1.10.23: Gibbs Energies- Salt Solutions- Excess Gibbs Energies

A given salt solution contains a single salt \(j\) which completely dissociates to form \(ν\) moles of ions from one mole of salt. Then the chemical potential of the salt \(j\) in aqueous solution at temperature \(\mathrm{T}\) and pressure \(\mathrm{p}\) is given by equation (a).

\[\mu_{j}(a q)=\mu_{j}^{0}(a q)+v \, R \, T \, \ln \left(Q \, m_{j} \, \gamma_{\pm} / m^{0}\right)+\int_{p^{0}}^{p} V_{j}^{\infty}(a q) \, d p\]

Here \(\mu_{\mathrm{j}}^{0}(\mathrm{aq})\) is the chemical potential of the salt in solution at the same temperature and the standard pressure where molality \(\mathrm{m}_{\mathrm{j}}=1 \mathrm{~mol} \mathrm{~kg}^{-1}\) and mean ionic activity coefficient \(\gamma_{\pm} = 1\). The chemical potential of water in aqueous solution at temperature \(\mathrm{T}\) and pressure \(\mathrm{p}\) is given by equation (b) where \(\phi\) is the practical osmotic coefficient [1-3],

\[\mu_{1}(\mathrm{aq})=\mu_{1}^{*}(\ell)-\mathrm{v} \, \phi \, \mathrm{R} \, \mathrm{T} \, \mathrm{M}_{1} \, \mathrm{m}_{\mathrm{j}}+\int_{\mathrm{p}}^{\mathrm{p}} \mathrm{V}_{1}^{*}(\ell) \, \mathrm{dp}\]

If we confine our attention to the properties of solutions at ambient pressure (which is very close to the standard pressure) then we can ignore the integrals in equations (a) and (b). Hence the Gibbs energy of the solution at the same \(\mathrm{T}\) and \(\mathrm{p}\) prepared using \(1 \mathrm{~kg}\) of water is given by equation (c).

\[\begin{aligned}
\mathrm{G}\left(\mathrm{aq} ; \mathrm{w}_{1}=1 \mathrm{~kg}\right)=\left(1 / \mathrm{M}_{1}\right) \,\left[\mu_{1}^{*}(\ell)-\mathrm{v} \, \phi \, \mathrm{R} \, \mathrm{T} \, \mathrm{M}_{1} \, \mathrm{m}_{\mathrm{j}}\right] \\
&+\mathrm{m}_{\mathrm{j}} \,\left[\mu_{\mathrm{j}}^{0}(\mathrm{aq})+\mathrm{v} \, \mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{Q} \, \mathrm{m}_{\mathrm{j}} \, \gamma_{\pm} / \mathrm{m}^{0}\right)\right]
\end{aligned}\]

As for solutions containing neutral solutes we cannot put a number value to \(\mathrm{G}\left(\mathrm{aq} ; \mathrm{w}_{1}=1 \mathrm{~kg}\right)\). If the properties of this salt solution are in fact ideal (in a thermodynamic sense) then \(\mathrm{G}\left(\mathrm{aq} ; \mathrm{w}_{1}=1 \mathrm{~kg}: \mathrm{id}\right)\) is given by equation (d).

\[\begin{aligned}
\mathrm{G}\left(\mathrm{aq} ; \mathrm{id} ; \mathrm{w}_{1}=1 \mathrm{~kg}\right)=&\left(1 / \mathrm{M}_{1}\right) \,\left[\mu_{1}^{*}(\ell)-\mathrm{v} \, \mathrm{R} \, \mathrm{T} \, \mathrm{M}_{\mathrm{l}} \, \mathrm{m}_{\mathrm{j}}\right] \\
&+\mathrm{m}_{\mathrm{j}} \,\left[\mu_{\mathrm{j}}^{0}(\mathrm{aq})+\mathrm{v} \, \mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{Q} \, \mathrm{m}_{\mathrm{j}} \, \gamma_{\pm} / \mathrm{m}^{0}\right)\right]
\end{aligned}\]

Hence in the case where \(j = \mathrm{~NaCl}\), \(ν = 2\) and \(\mathrm{Q} = 1\). In the next stage we use differences between \(\mathrm{G}\left(\mathrm{aq} ; \mathrm{w}_{1}=1 \mathrm{~kg}\right)\) and \(\mathrm{G}\left(\mathrm{aq} ; \mathrm{id} ; \mathrm{w}_{1}=1 \mathrm{~kg}\right)\) to define excess Gibbs energies for a solution prepared using \(1 \mathrm{~kg}\) of water. Then

\[\mathrm{G}^{\mathrm{E}}=\mathrm{G}\left(\mathrm{aq} ; \mathrm{w}_{1}=1 \mathrm{~kg}\right)-\mathrm{G}\left(\mathrm{aq} ; \mathrm{id} ; \mathrm{w}_{1}=1 \mathrm{~kg}\right)\]

For salt \(j\),

\[\mathrm{G}^{\mathrm{E}}=\mathrm{V} \, \mathrm{m}_{\mathrm{j}} \, \mathrm{R} \, \mathrm{T} \,\left[1-\phi+\ln \left(\gamma_{\pm}\right)\right]\]

According to the Gibbs-Duhem for a solution at constant temperature and constant pressure,

\[\mathrm{n}_{1} \, \mathrm{d} \mu_{1}(\mathrm{aq})+\mathrm{n}_{\mathrm{j}} \, \mathrm{d} \mu_{\mathrm{j}}(\mathrm{aq})=0\]

Hence for salt \(j\),

\[\begin{aligned}
&\left(1 / \mathrm{M}_{1}\right) \, \mathrm{d}\left[\mu_{1}^{*}(\ell)-\mathrm{v} \, \phi \, \mathrm{R} \, \mathrm{T} \, \mathrm{M}_{1} \, \mathrm{m}_{\mathrm{j}}\right] \\
&\quad+\mathrm{m}_{\mathrm{j}} \, \mathrm{d}\left[\mu_{\mathrm{j}}^{0}(\mathrm{aq})+\mathrm{v} \, \mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{Q} \, \mathrm{m}_{\mathrm{j}} \, \gamma_{\pm} / \mathrm{m}^{0}\right)\right]=0
\end{aligned}\]

We are concerned with the dependence of chemical potential on the molality of salt. Thus the amount of solvent and, for the salt, both \(\mathrm{Q}\) and \(v\) are fixed. Hence

\[\mathrm{d}\left[-\phi \, \mathrm{m}_{\mathrm{j}}\right]+\mathrm{m}_{\mathrm{j}} \, \mathrm{d}\left[\ln \left(\mathrm{m}_{\mathrm{j}} \, \gamma_{\pm} / \mathrm{m}^{0}\right)\right]=0\]

Hence,

\[-\phi \, d m_{j}-m_{j} \, d \phi+m_{j} \, d m_{j} / m_{j}+m_{j} \, d \ln \left(\gamma_{t}\right)=0\]

Then,

\[(\phi-1) \, d m_{j}+m_{j} \, d \phi=m_{j} \, d \ln \left(\gamma_{\pm}\right)\]

Hence we obtain an equation for \(\ln \left(\gamma_{\pm}\right)\) in terms of the dependence of \((\phi - 1)\) on molality bearing in mind that \(\ln \left(\gamma_{\pm}\right)\) equals zero and \(\phi\) equals 1 at ‘\(\mathrm{m}_{j} = 0\)’.

\[\ln \left(\gamma_{\pm}\right)=(\phi-1)+\int_{0}^{m_{j}}(\phi-1) \, d \ln \left(m_{j}\right)\]

From equation (f), the dependence of \(\mathrm{G}^{\mathrm{E}}\) on \(\mathrm{m}_{j}\) is given by equation (m).

\[\begin{aligned}
&(1 / \mathrm{V} \, \mathrm{R} \, \mathrm{T}) \, \mathrm{dG}^{\mathrm{E}} / \mathrm{dm}_{\mathrm{j}} \\
&=\left[1-\phi+\ln \left(\gamma_{\pm}\right)\right]-\mathrm{m}_{\mathrm{j}} \,\left(\mathrm{d} \phi / \mathrm{dm}_{\mathrm{j}}\right)+\mathrm{m}_{\mathrm{j}} \, \mathrm{d} \ln \left(\gamma_{\pm}\right) / \mathrm{dm}_{\mathrm{j}}=0
\end{aligned}\]

But according to equation (i),

\[-\phi-\mathrm{m}_{\mathrm{j}} \, \mathrm{d} \phi / \mathrm{dm}_{\mathrm{j}}+1+\mathrm{m}_{\mathrm{j}} \, \mathrm{d} \ln \left(\gamma_{\pm}\right) / \mathrm{dm}_{\mathrm{j}}=0\]

Hence[4],

\[\ln \left(\gamma_{\pm}\right)=(1 / \mathrm{V} \, \mathrm{R} \, \mathrm{T}) \, \mathrm{dG}^{\mathrm{E}} / \mathrm{dm}_{\mathrm{j}}\]

Footnotes

[1] Compilation of Data for polyvalent electrolytes; R. N. Goldberg, B. R. Staples, R. L. Nuttall and R. Arbuckle, NBS Special Publication 485, 1977.

[2] Thermal Properties of Aqueous Univalent –Univalent Electrolytes, V.B.Parker, NBS, 2, 1965.

[3] J.-L. Fortier and J. E. Desnoyers, J. Solution Chem.,1976,5,297.

[4]

\[\ln \left(\gamma_{\pm}\right)=\left[\frac{1}{[1] \,\left[\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right] \,[\mathrm{K}]}\right] \,\left[\frac{\mathrm{J} \mathrm{kg}^{-1}}{\mathrm{~mol} \mathrm{~kg}^{-1}}\right]=[1]\]