# 12.5 Solid-Liquid Equilibria

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How does the freezing point of the liquid mixture vary with composition? We divide both sides of Eq. 12.5.12 by $$T$$ and take differentials: \begin{gather} \s{ -a\dif(\mu\A/T)-b\dif(\mu\B/T)+\dif[\mu\solid /T] = 0 } \tag{12.5.13} \cond{(phase equilibrium)} \end{gather} The pressure is constant. Then $$\mu\A/T$$ and $$\mu\B/T$$ are functions of $$T$$ and $$x\A$$, and $$\mu\solid /T$$ is a function only of $$T$$. We find expressions for the total differentials of these quantities at constant $$p$$ with the help of Eq. 12.1.3: $$\dif(\mu\A/T) = -\frac{H\A}{T^2}\dif T + \frac{1}{T}\Pd{\mu\A}{x\A}{T,p} \dx\A \tag{12.5.14}$$ $$\dif(\mu\B/T) = -\frac{H\B}{T^2}\dif T + \frac{1}{T}\Pd{\mu\B}{x\A}{T,p}\dx\A \tag{12.5.15}$$ $$\dif[\mu\solid /T] = -\frac{H\m\solid }{T^2}\dif T \tag{12.5.16}$$ When we substitute these expressions in Eq. 12.5.13 and solve for $$\dif T/\dx\A$$, setting $$T$$ equal to $$T\subs{f}$$, we obtain $$\frac{\dif T\subs{f}}{\dx\A} = \frac{T\subs{f}}{aH\A+bH\B-H\m\solid } \left[ a\Pd{\mu\A}{x\A}{T,p}+b\Pd{\mu\B}{x\A}{T,p} \right] \tag{12.5.17}$$

The quantity $$aH\A+bH\B-H\m\solid$$ in the denominator on the right side of Eq. 12.5.17 is $$\Delsub{sol}H$$, the molar differential enthalpy of solution of the solid compound in the liquid mixture. The two partial derivatives on the right side are related through the Gibbs–Duhem equation $$x\A\dif\mu\A+x\B\dif\mu\B=0$$ (Eq. 9.2.27), which applies to changes at constant $$T$$ and $$p$$. We rearrange the Gibbs–Duhem equation to $$\dif\mu\B=-(x\A/x\B)\dif\mu\A$$ and divide by $$\dx\A$$: $$\Pd{\mu\B}{x\A}{T,p} = -\frac{x\A}{x\B} \Pd{\mu\A}{x\A}{T,p} \tag{12.5.18}$$ Making this substitution in Eq. 12.5.17, we obtain the equation $$\frac{\dif T\subs{f}}{\dx\A} = \frac{x\A T\subs{f}}{\Delsub{sol}H} \left( \frac{a}{x\A}-\frac{b}{x\B} \right)\Pd{\mu\A}{x\A}{T,p} \tag{12.5.19}$$ which can also be written in the slightly rearranged form $$\frac{\dif T\subs{f}}{\dx\A} = \frac{b T\subs{f}}{\Delsub{sol}H} \left( \frac{a}{b}-\frac{x\A}{1-x\A} \right)\Pd{\mu\A}{x\A}{T,p} \tag{12.5.20}$$

Suppose we heat a sample of the solid compound to its melting point to form a liquid mixture of the same composition as the solid. The molar enthalpy change of the fusion process is the molar enthalpy of fusion of the solid compound, $$\Delsub{fus}H$$, a positive quantity. When the liquid has the same composition as the solid, the dissolution and fusion processes are identical; under these conditions, $$\Delsub{sol}H$$ is equal to $$\Delsub{fus}H$$ and is positive.

Equation 12.5.20 shows that the slope of the freezing-point curve, $$T\subs{f}$$ versus $$x\A$$, is zero when $$x\A/(1-x\A)$$ is equal to $$a/b$$, or $$x\A=a/(a+b)$$; that is, when the liquid and solid have the same composition. Because $$\pd{\mu\A}{x\A}{T,p}$$ is positive, and $$\Delsub{sol}H$$ at this composition is also positive, we see from the equation that the slope decreases as $$x\A$$ increases. Thus, the freezing-point curve has a maximum at the mixture composition that is the same as the composition of the solid compound. This conclusion applies when both components of the liquid mixture are nonelectrolytes, and also when one component is an electrolyte that dissociates into ions.

Now let us assume the liquid mixture is an ideal liquid mixture of nonelectrolytes in which $$\mu\A$$ obeys Raoult’s law for fugacity, $$\mu\A=\mu\A^*+RT\ln x\A$$. The partial derivative $$\pd{\mu\A}{x\A}{T,p}$$ then equals $$RT/x\A$$, and Eq. 12.5.19 becomes $$\frac{\dif T\subs{f}}{\dx\A} = \frac{RT\subs{f}^2}{\Delsub{sol}H} \left( \frac{a}{x\A}-\frac{b}{x\B} \right) \tag{12.5.21}$$ By making the approximations that $$\Delsub{sol}H$$ is independent of $$T$$ and $$x\A$$, and is equal to $$\Delsub{fus}H$$, we can separate the variables and integrate as follows: $$\int_{T'\subs{f}}^{T''\subs{f}}\frac{\dif T\subs{f}}{T\subs{f}^2} = \frac{R}{\Delsub{fus}H}\left( \int_{x'\A}^{x''\A}\frac{a}{x\A}\dx\A +\int_{x'\B}^{x''\B}\frac{b}{x\B}\dx\B \right) \tag{12.5.22}$$ (The second integral on the right side comes from changing $$\dx\A$$ to $$-\dx\B$$.) The result of the integration is \begin{gather} \s{ \frac{1}{T'\subs{f}} = \frac{1}{T''\subs{f}} + \frac{R}{\Delsub{fus}H} \left( a\ln\frac{x''\A}{x'\A}+b\ln\frac{x''\B}{x'\B} \right) } \tag{12.5.23} \cond{(ideal liquid mixture in} \nextcond{equilibrium with solid} \nextcond{compound, $$\Delsub{sol}H{=}\Delsub{fus}H$$)} \end{gather}

Figure 12.7 Solid curve: freezing-point curve of a liquid melt of Zn and Mg that solidifies to the solid compound Zn$$_2$$Mg (Rodney P. Elliot, Constitution of Binary Alloys, First Supplement, McGraw-Hill, New York, 1965, p. 603). The curve maximum (open circle) is at the compound composition $$x''\subs{Zn}=2/3$$ and the solid compound melting point $$T''\subs{f}=861\K$$. Dashed curve: calculated using Eq. 12.5.23 with $$\Delsub{fus}H = 15.8\units{kJ mol\(^{-1}$$}\).

Let $$T'\subs{f}$$ be the freezing point of a liquid mixture of composition $$x'\A$$ and $$x'\B=1-x'\A$$, and let $$T''\subs{f}$$ be the melting point of the solid compound of composition $$x''\A=a/(a+b)$$ and $$x''\B=b/(a+b)$$. Figure 12.7 shows an example of a molten metal mixture that solidifies to an alloy of fixed composition. The freezing-point curve of this system is closely approximated by Eq. 12.5.23.

## 12.5.5 Solubility of a solid electrolyte

Consider an equilibrium between a crystalline salt (or other kind of ionic solid) and a solution containing the solvated ions: $\tx{M$$_{\nu_+}$$X$$_{\nu_-}$$(s)} \arrows \nu_+\tx{M$$^{z_+}$$(aq)} + \nu_-\tx{X$$^{z_-}$$(aq)}$ Here $$\nu_+$$ and $$\nu_-$$ are the numbers of cations and anions in the formula unit of the salt, and $$z_+$$ and $$z_-$$ are the charge numbers of these ions. The solution in equilibrium with the solid salt is a saturated solution. The thermodynamic equilibrium constant for this kind of equilibrium is called a solubility product, $$K\subs{s}$$.

We can readily derive a relation between $$K\subs{s}$$ and the molalities of the ions in the saturated solution by treating the dissolved salt as a single solute substance, B. We write the equilibrium in the form B$$^*$$(s)$$\arrows$$B(sln), and write the expression for the solubility product as a proper quotient of activities: $$K\subs{s} = \frac{a\mbB}{a\B^*} \tag{12.5.24}$$ From Eq. 10.3.16, we have $$a\mbB= \G\mbB \g_{\pm}^\nu(m_+/m\st)^{\nu_+}(m_-/m\st)^{\nu_-}$$. This expression is valid whether or not the ions M$$^{z_+}$$ and X$$^{z_-}$$ are present in solution in the same ratio as in the solid salt. When we replace $$a\mbB$$ with this expression, and replace $$a\B^*$$ with $$\G\B^*$$ (Table 9.5), we obtain $$K\subs{s} = \left(\frac{\G\mbB}{\G\B^*}\right) \g_{\pm}^\nu \left( \frac{m_+}{m\st} \right)^{\nu_+} \left( \frac{m_-}{m\st} \right)^{\nu_-} \tag{12.5.25}$$ where $$\nu=\nu_+ + \nu_-$$ is the total number of ions per formula unit. $$\g_{\pm}$$ is the mean ionic activity coefficient of the dissolved salt in the saturated solution, and the molalities $$m_+$$ and $$m_-$$ refer to the ions M$$^{z_+}$$ and X$$^{z_-}$$ in this solution.

The first factor on the right side of Eq. 12.5.25, the proper quotient of pressure factors for the reaction B$$^*$$(s)$$\ra$$B(sln), will be denoted $$\G\subs{r}$$ (the subscript “r” stands for reaction). The value of $$\G\subs{r}$$ is exactly $$1$$ if the system is at the standard pressure, and is otherwise approximately $$1$$ unless the pressure is very high.

If the aqueous solution is produced by allowing the salt to dissolve in pure water, or in a solution of a second solute containing no ions in common with the salt, then the ion molalities in the saturated solution are $$m_+=\nu_+m\B$$ and $$m_-=\nu_-m\B$$ where $$m\B$$ is the solubility of the salt expressed as a molality. Under these conditions, Eq. 12.5.25 becomes \begin{gather} \s{ K\subs{s} = \G\subs{r} \g_{\pm}^{\nu} \left(\nu_+^{\nu_+}\nu_-^{\nu_-}\right) \left( \frac{m\B}{m\st} \right)^{\nu} } \tag{12.5.26} \cond{(no common ion)} \end{gather} We could also have obtained this equation by using the expression of Eq. 10.3.10 for $$a\mbB$$.

If the ionic strength of the saturated salt solution is sufficiently low (i.e., the solubility is sufficiently low), it may be practical to evaluate the solubility product with Eq. 12.5.26 and an estimate of $$\g_{\pm}$$ from the Debye–Hückel limiting law (see Prob. 12.19). The most accurate method of measuring a solubility product, however, is through the standard cell potential of an appropriate galvanic cell (Sec. 14.3.3).

Since $$K\subs{s}$$ is a thermodynamic equilibrium constant that depends only on $$T$$, and $$\G\subs{r}$$ depends only on $$T$$ and $$p$$, Eq. 12.5.26 shows that any change in the solution composition at constant $$T$$ and $$p$$ that decreases $$\g_{\pm}$$ must increase the solubility. For example, the solubility of a sparingly-soluble salt increases when a second salt, lacking a common ion, is dissolved in the solution; this is a salting-in effect.

Equation 12.5.25 is a general equation that applies even if the solution saturated with one salt contains a second salt with a common ion. For instance, consider the sparingly-soluble salt M$$_{\nu_+}$$X$$_{\nu_-}$$ in transfer equilibrium with a solution containing the more soluble salt M$$_{\nu'_+}$$Y$$_{\nu'_-}$$ at molality $$m\C$$. The common ion in this example is the cation M$$^{z_+}$$. The expression for the solubility product is now \begin{gather} \s{ K\subs{s} = \G\subs{r} \g_{\pm}^{\nu} (\nu_+m\B+\nu'_+m\C)^{\nu_+}(\nu_-m\B)^{\nu_-}/(m\st)^{\nu} } \tag{12.5.27} \cond{(common cation)} \end{gather} where $$m\B$$ again is the solubility of the sparingly-soluble salt, and $$m\C$$ is the molality of the second salt. $$K\subs{s}$$ and $$\G\subs{r}$$ are constant if $$T$$ and $$p$$ do not change, so any increase in $$m\C$$ at constant $$T$$ and $$p$$ must cause a decrease in the solubility $$m\B$$. This is called the common ion effect.

From the measured solubility of a salt in pure solvent, or in an electrolyte solution with a common cation, and a known value of $$K\subs{s}$$, we can evaluate the mean ionic activity coefficient $$\g_{\pm}$$ through Eq. 12.5.26 or 12.5.27. This procedure has the disadvantage of being limited to the value of $$m\B$$ existing in the saturated solution.

We find the temperature dependence of $$K\subs{s}$$ by applying Eq. 12.1.12: $$\frac{\dif\ln K\subs{s}}{\dif T} = \frac{\Delsub{sol,B}H\st}{RT^2} \tag{12.5.28}$$ At the standard pressure, $$\Delsub{sol,B}H\st$$ is the same as the molar enthalpy of solution at infinite dilution, $$\Delsub{sol,B}H^{\infty}$$.