1.11.4: Gibbs-Duhem Equation- Aqueous Salt Solutions- Salt and Solvent- Debye-Huckel Limiting Law

A given aqueous salt solution at temperature \(\mathrm{T}\) and pressure \(\mathrm{p}\) (which is close to the standard pressure) is prepared using water (\(1 \mathrm{~kg}\)) and \(\mathrm{m}_{\mathrm{j}}\) moles of a 1:1 sa }\) and the thermodynamic properties of the solution are ideal.

\[\mu_{\mathrm{j}}(\mathrm{aq})=\mu_{\mathrm{j}}^{0}(\mathrm{aq})+2 \, \mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{m}_{\mathrm{j}} \, \gamma_{\pm} / \mathrm{m}^{0}\right) \nonumber \]

At all \(\mathrm{T}\) and \(\mathrm{p}\),

\[\operatorname{limit}\left(\mathrm{m}_{\mathrm{j}} \rightarrow 0\right) \gamma_{\pm}=1 \nonumber \]

The chemical potential of solvent, water \(\mu_{1}(\mathrm{aq})\) is related to salt molality \(\mathrm{m}_{\mathrm{j}}\) by equation (c) where \(\mu_{1}^{*}(\ell)\) is the chemical potential of water(\(\ell\)) at the same \(\mathrm{T}\) and \(\mathrm{p}\) and \(\phi\) is the molal osmotic coefficient.

\[\mu_{1}(\mathrm{aq})=\mu_{1}^{*}(\ell)-2 \, \phi \, \mathrm{R} \, \mathrm{T} \, \mathrm{M}_{1} \, \mathrm{m}_{\mathrm{j}} \nonumber \]

At all \(\mathrm{T}\) and \(\mathrm{p}\),

\[\operatorname{limit}\left(\mathrm{m}_{\mathrm{j}} \rightarrow 0\right) \phi=1\)

For a solution at fixed \(\mathrm{T}\) and \(\mathrm{p}\), the Gibbs-Duhem equation relates \(\mu_{1}(\mathrm{aq})\) and \(\mu_{j}(\mathrm{aq})\) using equation (e).

\[\left(1 / M_{1}\right) \, d \mu_{1}(a q)+m_{j} \, d \mu_{j}(a q)=0 \nonumber \]

Therefore

\[\begin{aligned}
&\left(1 / \mathrm{M}_{1}\right) \, \mathrm{d}\left[\mu_{1}^{*}(\ell)-2 \, \phi \, \mathrm{R} \, \mathrm{T} \, \mathrm{M}_{1} \, \mathrm{m}_{\mathrm{j}}\right] \\
&+\mathrm{m}_{\mathrm{j}} \, \mathrm{d}\left[\mu_{\mathrm{j}}^{0}(\mathrm{aq})+2 \, \mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{m}_{\mathrm{j}} \, \gamma_{\pm} / \mathrm{m}^{0}\right)\right]=0
\end{aligned} \nonumber \]

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

At ‘\(\mathrm{m}_{\mathrm{j}} =0\)’, \(\phi\) is unity. Therefore integration [1] of equation (g) yields equation (h)

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

We have an equation for \(\phi\) in terms of \(\mathrm{m}_{\mathrm{j}}\) and \(\gamma_{\pm}\). Equation (h) signals the limit of the thermodynamic analysis. To make progress we need an equation for \(\gamma_{\pm}\) in terms of \(\mathrm{m}_{\mathrm{j}}\). The Debye-Huckel Limiting Law provides such an equation having the form shown in equation (i) where \(\mathrm{S}_{\gamma}\) is positive and a function of temperature, pressure and relative permittivity of the solvent. Thus

\[\ln \left(\gamma_{\pm}\right)=-S_{\gamma} \,\left(m_{j} / m^{0}\right)^{1 / 2} \nonumber \]

From equation (h),

\[\phi=1-\frac{S_{\gamma}}{\left(m^{0}\right)^{1 / 2} \, m_{j}} \, \int_{0}^{m(j)} m_{j} \, d\left(m_{j}\right)^{1 / 2} \nonumber \]

Hence [2],

\[\phi=1-\left(\mathrm{S}_{\gamma} / 3\right) \,\left(\mathrm{m}_{\mathrm{j}} / \mathrm{m}^{0}\right)^{1 / 2}\)

We examine the impact of equations (i) and (k) on the chemical potentials of solute and solvent. From equation (c),

\[\mu_{1}(\mathrm{aq})-\mu_{1}^{*}(\ell)=-2 \, \mathrm{R} \, \mathrm{T} \, \mathrm{M}_{1} \, \mathrm{m}_{\mathrm{j}} \,\left[1-\left(\mathrm{S}_{\gamma} / 3\right) \,\left(\mathrm{m}_{\mathrm{j}} / \mathrm{m}^{0}\right)^{1 / 2}\right] \nonumber \]

For a salt solution having thermodynamic properties which are ideal,

\[\mu_{1}(\mathrm{aq} ; \mathrm{id})-\mu_{1}^{*}(\ell)=-2 \, \mathrm{R} \, \mathrm{T} \, \mathrm{M}_{1} \, \mathrm{m}_{\mathrm{j}} \nonumber \]

Then,

\[\mu_{1}(\mathrm{aq})-\mu_{1}(\mathrm{aq} ; \mathrm{id})=2 \, \mathrm{R} \, \mathrm{T} \, \mathrm{M}_{1} \,\left(\mathrm{S}_{\gamma} / 3\right) \,\left(\mathrm{m}^{0}\right)^{-1 / 2} \,\left(\mathrm{m}_{\mathrm{j}}\right)^{3 / 2} \nonumber \]

But \(\mathrm{S}_{\gamma}\) is positive. Hence \(\left[\mu_{1}(\mathrm{aq})-\mu_{1}(\mathrm{aq} ; \mathrm{id})\right]\) is positive so that in terms of the DHLL the chemical potential of water is raised above that in the corresponding solution having ideal thermodynamic properties.

For the solute, equation (a) requires that \(\mu_{j}(\mathrm{aq})\) is given by equation (o).

\[\mu_{\mathrm{j}}(\mathrm{aq})=\mu_{\mathrm{j}}^{0}(\mathrm{aq})+2 \, \mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{m}_{\mathrm{j}} / \mathrm{m}^{0}\right)+2 \, \mathrm{R} \, \mathrm{T} \, \ln \left(\gamma_{\pm}\right) \nonumber \]

Or,

\[\mu_{\mathrm{j}}(\mathrm{aq} ; \mathrm{id})=\mu_{\mathrm{j}}^{0}(\mathrm{aq})+2 \, \mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{m}_{\mathrm{j}} / \mathrm{m}^{0}\right) \nonumber \]

Then using equation (i),

\[\mu_{\mathrm{j}}(\mathrm{aq})-\mu_{\mathrm{j}}(\mathrm{aq} ; \mathrm{id})=-2 \, \mathrm{R} \, \mathrm{T} \, \mathrm{S}_{\mathrm{\gamma}} \,\left(\mathrm{m}_{\mathrm{j}} / \mathrm{m}^{0}\right)^{1 / 2} \nonumber \]

According therefore to the DHLL, salt j in a real solution is stabilised relative to that in an ideal solution. In other words according to the DHLL the salt is stabilised whereas the solvent is destabilised, the impact of ion-ion interactions on the Gibbs energy of a solution is moderated.

Footnotes

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

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

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

Or, \(\phi=1+\frac{1}{m_{j}} \, \int_{0}^{m(j)} m_{j} \, d \ln \left(\gamma_{\pm}\right)\)

[2] Put \(\left(m_{j}\right)^{1 / 2}=x\); \(\int_{0}^{m(j)} x^{2} \, d x=x^{3} / 3=\left(m_{j}\right)^{3 / 2} / 3\)