Electrolyte solutions
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- Written by: Germán Fernández
- Category: electrolyte solutions
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An electrolyte is a substance that produces ions in solution, which is evidenced by the fact that the solution is electrically conductive.
Due to the long-range intermolecular forces between ions in the solution, the use of activity coefficients when treating electrolyte solutions is essential, even for very dilute solutions.
Let the electrolyte be:
\begin{equation} M{\nu_{+}}X_{\nu_{-}}(s) \rightarrow \nu_{+}M_{ac}^{Z_{+}}+\nu_{ -}X_{ac}^{Z_{-}} \end{equation}
Let's compare this equation with the dissociation of barium nitrate. \begin{equation} Ba(NO_3)_2 \rightarrow Ba_{ac}^{2+}+2NO_{3}^{-} \end{equation} Where $M=Ba, X=NO_3, \nu_{+} =1, \nu_{-}=2, Z_{+}=+2, Z_{-}=-1$
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- Written by: Germán Fernández
- Category: electrolyte solutions
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The chemical potential of the positive ions in the solution is given by: \begin{equation} \mu_{+}=\left(\frac{\partial G}{\partial n_+}\right)_{T,P, n_{j\neq +}} \end{equation} We cannot measure $\mu_{+}$ or $\mu_{-}$ since we cannot add only negative or positive ions to the solution.
Therefore, the chemical potential of the electrolyte as a whole is defined. \begin{equation} \mu_{i}=\left(\frac{\partial G}{\partial n_i}\right)_{T,P,n_A} \end{equation} where $n_i$ represents the moles of dissolved electrolyte.
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- Written by: Germán Fernández
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For $n_i$ moles of an electrolyte of the type $M_{\nu_{+}}X_{\nu_{-}}$ we have:
\begin{equation} n_+=\nu_{+}n_i-n_{PI}\ ;\; \Rightarrow\;\;\; dn_+=\nu_{+}dn_i-dn_{PI} \end{equation}
\begin{equation} n_{-}=\nu_{-}n_i-n_{PI}\;\;\;\Rightarrow\; \;\; dn_{-}=\nu_{-}dn_i-dn_{PI} \end{equation}
Writing the Gibbs equation for dG in open systems
\begin{equation} dG=-SdT+VdP+\sum_{i}\mu_{ i}dn_{i}=-SdT+VdP+\mu_Adn_A+\mu_+dn_+ + \mu_-dn_- + \mu_{PI}dn_{PI} \end{equation}
Substituting $dn_-$ and $dn_+$ into this last equation
\begin{equation} dG=-SdT+VdP+\mu_Adn_A+\mu_+\left(\nu_+dn_i-dn_{PI}\right)+\mu_-\left(\mu_-dn_i-dn_{PI} \right)+\mu_{PI}dn_{PI} \end{equation}
Read more: Relationship between $\mu_{+}$,$\mu_{-}$ and $\mu_{i}$
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- Written by: Germán Fernández
- Category: electrolyte solutions
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Dissociating $n_i$ moles $M_{\nu_+}X_{\nu_-}$ we get: \begin{equation} n_+=\nu_+n_i\;\;\;\Rightarrow \;\;\; m_+=\nu_+ m_i \end{equation} \begin{equation} n_-=\nu_- n_i\;\;\;\Rightarrow \;\;\; m_-=\nu_- m_i \end{equation} Substituting into equation (61) \begin{equation} \mu_i=\mu_{i}^{0}+RTln\left[\gamma_{\pm}^{\ nu}\left(\frac{\nu_{+}m_i}{m^0}\right)^{\nu_+}\left(\frac{\nu_-m_i}{m^0}\right)^{ \nu_-}\right] \end{equation} Grouping Terms \begin{equation} \mu_i=\mu_{i}^{0}+RTln\left[\gamma_{\pm}^{\nu}\nu_{ \pm}^{\nu}\left(\frac{m_i}{m^0}\right)^{\nu}\right] \end{equation} Applying properties of natural logarithms. \begin{equation} \mu_i=\mu_{i}^{0}+\nu RTln\left[\gamma_{\pm}\nu_{\pm}\left(\frac{m_i}{m^{0} }\right)\right] \end{equation}
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- Written by: Germán Fernández
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Let $n_i$ be moles of $M_{\nu_+}X_{\nu_-}$ with which the solution is prepared.
$\alpha$ represents the fraction of $M^{Z_+}$ that does not associate (remains free in solution).
Under these conditions in the solution we have: \begin{eqnarray} n_+ & = & \alpha\nu_{+}n_i\\ n_{PI} & = & \nu_{-}n_i-n_+ =\nu_+n_i- \alpha\nu_{+}n_i=\nu_{+}n_i(1-\alpha)\\ n_- & = & \nu_-n_i-n_{PI}=\nu_-n_i-\nu_+n_i(1- \alpha)=\left[\nu_--\nu_+(1-\alpha)\right]n_i \end{eqnarray} Dividing by the kg of solvent gives the molalities. \begin{eqnarray} m_+ & = & \alpha \nu_+ m_i\\ m_- & = & \left[\nu_- - \nu_+(1-\alpha)\right]m_i \end{eqnarray} Substituting into equation (61) \begin{equation} \mu_i =\mu_{i}^{0} + RTln\left[\gamma_{\pm}^{\nu}\left(\frac{\alpha\nu_+m_i }{m^0}\right)^{\nu_+}\left(\frac{[\nu_--\nu_+(1-\alpha)]m_i}{m^0}\right)^{\nu_ -}\right] \end{equation}
Read more: Chemical potential of an electrolyte with ionic association