Tadeusz Górecki Ionic Equilibria Complex Formation · Tadeusz Górecki Ionic Equilibria Page 106...
Transcript of Tadeusz Górecki Ionic Equilibria Complex Formation · Tadeusz Górecki Ionic Equilibria Page 106...
Tadeusz Górecki Ionic Equilibria
Page 106
Complex Formation
Complex - any species in solution formed by the combination of two or more
simpler species, which can also exist independently in the solution.
Complex formation constant:
CdClClCd 2
]][[
][21
ClCd
CdClK
General formula: MLn, where n - coordination number.
Water is usually omitted in formulas (e.g. we write 2Cu instead of
2
42 )( OHCu or 2
62 )( OHCu )
Overall formation constants, i.e.:
]][[
][21
ClCd
CdCl
22
22
]][[
][
ClCd
CdCl etc.
Presenting equilibrium data:
Species concentration [MLn] vs. ligand concentration [L]
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Distribution diagrams (
Total
nn
M
ML
][
][ )
Compact distribution diagrams
Log plots of the distribution diagrams or equilibrium diagrams
Distribution diagrams
TCd
Cd
][
][ 2
0
TCd
CdCl
][
][1
TCd
CdCl
][
][ 22
TCd
CdCl
][
][ 33
TCd
CdCl
][
][ 2
44
][][][][][][ 2
432
2 CdClCdClCdClCdClCdCd T
]][[][ 2
1
ClCdCdCl
22
22 ]][[][ ClCdCdCl
32
33 ]][[][ ClCdCdCl
42
4
2
4 ]][[][ ClCdCdCl
Thus
4
4
3
3
2
21
0][][][][1
1
ClClClCl
4
4
3
3
2
21
11
][][][][1
][
ClClClCl
Cl
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4
4
3
3
2
21
2
22
][][][][1
][
ClClClCl
Cl
4
4
3
3
2
21
3
33
][][][][1
][
ClClClCl
Cl
4
4
3
3
2
21
4
44
][][][][1
][
ClClClCl
Cl
Note: [Cd2+
] cancels out, thus the fractional distribution does not depend in
this case on the metal concentration!
Compare with expressions for polyprotic acids, e.g.:
4321321
2
21
3
1
4
4
0][][][][
][
aaaaaaaaaa KKKKHKKKHKKHKH
H
Distribution diagram for Cd2+
complexes with Cl- at I = 3 ( nlog 1.5, 2.2,
2.3 and 1.6, respectively):
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 2 4
pCl
Alp
ha
0
12
3
4
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In logarithmic scale:
-5
-4.5
-4
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0
-0.5 0.5 1.5 2.5
pCl
log
alp
ha
0
1
2
3
4
Equilibrium diagram ( MCd T
410][ ):
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
-0.5 0.5 1.5 2.5 3.5
pCl
log
C
[Cd2+
]
[CdCl+]
[CdCl2][CdCl3
-]
[CdCl42-
]
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Compact distribution diagram (also called in this case "cumulative
distribution diagram") -- plot of 0 , 10 , etc.:
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-0.5 1.5 3.5
pCl
[Cd2+
][CdCl
+]
[CdCl2]
[CdCl3-]
[CdCl42-
]
When only the total ligand concentration is known, we need to include mass
balances.
Example: calculate the concentrations of all species in a solution containing
0.01 M CdCl2 (salt). pH is adjusted so that no Cd(OH)2 is formed.
Mass balances:
][][][][][01.0][ 2
432
2 CdClCdClCdClCdClCdCd T
][4][3][2][][02.0][ 2
432
CdClCdClCdClCdClClCl T
Which terms are significant? Let's assume (incorrectly, of course) that
complex formation is slight, so that:
01.0][ 2 Cd and 02.0][ Cl
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In this case:
0191.002.001.010]][[][ 98.12
1 ClCdCdCl
0016.0]][[][ 22
22 ClCdCdCl
532
33 102]][[][ ClCdCdCl
842
4
2
4 1002.8]][[][ ClCdCdCl
Thus, last two terms can be neglected.
][][][01.0][ 2
2 CdClCdClCdCd T
][2][][02.0][ 2CdClCdClClCl T
using expressions:
2
21
2
][][1
01.0][
ClClCd
and
02.0][2][][][][ 2
21
2 ClClCdClCl T
Substitution of the expression for [Cd2+
] into the mass balance on [Cl] yields
after rearrangement:
02.0])[01.01(][][ 1
2
1
3
2 ClClCl
Solution: 01385.0][ Cl , 00417.0][ 2 Cd
Concentrations of the remaining species can be easily calculated from the
overall formation constants.
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Ionic strength
12
0
1][][
][
ClCd
CdCl
logloglogloglog 0
11
Davies eq.:
I
I
Izz 2.0
15.0log 2
thus
log4log
and
log4loglog 0
11
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2
0
2222
020
2][][
][
ClCd
CdCl
I1.0log 0 , thus I1.0log6loglog 0
22
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33332
30
3][][
][
ClCd
CdCl
log6loglog 0
33
44442
2
40
4][][
][
ClCd
CdCl
log4loglog 0
44
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Complex formation effect on solubility
Complex formation increases solubility by binding one of the ions (removing
it from the solution).
Salt MX dissolved in a solution of ligand Y:
0]][[ sKXM
]][[][ 1
YMKMY
Mass balance on M:
...)][1][...][][][ 1 YKMMYMM T
][][ 0
X
KM s
...][1][
][ 10
YK
X
KM s
T
[M]T – solubility, equal to sum of all the forms of M in solution
Quite often, X- and Y
- can be the same species. In such cases, an increase in
[X-] at first causes a decrease in solubility (common ion effect), followed by
an increase in solubility due to complex formation.
AgCl in excess Cl-:
][][][][][][ 3
4
2
32
AgClAgClAgClAgClAgAg T
4
4
3
3
2
210 ][][][][1][
][
ClClClCl
Cl
KAg s
T
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-10
-9
-8
-7
-6
-5
-4
-3
-2
0 1 2 3 4 5 6
pCl
log
(so
lub
ilit
y)
Ks0 only
Complexes, I = 0
Complexes, including
ionic strength
Hydrolysis of metal ions
pH is the master variable. Formation of insoluble oxides or hydroxides needs
to be taken into account.
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Example: Hg-OH system. Hg2+
reacts with water to form hydroxide
complexes HgOH+ and Hg(OH)2. In concentrated solutions, HgO is formed.
Distribution diagram in dilute solutions (<10 –4
M):
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 1 2 3 4 5 6
pH
n
0
1
2
Hg2+
HgOH+
Hg(OH)2
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2
21
0][][1
1
OHOH
2
21
11
][][1
][
OHOH
OH
2
21
2
22
][][1
][
OHOH
OH
Apply only to unsaturated solutions.
If precipitate is present, an additional equilibrium must be fulfilled:
22
0 ]][[ OHHgKs
For unsaturated solutions, [Hg2+
] is determined from the total dissolved
concentration of Hg ([Hg]T) and 0 . In saturated solutions, [Hg2+
] is
determined from the solubility product, and the total dissolved concentration
of Hg can then be calculated from 0 .
Example: is 1 mM Hg(II) solution at pH = 1 saturated with respect to HgO?
mMHgHg T 99.0][][ 0
2
At pH of 1, the ion product is:
29213322 10)10)(10(]][[ OHHg
44.25
0 10sK
Thus, the solution is unsaturated.
Example: what is the species distribution for the same [Hg]T concentration at
pH = 3.0?
THgHg ][][ 0
2
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THgHgOH ][][ 1
THgOHHg ][])([ 22
Let us first assume that all Hg(II) is in the dissolved form, i.e. [Hg]T = 1 mM:
mMHgHg T 49.0][][ 0
2
On the other hand, the solubility product yields:
mMOH
KHg s 36.0
][][
2
02
Thus, the solution is saturated, and we have to use the latter number to
calculate the total concentration of dissolved Hg:
mMHg
Hg satsolT 737.0
49.0
36.0][][
0
2
)(
From the original 1 mM, 0.737 mM remains in solution, and 0.263 mM
precipitates as HgO.
Remaining species:
mMHgHgOH solT 145.0][][ )(1
mMHgOHHg solT 229.0][])([ )(22
Complete distribution diagram for 1 mM solution:
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0
0.0001
0.0002
0.0003
0.0004
0.0005
0.0006
0.0007
0.0008
0.0009
0.001
0 1 2 3 4 5 6
pH
Co
ncen
trati
on
[Hg2+
]
[HgOH+]
[Hg(OH)2]
HgO(s)
[Hg]dissolved
1 M solution, logarithmic diagram:
-14
-12
-10
-8
-6
-4
-2
0
0 1 2 3 4 5 6
pH
log
co
ncen
trati
on
[Hg2+
]
[HgOH+]
[Hg3(OH)33+
]
[Hg]dissolved
[Hg(OH)2]
[Hg2OH3+
]
[Hg(OH)3-]
Saturation
point
Polynuclear complexes reach their greatest influence at around pH = 1.2,
where HgO starts to form. Equilibria:
Tadeusz Górecki Ionic Equilibria
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7.10
22
3
2 10][][
])([
OHHg
OHHg
6.35
332
3
33 10][][
])([
OHHg
OHHg
At high pH, Hg(OH)3- becomes an important species:
-12
-10
-8
-6
-4
-2
0
5 6 7 8 9 10 11 12 13 14
pH
log
co
ncen
trati
on
[Hg2+
][HgOH
+]
[Hg(OH)2]
[Hg(OH)3-]
[OH-]
Al-OH system
Aluminum forms a series of mononuclear and polynuclear hydroxide
complexes. At higher pH, Al(OH)3 precipitates.
Mass balance for unsaturated solution:
])([])([])([][][][ 432
23 OHAlOHAlOHAlAlOHAlAl T
])([13])([3])([2 7
3213
5
43
4
22
OHAlOHAlOHAl
]][[
][3
2
1
OHAl
AlOH hence ]][[][ 3
1
2 OHAlAlOH etc.
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4
4
3
3
2
21
3 ][][][][1]([][ OHOHOHOHAlAl T
423
34
23
22 ][][3]][[2 OHAlOHAl
)][][13 32123
32,13
OHAl
Den
AlAl T
unsat
][][ 3
4
4
3
3
2
21 ][][][][1 OHOHOHOHDen
423
34
23
22 ][][3]][[2 OHAlOHAl
32123
32,13 ][][13 OHAl
[Al3+
] is present in the denominator, thus it cannot be calculated directly
(iteration necessary).
In saturated solutions:
3
03
][][
OH
KAl s
sat
Procedure: Find the pH at which precipitation of Al(OH)3 begins (i.e.
substitute Ks0/[OH-]
3 for [Al
3+]unsat and solve the resulting equation). Use
[Al3+
]unsat for pH values lower than the precipitation pH, and [Al3+
]sat for
higher pH. Alternatively, for each pH value calculate both [Al3+
]unsat and
[Al3+
]sat. Pick the smaller of the two values at each pH for the calculations.
Mononuclear complexes are often sufficient to describe the system.
Al13(OH)327+
( 5.336log 32,13 ) is well documented (precursor of
amorphous Al(OH)3), but it forms only very slowly at room temperature.
Thus, it is usually not very important.
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Equilibrium diagram for [Al]T = 0.5 M solution (I = 1 M):
31.8log 1 20log 22
2.16log 2 1.41log 34
3.24log 3 5.33log 0 sK
5.29log 4
-14
-12
-10
-8
-6
-4
-2
0
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
pH
Lo
g c
on
cen
trati
on
log [Al]T (sol)
log [Al3+]
log [AlOH]
log [Al2(OH)2]
log [Al3(OH)4]
log [Al(OH)2]
log [Al(OH)3]
log [Al(OH)4]
Saturation point
Fe-OH system
Fe3+
is very common. Soluble in acids and slightly soluble in bases. At
intermediate pH, largely precipitated as amorphous Fe(OH)3, FeOOH
(goethite) or Fe2O3 (hematite). The solubility products of these three species
have the same form, but different values (see Table 7.2). We will assume the
solid phase to be Fe(OH)3.
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Fe3+
forms 4 mononuclear and 2 polynuclear complexes. Polynuclear
Fe2OH24+
can be the dominant species at high concentrations in acidic
solutions.
Mass balance for unsaturated solution:
])([])([])([][][][ 432
23 OHFeOHFeOHFeFeOHFeFe T
])([3])([2 5
43
4
22
OHFeOHFe
Den
FeFe T
unsat
][][ 3
4
4
3
3
2
21 ][][][][1 OHOHOHOHDen
423
34
23
22 ][][3]][[2 OHFeOHFe
[Fe3+
] is present in the denominator, thus [Fe3+
]unsat cannot be calculated
directly (iteration necessary).
In saturated solutions:
3
03
][][
OH
KFe s
sat
Procedure is exactly the same as before. Once we have [Fe3+
], we can
calculate the concentrations of the remaining species at a given pH, e.g.:
223
22
4
2 ][][)(2
OHFeOHFe etc.
At higher pH, Fe(OH)3 starts to precipitate, and:
3
03
][][
OH
KFe s
sat
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Equilibrium diagram for [Fe]T = 0.5 M solution (I = 3 M):
2.11log 1 4.25log 22
1.22log 2 51log 34
28log 3 6.38log 0 sK
33log 4
-14
-12
-10
-8
-6
-4
-2
0
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
pH
Lo
g c
on
cen
trati
on
log [Fe]T (sol)
log [Fe3+]
log [FeOH]
log [Fe2(OH)2]
log [Fe3(OH)4]
log [Fe(OH)2]
log [Fe(OH)3]
log [Fe(OH)4]Saturation
point
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Acid mine drainage
In the presence of other complex-forming species, additional equilibria must
be taken into account.
HSOFeOHOsFeS 2
4
3
222 22
1
4
15)(
HSOsOHFeOHHSOFe 42)()(32 2
432
2
4
3
Both reactions increase acidity.
Mass balance on Fe:
])([])([])([][][][ 432
23 OHFeOHFeOHFeFeOHFeFe T
])([][])([3])([2 244
5
43
4
22
SOFeFeSOOHFeOHFe
Substituting equilibrium expressions:
4
4
3
3
2
21
3 ][][][][1]([][ OHOHOHOHFeFe T
)][][][][3]][[2 22
42
2
41
423
34
23
22
SOSOOHFeOHFe SS
Den
FeFe T
unsat
][][ 3
Mass balance on sulphur:
][])([])([][][ 4244
2
44
HSOSOFeSOFeSOSO T
a
SS
T
K
HSOFeFe
SOSO
][]][[][1
][][
2
4
3
2
3
1
42
4
Initial values of [Fe3+
] and [SO42-
] must be guessed and then refined through
iteration. Constraints in unsaturated solution: [Fe3+
] [Fe3+
]T and [SO42-
]
[SO42-
]T. In saturated solution: [Fe3+
] Ks0/[OH-]3.