1.15: Solution Equilibria and Colligative Properties

Statistical Mechanics of Solvation: Solvation Free Energies

We consider a solvent with coordinates \(\textbf{r}_1^\text{(a)}, \ldots, \textbf{r}_{N_a}^\text{(a)}\) in to which a solute with coordinates \(\textbf{r}_1^\text{(b)}, \ldots, \textbf{r}_{N_b}^\text{(b)}\). The total number of particles in the system is \(N = N_a + N_b\). Let \(\textbf{r}^\text{(a)}\) denote the full set of solvent coordinates, \(\textbf{r}^\text{(b)}\) denote the full set of solute coordinates, and \(\textbf{r} = (\textbf{r}^\text{(a)}, \textbf{r}^\text{(b)})\) denote the full set of coordinates in the system. The pure system of each system, before the solute is introduced into the solvent, is described by a potential energy

\[U^* (\textbf{r}) = U_a (\textbf{r}^\text{(a)}) + U_b (\textbf{r}^\text{(b)}) \label{16.1} \]

When the solute is introduced into the solvent, we change the thermodynamic state of the system, and we change the mechanical state as well. Hence, the potential energy for the solution is of the form

\[U (\textbf{r}) = U^* (\textbf{r}) + U_{ab} (\textbf{r}^\text{(a)}, \textbf{r}^\text{(b)}) \label{16.2} \]

where \(U_{ab}\) is an interaction potential between solvent and solute, which we can take to be a pair potential

\[U_{ab} (\textbf{r}^\text{(a)}, \textbf{r}^\text{(b)}) = \sum_{i=1}^{N_a} \sum_{j=1}^{N_b} u_{ab} \left( \left| \textbf{r}_1^\text{(a)} - \textbf{r}_j^\text{(b)} \right| \right) \label{16.3} \]

In order to affect the change of thermodynamic state so that we can compute the solvation free energy, we define an arbitrary path via the introduction of a coupling parameter \(\lambda\) that slowly, reversibly switches on the interaction between solvent and solute:

\[\overset{\sim}{U} (\textbf{r}; \lambda) = (1 - \lambda) U^* (\textbf{r}) + \lambda U (\textbf{r}) \label{16.4} \]

Note that \(\overset{\sim}{U} (\textbf{r};, 0) = U^* (\textbf{r})\) is the potential of the pure substances, and \(\overset{\sim}{U} (\textbf{r}, 1) = U (\textbf{r})\) is the potential energy for the solution. Here \(\lambda \in [0, 1]\) is a continuous parameter that we slowly switch from \(0\) to \(1\), thereby slowly solvating the solute in the solvent.

Let \(A (\lambda)\) denote the Helmholtz free energy for \(\overset{\sim}{U} (\textbf{r};, \lambda)\) at a particular value of \(\lambda\). The solvation Helmholtz free energy is then given by the relation

\[\Delta_\text{solv} A = A(1) - A(0) = \int_0^1 \dfrac{dA}{d \lambda} d \lambda \label{16.5} \]

Note that

\[\begin{align} A(\lambda) &= -\dfrac{1}{\beta} \: \text{ln} \: Z(\lambda) \\ Z(\lambda) &= \int d \textbf{r} \: e^{-\beta \overset{\sim}{U} (\textbf{r}; \lambda)} \\ \dfrac{dA}{d \lambda} &= -\dfrac{1}{\beta Z(\lambda)} \dfrac{dZ}{d \lambda} \\ &= \dfrac{1}{Z(\lambda)} \int d \textbf{r} \: \dfrac{ \partial \overset{\sim}{U}}{\partial \lambda} e^{-\beta \overset{\sim}{U} (\textbf{r}; \lambda)} \end{align} \nonumber \]

But

\[\dfrac{\partial \overset{\sim}{U}}{\partial \lambda} = -U^* (\textbf{r}) + U^* (\textbf{r}) + U_{ab} (\textbf{r}^\text{(a)}, \textbf{r}^\text{(b)}) = U_{ab} (\textbf{r}^\text{(a)}, \textbf{r}^\text{(b)}) \label{16.6} \]

Therefore,

\[\dfrac{dA}{d \lambda} = \dfrac{1}{Z(\lambda)} \int d \textbf{r} \: U_{ab} (\textbf{r}^\text{(a)}, \textbf{r}^\text{(b)}) e^{-\beta \overset{\sim}{U} (\textbf{r}; \lambda)} \label{16.7} \]

Now, define the radial distribution function at a given value of \(\lambda\) as

\[\overset{\sim}{g}_{ab} (r; \lambda) = \dfrac{N_a V}{\rho_a Z(\lambda)} \int_{\left| \textbf{r}_1^\text{(a)} - \textbf{r}_1^\text{(b)} \right| = r} d \textbf{r} \: e^{-\beta \overset{\sim}{U} (\textbf{r}; \lambda)} \label{16.8} \]

Then

\[\begin{align} \dfrac{dA}{d \lambda} &= \dfrac{1}{Z(\lambda)} N_a N_b \int d \textbf{r} \: u_{ab} \left( \left| \textbf{r}_1^\text{(a)} - \textbf{r}_1^\text{(b)} \right| \right) e^{-\beta \overset{\sim}{U} (\textbf{r}; \lambda)} \\ &= N_b \int d \textbf{r}_1^\text{(a)} d \textbf{r}_1^\text{(b)} u_{ab} \left( \left| \textbf{r}_1^\text{(a)} - \textbf{r}_1^\text{(b)} \right| \right) \left[ \dfrac{N_a}{Z(\lambda)} \int d \textbf{r}' \: e^{-\beta \overset{\sim}{U} (\textbf{r}; \lambda)} \right] \\ &= N_b \int d \textbf{R} \: d \textbf{r} \: u_{ab} ( \left| \textbf{r} \right| ) \left[ \dfrac{N_a}{Z(\lambda)} \int d \textbf{r}' \: e^{-\beta \overset{\sim}{U} (\textbf{r}; \lambda)} \right] \\ &= 4 \pi N_b V \int_0^\infty dr \: r^2 \: u_{ab}(r) \left[ \dfrac{N_a}{Z(\lambda)} \int d \textbf{r}' \: e^{-\beta \overset{\sim}{U} (\textbf{r}; \lambda)} \right] \\ &= 4 \pi N_b \rho_a \int_0^\infty dr \: r^2 \: u_{ab}(r) \left[ \dfrac{N_a V}{\rho_a Z(\lambda)} \int_{\left| \textbf{r} \right| = r} d \textbf{r} \: e^{-\beta \overset{\sim}{U} (\textbf{r}; \lambda)} \right] \\ &= 4 \pi N_b \rho_a \int_0^\infty dr \: r^2 \: u_{ab}(r) \: \overset{\sim}{g}_{ab} (r; \lambda) \end{align} \nonumber \]

We now integrate over \(\lambda\) to obtain the solvation free energy

\[\Delta_\text{solv} A = 4 \pi N_b \rho_a \int_0^\infty dr \: r^2 \: u_{ab} (r) \int_0^1 d \lambda \: \overset{\sim}{g}_{ab} (r; \lambda) \label{16.9} \]

Now define

\[g_{ab}(r) = \int_0^1 d \lambda \: \overset{\sim}{g}_{ab} (r; \lambda) \label{16.10} \]

which is the solvation radial distribution function and measures the work to bring together solute and solvent particles at a distance \(r\) from each other. Then,

\[\Delta_\text{solv} A = 4 \pi N_b \rho_a \int_0^\infty dr \: r^2 \: u_{ab} (r) \: g_{ab}(r) \label{16.11} \]

Now, we compute the solvation chemical potential as the solvation free energy per solute particle

\[\Delta_\text{solv} \mu = \dfrac{\partial \Delta_\text{solv} A}{\partial N_b} = 4 \pi \rho_a \int_0^\infty dr \: r^2 \: u_{ab} (r) \: g_{ab} (r) \label{16.12} \]

Finally, then, the molar solvation free energy is simply obtained by multiplying this by Avogadro’s number \(N_A\):

\[\Delta_\text{solv} \mu = \dfrac{\partial \Delta_\text{solv} A}{\partial N_b} = 4 \pi \rho N_A \int_0^\infty dr \: r^2 \: u_{ab} (r) \: g_{ab} (r) \label{16.13} \]

Chemical Potentials in Solution-Vapor Equilibria

When treating the vapor as an ideal gas, it is relevant to ask what the Gibbs free energy of the vapor phase, and therefore, the chemical potential are. Recall that the partition function of an ideal gas of just one component is

\[Q(N, V, T) = \dfrac{1}{N!} \left( \dfrac{V}{\lambda^3} \right)^N \label{16.14} \]

where \(\lambda = \sqrt{\beta h^2/2 \pi m}\). The Gibbs free energy, however, comes from the isobaric partition function \(\Delta (N, P, T)\), given by

\[\begin{align} \Delta (N, P, T) &= \int_0^\infty dV \: e^{-\beta PV} Q(N, V, T) \\ &= \dfrac{1}{N! \lambda^{3N}} \int_0^\infty e^{-\beta PV} V^N dV \\ &= \dfrac{1}{(\beta P)^{N+1} \lambda^{3N}} \\ &\approx \dfrac{1}{(\beta P)^N \lambda^{3N}} \end{align} \label{16.15} \]

Therefore, the Gibbs free energy is

\[G(N, P, T) = -\dfrac{1}{\beta} \text{ln} \: \Delta (N, P, T) = N \: \text{K} \cdot \text{kg} \cdot \text{mol}^{-1} \: T \: \text{ln} \: \beta + N \: \text{K} \cdot \text{kg} \cdot \text{mol}^{-1} \: T \: \text{ln} \: P + 3N \: \text{K} \cdot \text{kg} \cdot \text{mol}^{-1} \: T \: \text{ln} \: \lambda \nonumber \]

Now suppose we have a process in which we change \(P_1\) to \(P_2\) in a gas at fixed \(N\) and \(T\). The change in Gibbs free energy is

\[\Delta G = G(N, P_2, T) - G(N, P_1, T) = N \: \text{K} \cdot \text{kg} \cdot \text{mol}^{-1} \: T \: \text{ln} \: \left( \dfrac{P_2}{P_1} \right) = nRT \: \text{ln} \: \left( \dfrac{P_2}{P_1} \right) \nonumber \]

Thus, the molar Gibbs free energy is

\[\Delta \overset{\sim}{G} = \dfrac{\Delta G}{n} = \mu = RT \: \text{ln} \: \left( \dfrac{P_2}{P_1} \right) \nonumber \]

which is also the chemical potential of the gas.

In thermodynamics, we can only define changes relative to an arbitrary reference, so we actually need to define that reference. This reference is known as the standard state. For gases, the standard state is a pressure of \(1 \: \text{atm}\) at some specified temperature, assuming ideal-gas behavior, i.e., that \(PV = nRT\). The pressure of the standard state is denoted \(P^\text{o}\), and it is just a constant having the value of \(1 \: \text{atm}\). If we now consider a change from the standard state to an arbitrary pressure \(P\), the change in the Gibbs free energy is then given by

\[\Delta \overset{\sim}{G} = \Delta \overset{\sim}{G^\text{o}} (T) + RT \: \text{ln} \: \left( \dfrac{P}{P^\text{o}} \right) \nonumber \]

and the chemical potential, which is the same thing, is

\[\mu = \mu^\text{o} (T) + RT \: \text{ln} \: \left( \dfrac{P}{P^\text{o}} \right) \nonumber \]

Let us denote a general component in our system as \(j\), i.e., \(j = A, B\) in a two-component solution. For each component in the vapor phase,

\[\mu_j^\text{(vap)} = \mu_j^\text{o} + RT \: \text{ln} \: \left( \dfrac{P_j}{P_j^\text{o}} \right) \label{16.16} \]

However, in equilibrium \(\mu_j^\text{(vap)} = \mu_j^\text{(soln)}\). Therefore, if we take \(P_j^\text{o} = 1 \: \text{atm}\), then

\[\mu_j^\text{(vap)} = \mu_j^\text{(soln)} = \mu_h^\text{o} + RT \: \text{ln} \: P_j \label{16.17} \]

An equivalent expression for a pure sample of component \(j\) in an equilibrium between its liquid and vapor would appear as

\[\mu_j^* \text{(liquid)} = \mu_j^* \text{(vapor)} = \mu_j^\text{o} + RT \: \text{ln} \: P_j^* \label{16.18} \]

Subtracting the two expressions yields

\[ \mu_j^\text{(soln)} = \mu_j^* \text{(liquid)} + RT \: \text{ln} \left( \dfrac{P_j}{P_j^*} \right) \label{16.19} \]

Moreover,

\[\begin{array}{rcl} \mu_j^\text{(soln)} & = & \mu_j^\text{(vap)} = \mu_j^\text{o} + RT \: \text{ln} \: (x_j P) \\ & = & \mu_j^\text{o} + RT \: \text{ln} \: x_j + RT \: \text{ln} \: P \end{array} \label{16.20} \]

Thus, for each of the two components,

\[\begin{align} \mu_1^\text{(soln)} &= \mu_1^\text{(vap)} = \mu_1^\text{o} + RT \: \text{ln} \: x_1 + RT \: \text{ln} \: P \\ \mu_2^\text{(soln)} &= \mu_2^\text{(vap)} = \mu_2^\text{o} + RT \: \text{ln} \: x_2 + RT \: \text{ln} \: P \\ \mu_2^\text{(soln)} - \mu_1^\text{(soln)} &= \mu_2^\text{o} - \mu_1^\text{o} + RT \: \text{ln} \left( \dfrac{x_2}{x_1} \right) \end{align} \label{16.21} \]

This shows that the difference of chemical potentials of the two components in solution is related to the logarithms of the ration of the mole fractions. Due to the equilibrium condition

\[\mu_2^\text{(vap)} - \mu_1^\text{(vap)} = \mu_2^\text{o} - \mu_1^\text{o} + RT \: \text{ln} \: \left( \dfrac{x_2}{x_1} \right) \label{16.22} \]

which relates the chemical potential difference to the same mole fraction ratio (which we recall is the ratio of gas-phase mole fractions); the same mole fraction ratio must be the same in solution in order to determine the solution-phase chemical potential difference.

Non-Ideal Solutions and Activities

We showed that the solution chemical potential is

\[\mu_j^\text{(soln)} = \mu_j^* \text{(liquid)} + RT \: \text{ln} \left( \dfrac{P_j}{P_j^*} \right) \label{16.23} \]

In an ideal solution \(P_j = x_j P_j^*\), in which case

\[\mu_j^\text{(soln)} = \mu_j^* \text{(liquid)} +RT \: \text{ln} \: x_j \label{16.24} \]

In a non-ideal solution, we can fit the behavior of \(P_1\) vs. \(x_1\) as

\[P_1 = x_1 P_1^* e^{b(1 - x_1)^2 + c(1 - x_1)^3} \label{16.25} \]

and

\[\mu_1^\text{(soln)} = \mu_1^* \text{(liquid)} + RT \: \text{ln} \: x_1 + RTb(1 - x_1)^2 + RTc(1 - x_1)^3 \label{16.26} \]

The second and third and fourth terms on the right of this expression measure the deviation from ideal behavior, and we see that as \(x_1 \rightarrow 1\),

\[\mu_1^\text{(soln)} \rightarrow \mu_1^* \text{(liquid)} + RT \: \text{ln} \: x_1 \label{16.27} \]

Let us define

\[a_j = \dfrac{P_j}{P_j^*} \label{16.28} \]

which is called the activity. For an ideal solution, \(a_j = x_j\). Generally, however, \(P_j/P_j^* \neq x_j\) but is rather some function \(f(x_1)\) of \(x_1\). Thus, in the non-ideal case, we can write

\[\mu_j^\text{(soln)} = \mu_j^* \text{(liquid)} + RT \: \text{ln} \: a_j \label{16.29} \]

The ratio

\[\gamma_j = \dfrac{a_j}{x_j} \label{16.30} \]

is known as the activity coefficient and is equal to \(1\) for ideal solutions. Activities can be defined for any state of a substance, i.e., solutions, dissolved species, gases, liquids, solids,...

Characterizing Solid-Liquid Solutions and Dissolved Species

Solid-liquid solutions involved the dissolution of a solid substance in a solvent leading to a homogeneous mixture. We will discuss the microscopic nature of dissolved species later in the lecture. Before we get into that, however, let us discuss some basic measures for characterizing the composition of such a solution.

The most common measure of solution composition is the molarity defined as

\[\text{Molarity} = \dfrac{\text{moles of solute}}{\text{liters of solvent}} \label{16.31} \]

and is denoted \(M\). A second measure is known as the molality defined as

\[\text{molality} = \dfrac{\text{moles of solute}}{\text{kilograms of solvent}} \label{16.32} \]

Example \(\PageIndex{1}\):

A solution has a solvent (component \(1\)) and \(N - 1\) dissolved species indexed \(k = 2, \ldots, N\). Let \(x_1\) be the mole fraction of the solvent and \(x_2, \ldots, x_N\) be the mole fractions of each of the dissolved solute species. Let \(M_1\) be the molar mass of the solvent. Show that

\[x_1 = \dfrac{1}{1 + \dfrac{M_1 m}{1000}}, \: \: \: \: \: \: \: x_k = \dfrac{M_1 m_k/1000}{1 + \dfrac{M_1 m}{1000}} \label{16.33} \]

where \(m_k\) is the molality of the \(k^\text{th}\) component and \(m\) is the total molality of the solution, i.e.,

\[m_k = \dfrac{\text{moles of component} \: k}{\text{kilograms of solvent}}, \: \: \: \: \: \: \: m = \sum_{k=2}^N m_k \label{16.34} \]

Solution

Start with the definition of the mole fractions. For component, the definition is

\[x_1 = \dfrac{n_1}{n_1 + \Sigma_{k=2}^N n_k} \nonumber \]

Divide top and bottom by \(n_1\) to give

\[x_1 = \dfrac{1}{1 + \Sigma_{k=2}^N \dfrac{n_k}{n_1}} \nonumber \]

The number of moles of solvent is

\[\begin{align} n_1 &= \dfrac{\text{mass of solvent in grams}}{M_1} \\ &= \dfrac{1000 \text{(mass of solvent in kilograms)}}{M_1} \\ &= \dfrac{1000 \mu_1}{M_1} \end{align} \nonumber \]

where \(\mu_1\) is the mass of solvent in kilograms. Thus, \(x_1\) becomes

\[\begin{align} x_1 &= \dfrac{1}{1 + \dfrac{M_1}{1000} \sum_{k=2}^N \dfrac{n_k}{\mu_1}} \\ &= \dfrac{1}{1 + \dfrac{M_1 m}{1000}} \end{align} \nonumber \]

where \(n_k/\mu_1\) is the molality of component \(k\), and \(m = \Sigma_{k=2}^N m_k\). For component \(k\),

\[n_k = \dfrac{n_k}{n_1 + \sum_{j=2}^N n_j} = \dfrac{n_k/n_1}{1 + \sum_{j=2}^N \dfrac{n_j}{n_1}} \nonumber \]

Use the fact that \(n_1 = 1000 \mu_1/M_1\). Substituting this in, we obtain

\[n_k = \dfrac{\dfrac{M_1}{1000} \dfrac{n_k}{\mu_1}}{1 + \dfrac{M_1}{1000} \sum_{j=2}^N \dfrac{n_j}{\mu_1}} \nonumber \]

Again, since \(n_j/\mu_1 = m_j\), and \(m = \Sigma_{j=2}^N m_j\), we finally obtain

\[x_k = \dfrac{M_1 m_k/1000}{1 + \dfrac{M_1 m}{1000}} \nonumber \]

Example \(\PageIndex{2}\):

A solution is prepared by dissolving \(22.4 \: \text{g}\) of \(MgCl_2\) in \(0.2 \: \text{L}\) of water. The resulting solution has a density of \(1.098 \: \text{g/cm}^3\). Calculate the mole fraction, molarity, and molality of \(MgCl_2\) in the solution.

Solution

The molar mass of \(MgCl_2\) is \(95.22 \: \text{g/mol}\). Thus, the moles of \(MgCl_2\) are

\[\text{moles of} \: MgCl_2 = \dfrac{22.4 \: \text{g}}{95.22 \: \text{g/mol}} = 0.235 \: \text{moles} \nonumber \]

In addition

\[\text{mass of water} = (0.2 \: \text{L}) \left( \dfrac{1000 \: \text{cm}^3}{1.0 \: \text{L}} \right) (1.00 \: \text{g/cm}^3) = 200 \: \text{g} \nonumber \]

The molar mass of water is \(18.02 \: \text{g/mol}\). Hence, the moles of water are

\[\text{moles of water} = \dfrac{200 \: \text{g}}{18.02 \: \text{g/mol}} = 11.1 \: \text{moles} \nonumber \]

The mole fraction of \(MgCl_2\), which is component \(2\), becomes

\[x_2 = \dfrac{0.235 \: \text{moles}}{(11.1 + 0.235) \: \text{moles}} = 0.0207 \: \text{moles} \nonumber \]

The total solution volume is

\[\begin{align} \text{solution volume} &= \dfrac{\text{mass of solution}}{\text{density of solution}} \\ &= \dfrac{22.4 \: \text{g} + 200 \: \text{g}}{1.098 \: \text{g/cm}^3} \\ &= 204 \: \text{cm}^3 \\ &= 0.204 \: \text{L} \end{align} \nonumber \]

Thus, the molarity of \(MgCl_2\) is

\[\text{Molarity} = \dfrac{0.235 \: \text{moles}}{0.204 \: \text{L}} = 1.15 \: \text{M} \nonumber \]

Finally, the molality is

\[\text{molality} = \dfrac{0.235 \: \text{moles}}{0.2 \: \text{kg}} = 1.18 \: \text{mol/kg} = 1.18 \: \text{m} \nonumber \]

Freezing Point Depression

At the freezing point, the solid solvent is in equilibrium with the liquid solvent. This requires

\[\mu_1^\text{(solid, soln)} (P, T) = \mu_1^\text{(liq, soln)} (P, T) \label{16.35} \]

where component \(1\) is the solvent. Recall, however, that

\[\mu_1^\text{(liq, soln)} = \mu_1^* \text{(pure liquid)} + RT \: \text{ln} \: a_1 = \mu_1^\text{(solid, soln)} \label{16.36} \]

Thus,

\[\text{ln} \: a_1 = \dfrac{\mu_1^\text{(solid, soln)} - \mu_1^* \text{(pure liquid)}}{RT} \label{16.37} \]

How does \(a_1\) depend on \(T\)? Take the temperature derivative of both sides:

\[\dfrac{\partial \: \text{ln} \: a_1}{\partial T} = \dfrac{1}{R} \left[ \dfrac{\partial (\mu_1^\text{(solid, soln)}/T)}{\partial T} - \dfrac{\partial (\mu_1^* \text{(pure liquid)}/T)}{\partial T} \right] \label{16.38} \]

For a general chemical potential, use the fact that \(G(n, P, T) = \mu (P, T) = H - TS\),

\[\begin{align} \dfrac{G}{T} &= \dfrac{H}{T} - S = \dfrac{\mu}{T} n \\ \dfrac{\partial (G/T)}{\partial T} &= \mu \: \partial (\mu/T)/\partial T = -\dfrac{H}{T^2} + \dfrac{1}{T} \left( \dfrac{\partial H}{\partial T} \right) - \dfrac{\partial S}{\partial T} \end{align} \label{16.39} \]

However, remember that

\[\dfrac{\partial H}{\partial T} = C_P (T) \label{16.40} \]

and, under isobaric conditions,

\[\begin{align} S &= \int \dfrac{dQ_\text{rev}}{T} \\ &= \int \dfrac{dH}{T} \\ &= \int \dfrac{C_P (T) \: dT}{T} \\ dS &= \dfrac{C_P (T) \: dT}{T} \\ \dfrac{\partial S}{\partial T} &= \dfrac{C_P (T)}{T} \end{align} \label{16.41} \]

Therefore,

\[\begin{align} \dfrac{\partial (G/T)}{\partial T} &= n \partial (\mu/T)/\partial T = -\dfrac{H}{T^2} + \dfrac{C_P (T)}{T} - \dfrac{C_P}{T} \\ &= -\dfrac{H}{T^2} \end{align} \label{16.42} \]

Applying Equation 16.42 to the molar Gibbs free energy \(\mu\), we obtain

\[\dfrac{\partial (\mu/T)}{\partial T} = -\dfrac{\overset{\sim}{H}}{T^2} \label{16.43} \]

where \(\overset{\sim}{H}\) is the molar enthalpy. Thus, for the solvent

\[\dfrac{\partial (\mu_1/T)}{\partial T} = -\dfrac{\overset{\sim}{H_1}}{T^2} \label{16.44} \]

We can use these results to determine the activity and ultimately the amount of freezing point depression. If we differentiate Equation 16.37 with respect to temperature and apply the above result, we obtain

\[\begin{align} \dfrac{\partial \: \text{ln} \: a_1}{\partial T} &= \dfrac{1}{R} \left[ \dfrac{\partial (\mu_1^\text{(solid, soln)}/T)}{\partial T} - \dfrac{\partial (\mu_1^* \text{(pure liq)}/T)}{\partial T} \right] \\ &= -\dfrac{\overset{\sim}{H_1}^\text{(solid, soln)} - \overset{\sim}{H_1^*} \text{(pure liq)}}{RT^2} \\ &= \dfrac{\overset{\sim}{H_1} \text{(pure liq)} - \overset{\sim}{H_1}^\text{(solid, soln)}}{RT^2} \end{align} \label{16.45} \]

Now we assume that, in a dilute solution, the solvent dominates, so that \(\overset{\sim}{H_1^*} \text{(pure liq)} \approx \overset{\sim}{H_1}^\text{(liquid, soln)}\). Thus,

\[\begin{align} \dfrac{\partial \: \text{ln} \: a_1}{\partial T} &= \dfrac{\overset{\sim}{H_1}^\text{(liquid, soln)} - \overset{\sim}{H}^\text{(solid, soln)}}{RT^2} \\ &= \dfrac{\Delta_\text{fus} \overset{\sim}{H_1}}{RT^2} \end{align} \label{16.46} \]

In order to see if there is a change in the freezing point, we need to integrate both sides from \(T_\text{fus}^*\), the freezing point of the pure solvent, to \(T_\text{fus}\), the freezing point of the solution. This gives

\[\text{ln} \: a_1 = \int_{T_\text{fus}^*}^{T_\text{fus}} \dfrac{\Delta_\text{fus} \overset{\sim}{H_1}}{RT^2} dT \label{16.47} \]

Let us assume that the solution is dilute so that \(a_1 \approx x_1\), \(\text{ln} \: a_1 \approx \text{ln} \: x_1 = \text{ln}(1-x_2) \approx -x_2\), where \(x_2 \ll 1\). We then have

\[\begin{align} -x_2 &= \left. -\dfrac{\Delta_\text{fus} \overset{\sim}{H_1}}{RT} \right|_{T_\text{fus}^*}^{T_\text{fus}} \\ &= \dfrac{\Delta_\text{fus} \overset{\sim}{H_1}}{R} \left[ \dfrac{1}{T_\text{fus}^*} - \dfrac{1}{T_\text{fus}} \right] \\ -x_2 &\approx \dfrac{\Delta_\text{fus} \overset{\sim}{H_1}}{R} \left[ \dfrac{T_\text{fus} - T_\text{fus}^*}{T_\text{fus} T_\text{fus}^*} \right] \end{align} \label{16.48} \]

Since \(\Delta \overset{\sim}{H}_\text{fus} > 0\), it follows that \(T_\text{fus} < T_\text{fus}^*\), hence, for the solution, the freezing point is lowered.

Previously, we showed that

\[x_2 = \dfrac{M_1 m_2/1000}{1 + M_1 m/1000} \label{16.49} \]

For a two-component solution \(m_2 = m\), hence this becomes

\[x_2 = \dfrac{M_1 m/1000}{1 + M_1 m/1000} \label{16.50} \]

which we can write as

\[\begin{align} x_2 &= \dfrac{m}{m + \dfrac{1000}{M_1}} \\ &\approx \dfrac{m}{1000/M_1} \\ &= \dfrac{M_1 m}{1000} \end{align} \label{16.51} \]

where we have assumed that \(m \ll 1000/M_1\). Now, let \(\Delta T_\text{fus} = T_\text{fus}^* - T_\text{fus}\). Hence, we can write

\[x_2 = \dfrac{\Delta_\text{fus} \overset{\sim}{H_1}}{R} \dfrac{\Delta T_\text{fus}}{T_\text{fus} T_\text{fus}^*} \approx \dfrac{M_1 m}{1000} \label{16.52} \]

Solving for the change in the freezing temperature \(\Delta T_\text{fus}\), we obtain

\[\Delta T_\text{fus} \approx \dfrac{T_\text{fus} T_\text{fus}^* R}{\Delta_\text{fus} \overset{\sim}{H_1}} \dfrac{M_1}{1000} m \equiv K_f m \label{16.53} \]

where \(K_f\) is a constant known as the freezing-point depression constant.

By a similar argument, it can also be shown that the boiling point of the solution is elevated, i.e.,

\[\Delta T_\text{vap} = K_b m \label{16.54} \]

Boiling point elevation and freezing point depression can be used to determine molar masses. The following examples illustrate how this is done:

Example \(\PageIndex{3}\)

When \(5.50 \: \text{g}\) of biphenyl (\(C_{12} H_{10}\)) is dissolved in \(100.0 \: \text{g}\) of benzene, the boiling point increases by \(0.903^\text{o}C\). When \(6.3 \: \text{g}\) of an unknown hydrocarbon is dissolved in \(150.0 \: \text{g}\) of benzene, the boiling point of the solution increases by \(0.597^\text{o}C\). What is the molar mass of the solute?

Solution

\[\text{moles of biphenyl} = \dfrac{5.5 \: \text{g}}{154.2 \: \text{g/mol}} = 0.0357 \: \text{mol} \nonumber \]

\[\text{molality of solution} = \dfrac{\text{moles of solute}}{\text{kg of solvent}} = \dfrac{0.0357 \: \text{mol}}{0.1 \: \text{kg}} = 0.357 \: \text{mol/kg} \nonumber \]

Therefore, we can determine the constant \(K_b\):

\[K_b = \dfrac{\Delta T_b}{m} = \dfrac{0.903 \: \text{K}}{0.357 \: \text{mol/kg}} = 2.53 \: \text{K} \cdot \text{kg} \cdot \text{mol}^{-1} \nonumber \]

Now that we know the value of \(K_b\) for benzene, we can use it to determine the molar mass of the unknown solute. Given the increase in the boiling temperature, we solve for the molality of the unknown solution:

\[m= \dfrac{\Delta T_b}{K_b} = \dfrac{0.597 \: \text{K}}{2.53 \: \text{K} \cdot \text{kg} \cdot \text{mol}^{-1}} = 0.236 \: \text{mol/kg} \nonumber \]

From this, we can determine the number of moles of the solute:

\[\text{moles of solute} = \text{molality of solution} \times \text{kilograms of solvent} = (0.236 \: \text{mol/kg})(0.150 \: \text{kg}) = 0.0354 \: \text{mol} \nonumber \]

Then, given the number of grams of solute, the molar mass is

\[M= \dfrac{6.3 \: \text{g}}{0.0354 \: \text{mol}} = 178 \: \text{g/mol} \nonumber \]

Example \(\PageIndex{4}\)

When \(0.494 \: \text{g}\) of \(K_3 Fe (CN)_6\) is dissolved in \(100.0 \: \text{g}\) of water, the freezing point is found to be \(-0.1116^\text{o}C\). What is the dissociation reaction for \(K_3 Fe (CN)_6\)? The \(K_f\) for water is \(1.86 \: \text{kg} \cdot \text{K/mol}\).

Solution

We first calculate the molality of the solute:

\[m = \dfrac{0.1116 \: \text{K}}{1.86 \: \text{kg} \cdot \text{K/mol}} = 0.06 \: \text{mol/kg} \nonumber \]

Thus, we have

\[(0.06 \: \text{mol/kg}) (0.1 \: \text{kg}) = 0.006 \: \text{mol} \nonumber \]

moles of solute. The molar mass of \(K_3 Fe (CN)_6\) is \(329.25 \: \text{g/mol}\). This gives

\[\dfrac{0.494 \: \text{g}}{329.25 \: \text{g/mol}} = 0.0015 \: \text{mol} \nonumber \]

of \(K_3 Fe (CN)_6\). We know, however, that this solute dissociates into ions, and there must be \(0.0015 \times 4 = 0.006\) moles of ions in the solution. Accordingly, the dissociation reaction must be

\[K_3 Fe (CN)_6 \longrightarrow 3K^+ \text{(aq)} + \left[ Fe (CN)_6 \right]^{3-} \text{(aq)} \nonumber \]

which gives \(4\) moles of ions per \(1\) mole of undissociated solute.