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 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.

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.