# 16.13: Colligative Properties - Solubility of a Solute in an Ideal Solution


Although the result has few practical applications, we can also use these ideas to calculate the solubility of a solid solute in an ideal solution. The arguments are similar to those we used to estimate the freezing-point depression of a solution. The freezing point of a solution is the temperature at which the solution is in equilibrium with its pure-solid solvent. The solubility of a solute is the mole fraction of the solute in a solution that is at equilibrium with pure-solid solute. In this analysis, we assume that the solid phase is pure solute. Our analysis does not apply to a solid solution in equilibrium with a liquid solution. The properties of the solvent have no role in our description of the solid–ideal-solution equilibrium state. Consequently, our analysis produces a model in which the solubility of an ideal solute depends only on the properties of the solute; for a given solute, the ideal-solution solubility is the same in every solvent.

We specify the composition of the solution by the mole fractions of $$A$$ and $$B$$, again letting the solute be compound $$A$$. When we consider freezing-point depression, an $$A$$–$$B$$ solution of specified composition is in equilibrium with pure solid $$B$$, and we want to find the equilibrium temperature. When we consider solute solubility, the $$A$$–$$B$$ solution is in equilibrium with pure solid $$A$$ at a specified pressure, $$P^{\#}$$, and temperature, $$T$$; we want to find the equilibrium composition. Since pure solid $$A$$ is present, the temperature must be less than the melting point of pure $$A$$. We let the melting point of the pure solute be $$T_{FA}$$, at the specified system pressure.

The activity of pure solid $$A$$ and the system pressure are both constant; we have $$d{ \ln {\tilde{a}}_{A,\mathrm{solid}}=0\ }$$, $$dP=0$$, and

$d{\mu }_{A,\mathrm{soli}\mathrm{d}}=-{\overline{S}}^{\textrm{⦁}}_{A,\mathrm{solid}}dT$

In the saturated ideal solution in equilibrium with this solid, we have $${\tilde{a}}_{A,\mathrm{solution}}=y_A$$, $$dP=0$$, and

$d{\mu }_{A,\mathrm{sol}\mathrm{ution}}=-{\overline{S}}_{A,\mathrm{solution}}dT+RT\left(d{ \ln y_A\ }\right)$

The relationship $$d{\mu }_{A,\mathrm{sol}\mathrm{ution}}=d{\mu }_{A,\mathrm{solid}}$$ becomes

$-{\overline{S}}_{A,\mathrm{solution}}dT+RT\left(d{ \ln y_A\ }\right)=-{\overline{S}}^{\textrm{⦁}}_{A,\mathrm{solid}}dT$

and $d{ \ln y_A\ }=\left(\frac{\overline{S}_{A,\mathrm{solution}}-{\overline{S}}^{\textrm{⦁}}_{A,\mathrm{solid}}}{RT}\right)dT$

Now, $${\overline{S}}_{A,\mathrm{solution}}-{\overline{S}}^{\textrm{⦁}}_{A,\mathrm{solid}}$$ is the entropy change for the reversible (equilibrium) process in which one mole of pure solid $$A$$ dissolves in a very large volume of a saturated solution; the mole fraction of $$A$$ in this solution is constant at $$y_A$$. During this process, the pressure and temperature are constant at $$P^{\#}$$ and $$T$$. Letting the heat absorbed by the system during this process be $$q^{rev}_{P^{\#}}$$, we have

${\left({\overline{S}}_{A,\mathrm{solution}}-{\overline{S}}^{\textrm{⦁}}_{A,\mathrm{solid}}\right)}_{P^{\#},T}={q^{rev}_{P^{\#}}}/{T}$

The heat absorbed is also expressible as the difference between the partial molar enthalpy of $$A$$ in the solution and that of the pure solid; that is,

$q^{rev}_{P^{\#}}={\left({\overline{H}}_{A,\mathrm{solution}}-{\overline{H}}^{\textrm{⦁}}_{A,\mathrm{solid}}\right)}_{P^{\#},T}$

One of the properties of an ideal solution is that the enthalpy of mixing is zero. Thus, the partial molar enthalpy of $$A$$ in an ideal solution is independent of $$y_A$$, so that the partial molar enthalpy of $$A$$ in an ideal solution is the same as the partial molar enthalpy of pure liquid $$A$$; that is, $${\overline{H}}_{A,\mathrm{solution}}={\overline{H}}^{\textrm{⦁}}_{A,\mathrm{liquid}}$$, and

$q^{rev}_{P^{\#}}={\left({\overline{H}}_{A,\mathrm{solution}}-{\overline{H}}^{\textrm{⦁}}_{A,\mathrm{solid}}\right)}_{P^{\#},T}={\left({\overline{H}}^{\textrm{⦁}}_{A,\mathrm{liquid}}-{\overline{H}}^{\textrm{⦁}}_{A,\mathrm{solid}}\right)}_{P^{\#},T}={\left(\Delta_{\mathrm{fus}}H_A\right)}_{P^{\#},T}\approx {\left(\Delta_{\mathrm{fus}}H_A\right)}_{P^{\#},T_{FA}}$

Then,

$\left(\overline{S}_{A,\mathrm{solution}}-\overline{S}^{\textrm{⦁}}_{A,\mathrm{solid}}\right)_{P^{\#},T} \approx \frac{\left(\Delta_{\mathrm{fus}}H_A\right)_{P^{\#},T_{FA}}}{T}$

Dropping the subscript information and replacing the approximate equality, we have

$d{ \ln y_A\ }=\left(\frac{\Delta_{\mathrm{fus}}H_A}{RT^2}\right)dT$

At $$P^{\#}$$ and $$T_{FA}$$, $$\Delta_{\mathrm{fus}}H_A$$ is a property of pure $$A$$ and is independent of the solution composition. When the pure solid solute melts at $$T_{FA}$$, the solute mole fraction is unity in the liquid phase with which it is in equilibrium: At $$T_{FA}$$, $$y_A=1$$. At temperature $$T$$, $$y_A$$ is the solute mole fraction in the liquid-phase solution that is at equilibrium with the pure-solid solute. Integrating between the limits $$\left(1,T_{FA}\right)$$ and $$\left(y_A,T\right)$$, we have

$\int^{y_A}_1{d{ \ln y_A\ }}=\frac{\Delta_{\mathrm{fus}}H_A}{R}\int^T_{T_{FA}}{\frac{dT}{T^2}}$ and ${ \ln y_A\ }=\frac{-\Delta_{\mathrm{fus}}H_A}{R}\left(\frac{1}{T}-\frac{1}{T_{FA}}\right)$

For a given solute, $$\Delta_{\mathrm{fus}}H_A$$ and $$T_{FA}$$ are fixed and are independent of the characteristics of the solvent. The mole fraction of $$A$$ in the saturated solution depends only on temperature. Since $$\Delta_{\mathrm{fus}}H_A>0$$ and $$T<t_{fa}$$>, we find $${ \ln y_A\ }<0$$. Therefore, we find that $$y_A<1$$, as it must be. However, $$y_A$$ increases, with T, implying that the solubility of a solid increases as the temperature increases, as we usually observe.

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