Limiting molar conductivity of multiple solutes

Suppose there is a water solution at $\pu{25 °C}$ with $\ce{Na+},$ $\ce{Mg^2+}$ cations and $\ce{Cl-},$ $\ce{HCO3-}$ anions. Concentrations $[\ce{Na+}],$ $[\ce{Mg^2+}],$ $[\ce{Cl^-}]$ and $[\ce{HCO3-}]$ are known.


I cannot understand how to apply the Kohlrausch's law of the independent migration of ions here:


$$\Lambda\_\mathrm{m}^0 = \nu\_+\lambda\_+ + \nu\_-\lambda\_-$$


How this formula will transform for the solution with many kinds of ions?


Edit:


Electroneutrality condition holds, so:
$$[\ce{Na+}]+2 \cdot [\ce{Mg^2+}]=[\ce{Cl^-}]+[\ce{HCO3-}]$$ 
I've found the table with the limiting ionic conductivities $\lambda$ of each individual ion in water at $\pu{25 °C}$:


$\lambda\_{\ce{Na+}} = \pu{50 \cdot 10^{-4} S m2 mol-1}$


$\lambda\_{\ce{Mg^2+}} = \pu{106 \cdot 10^{-4} S m2 mol-1}$


$\lambda\_{\ce{Cl-}} = \pu{76 \cdot 10^{-4} S m2 mol-1}$


$\lambda\_{\ce{HCO3-}} = \pu{45 \cdot 10^{-4} S m2 mol-1}$


But I don't know what values for $\nu$ I should take.




I've found a answer for my question from [this](https://chemistry.stackexchange.com/questions/77349/how-can-i-calculate-the-molar-conductivity-of-a-solution-with-multiple-solutes) question.


The way to compute the conductivity of electrolyte with multiple ion types is given from the work Pawlowicz, Rich, ( 2008), Calculating the conductivity of natural waters, Limnol. Oceanogr. Methods, 6, [doi:10.4319/lom.2008.6.489](https://doi.org/10.4319/lom.2008.6.489).


For general case consider the system, which consist of $N\_+$ number of cation types, $N\_-$ number of anion types and $N\_{types} = N\_+ + N\_-$ total number of types, fully dissolved in solvent. This system can be viewed as a weighted sum of all possible pairwise combinations between cation and anion types. Then the conductivity $\sigma,\pu{[S m-1]}$ of the solution will be determined by:
$$\sigma = \sum\_{i=1}^{N\_+}\sum\_{j=1}^{N\_-}{\frac{c\_i^+ z\_i c\_j^- z\_j}{C\_{eq}} \Lambda\_{eqm, \, ij}}$$
where $c\_i^{\pm}$ - molar concentration of corresponding ion type, $\pu{[mol m-3]}=\pu{[mM]}$;


$z\_i$ - valency of corresponding atom;


$c\_i^{\pm} \cdot z\_i$ - equivalent ionic concentration of corresponding ion type $\pu{[mol m-3]}=\pu{[mM]}$;


$\Lambda\_{eqm, \, ij}$ - equivalent molar conductivity of binary subsystem of cation type $i$ and anion type $j$, $\pu{[S m^2 mol^{-1}]}$.


Equivalent ionic concentration $C\_{eq}$ is defined as:
$$C\_{eq} = \sum\_i^{N\_+}c\_i^{+} \cdot z\_i = \sum\_j^{N\_-}c\_j^{-} \cdot z\_j = \frac{1}{2}\sum\_k^{N\_{types}}c\_k^{\pm} \cdot z\_k$$


In general a binary solute dissociates according to formula:
$$\ce{A\_{\nu^+}B\_{\nu^-} -> {\nu^+}A^{z^+ +} + {\nu^-}B^{{z^-}-}}$$
where $\nu^+$ and $\nu^-$ are coprime numbers and represent the moles of ions for 1 mole of solute. It follows:
$$\nu^+ = z^-$$
$$\nu^- = z^+$$


Equivalent molar conductivity $\Lambda\_{eqm}$ of single binary electrolyte is defined as:
$$\Lambda\_{eqm} = \frac{\sigma}{c\nu\_+z^+} = \frac{\sigma}{c\nu\_-z^-}$$


For a binary solute, dissolved in water at normal pressure, $\pu{25 °C}$ and infinitely diluted, the law of the independent migration of ions can be written to find a limiting equivalent molar conductivity $\Lambda\_{eqm}^{0}$:
$$\Lambda\_{eqm}^{0} = (\frac{\nu\_+}{z^+ } \lambda\_{+}^{0} + \frac{\nu\_-}{z^- }\lambda\_{-}^{0}) = (\frac{z^-}{z^+ } \lambda\_{+}^{0} + \frac{z^+}{z^- }\lambda\_{-}^{0})$$
where $\lambda^{0}$ - limiting ionic conductivity of the ion, $\pu{[S m^2 mol^{-1}]}$.


Here interaction between ions is ignored.


Change of equivalent molar conductivity from changing of the solute equivalent concentration is determined by Debye-Hückel-Onsager equation of the form:
$$\Lambda\_{eqm} = \Lambda\_{eqm}^0 (1-A \sqrt{I}) - B \sqrt{I} = \Lambda\_{eqm}^0 K\_A - K\_B$$
where $A, \, B$ - Debye–Hückel–Onsager coefficients;


$I$ - stoichiometric ionic strength;
$$I = \frac{1}{2} \sum\_{k=1}^{N\_{types}}c\_i z\_i^2$$
$$ A = \frac{z^2eF^2}{3 \pi \eta}\left(\frac{2}{\varepsilon RT}\right)^{1/2}$$
$$ B = \frac{qz^3eF}{24 \pi \varepsilon RT}\left(\frac{2}{\varepsilon RT}\right)^{1/2}$$
where $\eta$ - viscosity of solvent, $\pu{[Pa s]}$;


$\varepsilon$ - dielectric permittivity of solvent;


$q$ - coefficient depends on $z^+/z^-$ of binary solute.


The expression for the conductivity of electrolyte (solvent - water at $\pu{25 °C}$ and normal pressure, ion pairng reduction factor is ignored ($\alpha\_{ij}=1$ for all $i,j$)) will be:
$$\sigma = \sum\_{i=1}^{N\_+}\sum\_{j=1}^{N\_-}{\frac{c\_i^+ z\_i c\_j^- z\_j}{C\_{eq}} (\frac{z\_j^-}{z\_i^+} \lambda\_i^0 \, K\_{A, \, ij} + \frac{z\_i^+}{z\_j^- } \lambda\_j^0 \, K\_{A, \, ij} - K\_{B, \,ij})}$$


In my case I have 2 cations and 2 anions, their pairwise combination give 4 different solutes, which are assumed to be completely dissociate in water. Relation between solute molar concentration and its ions molar concentrations can be established.


1. $\ce{NaCl -> Na+ + Cl-}$
2. $\ce{NaHCO3 -> Na+ + HCO3-}$
3. $\ce{MgCl2 -> Mg^2+ + 2Cl-}$
4. $\ce{Mg(HCO3)2 -> Mg^2+ + 2HCO3-}$


Conductivity of my electrolyte will be:
\begin{align\*}
\sigma &= \frac{1}{[\ce{Na+}]+2 \cdot [\ce{Mg^2+}]} \cdot \\
 &\cdot \Bigl(\ce{[Na+]} \ce{[Cl-]} K\_{A(\ce{NaCl})} ( \lambda\_{\ce{Na+}}^0 + \lambda\_{\ce{Cl-}}^0 - \frac{K\_{B(\ce{NaCl})}}{K\_{A(\ce{NaCl})}} )+\\
 &+ \ce{[Na+]} \ce{[HCO3-]} K\_{A(\ce{NaHCO3})} ( \lambda\_{\ce{Na+}}^0 + \lambda\_{\ce{HCO3-}}^0 - \frac{K\_{B(\ce{NaHCO3})}}{K\_{A(\ce{NaHCO3})}})+\\
 &+ 2\ce{[Mg^2+]} \ce{[Cl-]} K\_{A(\ce{MgCl2})} ( \frac{1}{2}\lambda\_{\ce{Mg^2+}}^0 + 2\lambda\_{\ce{Cl-}}^0 - \frac{K\_{B(\ce{MgCl2})}}{K\_{A(\ce{MgCl2})} } )+\\
 &+ 2\ce{[Mg^2+]} \ce{[HCO3-]} K\_{A(\ce{Mg(HCO3)2})} ( \frac{1}{2}\lambda\_{\ce{Mg^2+}}^0 + 2\lambda\_{\ce{HCO3-}}^0 - \frac{ K\_{B(\ce{Mg(HCO3)2})}} {K\_{A(\ce{Mg(HCO3)2})}})
 \Bigr) 
\end{align\*}


All the values here depend from given concentration of ions or from values found in tables. Hope it is correct, please, correct me if I'm wrong. 


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