In the previous chapter, we learned that systems tend to move toward lower Gibbs functions and that the chemical potential determines the direction of spontaneous change.
These ideas provide a powerful way to understand solubility.
Consider a solid solute placed in a liquid solvent. If dissolving the solute lowers the Gibbs function of the system, the solute will dissolve. If remaining as a solid produces the lower Gibbs function, the solute will remain undissolved.
The maximum solubility of a solute therefore corresponds to an equilibrium between the dissolved and undissolved forms of the solute.
Suppose a solid solute \(B\) is in equilibrium with a saturated solution.
At equilibrium, there can be no net transfer of solute between the solid and the solution.
Consequently, the chemical potentials of the two forms of the solute must be equal:
\[ \boxed{ \mu_B^{(s)} = \mu_B^{(soln)} } \]
This equation is the thermodynamic criterion for saturation.
If \[ \mu_B^{(s)} > \mu_B^{(soln)} \] the solute dissolves.
If \[ \mu_B^{(s)} < \mu_B^{(soln)} \] the solute precipitates from solution.
For an ideal solution, the chemical potential of the dissolved solute is
\[ \mu_B^{(soln)} = \mu_B^\circ + RT\ln x_B \]
where \(x_B\) is the mole fraction of the solute.
Applying the saturation condition gives
\[ \mu_B^{(s)} = \mu_B^\circ + RT\ln x_B \]
Rearranging,
\[ \boxed{ \ln x_B = \frac{ \mu_B^{(s)} - \mu_B^\circ } {RT} } \]
The difference between the chemical potentials of the solid and liquid forms is simply the Gibbs function of fusion:
\[ \Delta G_{\mathrm{fus}} = \mu_B^\circ - \mu_B^{(s)} \]
Therefore,
\[ \boxed{ \ln x_B = -\frac{\Delta G_{\mathrm{fus}}}{RT} } \]
The Gibbs function of fusion is temperature dependent.
Using the Gibbs-Helmholtz equation, it can be shown that
\[ \frac{d(\ln x_B)}{dT} = \frac{\Delta H_{\mathrm{fus}}} {RT^2} \]
Assuming that the enthalpy of fusion is approximately constant over the temperature range of interest, integration yields
\[ \boxed{ \ln x_B = -\frac{\Delta H_{\mathrm{fus}}}{R} \left( \frac{1}{T} - \frac{1}{T_f} \right) } \]
where \(T_f\) is the melting point of the pure solute.
This expression shows that solubility depends primarily on two properties:
Substances with low melting points or small enthalpies of fusion generally dissolve more readily than substances with high melting points and large enthalpies of fusion.
The equation also predicts that solubility usually increases with temperature, a result that is familiar from everyday experience.
Dissolution is fundamentally a competition between two phases of the same substance.
The solid phase tends to persist because of the strong interactions holding the crystal together. The solution phase tends to be favored because mixing lowers the chemical potential of the dissolved solute.
Saturation occurs when these two tendencies exactly balance.
Big picture: Solubility is determined by the condition that the chemical potential of the solute must be the same in the solid and solution phases. This equilibrium condition provides a thermodynamic explanation for why different substances exhibit different solubilities and why solubility often depends strongly on temperature.
Naphthalene has a melting point of
\[ T_f = 353\ \mathrm{K} \]
and an enthalpy of fusion of
\[ \Delta H_{\mathrm{fus}} = 19.0\ \mathrm{kJ\,mol^{-1}}. \]
Estimate the mole-fraction solubility of naphthalene in an ideal solvent at \(298\ \mathrm{K}\).
The solubility of a solid solute in an ideal solution is given by
\[ \ln x_B = -\frac{\Delta H_{\mathrm{fus}}}{R} \left( \frac{1}{T} - \frac{1}{T_f} \right) \]
Substituting the values:
\[ \ln x_B = -\frac {19000\ \mathrm{J\,mol^{-1}}} {8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}} \left( \frac{1}{298\ \mathrm{K}} - \frac{1}{353\ \mathrm{K}} \right) \]
This gives:
\[ \ln x_B = -\frac {19000\ \mathrm{J\,mol^{-1}}} {8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}} \left( 3.356\times10^{-3}\ \mathrm{K^{-1}} - 2.833\times10^{-3}\ \mathrm{K^{-1}} \right) \]
\[ \ln x_B = -2285.3\ \mathrm{K} \left( 5.23\times10^{-4}\ \mathrm{K^{-1}} \right) \]
\[ \ln x_B = -1.20 \]
Exponentiating both sides:
\[ x_B = e^{-1.20} \]
\[ x_B = 0.30 \]
\[ \boxed{ x_B = 0.30 } \]
Interpretation: At \(298\ \mathrm{K}\), the saturated solution is predicted to contain approximately 30 mol% naphthalene. The solubility is less than unity because the temperature is below the melting point of the pure solute, making the solid phase thermodynamically favorable. However, the entropy gained by dissolving some of the naphthalene is sufficient to stabilize a significant concentration in solution.
| Question 1 |
|---|
| Anthracene has a melting point of \[ T_f = 489\ \mathrm{K} \] and an enthalpy of fusion of \[ \Delta H_{\mathrm{fus}} = 28.8\ \mathrm{kJ\,mol^{-1}}. \] Estimate the mole-fraction solubility of anthracene in an ideal solvent at \(298\ \mathrm{K}\). |
Show AnswerUse \[ \ln x_B = -\frac{\Delta H_{\mathrm{fus}}}{R} \left( \frac{1}{T} - \frac{1}{T_f} \right) \] \[ \ln x_B = -\frac {28800\ \mathrm{J\,mol^{-1}}} {8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}} \left( \frac{1}{298\ \mathrm{K}} - \frac{1}{489\ \mathrm{K}} \right) \] \[ \ln x_B = -4.54 \] \[ x_B = e^{-4.54} = 0.0107 \] \[ \boxed{ x_B = 0.011 } \] Anthracene is much less soluble than naphthalene because it has both a higher melting point and a larger enthalpy of fusion. |
| Question 2 |
|---|
| A solute has a melting point of \[ T_f = 350\ \mathrm{K} \] and an enthalpy of fusion of \[ \Delta H_{\mathrm{fus}} = 12.0\ \mathrm{kJ\,mol^{-1}}. \] Estimate the mole-fraction solubility of the solute in an ideal solvent at \(320\ \mathrm{K}\). |
Show AnswerUsing \[ \ln x_B = -\frac{\Delta H_{\mathrm{fus}}}{R} \left( \frac{1}{T} - \frac{1}{T_f} \right) \] gives \[ \ln x_B = -\frac {12000\ \mathrm{J\,mol^{-1}}} {8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}} \left( \frac{1}{320\ \mathrm{K}} - \frac{1}{350\ \mathrm{K}} \right) \] \[ \ln x_B = -0.387 \] \[ x_B = e^{-0.387} = 0.679 \] \[ \boxed{ x_B = 0.679 } \] Because the temperature is close to the melting point of the solute, the solution can accommodate a much larger amount of dissolved solute before saturation is reached. |
Big picture: Solubility can be understood entirely in terms of chemical potential. A solute dissolves until the chemical potential of the dissolved species becomes equal to that of the solid phase. This equality condition provides a thermodynamic explanation for why different substances exhibit different solubilities and why solubility often depends strongly on temperature.