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CHAPTER 5. IONS IN SOLUTION
5.1
Metal Ions in Solution…………………………………………………………...... 1
5.1.1 Hydration and Hydrolysis of Metal Ions…………………………………… 1
5.1.2 Ionization of Water and Hydration of the Proton…………………………... 4
5.1.3 Polymerization and Precipitation…………………………………………… 5
5.1.4 Structures of Polycations and Polyoxoanions………………………………. 7
5.2
The Concept of Equilibrium Constant…………………………………………. 10
5.2.1 The Solubility Product…………………………………………………….. 10
5.2.2 The Free Energy of Reaction and the Equilibrium Constant……………… 11
5.2.3 Solubility Calculations……………………………………………………..14
5.2.4 Solubility of Sparingly Soluble Salts……………………………………… 15
5.2.5 Lattice Energy, Hydration Energy, and Solubility………………………... 17
5.3
Acids and Bases…………………………………………………………………... 18
5.3.1 Acidic and Basic Solutions………………………………………………... 18
5.3.2 Strong Acids and Bases…………………………………………………… 19
5.3.3 Weak Monoprotic Acids and Bases……………………………………….. 23
5.4
Structural Aspects of Acid-Base Strength…………………………………….... 25
5.4.1 Gas Phase Acidity…………………………………………………………. 25
5.4.2 Hydration Effects………………………………………………………….. 28
5.4.3 Strengths of Hydroxyl-Group Acids………………………………………. 29
5.5
Graphical Representations of Ionic Equilibria………………………………… 30
5.5.1 Logarithmic Concentration Diagrams…………………………………….. 30
5.5.2 Distribution Diagrams…………………………………………………….. 34
5.6
The Partial Charge Model and the Reactions of Metal Cations……………….34
5.6.1 The Acidity of Hydrated Metal Cations…………………………………... 34
5.6.2 The Mean Electronegativity of an Aqueous Solution……………………... 35
5.6.3 Hydrolysis of Metal Cations………………………………………………. 37
5.6.4 Polymerization…………………………………………………………….. 41
5.6.5 Precipitation……………………………………………………………….. 44
5.1. Metal Ions in Solution
5.1.1 Hydration and Hydrolysis of Metal Ions
In the aqueous processing of metals, we are concerned with two major questions:
(i) What happens to metal ions in solution? and (ii) What happens to a solid placed in an
aqueous environment? Let us for a moment focus our attention on the water molecules
which constitute the solvent phase of the aqueous solution. Two hydrogen atoms and an
oxygen atom come together to form a water molecule by sharing their outer electrons:
oo
oO o
o o
+
H +
H
=
o oH
o

oO o
o 
H
Each pair of shared electrons constitutes a covalent bond.
(5.1)
It can be seen from this
simplified picture that in a water molecule, the oxygen atom is surrounded by 8 outer
electrons, four of which are involved in covalent bonding and 4 of which are unshared. We
can add to this picture, the fact that oxygen has a higher affinity than hydrogen for
electrons. Thus in a water molecule, the electron cloud is denser around the oxygen
nucleus than around the hydrogen nuclei. This effect is acknowledged by ascribing a
dipolar structure to the water molecule:
H+
(-) O
H+
Thus like a magnet, the water molecule has a positive and a negative pole.
Now let us consider a metal ion immersed in this sea of dipoles. Let us take the
simplest case, i.e., the hypothetical situation of a metal ion Mz+ immersed in water with no
other ions being present.
Through its charge, this ion exerts some influence on the
surrounding water molecules. The metal ion is positively charged and by analogy with the
well known behavior of magnets, we would expect it to attract the negative pole of a water
molecule. This is called an electrostatic interaction:
H+
H

Mz+ + O-
Mz+
O
H+
(5.2)
H
Depending on the nature of the metal ion, several more water molecules may be attracted
towards it in a similar manner. Thus we say that metal ions are hydrated in aqueous
solution.
We would expect that this interaction will be most intense for the water
molecules which are nearest to the central metal ion.
These nearest neighbor water
molecules constitute the first or inner hydration sphere of the hydrated ion. The hydration
number is the number of such nearest neighbor water molecules, and is generally taken to
be equal to 6. Thus in aqueous solution, we can represent the dissolved metals ions as
M(H2O)6z+, e.g., Ni(H2O)62+, Al(H2O)63+, etc.
In addition to the attractive Mz+-O- interaction noted above, we must recognize that
a repulsive Mz+-H+ interaction also exists. In fact, we may go so far as to say that the
hydration of a metal ion really involves two simultaneous processes: (a) attraction of the
negative pole of the water molecule to the positively charged metal ion, and (b) repulsion
of the positive pole of the water molecule by the positive charge on the metal ion. These
two processes, in competition, can cause a splitting of the bound water molecule:
H+
-
M2+ + O
H+
M2+
H+
-
H+
M2+
O
-
O
H+
H+
Mz+  (OH)- + H+
(5.3)
The term hydrolysis describes processes such as Equation 5.3 in which the O-H
bonds of water are broken. The resulting OH--containing product is termed a hydroxo
metal complex. It is conventional practice to express the formation of a hydroxo complex
as indicated in Equation 5.4.
Mz+ + H2O = MOH(z-1)+ + H+
However, in view of our previous treatment of hydration, we can also write:
(5.4)
M(H2O)6z+ + H2O = M(H2O)5(OH)+ + H3O+
(5.5)
The hydrolysis process may not necessarily stop with the splitting of one water molecule; it
may involve the other water molecules in the inner hydration sphere. Thus we speak of
successive or stepwise hydrolysis:
M(H2O)62+
=
M(H2O)5(OH)+ + H+
M(H2O)5(OH)+
=
M(H2O)4(OH)2o + H+
M(H2O)4(OH)2o
=
M(H2O)3(OH)3- + H+
M(H2O)3(OH)3-
=
M(H2O)2(OH)42- + H+ etc.
(5.6)
5.1.2 Ionization of Water and Hydration of the Proton
A similar electrostatic process involving only water molecules, leads to the
ionization of water and the hydration of the proton. The ionization of water may be viewed
in terms of the transfer of a proton from a water molecule to a second water molecule,
yielding a hydrated hydrogen ion and a hydroxyl ion:
H
O+H
H
-
+H
O
-
H
O
+H
-
+H
O
-
+H
+H
H2O + H2O = H3O+ + OH-
O
O
+H
+H
+H
(5.7)
The species H3O+ is called the oxonium ion and it may be further hydrated to give
the trihydrated oxonium ion, H3O(H2O)3+:
H
H
O
H
H
O
O
H
H
H
H
O
H
In the above structure the dashed lines are used to emphasize the fact that the O-H bonds
between the surrounding water molecules and the oxonium protons are weaker than those
within the oxonium ion. The weaker bonds represent hydrogen bonds.
It is common practice to express the ionization of water without explicitly
indicating the hydration of the proton:
H2O = H+ + OH-
(5.8)
However there are situations where it is necessary to invoke the hydrated nature of the
proton.
5.1.3 Polymerization and Precipitation
The hydrolysis products encountered above may react further with each other to
give polymeric species:
M(H2O)5(OH)2+ + M(H2O)5(OH)2+ = [(H2O)4M-(OH)2-M(H2O)4]4+ + 2H2O
(5.9)
For example, Fe(H2O)5OH2+ can undergo the following dimerization reaction:
Fe(H2O)5OH2+ + Fe(H2O)5OH2+ =
OH
Fe(H2O)4]4+ + 2H2O
[(H2O)4 Fe
OH
(5.10)
In other words, the two OH groups form bridges between the two nuclei, i.e., the two Fe3+
ions.
These polynuclear species can in turn become hydrolyzed:
OH
OH
Fe(H2O)4]4+ = [(OH)(H2O)3Fe
[(H2O)4Fe
OH
Fe(H2O)3(OH)]2+ + 2H+ (5.11)
OH
The presence of this new hydrolyzed species can then lead to more polymerization:
OH
OH
Fe(H2O)3(OH)]2+ + [(OH)(H2O)3Fe
[(OH)(H2O)3Fe
OH
OH
OH
= [(OH) (H2O)3Fe
Fe(H2O)3(OH)]2+
OH
Fe(H2O)2
OH
OH
Fe(H2O)3(OH)]4+ + 2H2O (5.12)
(H2O)2Fe
OH
OH
Hydrolysis-polymerization reactions can proceed until finally the resulting
polymers become too large to remain soluble, i.e., they achieve colloidal dimensions.
When this stage is reached, we say we have precipitated a solid. In this hypothetical
system consisting of only one type of metal ion in water (i.e., in the absence of anions or
other metals ions), each polymerization reaction involves the elimination of water and
therefore ultimately we would expect metal hydroxides to be formed.
EXAMPLE 5.1 Hydrolysis reactions
For each of the polynuclear hydroxo species below, write down the corresponding chemical equation
that describes its formation via interaction of Mz+ and H2O.
(a)
Be6(OH)84+
(b) Zr4(OH)88+
Solution
(a) 6Be2+ + 8H2O = Be6(OH)84+ + 8H+
(c) Mg4(OH)44+
(d) Co2OH3+
(b) 4Zr4+ + 8H2O = Zr4(OH)88+ = 8H+
(c) 4Mg2+ + 4H2O = Mg4(OH)44+ + 4H+
(d) 2Co2+ + H2O = Co2OH3+ + H+
5.1.4 Structures of Polycations and Polyoxoanions
The species of interest here are those in which metal ions are linked by hydroxyl
(M-OH-M) and/or oxo (M-O-M) bridges. In the case of complexes based on M(II), M(III),
and M(IV) atoms, the hydroxyl bridge is used almost exclusively. Table 5.1 presents a
summary of the structural information on these species. (see Baes and Messmer, p. 420).
It can be seen that symmetrical structures are preferred. In these structures, up to six
cations are organized into groups of two, three, four, or six and the coordination number of
the hydroxyl ion may be 2 or 3. The structure of the square-planar M4(OH)8(H2O)16
complex is illustrated in Figure 5.1a (B&M, p. 157). Adjacent metal cations are linked by
a double bridge of hydroxyls. Each metal ion is eight-coordinate - through the oxygens of
four hydroxyls and four water molecules. The M6(OH)84+ and M6(OH)126+ structures are
illustrated respectively in Figures 5.1b and 5.1c. (See Baes & Mesmer, p.378) In addition
to the species listed in Table 5.1, Al(III) forms the complex Al13O4(OH)247+. This species
(Figure 5.1d; see Baes and Mesmer, p.118) consists of a central AlO45- tetrahedron
surrounded by twelve AlO6 octahedra; the octahedra are linked by shared edges.
In the case of the M(V) and M(VI) atoms, O2- bridges are prominent. The elements
which form these polyoxoanions come predominantly from the left-side of the d-block:
V(V), Nb(V), Ta(V), Cr(VI), Mo(VI), and W(VI). Table 5.2 shows that the polymerization
ranges from dimers to 19-mers. The typical structure consists of MO6 octahedra linked at
the corners or edges, as illustrated in Figure 5.2 (Huheey).
Table 5.1 Polynuclear hydroxo complexes (Baes and Mesmer, p. 420)
Species
Probable structure
= Mz+, o = OH-
Cation (source)
M2OH3+
M4(OH)88+
Be2+, Mn2+, Co2+, Ni2+,
Zn2+, Cd2+, Hg2+, Pb2+
Cu2+, Sn2+, UO22+, NpO22+, PuO22+,
VO2+,
Al3+ Sc3+, Ln3+, Ti3+, Cr3+
Th4+
Be2+, Hg2+
Sn2+, Pb2+
Al3+, Cr3+, Fe3+, In3+
UO22+, NpO22+, PuO22+
Sc3+, Y3+, Ln3+
Mg2+, Co2+, Ni2+, Cd2+
Pb2+
Zr4+, Th4+
M6(OH)84+
Be2+, Pb2+
M6(OH)126+
Bi3+
M2(OH2(2z-2)+
M3(OH)33+
M3(OH)4(3z-4)+
M3(OH)5(3z-5)+
M4(OH)44+
M4 square with eight OHions, one centered over and
under each edge.
M6 octahedron with eight
OH- ions centered on faces.
M6 octahedron with 12 OHions centered along edges.
Table 5.2 Polyoxoanions
M2
V2O74-
V(V)
M3
M4
V3O93-
V4O124-
M6
Nb6O198-
Ta(V)
Ta6O198-
M12
M19
Cr2O72Mo7O246-
Mo(VI)
W(VI)
M10
V10O286-
Nb(V)
Cr(VI)
M7
W6O192-
Mo19O594W12O4110-
See B&M under the various metal ions, Huheey, pp.755-764, SAL, pp.164-165)
(a)
(b)
(c)
(d)
Figure 5.1 Structures of polycations: (a) M4(OH)8(H2O)168+ , (b) M6(OH)84+
(c) M6(OH)126+ , (d) the Al13O4(OH)247+ polycation.
Figure 5.2 Evolution of the structures of polyoxoanions. (Huheey, p. 757)
5.2
5.2.1
The Concept of Equilibrium Constant
The Solubility Product
We learned from our hypothetical metal ion-water system above that successive
polymerization-hydrolysis reactions can result in the precipitation of a solid. When a solid,
let us say a metal hydroxide, M(OH)2(s), is placed in an aqueous solution, it will dissolve
according to a reaction such as:
M(OH)2(s) = M2+ + 2OH-
(5.13)
Actually, in view of the hydration of metal ions, it is more appropriate to write
M(OH)2 (s) + 6H2O = M(H2O)62+ + 2OH-
(5.14)
We would also expect further reactions to occur in the aqueous solution, e.g., hydrolysis,
and polymerization. As more and more solid dissolves, these aqueous phase interactions
also increase in extent until finally, the reverse process, i.e., precipitation, starts.
Thus, in effect, the dissolution process can be considered, ideally, in terms of a
reversible reaction in which the forward reaction will be dissolution and the reverse
reaction, precipitation. The criterion which is used to determine which of these two
reactions predominates under a given set of conditions, are called the solubility product.
The concentration solubility product for the metal hydroxide reaction (Equation 5.13) is
defined as:
Kso = [M2+] [OH-]2
(5.15)
where Kso is the solubility product constant and the square brackets represent
concentration. The solid M(OH)2 will precipitate out of solution when the product [M2+]
[OH-]2 exceeds the value of Kso. Alternatively, it is sometimes more convenient to write
the dissolution reaction in terms of the hydrogen ion:
M(OH)2(s) + 2H+ = M2+ + 2H2O
(5.16)
In this case, the solubility equilibrium is described by
*Kso = [M2+]/[H+]2
(5.17)
5.2.2
The Free Energy of Reaction and the Equilibrium Constant
Consider a general reaction,
aA + bB = cC + dD
(5.18)
For any arbitrary conditions (i.e., for a system not necessarily at equilibrium), the free
energy is given by
G = Go + RTln Q
(5.19)
where Go is the standard Gibbs free energy of the reaction and Q is the reaction quotient
and is given by
Q = {C}c{D}d/{A}a{B}b
(5.20)
where the symbol { } denotes activity.
At equilibrium, G = 0, and the equilibrium value of Q (i.e., Qeq) is termed the
equilibrium constant, K:
b
K = {C} ceq {D} deq /{A} aeq {B}eq
(5.21)
Go = -RT 1n Qeq = -RT 1n K
(5.22)
It follows therefore that
Application of Equation 5.22 to Equation 5.19 gives:
G = RT (1n Q - 1n K)
When
(5.23)
G = 0,
equilibrium has been attained and ln Qeq = 1n K
G < 0,
G > 0,
reactants form products simultaneously and ln Q < ln K
conversion of reactants to products is impossible; however the
reverse reaction is spontaneous and 1n Q > 1n K
The standard Gibbs free energy of reaction is a function of the standard free energies
of formation (Gof) of the reactants and products associated with the reaction. That is, for
the general reaction described by Equation 5.18, the standard Gibbs free energy (Go) is
given by:
o
o
o
o
Go = cGf (C) + dGf (D) - aGf (A) - bGf (B)
(5.24)
o
where Gf (C) is the standard free energy of formation of species C, etc.
In infinitely dilute solutions, the activity of a dissolved species equals its
concentration, i.e., {C} = [C], etc., and therefore under these conditions,
Q = [C]c [D]d/[A]a[B]b
(5.25)
In more concentrated solutions, however, enhanced ionic interactions lead to nonidealities
and the ionic activity generally deviates significantly from the concentration. Under these
circumstances the concentration and activity can be related through an activity coefficient
(), defined for species A as:
{A} = A [A]
(5.26)
The activity coefficient of a given species is a function of the total ionic concentration in
the electrolyte. Nonideality effects are treated in greater detail in Chapter 6. For the rest of
this chapter it is assumed that the aqueous solutions are infinitely dilute so that
concentration and activity equilibrium constants are numerically the same.
The quantities Kso and K *so defined above (see Equations 5.15 and 5.17) are
equilibrium constants. Referring to Equation 5.13 and recalling Equations 5.18 and 5.20,
we can write:
2 /{M(OH) (s)} = {M2+} {OH-} 2
Kso = {Mz+}eq{OH-} eq
2
eq
eq
eq
2
= [M2+]eq[OH-] eq
(5.27a)
(5.27b)
In equation 5.27a use is made of the fact that the activity of a pure substance is unity, i.e.,
{M(OH)2(s)} = 1; Equation 5.27b results from Equation 5.27a, upon applying the
assumption of infinitely dilute solution. The ionization of water, Equation 5.8 is also
characterized by an equilibrium constant, termed the ionization constant, Kw:
Kw = {H+}eq{OH-}eq/{H2O}eq = {H+}eq{OH-}eq
(5.28a)
= [H+]eq[OH-]eq (infinitely dilute condition) = 10-14 at 25°C
(5.28b)
In Equation 5.28a, the activity of water is indicated as unity since for an infinitely dilute
solution the solvent water may be considered to be a pure substance.
EXAMPLE 5.2 Calculation of equilibrium constants from free energy data
Based on the G of data (Wagman et al, The NBS Tables of Chemical Thermodynamic Properties,
1982) provided below, determine the free energies of reaction and the equilibrium constants for the following
reactions:
(a) 1/2O2 + H2 = H2O
(b) H+ + OH- = H2O
(c) CuO + 2H+ = Cu2+ + H2O
(d) Cu2+ + H2 = Cu + 2H+
Solution
First we must recall Equation 5.22:
o
o
logK = -Gr /2.303 RT = -Gr /(2.303) (8.3143 Jmol-1K-1) (298.15K)
= -G or /5.709, with G or in kJ mol-1
Note that this expression is valid only for 25°C, i.e., T = 298.15 K
(a)
o
G or = G of (H 2 O) - 1/2 G of (O 2 ) - Gf (H2)
= (-237.129)-1/2(0) - (0) = -237.129 kJ mol-1
logK = -G or /2.303 RT = -(-237.129)/5.709 = 41.54
(b)
o
G or = G of (H 2 O) - G of (H  ) -Gf (OH-)
= (-237.129) - (0) - (-157.244) = - 79.89 kJ mol-1
logK = - G or /2.303 RT = -(-79.89)/5.709 = 14.0
(c)
o
G or = G of (Cu 2 ) + G of (H 2 O) - G of (CuO ) - 2Gf (H+)
= (65.49) + (-237.129) - (-129.7) - 2 (0.) = - 41.94 kJ mol-1
o
logK = -Gr /2.303 RT = - (-41.94)/5.709 = 7.35
(d)
G or = G of (Cu ) + 2G of (H  ) - G of (Cu 2 ) - G of (H 2 )
= (0) + 2(0.) - (65.49) - (0.) = - 65.49 kJ mol-1
o
logK = -Gr /2.303 RT = - (-65.49)/5.709 = 11.47
5.2.3 Solubility Calculations
The equilibrium constant represents an important parameter which allows us to make
theoretical predictions and assess the likelihood of reactions. Suppose we wish to find the
equilibrium constant for the dissolution-precipitation reaction described by Equation 5.16.
This can be done by combining Equation 5.13 with Equation 5.8:
(-2)x Eq. 5.8:
M(OH)2(s) = M2+ + 2OH-
Kso
2H+ + 2OH- = 2H2O
1/Kw
(5.30)
M(OH)2(s) + 2H+ = M2+ + 2H2O
*Kso
(5.16)
(5.29)
2
The equilibrium constant for Equation 5.16 is given by:
2
*Kso = [M2+]eq/[H+]eq2 = [M2+]eq[OH-]2eq/[OH-]eq2[H+]eq2 = Kso/Kw
(5.31)
In general, for a hydroxide M(OH)z(s), the following relation holds,
z
*Kso = Kso/Kw
(5.32)
Let us consider Equation 5.16 again. Using the definition of the reaction quotient for
infinitely dilute solution, i.e., Equation 5.25, it follows that,
Q = {M2+}/{H+}2 = [M2+]/[H+]2
(5.33)
The equilibrium condition is then given by
logQeq = log*Kso
(5.34)
Therefore, from Equations 5.33 and 5.34,
log[M2+]eq = -2pHeq + log*Kso
(5.35)
where pH is defined as:
pH = - log {H+}
= - log [H+] (infinite dilution condition)
5.2.4
(5.36a)
(5.36b)
Solubility of Sparingly Soluble Salts
Consider the salt MyAz consisting of cations Mz+ and anions Ay-. When this salt is
present in a saturated solution of its ions, a chemical equilibrium exists:
MyAz(s) = yMz+ + zAy-
(5.37)
The equilibrium constant, i.e., solubility product, is given by
Kso = [Mz+]y [Ay-]z
(5.38)
where it is understood that the respective activities are equilibrium values.
According to Equation 5.37, dissolution of one mole of salt yields y moles of cations
and z moles of anions. Let C represent the solubility of the salt M yAz in pure water, e.g.,
the moles of salt which dissolve to give one liter of saturated solution. Then a mass
balance on the dissolved species gives,
[Mz+] = yC
[Ay-] = zC
(5.39)
(5.40)
Therefore, Equation 5.38 becomes
Kso = [yC]y [zC]z
(5.41)
EXAMPLE 5.3 Solubility of sparingly soluble salts
a)
Determine the solubility product of barium sulfate, BaSO 4, at 25°C given that the solubility of
this salt is 1.05 x 10-5 mol/liter.
b)
Calculate the solubility of calcium fluoride, CaF2, using the fact that the solubility product of this
salt is 4.0 x 10-11.
c)
On the basis of the results obtained in part (a), determine the solubility of BaSO 4 in 1.0 x 10-4
molar BaCl2 solution.
Solution
2(a) BaSO4(s) = Ba2+ + SO4
Dissolution of one mole of salt yields one mole of Ba2+ and one mole of SO 24 . Let C be the solubility of
the salt. Then it follows that:
2[Ba2+] = [SO4 ] = C = 1.05 x 10-5 mol/L
Thus
2Kso = [Ba2+] [SO4 ] = (1.05 x 10-5)2 = 1.10 x 10-10
(b) CaF2(s) = Ca2+ + 2FDissolution of one mol of salt yields one mole of Ca2+ and two moles of F-. Thus if C is the solubility of the
salt, then,
[Ca2+] = C and [F-] = 2C
Thus
Kso = [Ca2+][F-]2 = (C) (2C)2 = 4C3
Therefore
C = (Kso/4)1/3 = (4.0 x10-11/4)1/3 = 2.15 x 10-4 mol/L
(c) Let the solubility be C mol/L. Now the solution also contains 1.0 x 10 -4 mol/L BaCl2. Thus
[Ba2+] = (C + 1.0 x 10-4) mol/L
2Kso = 1.10 x 10-10 = [Ba2+] [SO4 ] = (C + 1.0 x 10-4) (C)
That is,
C2 + 1.0 x 10-4 C - 1.10 x 10-10 = 0
Applying the quadratic formula,
C
=
-1.0x10-4 +
(1.0x10-4)2 - 4(1)(-1.0x10-10)
2(1)
= (-1.0 x 10-4 + 1.02 x 10-4)/2 = 1.0 x 10-6 mol/L
5.2.5
Lattice Energy, Hydration Energy, and Solubility
Some insight into the energetics of the dissolution of ionic solids may be gained by
considering the thermodynamic cycle presented in Figure 5.3. The free energy change
associated with the disintegration of the solid MyAz(s) into isolated gaseous ions is
represented by GLo. The hydration of the gaseous ions when they are introduced into the
aqueous phase results in the free energy change Gho. The free energy of solution of the
solid is depicted as Gso. It follows from Figure 5.3 that
Gso = GLo + Gho
(5.42)
yMz+ (g) + zAy- (g)
 Gh

 GL
My Az (s)
 Gs
yMz+ (aq) + zAy- (aq)
Figure 5.3 Thermodynamic cycle illustrating the contributions to the standard free energy
of solution.
Recalling the Kapustinskii equation (Equation 4.8), etc., we can write GLo as:
GLo = K (y + z)(z+y-)/(r+ + r-) - 31.1 (y + z) kJ mol-1
(5.43)
K = 1.079 x 105 kJ mol-1 pm (see Johnson, p.48). Also an ionic model of ion hydration
(see Chapter 6) results in the following equation for the free energy of hydration:
Gho = -[B z2/(r + k)] + 7.9 kJ mol-1
(5.44)
where B = 6.8583 x 104 kJ mol-1 pm, r is the ionic crystal radius, k has a value of 80 pm
for cations and 17 for anions. (See SAL, pp.133-135; Huheey, pp.310-314; Johnson,
pp.116-)
5.3
5.3.1
Acids and Bases
Acidic and Basic Solutions
As indicated in Equation 5.28b the ionization constant of water has a value of 1014 at room temperature. Thus, in an equilibrium aqueous system containing only pure
water,
[H+] = [OH-] = Kw 1/2 = 10-7
(5.45)
pH = 7
(5.46)
or
Pure water is considered to be neutral and the solution is said to be an acidic
solution when
[H+] > [OH-] or pH < 7
(5.47)
On the other hand, the solution is considered a basic solution when
[H+] < [OH-] or pH > 7
(5.48)
Shifts in pH from the neutral value are obtained by introducing acids and bases
into the aqueous system. According to the Bronsted acid-base concept, an acid is a proton
donor while a base is a proton acceptor. That is, the ionization of an acid (HA) may be
written as:
HA = H+ + A-
(5.49)
In contrast, the ionization of a base (B) may be represented as:
B + H+ = BH+
(5.50)
EXAMPLE 5.4 The neutral pH of hydrothermal solutions
The ionization constant of water varies with temperature. Determine the pH of neutral solution at
elevated temperatures given the following data:
T(°C)
Kw
100
-12.26
200
-11.27
300
-11.39
Solution
For neutral pH, [H+] = [OH-] = Kw1/2 (Eq. 5.45). Therefore, the corresponding pH is given by pH = -(1/2)
log Kw. Accordingly, neutral pH has the following values: 6.13 (100 oC), 5.64 (200 oC), 5.70 (300 oC).
5.3.2
Strong Acids and Bases
A strong acid is one which dissociates completely to give protons and the
corresponding anions. Similarly a strong base is completely protonated to give the
corresponding ionic form. In contrast, weak acids and bases are only partially dissociated.
Table 5.3 presents a collection of several strong acids and bases.
Table 5.3 Ionization of Strong Acids and Bases
Name of Acid or Base
Chemical Formula
Ionization Reaction
Nitric acid HNO3
HNO3 = H+ + NO3-
Nitrate
Sulfuric acid
H2SO4
Hydrochloric acid
Hydrobromic acid
Hydroiodic acid
Perchloric acid
HCl
HBr
HI
HClO4
Lithium hydroxide
Sodium hydroxide
Potassium hydroxide
Rubidium hydroxide
Cesium hydroxide
LiOH
NaOH
KOH
RbOH
CsOH
H2SO4 = H+ + HSO4HCl = H+ + ClHBr = H+ + BrHI = H+ + IHClO4 = H+ + ClO4LiOH = Li+ + OHNaOH = Na+ + OHKOH = K+ + OHRbOH = Rb+ + OHCsOH = Cs+ + OH-
Anion____
pKa___
Bisulfate
-2
Chloride
Bromide
Iodide
Perchlorate
-7
-9
-11
-10
Hydroxyl
Hydroxyl
Hydroxyl
Hydroxyl
Hydroxyl
The pH of solutions of strong acids and bases can be calculated by combining the
ionization equilibrium of water with appropriate mass balances. Consider a solution
containing C mol/L of the strong acid HA. We wish to determine the pH of this solution.
In view of the ionization of water, the following condition must be satisfied:
[H+][OH-] = Kw
(5.51)
Let [H+]' represent the concentration of protons released into the solution by the
ionization of water. Then, since each proton produced from a water molecule also leads to
the release of a hydroxide ion,
[H+]' = [OH-]
(5.52)
Also, let [H+]" represent the concentration of protons associated with the dissociation of
HA. Then, since HA is a strong acid, there is complete dissociation, and therefore
[H+]" = [A-] = C
(5.53)
Thus, a mass balance on hydrogen ions may be written as:
[H+] = [H+]' + [H+ ]"
= [OH-] + C
(5.54a)
(5.54b)
Combination of Equations 5.51 and 5.54b, with elimination of [OH-], gives [H+]
as a quadratic equation:
[H+]2 - C[H+] - Kw = 0
(5.55)
Since [H+] has to be a positive number, the solution of Equation 5.55 is
[H+] = [C + (C2 + 4 Kw)1/2]/2
(5.56)
It can be seen from this expression that
[H+] ~ C
for C >> 2Kw1/2
(5.57a)
~ Kw1/2
for C << 2Kw1/2
(5.57b)
Thus, the proton concentration of a highly concentrated solution of a strong acid
corresponds to the analytical concentration of the acid. In contrast, the proton
concentration in a very dilute solution of a strong acid is the same as that of pure water.
EXAMPLE 5.5 Concentration dependence of the pH of a strong acid
Determine the pH of a strong acid solution with the following concentrations (mol/L): (a) 10 -4, (b) 10-7, (c)
10-10.
Solution
We recall that Kw = 10-14. Therefore, Kw1/2 = 10-7.
(a) C = 10-4 mol/L >> 2 Kw1/2. Thus, it follows from Equation 5.57 that [H+] = C = 10-4.
Therefore, pH = 4.
(b) Here C = 10-7 mol/L = Kw1/2. Therefore, we must use the complete expression, Equation 5.56.
We find [H+] = 1.62 x 10-7 mol/L. Therefore, pH = 6.8
(c) C = 10-10 mol/L << 2 Kw1/2. Accordingly, invoking Equation 5.58, pH = -log(Kw1/2) = 7.00
A similar approach may be taken for the determination of the pH of a strong base. As
+
-
already noted above, the dissociation of a water molecule releases one H ion and one OH .
Thus, if [OH-]' represents the concentration of hydroxide ions released into solution via the
dissociation of water, then it follows that:
-
+
[OH ]' = [H ]
(5.58a)
Consider a base M(OH)z, which dissociates as:
M(OH)z = Mz+ + zOH-
(5.58b)
For a strong base M(OH)z, there is complete ionization. Let the solution contain C mol/L
-
of the base. Let [OH ]" represent the concentration of hydroxide ions present in solution as
a result of the dissociation of the base. Then, since the dissociation of a mole of M(OH)z
produces a mole of M
z+
-
plus z moles of OH ,
[OH-]" = z[Mz+] = zC
(5.58c)
It follows, therefore, that the total hydroxide concentration is given by
[OH-] = [OH-]' + [OH-]" = [H+] + zC
(5.58d)
+
-
Recalling the ionization equilibrium of water, we can use the fact that [H ] = Kw/[OH ].
Thus, Equation 5.58d can be rewritten as:
[OH-] = (Kw/[OH-]) + zC
(5.58e)
That is,
-2
-
[OH ] - zC[OH ] - Kw = 0
(5.58f)
[OH-] = [zC + (z2C2 + 4Kw)1/2]/2
(5.58g)
Therefore,
The relevant limiting conditions are:
-
[OH ] ≈ zC, [H+] = Kw/zC for zC >> 2Kw1/2
+
≈Kw1/2, [H ] = Kw1/2 for zC << 2K w1/2
(5.58h)
(5.58i)
_________________________________________________________________________
EXAMPLE 5.6 Concentration dependence of the pH of a strong base
Repeat Ex.5.5 for the solution of a strong base MOH.
5.3.3
Weak Monoprotic Acids and Bases
As indicated above in Section 5.3.2, weak acids and bases are not completely
dissociated. Table 5.4 presents examples of weak monoprotic acids and bases, i.e., weak
acids which dissociate to give only one proton per molecule, and weak bases which can
only bind one proton. Acids carrying more than one ionizable proton (called polyprotic
acids) and bases capable of binding more than one proton are considered in Chapter 7.
Table 5.4 also includes equilibrium constants for the various dissociation reactions.
In the case of the acids, the greater the numerical value of log K, the greater the tendency
for the acid to ionize. Examination of the tabulated log K values therefore clearly indicates
that the designation "weak" is a relative term. For example, it will be noticed that log K = 1.92 for methylsulfonic acid whereas the corresponding value for phenol is -9.98. Thus,
the sulfonic acid is a much stronger acid than phenol. In the case of the amines, the greater
the numerical value of log K, the greater the tendency towards protonation, i.e., the
stronger the base. Therefore, here too, there is a range of basicities, with dimethylamine
(log K = 10.77) giving the highest and aniline (log K = 4.60) giving the lowest.
The pH of a solution of a weak acid or base may be determined by combining the
equilibrium expressions for the ionization of water, and the acid or base with appropriate
mass balances. Consider the weak acid HA which dissociates as
HA = H+ + A-
(5.59)
The corresponding equilibrium constant is given by:
K = [H+] [A-]/[HA]
(5.60)
A mass balance on the proton gives
[H+] = [H+]' + [H+]" = [OH-] + [A-]
(5.61)
Table 5.4 Ionization of Weak Monoprotic Acids and Bases
Name of Acid or Base
Chemical Formula
Ionization Reaction
log K
Methylsulfonic acid
CH3SO3H
CH3SO3H = H+ + CH3SO3
-1.92
Hydrogen sulfate ion
HSO4
2HSO4 = H+ + SO4
-1.99
Hydrofluoric acid
HF
HF = H+ + F-
-3.17
S
||
Diethyldithiocarbamic acid
(C2H5)2C-SH
(C2H5)2C(S)SH
= H+ + (C2H5)2C(S)S-
-3.42
Formic acid
HCOOH
HCOOH = H+ + COO-
-3.75
Acetic acid
CH3COOH
CH3COOH = H+ + CH3COO-
-4.76
Hydrogen hypochlorite
HOCl
HOCl = H+ + OCl-
-7.53
Hydrogen hypobromite
HOBr
HOBr = H+ + OBr-
-8.63
Hydrocyanic acid
HCN
HCN = H+ + CN-
-9.21
Phenol (Hydroxybenzene)
OH
O-
OH
O
O
= H+ +
-9.98
O
Hydrogen hypoiodite
HOI
HOI = H+ + OI-
-10.64
Dimethylamine
(CH3)2N
(CH3)2NH + H+
= (CH3)2NH4+
10.77
Methyl amine
CH3NH2
CH3NH2 + H+ = CH3NH3+
10.64
Trimethylamine
(CH3)3N
(CH3)3N + H+ = (CH3)3NH+
Ammonia
NH3
NH3 + H+ = NH4+
O
Pyridine
Aniline
O
N
O
- NH2
O
9.80
9.24
O
+
+
N +H = NH
5.23
- NH2 + H+ =
O - NH3+
4.60
where [H+]' and [H+]" respectively represent the protons released by the dissociation of the
water and acid molecules. If the acid is present in solution at an analytical concentration of
C, then a mass balance on the anion becomes:
C = [HA] + [A-]
(5.62)
where [HA] is the concentration of the residual unionized acid.
Combination of Equations 5.60 and 5.62 gives:
[A-] = C/(1 + [H+]/Ka)
(5.63)
[HA] = C([H+]/Ka)/(1 + [H+]/Ka)
(5.64)
Recalling the equilibrium relation for the ionization of water,
[OH-] = Kw/[H+]
(5.65)
Substitution of Equations 5.63 and 5.65 respectively for [A-] and [OH-] in Equation 5.61
gives the following results:
[H+] = Kw/[H+] + C/(1 + [H+]/Ka)
(5.66a)
C = (1 + ([H+]/Ka))([H+] - (Kw/[H+]))
(5.66b)
EXAMPLE 5.7 The pH of a weak acid solution
Determine the pH of a 10-3 mol/L solution of (a) hydrogen sulfate ion (pK a = 1.99), (b) formic acid
(pKa = 3.75), (c) hypochlorous acid (pKa = 7.53), (d) phenol (pKa = 9.98)
Solution
5.4
Structural Aspects of Acid-Base Strength
5.4.1 Gas-Phase Acidity
We saw above that the strength of a Bronsted acid, HA, is expressed by the acid
ionization constant:
HA(aq) + H2O = A-(aq) + H3O+(aq)
Ka
(5.67)
We found that some acids are strong (Ka > 1, pKa < 0; Table 5.3), while others are weak
(Ka <1, pKa > 0; Table 5.4). We now wish to address this question: what are the factors
that govern acid strength ?
A convenient starting point is to consider the feasibility of gas-phase proton transfer.
It is helpful to define a quantity termed the proton-gain enthalpy, Hpo, which is the
standard enthalpy of protonation for the following gas phase reaction:
A- + H+ = HA
Hpo (A-)
(5.68)
A related quantity is the proton affinity, Ap, defined as the negative of the proton-gain
enthalpy (i.e., Ap = -Hpo). The proton affinity is high when proton attachment is highly
favored, as reflected in a highly negative value of the proton-gain enthalpy. Proton affinity
values for selected bases are presented in Table 5.5 (Shriver, p.152).
We can use a thermodynamic cycle to discern the factors that contribute to the
magnitude of the proton affinity. On the basis of the thermodynamic cycle presented in
Figure 5.4, the proton-gain enthalpy can be expressed as:
Hpo (A-) = Ae(A) - I(H) -B(HA)
(5.69)
where Ae(A) = the electron affinity of A, I(H) = ionization energy of H, and B(HA) = the
H-A bond enthalpy. Alternatively, the proton afinity can be expressed as:
Ap = -Hpo = B(HA) + I(H) - Ae(A)
(5.70)
Equations 5.69 and 5.70 can be used to rationalize trends observed when one goes across
the period or down a group in the Periodic Table. Thus, on proceeding from left to right
across a period, the electron affinity increases, and this is expected to lower the proton
affinity. On the other hand, as one goes down a group, the bond dissociation enthalpy
decreases, and this should decrease the proton affinity.
Table 5.5 Proton affinities (Shriver, p.152)
Conjugate acid
Base
F-
HF
Ap/kJ mol-1
Ap/kJ mol-1
1553
1150
HCl
Cl
-
1393
1090
HBr
Br-
1353
1079
-
HI
I
1314
1068
CH4
CH3-
1741
1380
NH3
NH2-
1670
1351
PH3
PH2
-
1548
1283
H2O
OH-
1634
1188
-
CN
1476
1183
H2O
723
1130
NH3
865
1182
936
1160
HCN
+
H3O
NH4+
+
C6H5NH
C6H5N

o
*Ap = -Hp is the gas-phase proton affinity: Ap the effective proton affinity for the base in water.
Source: J. E. Bartmess and R. J. McIver, in Gas Phase Ion Chemistry, M. T. Bowers (ed.). Academic Press.
New York (1978).
A- (g) + H (g)
-I (H)
- eA e (A)
A (g) + H (g)
+ e-
-B (HA)
-
A (g) + H +(g)
HA (g)
 Hop (A- )
Figure 5.4
Thermodynamic cycle illustrating the contributions to the proton affinity.
EXAMPLE 5.8 Proton transfer to water in the gas phase
On the basis of the gas-phase proton affinity values presented in Table 5.5, determine the feasibility of proton
transfer from the various acids to water in the gas phase.
Solution
The proton transfer to water in the gas phase can be evaluated as shown below:

 

HA(g) = A-(g) + H+(g)
-Hpo (A-)
H2O(g) + H+(g) = H3O+(g)
Hpo (H2O)
__________________________________________
HA(g) + H2O(g) = A-(g) + H3O+(g)
(1)
Ho = Hpo (H2O) - Hpo (A-) = Ap (A-) - Ap (H2O)
(4)
(2)
(3)
According to Equation 4, in order for proton transfer from an acid HA to water to be favorable in the gas
phase (i.e., Ho < 0), we must have Ap(A-) < Ap (H2O). As can be seen from Table 5.5, none of the acids
listed satisfies this condition.
5.4.2 Hydration Effects
The conclusion obtained in Ex. 5.8 (which involves gas-phase proton transfer)
appears to contradict our everyday experience with aqueous solutions: HCl, for example, is
a strong acid that readily transfers protons to water molecules in aqueous solution. This
apparent contradiction is resolved by recognizing the key role played by hydration in
proton transfer in aqueous solution. To account for hydration effects, an effective proton
affinity, Ap', is defined on the basis of the enthalpy change associated with the gas-phase
transfer of a proton to a cluster of water molecules. Values of Ap' are also included in
Table 5.5, and these can be used to treat proton transfers in aqueous solution:
HA(aq) + H2O(aq) = A-(aq) + H3O+(aq)
Ho = Ap'(A-) - Ap'(H2O)
(5.71)
When the Ap and Ap' values in Table 5.5 are compared, it can be seen that for all the
bases above H2O, hydration decreases the proton affinity (Ap(A-) > Ap'(A-)). On the other
hand, for H2O, NH3, and C5H5N, hydration increases the proton affinity (Ap < Ap'). The
decrease in proton affinity means that hydration stabilizes A-. On the other hand, the
increase in proton affinity means that hydration stabilizes A (e.g., H3O+, NH4+, and
C5H5NH+). It is interesting to note that in these examples, the stabilized species are the
charged conjugate species (e.g., Cl- vs. HCl, NH4+ vs. NH3). These trends illustrate the
fact that the electrostatic interactions between water dipoles and ions will stabilize ions in
solution relative to the gas phase. Thus for a series of bases, we expect proton affinity to
increase with decrease in ionic size and with increase in ionic charge. Another factor
which influences proton affinity is hydrogen bonding between aqueous species and water
molecules.
_________________________________________________________________________
EXAMPLE 5.9 Trends in the basicity of ammonia and substituted amines
The ability of an ammonia molecule to accept a proton, i.e., its basicity, is related to the availability of
the lone electron-pair on the nitrogen atom (H3N: ->H+). Using this fact and the ability of H2O to serve as a
hydrogen-bond acceptor, as noted above, rationalize the following trends in the basicity of ammonia and
amines.
(a) How would one expect the basicity to change upon substituting one of the hydrogen atoms in
ammonia with (i) an electron-withdrawing group, (ii) an electron-donating group.
(b) Given pKb (MeNH2) = 3.36, pKb (NH3) = 4.74, pKb (NH2NH2) = 5.77, pKb (NH2OH) = 7.97,
determine whether the groups -Me, -NH2, and -OH are electron-donating or electron-withdrawing.
(c) On the basis of the finding in part (b), speculate about the trend in the gas-phase basicity of the
following methylamines: MeNH2, Me2NH, Me3N.
(d) Rationalize the following trends: pKb (NH3) = 4.74, pKb (MeNH2) = 3.36, pKb (Me2NH) =
3.29, pKb (Me3N) = 4.28
Solution
(a) An electron-withdrawing group will decrease the availability of the lone-pair electron and this
would be expected to decrease the proton affinity and therefore the basicity. On the other hand, an
electron-donating group will amplify the availability of the lone-pair electron, which should
register as an increase in basicity.
(b) On the basis of part (a), since substitution of -NH2 or -OH results in a decrease in basicity (an
increase in pKb), these groups must be electron-withdrawing. On the other hand, the presence of
the -Me group increases the basicity (a decreaase in pK b) and therefore this group must be
electron-donating.
(c) Since the -Me group was found above to be electron-donating, one would expect the basicity to
increase with successive substitution of this group, i.e., the trend NH 3 < MeNH2 < Me2NH <
Me3N.
(d) The trend expected from electron donation (part (c)) is satisfied with respect to the trend pK b
(NH3) > pKb (MeNH2) > pKb (Me2NH). The trend pKb (Me2NH) < pKb (Me3N) is, however,
unexpected. This reversal in basicity can be rationalized by considering the role of hydrogen
bonding in the solvation of the ammonium ions. As already noted above, hydration enhances
basicity. Thus, for the protonated amines, hydrogen bonding is mediated through the hydrogen
atoms, and consequently, the extent of hydrogen bonding and hence increased basicity is expected
to follow the order: MeNH2 > Me2NH > Me3N. This is the reverse of the trend based on electron
donation. Hence, the basicity goes through a maximum with increasing substitution.
________________________________________________________________________________________
5.4.3 Strengths of Hydroxyl -Group Acids
Hydroxyl-group acids are characterized by the presence of hydroxyl groups which
can deprotonate. These acids fall into three classes, i.e., aqua acids (e.g., M(OH2)nz+),
hydroxoacids (e.g., M(OH)z), and oxoacids (e.g., H2SO4). In an aqua acid, the hydroxyl
group comes from a water molecule attached to a metal ion. In the case of a hydroxoacid,
the hydroxyl group is attached directly to the central atom and no oxo group is bonded
directly to this atom. An oxoacid results when the central atom is bonded to both a
hydroxo group and an oxo group. The progression: aqua acid  hydroxoacid  oxoacid,
can be considered in terms of a consecutive deprotonation process:
H2O-M-OH2 (aqua acid) = HO-M-OH (hydroxoacid) + 2H+
(5.72)
HO-M-OH (hydroxoacid) = HO-M=O (oxoacid) + H+
(5.73)
Which elements are most likely to form aquoacids? Typically in aquo acids, the
central atom has a low oxidation number: atoms from the s-block, d-block, and leftside of
the p-block. In some cases, trends in the acidities of aquo acids can be correlated with an
electrostatic model. This is the case when the central atom of the aquo acids is derived
from s-block elements. These are also the elements which have the greatest tendency to
form ionic solids. Here the tendency to deprotonate is expected to increase with increase in
charge and decrease in size. Such an electrostatic model is less useful for the d-band and pblock metals, where there is significant covalent bonding.
Trends for oxoacids (see SAL, p.159)
5.5
Graphical Representations of Ionic Equilibria
5.5.1 Logarithmic Concentration Diagrams
Recall Equation 5.16:
M(OH)2(s) + 2H+ = M2+ + 2H2O
(5.16)
For a given pH, when [M2+] is less than the equilibrium value, it follows from Equations
5.33 and 5.34 that log Q < log K and Equation 5.16 proceeds from left to right, i.e.,
dissolution takes place. Under these conditions, it can be said that M2+ is more stable than
M(OH)2(s). On the other hand, when [M2+] is greater than the equilibrium value, the
reaction proceeds from right to left, i.e., precipitation occurs, and in this case M(OH)2(s) is
more stable than M2+. Thus for the M2+ - M(OH)2(s) pair, M2+ is the stable species when
log[M2+] < -2pH + log *Kso
(5.74)
while M(OH)2(s) is the stable species when
log[M2+] > -2pH + log *Kso
(5.75)
We saw previously that the concentration of metal ion, [M2+]eq, which is in
equilibrium with the metal hydroxide M(OH)2(s) is given by
log[M2+]eq = -2pHeq + log*Kso
(5.76)
y = mx + c
(5.77)
This equation is of the form
i.e., the equation of a straight line with a slope m = -2, and an intercept at log[M2+]eq = 0
given by
pHeq = (1/2) log*Kso
(5.78)
Thus, assuming a negative slope, it can be seen that on a log[M2+] versus pH plot, M2+ will
be stable to the left of this straight line (Equation 5.74) while M(OH)2(s) will be stable to
the right (Equation 5.75).
EXAMPLE 5.10 The solubility diagram for Fe(OH)3 in acidic solution
The following reaction occurs in the Fe(OH)3(s)-H2O system:
Fe(OH)3(s) + 3H+ = Fe3+ + 3H2O
where log *Kso = 3.3. Prepare a log[M] vs pH diagram that describes the solubility of Fe(OH) 3(s).
Solution
Following Equation 5.35, 5.76, and 5.77, the solubility relation is given by
log[Fe3+]eq = 3.3 -3pHeq
(1)
Furthermore, according to Equations 5.76 and 5.77, the straight line represented by Equation 1 can be drawn
as shown in Figure E.5.10. Finally, recalling the inequalities given by Equations 2.74 and 2.75, Fe 3+ is
stable to the left of this line, while Fe(OH)3(s) is stable to the right.
0
-1
-2
Fe(OH) 3 (s)
3+
log[Fe ] -3
Fe3+
-4
-5
-6
0
2
pH
4
6
Figure E.5.10 Solubility of Fe(OH)3(s), (neglecting hydrolysis).
EXAMPLE 5.11 Selective precipitation and selective dissolution
a)
You have been asked to investigate the method of selective-precipitation as a means of separating
metal ions from each other. A multicomponent solution containing Ag+, Zn2+, Ca2+ at 10-5
mol/liter each has been provided. The solution is initially acidic and you have a bottle of
concentrated NaOH (caustic soda) at your disposal. As you add the caustic soda solution
dropwise, (i) Which of the metals would you expect to precipitate first? (ii) What is the order in
which the remaining metals would precipitate? (iii) What would be the concentration of the other
metals when the last metal ion begins to precipitate?
b) In another experiment similar to that described above, you presented your technician with a
solution containing 1 mol/liter each of Mg2+, Cu2+, Ca2+, Fe3+. Unfortunately, you explained
your technique hurriedly to the technician and you came back to find that instead of adding caustic
soda dropwise, he added a large dose at once, thereby, precipitating all the ions from solution.
You decide then to apply the technique of selective dissolution(leaching) to separate the metal
ions in the precipitate from each other. In this method, you add sulfuric acid dropwise to an
aqueous slurry of the precipitate. (i) Which of the metals will dissolve first? (ii) Which will
dissolve last?
Relevant thermodynamic data are provided below:
Reaction
Fe(OH)3(s) = Fe3+ + 3OHCu(OH)2(s) = Cu2+ + 2OHZn(OH)2(s) = Zn2+ + 2OHMg(OH)2(s) = Mg2+ + 2OHCa(OH)2(s) = Ca2+ + 2OHAg(OH) (s) = Ag+ + OH-
logK(25°C)
-38.7
-19.0
-16.0
-9.2
-5.43
-7.8
It was assumed in Ex. 5.10 that the solids were in equilibrium with only hydrated
metal ions, i.e., we ignored the fact that metal ions may be hydrolyzed in aqueous solution.
In fact, to fully describe the solubility of solids, we need to take into account the possibility
that solids may be in equilibrium with hydrolyzed metal ions. Like the dissolutionprecipitation reactions discussed above, hydrolysis reactions are also characterized by
equilibrium constants. These constants can be combined with solubility product constants
to establish the solubility of an oxide or hydroxide over the entire acid to alkaline pH
range. This more complicated situation is addressed in Chapter 7.
Another kind of concentration diagram is that describing the dissociation of a weak
acid or base. For the acid HA, the effect of pH on the concentration ([A]) of HA or A - can
be depicted graphically in terms of a log [A] vs pH plot. Referring to Equations 5.63 and
5.64, the following limiting results may be obtained:
Low pH (i.e., [H+] >> K):
[A-] = CK/[H+]
(5.78)
log [A-] = log C + log K + pH
(5.79)
[HA] = C
(5.80)
log [HA] = log C
(5.81)
or
Also,
or
High pH (i.e., [H+] << K):
[A-] = C
or
(5.82)
log [A-] = log [C]
(5.83)
[HA] = C[H+]/K
(5.84)
log [HA] = log C - logK - pH
(5.85)
Also,
or
Figure 5.5 presents a schematic illustration of the above log [A] vs. pH relationships.
log[C]
[HA]
[A- ]
log[A]
pKa
pH
Figure 5.5 A schematic speciation diagram for the weak acid HA.
5.5.2 Distribution Diagrams
Let us return to the acid HA. Again Let C be the analytical concentration of the acid
in aqueous solution. Then we can define the following concentration fractions:
A = [A-]/C
HA = [HA]/C
(5.86)
(5.87)
A plot of A vs. pH is termed the dissociation curve. On the other hand a plot of HA vs.
pH is called the formation curve. These plots may also be designated as distribution
diagrams; a vertical line drawn through a given pH directly indicates the relative amounts
of the acid-derived species A- and HA. For example in the case of the HA vs. pH curve,
the region below the curve gives HA whereas that above the curve corresponds to A.
EXAMPLE 5.12 Distribution diagrams for weak bases
Using the logK data provided in Table 5.4, prepare distribution diagrams for the following:
(a) weak bases: ammonia, aniline, and dimethylamine.
(b) weak acids: formic acid, hydrogen hypochlorite, hydrogen sulfate ion, and phenol.
Solution
5.6 The Partial Charge Model and the Reactions of Metal Cations
5.6.1 The Acidity of Hydrated Metal Cations
In Section 2.4.3, the concept of electronegativity equalization was presented.
According to this concept, the shared electrons in a stable molecule are distributed among
the constituent atoms in such a manner that the electronegativity (i) of each bonded atom
attains the same magnitude as the mean electronegativity of the complex (). As discussed
in Section 2.4.3, this electronegativity equalization results in the acquisition of partial
charges by the constituent atoms. For a complex with total charge z, the partial charges on
the constituent atoms satisfy the condition:
z =  pi i
(5.89)
i
where pi is the number of moles of atom i in one mole of the complex. The partial charge
distribution in a molecule provides a powerful predictive tool for the aqueous chemistry of
metal cations.
EXAMPLE 5.13 Partial charge distribution in hydrated metal cations
z+
2+
3+
Determine the mean electronegativity () of the following M(OH)6 species: (a) Mg , (b) Al , (c)
5+
6+
Ti , (d) V , (e) W . For each hydrated species, determine the partial charges on the constituent atoms.
4+
Solution
(a)
2+
Mg(OH2)6 : From Table 2.9 (Allred-Rochow),
oMg = 1.23,  H = 2.20,  oO = 3.50. Also, for
o
the hydrated M(II) cation, z=2. Thus, using Equation 2.12h,


 



    

 

  
= [1.11 + 17.80 + 11.23 + 2.72]/[(0.90) + (8.09) + (3.21)]
= 32.86/12.2 = 2.69
Then, from Equation 2.12g,
H = ( - H )/1.36(H )
= (0.49)/2.02 = 0.24
o
o 1/2
= (2.69-2.20)/1.36(2.20)
1/2
O = (-O )/1.36(O ) = (2.69 - 3.50)/1.36 (3.50)
= -0.81/2.544 = -0.32
o
o 1/2
Mg = ( - Mg )/1.36(Mg )
= (1.46)/(1.51) = 0.97
o
o 1/2
1/2
= (2.69 - 1.23)/1.36 (1.23)
1/2
___________________________________________
It can be seen from Ex 5.13 that as the oxidation state of the metal cation increases
from +2(Mg) to +6(W), the partial charge on the metal cation (M) decreases dramatically.
This trend reflects the ability of the metal cation to draw electrons from the water
molecules through the M 2 bonds. Additionally, the results of Ex 5.13 indicate that
the partial charge on the proton (H) increases with increase in the charge on the metal
cation. In fact, in W(OH2)66+, the partial charge on H(H= 0.60) is higher than that on
W(W = 0.31). It will be recalled from Ex 2.13 that for the free water molecule, the partial
charge on the proton is H=0.2. This means that the O-H bond is weaker when a water
molecule is in the hydration sphere of a metal cation than when the water molecule is free.
Another way of expressing this fact is to say that a water molecule in the hydration sphere
of a metal ion is a stronger acid than a free water molecule in the bulk aqueous phase.
5.6.2 The Mean Electronegativity of an Aqueous Solution
The chemical potential (i) of a species i is given by
i = io + RTlnCi
(5.90a)
where io is the standard chemical potential, Ci is the concentration of the species, R is the
universal gas constant, and T is the absolute temperature. Thus, for the proton in aqueous
solution,
+
+
+
 (H ) = io (H ) + RTln[H ]
+
= io (H ) - 2.3RT pH
(5.90b)
(5.90c)
The electronegativity may be viewed as an electronic chemical potential.
Therefore, assuming a proportionality between  and (i.e., i = ki, where k is the
proportionality constant), we can write:
+
+
k(H ) = kio (H ) - 2.3RTk pH
(5.91a)
or
+
o
+
(H ) =  (H ) - 2.3RTkpH
(5.91b)
It follows from Equation 5.91b that when pH = 0,
+
o
+
(H ) =  (H )
(5.92)
In strongly acidic solutions, the proton is bihydrated and therefore, the species [H2O-HOH2]+ (i.e., H5O2+) can be chosen as the reference state for the hydrated proton.
That is, at pH 0,
+
o
+
+
(H ) =  (H ) = (H5O2 ) = 2.732
(5.93)
Using Equation 5.93 in Equation 5.91b gives
+
+
(H ) = (H5O2 ) - 2.3k RT pH
= 2.732 - 2.3k RT pH
(5.94)
Now, in water, the proton is highly delocalized and therefore, invoking the principle
of electronegativity equalization, we expect the electronegativity of the hydrated proton to
equal the mean electronegativity of the aqueous solution (w). Thus,
+
(H ) = w
(5.95)
At pH 7, the aqueous solution has no excess of hydrated proton or hydroxide ions.
Therefore, it is convenient to assume that at pH,
+
(H ) = w = (H2O) = 2.491
(5.96)
Combining Equations 5.94 and 5.96 gives:
2.3kRT = [2.732 - (H+)]/pH
= [2.732 - (H2O)]/(7)
= [2.732 - 2.491]/7 = 0.035
(5.97)
Thus, Equation 5.94 becomes:
(H+) = w = 2.732 - 0.035pH
5.6.3a Hydrolysis of Metal Cations
(5.98)
According to the Partial Charge Model (PCM), complex formation can be
rationalized in terms of electronegativity equalization. Let us consider the hydrolysis of a
metal cation. As discussed in Section 5.1, this refers to the reaction with water molecules
that leads to deprotonation and the formation of hydroxo or oxo metal complexes:
M(OH2)n
z+
+ hH2O = M(OH)h(OH2)n-h
(z-h)+
+
+ hH3O
(5.99)
(z-h)+
The complex on the right-hand-side of Equation 5.99 can be expressed as MOnH2n-h
.
z+
Then, it can be seen that the complex is: (a) an aquo-ion, M(OH2)n , when h=0, (b) an
(2n-z)-
oxo-ion, MOn
, when h = 2n, (c) an oxo-hydroxo complex, MOx(OH)n-x
when 0<h<2n and h>n, (d) a hydroxo-aquo complex, M(OH)h(OH2)n-h
0<h<2n and h<n, and (e) a hydroxo-complex, M(OH)n
(n+x-z)-
,
(z-h)+
, when
(n-z)-
, when 0<h<2n and h=n.
EXAMPLE 5.14a Hydrolysis of Cr(VI) and Cr(III) metal ions
The following hydrolysis products of Cr(VI) are known:
corresponding Cr(III) complexes are:
3+
Cr(OH 2)6 ,
o
-
2-
CrO 2(OH)2 , CrO3(OH) , CrO4 .
2+
Cr(OH)(OH2)5 ,
The
+
o
Cr(OH)2(OH2)4 , Cr(OH)3(OH2)3 ,
-
Cr(OH)4 . For each complex, determine the value of h and state whether it is an aquo-ion, oxo-ion, etc.
5.6.3b Acid Ionization of Hydrated Metal Ions
Interaction between the partial positive charges on the metal ion and the protons of
the bound H2O and OH- can lead to the deprotonation of a hydrated metal ion:

M(OH 2 ) 2n  hH 2 O  M(OH) h (OH 2 ) (zn hh)  hH 3 O 
(5.99)
When OH is negative, the attractive OH--H+ interaction will discourage deprotonation.
However, deprotonation will continue so long as OH is positive, i.e., proton transfer ceases
when OH=0.
EXAMPLE 5.14b Deprotonation of hydrated metal ions
a)
Show that the maximum extent of deprotonation is achieved when:
h = (z-nH - M)/(1-H)
b) Show that the mean electronegativity () of the stable complex is given by:
(1)
 = OH = 2.71
c)
(2)
Show from Equation 1 and 2 that:


h = 1.45z – 0.45n – 1.07(2.71 -  M )/( M )1/2
d) With the aid of Equation 3, speculate on the nature of the cationic species that are expected to exist at
very low pH in a dilute aqueous solution.
(3)
5.6.3c Basic Ionization of Oxo Metal Ions
z) 
Consider the anionic oxo species MO (2m
. The negative charges carried by the
m
oxygen atoms can lead to attractive interaction with the protons from the solvent water
molecules:
z) 
 z q)
MO (2m
+ qH2O = MOmH (2m
+ qOHm
q
(5.99b)
The protonation will proceed so long as H < 0. The limit is reached when H = 0.
EXAMPLE 5.14c Ionization of oxo metal ions
a)
With the aid of the partial charge model show that the stable protonated oxo species is given by:
q = 2m – z + mO + M
b) Show that the mean electronegativity of the stable species is given by:
(1)
 = 2.1
c)
(2)
On the basis of Equations 1 and 2, show that


q = 1.45m – z + 0.74(2.1 -  M )/( M )1/2
(3)
d) With the aid of Equation 3, speculate on the nature of the anionic species that are expected to exist at
high pH in a dilute aqueous solution.
(z-h)+
Based on Equation 5.89, it can be shown that for the species M(OH)h (OH2)n-h
,
the hydrolysis parameter, h, is given by:
h = (z - M - nO - 2nH)/(1 - H)
(5.100)
A relevant question then is this: How far can the deprotonation process proceed?
That is, what is the limiting value of the hydrolysis parameter, h, in the complex
(z-h)+
M(OH)h(OH2)n-h
? In order to address this question, we invoke the principle of
electronegativity equalization. That is, the proton exchange between the complex and the
aqueous solution will proceed up to the point where the mean electronegativity of the
complex () becomes equal to the mean electronegativity of the aqueous solution (w).
Thus,
 = w
(5.101)
We recall the expression for the partial charge:
o
o 1/2
i = ( - i )/1.36 (i )
(5.102)
An expression for w was derived above (Equation 5.98). Thus,
o
o 1/2
iw - i )/1.36 (i )
o
(5.103)
o 1/2
i = (2.732 - 0.035 pH - i )/1.36 (i )
o
(5.104)
o
Using H = 2.10 and O = 3.50 in Equation 5.104 leads to:
H = 0.321 - 0.0178 pH
(5.105)
O = -0.302 - 0.138 pH
(5.106)
Also, from Equation 5.104, the partial charge on the metal is given by:
o
o 1/2
M = (2.732 - 0.035 pH - M )/1.36(M )
= 2.01 (Mo)1/2 – 0.0257(Mo)-1/2 pH – 0.735(Mo)1/2
(5.107)
It follows from Equations 5.100, 5.105, 5.106 that:
h = [z - M - n(-0.302 - 0.0138 pH) -2n(0.321 - 0.0178 pH)]/[1 - (0.321 - 0.0178 pH)]
= [z - M + 0.302n + 0.0138n pH - 0.642n + 0.0356n pH]/[0.679 + 0.0178 pH]
h = [z - M - 0.34n + 0.0494n pH]/[0.679 + 0.0178 pH]
(5.108)
EXAMPLE 5.15 Hydrolysis of Si(IV)
(a)
Given that for Si(IV), Si = 1.74 and n = 4, show that application of the partial charge model
gives the pH dependence of the hydrolysis parameter (h) as:
o
h = (2.088 + 0.217 pH)/(0.679 + 0.018 pH)
(b)
Using the above expression determine the hydrolyzed Si(IV) complexes that can form in the pH
range of 0-14.
Solution
(a)
According to Equation 5.108,
h = [z - M - 0.34n + 0.0494n pH]/[0.679 + 0.0178 pH]
(5.108)
From Equation 5.107,
Si = (2.732 - 0.035 pH - Si )/1.36 (Si )
1/2
Si = (2.732 - 0.035 pH - 1.74)/1.36(1.74)
= (0.992 - 0.035 pH)/1.794
= 0.553 - 0.0195 pH
o
o 1/2
(1)
For Si(IV), z=4 and n=4. Therefore, inserting these values and Si from Equation 1 into Equation
5.108 gives:
h = [4 - 0.553 + 0.0195 pH - 0.34(4) + 0.0494(4) pH]/[0.679 + 0.0178 pH]
= (2.088 + 0.217 pH)/(0.679 + 0.0178 pH)
(b)
(2)
When pH = 0, Equation 2 gives:
h = 2.088/0.679 = 3.1
Correspondingly, for pH = 14,
h = [2.088 + 0.217(14)]/[0.679 + 0.0178(14)]
= 5.126/0.928 = 5.52
Thus, taking h as the nearest whole number, we find that in the pH = 0-14 range, h varies from 3 at
pH=0 to 6 at pH = 14. Therefore, summarizing,
h
3
4
5
6
(4-h)+
M(OH)h(OH2)4-h
+
Si(OH)3(OH2)
o
Si(OH)4
Si(OH)5(OH2)-1 (i.e., SiO(OH)3 )
22Si(OH)6(OH2)-2 (i.e., SiO2(OH)2 )
5.6.4 Polymerization
Polymerization via Olation. In the framework of the partial charge model, olation
may be visualized as a two-step process. First, the negative end of the M-OH group
interacts with the positive end of a hydrated metal cation, giving an intermediate complex.
Next, a water molecule is expelled from the intermediate complex:
M  OH  M   OH2  M  OH  M  OH2 
(5.109)
 M  OH  M  H2 O
It must be noted that the removal of a water molecule is necessary because hydrated metal
cations typically have the highest coordination number available to the M
z+
ion.
In order for polymerization to occur, a negative partial charge must exist on the OH
group in the MOH unit (i.e., OH < 0). Additionally, the positive partial charge on M (in
the MOH unit) must be sufficiently large. The available experimental data suggest M ≥ +
0.3. Figure 5.6 presents typical examples of the different types of hydroxo bridges that are
found in polynuclear hydroxo metal complexes.
Figure 5.6. Types of hydroxo bridges in polynuclear complexes.
EXAMPLE 5.16 Polymerization limit of hydroxo complexes
For each of the following dimeric complexes: (1) determine the partial charge (OH) on the OH group
of the monomer, (2) speculate on the feasibility of dimer formation, and (3) speculate on the feasibility of
further polymerization beyond the dimer.
OH
(a)
[(H2O)4 Cr
Cr (OH2)4]
4+
Ni (OH2)4]
4+
OH
OH
(b)
[(H2O)4 Ni
OH
Solution
(a)
For the monomer, [Cr(OH)(OH2)] , the partial charge on the hydroxyl group is OH = -0.02. Also, M
= 1.56. Thus, the two criteria for polymerization (i.e., OH<0, M ≥ + 0.3) are met. Therefore,
dimerization is feasible. For the dimer, OH = +0.01. Therefore, polymerization is not possible.
(b)
For the monomer, [Ni(OH)(OH2)5] , OH = -0.12. Accordingly dimerization is expected. For the
dimer, OH = -0.10. Therefore further polymerization is feasible. In fact, condensation of two dimers
4+
gives the tetramer [Ni4(OH)4(OH2)12] whose structure is depicted in Table 5.1. For the tetramer, OH
= +0.6 and therefore further polymerization beyond this is not possible.
2+
+
Polymerization via Oxolation. When metal complexes no longer have water
molecules as ligands, the olation pathway for polymerization is no longer available. In this
case, polymerization must proceed via oxolation.
The metal ions that undergo oxolation tend to be high-valent (z>4+) which form
oxo-hydroxo anions, [MOx(OH)m-x]
(m+x-z)-
. Two different pathways are available for
oxolation, depending on whether or not the coordination of the metal ion in the starting
complex is the highest attainable. If coordination is not saturated, then oxolation can lead
to expansion of the coordination:


O
M +
O
M 


 










In this case, there is no need to eliminate hydroxyl groups or water molecules.
If the metal coordination is fully saturated in the starting complex, then oxolation
requires the elimination of hydroxyl groups or water molecules:
-
+


M  OH +  M  OH   M  O  M  OH
H
 M  OH- +  M+  OH M  O  M  OH
H OH
H... OH
 M  O M  M  O  M  M  O  M  



It can be seen from the above steps that the first step involves OH( )M( 
interaction. Thus, a useful guide in ascertaining the feasibility of additional polymerization
is to consider the sign of the partial charge on the OH group.
EXAMPLE 5.17 Polarization of Cr(VI) complexes
Determine the extent to which the following Cr complexes can undergo polymerization:
o
(a) [CrO2(OH)2] , (b) [CrO3(OH)] .
Solution
(a)
The first stage of polymerization is dimer formation:
2[CrO2(OH)2]  [(HO)O2Cr-O-CrO2(OH)] + H2O
o
o
The partial charge on the OH groups in the dimer is determined as OH = +0.04. The positive partial
charge indicates polymerization beyond the dimer is not feasible.
(b)
The relevant dimerization reaction is:
-
2-
2[CrO3(OH)] = [Cr2O7] + H2O
In this case, after forming the dimer, no more hydroxyl groups are available for further polymerization.
Thus, polymerization ceases after the dimer forms.
________________________________________________________________________________________
5.6.5a Precipitation
Formation of a solid metal hydroxide/oxide phase may be viewed in terms of infinite
polymerization of neutral complexes (h=z) via the hydroxo groups. Complete olation
results in a metal hydroxide while complete oxolation gives a metal oxide. Formation of an
oxo bridge from an ol bridge involves the transfer of a proton from one hydroxo group to a
neighboring hydroxo group:
H OH
H... OH
H 
M O M  M OMMOM

It can be seen that this process produces a water molecule. The nature of the partial
charge on this water molecule provides a guide as to the relative importance of ol and oxo
groups in the solid product:
(a)
(H2O) < 0. The metal cation M() is attracted to the water molecule ().
Therefore, the water molecule is retained and a reversal of the proton
transfer occurs. In this case, oxolation is not favored.
(b)
(H2O)>0. The cation M() repels the water molecule. Elimination of the
water molecule is therefore feasible and oxolation is favored.
EXAMPLE 5.18 Olation vs. oxolation in metal metal (hydr)oxide precipitation
Starting with the following neutral soluble species, speculate on the likelihood that precipitation will
result in a hydroxide or oxide solid product:
o
o
Mn(OH)2(OH2)4 , Y(OH)3(OH2)3 ,
o
o
Fe(OH)3(OH2)3 , Zr(OH)4(OH2)4
Solution
Neutral Complex (H2O)
Predicted Solid Phase
Observed
Solid
Phase
o
Mn(OH)2(OH2)4
-0.02
Mn(OH)2
o
Y(OH)3(OH2)3
-0.05
Y(OH)3
o
Fe(OH)3(OH2)3
+0.03
Fe2O3
________________________________________________________________________________________
5.6.5b Precipitation (cont’d)
Precipitation occurs in aqueous solution when the pH reaches the corresponding point
of zero charge. At this pH, the charge on the metal species is zero or nearly zero and,
therefore, electrostatic repulsion is too weak to stop their collision. Following collision,
condensed species form through olation or oxolation. In either case, OH groups are

involved. From the acid side, cationic species M(OH)h(OH2) (zn hh) acquire additional OHgroups and decrease in charge, as the pH is raised. On the other hand, starting from a basic
 z q)
pH, anionic species MOnH (2n
can be condensed by lowering the pH. At the point of
q
zero charge where precipitation occurs, q = 2n-z.
Consider the neutral species (MOnH2n-z) at the point of zero charge. The condition for
the condensation of these species is that OH < 0. Therefore, the limit for condensation is
reached when OH = 0. At this limit the partial charge model gives:
M + nO + (2n-z)H = 0
(5.112)
Also, it follows from the condition OH = 0 that
 = OH = 2.71
(5.113)
EXAMPLE 5.19 The mean electronegativity of the neutral species (MOnH2n-z).
Consider the neutral species (MOnH2n-z). For the limiting condition OH = 0, determine:
a) the mean electronegativity OH
b) the electronegativity of the bonded O and H in OH
c) the mean electronegativity of the neutral species (MOnH2n-z)
d) the partial charges O and H
Solution
a)
Using Equation 2.12h, and recognizing that, for the condition OH = 0 the charge (z) on OH is
zero, gives:
OH = (OH)1/2 = [(3.50)(2.10)]1/2 = 2.71
b) Recalling the principle of electronegativity equalization, when two atoms combine to give a
stable molecule (in this case OH), the two partially charged atoms acquire the same
electronegativity. It follows, therefore, that:
O = H = OH = 2.71
c)
Again, according to the principle of electronegativity equalization, for the stable neutral
molecule (MOnH2n-z), the partially charged atoms acquire the same electronegativity.
Therefore, in view of part (b), and the fact that O, H, and M are atoms in the same stable
molecule, i.e., (MOnH2n-z),
complex = O = H = M = 2.71
d) Recalling Equation 2.12g,


i = ( -  i )/1.36( i )1/2

(2.12g)

Therefore, for  i =  O = 3.50,
o = (2.71 – 3.50)/1.36(3.50)1/2 = -0.310


Similarly, for  i =  H = 2.10,
H = (2.71 – 2.10)/1.36(2.10)1/2 = +0.310
Alternatively, we can argue that given that OH = 0 for the neutral species, it follows that H = -O =
0.310.
e)
We can express the neutral molecule (MOnH2n-z) as (M(OH)nHn-z). It follows, therefore, that
M + nOH + (n-z)H = 0
But OH = 0. Thus,
M = (z-n)H
f)
From Equation 2.12g,


M = ( -  M )/1.36( M )1/2
Thus,


 M + 1.36M( M )1/2 -  = 0
Solution of this quadratic equation gives

( M )1/2 = {-1.36M + [(1.36M)2 + 4]}/2
Making the substitutions M = (z-n)H, H = 0.310,  = 2.71 gives

( M )1/2 = 0.21(n-z) + [2.71 + 0.044(z-n)2]1/2
5.6.6 Alkoxide Reactions
Three important reactions of metal alkoxides (M(OR)n) are hydrolysis (Equation
5.114a), condensation (Equation 5.114b), and chemical modifications (Equation 5.114c)
M  OR + HO  H  M  OH + ROH
(5.114a)
M  OR + HO  M  M  O M + ROH
(5.114b)
M  OR + HO  R  M  O  R + ROH
(5.114c)
All these reactions can be viewed in terms of the following three-step process:
a) A nucleophilic addition of an OH group to the metal atom:
+
X  O + M  OR

H
X  O  M  OR

H
(5.115)
This step is feasible provided OH < 0 and M > 0.
b) A transition state during which a proton from the entering ligand (XOH) attaches to
an oxygen of a neighboring OR group:
-
XOMOR

+H
XO  M  O  R

H
(5.116)
The requirement here is that H > OR.
c) Release of the protonated group (ROH):
+ +
XO  M  O  R

H
XO  M + ROH
(5.117)
Removal of the ROH group requires that ROH > 0.
Thus, consideration of the partial charge distribution leads to the following conditions:
M > 0, H > OR, and ROH > 0
EXAMPLE 5.20 Correlation between M and the hydrolysis rate of silicon alkoxides
The hydrolysis rate of silicon alkoxides (Si(OR) 4) were determined by Aelion and Akerman (
following results were obtained:
R
kx102 M-1s-1[H+]-1
Ethyl
5.1
Butyl
1.9
). The
Hexyl
0.83
Rationalize these results with the aid of the partial charge model. What will be the effect of an increasingly
positive partial charge on Si?
Solution
As noted above, the first step in the alkoxide hydrolysis reaction involves the nucleophilic addition of the
OH group to the metal ion (Equation 5.115). The requirement M > 0 suggests that increasing the positive
partial charge on the metal atom should enhance the reaction rate.
Si(OR)4 can be expressed as Si(OCnH2n+1)4 or SiC4nH8n+4O4. Using Equations 2.12h and 2.12g, the mean
electronegativities and partial charges can be calculated, as summarized in Table E5.20.
 1/ 2
 1/ 2
  [( Si
)
 4n( C )1 / 2  (8n  4)( H )1 / 2  4( O )1 / 2 ] /[1 /( Si
)
 4n /( C )1 / 2  (8n  4) /( H )1 / 2  4 /( O )1 / 2 ]
 i  (   i ) / 1.36 ( i )1/ 2
TABLE E5.20
R
Ethyl
n
2

M
OR
ROH
H
kx102 M-1s-1[H+]-1
5.1
Butyl
Hexyl
4
6
1.9
0.83
EXAMPLE 5.21 Comparison of the hydrolysis rates of titanium and silicon alkoxides
Titanium alkoxides are known to hydrolyze more rapidly than the corresponding silicon precursors.
Rationalize this observation with the aid of the partial charge model. Show that for the alkoxides M(OR) 4
this observation should be expected at least up to R = hexyl.
Solution
Based on Equations 2.12h and 2.12g, the mean electronegativities and the partial charges can be calculated,
as summarized in Table E5.21. It can be seen that in all cases, the positive charge on titanium is
approximately twice that on silicon. Thus, recalling Equation 5.115 above, we should expect the rate of
hydrolysis of titanium alkoxides to be faster than that of the corresponding silicon compounds.
TABLE E5.21
EXAMPLE 5.22 Alcohol vs. water elimination in the condensation of metal alkoxides
In principle, the condensation of the hydrolyzed species M(OR) 3(OH) can occur via the elimination of a
water molecule (Equation 5.114a) or an alcohol molecule (Equation 5.114b):
M(OR)3OH + HO  M(OR)3
M(OR)3OH + RO  M(OR)2(OH)
(RO)3M  O  M(OR)3 + H2O
(RO)3M  O  M(OR)2OH + ROH
(1)
(2)
Refer to the three-step process for alkoxide reactions. Consider the case where R = Pr = C 3H7.
Show that for both Si and Ti, the conditions for steps 1 and 2 are satisfied (i.e., OH < 0, M > 0 (Step
1); H > 0 (Step 2).
b) Determine whether the condensation reaction will favor alcohol or water elimination.
a)
Solution
According to the three-step mechanism for alkoxide reactions, the first step is nucleophilic addition
of an OH group to a metal atom:
Step 1a:
- +
(RO)3M  O + M  (OR)3
 
H OH
(RO)3M  O  M  (OR)3


H OH
Required: OH < 0, M > 0.
Step 2a:
(RO)3M  O  M  (OR)3


H
OH
+
Required: H > OH.
(RO)3MO  M  (OR)3

OH2
Step 3a:
(RO)3M  O  M(OR)3

Required:  H2O > 0.
OH2
(RO)3M O  M(OR)3 + H2O
The first step (Step 1b) of the alcohol elimination pathway involves nucleophilic OH addition.
Step 1b:
- +
(RO)3M  O + M  (OR)2(OH)
 
H OR
(RO)3M O  M  (OR)2(OH)


H
OR
Required: OH < 0, M > 0.
Step 2b:
(RO)3M  O  M  (OR)2(OH)


H
OR
+ -
(RO)3M  O  M  (OR)2(OH)

HOR
Required: H > OR
Step 3b:
(RO)3M  O  M  (OR)2(OH)

HOR
(RO)3M  O  M(OR)2(OH) + ROH
Required: ROH > 0
Compound

M
OH
H
H O
2
ROH
Si(OPr)3(OH)
Ti(OPr)3(OH)
Si2(OPr)6(OH)2
Ti2(OPr)6(OH)2
5.6.7 Metal Complexation
A monodentate monovalent anion X- can form a complex with a hydrated metal ion
via a ligand substitution (anation) reaction in which a water molecule is released:
M(OH2) zn + X- = [M(X)(OH2)n-1](z-1) + H2O
(5.118)
Consider the complex [M(X)(OH2)n-1](z-1). In order for this complex to remain stable in
aqueous solution, it must withstand: (a) the tendency towards ionic dissociation (Equation
5.119) caused by the high dielectric constant of the solvent water, and (b) the tendency
towards hydrolytic dissociation (Equation 5.120):


[M(OH)h(X)(OH2)n-h-1] (z h 1) + H2O = [M(OH)h(OH2)n-h] (zh) + X[M(OH)h(X)(OH2)n-h-1] (z h 1)
(5.119)

= [M(OH)h+1(HX)(OH2)n-h-2] (z h 1)

(5.120)
= [M(OH)h+1(OH2)n-h-2] (zh 1) + HX
The stability of the M  X bond, as related to Equations 5.119 and 5.120, can be
assessed with the help of the partial charge model. In the case of the tendency towards
ionic dissociation (Equation 5.119), the electronegativity of X- can be compared with that
for H2O( = 2.49). If X- is more electronegative (i.e., X >  H2O ) electron transfer from M
to X occurs, the negative charge on X increases (X < -1), and the M X bond becomes
more ionic. The result is enhanced ionization. On the other hand, if x <  H2O , electrons
will be withdrawn from X-, thereby decreasing the negative charge on X (X > -1). The
result is a more covalent M  X bond and a decreased tendency towards ionic
dissociation.
Considering the tendency towards hydrolytic dissociation (Equation 5.120), we
focus on the intermediate species [M(OH)h+1(HX)(OH2)n-h-2](z-h-1). The condition HX < 0
discourages the release of HX and, therefore, the M  X bond is stable. On the other
hand, HX > 0 favors the release of HX and, therefore, hydrolytic dissociation.
EXAMPLE 5.23 Complexation of Fe3+ by bidentate monovalent anions
Determine whether the following monovalent anions can complex the hydrated ferric ion
3




Fe(OH2) 6 : ClO 4 , NO 3 , HSO 4 , H2PO 4 , and CH3COO-. Assume that these ions are bidentate ligands.
Solution
Since these anions are bidentate, they can displace two water molecules to give complexes of the
type [Fe(X)(OH2)4]2+. The corresponding destabilization reactions can be written as:
Ionic dissociation
[Fe(X)(OH2)4]2+ + 2H2O = [Fe(OH2)6]3+ + X-
Hydrolytic dissociation
[Fe(OH)(HX)(OH2)3]2+ = [Fe(OH)(OH2)5]2+ + HX
Table E5.23 presents tabulation of the following quantities: X (the mean electronegativity of anion X-), x
(the partial charge of X in [Fe(X)(OH2)4]2+), and HX (the partial charge of HX in [Fe(OH)(HX)(OH2)4]2+.
Consider an anion Xx-. Define the change in charge x as
x = charge on bonded anion - charge on free anion = x – (-x) = x + x
For a monovalent anion, x = 1 and therefore we get:
x = 1 + x
When x < -1, x < 0, and anion complexation is not feasible. On the other hand, when x > -1, x > 0, and
anion complexation is feasible.
Table E5.23
-
Anion (X )

ClO 4

NO 3

HSO 4

HCO 3
CH3COO-
X
x
x
HX
FURTHER READING
1.
J. N. Butler, Ionic Equilibria:
Reading, MA, 1964.
A Mathematical Approach, Addison-Wesley,
2.
W. Stumm and J. J. Morgan, Aquatic Chemistry, 2nd ed., Wiley, New York, 1981.
3.
D. A. Johnson, Some Thermodynamic Aspects of Inorganic Chemistry, 2nd ed.,
Cambridge, New York, 1982.
4.
A. F. Wells, Structural Inorganic Chemistry., 5th ed., Oxford, New York, 1984.
5.
L. Pauling, The Nature of the Chemical Bond, 3rd. ed., Cornell University Press,
Ithaca, New York, 1960.
6.
J. E. Huheey, E.A. Keiter, and R. L. Keiter, Inorganic Chemistry. Principles of
Structure and Reactivity, 4th ed., HarperColins 1993.
7.
A. G. Sharpe, Inorganic Chemistry, 3rd. ed., Longman, London, 1992.
8.
D. E. Shriver, P. W. Atkins, and C. H. Langford, Inorganic Chemistry, Freeman,
New York, 1990.
9.
B. Webster, Chemical Bonding Theory, Blackwell, Oxford, 1990.
10.
F. A. Cotton and G. Wilkinson, Advanced Inorganic Chemistry, 5th ed., Wiley New
York, 1988.
11.
M. Henry, J. P. Jolivet, and J. Livage, "Aqueous Chemistry of Metal Cations:
Hydrolysis, Condensation and Complexation", Structure and Bonding, 77, 153-206
(1992).
12.
J. Livage, M. Henry, and C. Sanchez, "Sol-Gel Chemistry of Transition Metal
Oxides", Prog. Solid Stage Chem., 18, 259-341 (1988).
13.
J. P. Jolivet, Metal Oxide Chemistry and Synthesis. From Solution to Solid State,
Wiley, New York, 2000.