85 nevertheless.

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• Estimate the concentration of all species in a
  solution of ST 0.02 M and saturated with
  respect to Ag2S as a function of pH (in other
  words, calculate a solubility diagram).
• AgT Ag AgHS0 Ag(HS)2-
  2Ag2S3H22-
• Ks0 Ag2S2-, but S2- ?2ST so
• Ks0 Ag2 ?2ST

Ag HS- ? AgHS0 log K1 13.3 AgHS0 HS- ?
Ag(HS)2- log K2 3.87 Ag2S(s) 2HS- ?
Ag2S3H22- log Ks3 -4.82
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3
2
4
5
1
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• Region 1 AgHS0 and H2S0 are predominant
• Ag2S(s) H2S0 ? 2AgHS0
• log AgHS0 1/2log H2S0 1/2log K

Region 2 Ag(HS)2- and H2S0 are
predominant Ag2S(s) 3H2S0 ? 2Ag(HS)2- 2H log
Ag(HS)2- 3/2log H2S0 1/2log K pH
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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

Region 4 Ag2S3H22- and HS- are
predominant Ag2S(s) 2HS- ? Ag2S3H22- log
Ag2S3H22- 2log HS- log K
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Region 5 Ag2S3H22- and S2- are
predominant Ag2S(s) 2S2- 2H ? Ag2S3H22- log
Ag2S3H22- 2log S2- log K - 2pH
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THE CHELATE EFFECT

• Multidentate ligands are much stronger complex
  formers than monodentate ligands.
• Chelates remain stable even at very dilute
  concentrations, whereas monodentate complexes
  tend to dissociate.

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WHAT IS THE CAUSE OF THE CHELATE EFFECT?

• ?Gro ?Hro - T?Sr0
• 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 4NH30 ? Cu(NH3)42 4H2O
• Cu(H2O)42 N4 ? Cu(N4)2 4H2O
• The second reaction results in a greater increase
  in ?Sr0.

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Figure 6-11 from Stumm and Morgan. Effect of
dilution on degree of complexation.
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Figure 6-12a from Stumm Morgan. Complexing of
Fe(III). The degree of complexation is expressed
as ?pFe for various ligands at a concentration of
10-2 M. The complexing effect is highly
pH-dependent because of the competing effects of
H and OH- at low and high pH, respectively.
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Figure 6-12b from Stumm Morgan. Chelation of
Zn(II).
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METAL-ION BUFFERS

• Analogous to pH buffers. Consider
• Me L ? MeL

If we add MeL and L in approximately equal
quantities, Me will be maintained approximately
constant unless a large amount of additional
metal or ligand is added. If MeL L, then
pMe pK!
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• Example Calculate Ca2 of a solution with the
  composition - 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.

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Equations (i) and (ii) must be solved by trial
and error. We know pH so we can calculate ?4
directly. We can then assume that ?HiY4-i ? YT
- CaT. This permits us to 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- 6.05x10-9 M H3Y-
3.07x10-5 M H2Y2- 8.8x10-3 M HY3-
8.21x10-4 M H4Y0 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

Log S is a statistical factor. For example, the
probability of forming MAB relative to MA2 and
MB2 is S 2 because there are two distinct ways
of forming MAB, i.e., A-M-B and B-M-A. The
probability of forming MA2B relative to MA3 and
MB3 is S 3.
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In simple cases we can use the following formula

• In general, mixed complexes usually only
  predominate under a very restricted set of
  conditions.

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Figure 6-15 from Stumm and Morgan. Predominance
of Hg(II) species as a function of pCl and pH. In
seawater, HgCl42- 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
  by EDTA.

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• But, let us combine the two above expressions to
  get
• CaY2- Fe3 ? FeY- Ca2 log Kexchange 14.4

Thus, the relative importance of the two EDTA
complexes depends also on the ratio of calcium to
iron in solution. For an exact solution to this
problem, we also need to consider the species
FeYOH and FeY(OH)2.
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Figure 6.17a from Stumm Morgan. Competitive
effect of Ca2 on complexation of Fe(III) with
EDTA. Fe(OH)3(s) precipitates at pH gt 8.6.
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Figure 6.17b from Stumm Morgan. Competitive
effect of Ca2 on complexation of Fe(III) with
EDTA.
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Figure 6.17c from Stumm Morgan. Competitive
effect of Ca2 on complexation of Fe(III) with
citrate.