Complex Formation
Complex - any ______ in solution formed by the combination of __________
____________, which can also exist _______________ in the solution.
Complex formation constant:
+
−
+
+ Cl ↔ CdCl
Cd
2] ][
[
] [
1 2+ −
=
+Cl Cd
K CdCl
General formula: MLn, where n - _________________.
Water is usually _________________ (e.g. we write ______ instead of __________ or ____________)
_______________ constants, i.e.:
] ][
[
] [
1 2+ −
=
+Cl Cd
β CdCl
2 2
2
2
[ ][ ]
] [
−
=
+Cl Cd
β CdCl
etc.Presenting equilibrium data:
• ________ concentration _______ vs. _______ concentration ____
• _____________ diagrams (
Total n
n
M
ML ] [
]
= [
α
)• __________________ diagrams
• ________ of the ___________ diagrams or ____________ diagrams
Distribution diagrams
Cd
TCd ] [
]
[
20
=
+α
Cd
TCdCl ] [
] [
1
=
+α
Cd
TCdCl ] [
]
[
22
= α
Cd
TCdCl ] [
]
[
33
=
−α
Cd
TCdCl ] [
] [
424
=
−α
] [
] [
] [
] [
] [
]
[ Cd
T= Cd
2++ CdCl
++ CdCl
2+ CdCl
3−+ CdCl
42−]
][
[ ]
[ CdCl
+= β
1Cd
2+Cl
−2 2
2
2
] [ ][ ]
[ CdCl = β Cd
+Cl
−3 2
3
3
] [ ][ ]
[ CdCl
−= β Cd
+Cl
−4 2
4 2
4
] [ ][ ]
[ CdCl
−= β Cd
+Cl
−Thus
4 4
3 3
2 2
1
0
1 [ ] [ ] [ ] [ ]
1
−
−
−
−
+ + +
= +
Cl Cl
Cl
Cl β β β
α β
4 3
2 1
1
1 [ ] [ ] [ ] [ ]
] [
−
−
−
−
−
+ +
+
= +
Cl Cl
Cl Cl
Cl
β β
β β
α β
4 4
3 3
2 2
1
2 2
2
1 [ ] [ ] [ ] [ ]
] [
−
−
−
−
−
+ +
+
= +
Cl Cl
Cl Cl
Cl
β β
β β
α β
4 4
3 3
2 2
1
3 3
3
1 [ ] [ ] [ ] [ ]
] [
−
−
−
−
−
+ +
+
= +
Cl Cl
Cl Cl
Cl
β β
β β
α β
4 4
3 3
2 2
1
4 4
4
1 [ ] [ ] [ ] [ ]
] [
−
−
−
−
−
+ +
+
= +
Cl Cl
Cl Cl
Cl
β β
β β
α β
Note: [Cd2+] __________, thus the fractional distribution _____________ in this case on the _________________!
Compare with
α
expressions for polyprotic acids, e.g.:4 3 2 1 3
2 1 2
2 1 3
1 4
4
0 [ ] [ ] [ ] [ ]
] [
a a a a a
a a a
a
a H K K H K K K H K K K K
K H
H
+ +
+
= + + + + + +
α
Distribution diagram for Cd2+ complexes with Cl- at I = 3 (
log β
n=
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
Alpha
α0
α1
α2
α3
α4
In ________________:
-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 alpha
α0
α1
α2
α3
α4
Equilibrium diagram (
[ Cd ]
T= 10
−4M
):-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-
] [CdCl4
2-]
____________________ diagram (also called in this case "___________
____________ diagram") -- plot of
α
0,α +
0α
1, 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
-] [CdCl4
2-]
When only the _____________________ is known, we need to include ____
___________.
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.
____ balances:
] [
] [
] [
] [
] [
01 . 0 ]
[ Cd
T= = Cd
2++ CdCl
++ CdCl
2+ CdCl
3−+ CdCl
42−] [
4 ] [
3 ] [
2 ] [
] [ 02 . 0 ]
[ Cl
− T= = Cl
−+ CdCl
++ CdCl
2+ CdCl
3−+ CdCl
42−Which terms are significant? Let's assume (_____________________) that complex formation is ______, so that:
01 . 0 ]
[ Cd
2+=
and[ Cl
−] = 0 . 02
In this case:
=
⋅
⋅
=
=
+ −+
] [ ][ ] 10 0 . 01 0 . 02
[ CdCl β
1Cd
2Cl
1.98=
=
2 2+ − 22
] [ ][ ]
[ CdCl β Cd Cl
=
=
+ −− 2 3
3
3
] [ ][ ]
[ CdCl β Cd Cl
=
=
+ −− 2 4
4 2
4
] [ ][ ]
[ CdCl β Cd Cl
Thus, _______ terms can be neglected.
] [
] [
] [
01 . 0 ]
[ Cd
T= = Cd
2++ CdCl
++ CdCl
2] [
2 ] [
] [ 02 . 0 ]
[ Cl
T= = Cl
−+ CdCl
++ CdCl
2using
α
expressions:2 2
1 2
] [ ]
[ 1
01 . ] 0
[
+ − −+
= +
Cl Cd Cl
β β
and
( [ ] 2 [ ] ) 0 . 02
] [
] [ ]
[ Cl
T= Cl
−+ Cd
2+β
1Cl
−+ β
2Cl
− 2=
Substitution of the expression for _____ into the _________________ yields after rearrangement:
=
⋅
− +
+
− −−
] [ ] ( 1 0 . 01 )[ ]
[
3 1 2 12
Cl β Cl β Cl
β
Solution:
[ Cl
−] = __________
,[ Cd
2+] = __________
Concentrations of the remaining species can be easily calculated from the ____________________________.
Ionic strength
− + +
+
−
− + +
+ + +
=
= γ γ
β γ γ
γ
β
2γ
10
1
[ ] [ ]
] [
Cl Cd
CdCl
+
− +
+
+ −
+
= β γ γ γ
β log log log log
log
1 10Davies eq.:
−
− +
= I
I z I
z
0 . 2
5 1 . 0
log γ
2thus
− +
= γ γ
+ +
+
= γ
γ 4 log log
and
1
=
log β
2 0 2 2
2 2
0 0 2
2
[ ] [ ]
] [
− + +
−
− + +
+
=
= γ γ
β γ γ
γ β γ
Cl Cd
CdCl
I 1 . 0
log γ
0=
, thuslog β
2=
3 3 3
3 2
0 3
3
[ ] [ ]
] [
− + +
−
−
− + +
+ − −
=
= γ γ
β γ γ
γ β γ
Cl Cd
CdCl
3
= log β
4 4 4
4 2
2 0 4
4
[ ] [ ]
] [
− + +
=
−
− + +
+ − =
=
= γ γ
β γ γ
γ β γ
Cl Cd
CdCl
4
=
log β
Complex formation effect on solubility
Complex formation ________________ by __________________ (_______
_________________).
Salt MX dissolved in a solution of ligand Y:
]
0][
[ M
+X
−= K
s] ][
[ ]
[ MY = K
1M
+Y
−____ balance on M:
( 1 [ ] ...) )
] [
...
] [
] [ ]
[ M
T= M
++ MY + = M
++ K
1Y
−+
] ] [
[
+=
−0X M K
s( 1 [ ] ... )
] ] [
[ =
−0+ K
1Y
−+ X
M
TK
s[M]T – solubility, equal to ___________________ of M in solution
Quite often, X- and Y- can be the same species. In such cases, an increase in [X-] at first causes ___________________ (__________ effect), followed by an ________________ due to ________________.
AgCl in excess Cl-:
] [
] [
] [
] [
] [
]
[ Ag
T= Ag
++ AgCl + AgCl
2−+ AgCl
32−+ AgCl
43−(
4 4)
3 3
2 2
1
0
1 [ ] [ ] [ ] [ ]
] ] [
[ =
−+ Cl
−+ Cl
−+ Cl
−+ Cl
−Cl
Ag
TK
sβ β β β
-10 -9 -8 -7 -6 -5 -4 -3 -2
0 1 2 3 4 5 6
pCl
log (solubility)
Ks0 only Complexes, I = 0
Complexes, including ionic strength
Hydrolysis of metal ions
pH is the _____________. Formation of ________________________ needs to be taken into account.
Example: Hg-OH system. Hg2+ _____________to form ________ complexes HgOH+ and Hg(OH)2. In concentrated solutions, ____ is formed.
Distribution diagram in __________ (_____ 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 ααα
αn
α0
α1
α2
Hg2+
HgOH+
Hg(OH)2
2 2
1
0
1 [ ] [ ]
1
−
−
+
= +
OH OH β
α β
2 2
1
1
1
1 [ ] [ ]
] [
−
−
−
+
= +
OH OH
OH β β
α β
2 2
1
2 2
2
1 [ ] [ ]
] [
−
−
−
+
= +
OH OH
OH β β
α β
Apply only to _____________ solutions.
If ___________ is present, an ___________________ must be fulfilled:
2 2
0
= [ Hg
+][ OH
−] K
sFor _____________ solutions, [Hg2+] is determined from the _____________
______________ of Hg ([Hg]T) and
α
0. In ________ solutions, [Hg2+] is determined from the ______________, and the _______________________of Hg can then be calculated from ___.
Example: is 1 mM Hg(II) solution at pH = 1 saturated with respect to HgO?
=
+
=
Hg
THg ] [ ]
[
2α
0At pH of 1, the ion product is:
=
=
− −−
+ 2 3 13 2
2
][ ] ( 10 )( 10 )
[ Hg OH
0
= K
sThus, the solution is _____________.
Example: what is the species distribution for the same [Hg]T concentration at pH = 3.0?
Hg
THg ] [ ]
[
2+= α
0Hg
THgOH ] [ ]
[
+= α
1Hg
TOH
Hg ( ) ] [ ]
[
2= α
2Let us first assume that all Hg(II) is in the _______ form, i.e. ____________:
=
+
=
Hg
THg ] [ ]
[
2α
0On the other hand, the __________________ yields:
=
=
−+
2 2 0
] ] [
[ OH
Hg K
sThus, the solution is ____________, and we have to use the ___________ to calculate the total concentration of ___________ Hg:
=
=
+0 2 )
(
] ] [
[
T solα
satHg Hg
From the original 1 mM, ____ mM __________________, and _____ mM ____________________.
Remaining species:
=
+
=
) (
1
[ ]
]
[ HgOH α Hg
T sol=
=
2 ( )2
] [ ]
) (
[ Hg OH α Hg
T solComplete distribution diagram for 1 mM solution:
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
Concentration
[Hg2+]
[HgOH+]
[Hg(OH)2] HgO(s) [Hg]dissolved
1 M solution, _____________ diagram:
-14 -12 -10 -8 -6 -4 -2 0
0 1 2 3 4 5 6
pH
log concentration
[Hg2+] [HgOH+]
[Hg3(OH)3 3+] [Hg]dissolved
[Hg(OH)2]
[Hg2OH3+]
[Hg(OH)3-
] Saturation
point
____________ complexes reach their greatest influence at around pH = ___, where ______________. Equilibria:
7 . 10 2
2
3
2
10
] [
] [
] ) (
[
+ +−=
OH Hg
OH
Hg
35.63 3
2
3 3
3
10
] [
] [
] ) (
[
+ +−=
OH Hg
OH Hg
At ________, 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 concentration
[Hg2+]
[HgOH+] [Hg(OH)2]
[Hg(OH)3 -] [OH-]
Al-OH system
Aluminum forms a series of _____________ and ____________________
complexes. At higher pH, _____________________.
____ balance for ____________ solution:
+ +
+ +
+
= [
+] [
+] [ ( )
+] [ ( ) ] [ ( )
−] ]
[ Al
TAl
3AlOH
2Al OH
2Al OH
3Al OH
4] ) (
[ 13 ] ) (
[ 3 ] ) (
[
2
2 42++
3 54++
13 732++ Al OH Al OH Al OH
] ][
[
] [
3
2
1 + −
=
+OH Al
β AlOH
hence[ AlOH
2+] = β
1[ Al
3+][ OH
−]
etc.+ +
+ +
+
= [
3+]( 1
1[
−]
2[
−]
2 3[
−]
3 4[
−]
4]
[ Al
TAl β OH β OH β OH β OH
+ +
+ 2 β
22[ Al
3+][ OH
−]
23 β
34[ Al
3+]
2[ OH
−]
4)
] [
] [
13
13,32 3+ 12 − 32+ β Al OH
Den Al
unsat[ Al ]
T]
[
3+=
+ +
+ +
+
= 1
1[ OH
−]
2[ OH
−]
2 3[ OH
−]
3 4[ OH
−]
4Den β β β β
+ +
+ 2 β
22[ Al
3+][ OH
−]
23 β
34[ Al
3+]
2[ OH
−]
432 12
3 32 ,
13
[ ] [ ]
13
+ −+ β Al OH
_____ is present in the ____________, thus it ______ be ________________
(__________ necessary).
In __________ solutions:
3 3 0
] ] [
[
+=
−OH Al
satK
sProcedure: Find the pH at which ______________________ begins (i.e.
substitute _________ for ________ and solve the resulting equation). Use __________ for pH values lower than the precipitation pH, and ______ for higher pH. Alternatively, for each pH value calculate both __________ and _______. Pick the _______of the two values at each pH for the calculations.
____________ complexes are often sufficient to describe the system.
Al13(OH)32
7+ (
log β
13,32= 336 . 5
) is well documented (precursor ofamorphous Al(OH)3), but it forms only ___________ at room temperature.
Thus, it is usually _________________.
Equilibrium diagram for [Al]T = 0.5 M solution (I = 1 M):
31 . 8
log β
1= log β
22= 20
2 . 16
log β
2= log β
34= 41 . 1
3 . 24
log β
3= log K
s0= − 33 . 5
5 . 29 log β
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
Log concentration
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 ______ and slightly soluble in ______. At ___________ pH, largely __________ as amorphous Fe(OH)3, α−FeOOH (goethite) or Fe2O3 (hematite). The solubility products of these three species have the _________, but ____________ (see Table 7.2). We will assume the solid phase to be ________.
Fe3+ forms 4 ____________ and 2 ___________ complexes. ____________
Fe2OH24+ can be the dominant species at __________________ in ______
solutions.
_____ balance for _____________ solution:
+ +
+ +
+
= [
+] [
+] [ ( )
+] [ ( ) ] [ ( )
−] ]
[ Fe
TFe
3FeOH
2Fe OH
2Fe OH
3Fe OH
4] ) (
[ 3 ] ) (
[
2
2 42++
3 54++ Fe OH Fe OH
Den Fe
unsat[ Fe ]
T]
[
3+=
+ +
+ +
+
= 1
1[ OH
−]
2[ OH
−]
2 3[ OH
−]
3 4[ OH
−]
4Den β β β β
4 2
3 34 2
3
22
[ ][ ] 3 [ ] [ ]
2
+ −+
+ −+ β Fe OH β Fe OH
[Fe3+] is present in the _____________, thus [Fe3+]unsat ______ be _________
_______ (________ necessary).
In __________ solutions:
3 3 0
] ] [
[
+=
−OH Fe
satK
sProcedure 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.:
2 2
3 22 4
2
( ) [ ] [ ]
2
− +
+
= Fe OH
OH
Fe β
etc.At _________, Fe(OH)3 starts to precipitate, and:
3 3 0
] ] [
[
+=
−OH
Fe
satK
sEquilibrium diagram for [Fe]T = 0.5 M solution (I = 3 M):
2 . 11
log β
1= log β
22= 25 . 4
1 . 22
log β
2= log β
34= 51
28
log β
3= log K
s0= − 38 . 6
33 log β
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
Log concentration
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
Acid mine drainage
In the presence of other complex-forming species, _________________ must be taken into account.
+
−
+
+ +
→ +
+ O H O Fe SO H
s
FeS
2 2 2 32
422 1 4
) 15 (
+
− +
−
+
+ SO + H + H O ↔ Fe OH s + SO + H
Fe
32
423
2( )
3( ) 2
424
Both reactions _________________.
____ balance on Fe:
+ +
+ +
+
= [
+] [
+] [ ( )
+] [ ( ) ] [ ( )
−] ]
[ Fe
TFe
3FeOH
2Fe OH
2Fe OH
3Fe OH
4] ) (
[ ] [
] ) (
[ 3 ] ) (
[
2
2 42++
3 54++
4++
4 2−+ Fe OH Fe OH FeSO Fe SO
Substituting _______________________:
+ +
+ +
+
= + − − − 4 − 4
3 3
2 2
1
3 ](1 [ ] [ ] [ ] [ ]
[ ]
[Fe T Fe β OH β OH β OH β OH
) ] [
] [
] [
] [
3 ] ][
[
2 22 3+ − 2 + 34 3+ 2 − 4 + 1 42− + 2 42− 2 + β Fe OH β Fe OH βS SO βS SO
Den Fe
unsat[ Fe ]
T]
[
3+=
____ balance on sulphur:
] [
] ) (
[ ] ) (
[ ] [
]
[ SO
4 T= SO
42−+ Fe SO
4 ++ Fe SO
4 2−+ HSO
4−a S
S
T
K SO H
Fe Fe
SO SO
] ] [
][
[ ]
[ 1
] ] [
[
2 4 3
2 3
1 2 4
4 +
− +
+
−
+ +
+
=
β β
Initial values of [Fe3+] and [SO42-] must be _______ and then ______ through __________. Constraints in _________ solution: [Fe3+] ≤ [Fe3+]T and [SO4
2-]
≤ [SO4
2-]T. In __________ solution: [Fe3+] ≤ Ks0/[OH-]3.
