Representation of Electrical Conductances by the Quint–Viallard Conductivity Equation. Part 6. Unsymmetrical 2:1 Type “Complex Ions” Electrolytes: Cadmium Bromide and Cadmium Iodide

Abstract

Systematic and precise measurements of electrical conductivities of dilute aqueous solutions of cadmium bromide and cadmium iodide were performed from 15 to 35 °C. The conductances were interpreted in terms of a molecular model that includes a mixture of two 1:1 and 2:1 electrolytes. The molar limiting conductances of $\lambda$ ⁰(CdX⁺, T) ions, the equilibrium constants of formation of CdX⁺ complexes K(T) and the corresponding standard thermodynamic functions were evaluated using the Quint–Viallard conductance equations, the Debye–Hückel equations for activity coefficients and the mass action equations. Mathematical representations are also presented to an extension of molecular models, to the simultaneous formation of MeX⁺ and MeX₂ species and to the possibility of hydrolysis reactions with formation of Me(OH)⁺ ions.

Appendix: Chemical Equilibria and Conductance Equations in Cases of Formation of Neutral Species and Hydrolysis

An extension of molecular model to formation in solution of the neutral MeX₂ species is governed by the following chemical reactions:

$$ \begin{aligned} {\text{MeX}}_{ 2} \to {\text{Me}}^{{ 2 { + }}} + 2 {\text{X}}^{ - } \hfill \\ {\text{Me}}^{{ 2 { + }}} + {\text{X}}^{ - } \mathop \rightleftharpoons \limits^{{K_{ 1} }} {\text{MeX}}^{ + } \hfill \\ {\text{MeX}}^{ + } + {\text{X}}^{ - } \mathop \rightleftharpoons \limits^{{K_{ 2} }} {\text{MeX}}_{ 2} \hfill \\ \end{aligned} $$

(16)

and in terms of the mass-action equations:

$$ \begin{aligned} K_{ 1} = \frac{{\left[ {{\text{MeX}}^{ + } } \right]}}{{\left[ {{\text{Me}}^{{ 2 { + }}} } \right]\left[ {{\text{X}}^{ - } } \right]}}\frac{{f_{{{\text{MeX}}^{ + } }} }}{{f_{{{\text{Me}}^{{ 2 { + }}} }} f_{{{\text{X}}^{ - } }} }} = \frac{{\left[ {{\text{MeX}}^{ + } } \right]}}{{\left[ {{\text{Me}}^{{ 2 { + }}} } \right]\left[ {{\text{X}}^{ - } } \right]}}F_{1} (c) \hfill \\ K_{ 2} = \frac{{\left[ {{\text{MeX}}_{ 2} } \right]}}{{\left[ {{\text{MeX}}^{ + } } \right]\left[ {{\text{X}}^{ - } } \right]}}\frac{{f_{{{\text{MeX}}_{ 2} }} }}{{f_{{{\text{MeX}}^{ + } }} f_{{{\text{X}}^{ - } }} }} = \frac{{\left[ {{\text{MeX}}_{ 2} } \right]}}{{\left[ {{\text{MeX}}^{ + } } \right]\left[ {{\text{X}}^{ - } } \right]}}F_{2} (c). \hfill \\ \end{aligned} $$

(17)

Denoting the fractions of Me²⁺ and MeX⁺ ions in solution as α and β, respectively, from the charge and material balance we have:

$$ \begin{aligned} \left[ {{\text{Me}}^{{ 2 { + }}} } \right] = c\alpha \hfill \\ \left[ {{\text{MeX}}^{ + } } \right] = c\beta \hfill \\ \left[ {{\text{X}}^{ - } } \right] = c(2\alpha + \beta ) \hfill \\ \left[ {{\text{MeX}}_{ 2} } \right] = c(1 - \alpha - \beta ) \hfill \\ I = c(3\alpha + \beta ). \hfill \\ \end{aligned} $$

(18)

Introducing Eq. 18 into Eq. 17, the chemical equilibrium equations take the form

$$ \begin{aligned} K_{ 1} = \frac{\beta }{{c{\kern 1pt} \alpha (2\alpha + \beta )}}F_{1} (c) \hfill \\ K_{2} = \frac{(1 - \alpha - \beta )}{{c{\kern 1pt} \beta (2\alpha + \beta )}}F_{2} (c). \hfill \\ \end{aligned} $$

(19)

For every concentration c, the fractions α and β can be successively evaluated by an iterative solution of two quadratic equations:

$$ \begin{aligned} \alpha = \frac{{ - \beta {\kern 1pt} Q_{1} (c) + \sqrt {{\kern 1pt} {\kern 1pt} [\beta {\kern 1pt} Q_{1} (c)]^{2} + 8\beta {\kern 1pt} Q_{1} (c)} }}{{4{\kern 1pt} {\kern 1pt} Q_{1} (c)}} \hfill \\ \beta = \frac{{ - {\kern 1pt} {\kern 1pt} [1 + 2\alpha {\kern 1pt} Q_{2} (c)] + \sqrt {{\kern 1pt} {\kern 1pt} [1 + 2\alpha {\kern 1pt} Q_{2} (c)]^{2} + 4(1 - \alpha )Q_{2} (c)} }}{{2Q_{2} (c)}} \hfill \\ \end{aligned} $$

(20)

where

$$ \begin{aligned} Q_{1} (c) = \frac{{K_{1} c}}{{F_{1} (c)}} \hfill \\ Q_{2} (c) = \frac{{K_{2} c}}{{F_{2} (c)}}. \hfill \\ \end{aligned} $$

(21)

Molar conductance of electrolyte Λ(c, T) in this case from Eq. 1 is:

$$ \varLambda (c,T) = 2\alpha {\kern 1pt} {\kern 1pt} \lambda_{{{\text{Me}}^{{ 2 { + }}} }} (c,T) + \beta {\kern 1pt} {\kern 1pt} \lambda_{{{\text{MeX}}^{ + } }} (c,T) + (2\alpha + \beta ){\kern 1pt} {\kern 1pt} \lambda_{{{\text{X}}^{ - } }} (c,T) $$

(22)

and in terms of the ion pairs (Me²⁺ + X⁻) and (MeX⁺ + X⁻) is:

$$ \begin{aligned} \varLambda (c,T) = 2\alpha {\kern 1pt} \varLambda_{1} (c,T) + \beta {\kern 1pt} \varLambda_{2} (c,T) \hfill \\ \varLambda_{1} (c,T) = {\kern 1pt} {\kern 1pt} \left[ {\lambda_{{{\text{Me}}^{{ 2 { + }}} }} (c,T) + \lambda_{{{\text{X}}^{ - } }} (c,T)} \right] \hfill \\ \varLambda_{2} (c,T) = {\kern 1pt} {\kern 1pt} {\kern 1pt} \left[ {{\kern 1pt} \lambda_{{{\text{MeX}}^{ + } }} (c,T) + {\kern 1pt} {\kern 1pt} \lambda_{{{\text{X}}^{ - } }} (c,T)} \right]. \hfill \\ \end{aligned} $$

(23)

Thus, essentially the form of conductivity equations remains the same as in Eq. 11, but the appearance of neutral MeX₂ species reduces the total number of ions in solution.

In the simplest case of hydrolysis process, the following reactions occur:

$$ \begin{aligned} {\text{MeX}}_{ 2} \to {\text{Me}}^{{ 2 { + }}} + 2 {\text{X}}^{ - } \hfill \\ {\text{Me}}^{{ 2 { + }}} + {\text{X}}^{ - } \mathop \rightleftharpoons \limits^{K} {\text{MeX}}^{ + } \hfill \\ {\text{Me}}^{{ 2 { + }}} + {\text{OH}}^{ - } \mathop \rightleftharpoons \limits^{{K_{\text{h}} }} {\text{MeOH}}^{ + } \hfill \\ {\text{H}}_{ 2} {\text{O}}\mathop \rightleftharpoons \limits^{{K_{\text{w}} }} {\text{H}}^{ + } + {\text{OH}}^{ - } . \hfill \\ \end{aligned} $$

(24)

The mass-action equations of the complexation and hydrolysis processes are:

$$ \begin{aligned} K = \frac{{\left[ {{\text{MeX}}^{ + } } \right]}}{{\left[ {{\text{Me}}^{{ 2 { + }}} } \right]\left[ {{\text{X}}^{ - } } \right]}}\frac{{f_{{{\text{MeX}}^{ + } }} }}{{f_{{{\text{Me}}^{{ 2 { + }}} }} f_{{{\text{X}}^{ - } }} }} = \frac{{\left[ {{\text{MeX}}^{ + } } \right]}}{{\left[ {{\text{Me}}^{{ 2 { + }}} } \right]\left[ {{\text{X}}^{ - } } \right]}}F(c) \hfill \\ K_{\text{h}} = \frac{{\left[ {{\text{MeOH}}^{ + } } \right]}}{{\left[ {{\text{Me}}^{{ 2 { + }}} } \right]\left[ {{\text{OH}}^{ - } } \right]}}\frac{{f_{{{\text{MeOH}}^{ + } }} }}{{f_{{{\text{Me}}^{{ 2 { + }}} }} f_{{{\text{OH}}^{ - } }} }} = \frac{{\left[ {{\text{MeOH}}^{ + } } \right]}}{{\left[ {{\text{Me}}^{{ 2 { + }}} } \right]\left[ {{\text{OH}}^{ - } } \right]}}F_{\text{h}} (c) \hfill \\ K_{\text{w}} = \left[ {{\text{H}}^{ + } } \right]\left[ {{\text{OH}}^{ - } } \right]f_{{{\text{H}}^{ + } }} f_{{{\text{OH}}^{ - } }} = \left[ {{\text{H}}^{ + } } \right]\left[ {{\text{OH}}^{ - } } \right]F_{\text{w}} (c). \hfill \\ \end{aligned} $$

(25)

In this case, from the material and charge balance we have:

$$ \begin{aligned} \left[ {{\text{Me}}^{{ 2 { + }}} } \right] = c\alpha \hfill \\ \left[ {{\text{MeX}}^{ + } } \right] = c\beta \hfill \\ \left[ {{\text{MeOH}}^{ + } } \right] = c(1 - \alpha - \beta ) \hfill \\ \left[ {{\text{X}}^{ - } } \right] = c(2 - \beta ) \hfill \\ \left[ {{\text{OH}}^{ - } } \right] = c\gamma \hfill \\ \left[ {{\text{H}}^{ + } } \right] = c(1 - \alpha - \beta + \gamma ) \hfill \\ I = c(2 + \alpha - \beta + \gamma ) \hfill \\ \end{aligned} $$

(26)

and therefore

$$ \begin{aligned} K = \frac{\beta }{{c{\kern 1pt} \alpha (2 - \beta )}}F(c) \hfill \\ K_{\text{h}} = \frac{(1 - \alpha - \beta )}{{c{\kern 1pt} \alpha {\kern 1pt} \gamma }}F_{\text{h}} (c) \hfill \\ K_{\text{w}} = c^{2} (1 - \alpha - \beta + \gamma )\gamma {\kern 1pt} F_{\text{w}} (c). \hfill \\ \end{aligned} $$

(27)

This gives the set of three equations that should be solved by an iterative procedure:

$$ \begin{aligned} \alpha = \frac{1 - \beta }{{1 + \gamma {\kern 1pt} Q_{\text{h}} (c)}} \hfill \\ \beta = \frac{{2\alpha {\kern 1pt} Q(c)}}{{1 + \alpha {\kern 1pt} Q(c)}} \hfill \\ \gamma = \frac{{ - (1 - \alpha - \beta ) + \sqrt {(1 - \alpha - \beta )^{2} + 4Q_{\text{w}} (c)} }}{2} \hfill \\ \end{aligned} $$

(28)

where

$$ \begin{aligned} Q(c) = \frac{Kc}{F(c)} \hfill \\ Q_{\text{h}} (c) = \frac{{K_{\text{h}} c}}{{F_{\text{h}} (c)}} \hfill \\ Q_{\text{w}} (c) = \frac{{K_{\text{w}} c^{2} }}{{F_{\text{w}} (c)}}. \hfill \\ \end{aligned} $$

(29)

The ionic contributions to the molar conductance Λ(c, T) are given by:

$$ \begin{aligned} \varLambda (c,T) = 2\alpha {\kern 1pt} {\kern 1pt} \lambda_{{{\text{Me}}^{{ 2 { + }}} }} (c,T) + \beta {\kern 1pt} {\kern 1pt} \lambda_{{{\text{MeX}}^{ + } }} (c,T) + (1 - \alpha - \beta )\lambda_{{{\text{MeOH}}^{ + } }} (c,T) \hfill \\ \quad \quad \quad \quad + (2 - \beta ){\kern 1pt} {\kern 1pt} \lambda_{{{\text{X}}^{ - } }} (c,T) + (1 - \alpha - \beta + \lambda )\lambda_{{{\text{H}}^{ + } }} (c,T) + \gamma {\kern 1pt} \lambda_{{{\text{OH}}^{ - } }} (c,T). \hfill \\ \end{aligned} $$

(30)

In order to apply the Quint–Viallard conductance equation, the following conductances of six pairs of ions should be introduced:

$$ \begin{aligned} \varLambda_{1} (c,T) = {\kern 1pt} {\kern 1pt} \left[ {\lambda_{{{\text{Me}}^{{ 2 { + }}} }} (c,T) + \lambda_{{{\text{X}}^{ - } }} (c,T)} \right] \hfill \\ \varLambda_{2} (c,T) = {\kern 1pt} {\kern 1pt} \left[ {\lambda_{{{\text{Me}}^{{ 2 { + }}} }} (c,T) + \lambda_{{{\text{OH}}^{ - } }} (c,T)} \right] \hfill \\ \varLambda_{3} (c,T) = {\kern 1pt} {\kern 1pt} {\kern 1pt} \left[ {{\kern 1pt} \lambda_{{{\text{MeX}}^{ + } }} (c,T) + {\kern 1pt} {\kern 1pt} \lambda_{{{\text{X}}^{ - } }} (c,T)} \right] \hfill \\ \varLambda_{4} (c,T) = {\kern 1pt} {\kern 1pt} {\kern 1pt} \left[ {{\kern 1pt} \lambda_{{{\text{MeX}}^{ + } }} (c,T) + {\kern 1pt} {\kern 1pt} \lambda_{{{\text{OH}}^{ - } }} (c,T)} \right] \hfill \\ \varLambda_{5} (c,T) = {\kern 1pt} {\kern 1pt} {\kern 1pt} \left[ {{\kern 1pt} \lambda_{{{\text{H}}^{ + } }} (c,T) + {\kern 1pt} {\kern 1pt} \lambda_{{{\text{OH}}^{ - } }} (c,T)} \right] \hfill \\ \varLambda_{6} (c,T) = {\kern 1pt} {\kern 1pt} {\kern 1pt} \left[ {{\kern 1pt} \lambda_{{{\text{H}}^{ + } }} (c,T) + {\kern 1pt} {\kern 1pt} \lambda_{{{\text{X}}^{ - } }} (c,T)} \right]. \hfill \\ \end{aligned} $$

(31)

Arranging (31) with using (30), the molar conductance of solutions is given by:

$$ \begin{aligned} \varLambda (c,T) = \alpha {\kern 1pt} \varLambda_{1} (c,T) + \alpha {\kern 1pt} \varLambda_{2} (c,T) + \beta {\kern 1pt} {\kern 1pt} \varLambda_{3} (c,T) + (1 - \alpha - \beta ){\kern 1pt} {\kern 1pt} \varLambda_{4} (c,T) \hfill \\ \quad \quad \quad \quad +( - 1 + \beta + \gamma ){\kern 1pt} {\kern 1pt} \varLambda_{5} (c,T) + (2 - \alpha - 2\beta ){\kern 1pt} {\kern 1pt} \varLambda_{6} (c,T). \hfill \\ \end{aligned} $$

(32)