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) $$\mu_i =\mu_{i}^{0} + RTln\left[\gamma_{\pm}^{\nu}\left(\frac{\alpha\nu_+m_i }{m^0}\right)^{\nu_+}\left(\frac{[\nu_--\nu_+(1-\alpha)]m_i}{m^0}\right)^{\nu_ -}\right]$$

Grouping Terms $$\mu_i =\mu_{i}^{0} + RTln\left[\gamma_{\pm}^{\nu}\alpha^{\ nu_+}\nu_{\pm}^{\nu}\left[1-(1-\alpha)\frac{\nu_+}{\nu_-}\right]^{\nu_-}\left(\ frac{m_i}{m^0}\right)^{\nu}\right]$$ Dividing all exponents by $\nu$ $$\mu_i =\mu_{i}^{0 } + \nu RTln\left[\gamma_{\pm}\alpha^{\nu_+/\nu}\nu_{\pm}\left[1-(1-\alpha)\frac{\nu_+}{ \nu_-}\right]^{\nu-/\nu}\left(\frac{m_i}{m^0}\right)\right]$$ Calling $\gamma_i=\alpha^{\ nu_+/\nu}\left[1-(1-\alpha)\frac{\nu_+}{\nu}\right]^{\nu_-/\nu}\gamma_{\pm}$, the potential chemical of an electrolyte with ionic association we have: $$\mu_ i=\mu_{i}^{0}+\nu RTln\left[\gamma_i \nu_{\pm}\frac{m_i}{m^0}\right]$$

$\gamma_i$: Stoichiometric activity coefficient on the molality scale.

$\gamma_{\pm}$: Mean ionic activity coefficient on the molality scale.