chapter 7 complex formation equilibria
7.1.Introduction………………………………………………………………………………...1
7.2.Speciation and Distribution……………………………………………………………….1
7.2.1Metal-Ligand Systems…………………………………………………………..…..1
7.2.2Speciation and Distribution Diagrams…………………………………………..…..5
7.2.3Polynuclear Hydroxo Complexes…………………………………………………13
7.2.4Polyprotic Complexes……………………………………………………………...15
7.3Speciation and the Solubility of Metal Salts…………………………………………….20
7.3.1Complex Formation and Solubility………………………………………………..20
7.3.2Anion Speciation and Solubility…………………………………………………...21
7.3.3Metal Ion Speciation and Solubility……………………………………………….. 24
7.1.Introduction
Complexation reactions of a metal ion can be considered in terms of the displacement of anionic water molecules within the hydration sphere by anionic or neutral ligands:
M(H2O)+ L = M(H2O)5Lz+ + H2O (7.1)
where L is a neutral ligand in this case. It can be seen from the above expression that the hydrated metal ion also constitutes a complex, with the water molecules as the ligands. Similarly the hydroxyl ions associated with the hydrolysis products encountered in Chapter 5, are also ligands.
The structural aspects of metal-ligand complexes were treated in Chapter 3. In the present chapter we direct our attention to the quantitative description of the equilibrium reactions involving metal complexes. We shall also consider the reactions of protolyzable species such, as polyprotic acids (HnA) and polyamines (e.g., hydrazine, NH2-NH2). The reactions of interest here include those between aqueous species, as well as reactions between aqueous species and solid compounds.
7.2Speciation and Distribution
7.2.1Metal-Ligand Systems
It is convenient to use a short-hand expression where the water molecules are not explicitly indicated:
Mz+ + L = MLz+(7.2)
This convention is adopted in the rest of our discussion. The following are examples of complexation reactions:
Ag+ + OH-=AgOH(aq) (hydroxo complex)(7.3a)
Ni2+ + 6NH3=Ni(NH3)(7.3b)
Co2+ + 4Cl-=CoCl(7.3c)
Au+ + 2CN- =Au(CN)(7.3d)
It will be noticed that in the above equations, a variety of ligand numbers is indicated ranging from one hydroxyl ligand for AgOH(aq) to six ammonia ligands for Ni(NH3). In general the formation of complexes proceeds in a step-wise manner, with each consecutive reaction governed by its own equilibrium constant:
Mz+ + L= Mz+K1(7.4a)
MLz+ + L= MLK2(7.4b)
ML+ L= MLK3, etc(7.4c)
where
Ki = [ML]/[ML]L](7.5)
The constants K1, K2... etc are called stability constants, in recognition of the fact that they provide a measure of the stability of the respective complexes. Thus a large equilibrium constant indicates a highly stable complex. Corresponding to the consecutive stability constants, Ki, are overall stability constants,i, defined as:
Mz+ + L=MLz+ 1(7.6a)
Mz+ + 2L =ML2(7.6b)
Mz+ + 3L =ML3(7.6c)
where
i = K1K2...Ki= [ML]/[Mz+][L]i(7.6)
Table 7.1 provides a collection of selected stability constants for various complexes. Several factors account for the differences in stability constants, among which are (a) properties of the central metal ion (ionic size, ionic charge, electronegativity), (b) properties of the ligand (size, ionic charge, basic character, extent of chelation, steric effects, size of chelate ring).
Table 7.1 Stability constants for selected metal-ligand complexes
Knowledge of the relevant equilibrium or stability constants allows us to calculate the concentrations of all the individual complexes for any given solution conditions; the resulting set of concentrations is called the speciation of the aqueous phase for those particular conditions. Consider a system consisting of a metal ion Mz+ and a ligand L, with overall stability constants i as indicated by Equations 7.4a-c, and 7.6. The concentrations of the various complexes may be written as:
[MLz+] = 1[L] [Mz+](7.7a)
[ML] = 2[L]2[Mz+](7.7b)
[ML] = i[L]i[Mz+](7.7c)
[ML] = n[L]n[Mz+](7.7d)
If conditions are such that the metal ion is not significantly hydrolyzed, then the mass balance on metal is:
CM = [Mz+] + [MLz+] + [ML] + ..... + [MLiz+] +...
= [Mz+] + i[L][Mz+] + 2[L]2[Mz+] + ..... +i[L]i[Mz+] + .....
Similarly a mass balance on the ligand is given by:
CL = [L] + [MLz+] + 2[ML] + ... + i[ML] + ... + n[ML]
= [L] + i[L][Mz+] + 22[L]2[Mz+] + ... + ii[L]i[Mz+ ] + ... + nn[L]n[Mz+]
A typical kind of problem encountered in aqueous systems is the following: Given total dissolved metal and ligand concentrations (i.e., CM and CL), determine the speciation; i.e., determine the equilibrium concentrations [Mz+], [L], [ML]. It can be seen from Equations 7.7a-d that the concentrations of the various complexes can be readily determined once the free metal ion and free ligand concentrations (i.e., [Mz+ and [L]) are known. The trick is how to determine [Mz+] and [L] given CM and CL.
We begin with a simple system. Only one complex forms, i.e., the monocomplex MLz+. Furthermore, let us assume that there is excess ligand. That is, CM < CL. Then the mass balance on metal (Equation 7.8) becomes:
CM = [Mz+] + [MLz+](7.10)
The mass balance on ligand (Equation 7.9) now simplifies to:
CL = [L] + [MLz+](7.11)
It follows from Equations 7.7a and 7.10 that:
[Mz+] = CM/{1 + 1[L]}(7.12)
[MLz+] = 1[L]CM/{1 + 1[L]}(7.13)
Substitution of Equation 7.13 in Equation 7.11 gives the following quadratic equation:
[L]2 + (1 + 1CM - 1CL)[L] - CL = 0(7.14)
Equation 7.14 can be solved readily with the quadratic formula.
The above analysis is greatly simplified for the special case where the ligand is present in excess. That is CL > CM. Under these conditions the chloride mass balance becomes:
CL = [L] (7.15)
Therefore [Mz+] and [MLz+] are readily determined by substituting CL for [L] in Equations 7.12 and 7.13.
Example 7.1 Speciation in the Systems Mg-F-H2O, Sr-F-H2O, and Ba-F-H2O
The following data are provided for the M-F-H2O systems:
Mg2+ + F- = MgF+logK = 1.32, I = 1.0
Sr2+ + F- = SrF+logK = 0.1, I = 1.0
Ba2+ + F- = BaF+logK = -0.3, I = 1.0
For each M-F-H2O system, determine the speciation of a solution containing 0.1 mol/L fluoride and 1.0 x 10-4 mol/L metal.
Solution
Here CL = 10-1 mol/L > CM = 10-4 mol/L. Therefore it follows that [L] = CL = 10-1 mol/L and from Equations 7.12 and 7.13,
[M2+] = (10-4)/{1 + K (10-1)}
[MF+] = K(10-1)(10-4)/{1 + K (10-1)} = 10-5 K/(1 + 10-1K)
(i)For Mg(II), logK = 1.32, i.e. K=20.9. Thus
[Mg2+] = (10-4)/{1 + (20.9)(10-1)} = 3.24 x 10-5 mol/L.
From mass balance (see Equation 4.20),
[MgF+] = CMg - [Mg2+] = 1.0 x10-4 - 3.24 x 10-5 = 6.76 x 10-5 mol/L.
(ii)For Sr(II), logK = 0.1, i.e. K=1.26:
[Sr2+] = (10-4)/{1 + (1.26)(10-1)} = 8.88 x 10-5 mol/L.
[SrF+] = 1.12 x 10-5 mol/L.
(iii)For Ba(II), logK = -0.3, i.e. K = 0.50
[Ba2+] = (10-4)/{1 + (0.50)(10-1)} = 9.52 x 10-5 mol/L.
[BaF+] = 0.48 x 10-5 mol/L.
7.2.2Speciation and Distribution Diagrams
In Section 5.5 the concept of graphical representation of ionic equilibria was introduced. We now present further elaboration. The analysis here is complicated by the effects of stepwise complexation. However the basic approach is similar to that developed in Chapter 5. Referring to the metal-ligand system discussed in Section 7.2.1, we can define the following fractions:
i = [ML]/CM(7.16)
where 0 i n. It follows from Equations 7.8 and 7.16 that
Also from Equations 7.7a-d, 7.16 and 7.17,
i = i[Li]io(7.18)
where 1 i n.
The i values can be displayed graphically in a variety of ways. A speciation diagram consists of curves of i vs log[L] or vs pL (= -log[L]). A distribution diagram presents plots of fj vs log[L] or vs pL, where
and 0 j n. A formation curve provides a plot of the average number of ligands () bound per central metal ion as a function of log[L] or pL. The quantity is defined as
= (CL - [L])/CM(7.20)
It follows from Equations 7.9, 7.18, and 7.20 that
Another kind of plot is that involving a plot of the concentration of free ligand as a function of the analytical concentration of dissolved metal. The appropriate expression needed to prepare this plot is readily obtained by rearranging Equation 7.20 as:
[L]/CM = (CL/CM) - (7.22)
Finally, by following a procedure similar to that developed in Section 5.5.1, logarithmic concentration diagrams can be prepared
EXAMPLE 7.2 Speciation diagrams for the systems Mg-, Sr-, and Ba-F-H2O
Using the data presented in Ex.7.1, develop (a) speciation and (b) distribution diagrams for the Mg-F-H2O, Sr-F-H2O and Ba-F-H2O systems.
Solution
According to Equations 7.12 and 7.17,
o = [Mz+]/CM = 1/{1 + 1[L]}(1)
Similarly, from Equations 7.13 and 7.16,
1 = [MLz+]/CM = 1 [L]/{1 + 1 [L]}(2)
It follows from Equations 1 and 2 that o = 1 when
[L] = 1/1(3)
Under conditions where the total concentration of dissolved ligand, CL, is relatively high, i.e., CL > CM, [L] in Equations 1-3 can be replaced by CL. Thus in the case of the Mg-F-H2O system,
o = [Mg2+ ]/CMg = 1/{1 + 20.9 CF}(4)
1 = [MgF+]/CMg = 20.9CF/{1 + 20.9CF}(5)
where CF is the total concentration of dissolved fluoride under conditions where CF > CMg. Similar expressions may be derived for Sr and Ba.
The relevant diagrams are presented in Figures E7.2 a, b, and c respectively for the systems Mg-, Sr-, and Ba-F-H2O. It can be seen that the fluoride concentration where o = 1 increases in the order Mg < Sr < Ba. That is, MgF+, with the highest stability constant requires the lowest fluoride ion concentration to stabilize the corresponding fluorocomplex.
Figure E7.2Speciation diagrams for the systems (a) Mg-F-H2O, (b) Sr-F-H2O, and
(c) Ba-F-H2O.
EXAMPLE 7.3 Speciation and distribution diagrams for the Ag(I) - NH3-H2O system
The stability constants of Ag(I) ammine complexes have been determined at 25°C and an ionic strength of 2.0 mol/L as:
Ag+ + NH3 = AgNH3+log1 = 3.26(1)
Ag+ + 2NH3 = Ag(NH3)log2 = 7.20(2)
With the aid of these data generate the relevant speciation and distribution diagrams.
Solution
According to Equations 1 and 2,
[AgNH3+] = 1[NH3][Ag+](3)
[Ag(NH3)2+] = 2[NH3]2[Ag+](4)
Thus the total dissolved silver concentration, CM, is given by
CM = [Ag+] + [AgNH3+] + [Ag(NH3]2+
= [Ag+]{1 + 1[NH3] + 2[NH3]2}(5)
It follows from Equation 5 that
o = [Ag+]/CM = 1/{1 + 1[NH3} + 2{NH3]2}(6)
Also the respective fractions of the ammine complexes are given by:
1 = [AgNH3+]/CM = 1[NH3]o(7)
2 = [Ag(NH3)2+]/CM = 2[NH3]2o(8)
Using the stability constants provided, and Equations 6, 7, and 8, values of o, 1, and 2 can be determined as a function of [NH3]. Note that [NH3] is not the total ammonia concentration, rather it represents the concentration of free (i.e. unbound) ammonia. Figure E7.3a presents the resulting speciation diagram. Recalling Equation 7.19, the corresponding distribution diagram is obtained by evaluating the functions:
The resulting distribution diagram is shown in Figure E7.3b.
Figure E7.3: The Ag-NH3-H2O system: (a) Speciation diagram; (b) Distribution diagram.
Example 7.4 Speciation Diagram for the Ag- OH- -H2O System
Prepare a speciation diagram (i vs pH) for the Ag-OH-H2O system, given the following data:
Ag+ + H2O = AgOH(aq) + H+log*1 = - 12.0(1)
Ag+ + 2H2O = Ag(OH)+ 2H+log*2 = - 24.0 (2)
Solution
If follows from Equations 1 and 2 that
*1 = [AgOH(aq)][H+]/[Ag+](3)
*2 = [Ag(OH)][H+]2/[Ag+](4)
Thus
[AgOH(aq)] = (*1/[H+])[Ag+](5)
[Ag(OH)] = (*2/[H+]2)[Ag+](6)
Accordingly we can write the mass balance on total dissolved silver as:
CM = [Ag+] + [AgOH(aq)] + [Ag(OH)]
= [Ag+] {1 + (*1/[H+]) + (*2/[H+]2)}(7)
Therefore
o = [Ag+]/CM = 1/{1 + (*1/[H+]) + (*2/[H+]2)}(8)
Also from Equations 5, 6, and 8,
1 = [AgOH(aq)]/CM = (*1/[H+])o(9)
2 = [Ag(OH)]/CM = (*2/[H+]2)o(10)
Figure E7.4 presents the required speciation diagram. It can be seen that as pH increases, the predominant form of dissolved silver changes in the sequence.
Ag+ AgOH(aq) Ag(OH).
Figure E7.4 Speciation diagram for the Ag(I)-OH--H2O system
example 7.5 speciation Diagram for the Zn-OH--H2O System
Given the following equilibrium data, calculate as a function of pH, the concentrations of all species present in a 10-3M Zn(NO3)2 solution.
Zn2+ + OH-=ZnOH+log 1= 5.0
Zn2+ + 2OH-=Zn(OH)2(aq)log 2= 11.1
Zn2+ + 3OH-=Zn(OH)log3= 13.9
Zn2++ 4OH-=Zn(OH)log4= 15.1
Solution
Let us obtain equilibrium expressions in terms of H+:
Zn2+ + OH-=ZnOH+log 1 = 5.0
H2O=H+ + OH-log Kw =-14.0
Zn2+ + H2O=ZnOH+ + H+log*1 = -9.0
*1=[ZnOH+][H+]/[Zn2+] (1a)
[ZnOH+]=(*1/[H+])[Zn2+] (1b)
Zn2+ + 2OH-=Zn(OH)2(aq)log2 = 11.1
2H2O=2H+ + 2OH-2logKw = -28.0
Zn2+ + 2H2O=Zn(OH)2(aq) + 2H+log*2 = -16.9
=[Zn(OH)2(aq)][H+]2/[Zn2+] (2a)
[Zn(OH)2(aq)] = (*2/[H+]2)[Zn2+] (2b)
Zn2+ + 3OH-=Zn(OH)log 3 = 13.9
3H2O= 3H+ + 3OH-3 log Kw = -42.0
Zn2+ + 3H2O=Zn(OH)+ 3H+log*3 = 28.1
*3=[Zn(OH)][H+]3/[Zn2+] (3a)
[Zn(OH)]= (*K3/[H+]3)[Zn2+] (3b)
Zn2+ + 4OH-= Zn(OH)log 4 = 15.1
4H2O =4H+ + 4OH-4 log Kw = - 56.0
Zn2+ + 4H2O=Zn(OH)+ 4H+log*4 = -40.9
*4=[Zn(OH)][H+]4/[Zn2+] (4a)
[Zn(OH)]=(*K4/[H+]4) [Zn2+] (4b)
The aqueous complexes must satisfy the following mass balance on total dissolved zinc:
[Zn]T = [Zn2+] + [ZnOH+] + [Zn(OH)2(aq)] + [Zn(OH)] + [Zn(OH)]
i.e.[Zn]T = [Zn2+] (1 + *1/[H+] + *2/[H+]2 + *3/[H+]3 + *4/[H+]4)
We can define the following distribution ratios:
o = [Zn2+]/[Zn]T=1/(1+*1/[H+] + *2/[H+]2 + *3/[H+]3 + *4/[H+]4)
1 = [ZnOH+]/[Zn]T=(*1/[H+])[Zn2+]/[Zn]T = (*1/[H+])0
2 =[Zn(OH)2(aq)]/[Zn]T= (*2/[H+]2)0
3 = [Zn(OH)]/[Zn]T=(*3/[H+]3)0
4 = [Zn(OH)]/[Zn]T=(*4/[H+]4)0
Now, log*1 = -9.0, therefore *1 = 10-9. Similarly, *2 = 1.26 x 10-17,
Figure E7.5. Speciation diagram for the Zn-OH--H2O system.
= 7.94 x 10-29 and = 1.26 x 10-41. Let
A = *1/[H+] = 10-9/[H+]
B = *2/[H+]2 = 1.26 x 10-17/[H+]2
C = *3/[H+]3 = 7.9x 10-29/[H+]3
D = *4/[H+]4 = 1.26 x 10-41/[H+]4
Then,
0 = 1/(1 + A + B + C + D)
1 = Ao, = Bo, 3 = Co, 4 = Do
These expressions may be used to calculate 0, 1, 2, 3 and 4 as a function of pH. The resulting speciation diagram is presented in Figure E7.5.
7.2.3Polynuclear Hydroxo Complexes
In Chapter 5.1.3 it was found that hydroxo complexes can combine to form dimers and higher aggregates. Such aggregates are called polynuclear complexes and the hydroxyl ion is termed a bridging ligand. A molecule or ion can serve as a bridging ligand if it contains an atom that possesses at least two pairs of free electrons. For the M2+, M3+, and M4+ cations, the hydroxyl ion is the most common bridging ligand. For the higher valent cations (M5+, M6+), polyanions form that typically involve O2- bridges. A summary of the structures of the polynuclear hydroxo complexes was presented above in Table 5.1. Table 7.2 presents stability constants and solubility products for selected M-OH--H2O systems.
Consider the case where a metal cation Mz+ forms the hydroxo species Mx(OH)y(xz-y):
xMz+ + yOH- = Mx(OH)xy(7.23)
where
xy = [Mx(OH)]/[Mz+]x [OH-]y(7.24)
Equation 7.23 can be expressed alternatively in terms of protons as:
xMz+ + yH2O = Mx(OH)+ yH+(7.25)
where
*xy = [Mx(OH)][H+]y[Mz+]x(7.26)
Table 7.2 Stability Constants and Solubility Products for Selected Metal-Water Systems
ReactionlogK(I=3.0)logK(I=0)
Gd3+ + OH- = GdOH2+ 5.0 -
2Gd3+ + 2OH- = Gd2(OH) 14.13 -
Gd(OH)3(s) = Gd2+ + 3OH- - -25.7
Mg2+ + OH- = MgOH+ 1.85 2.58
4Mg2+ + 4OH- = Mg4(OH) 16.93 16.3
Mg(OH)2(s) = Mg2+ + 2OH- - -11.15
V3+ + OH- = VOH2+ 11.1 11.7
2V3+ + 2OH- = V2(OH) 23.43 24.2
2V3+ + 3OH- = V2(OH) 34.5 -
V(OH)3(s) = V3+ + 3OH- - -34.4
VO2+ + OH- = VO(OH)+ 7.9 8.3
VO2+ + 2OH- = VO(OH)2(aq) 18.31 -
2VO2+ + 2OH- = (VO)2(OH) 28.35 21.3
VO(OH)2(s) = VO2+ + 2OH- - -23.5
Y3+ + OH- = YOH2+ 5.1 6.3
2Y3+ + 2OH- = Y2(OH) 14.06 13.8
3Y3+ + 5OH- = Y3(OH) 37.1 38.4
Y(OH)3(s) = Y3+ + 3OH- - -23.2
La3+ + OH- = LaOH2+ 4.1 5.5
2La3+ + OH- = La2OH5+ (4.2) -
5La3+ + 9OH- = La5(OH) 56.2 54.8
La(OH)3(s) = La3+ + 3OH- - -20.7
Recall that the equilibria involving the monomeric species are described by
Mz+ + iH2O = M(OH)+ iH+(7.27)
where
*i = [M(OH)][H+]i/[Mz+](7.28)
It follows from Equations 7.26 and 7.28 that the mass balance on total dissolved metal is given by:
Thus the fraction of total dissolved metal present as the hydrated metal ion Mz+ can be expressed as:
With the aid of Equation 4.37, the corresponding fractions for the other dissolved species can be determined, i.e.,
= (*i/[H+]i) o(7.31b)
xy= [Mx(OH)]/CM = xxy/[H+]y) [Mz+]x/CM(7.32a)
= x(*xy/[H+]y) [Mz+]x-1 o(7.32b)
It can be seen from Equations 7.30-7.32 that the concentration fractions are functions of [H+] as well as [Mz+]. This is in contrast to the case of systems that are free of polynuclear hydroxo complexes where there is no Mz+ dependence (see Examples 7.4 and 7.5).
7.2.4Polyprotic Complexes
A polyprotic acid can ionize with release of two or more protons. Thus in the case of a triprotic acid, H3A, the stepwise ionization can be represented as:
H3A(aq) = H2A- + H+ K1(7.33a)
H2A- = HA2- + H+K2(7.33b)
HA2- = A3- + H+K3(7.33c)
Table 7.3 Ionization Constants of Selected Polyprotic Acids
CompoundChemical FormulapK1pK2pK3pK4
O
||
Carbonic acid (H2A)H2CO3; HO-C-OH 6.35 10.33 - -
O O
|| ||
Oxalic acid (H2A)H2C2O4; HO-C-C-OH1.25 4.27 - -
OO
|| ||
Tartaric acid (H2A)HO-C-CH-CH-C-OH 3.17 4.91 - -
| |
HOOH
OOHO
|| | ||
Citric acid (H3A) HO-C-CH2-C-CH2-C-OH 3.13 4.76 6.40 -
|
O = C - OH
Ethylene diamineHOOC-CH2CH2COOH 1.99 2.67 6.16 10.26
tetra-acetic acid
(EDTA)(H4A)N-(CH2)2-N (Butler, p. 209)
HOOC-CH2CH2-COOH
Hydrogen phosphiteHO
(Phosphorous acid) (H2A)H — P = O 1.5 6.79 - -
HO
Hydrogen phosphateHO
(Phosphoric acid) (H3A)HO — P = O 2.15 7.20 12.35 -
HO
OO
|| ||
Hydrogen diphosphateHO-P-O-P-OH 0.8 2.2 6.70 9.40
(Pyrophosphoric acid) (H4A)| |
HOOH
Hydrogen arsenateH3AsO4 2.24 6.96 11.50 -
(Arsenic acid) (H3A)
Hydrogen sulfide (H2A)H2S 7.02 13.9 - -
Hydrogen thiosulfate (H2A)H2S2O3 0.6 1.6 - -
O
||
Hydrogen sulfiteH2SO3;HO-S-OH 1.91 7.18 - -
(Sulfurous acid) (H2A)
Hydrogen selenide (H2A)H2Se 3.89 15.0 - -
Data from Martell and Smith, Critical Stability Constants
In the mathematical analysis of ionic equilibria in systems containing polyprotic acids, it is convenient to view the anion as the central ion and the proton as the ligand. It can be seen by comparing Equations 7.4a-c and 7.33a-c that the successive protonation of the central anion is mathematically similar to the stepwise addition of a ligand to a central metal ion. Table 7.3 presents a selection of polyprotic acids with their corresponding stability constants.
It is useful to include protolyzable acids and polyamines under this section since ionic equilibria involving these compounds are described by a mathematical formalism similar to that for polyprotic acids. Stability constants for selected protolyzable acids and polyamines are presented in Table 7.4. To illustrate the ionization of these compounds, hydrazine (NH2-NH2) may be represented as A and the protonated products as HA+ and H2A2+; the corresponding ionization equilibria may be depicted as:
H2A2+ = HA+ + H+K1(7.34a)
HA+ = A + H+K2(7.34b)
Similarly salicylaldoxime may be represented as H2A with the following ionized species A2-, HA- and H3A+:
H3A+ = H2A + H+K1(7.35a)
H2A = HA- + H+K2(7.35b)
HA- = A2- + H+K3(7.35c)
Table 7.4 Ionization Constants for Selected Protolyzable Acids and Polyamines
CompoundChemical FormulapK1pK2pK3
Salicylaldoxine (H2A, H3A+)O -C-H O1.379.1812.11
| ||
OHNOH
8-Hydroxyquinoline (Oxine;O O 4.919.81-
HA, H2A+) |N
OH
Ethylenediamine (en;NH2CH2CH2NH26.859.93-
A, H2A2+)
Hydrazine (A, H2A2+)NH2-NH2 -0.97.98-
Data from Smith and Martell, Critical Stability Constants
Following a procedure similar to that presented for metal complexes, speciation and distribution diagrams can be developed for polyprotic acids. As an illustration, Equations 7.33a - 7.33c can be rewritten in terms of overall ionization reactions as:
H3A(aq) = H+ + H2A-K1' = K1(7.36a)
H3A(aq) = 2H+ + HA2-K2' = K1K2(7.36b)
H3A(aq) = 3H+ + A3-K3' = K1K2K3(7.36c)
It follows from Equations 7.36a - 7.36c that:
[H2A-]/[H3A(aq)] = K1'/[H+](7.37a)
[HA2-]/[H3A(aq)] = K2'/[H+]2(7.37b)
[A3-]/[H3A(aq)] = K3'/[H+]3(7.37c)
A mass balance on total dissolved acid can be written as:
CA = [H3A(aq)] + [H2A-] + [HA2-] + [A3-](7.38)
The fraction of the total dissolved acid that exists in the form of the undissociated neutral molecule H3A(aq) is given by
3 = [H3A(aq)]/CA(7.39)
Equations 7.38 and 7.39 may be combined to give:
1/3 = CA/[H3A(aq)] = 1 + [H2A-]/[H3A(aq)] + [HA2-]/[H3A(aq)] + [A3-]/[H3A(aq)](7.40)
Further, substitution of Equations 7.37a - 7.37c in Equation 7.40 gives 3 as:
3 = [H3A(aq)]/CA = 1/{1 + K1'/[H+] + K2'/[H+]2 + K3'/[H+]3}(7.41)
Once 3 is known, the remaining fractions are readily determined. From Equations 7.37a and 7.38,
2 = [H2A-]/CA = ([H2A-]/[H3A(aq)])([H3A(aq)]/CA)
= (K1'/[H+]) 3(7.42a)
Similarly, from Equations 7.37b and 7.38,
1 = [HA2-]/CA = (K2'/[H+]2) 3(7.42b)
Finally, from Equations 7.35c and 7.36,
o = [A3-]/CA = (K3'/[H+]3)3(7.42c)
______
EXAMPLE 7.6 Speciation and distribution diagrams for arsenic acid.
Prepare speciation and distribution diagrams for arsenic acid, using the ionization constants provided in Table 7.3.
Solution
According to the data presented in Table 7.3,
H3AsO4(aq) = H+ + H2AsOlogK1 = -2.24(1)
H2AsO= H+ + HAsOlogK2 = -6.96(2)
HAsO= H+ + AsOlogK3 = -11.50(3)
Figure E7.6: (a) Speciation diagram for arsenic acid. (b) Distribution diagram for arsenic acid.
Comparing Equations 1-3 with Equations 7.33a - 7.33c, it follows from Equations 7.36a - 7.36c that:
K1' = K1 = 5.75 x 10-3(4)
K2' = K1K2 = 6.31 x 10-10(5)
K3' = K1K2K3 = 2.0 x 10-21(6)
Inserting these numerical values of K1', K2' and K3' into Equations 7.41 to 7.42c, the fractions o, 1, 2, and 3 can be determined as a function of pH. The resulting speciation and distribution diagrams are presented in Figures E7.6a and E7.6b respectively.
______
7.3.Speciation and the Solubility of Metal Salts
7.3.1Complex Formation and Solubility
In Sections 5.2 and 5.4 the solubility of oxides and hydroxides was discussed without considering the effects of hydroxo complexes. Similarly the discussion of the solubility of metal salts in Chapter 5 ignored the effects of metal complexes as well as the acid-base reactions of the anions of the salts. We now incorporate these effects into the analysis of solubility equilibria.
In connection with the effects of anion speciation on the solubility of a salt MA, we can distinguish between three different situations respectively, where (a) Az- is the anion of a strong acid, (b) Az- is the anion of a monoprotic acid, and (c) Az- is the anion of a polyprotic acid. In case (a) there is no need to consider anion speciation. On the other hand, in connection with metal complexation, we can envision the following different situations: (a) Mz+ is uncomplexed, (b) Mz+ is complexed by the anion of the salt (i.e. A-), and (c) Mz+ is complexed by a foreign ligand. Here case (b) includes the situation where the salt is a metal hydroxide and therefore the resulting complex is a hydroxo species.
7.3.2 Anion Speciation and Solubility
Consider a metal salt MA which dissolves to give Mz+ and Az- ions:
MA (s) = Mz+ (aq) + Az- (aq) Ks0(7.43)
Suppose Az- is the anion of a weak acid. Then protonation reactions contribute to the overall equilibria:
Az- + H+ = HA(1-z)K1(7.44)
HA(1-z) + H+ = H2A(2-z)K2(7.45)
......
Hn-1A(n-1-z) + H+ = HnA(n-z)Kn(7.46)
It follows from Equations 7.44 - 7.46 that
[HA(1-z)]=K1 [H+] [Az-]=1 [H+] [Az-](7.47)
[H2A(2-z)]=K1K2 [H+]2[Az-]=2 [H+]2 [Az-](7.48)
......
[HnA(n-z)]=K1K2..... Kn[H+]n[Az-]=n [H+]n [Az-](7.49)
If the solubility of the salt is S, then the relevant mass balances are:
[Mz+] =S(7.50)
[Az-] +[HA(1-z)] + ...... + [HnA (n-z)] = S(7.51)
Inserting Equations 7.47-7.49 in Equation 7.51 gives:
Now, the solubility product, Ks0 is given by
Ks0 = [Mz+] [Az-](7.53)
Therefore, by combining Equations 7.50, 7.52, and 7.53, we get
That is,
It can be seen from Equation 7.55 that as a result of the protonation of the anion, the solubility of the salt increases with increase in the acidity of the aqueous solution.
EXAMPLE 7.7 Solubility of the Salt of a Weak Acid
Ignoring metal ion hydrolysis and the possible formation of insoluble metal hydroxide determine the pH dependence of the solubility of the following metal salts: (a) CdCO3(s), (b) CaSO3(s). The relevant equilibrium data are provided below.
CdCO3(s)=Cd2+ + COlogKs0= -13.74
CO+ H+=HCOlogK1= 10.33
HCO+ H+=H2CO3 (aq)logK2= 6.35
CaSO3(s)=Ca2+ + SOlogKs0= -6.5
SO+ H+=HSOlogK1= 7.18
HSO+ H+=H2SO3 (aq)logK2= 1.91
Solution
(a)Let the solubility of CdCO3(s) be S. Then following Equations 7.50 and 7.51, the following mass balances can be written:
[Cd2+] = S(1)
[CO32-] + [HCO3-] + [H2CO3(aq)] = S(2)
Also from Equations 7.47 and 7.48, and the given thermodynamic data,
[HCO3-] = 1[H+][CO32-] (3)
[H2CO3(aq)] = 2[H+]2[CO32-](4)
where 1 = K = 2.14 x 1010 and 2 = K1K2 = 4.79 x 1016.
Combining Equations 2-4 gives:
S = [CO32-]{1 + 1[H+] + 2[H+]2}(5)
Thus
[CO32-] = S/{1 + 1[H+] + 2[H+]2}(6)
Also, the solubility product is given by
Ks0 = [Cd2+][CO32-](7)
It follows therefore from Equations 1, 6 and 7 that
Ks0 = S2/{1 + 1[H+] + 2[H+]2}(8)
Thus
S = Ks01/2 {1 + 1[H+] + 2[H+]2}1/2(9)
In very acidic solution, Equation 9 can be approximated as:
S ≈ Ks01/221/2 [H+]2/2(10)
That is,
logS ≈ log(Ks01/22) - 2pH(11)
For intermediate acidities,
S ≈ Ks01/21[H+](12)
or,
logS ≈ log(Ks01/21) - pH(13)
Finally, for relatively low acidities,
S ≈ Ks01/2(14)
and therefore,
logS ≈ 1/2 logKs0(15)
Figure E7.7 presents a plot of logS versus pH. In accordance with Equations 11, 13, and 15, it can be seen that at low pH the solubility decreases rapidly with pH but becomes independent of pH at relatively high pH.
Figure E7.7 Solubility of CdCO3(s) as a function of pH.
7.3.3 Metal Ion Speciation and Solubility
Let us first consider the case where the insoluble salt is MA and metal ion (Mz+) is complexed by the anion (Az-) of the salt. To simplify our analysis, we will assume that Az- is the anion of a strong acid, i.e. we do not need to consider protolytic reactions of the anion. The complexation reactions of the metal ion may described thus:
Mz++Az-=MA(aq)1(7.56)
Mz++2Az-=MA2(7.57)
......
Mz++nAz-=MAn(7.58)
where
[MA(aq)]=1[Az-] [Mz+](7.59)
[MA]=2[Az-]2 [Mz+](7.60)
......
[MA]=n[Az-]n [Mz+](7.61)
The relevant mass balances are:
[Mz+] + [MA(aq)] + ...... + [MA] = S(7.62)
It follows from Equations 7.59-7.62 that
The solubility product (Equation 7.53) is then given by
Equation 7.64 may be rearranged to give the solubility as:
It is instructive to consider two limiting forms of Equation 7.65. When [Az-]is relatively low, the solubility expression becomes:
S = Ks0/[Az-](7.66)
Thus at low anion concentration, the solubility decreases with increasing anion concentration. On the other hand, when [Az-] is relatively high, Equation 7.65 reduces to:
It can be seen from Equation 7.67 that if n=1, then the solubility is given by
S = Ks0 1(7.68)
That is, the solubility becomes independent of the anion concentration. In contrast, when n is greater than one, Equation 7.67 indicates that in this case the solubility increases with increase in the anion concentration. It follows therefore that when the anion of the salt forms complexes with the metal ion of the insoluble salt such that n > 1, then a minimum occurs in the curve of solubility versus anion concentration.