3. Solubility product(s) for MnO2

HgS ¼ Hg þ S ðKsp1 ¼ ½Hg�½S�Þ ð4Þ

HgI2 ¼ Hg þ I2ðKsp1 ¼ ½Hg�½I2�Þ ð5Þ

<sup>4</sup> <sup>þ</sup> H2S <sup>¼</sup> HgS <sup>þ</sup> 4Cl–<sup>1</sup> <sup>þ</sup> 2Hþ<sup>1</sup> <sup>ð</sup>7<sup>Þ</sup>

, pKsp ¼ 28:55Þ ð6Þ

<sup>4</sup> þ H2S ð7aÞ

<sup>2</sup> , ðpKsp<sup>2</sup> ¼ 17:33Þ ð8Þ

], Ksp (Eq. (3)).

�½NHþ<sup>1</sup>

�½NH3�½HPO�<sup>2</sup>

<sup>4</sup> �½PO�<sup>3</sup>

<sup>4</sup> �Þ ð9Þ

<sup>4</sup> � ¼ KspK1N=K3PÞ ð10Þ

; then pKsp1 ¼ 20:80:

, 1/A = RT/

is far more favored from thermodynamic viewpoint; nonetheless, the solubility product (Ksp) for HgS is commonly formulated on the basis of reaction (3). We obtain pKsp1 = pKsp – 2A

Equilibrium constants are usually formulated for the simplest reaction notations. However, in this respect, Eq. (4) is simpler than Eq. (3). Moreover, we are "accustomed" to apply solubility products with ions (cations and anions) involved, but this custom can easily be overthrown. A similar remark may concern the notation referred to elementary dissociation of mercuric

<sup>ð</sup>Ksp ¼ ½Hgþ<sup>2</sup>

The species in the expression for solubility products do not predominate in real chemical systems, as a rule. However, the precipitation of HgS from acidified (HCl) solution of mercury

Eq. (7) can be applied to formulate the related solubility product, Ksp2, for HgS. To be online with customary requirements put on the solubility product formulation, Eq. (7) should be

HgS <sup>þ</sup> 4Cl–<sup>1</sup> <sup>þ</sup> 2Hþ<sup>1</sup> <sup>¼</sup> HgCl�<sup>2</sup>

<sup>4</sup> �½H2S�

The solubility product for MgNH4PO4 can be formulated on the basis of reactions:

, [H2S] = 1020.0[H+1]

<sup>4</sup> <sup>þ</sup> PO�<sup>3</sup>

<sup>4</sup> <sup>ð</sup>Ksp1 ¼ ½Mgþ<sup>2</sup>

2 [S–<sup>2</sup>

<sup>4</sup> <sup>ð</sup>Ksp ¼ ½Mgþ<sup>2</sup>

salt with H2S solution can be presented in terms of predominating species; we have

�½I �1 � 2

(E01�E02), where <sup>E</sup><sup>01</sup> = 0.850 V for Hg+2 + 2e–<sup>1</sup> <sup>=</sup> Hg, <sup>E</sup><sup>02</sup> <sup>=</sup> –0.48 V for <sup>S</sup> + 2e–<sup>1</sup> = S–<sup>2</sup>

F�ln10, A = 16.92 for 298 K; then pKsp1 = 7.4.

96 Descriptive Inorganic Chemistry Researches of Metal Compounds

we obtain pKsp1 = pKsp – 2A(E01–E03), where

where I2 denotes a soluble form of iodine in a system. From

HgI2 <sup>¼</sup> Hgþ<sup>2</sup> <sup>þ</sup> 2I�<sup>1</sup>

HgCl�<sup>2</sup>

<sup>K</sup>sp2 <sup>¼</sup> <sup>½</sup>HgCl�<sup>2</sup>

½Cl�<sup>1</sup> � 4 ½H<sup>þ</sup><sup>1</sup> �

] 4

Applying the law of mass action to Eq. (7a), we have

] = 1015.07[Hg+2][Cl–<sup>1</sup>

MgNH4PO4 <sup>¼</sup> Mgþ<sup>2</sup> <sup>þ</sup> NH3 <sup>þ</sup> HPO�<sup>2</sup>

MgNH4PO<sup>4</sup> <sup>¼</sup> Mgþ<sup>2</sup> <sup>þ</sup> NHþ<sup>1</sup>

<sup>E</sup><sup>01</sup> <sup>¼</sup> <sup>0</sup>:850 V for Hg<sup>þ</sup><sup>2</sup> <sup>þ</sup> 2e�<sup>1</sup> <sup>¼</sup> Hg, E<sup>03</sup> <sup>¼</sup> <sup>0</sup>:621 V for I2 <sup>þ</sup> 2e�<sup>1</sup> <sup>¼</sup> 2I�<sup>1</sup>

iodide precipitate

rewritten into the form

where [HgCl4

–2

The scheme presented above cannot be extended to all oxides. For example, one cannot recommend the formulation of this sequence for MnO2, i.e., MnO2 + 2H2O = Mn(OH)4 ) Mn(OH)4 = Mn+4 + 4OH–<sup>1</sup> ) <sup>K</sup>sp0 = [Mn+4][OH–<sup>1</sup> ] 4 ; Mn+4 ions do not exist in aqueous media, and MnO2 is the sole Mn(+4) species present in such systems. In effect, Ksp0 for MnO2 is not known in the literature, compare with Ref. [32]. However, the Ksp for MnO2 can be formally calculated according to an unconventional approach, based on the disproportionation reaction

$$5\text{MnO}\_2 + 4\text{H}^+ = 2\text{MnO}\_4^{-1} + 3\text{Mn}^{+2} + \text{H}\_2\text{O} \tag{12}$$

reverse to the symproportionation reaction 2MnO4 �<sup>1</sup> + 3Mn+2 + H2O=5MnO2 + 4H+1. The Ksp = Ksp1 value can be found there on the basis of E<sup>01</sup> and E<sup>02</sup> values [33], specified for reactions:

$$\text{MnO}\_4^{-1} + 4\text{H}^+ + 3\text{e}^{-1} = \text{MnO}\_2 + 2\text{H}\_2\text{O}(E\_{01} = 1.692\text{ V})\tag{13}$$

$$\text{MnO}\_2 + 4\text{H}^+ + 2\text{e}^- = \text{Mn}^{+2} + 2\text{H}\_2\text{O} (\text{E}\_{02} = 1.228 \text{ V}) \tag{14}$$

Eqs. (13) and (14) are characterized by the equilibrium constants:

$$K\_{\rm e1} = \frac{[\text{MnO}\_2][\text{H}\_2\text{O}]^2}{[\text{MnO}\_4^{-1}][\text{H}^+]^4[\text{e}^{-1}]^3}, \qquad K\_{\rm e2} = \frac{[\text{Mn}^{+2}][\text{H}\_2\text{O}]^2}{[\text{MnO}\_2][\text{H}^+]^4[\text{e}^{-1}]^2} \tag{15}$$

defined on the basis of mass action law (MAL) [14], where logKe1 = 3�A�E01, logKe2 = 2�A�E02, A = 16.92. From Eqs. (13) and (14), we get

$$2\mathbf{MnO}\_2 + 4\mathbf{H}\_2\mathbf{O} + 3\mathbf{MnO}\_2 + 12\mathbf{H}^{+1} + 6\mathbf{e}^{-1} = 2\mathbf{MnO}\_4^{-1} + 8\mathbf{H}^{+1} + 6\mathbf{e}^{-1} + 3\mathbf{Mn}^{+2} + 6\mathbf{H}\_2\mathbf{O} \tag{16}$$

Assuming [MnO2] = 1 and [H2O] = 1 on the stage of the Ksp1 formulation for reaction (16), equivalent to reaction (12), we have

$$K\_{\rm sp1} = \frac{[\rm MnO\_4^{-1}]^2 [\rm Mn^{+2}]^3}{[\rm H^{+1}]^4} \tag{17}$$

and then

$$K\_{\rm sp1} = (K\_{\rm e2})^3 \cdot (K\_{\rm e1})^{-2} \tag{18}$$

$$p\text{K}\_{\text{sp1}} = 3\log\text{K}\_{\text{e2}} - 2\log\text{K}\_{\text{el}} = 6A(\text{E}\_{01} - \text{E}\_{02}) = 6 \cdot 16.92 \cdot (1.692 - 1.228) = 47.11 \tag{19}$$

The solubility products with MnO2 involved can be formulated on the basis of other reactions. For example, addition of

$$\mathbf{M} \mathbf{n}^{+2} = \mathbf{M} \mathbf{n}^{+3} + \mathbf{e}^{-1} \tag{20}$$

to Eq. (14) gives

Solubility Products and Solubility Concepts http://dx.doi.org/10.5772/67840 99

$$\mathbf{M}\mathbf{n}\mathbf{O}\_2 + 4\mathbf{H}^{+1} + 2\mathbf{e}^{-1} + \mathbf{M}\mathbf{n}^{+2} = \mathbf{M}\mathbf{n}^{+2} + 2\mathbf{H}\_2\mathbf{O} + \mathbf{M}\mathbf{n}^{+3} + \mathbf{e}^{-1} \tag{21}$$

Multiplication of Eq. (21) by 3, and then addition to Eq. (13a)

$$\mathbf{MnO\_2} + 2\mathbf{H\_2O} = \mathbf{MnO\_4^{-1}} + 4\mathbf{H^{+1}} + 3\mathbf{e^{-1}}$$

(reverse to Eq. (13)) gives the equation

$$\begin{aligned} &3\mathbf{M}\mathbf{n}\mathbf{O}\_2 + 12\mathbf{H}^{+1} + 6\mathbf{e}^{-1} + 3\mathbf{M}\mathbf{n}^{+2} + \mathbf{M}\mathbf{n}\mathbf{O}\_2 + 2\mathbf{H}\_2\mathbf{O} \\ &= 3\mathbf{M}\mathbf{n}^{+2} + 6\mathbf{H}\_2\mathbf{O} + 3\mathbf{M}\mathbf{n}^{+3} + 3\mathbf{e}^{-1} + \mathbf{M}\mathbf{n}\mathbf{O}\_4^{-1} + 4\mathbf{H}^{+1} + 3\mathbf{e}^{-1} \end{aligned} \tag{22}$$

and its equivalent form, obtained after simplifications,

$$4\text{MnO}\_2 + 8\text{H}^+ = 3\text{Mn}^{+3} + \text{MnO}\_4^{-1} + 4\text{H}\_2\text{O} \tag{22a}$$

Eq. (22) and then Eq. (22a) is characterized by the solubility product

$$K\_{\rm sp2} = \frac{[\rm MnO\_4^{-1}][\rm Mn^{+3}]^3}{[\rm H^{+1}]^8} = (K\_{\rm e2})^3 \cdot (K\_{\rm e3})^{-3} \cdot (K\_{\rm e1})^{-1} \tag{23}$$

where

Mn(OH)4 = Mn+4 + 4OH–<sup>1</sup> ) <sup>K</sup>sp0 = [Mn+4][OH–<sup>1</sup>

98 Descriptive Inorganic Chemistry Researches of Metal Compounds

reverse to the symproportionation reaction 2MnO4

Eqs. (13) and (14) are characterized by the equilibrium constants:

<sup>4</sup> �½H<sup>þ</sup><sup>1</sup> � 4 ½e �1 �

<sup>K</sup>e1 <sup>¼</sup> <sup>½</sup>MnO2�½H2O�

<sup>2</sup>MnO2 <sup>þ</sup> 4H2O <sup>þ</sup> <sup>3</sup>MnO2 <sup>þ</sup> 12Hþ<sup>1</sup> <sup>þ</sup> 6e�<sup>1</sup> <sup>¼</sup> 2MnO�<sup>1</sup>

½MnO�<sup>1</sup>

MnO�<sup>1</sup>

A = 16.92. From Eqs. (13) and (14), we get

equivalent to reaction (12), we have

and then

For example, addition of

to Eq. (14) gives

] 4

and MnO2 is the sole Mn(+4) species present in such systems. In effect, Ksp0 for MnO2 is not known in the literature, compare with Ref. [32]. However, the Ksp for MnO2 can be formally calculated according to an unconventional approach, based on the disproportionation reaction

Ksp = Ksp1 value can be found there on the basis of E<sup>01</sup> and E<sup>02</sup> values [33], specified for reactions:

defined on the basis of mass action law (MAL) [14], where logKe1 = 3�A�E01, logKe2 = 2�A�E02,

Assuming [MnO2] = 1 and [H2O] = 1 on the stage of the Ksp1 formulation for reaction (16),

<sup>4</sup> � 2 ½Mn<sup>þ</sup><sup>2</sup> � 3

½H<sup>þ</sup><sup>1</sup> �

<sup>3</sup> � ðKe1<sup>Þ</sup>

pKsp1 ¼ 3logKe2 � 2logKe1 ¼ 6AðE<sup>01</sup> � E02Þ ¼ 6 � 16:92 � ð1:692 � 1:228Þ ¼ 47:11 ð19Þ

The solubility products with MnO2 involved can be formulated on the basis of other reactions.

<sup>K</sup>sp1 <sup>¼</sup> <sup>½</sup>MnO�<sup>1</sup>

Ksp1 ¼ ðKe2Þ

<sup>4</sup> <sup>þ</sup> 4Hþ<sup>1</sup> <sup>þ</sup> 3e�<sup>1</sup> <sup>¼</sup> MnO2 <sup>þ</sup> 2H2OðE<sup>01</sup> <sup>¼</sup> <sup>1</sup>:692 VÞ ð13<sup>Þ</sup>

MnO2 <sup>þ</sup> 4Hþ<sup>1</sup> <sup>þ</sup> 2e�<sup>1</sup> <sup>¼</sup> Mnþ<sup>2</sup> <sup>þ</sup> 2H2OðE<sup>02</sup> <sup>¼</sup> <sup>1</sup>:228 VÞ ð14<sup>Þ</sup>

<sup>3</sup> , Ke2 <sup>¼</sup> <sup>½</sup>Mn<sup>þ</sup><sup>2</sup>

<sup>5</sup>MnO2 <sup>þ</sup> 4Hþ<sup>1</sup> <sup>¼</sup> 2MnO�<sup>1</sup>

2

; Mn+4 ions do not exist in aqueous media,

<sup>4</sup> <sup>þ</sup> 3Mnþ<sup>2</sup> <sup>þ</sup> H2O <sup>ð</sup>12<sup>Þ</sup>

�<sup>1</sup> + 3Mn+2 + H2O=5MnO2 + 4H+1. The

�½H2O� 2

� 4 ½e �1 �

<sup>4</sup> <sup>þ</sup> 8Hþ<sup>1</sup> <sup>þ</sup> 6e�<sup>1</sup> <sup>þ</sup> 3Mnþ<sup>2</sup> <sup>þ</sup> 6H2O

<sup>4</sup> ð17Þ

�<sup>2</sup> <sup>ð</sup>18<sup>Þ</sup>

Mnþ<sup>2</sup> <sup>¼</sup> Mnþ<sup>3</sup> <sup>þ</sup> <sup>e</sup>�<sup>1</sup> <sup>ð</sup>20<sup>Þ</sup>

<sup>2</sup> ð15Þ

ð16Þ

½MnO2�½H<sup>þ</sup><sup>1</sup>

$$K\_{\rm e3} = \frac{[\rm Mn^{+2}]}{[\rm Mn^{+3}][e^{-1}]} \tag{24}$$

for Mn+3 + e�<sup>1</sup> = Mn+2 (E<sup>03</sup> = 1.509 V) (reverse to Eq. (20)), logKe3 <sup>=</sup> <sup>A</sup>�E03. Then

$$pK\_{\rm sp2} = 3A \cdot (E\_{01} - 2E\_{02} + E\_{03}) + 37.82 \tag{25}$$

Formulation of Ksp<sup>i</sup> for other combinations of redox and/or nonredox reactions is also possible. This way, some derivative solubility products are obtained. The choice between the "output" and derivative solubility product values is a matter of choice. Nevertheless, one can choose the Ksp3 value related to the simplest expression for the solubility product Ksp3 = [Mn+2][MnO4 �2 ] involved with reaction 2MnO2 = Mn+2 + MnO4 �2 .

As results from calculations, the low Ksp<sup>i</sup> (i = 1,2,3) values obtained from the calculations should be crossed, even in acidified solution with the related manganese species presented in Figure 1. In the real conditions of analysis, at C<sup>a</sup> = 1.0 mol/L, the system is homogeneous during the titration, also after crossing the equivalence point, at Φ = Φeq > 0.2; this indicates that the corresponding manganese species form a metastable system [34], unable for the symproportionation reactions.

Figure 1. The log[Xi] versus Φ relationships for different manganese species Xi, plotted for titration of V<sup>0</sup> = 100 mL solution of FeSO4 (C<sup>0</sup> = 0.01 mol/L) + H2SO4 (C<sup>a</sup> = 1.0 mol/L) with V mL of C = 0.02 mol/L KMnO4; Φ = CV/(C<sup>0</sup>V0). The species Xi are indicated at the corresponding lines.

#### 4. Calculation of solubility

In this section, we compare two options applied to the subject in question. The first/criticized option, met commonly in different textbooks, is based on the stoichiometric considerations, resulting from dissociation of a precipitate, characterized by the solubility product Ksp value, and considered a priori as an equilibrium solid phase in the system in question; the solubility value obtained this way will be denoted by s\* [mol/L]. The second option, considered as a correct resolution of the problem, is based on full physicochemical knowledge of the system, not limited only to Ksp value (as in the option 1); the solubility value thus obtained is denoted as s [mol/L]. The second option fulfills all requirements expressed in GATES and involved with basic laws of conservation in the systems considered. Within this option, we check, among others, whether the precipitate is really the equilibrium solid phase. The results (s\* , s) obtained according to both options (1 and 2) are compared for the systems of different degree of complexity. The unquestionable advantages of GATES will be stressed this way.

#### 4.1. Formulation of the solubility s\*

The solubility s\* will be calculated for a pure precipitate of: (1o ) AaBb or (2o ) AaBbCc type, when introduced into pure water. Assuming [A]=a∙s \* and [B]=b∙s \* , from Eq. (1), we have

$$\mathbf{s}^\* = \left(\frac{K\_{\rm sp}}{\mathbf{a}^\mathbf{a} \cdot \mathbf{b}^\mathbf{b}}\right)^{1/(\mathbf{a}+\mathbf{b})} \tag{26}$$

and assuming [A]=a∙s \* , [B]=b∙s \* , [C]=c∙s \* , from Eq. (2), we have

$$\mathbf{s}^\* = \left(\frac{K\_{\rm sp}}{\mathbf{a}^\mathbf{a} \cdot \mathbf{b}^\mathbf{b} \cdot \mathbf{c}^c}\right)^{1/(\mathbf{a}+\mathbf{b}+\mathbf{c})} \tag{27}$$

As a rule, the formulas (26) and (27) are invalid for different reasons, indicated in this chapter. This invalidity results, among others, from inclusion of the simplest/minor species in Eq. (26) or (27) and omission of hydroxo-complexes + other soluble complexes formed by A, and protocomplexes + other soluble complexes, formed by B. In other words, not only the species entering the expression for the related solubility product are present in the solution considered. Then the concentrations: [A], [B] or [A], [B], and [C] are usually minor species relative to the other species included in the respective balances, considered from the viewpoint of GATES [8].

#### 4.2. Dissolution of hydroxides

4. Calculation of solubility

species Xi are indicated at the corresponding lines.




log[Xi

]

100 Descriptive Inorganic Chemistry Researches of Metal Compounds



**MnSO4**

**Mn+2**

**Mn(OH) +2**

**Mn+3**

0

4.1. Formulation of the solubility s\*

In this section, we compare two options applied to the subject in question. The first/criticized option, met commonly in different textbooks, is based on the stoichiometric considerations, resulting from dissociation of a precipitate, characterized by the solubility product Ksp value, and considered a priori as an equilibrium solid phase in the system in question; the solubility value obtained this way will be denoted by s\* [mol/L]. The second option, considered as a correct resolution of the problem, is based on full physicochemical knowledge of the system, not limited only to Ksp value (as in the option 1); the solubility value thus obtained is denoted as s [mol/L]. The second option fulfills all requirements expressed in GATES and involved with basic laws of conservation in the systems considered. Within this option, we check, among

Figure 1. The log[Xi] versus Φ relationships for different manganese species Xi, plotted for titration of V<sup>0</sup> = 100 mL solution of FeSO4 (C<sup>0</sup> = 0.01 mol/L) + H2SO4 (C<sup>a</sup> = 1.0 mol/L) with V mL of C = 0.02 mol/L KMnO4; Φ = CV/(C<sup>0</sup>V0). The

0.0 0.1 0.2 0.3 0.4 F

**MnO4 - 1**

**MnO4 - 1**

**Mn(OH) +2**

**MnO4 - 2**

**Mn(OH) +1**

according to both options (1 and 2) are compared for the systems of different degree of

, s) obtained

) AaBbCc type,

) AaBb or (2o

, from Eq. (1), we have

\*

\* and [B]=b∙s

others, whether the precipitate is really the equilibrium solid phase. The results (s\*

complexity. The unquestionable advantages of GATES will be stressed this way.

The solubility s\* will be calculated for a pure precipitate of: (1o

when introduced into pure water. Assuming [A]=a∙s

We refer first to the simplest two-phase systems, with insoluble hydroxides as the solid phases. In all instances, s\* denotes the solubility obtained from stoichiometric considerations, whereas s relates to the solubility calculated on the basis of full/attainable physicochemical knowledge related to the system in question where, except the solubility product (Ksp), other physicochemical data are also involved.

Applying formula (26) to hydroxides (B = OH�<sup>1</sup> ): Ca(OH)2 (pKsp1 = 5.03) and Fe(OH)3 (pKsp2 = 38.6), we have [35]

$$\mathbf{Ca(OH)}\_{2} = \mathbf{Ca^{+2}} + 2\mathbf{OH^{-1}}\\(K\_{\mathrm{sp1}} = [\mathrm{Ca^{+2}}][\mathrm{OH^{-1}}]^2, \ \mathrm{s^\*} = (K\_{\mathrm{sp1}}/4)^{1/3} = 0.0133 \text{ mol/L} \qquad (28)$$

$$\mathrm{Fe(OH)}\_{3} = \mathrm{Fe^{+3}} + \mathrm{3OH^{-1}}\\(\mathrm{K\_{sp2}} = [\mathrm{Fe^{+3}}][\mathrm{OH^{-1}}]^3, \mathrm{ s}^\* = (\mathrm{K\_{sp2}}/27)^{1/4} = 0.98 \times 10^{-10} \mathrm{mol/L} \,\mathrm{s} \,\tag{29}$$

respectively. However, Ca+2 and Fe+3 form the related hydroxo-complexes: [CaOH+1] = 101.3�[Ca+2][OH�<sup>1</sup> ] and: [FeOH+2] = 1011.0�[Fe+3][OH�<sup>1</sup> ], [Fe(OH)2 +1] = 1021.7�[Fe+3][OH�<sup>1</sup> ] 2 ; [Fe2(OH)2 +4] = 1025.1�[Fe+3] 2 [OH�<sup>1</sup> ] <sup>2</sup> [31]. The corrected expression for the solubility of Ca(OH)2 is as follows

$$\mathbf{s} = [\mathbf{C}\mathbf{a}^{+2}] + [\mathbf{C}\mathbf{a}\mathbf{O}\mathbf{H}^{+1}] \tag{30}$$

Inserting [Ca+2] = Ksp1/[OH�<sup>1</sup> ] <sup>2</sup> and [OH�<sup>1</sup> ] = KW/[H+1], [H+1] = 10�pH (pK<sup>W</sup> = 14.0 for ionic product of water, KW) into the charge balance

$$\mathbf{2[Ca^{+2}]} + [\mathbf{CaOH^{+1}}] + [\mathbf{H^{+1}}] - [\mathbf{OH^{-1}}] = \mathbf{0} \tag{31}$$

we get, by turns,

$$2 \cdot 10^{-5.03} / [\text{OH}^{-1}]^2 + 10^{1.3} \cdot 10^{-5.03} / [\text{OH}^{-1}][\text{H}^{+1}] - [\text{OH}^{-1}] = 0$$

$$\Rightarrow 2 \cdot 10^{-5.03 + 28 - 2 \text{pH}} + 10^{1.3} \cdot 10^{-5.03 + 14 - \text{pH}} + 10^{-\text{pH}} - 10^{\text{pH}-14} = 0$$

$$\text{y} (\text{pH}) = 2 \cdot 10^{22.97 - 2 \text{pH}} + 10^{10.17 - \text{pH}} + 10^{-\text{pH}} - 10^{\text{pH}-14} = 0 \tag{32}$$

where pH ¼ �log½H<sup>þ</sup><sup>1</sup> �. Applying the zeroing procedure to Eq. (30), we get pH0 = 12.453 (Table 1), where: [Ca+2] = 0.0116, [CaOH+1] = 0.00656, s = 0.0182 mol/L (Eq. (28)). As we see, [CaOH+1] is comparable with [Ca+2], and there are none reasons to omit [CaOH+1] in Eq. (28).

The alkaline reaction in the system with Ca(OH)2 results immediately from Eq. (29): [OH�<sup>1</sup> ] – [H+1]=2½Caþ<sup>2</sup> �þ½CaOH<sup>þ</sup><sup>1</sup> � > 0:

Analogously, for the system with Fe(OH)3, we have the charge balance

$$2[\text{Fe}^{+3}] + 2[\text{FeOH}^{+2}] + [\text{Fe(OH)}\_{2}^{+1}] + 4[\text{Fe}\_{2}(\text{OH})\_{2}^{+4}] + [\text{H}^{+1}] - [\text{OH}^{-1}] = 0 \tag{33}$$

and then

$$\text{y(pH)} = 3 \cdot 10^{3.4 - 3 \text{pH}} + 2 \cdot 10^{0.4 - 2 \text{pH}} + 10^{-2.9 - \text{pH}} + 4 \cdot 10^{3.9 - 4 \text{pH}} + 10^{-\text{pH}} \text{--} 10^{\text{pH}-14} = 0 \qquad (34)$$

Eq. (32) zeroes at pH0 = 7.0003 (Table 2), where the value

$$\mathbf{s} = \mathbf{[Fe^{+3}]} + \mathbf{[FeOH^{+2}]} + \mathbf{[Fe(OH)\_2^{+1}]} + 2\mathbf{[Fe\_2(OH)\_2^{+4}]} \tag{35}$$

is close to s ffi [Fe(OH)2 +1] = 10–9.9. Alkaline reaction for this system, i.e., [OH�<sup>1</sup> ] > [H+1], results immediately from Eq. (30), and pH0 = 7.0003 (>7).

At pH = 7, Fe(OH)2 +1 (not Fe+3) is the predominating species in the system, [Fe(OH)2 +1]/[Fe+3] = 1021.7–<sup>14</sup> = 5�107 , i.e., the equality/assumption s\* = [Fe+3] is extremely invalid. Moreover, the value [OH�<sup>1</sup> ]=3�s \* = 2.94�10–<sup>10</sup> = 10–9.532, i.e., pH = 4.468; this pH-value is contradictory with the inequality [OH�<sup>1</sup> ] > [H+1] resulting from Eq. (31). Similarly, extremely invalid result was


Table 1. Zeroing the function (30) for the system with Ca(OH)2 precipitate introduced into pure water (copy of a fragment of display).


<sup>2</sup>½Caþ<sup>2</sup>

�

<sup>2</sup> � <sup>10</sup>�5:<sup>03</sup>=½OH�<sup>1</sup>

102 Descriptive Inorganic Chemistry Researches of Metal Compounds

�þ½CaOH<sup>þ</sup><sup>1</sup>

� þ <sup>2</sup>½FeOH<sup>þ</sup><sup>2</sup>

<sup>3</sup>½Feþ<sup>3</sup>

is close to s ffi [Fe(OH)2

]=3�s

At pH = 7, Fe(OH)2

the inequality [OH�<sup>1</sup>

1021.7–<sup>14</sup> = 5�107

fragment of display).

value [OH�<sup>1</sup>

� > 0:

Eq. (32) zeroes at pH0 = 7.0003 (Table 2), where the value

<sup>s</sup> ¼ ½Feþ<sup>3</sup>

immediately from Eq. (30), and pH0 = 7.0003 (>7).

pH y(pH) [OH�<sup>1</sup>

Analogously, for the system with Fe(OH)3, we have the charge balance

�þ½FeðOHÞ

�þ½FeOH<sup>þ</sup><sup>2</sup>

we get, by turns,

where pH ¼ �log½H<sup>þ</sup><sup>1</sup>

[H+1]=2½Caþ<sup>2</sup>

and then

�þ½CaOH<sup>þ</sup><sup>1</sup>

�þ½Hþ<sup>1</sup>

<sup>2</sup> <sup>þ</sup> 101:<sup>3</sup> � <sup>10</sup>�5:<sup>03</sup>=½OH�<sup>1</sup>

) <sup>2</sup> � <sup>10</sup>�5:03þ28�2pH <sup>þ</sup> 101:<sup>3</sup> � <sup>10</sup>�5:03þ14�pH <sup>þ</sup> <sup>10</sup>�pH–10pH�<sup>14</sup> <sup>¼</sup> <sup>0</sup>

(Table 1), where: [Ca+2] = 0.0116, [CaOH+1] = 0.00656, s = 0.0182 mol/L (Eq. (28)). As we see, [CaOH+1] is comparable with [Ca+2], and there are none reasons to omit [CaOH+1] in Eq. (28).

The alkaline reaction in the system with Ca(OH)2 results immediately from Eq. (29): [OH�<sup>1</sup>

þ1

<sup>2</sup> � þ 4½Fe2ðOHÞ

<sup>y</sup>ðpHÞ ¼ <sup>3</sup> � 103:4�3pH <sup>þ</sup> <sup>2</sup> � 100:4�2pH <sup>þ</sup> <sup>10</sup>�2:9�pH <sup>þ</sup> <sup>4</sup> � 103:9�4pH <sup>þ</sup> <sup>10</sup>�pH–10pH�<sup>14</sup> <sup>¼</sup> <sup>0</sup> <sup>ð</sup>34<sup>Þ</sup>

�þ½FeðOHÞ

12.451 0.000377 0.02825 0.01169 0.006592 12.452 0.000193 0.02831 0.01164 0.006577 12.453 8.30E-06 0.02838 0.01159 0.006561 12.454 �0.000176 0.02844 0.01153 0.006546 12.455 �0.000359 0.02851 0.01148 0.006531

Table 1. Zeroing the function (30) for the system with Ca(OH)2 precipitate introduced into pure water (copy of a

+1] = 10–9.9. Alkaline reaction for this system, i.e., [OH�<sup>1</sup>

+1 (not Fe+3) is the predominating species in the system, [Fe(OH)2

þ1

, i.e., the equality/assumption s\* = [Fe+3] is extremely invalid. Moreover, the

\* = 2.94�10–<sup>10</sup> = 10–9.532, i.e., pH = 4.468; this pH-value is contradictory with

] > [H+1] resulting from Eq. (31). Similarly, extremely invalid result was

�–½OH�<sup>1</sup>

�½Hþ<sup>1</sup>

<sup>y</sup>ðpHÞ ¼ <sup>2</sup> � 1022:97�2pH <sup>þ</sup> 1010,17�pH <sup>þ</sup> <sup>10</sup>�pH–10pH�<sup>14</sup> <sup>¼</sup> <sup>0</sup> <sup>ð</sup>32<sup>Þ</sup>

�. Applying the zeroing procedure to Eq. (30), we get pH0 = 12.453

þ4 <sup>2</sup> �þ½Hþ<sup>1</sup>

<sup>2</sup> � þ 2½Fe2ðOHÞ

��½OH�<sup>1</sup>

� ¼ 0 ð31Þ

] –

� ¼ 0 ð33Þ

<sup>2</sup> � ð35Þ

] > [H+1], results

+1]/[Fe+3] =

� ¼ 0

�–½OH�<sup>1</sup>

þ4

] [Ca+2] [CaOH+1]

Table 2. Zeroing the function (32) for the system with Fe(OH)3 precipitate introduced into pure water (copy of a fragment of display).

obtained in Ref. [36], where the strong hydroxo-complexes were totally omitted, and weak chloride complexes of Fe+3 ions were included into considerations.

Taking only the main dissociating species formed in the solution saturated with respect to Fe(OH)3, we check whether the reaction Fe(OH)3 = Fe(OH)2 +1 + OH�<sup>1</sup> with Ksp1 = [Fe(OH)2 +1] [OH�<sup>1</sup> ] = 1021.7�10–38.6 = 10–16.9 can be used for calculation of solubility s<sup>0</sup> ¼ ðKsp1<sup>Þ</sup> <sup>1</sup>=<sup>2</sup> for Fe(OH)3; the answer is also negative. Simply, the main part of OH�<sup>1</sup> ions originates here from dissociation of water, where the precipitate has been introduced, and then Fe(OH)2 +1 and OH�<sup>1</sup> differ significantly. As we see, the diversity in Ksp value related to a precipitate depends on its dissociation reaction notation, which disqualifies the calculation of s\* based solely on the Ksp value. This fact was not stressed in the literature issued hitherto.

Concluding, the application of the option 1, based on the stoichiometry of the reaction (29), leads not only to completely inadmissible results for s<sup>+</sup> , but also to a conflict with one of the fundamental rules of conservation obligatory in electrolytic systems, namely the law of charge conservation.

Similarly, critical/disqualifying remarks can be related to the series of formulas considered in the chapter [37], e.g., Ksp = 27(s\* ) <sup>4</sup> for precipitates of A3B and AB3 type, and Ksp = 108(s\* ) <sup>5</sup> for A2B3 and A3B2. For Ca5(PO4)3OH, the formula Ksp = 84375(s\* ) <sup>9</sup> (!) was applied [38].

As a third example let us take a system, where an excess of Zn(OH)2 precipitate is introduced into pure water. It is usually stated that Zn(OH)2 dissociates according to the reaction

$$\text{Zn}(\text{OH})\_2 = \text{Zn}^{+2} + 2\text{OH}^{-1} \tag{36}$$

applied to formulate the expression for the solubility product

$$K\_{\rm sp3} = [\text{Zn}^{+2}][\text{OH}^{-1}]^2 (pK\_{\rm sp3} = 15.0) \tag{37}$$

The soluble hydroxo-complexes Zn(OH)i +2�<sup>i</sup> (i=1,…,4), with the stability constants, K<sup>i</sup> OH, expressed by the values logKi OH = 4.4, 11.3, 13.14, 14.66, are also formed in the system in question. The charge balance (ChB) has the form

$$-2[\text{Zn}^{+2}] + [\text{ZnOH}^{+1}] - [\text{Zn(OH)}\_3^{-1}] - 2[\text{Zn(OH)}\_4^{-2}] + [\text{H}^{+1}] - [\text{OH}^{-1}] = 0\tag{38}$$

$$\text{i.e., } 2 \cdot 10^{-15} \text{[OH}^{-1}]^2 + 10^{4.4} \cdot 10^{-15} \text{[OH}^{-1}] - 10^{13.14} \cdot 10^{-15} \text{[OH}^{-1}] - 2 \cdot 10^{14.66} \cdot 10^{-15} \text{[OH}^{-1}]^2 = 0$$

$$\text{y} (\text{pH}) = 2 \cdot 10^{13 - 2 \text{pH}} + 10^{3.4 - \text{pH}} \text{-} 10^{-15.86 + \text{pH}} - 2 \cdot 10^{-28.34 + 2 \text{pH}} + 10^{-\text{pH}} \text{-} 10^{\text{pH}-14} = 0 \qquad (39)$$

The function (39) zeroes at pH0 = 9.121 (see Table 3). The basic reaction of this system is not immediately stated from Eq. (38) (there are positive and negative terms in expression for [OH�<sup>1</sup> ] � [H+1]). The solubility s value

$$\mathbf{s} = \mathbf{[Zn^{+2}]} + \mathbf{[ZnOH^{+1}]} + \mathbf{[Zn(OH)\_2]} + \mathbf{[Zn(OH)\_3^{-1}]} + \mathbf{[Zn(OH)\_4^{-2}]} = 2.07 \cdot 10^{-4} \text{ s}$$

calculated at this point is different from s\* = (Kso3/4)1/3 = 6.3�10�<sup>6</sup> , and [OH�<sup>1</sup> ]/[Zn+2] 6¼ 2; such incompatibilities contradict application of this formula.

#### 4.3. Dissolution of MeL2-type salts

Let us refer now to dissolution of precipitates MeL2 formed by cations Me+2 and anions L�<sup>1</sup> of a strong acid HL, as presented in Table 4. When an excess of MeL2 is introduced into pure water, the concentration balances and charge balance in two-phase system thus formed are as follows:


Table 3. Zeroing the function (39) for the system with Zn(OH)2 precipitate introduced into water; pK<sup>W</sup> = 14.


Table 4. logKi OH and logKi values for the stability constants Ki and Kj of soluble complexes Me(OH)<sup>i</sup> +2-i and MeL<sup>j</sup> +2-j and pKsp values for the precipitates MeL2; [MeL<sup>i</sup> +2-i ]=Ki[Me+2][L�<sup>1</sup> ] i , Ksp = [Me+2][L�<sup>1</sup> ] 2 .

$$\left[\text{MeL}\_2\right] + \left[\text{Me}^{\cdot+2}\right] + \sum\_{i=1}^{l} \left[\text{Me}(\text{OH})\_i^{\cdot+2-i}\right] + \sum\_{j=1}^{l} \left[\text{MeL}\_j^{\cdot+2-j}\right] = \text{C}\_{\text{Me}}\tag{40}$$

$$\mathbf{2}[\mathbf{MeL}\_2] + [\mathbf{L}^{-1}] + \sum\_{j=1}^{J} j[\mathbf{MeL}\_j^{+2-j}] = \mathbf{C\_L} \tag{41}$$

$$\left[\text{H}^{+1}\right]-\left[\text{OH}^{-1}\right]+2\left[\text{Me}^{+2}\right]+\sum\_{i=1}^{l}(2-i)\left[\text{Me}(\text{OH})\_{i}^{+2-i}\right]+\sum\_{j=1}^{l}(2-j)\left[\text{MeL}\_{j}^{+2-j}\right]-\left[\text{L}^{-1}\right]=0\quad(42)$$

where [MeL2] denotes the concentration of the precipitate MeL2. At C<sup>L</sup> = 2CMe, we have

$$\text{L[Me}^{\cdot+2}] + 2\sum\_{i=1}^{I} [\text{Me}(\text{OH})\_i^{\cdot+2-i}] + \sum\_{j=1}^{J} (2-j)[\text{MeL}\_j^{\cdot+2-j}] = [\text{L}^{-1}] \tag{43}$$

From Eqs. (40) and (41)

<sup>2</sup>½Znþ<sup>2</sup>

]

i.e., 2�10�15/[OH�<sup>1</sup>

<sup>s</sup> ¼ ½Znþ<sup>2</sup>

4.3. Dissolution of MeL2-type salts

Me+2 MeOH+1 Me(OH)2 Me(OH)3

OH logK<sup>2</sup>

pKsp values for the precipitates MeL2; [MeL<sup>i</sup>

Hg+2 10.3 21.7 21.2 I

Pb+2 6.9 10.8 13.3 I

OH logK<sup>3</sup>

logK<sup>1</sup>

Table 4. logKi

[OH�<sup>1</sup>

follows:

pH [OH�<sup>1</sup>

�þ½ZnOH<sup>þ</sup><sup>1</sup>

104 Descriptive Inorganic Chemistry Researches of Metal Compounds

] � [H+1]). The solubility s value

�þ½ZnOH<sup>þ</sup><sup>1</sup>

<sup>2</sup> + 104.4�10�15/[OH�<sup>1</sup>

calculated at this point is different from s\* = (Kso3/4)1/3 = 6.3�10�<sup>6</sup>

] [Zn+2] [ZnOH+1] [Zn(OH)2] [Zn(OH)3

incompatibilities contradict application of this formula.

�–½ZnðOHÞ

�1

�þ½ZnðOHÞ2�þ½ZnðOHÞ

<sup>3</sup> � � 2½ZnðOHÞ

<sup>y</sup>ðpHÞ ¼ <sup>2</sup> � <sup>10</sup><sup>13</sup>�2pH <sup>þ</sup> 103:4�pH–10�15:86þpH–<sup>2</sup> � <sup>10</sup>�28:34þ2pH <sup>þ</sup> <sup>10</sup>�pH–10pH�<sup>14</sup> <sup>¼</sup> <sup>0</sup> <sup>ð</sup>39<sup>Þ</sup>

The function (39) zeroes at pH0 = 9.121 (see Table 3). The basic reaction of this system is not immediately stated from Eq. (38) (there are positive and negative terms in expression for

Let us refer now to dissolution of precipitates MeL2 formed by cations Me+2 and anions L�<sup>1</sup> of a strong acid HL, as presented in Table 4. When an excess of MeL2 is introduced into pure water, the concentration balances and charge balance in two-phase system thus formed are as

9.118 1.3122E-05 5.8076E-06 1.9143E-06 0.0002 1.8113E-07 7.8705E-11 2.2702E-07 0.00020743 9.119 1.3152E-05 5.7810E-06 1.9099E-06 0.0002 1.8155E-07 7.9068E-11 1.3858E-07 0.00020740 9.120 1.3183E-05 5.7544E-06 1.9055E-06 0.0002 1.8197E-07 7.9433E-11 5.0322E-08 0.00020737 9.121 1.3213E-05 5.7280E-06 1.9011E-06 0.0002 1.8239E-07 7.9800E-11 �3.7750E-08 0.00020734 9.122 1.3243E-05 5.7016E-06 1.8967E-06 0.0002 1.8281E-07 8.0168E-11 �1.2564E-07 0.00020731 9.123 1.3274E-05 5.6755E-06 1.8923E-06 0.0002 1.8323E-07 8.0538E-11 �2.1335E-07 0.00020728

] – 1013.14�10�15∙[OH�<sup>1</sup>

�2 <sup>4</sup> �þ½H<sup>þ</sup><sup>1</sup>

�1

�1

] [Zn(OH)4

�2

�<sup>1</sup> L�<sup>1</sup> MeL+1 MeL2 MeL3

, Ksp = [Me+2][L�<sup>1</sup>

OH and logKi values for the stability constants Ki and Kj of soluble complexes Me(OH)<sup>i</sup>

] i

]=Ki[Me+2][L�<sup>1</sup>

Table 3. Zeroing the function (39) for the system with Zn(OH)2 precipitate introduced into water; pK<sup>W</sup> = 14.

+2-i

OH logK<sup>1</sup> logK<sup>2</sup> logK<sup>3</sup> logK<sup>4</sup> pKsp

] 2 .

�<sup>1</sup> 12.87 23.82 27.60 29.83 28.54

�<sup>1</sup> 1.26 2.80 3.42 3.92 8.98 Cl�<sup>1</sup> 1.62 2.44 2.04 1.0 4.79

<sup>3</sup> �þ½ZnðOHÞ

��½OH�<sup>1</sup>

�2

, and [OH�<sup>1</sup>

<sup>4</sup> � ¼ <sup>2</sup>:<sup>07</sup> � <sup>10</sup>�<sup>4</sup>

�<sup>1</sup> MeL4

+2-i

and MeL<sup>j</sup>

+2-j and

�<sup>2</sup> MeL2

] – <sup>2</sup>�1014.66�10�15∙[OH�<sup>1</sup>

� ¼ 0 ð38Þ

] <sup>2</sup> = 0

]/[Zn+2] 6¼ 2; such

] y(pH) s [mol/L]

$$\alpha = [\text{H}^{+1}] - [\text{OH}^{-1}] = \sum\_{i=1}^{I} i [\text{Me}(\text{OH})]^{+2-i}\_{i} \tag{44}$$

i.e., reaction of the solution is acidic, [H+1] > [OH�<sup>1</sup> ]. Applying the relations for the equilibrium constants:

[Me+2][L�<sup>1</sup> ] <sup>2</sup> = Ksp, [Me(OH)<sup>i</sup> +2�i ] = Ki OH[Me+2][OH�<sup>1</sup> ] <sup>i</sup> (i = 1,…, I), [MeL<sup>j</sup> +2�j ] = Kj[Me+2][L�<sup>1</sup> ] j (j = 1,…, J)

from Eqs. (43) and (44) we have

$$2[\mathbf{M}\mathbf{e}^{+2}]^{3/2} \cdot (1 + (1 + \sum\_{i=1}^{l} \mathbf{x}\_i) + \mathbf{K}\_{\mathbf{sp}} ^{1/2} \cdot [\mathbf{M}\mathbf{e}^{+2}] \cdot \sum\_{j=1}^{l} (2 - j) \mathbf{K}\_j [\mathbf{L}^{-1}] - \mathbf{K}\_{\mathbf{sp}} ^{1/2} = \mathbf{0} \tag{45}$$

where

$$\begin{aligned} [\text{Me}^{+2}] &= \frac{\alpha}{\sum\_{i=1}^{l} i \cdot \mathbf{x}\_{i}}; \alpha = [\text{H}^{+1}] - [\text{OH}^{-1}] = 10^{-\text{pH}} - 10^{\text{pH}-\rho \text{K}\_{\text{W}}} ; [\text{L}^{-1}] \\ &= \left(\frac{\text{K}\_{\text{op}}}{[\text{Me}^{+2}]}\right)^{1/2} ; \mathbf{x}\_{i} = \text{K}\_{i}^{\text{OH}} \cdot (\text{K}\_{\text{W}} / [\text{H}^{+1}])^{i} \end{aligned}$$

In particular, for I = 3, J =4(Table 4), we have

$$-2\cdot \left(1+\sum\_{i=1}^{3} \mathbf{x}\_{i}\right) \cdot \frac{[\text{Me}]^{2}}{K\_{\text{sp}}^{1/2}} + K\_{\text{I}} \cdot [\text{Me}]^{3/2} - (K\_{\text{3}} \cdot K\_{\text{sp}} + 1) \cdot [\text{Me}]^{1/2} - 2 \cdot K\_{\text{4}} \cdot K\_{\text{sp}} {}^{3/2} = 0 \qquad (46)$$

Applying the zeroing procedure to Eq. (46) gives the pH = pH0 of the solution at equilibrium. At this pH0 value, we calculate the concentrations of all species and solubility of this precipitate recalculated on sMe and sL. When zeroing Eq. (46), we calculate pH = pH0 of the solution in equilibrium with the related precipitate. The solubilities are as follows:

$$\mathbf{s} = \mathbf{s}\_{\text{Me}} = [\mathbf{Me}^{+2}] + \sum\_{i=1}^{I} [\mathbf{Me}(\mathbf{OH})\_i^{+2-i}] + \sum\_{j=1}^{J} [\mathbf{MeL}\_j^{+2-j}] \tag{47}$$

$$\mathbf{s} = \mathbf{s}\_{\mathbb{L}} = [\mathbf{L}^{-1}] + \sum\_{j=1}^{4} j[\mathbf{MeL}\_{j}^{+2-j}] \tag{48}$$

The calculations of sMe and s<sup>L</sup> for the precipitates specified in Table 4 can be realized with use of Excel spreadsheet, according to zeroing procedure, as suggested above (Table 1).

For PbI2: pH0 = 5.1502, sPb = 6.5276∙10�<sup>4</sup> , sI = 1.3051∙10�<sup>3</sup> , see Table 6. The difference between sI and 2sPb = 1.3055∙10�<sup>3</sup> results from rounding the pH0-value.

For HgI2: pH0 = 6.7769, sHg = 1.91217∙10�<sup>5</sup> , sI = 3.82435∙10�<sup>5</sup> , see Table 7. The difference between s<sup>I</sup> and 2sHg = 3.82434∙10�<sup>5</sup> results from rounding the pH-value. The concentration [HgI2] = K2Ksp = 1.90546∙10�<sup>5</sup> is close to the sHg value. For comparison, 4(s\* ) <sup>3</sup> = Ksp ⟹ s \* = 1.93∙10�10, i.e., s\* /s ≈ 10�<sup>5</sup> .


Table 5. Fragment of display for PbCl2.


Table 6. Fragment of display for PbI2.


Table 7. Fragment of display for HgI2.

<sup>2</sup> � <sup>1</sup> <sup>þ</sup><sup>X</sup>

3

i¼1 xi !

� ½Me� 2

For PbI2: pH0 = 5.1502, sPb = 6.5276∙10�<sup>4</sup>

1.93∙10�10, i.e., s\*

For HgI2: pH0 = 6.7769, sHg = 1.91217∙10�<sup>5</sup>

/s ≈ 10�<sup>5</sup> .

pH [Pb+2] [PbOH+1] [Pb(OH)2] [Pb(OH)3

Table 5. Fragment of display for PbCl2.

Table 6. Fragment of display for PbI2.

pH [Pb+2] [PbOH+1] [Pb(OH)2] [Pb(OH)3

Ksp

106 Descriptive Inorganic Chemistry Researches of Metal Compounds

<sup>1</sup>=<sup>2</sup> þ K<sup>1</sup> � ½Me�

<sup>s</sup> <sup>¼</sup> <sup>s</sup>Me ¼ ½Me<sup>þ</sup><sup>2</sup>

and 2sPb = 1.3055∙10�<sup>3</sup> results from rounding the pH0-value.

equilibrium with the related precipitate. The solubilities are as follows:

� þ<sup>X</sup> I

<sup>s</sup> <sup>¼</sup> <sup>s</sup><sup>L</sup> ¼ ½L�<sup>1</sup>

[HgI2] = K2Ksp = 1.90546∙10�<sup>5</sup> is close to the sHg value. For comparison, 4(s\*

�1

i¼1

of Excel spreadsheet, according to zeroing procedure, as suggested above (Table 1).

<sup>3</sup>=<sup>2</sup> � ðK<sup>3</sup> � <sup>K</sup>sp <sup>þ</sup> <sup>1</sup>Þ�½Me�

Applying the zeroing procedure to Eq. (46) gives the pH = pH0 of the solution at equilibrium. At this pH0 value, we calculate the concentrations of all species and solubility of this precipitate recalculated on sMe and sL. When zeroing Eq. (46), we calculate pH = pH0 of the solution in

½MeðOHÞ<sup>i</sup>

� þ<sup>X</sup> 4

The calculations of sMe and s<sup>L</sup> for the precipitates specified in Table 4 can be realized with use

, sI = 1.3051∙10�<sup>3</sup>

between s<sup>I</sup> and 2sHg = 3.82434∙10�<sup>5</sup> results from rounding the pH-value. The concentration

4.5343 0.010749606 2.92208E-05 7.94315E-11 8.59592E-18 0.017405892 0.004466836 6.90723E-05 2.44685E-07 0.038842191 0.000138249 4.5344 0.010744657 2.92141E-05 7.94315E-11 8.5979E-18 0.017401884 0.004466836 6.90882E-05 2.44798E-07 0.038851136 7.7139E-05 4.5345 0.01073971 2.92074E-05 7.94315E-11 8.59988E-18 0.017397878 0.004466836 6.91041E-05 2.44911E-07 0.038860083 1.60945E-05 4.5346 0.010734765 2.92007E-05 7.94315E-11 8.60186E-18 0.017393872 0.004466836 6.912E-05 2.45023E-07 0.038869032 -4.48848E-05 4.5347 0.010729823 2.91939E-05 7.94315E-11 8.60384E-18 0.017389867 0.004466836 6.91359E-05 2.45136E-07 0.038877983 -0.000105799


] [PbI+1] [PbI2] [PbI3

5.15 0.000630817 7.07789E-06 7.94152E-11 3.54735E-17 1.47894E-05 6.60693E-07 3.54853E-09 1.44576E-11 0.001288393 0.000138249 5.1501 0.000630527 7.07626E-06 7.94152E-11 3.54816E-17 1.4786E-05 6.60693E-07 3.54935E-09 1.44643E-11 0.001288689 7.7139E-05 5.1502 0.000630236 7.07463E-06 7.94152E-11 3.54898E-17 1.47826E-05 6.60693E-07 3.55016E-09 1.44709E-11 0.001288986 1.60945E-05 5.1503 0.000629946 7.073E-06 7.94152E-11 3.5498E-17 1.47792E-05 6.60693E-07 3.55098E-09 1.44776E-11 0.001289283 -4.48848E-05 5.1504 0.000629656 7.07137E-06 7.94152E-11 3.55061E-17 1.47758E-05 6.60693E-07 3.5518E-09 1.44843E-11 0.00128958 -0.000105799

j¼1

þ2�i

j½MeL<sup>j</sup>

, sI = 3.82435∙10�<sup>5</sup>

� þ<sup>X</sup> J

þ2�j

j¼1

�1 ] [PbCl4 �2 ] [Cl�<sup>1</sup>

�1 ] [PbI4 �2 ] [I�<sup>1</sup>

½MeL<sup>j</sup>

þ2�j

<sup>1</sup>=<sup>2</sup> � <sup>2</sup> � <sup>K</sup><sup>4</sup> � <sup>K</sup>sp

<sup>3</sup>=<sup>2</sup> <sup>¼</sup> <sup>0</sup> <sup>ð</sup>46<sup>Þ</sup>

� ð47Þ

� ð48Þ

, see Table 6. The difference between sI

, see Table 7. The difference

)

<sup>3</sup> = Ksp ⟹ s

] y

] y

\* =

#### 4.4. Dissolution of CaCO3 in the presence of CO2

The portions 0.1 g of calcite CaCO3 (M = 100.0869 g/mol, d = 2.711 g/cm3 ) are inserted into 100 mL of: pure water (task A) or aqueous solutions of CO2 specified in the tasks: B1, B2, B3, and equilibrated. Denoting the starting (t = 0) concentrations [mol/L]: Co for CaCO3 and CCO2 for CO2 in the related systems, on the basis of equilibrium data collected in Table 8:

(A) we calculate pH = pH01 and solubility s = s(pH01) of CaCO3 at equilibrium in the system;

(B1) we calculate pH = pH02 and solubility s = s(pH02) of CaCO3 in the system, where CCO2 refers to saturated (at 25 <sup>o</sup> C) solution of CO2, where 1.45 g CO2 dissolves in 1 L of water [39].

(B2) we calculate minimal CCO2 in the starting solution needed for complete dissolution of CaCO3 in the system and the related pH = pH03 value, where s = s(pH03)=C<sup>o</sup> ;

(B3) we plot the logsCa versus V, pH versus V and logsCa versus pH relationships for the system obtained after addition of V mL of a strong base MOH (C<sup>b</sup> = 0.1) into V<sup>0</sup> = 100 mL of the system with CaCO3 presented in (B1). The quasistatic course of the titration is assumed.

The volume 0.1/2.711 = 0.037 cm3 of introduced CaCO3 is negligible when compared with V<sup>0</sup> at the start (t = 0) of the dissolution. Starting concentration of CaCO3 in the systems: A, B1, B2, B3 is C<sup>o</sup> = (0.1/100)/0.1 = 10<sup>2</sup> mol/L. At t > 0, concentration of CaCO3 is co mol/L. The balances are as follows:


Table 8. Equilibrium data.

$$\mathbf{C}^{\diamond} = \mathbf{c}^{\diamond} + [\mathbf{C}\mathbf{a}^{+2}] + [\mathbf{C}\mathbf{a}\mathbf{O}\mathbf{H}^{+1}] + [\mathbf{C}\mathbf{a}\mathbf{H}\mathbf{C}\mathbf{O}\_3^{+1}] + [\mathbf{C}\mathbf{a}\mathbf{C}\mathbf{O}\_3] \text{ (for } \mathbf{A}, \mathbf{B1}, \mathbf{B2}, \mathbf{B3}) \tag{49}$$

$$\mathbf{C}^{\diamond} = \mathbf{c}^{\diamond} + \left[ \mathbf{CaHCO\_3^{+1}} \right] + \left[ \mathbf{CaCO\_3} \right] + \left[ \mathbf{H\_2CO\_3} \right] + \left[ \mathbf{HCO\_3^{-1}} \right] + \mathbf{CO\_3^{-2}} \left( \text{for A} \right) \tag{50}$$

$$\text{C}^{\text{o}} + \text{C}\_{\text{CO2}} = \text{c}^{\text{o}} + [\text{CaHCO}\_{3}^{+}] + [\text{CaCO}\_{3}] + [\text{H}\_{2}\text{CO}\_{3}] + [\text{HCO}\_{3}^{-1}] + [\text{CO}\_{3}^{-2}] (\text{for B1, B2, B3}) \tag{51}$$

$$\left[\text{H}^{+}\text{I}\right]-\left[\text{OH}^{-1}\right]+2\left[\text{Ca}^{+2}\right]+\left[\text{CaOH}^{+1}\right]+\left[\text{CaHCO}\_{3}^{+1}\right]-\left[\text{HCO}\_{3}^{-1}\right]-2\left[\text{CO}\_{3}^{-2}\right]=0\left(\text{for A, B1, B2}\right)\tag{52}$$

$$\left[\text{Ca}^{+2}\right] + \left[\text{CaOH}^{+1}\right] = \left[\text{H}\_2\text{CO}\text{s}\right] + \left[\text{HCO}\_3^{-1}\right] + \left[\text{CO}\_3^{-2}\right] \tag{53}$$

$$\begin{aligned} \left[\text{Ca}^{+2}\right] &= \left[\text{CO}\_{3}^{-2}\right] \cdot f\_{1} \Rightarrow \left[\text{Ca}^{+2}\right] = 10^{-4.24} \cdot \left(f\_{1}/f\_{2}\right)^{0.5}; \left[\text{CO}\_{3}^{-2}\right] = 10^{-4.24} \cdot \left(f\_{2}/f\_{1}\right)^{0.5}; \left[\text{CaOH}^{+1}\right] \\ &= 10^{\text{pH}-16.94} \cdot \left(f\_{1}/f\_{2}\right)^{0.5}; \\ \left[\text{CaCO}\_{3}\right] &= 10^{-5.26}; \left[\text{CaHCO}\_{3}^{+1}\right] = 10^{2.96-\text{pH}}; \left[\text{HCO}\_{3}^{-1}\right] = 10^{6.09-\text{pH}} \cdot \left(f\_{2}/f\_{1}\right)^{0.5}; \left[\text{H}^{+1}\right] \\ &= 10^{-\text{pH}}; \left[\text{OH}^{-1}\right] = 10^{\text{pH}-14}. \end{aligned}$$

$$\begin{split} z = z(\text{pH}) &= 10^{-\text{pH}} - 10^{\text{pH}-14} + 2 \cdot 10^{-4.24} \cdot \left( f\_1 / f\_2 \right)^{0.5} + 10^{\text{pH}-16.94} \cdot \left( f\_1 / f\_2 \right)^{0.5} \\ &+ 10^{2.96-\text{pH}} - 10^{6.09-\text{pH}} \cdot \left( f\_2 / f\_1 \right)^{0.5} - 2 \cdot 10^{-4.24} \cdot \left( f\_2 / f\_1 \right)^{0.5} \end{split} \tag{54}$$

$$\mathbf{s} = [\mathbf{C}\mathbf{a}^{+2}] + [\mathbf{C}\mathbf{a}\mathbf{O}\mathbf{H}^{+1}] + [\mathbf{C}\mathbf{a}\mathbf{H}\mathbf{C}\mathbf{O}\_{3}^{+1}] + [\mathbf{C}\mathbf{a}\mathbf{C}\mathbf{O}\_{3}] \tag{55}$$

$$\mathbf{a} = 10^{-4.24} \cdot \left( f\_1 / f\_2 \right)^{0.5} + 10^{\text{pH} - 16.94} \cdot \left( f\_1 / f\_2 \right)^{0.5} + 10^{2.96 - \text{pH}} + 10^{-5.26} \tag{55a}$$

$$\begin{aligned} &\text{[H}\_2\text{CO}\_3\text{]} + \text{[HCO}\_3^{-1}] + \text{[CO}\_3^{-2}] - \text{([Ca}^{+2}] + \text{[CaOH}^{+1}]) \\ &= \text{C}\_{\text{CO2}} \Rightarrow \text{[CO}\_3^{-2}] \cdot f\_1 \text{--} [\text{Ca}^{+2}] \cdot f\_2 - \text{C}\_{\text{CO2}} = 0 \Rightarrow [\text{Ca}^{+2}]^2 \cdot f\_2 + \text{C}\_{\text{CO2}} \cdot [\text{Ca}^{+2}] - \text{K}\_{\text{sp}} \cdot f\_1 = 0 \end{aligned}$$

In this case,

<sup>C</sup><sup>o</sup> <sup>¼</sup> co þ ½Caþ<sup>2</sup>

<sup>C</sup><sup>o</sup> <sup>þ</sup> CCO2 <sup>¼</sup> co þ ½CaHCO<sup>þ</sup><sup>1</sup>

��½OH�<sup>1</sup>

� �½OH�<sup>1</sup>

• For (A)

<sup>½</sup>Caþ<sup>2</sup>

where [M+1] = CbV/(V0+V).

�<sup>f</sup> <sup>2</sup> ¼ ½CO�<sup>2</sup>

<sup>½</sup>CaCO3� ¼ <sup>10</sup>�5:<sup>26</sup>; <sup>½</sup>CaHCO<sup>þ</sup><sup>1</sup>

<sup>s</sup> ¼ ½Caþ<sup>2</sup>

<sup>¼</sup> <sup>10</sup>�4:<sup>24</sup> � ð<sup>f</sup> <sup>1</sup>=<sup>f</sup> <sup>2</sup><sup>Þ</sup>

We have s = s(pH = pH01) = 1.159�10�<sup>4</sup> mol/L.

Subtraction of Eq. (49) from Eq. (51) gives

<sup>¼</sup> <sup>10</sup>pH�16:<sup>94</sup> � ð<sup>f</sup> <sup>1</sup>=<sup>f</sup> <sup>2</sup><sup>Þ</sup>

<sup>¼</sup> <sup>10</sup>�pH; <sup>½</sup>OH�<sup>1</sup>

• For (B1)

<sup>½</sup>Hþ<sup>1</sup>

<sup>½</sup>Hþ<sup>1</sup>

<sup>C</sup><sup>o</sup> <sup>¼</sup> co þ ½CaHCO<sup>þ</sup><sup>1</sup>

108 Descriptive Inorganic Chemistry Researches of Metal Compounds

� þ <sup>2</sup>½Caþ<sup>2</sup>

From Eqs. (49) and (50), we have

<sup>½</sup>Caþ<sup>2</sup>

<sup>3</sup> � � <sup>f</sup> <sup>1</sup> ) ½Caþ<sup>2</sup>

0:5 ;

� ¼ 10pH�<sup>14</sup>:

�þ½Mþ<sup>1</sup>

�þ½CaOH<sup>þ</sup><sup>1</sup>

�þ½CaOH<sup>þ</sup><sup>1</sup>

�þ½CaOH<sup>þ</sup><sup>1</sup>

� þ <sup>2</sup>½Caþ<sup>2</sup>

�þ½CaHCO<sup>þ</sup><sup>1</sup>

<sup>3</sup> �þ½CaCO3�þ½H2CO3�þ½HCO�<sup>1</sup>

�þ½CaHCO<sup>þ</sup><sup>1</sup>

�þ½CaOH<sup>þ</sup><sup>1</sup>

<sup>3</sup> �þ½CaCO3�þ½H2CO3�þ½HCO�<sup>1</sup>

<sup>3</sup> ��½HCO�<sup>1</sup>

�þ½CaHCO<sup>þ</sup><sup>1</sup>

�¼½H2CO3�þ½HCO�<sup>1</sup>

0:5

; <sup>½</sup>CO�<sup>2</sup>

<sup>3</sup> � ¼ 106:09�pH � ðf2=f1<sup>Þ</sup>

<sup>0</sup>:<sup>5</sup> � <sup>2</sup> � <sup>10</sup>�4:<sup>24</sup> � ð<sup>f</sup> <sup>2</sup>=<sup>f</sup> <sup>1</sup><sup>Þ</sup>

Considering the solution saturated with respect to CaCO3 and denoting: f<sup>1</sup> = 1016.71�2pH +

and applying the zeroing procedure to the function (54), we find pH01 = 9.904, at z = z

1010.33�pH + 1, f<sup>2</sup> = 1 + 10pH�12.7, from Eq. (53) and Table 1, we have the relations:

� ¼ <sup>10</sup>�4:<sup>24</sup> � ð<sup>f</sup> <sup>1</sup>=<sup>f</sup> <sup>2</sup><sup>Þ</sup>

<sup>3</sup> � ¼ <sup>10</sup><sup>2</sup>:96�pH; <sup>½</sup>HCO�<sup>1</sup>

Inserting them into the charge balance (52), rewritten into the form

(pH01) = 0. The solubility s = s(pH) of CaCO3, resulting from Eq. (49), is

�þ½CaHCOþ<sup>1</sup>

<sup>0</sup>:<sup>5</sup> <sup>þ</sup> 10pH�16:<sup>94</sup> � ð<sup>f</sup> <sup>1</sup>=<sup>f</sup> <sup>2</sup><sup>Þ</sup>

<sup>z</sup> <sup>¼</sup> <sup>z</sup>ðpHÞ ¼ <sup>10</sup>�pH � 10pH�<sup>14</sup> <sup>þ</sup> <sup>2</sup> � <sup>10</sup>�4:<sup>24</sup> � ð<sup>f</sup> <sup>1</sup>=<sup>f</sup> <sup>2</sup><sup>Þ</sup>

�þ½CaOHþ<sup>1</sup>

<sup>þ</sup>102:96�pH–10<sup>6</sup>:09�pH � ð<sup>f</sup> <sup>2</sup>=<sup>f</sup> <sup>1</sup><sup>Þ</sup>

<sup>3</sup> �þ½CaCO3� ðfor A, B1, B2, B3Þ ð49Þ

<sup>3</sup> � � <sup>2</sup>½CO�<sup>2</sup>

<sup>3</sup> � ðfor AÞ ð50Þ

<sup>3</sup> �ðfor B1, B2, B3Þ ð51Þ

<sup>3</sup> � ¼ 0ðfor A, B1, B2Þ

<sup>3</sup> � ð53Þ

0:5

0:5

<sup>0</sup>:<sup>5</sup> ð54Þ

; <sup>½</sup>CaOH<sup>þ</sup><sup>1</sup>

�

<sup>3</sup> � ¼ 0ðfor B3Þ

ð52Þ

ð52aÞ

<sup>3</sup> � þ CO�<sup>2</sup>

<sup>3</sup> �þ½CO�<sup>2</sup>

<sup>3</sup> � � <sup>2</sup>½CO�<sup>2</sup>

<sup>3</sup> ��½HCO�<sup>1</sup>

<sup>3</sup> �þ½CO�<sup>2</sup>

<sup>3</sup> � ¼ <sup>10</sup>�4:<sup>24</sup> � ð<sup>f</sup> <sup>2</sup>=<sup>f</sup> <sup>1</sup><sup>Þ</sup>

<sup>0</sup>:<sup>5</sup> <sup>þ</sup> 10pH�16:<sup>94</sup> � ð<sup>f</sup> <sup>1</sup>=<sup>f</sup> <sup>2</sup><sup>Þ</sup>

<sup>3</sup> �þ½CaCO3� ð55Þ

<sup>0</sup>:<sup>5</sup> <sup>þ</sup> <sup>10</sup><sup>2</sup>:96�pH <sup>þ</sup> <sup>10</sup>�5:<sup>26</sup> <sup>ð</sup>55a<sup>Þ</sup>

0:5 ; <sup>½</sup>Hþ<sup>1</sup> �

$$\left[\text{Ca}^{+2}\right] = \frac{\sqrt{\left(\text{C}\_{\text{CO}\_2}\right)^2 + 4 \cdot \text{K}\_{\text{sp}} \cdot f\_1 \cdot f\_2} - \text{C}\_{\text{CO}\_2}}{2 \cdot f\_2} \tag{56}$$

where CCO2 = 1.45/44 = 0.0329 mol/L. Eq. (55) has the form

$$\mathbf{s} = [\mathbf{C}\mathbf{a}^{+2}] \cdot f\_2 + 10^{2.96 - \text{pH}} + 10^{-5.26} \tag{57}$$

and the charge balance is transformed into the zeroing function

$$z = z(\text{pH}) = 10^{-\text{pH}} - 10^{\text{pH}-14} + [\text{Ca}^{+2}] \cdot (2 + 10^{\text{pH}-12.7}) + 10^{2.96-\text{pH}} - [\text{CO}\_3^{-2}] \cdot (10^{10.33-\text{pH}} + 2) \tag{58}$$

where [CO3 �2 ] = 10-8.48/[Ca+2], and [Ca+2] is given by Eq. (56). Eq. (58) zeroes at pH = pH02 = 6.031. Then from Eq. (57) we calculate s = s(pH02) = 6.393∙10�<sup>3</sup> mol/L, at pH = pH02 = 6.031.

• For (B2)

At pH = pH03, where c<sup>o</sup> = 0, i.e., s = C<sup>o</sup> , the solution (a monophase system) is saturated toward CaCO3, i.e., the relation [Ca+2][CO3 �2 ] = Ksp is still valid. Applying Eqs. (56) and (57), we find pH values zeroing Eq. (58) at different, preassumed CCO2 values. Applying these pH-values in Eq. (57), we calculate the related s = s(pH, CCO2 ) values (Eq. (57), Table 9). Graphically, CCO2 = 0.100 is found at pH03 = 5.683, as the abscissa of the point of intersection of the lines: s = s(pH) and s = Co = 0.01. Table 9 shows other, preassumed s=Co values.

• For (B3)

We apply again the formulas used in (B1) and (B2), and the charge balance (Eq. (52a)), which is transformed there into the function


Table 9. The set of points used for searching the CCO2 value at s = Co = 0.01; at this point, we have pH03 = 5.683.

Figure 2. Graphical presentation of the data considered in (b3): (a) pH versus V, (b) log sCa versus V, (c) log sCa versus pH relationships.

$$\begin{split} z = z(\text{pH}, \text{V}) &= 10^{-\text{pH}} - 10^{\text{pH}-14} + \text{C}\_{\text{b}}V/(V\_0 + V) + [\text{Ca}^{+2}] \cdot (2 + 10^{\text{pH}-12.7}) \\ &+ 10^{2.96-\text{pH}} - [\text{CO}\_3^{-2}] \cdot (10^{10.33-\text{pH}} + 2) \end{split} \tag{59}$$

applied for zeroing purposes, at different V values. The data thus obtained are presented graphically in Figures 2a–c. The data presented in the dynamic solubility diagram (Figure 2b), illustrating the solubility changes affected by pH changes (Figure 2a) resulting from addition of a base, MOH; Figure 2c shows a synthesis of these changes. Solubility product of Ca(OH)2 is not crossed in this system.

#### 5. Nonequilibrium solid phases in aqueous media

Some solids when introduced into aqueous media (e.g., pure water) may appear to be nonequilibrium phases in these media.

#### 5.1. Silver dichromate (Ag2Cr2O7)

The equilibrium data related to the system, where Ag2Cr2O7 is introduced into pure water, were taken from Refs. [33, 40, 41], and presented in Table 10. A large discrepancy between pKsp2 values (6.7 and 10) in the cited literature is taken here into account. We prove that Ag2Cr2O7 changes into Ag2CrO4.

On the dissociation step, each dissolving molecule of Ag2Cr2O7 gives two ions Ag+1 and 1 ion Cr2O7 �2 , where two atoms of Cr are involved; in the contact with water, these ions are hydrolyzed, to varying degrees. In the initial step of the dissolution, before the saturation of the solution with respect to an equilibrium solid phase (not specified at this moment), we can write the concentration balances


Table 10. Physicochemical equilibrium data relevant to the Ag2Cr2O7 + H2O system (pK = �logK), at "room" temperatures.

<sup>z</sup> <sup>¼</sup> <sup>z</sup>ðpH, <sup>V</sup>Þ ¼ <sup>10</sup>�pH � 10pH�<sup>14</sup> <sup>þ</sup> CbV=ðV<sup>0</sup> <sup>þ</sup> <sup>V</sup>Þþ½Caþ<sup>2</sup>

applied for zeroing purposes, at different V values. The data thus obtained are presented graphically in Figures 2a–c. The data presented in the dynamic solubility diagram (Figure 2b), illustrating the solubility changes affected by pH changes (Figure 2a) resulting from addition of a base, MOH; Figure 2c shows a synthesis of these changes.

(a) (b) (c)

Figure 2. Graphical presentation of the data considered in (b3): (a) pH versus V, (b) log sCa versus V, (c) log sCa versus pH

Some solids when introduced into aqueous media (e.g., pure water) may appear to be

The equilibrium data related to the system, where Ag2Cr2O7 is introduced into pure water, were taken from Refs. [33, 40, 41], and presented in Table 10. A large discrepancy between pKsp2 values (6.7 and 10) in the cited literature is taken here into account. We prove that Ag2Cr2O7

On the dissociation step, each dissolving molecule of Ag2Cr2O7 gives two ions Ag+1 and 1 ion

hydrolyzed, to varying degrees. In the initial step of the dissolution, before the saturation of the solution with respect to an equilibrium solid phase (not specified at this moment), we can

, where two atoms of Cr are involved; in the contact with water, these ions are

<sup>þ</sup>10<sup>2</sup>:96�pH � ½CO�<sup>2</sup>

Solubility product of Ca(OH)2 is not crossed in this system.

5. Nonequilibrium solid phases in aqueous media

nonequilibrium phases in these media.

110 Descriptive Inorganic Chemistry Researches of Metal Compounds

5.1. Silver dichromate (Ag2Cr2O7)

write the concentration balances

changes into Ag2CrO4.

Cr2O7 �2

relationships.

��ð<sup>2</sup> <sup>þ</sup> 10pH�12:<sup>7</sup>

<sup>3</sup> ��ð1010:33�pH <sup>þ</sup> <sup>2</sup><sup>Þ</sup> <sup>ð</sup>59<sup>Þ</sup>

Þ

$$2[\mathbf{Ag\_2Cr\_2O\_7}] + [\mathbf{Ag^{+1}}] + [\mathbf{AgOH}] + [\mathbf{Ag(OH)\_2^{-1}}] + [\mathbf{Ag(OH)\_3^{-2}}] = 2\mathbf{C\_0} \tag{60}$$

$$2[\mathbf{Ag}\_2\mathbf{Cr}\_2\mathbf{O}\_7] + [\mathbf{H}\_2\mathbf{CrO}\_4] + [\mathbf{HClO}\_4^{-1}] + [\mathbf{CrO}\_4^{-2}] + 2[\mathbf{HClCr}\_2\mathbf{O}\_7^{-1}] + 2[\mathbf{Cr}\_2\mathbf{O}\_7^{-2}] = 2\mathbf{C}\_0 \tag{61}$$

where 2C0 is the total concentration of the solid phase in the system, at the moment (t = 0) of introducing this phase into water, [Ag2Cr2O7] is the concentration of this phase at a given moment of the intermediary step. As previously, we assume that addition of the solid phase (here: Ag2Cr2O7) does not change the volume of the system in a significant degree, and that Ag2Cr2O7 is added in a due excess, securing the formation of a solid (that is not specified at this moment), as an equilibrium solid phase. The balances in Eqs. (60) and (61) are completed by the charge balance

$$\begin{aligned} \left[\text{H}^{+}\right]-\left[\text{OH}^{-1}\right]+\left[\text{Ag}^{+}\right]-\left[\text{Ag}(\text{OH})\_{2}^{-1}\right]-2\left[\text{Ag}(\text{OH})\_{3}^{-2}\right]-\left[\text{HCrO}\_{4}^{-1}\right] \\ -2\left[\text{CrO}\_{4}^{-2}\right]-\left[\text{HCr}\_{2}\text{O}\_{7}^{-1}\right]-2\left[\text{Cr}\_{2}\text{O}\_{7}^{-2}\right]=0 \end{aligned} \tag{62}$$

used, as previously, to formulation of the zeroing function, y = y(pH), and the set of relations for equilibrium data specified in Table 10. From these relations, we get

$$\begin{aligned} [\text{H}\_2\text{CrO}\_4] = 10^{7.3-2\text{pH}} \cdot [\text{CrO}\_4^{-2}]; \qquad [\text{HCrO}\_4^{-1}] = 10^{6.5-\text{pH}} \cdot [\text{CrO}\_4^{-2}];\\ [\text{HCr}\_2\text{O}\_7^{-1}] = 10^{14.59-3\text{pH}} \cdot [\text{CrO}\_4^{-2}]^2; \end{aligned} \tag{63}$$

$$[\text{Cr}\_2\text{O}\_7\text{}^{-2}] = 10^{14.52 - 2\text{pH}} \cdot [\text{CrO}\_4\text{}^{-2}]^2 \tag{63a}$$

Denoting by 2c0 (< 2C0) the total concentration of dissolved Ag and Cr species formed, in a transition stage, from Ag2Cr2O7, we can write

$$\left[\text{Ag}^{+1}\right] + \left[\text{AgOH}\right] + \left[\text{Ag}(\text{OH})\_2^{-1}\right] + \left[\text{Ag}(\text{OH})\_3^{-2}\right] = 2\text{c}\_0\tag{64}$$

$$[\text{H}\_2\text{CrO}\_4] + [\text{HCrO}\_4^{-1}] + [\text{CrO}\_4^{-2}] + 2[\text{HCr}\_2\text{O}\_7^{-1}] + 2[\text{Cr}\_2\text{O}\_7^{-2}] = 2\text{c}\_0\tag{65}$$

From Table 10 and formulas (63)–(65) we get the relations:

$$\text{(a)}\ [\text{Ag}^{+}] = 2\text{c}\_{0}/\text{g}\_{0}\text{:}\ 2\text{g}\_{2}[\text{CrO}\_{4}^{-2}]^{2} + \text{g}\_{1}[\text{CrO}\_{4}^{-1}] - 2\text{c}\_{0} = 0 \Rightarrow \\ \text{(b)}\ [\text{CrO}\_{4}^{-2}] = \frac{\left(g\_{1}^{2} + 16 \cdot \text{c}\_{0}\text{g}\_{2}\right)^{0.5} - g\_{1}}{4 \cdot g\_{2}} \tag{66}$$

where g<sup>0</sup> = 1 + 10pH�11.7 + 102pH�24.4 + 103pH�37.2; g<sup>1</sup> = 107.3�2pH + 106.5�pH + 1; g<sup>2</sup> = 1014.59�3pH + 1014.52�2pH. Applying them in Eq. (62), we get the zeroing function

$$\mathbf{y} = \mathbf{y}(\mathbf{pH}) = 10^{-\text{pH}} \mathbf{-10^{pH-14}} + \mathbf{g}\_3 \cdot [\mathbf{Ag}^{+1}] - \mathbf{g}\_4 \cdot [\mathbf{CrO}\_4^{-2}] - \mathbf{g}\_5 \cdot [\mathbf{CrO}\_4^{-2}]^2 \tag{67}$$

where g<sup>3</sup> = 1 – 102pH�24.4 – 2∙103pH�37.2; g4 = 106.5�pH + 2; g5 = 1014.59�3pH + 2∙1014.52�2pH, and [Ag+1] and [CrO4 �2 ] are defied above, as functions of pH.

The calculation procedure, realizable with use of Excel spreadsheet, is as follows. We assume a sequence of growing numerical values for 2c0. At particular 2c0 values, we calculate pH = pH(2c0) value zeroing the function (67), and then calculate the values of the products: q<sup>1</sup> = [Ag+1] 2 [CrO4 �2 ]/Ksp1 and q<sup>2</sup> = [Ag+1] 2 [Cr2O7 �2 ]/Ksp2, where: [Ag+1], [CrO4 �2 ], and [Cr2O7 �2 ] are presented above (Eqs. (66a), (66b) and (63a), resp.), pKsp1 = 11.9, pKsp2 = 6.7. As results from Figure 3, where logq<sup>1</sup> and logq<sup>2</sup> are plotted as functions of 2c0; logq<sup>1</sup> = 0 ⇔ q<sup>1</sup> = 1 ⇔ [Ag+1] 2 [CrO4 �2 ] = Ksp1 at lower 2c0 value, whereas logq<sup>2</sup> < 0 ⇔ q<sup>2</sup> < 1 ⇔ [Ag+1] 2 [Cr2O7 �2 ] < Ksp2, both for pK<sup>2</sup> = 6.7 and 10, cited in the literature. The x1=1

Figure 3. The convergence of logq1 and logq2 to 0 value; Ksp1 is attained at lower 2c0 value.

value is attained at 2c0 = 3.5∙10�<sup>4</sup> ⟹ c0 = 1.75∙10�<sup>4</sup> ; then Ag2CrO4 precipitates as the new solid phase, i.e., total depletion of Ag2Cr2O7 occurs. It means that Ag2Cr2O7 is not the equilibrium solid phase in this system. This fact was confirmed experimentally, as stated in [42], i.e., Ag2Cr2O7 is transformed into Ag2CrO4 upon boiling with H2O; at higher temperatures, this transformation proceeds more effectively. Concluding, the formula s\* = (Ksp2/4)1/3 applied for Ksp2 = [Ag+1] 2 [Cr2O7 �2 ] is not "the best answer," as stated in Ref. [43].

The system involved with Ag2CrO4 was also considered in context with the Mohr's method of Cl�<sup>1</sup> determination [44–46]. As were stated there, the systematic error in Cl�<sup>1</sup> determining according to this method, expressed by the difference between the equivalence (eq) volume (Veq = C0V0/C) and the volume Vend corresponding to the end point where the Ksp1 for Ag2CrO4 is crossed, equals to

$$V\_{\rm eq} - V\_{\rm end} = \frac{K\_{\rm sp}}{\rm C} \cdot \left(\frac{\rm C\_{01}V\_0}{K\_{\rm sp1}}\right)^{0.5} \cdot \left(V\_0 + V\_{\rm end}\right)^{0.5} - \frac{1}{\rm C} \cdot \left(\frac{K\_{\rm sp1}}{\rm C\_{01}V\_0}\right)^{0.5} \cdot \left(V\_0 + V\_{\rm end}\right)^{1.5}$$

where Ksp = [Ag+1][Cl�<sup>1</sup> ] (pKsp = 9.75), V<sup>0</sup> is the volume of titrant with NaCl (C0)+K2CrO4 (C01) titrated with AgNO3 (C) solution; Vend = Veq at C01 = (1 + Vend/V0)∙Ksp1/Ksp.

All calculations presented above were realized using Excel spreadsheets. For more complex nonequilibrium two-phase systems, the use of iterative computer programs, e.g., ones offered by MATLAB [8, 47], is required. This way, the quasistatic course of the relevant processes under isothermal conditions can be tested [48].

#### 5.2. Dissolution of struvite

½Ag<sup>þ</sup><sup>1</sup>

112 Descriptive Inorganic Chemistry Researches of Metal Compounds

� ¼ 2c0=g0; <sup>2</sup>g2½CrO�<sup>2</sup>

�2

the products: q<sup>1</sup> = [Ag+1]

], and [Cr2O7

of 2c0; logq<sup>1</sup> = 0 ⇔ q<sup>1</sup> = 1 ⇔ [Ag+1]

2 [Cr2O7 �2

ðaÞ ½Ag<sup>þ</sup><sup>1</sup>

[Ag+1] and [CrO4

q<sup>2</sup> < 1 ⇔ [Ag+1]

[CrO4 �2 ½H2CrO4�þ½HCrO�<sup>1</sup>

From Table 10 and formulas (63)–(65) we get the relations:

<sup>4</sup> �

1014.52�2pH. Applying them in Eq. (62), we get the zeroing function

<sup>y</sup> <sup>¼</sup> <sup>y</sup>ðpHÞ ¼ <sup>10</sup>�pH–10pH�<sup>14</sup> <sup>þ</sup> <sup>g</sup><sup>3</sup> � ½Ag<sup>þ</sup><sup>1</sup>

2 [CrO4 �2

�2

�þ½AgOH�þ½AgðOHÞ

<sup>4</sup> �þ½CrO�<sup>2</sup>

<sup>2</sup> <sup>þ</sup> <sup>g</sup>1½CrO�<sup>1</sup>

] are defied above, as functions of pH.

2 [CrO4 �2

Figure 3. The convergence of logq1 and logq2 to 0 value; Ksp1 is attained at lower 2c0 value.

�1

<sup>4</sup> � þ <sup>2</sup>½HCr2O�<sup>1</sup>

where g<sup>0</sup> = 1 + 10pH�11.7 + 102pH�24.4 + 103pH�37.2; g<sup>1</sup> = 107.3�2pH + 106.5�pH + 1; g<sup>2</sup> = 1014.59�3pH +

where g<sup>3</sup> = 1 – 102pH�24.4 – 2∙103pH�37.2; g4 = 106.5�pH + 2; g5 = 1014.59�3pH + 2∙1014.52�2pH, and

The calculation procedure, realizable with use of Excel spreadsheet, is as follows. We assume a sequence of growing numerical values for 2c0. At particular 2c0 values, we calculate pH = pH(2c0) value zeroing the function (67), and then calculate the values of

11.9, pKsp2 = 6.7. As results from Figure 3, where logq<sup>1</sup> and logq<sup>2</sup> are plotted as functions

]/Ksp1 and q<sup>2</sup> = [Ag+1]

<sup>2</sup> �þ½AgðOHÞ

<sup>4</sup> �–2c0 <sup>¼</sup> <sup>0</sup> ) ðbÞ ½CrO�<sup>2</sup>

� –g<sup>4</sup> � ½CrO�<sup>2</sup>

2 [Cr2O7 �2

] are presented above (Eqs. (66a), (66b) and (63a), resp.), pKsp1 =

] < Ksp2, both for pK<sup>2</sup> = 6.7 and 10, cited in the literature. The x1=1

�2

<sup>7</sup> � þ <sup>2</sup>½Cr2O�<sup>2</sup>

<sup>4</sup> � ¼ <sup>ð</sup>g<sup>2</sup>

<sup>4</sup> � � <sup>g</sup>5�½CrO�<sup>2</sup>

] = Ksp1 at lower 2c0 value, whereas logq<sup>2</sup> < 0 ⇔

<sup>3</sup> � ¼ 2c0 ð64Þ

<sup>1</sup> þ 16 � c0g2Þ

<sup>4</sup> �

]/Ksp2, where: [Ag+1],

4 � g<sup>2</sup>

<sup>0</sup>:<sup>5</sup> � <sup>g</sup><sup>1</sup>

<sup>2</sup> <sup>ð</sup>67<sup>Þ</sup>

ð66Þ

<sup>7</sup> � ¼ 2c0 ð65Þ

The fact that NH3 evolves from the system obtained after leaving pure struvite pr1 in contact with pure water, e.g., on the stage of washing this precipitate, has already been known at the end of nineteenth century [49]. It was noted that the system obtained after mixing magnesium, ammonium, and phosphate salts at the molar ratio 1:1:1 gives a system containing an excess of ammonium species remaining in the solution and the precipitate that "was not struvite, but was probably composed of magnesium phosphates" [50]. This effect can be explained by the reaction [20]

$$\text{CMgNH}\_4\text{PO}\_4 = \text{Mg}\_3(\text{PO}\_4)\_2 + \text{HPO}\_4^{-2} + \text{NH}\_3 + 2\text{NH}\_4^{+1} \tag{68}$$

Such inferences were formulated on the basis of X-ray diffraction analysis, the crystallographic structure of the solid phase thus obtained. It was also stated that the precipitation of struvite requires a significant excess of ammonium species, e.g., Mg:N:P = 1:1.6:1. Struvite (pr1) is the equilibrium solid phase only at a due excess of one or two of the precipitating reagents. This remark is important in context with gravimetric analysis of magnesium as pyrophosphate. Nonetheless, also in recent times, the solubility of struvite is calculated from the approximate formula s\* = (Ksp1) 1/3 based on an assumption that it is the equilibrium solid phase in such a system.

Struvite is not the equilibrium solid phase also when introduced into aqueous solution of CO2 (CCO2 , mol/L), modified (or not) by free strong acid HB (Ca, mol/L) or strong base MOH (Cb, mol/L).

The case of struvite requires more detailed comments. The reaction (68) was proved theoretically [20], on the basis of simulated calculations performed by iterative computer programs, with use of all attainable physicochemical knowledge about the system in question. For this purpose, the fractions

$$\begin{split} q\_1 &= [\mathbf{M}\mathbf{g}^{+2}][\mathbf{N}\mathbf{H}\_4^{+1}][\mathbf{P}\mathbf{O}\_4^{-3}]/K\_{\mathrm{sp}1}, q\_2 = [\mathbf{M}\mathbf{g}^{+2}]^3[\mathbf{P}\mathbf{O}\_4^{-3}]^2/K\_{\mathrm{sp}2}, q\_3 \\ &= [\mathbf{M}\mathbf{g}^{+2}][\mathbf{H}\mathbf{P}\mathbf{O}\_4^{-2}]/K\_{\mathrm{sp}3}, q\_4[\mathbf{M}\mathbf{g}^{+2}][\mathbf{O}\mathbf{H}^{-1}]^2/K\_{\mathrm{sp}4} \end{split} \tag{69}$$

were calculated for: pr1 = MgNH4PO4 (pKsp1 = 12.6), pr2 = Mg3(PO4)2 (pKsp2 = 24.38), pr3 = MgHPO4 (pKsp3 = 5.5), pr4 = Mg(OH)2 (pKsp4 = 10.74) and are presented in Figure 4, at an initial concentration of pr1, equal C0 = [pr1]t=0 = 10�<sup>3</sup> mol/L (pC<sup>0</sup> = (ppr1)t=0 = 3); ppr1 = �log [pr1]. As we see, the precipitation of pr2 (Eq. (68)) starts at ppr1 = 3.088; other solubility products are not crossed. The changes in concentrations of some species, resulting from dissolution of pr1, are indicated in Figure 5, where s is defined by equation [20]

$$\begin{split} \mathbf{s} = \mathbf{s}\_{\text{Mg}} &= [\mathbf{Mg}^{+2}] + [\mathbf{MgOH}^{+1}] + [\mathbf{MgH}\_{2}\mathbf{PO}\_{4}^{+1}] + [\mathbf{MgHPO}\_{4}] + [\mathbf{MgPO}\_{4}^{-1}] \\ &+ [\mathbf{MgNH}\_{3}^{+2}] + [\mathbf{Mg(NH\_3)}\_{2}^{+2}] + [\mathbf{Mg(NH\_3)}\_{3}^{+2}] \end{split} \tag{70}$$

involving all soluble magnesium species are identical in its form, irrespective of the equilibrium solid phase(s) present in this system. Moreover, it is stated that pH in the solution equals

Figure 4. Plots of logqi versus ppr1 = �log[pr1] relationships, at (ppr1)t=0 = 3; i = 1,2,3,4 refer to pr1, pr2, pr3 and pr4, respectively.

Figure 5. The speciation curves for indicated species resulting from dissolution of pr1 at (ppr1)t=0 = 3.

Struvite is not the equilibrium solid phase also when introduced into aqueous solution of CO2 (CCO2 , mol/L), modified (or not) by free strong acid HB (Ca, mol/L) or strong base MOH (Cb,

The case of struvite requires more detailed comments. The reaction (68) was proved theoretically [20], on the basis of simulated calculations performed by iterative computer programs, with use of all attainable physicochemical knowledge about the system in question. For this

<sup>4</sup> �=Ksp3, q4½Mgþ<sup>2</sup>

were calculated for: pr1 = MgNH4PO4 (pKsp1 = 12.6), pr2 = Mg3(PO4)2 (pKsp2 = 24.38), pr3 = MgHPO4 (pKsp3 = 5.5), pr4 = Mg(OH)2 (pKsp4 = 10.74) and are presented in Figure 4, at an initial concentration of pr1, equal C0 = [pr1]t=0 = 10�<sup>3</sup> mol/L (pC<sup>0</sup> = (ppr1)t=0 = 3); ppr1 = �log [pr1]. As we see, the precipitation of pr2 (Eq. (68)) starts at ppr1 = 3.088; other solubility products are not crossed. The changes in concentrations of some species, resulting from

<sup>4</sup> �=Ksp1, q<sup>2</sup> ¼ ½Mgþ<sup>2</sup>

�þ½MgH2PO<sup>þ</sup><sup>1</sup>

<sup>2</sup> �þ½MgðNH3Þ

þ2

involving all soluble magnesium species are identical in its form, irrespective of the equilibrium solid phase(s) present in this system. Moreover, it is stated that pH in the solution equals

Figure 4. Plots of logqi versus ppr1 = �log[pr1] relationships, at (ppr1)t=0 = 3; i = 1,2,3,4 refer to pr1, pr2, pr3 and pr4,

� 3 ½PO�<sup>3</sup> <sup>4</sup> � 2

�½OH�<sup>1</sup> � 2 =Ksp4 =Ksp2, q<sup>3</sup>

<sup>4</sup> �þ½MgHPO4�þ½MgPO�<sup>1</sup>

<sup>3</sup> � <sup>ð</sup>70<sup>Þ</sup>

þ2

ð69Þ

<sup>4</sup> �

mol/L).

respectively.

purpose, the fractions

<sup>q</sup><sup>1</sup> ¼ ½Mgþ<sup>2</sup>

114 Descriptive Inorganic Chemistry Researches of Metal Compounds

<sup>s</sup> <sup>¼</sup> sMg ¼ ½Mgþ<sup>2</sup>

þ½MgNHþ<sup>2</sup>

¼ ½Mgþ<sup>2</sup>

�½NHþ<sup>1</sup>

�½HPO�<sup>2</sup>

dissolution of pr1, are indicated in Figure 5, where s is defined by equation [20]

�þ½MgOH<sup>þ</sup><sup>1</sup>

<sup>3</sup> �þ½MgðNH3Þ

<sup>4</sup> �½PO�<sup>3</sup>

ca. 9–9.5 (Figure 6); this pH can be affected by the presence of CO2 from air. Under such conditions, NH4 +1 and NH3 occur there at comparable concentrations [NH4 +1] ≈ [NH3], but [HPO4 �2 ]/[PO4 �3 ] = 1012.36�pH ≈ 10<sup>3</sup> . This way, the scheme (10) would be more advantageous, provided that struvite is the equilibrium solid phase; but it is not the case, see Eq. (68). The reaction (68) occurs also in the presence of CO2 in water where struvite was introduced.

Figure 6. The pH versus log[pr2] relationship; pr2 = Mg3(PO4)2, at [ppr1]t=0 = 3. The numbers at the corresponding lines indicate pCO2 ¼ �logCCO2 values; pCO2 ¼ ∞ ⇔ CCO2 = 0.

Figure 7. The speciation curves for indicated species Xi zi , resulting from dissolution of pr1 = MgNH4PO4, at (pC0, pCO2, pCb) = (2, 4, 2); s<sup>0</sup> is defined by Eq. (71).

After introducing struvite pr1 (at pC0 = [ppr1]t=0 = 2) into alkaline (C<sup>b</sup> = 10�<sup>2</sup> mol/L KOH, pCb = 2) solution of CO2 (pCO2 = 4), the dissolution is more complicated and proceeds in three steps, see Figure 7.

In step 1, pr4 precipitates first, pr1 + 2OH�<sup>1</sup> = pr4 + NH3 + HPO4 �2 , nearly from the very start of pr1 dissolution, up to ppr1 = 2.151, where Ksp2 is attained. Within step 2, the solution is saturated toward pr2 and pr4. In this step, the reaction expressed by the notation 2pr1 + pr4 = pr2 + 2NH3 + 2H2O occurs up to total depletion of pr4 (at ppr1 = 2.896). In this step, the reaction 3pr1 + 2OH�<sup>1</sup> = pr2 + 3NH3 + HPO4 �<sup>1</sup> + 2H2O occurs up to total depletion of pr1, i.e., the solubility product Ksp1 for pr1 is not crossed. The curve s<sup>0</sup> (Figure 7) is related to the function

$$\mathbf{s'} = \mathbf{s} + [\mathbf{M}\mathbf{g}\mathbf{H}\mathbf{C}\mathbf{O}\_3^{+1}] + [\mathbf{M}\mathbf{g}\mathbf{C}\mathbf{O}\_3] \tag{71}$$

where s is expressed by Eq. (70).

#### 6. Solubility of nickel dimethylglyoximate

The precipitate of nickel dimethylglyoximate, NiL2, has soluble counterpart with the same formula, i.e., NiL2, in aqueous media. If NiL2 is in equilibrium with the solution, concentration of the soluble complex NiL2 assumes constant value: [NiL2] = K2∙[Ni2+][L�] <sup>2</sup> = K2∙Ksp, where K<sup>2</sup> = 1017.24, Ksp = [Ni2+][L�] <sup>2</sup> = 10�23.66 [14, 17, 18], and then [NiL2] = 10�6.42 (i.e., log[NiL2] = �6.42). The concentration [NiL2] is the constant, limiting component in expression for solubility s = sNi of nickel dimethylglyoximate, NiL2. Moreover, it is a predominant component in

Figure 8. Solubility curves for nickel dimethylglyoximate NiL2 in (a) ammonia, (b) acetate+ammonia, and (c) citrate +acetate+ ammonia media at total concentrations [mol/L]: CNi = 0.001, C<sup>L</sup> = 0.003, C<sup>N</sup> = 0.5, CAc = 0.3, CCit = 0.1 [14].

expression for s in alkaline media, see Figure 8. This pH range involves pH of ammonia buffer solutions, where NiL2 is precipitated from NiSO4 solution during the gravimetric analysis of nickel; the expression for solubility

After introducing struvite pr1 (at pC0 = [ppr1]t=0 = 2) into alkaline (C<sup>b</sup> = 10�<sup>2</sup> mol/L KOH, pCb = 2) solution of CO2 (pCO2 = 4), the dissolution is more complicated and proceeds in three steps, see

of pr1 dissolution, up to ppr1 = 2.151, where Ksp2 is attained. Within step 2, the solution is saturated toward pr2 and pr4. In this step, the reaction expressed by the notation 2pr1 + pr4 = pr2 + 2NH3 + 2H2O occurs up to total depletion of pr4 (at ppr1 = 2.896). In this step, the

the solubility product Ksp1 for pr1 is not crossed. The curve s<sup>0</sup> (Figure 7) is related to the

The precipitate of nickel dimethylglyoximate, NiL2, has soluble counterpart with the same formula, i.e., NiL2, in aqueous media. If NiL2 is in equilibrium with the solution, concentration

�6.42). The concentration [NiL2] is the constant, limiting component in expression for solubility s = sNi of nickel dimethylglyoximate, NiL2. Moreover, it is a predominant component in

<sup>2</sup> = 10�23.66 [14, 17, 18], and then [NiL2] = 10�6.42 (i.e., log[NiL2] =

<sup>0</sup> <sup>¼</sup> <sup>s</sup> þ ½MgHCO<sup>þ</sup><sup>1</sup>

of the soluble complex NiL2 assumes constant value: [NiL2] = K2∙[Ni2+][L�]

�2

zi , resulting from dissolution of pr1 = MgNH4PO4, at (pC0, pCO2,

�<sup>1</sup> + 2H2O occurs up to total depletion of pr1, i.e.,

<sup>3</sup> �þ½MgCO3� ð71Þ

, nearly from the very start

<sup>2</sup> = K2∙Ksp, where

In step 1, pr4 precipitates first, pr1 + 2OH�<sup>1</sup> = pr4 + NH3 + HPO4

s

6. Solubility of nickel dimethylglyoximate

reaction 3pr1 + 2OH�<sup>1</sup> = pr2 + 3NH3 + HPO4

Figure 7. The speciation curves for indicated species Xi

116 Descriptive Inorganic Chemistry Researches of Metal Compounds

pCb) = (2, 4, 2); s<sup>0</sup> is defined by Eq. (71).

where s is expressed by Eq. (70).

K<sup>2</sup> = 1017.24, Ksp = [Ni2+][L�]

Figure 7.

function

$$\mathbf{s} = \mathbf{s}\_{\text{Ni}} = [\text{Ni}^{+2}] + [\text{NiO}\text{H}^{+1}] + [\text{NiSO}\_4] + \sum\_{i=1}^{6} [\text{Ni}(\text{NH}\_3)\_i^{+2}] + [\text{NiI}\_2] \tag{72}$$

The effect of other, e.g., citrate (Cit) and acetate (Ac) species as complexing agents can also be considered for calculation purposes, see the lines b and c in Figure 8. The presence of citrate does not affect significantly the solubility of NiL2 in ammonia buffer media, i.e., at pH ≈ 9, where sNi ffi [NiL2].

Calculations of s = sNi were made at CNi = 0.001 mol/L and C<sup>L</sup> = 0.003 mol/L HL, i.e., at the excessive HL concentration equal C<sup>L</sup> – 2CNi = 0.001 mol/L. Solubility of HL in water, equal 0.063 g HL/100 mL H2O (25o C) [51], corresponds to concentration 0.63/116.12 = 0.0054 mol/L of the saturated HL solution, 0.003 < 0.0054. Applying higher CL values needs the HL solution in ethanol, where HL is fairly soluble. However, the aqueous-ethanolic medium is thus formed, where equilibrium constants are unknown. To avoid it, lower CNi and C<sup>L</sup> values were applied in calculations. The equilibrium data were taken from Ref. [31].

The soluble complex having the formula identical to the formula of the precipitate occurs also in other, two-phase systems. In some pH range, concentration of this soluble form is the dominant component of the expression for the solubility s. As stated above, such a case occurs for NiL2. Then one can assume the approximation

$$\mathbf{s} = \mathbf{K\_2} \mathbf{K\_{sp}} \tag{73}$$

Similar relationship exists also for other precipitates. By differentiation of Eq. (73) with respect to temperature Tat p = const, and application of van't Hoff's isobar equation for K<sup>2</sup> and Ksp, we obtain

$$\frac{1}{\text{s}} \cdot \left(\frac{\partial \text{s}}{\partial T}\right)\_p = \frac{1}{RT^2} \cdot \left(\Delta G\_1^o + \Delta G\_2^o\right) \tag{74}$$

where

$$\Delta G\_1^\circ = RT^2 \cdot \left(\frac{\partial \ln K\_y}{\partial T}\right)\_p \quad \text{and} \quad \Delta G\_2^\circ = RT^2 \cdot \left(\frac{\partial \ln K\_2}{\partial T}\right)\_p$$

Because, as a rule,

$$\left(\frac{\partial K\_p}{\partial T}\right)\_p > 0 \quad \text{and} \quad \left(\frac{\partial K\_2}{\partial T}\right)\_p < 0$$

then ΔG<sup>o</sup> <sup>1</sup> > 0 and ΔG<sup>o</sup> <sup>2</sup> < 0, and Eq. (74) can be rewritten into the form

$$\frac{1}{\text{s}} \cdot \left(\frac{\partial \text{s}}{\partial T}\right)\_p = \frac{1}{RT^2} \cdot \left(|\Delta G\_1^o| - |\Delta G\_2^o|\right) \tag{75}$$

If <sup>j</sup>ΔG<sup>o</sup> <sup>1</sup><sup>j</sup> <sup>≈</sup> <sup>j</sup>ΔG<sup>o</sup> <sup>2</sup>j within the temperature range (T0, T), the value of s is approximately constant. Let T<sup>0</sup> denote the room temperature (at which,as a rule—all the equilibrium constants are determined) and T 6¼ T<sup>0</sup> is the temperature at which the precipitate is filtered and washed. In this case, the solubility s and then theoretical accuracy of gravimetric analysis does not change with temperature.

#### 7. Calculation of solubility in dynamic redox systems

#### 7.1. Preliminary information

The redox system presented in this section is resolvable according to generalized approach to redox systems (GATES), formulated by Michałowski (1992) [8]. According to GATES principles, the algebraic balancing of any electrolytic system is based on the rules of conservation of particular elements/cores Yg (g = 1,…, G), and on charge balance (ChB), expressing the rule of electroneutrality of this system; the terms element and core are then distinguished. The core is a cluster of elements with defined composition (expressed by its chemical formula) and external charge that remains unchanged during the chemical process considered, e.g., titration. For ordering purposes, we assume: Y<sup>1</sup> = H, Y<sup>2</sup> = O,…. For modeling purposes, the closed systems, composed of condensed phases separated from its environment by diathermal (freely permeable by heat) walls, are considered; it enables the heat exchange between the system and its environment. Any chemical process, such as titration, is carried out under isothermal conditions, in a quasistatic manner; constant temperature is one of the conditions securing constancy of equilibrium constants values. An exchange of the matter (H2O, CO2, O2,…) between the system and its environment is thus forbidden, for modeling purposes. The elemental/core balance F(Yg) for the g-th element/core (Yg) (g = 1,…, G) is expressed by an equation interrelating the numbers of Ygatoms or cores in components of the system with the numbers of Yg-atoms/cores in the species of the system thus formed; we have F(H) for Y<sup>1</sup> = H, F(O) for Y<sup>2</sup> = O, etc.

The key role in redox systems is due to generalized electron balance (GEB) concept, discovered by Michałowski as the Approach I (1992) and Approach II (2006) to GEB; both approaches are equivalent:

1 s � <sup>∂</sup><sup>s</sup> ∂T 

<sup>1</sup> <sup>¼</sup> RT<sup>2</sup> � <sup>∂</sup>lnKsp

∂Ksp ∂T p

1 s � <sup>∂</sup><sup>s</sup> ∂T 

7. Calculation of solubility in dynamic redox systems

the system thus formed; we have F(H) for Y<sup>1</sup> = H, F(O) for Y<sup>2</sup> = O, etc.

ΔG<sup>o</sup>

118 Descriptive Inorganic Chemistry Researches of Metal Compounds

where

then ΔG<sup>o</sup>

If <sup>j</sup>ΔG<sup>o</sup>

Because, as a rule,

<sup>1</sup><sup>j</sup> <sup>≈</sup> <sup>j</sup>ΔG<sup>o</sup>

with temperature.

7.1. Preliminary information

<sup>1</sup> > 0 and ΔG<sup>o</sup>

p <sup>¼</sup> <sup>1</sup>

∂T 

> p <sup>¼</sup> <sup>1</sup>

p

RT<sup>2</sup> � ðΔG<sup>o</sup>

and ΔGo

∂T p < 0

<sup>2</sup>j within the temperature range (T0, T), the value of s is approximately constant.

1j�jΔGo

> 0 and <sup>∂</sup>K<sup>2</sup>

<sup>2</sup> < 0, and Eq. (74) can be rewritten into the form

RT<sup>2</sup> � ðjΔG<sup>o</sup>

Let T<sup>0</sup> denote the room temperature (at which,as a rule—all the equilibrium constants are determined) and T 6¼ T<sup>0</sup> is the temperature at which the precipitate is filtered and washed. In this case, the solubility s and then theoretical accuracy of gravimetric analysis does not change

The redox system presented in this section is resolvable according to generalized approach to redox systems (GATES), formulated by Michałowski (1992) [8]. According to GATES principles, the algebraic balancing of any electrolytic system is based on the rules of conservation of particular elements/cores Yg (g = 1,…, G), and on charge balance (ChB), expressing the rule of electroneutrality of this system; the terms element and core are then distinguished. The core is a cluster of elements with defined composition (expressed by its chemical formula) and external charge that remains unchanged during the chemical process considered, e.g., titration. For ordering purposes, we assume: Y<sup>1</sup> = H, Y<sup>2</sup> = O,…. For modeling purposes, the closed systems, composed of condensed phases separated from its environment by diathermal (freely permeable by heat) walls, are considered; it enables the heat exchange between the system and its environment. Any chemical process, such as titration, is carried out under isothermal conditions, in a quasistatic manner; constant temperature is one of the conditions securing constancy of equilibrium constants values. An exchange of the matter (H2O, CO2, O2,…) between the system and its environment is thus forbidden, for modeling purposes. The elemental/core balance F(Yg) for the g-th element/core (Yg) (g = 1,…, G) is expressed by an equation interrelating the numbers of Ygatoms or cores in components of the system with the numbers of Yg-atoms/cores in the species of

<sup>1</sup> <sup>þ</sup> <sup>Δ</sup>G<sup>o</sup>

<sup>2</sup> <sup>¼</sup> RT<sup>2</sup> � <sup>∂</sup>lnK<sup>2</sup>

∂T p

<sup>2</sup>Þ ð74Þ

<sup>2</sup>jÞ ð75Þ

$$\text{Therefore, Approximate II to GEB} \leftrightarrow \text{Approach I to GEB} \tag{76}$$

GEB is fully compatible with charge balance (ChB) and concentration balances F(Yg), formulated for different elements and cores. The primary form of GEB, pr-GEB, obtained according to Approach II to GEB is the linear combination

$$\text{pr}-\text{GEB}=\text{2}\cdot F(\text{O}) - F(\text{H})\tag{77}$$

Both approaches (I and II) to GEB were widely discussed in the literature [7–12, 14, 15, 17, 18, 34, 52–74], and in three other chapters in textbooks [75–79] issued in 2017 within InTech. The GEB is perceived as a law of nature [9, 10, 17, 67, 71, 73, 74], as the hidden connection of physicochemical laws, as a breakthrough in the theory of electrolytic redox systems. The GATES refers to mono- and polyphase, redox, and nonredox, equilibrium and metastable [20, 21–23, 78, 79] static and dynamic systems, in aqueous, nonaqueous, and mixedsolvent media [69, 72], and in liquid-liquid extraction systems [53]. Summarizing, Approach II to GEB needs none prior information on oxidation numbers of all elements in components forming a redox system and in the species in the system thus formed. The Approach I to GEB, considered as the "short" version of GEB, is useful if all the oxidation numbers are known beforehand; such a case is obligatory in the system considered below. The terms "oxidant" and "reductant" are not used within both approaches. In redox systems, 2∙F(O) – F(H) is linearly independent on CHB and F(Yg) (g ≥ 3,…, G); in nonredox systems, 2∙F(O) – F(H) is dependent on those balances. This property distinguishes redox and nonredox systems of any degree of complexity. Within GATES, and GATES/GEB in particular, the terms: "stoichiometry," "oxidation number," "oxidant," "reductant," "equivalent mass" are considered as redundant, old-fashioned terms. The term "mass action law" (MAL) was also replaced by the equilibrium law (EL), fully compatible with the GATES principles. Within GATES, the law of charge conservation and law of conservation of all elements of the system tested have adequate importance/ significance.

A detailed consideration of complex electrolytic systems requires a collection and an arrangement of qualitative (particular species) and quantitative data; the latter ones are expressed by interrelations between concentrations of the species. The interrelations consist of material balances and a complete set of expressions for equilibrium constants. Our further considerations will be referred to a titration, as a most common example of dynamic systems. The redox and nonredox systems, of any degree of complexity, can be resolved in analogous manner, without any simplifications done, with the possibility to apply all (prior, preselected) physicochemical knowledge involved in equilibrium constants related to a system in question. This way, one can simulate (imitate) the analytical prescription to any process that may be realized under isothermal conditions, in mono- and two-phase systems, with liquid-liquid extraction systems included.

#### 7.2. Solubility of CuI in a dynamic redox system

The system considered in this section is related to iodometric, indirect analysis of an acidified (H2SO4) solution of CuSO4 [14, 64]. It is a very interesting system, both from analytical and physicochemical viewpoints. Because the standard potential E<sup>0</sup> = 0.621 V for (I2, I�<sup>1</sup> ) exceeds E<sup>0</sup> = 0.153 V for (Cu+2, Cu+1), one could expect (at a first sight) the oxidation of Cu+1 by I2. However, such a reaction does not occur, due to the formation of sparingly soluble CuI precipitate (pKsp = 11.96).

This method consists of four steps. In the preparatory step (step 1), an excess of H2SO4 is neutralized with NH3 (step 1) until a blue color appears, which is derived from Cu(NH3)<sup>i</sup> +2 complexes. Then the excess of CH3COOH is added (step 2), to attain a pH ca. 3.6. After subsequent introduction of an excess of KI solution (step 3), the mixture with CuI precipitate and dissolved iodine formed in the reactions: 2Cu+2 + 4I�<sup>1</sup> = 2CuI + I2, 2Cu+2 + 5I�<sup>1</sup> = 2CuI + I3 �1 is titrated with Na2S2O3 solution (step 4), until the reduction of iodine: I2 + 2S2O3 �<sup>2</sup> = 2I�<sup>1</sup> + S4O6 �2 , I3 �<sup>1</sup> + 2S2O3 �<sup>2</sup> = 3I�<sup>1</sup> + S4O6 �<sup>2</sup> is completed; the reactions proceed quantitatively in mildly acidic solutions (acetate buffer), where the thiosulfate species are in a metastable state. In strongly acidic media, thiosulfuric acid disproportionates according to the scheme H2S2O3 = H2SO3 + S [80].

#### 7.3. Formulation of the system

We assume that V mL of C mol/L Na2S2O3 solution is added into the mixture obtained after successive addition of: V<sup>N</sup> mL of NH3 (C1) (step 1), VAc mL of CH3COOH (C2) (step 2), VKI mL of KI (C3) (step 3), and V mL of Na2S2O3 (C) (step 4) into V<sup>0</sup> mL of titrand D composed of CuSO4 (C0)+H2SO4 (C01). To follow the changes occurring in particular steps of this analysis, we assume that the corresponding reagents in particular steps are added according to the titrimetric mode, and the assumption of the volumes additivity is valid.

In this system, three electron-active elements are involved: Cu (atomic number ZCu = 29), I (ZI = 53), S (ZS = 16). Note that sulfur in the core SO4 �<sup>2</sup> is not involved here in electron-transfer equilibria between S2O3 �<sup>2</sup> and S4O6 �2 ; then the concentration balance for sulfate species can be considered separately.

The balances written according to Approach I to GEB, in terms of molar concentrations, are as follows:

• Generalized electron balance (GEB)

<sup>ð</sup>ZCu–2Þð½Cuþ<sup>2</sup> �þ½CuOH<sup>þ</sup><sup>1</sup> �þ½CuðOHÞ2�þ½CuðOHÞ �1 <sup>3</sup> �þ½CuðOHÞ �2 <sup>4</sup> �þ½CuNHþ<sup>2</sup> <sup>3</sup> �þ½CuðNH3Þ þ2 <sup>2</sup> � þ½CuðNH3Þ þ2 <sup>3</sup> �þ½CuðNH3Þ þ2 <sup>4</sup> �þ½CuCH3COO<sup>þ</sup><sup>1</sup> �þ½CuðCH3COOÞ2�Þ þ ðZCu � <sup>2</sup> <sup>þ</sup> ZI � <sup>5</sup>Þ½CuIO<sup>þ</sup><sup>1</sup> <sup>3</sup> � þðZCu � <sup>1</sup>Þð½Cuþ<sup>1</sup> �þ½CuNHþ<sup>1</sup> <sup>3</sup> �þ½CuðNH3Þ þ1 <sup>2</sup> �Þ þ ðZCu <sup>þ</sup> ZIÞ½CuIð<sup>s</sup>Þ�þðZCu <sup>þ</sup> 2ZI <sup>þ</sup> <sup>1</sup>Þ½CuI�<sup>1</sup> <sup>2</sup> � þðZI þ 1Þ½I �1 �þð3ZI þ 1Þ½I �1 <sup>3</sup> � þ 2ZIð½I2� þ <sup>a</sup> � ½I2ð<sup>s</sup>Þ�Þ þ ðZI � <sup>1</sup>Þð½HIO�þ½IO�<sup>1</sup> �Þ þ ðZI � <sup>5</sup>Þð½HIO3�þ½IO�<sup>1</sup> <sup>3</sup> �Þ þðZI � <sup>7</sup>Þð½H5IO6�þ½H4IO�<sup>1</sup> <sup>6</sup> �þ½H3IO�<sup>2</sup> <sup>6</sup> �Þ þ <sup>2</sup> � ðZS � <sup>2</sup>Þð½H2S2O3�þ½HS2O�<sup>1</sup> <sup>3</sup> �Þ þ ½S2O�<sup>2</sup> <sup>3</sup> �Þ <sup>þ</sup> <sup>4</sup> � ðZS–2:5Þ½S4O�<sup>2</sup> <sup>6</sup> �þðZCu–<sup>1</sup> <sup>þ</sup> <sup>2</sup> � ðZS–2ÞÞ½CuS2O�<sup>1</sup> <sup>3</sup> �þðZCu–1 þ 4 � ðZS–2ÞÞ½CuðS2O3Þ �3 <sup>2</sup> �þðZCu–1þ 6 � ðZS–2ÞÞ½CuðS2O3Þ �5 <sup>3</sup> � � ððZCu � 2ÞC0V0 þ ðZI þ 1ÞC3VKI þ 2 � ðZS � 2ÞCVÞ= ðV0þVNþVAcþVKIþVÞ ¼ 0

ð78Þ
