Skip to main content
Chemistry LibreTexts

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

  • Page ID
  • 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


    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,


    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

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


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

    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.

    • Was this article helpful?