Tadeusz Górecki Ionic Equilibria Complex Formation · Tadeusz Górecki Ionic Equilibria Page 106 Complex Formation Complex - any species in solution formed by the combination of

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]


Page 107

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


Page 108

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


Page 109

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-

]


Page 110

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


Page 111

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.


Page 112

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


Page 113

2

0

2222

020

2][][

][

ClCd

CdCl

I1.0log 0 , thus I1.0log6loglog 0

22


Page 114

33332

30

3][][

][

ClCd

CdCl

log6loglog 0

33

44442

2

40

4][][

][

ClCd

CdCl

log4loglog 0

44


Page 115

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


Page 116

-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.


Page 117

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


Page 118

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


Page 119

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:


Page 120

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:


Page 121

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.


Page 122

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.


Page 123

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.


Page 124

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


Page 125

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


Page 126

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.

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Tadeusz Górecki Ionic Equilibria Complex Formation · Tadeusz Górecki Ionic Equilibria Page 106 Complex Formation Complex - any species in solution formed by the combination of