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Background

The miscibility, or solubility, of one liquid in another is both difficult to predict and difficult to explain, yet if we are willing to give up theoretical rigor, we can make some useful predictions by simply invoking "like dissolves like" (Padias p. 40). This phrase means liquids usually mix when their constituent molecules are electronically similar. Two polar liquids should mix, so should two nonpolar liquids. An "unlike" combination of polar and nonpolar liquids, however, should resist mixing.

You will get to investigate the application and limits of "like dissolves like" in this experiment by measuring the miscibility of water and several organic liquids, including eight alcohols: methanol, ethanol, 1- and 2-propanol, 1- and 2-butanol, 2-methyl-2-propanol (tert-butyl alcohol), and 3-methyl-1-butanol (isopentyl alcohol). Alcohols contain a polar hydroxyl group (-OH) just like water, but they also contain a nonpolar alkyl group (-R) so it is not at all obvious whether alcohols are "like" or "unlike" water. Only experimental measurements can tell you which part of the alcohol, the polar OH or nonpolar R group, determines its solubility in water.

Thermodynamics of mixing

"Like dissolves like" is a powerful and easy-to-remember tool, but it is defective as an explanation for mixing. It seems to suggest that "like" molecules have some special affinity for each other and this is responsible for their mixing. It also seems to suggest that "unlike" molecules somehow repel each other. Both ideas are incorrect.

To see why, we need to examine mixing from first a molecular perspective and then a thermodynamic perspective. The molecular perspective states that when two liquids come into contact, the molecules in each liquid jostle around and mix with each other until a dynamic equilibrium is achieved. We say the equilibrium is dynamic because we know that the molecules continue to jostle around, but we also say that the system is at equilibrium because the degree of mixing is no longer changing.

Thermodynamics tells us that the equilibrium state differs from the initial unmixed state by a change in free energy, G. More specifically, the degree of mixing found at equilibrium state is the one that minimizes G for the entire system (defined here as all of the molecules in the two liquids). To see how we work with a quantity like G, let’s assume that mixing occurs when the two liquids are placed in contact, i.e., the system starts off unmixed with free energy G_unmixed and then reaches equilibrium as a mixture with free energy G_mixed. The change in free energy is given by ΔG_mixing, where:

ΔG_mixing = G_mixed - G_unmixed

Since we assumed mixing occurs, and we know that the equilibrium state has the lowest possible free energy, G_mixed < G_unmixed. Or, ΔG_mixing < 0. This last result should be a familiar one from Chem 101/102. All spontaneous or favorable changes require a decrease in free energy.

Now, to identify factors that affect free energy, we invoke another relationship:

G = H - TS

where H represents enthalpy, the energy that is mainly associated with chemical bonds and intermolecular interactions, and S represents entropy, a measure of energy disorder. Combining above equations, we obtain:

ΔG_mixing = G_mixed - G_unmixed

= (H_mixed - TS_mixed) - (H_unmixed - TS_unmixed)

= (H_mixed - H_unmixed) - T(S_mixed - S_unmixed)

= ΔH_mixing - TΔS_mixing

(Notice that ΔX always means X_{final state} - X_{initial state}, or in the case of mixing, X_mixed - X_unmixed.)

The bottom equation tells us that there are two ways to make mixing favorable, i.e., to make ΔG_mixing < 0. Mixing is favored if it lowers the system's enthalpy, ΔH_mixing < 0, or if it raises the system's entropy, ΔS_mixing > 0. Of course, mixing could do both, lower enthalpy and raise entropy, but this isn't necessary. As long as one condition is satisfied, and that condition represents the dominant effect, mixing will occur. Unfortunately, phenomena that depend on two independent phenomena are inherently difficult to predict and rationalize, and solubility is no exception. There is no easy way to make solubility predictions, but we can learn a bit more by taking a closer look at the factors that affect ΔH_mixing and ΔS_mixing.

Enthalpy of mixing. The role of enthalpy in the mixing of two arbitrary liquids, A and B, can be clarified by visualizing the mixing of A and B as a three-stage process:

1. the molecules in liquid A separate to make gaseous A (vaporization of A)
2. the molecules in liquid B separate to make gaseous B (vaporization of B)
3. the two gases condense together to make "equilibrated" liquid AB (condensation of AB; exactly the reverse of AB vaporization)

The overall enthalpy change for mixing is just the sum of the enthalpy changes associated with each step:

ΔH_mixing = ΔH_{vaporization A} + ΔH_{vaporization B} + ΔH_{condensation AB}

Believe it or not, this equation simplifies our analysis. First, vaporization and condensation are really the same process, so by simply changing the last addition to a subtraction, we can write ΔH_mixing as a combination of three vaporization energies.

ΔH_mixing = ΔH_{vaporization A} + ΔH_{vaporization B} - ΔH_{vaporization AB}

Second, vaporization is a relatively simple process to understand. A liquid vaporizes into a gas when the kinetic energy of the molecules in the liquid becomes large enough to overcome the attractive intermolecular forces holding the molecules in the liquid. The stronger the forces between molecules, the greater the energy needed to overcome them.

Many kinds of intermolecular forces exist (Padias, p. 38-40), but we will consider only two here: hydrogen bonds (Loudon, p. 336-339, 342) and van der Waals forces (also known as dispersion or London forces, Loudon, p. 71).

The importance of hydrogen bonds can be seen in the following image. (This image is identical to the methanol-water mixture seen on the experiment's overview page except that the atoms in the present image have been rendered in "tube" format for clarity.) The long H ---- O contact in each hydrogen bond has been highlighted as a dashed yellow line and you can easily see that nearly every hydroxyl (-OH) group engages in at least one hydrogen bond. Many molecules engage in multiple hydrogen bonds.

tube molecular model of a methanol (CH₃OH)-water (H₂O) solution

Van der Waals forces are more general; they occur anywhere two molecules come into close contact. Therefore, as a first approximation, we can view a molecule's ability to create these forces as a characteristic of the molecule's surface area. Larger molecules with larger surfaces will generate stronger van der Waals forces per molecule.

All molecules generate van der Waals forces, but water and alcohols can also form hydrogen bonds, which are much stronger. This kind of thinking is entirely too simple-minded, however, because it neglects the role of surface area. The following image of "liquid" 1-butanol shows how van der Waals forces take over when a molecule has a large nonpolar surface. Only a few hydrogen bonds (dashed yellow lines) can be detected in the model, while van der Waals forces (not shown) exist between every pair of neighboring molecules.

tube molecular model of liquid 1-butanol (CH₃CH₂CH₂CH₂OH)

Now that we have some idea of the forces that operate in water (mainly hydrogen bonding) and alcohols (a combination of hydrogen bonding and van der Waals forces), we can try to make sense of vaporization enthalpies and how they influence mixing.

For our first case, consider the mixing of two identical liquids A and B. That is, consider a situation where we have split a liquid between two containers labeled "A" and "B", and we will now combine the liquids in a single vessel. Obviously the "A" and "B" molecules, being identical, will mix, a dramatic confirmation of the "like dissolves like" scenario. From the standpoint of vaporization enthalpies, though, the energy needed to vaporize a given quantity of "A", "B", or "AB", would be exactly the same. The overall enthalpy of mixing might be slightly negative because there are more molecules (and more intermolecular interactions) in the AB mixture, but this minor enthalpy change hardly explains why mixing is strongly favored.

Now consider a scenario in which liquids A and B are very "unlike", as in the proverbial "oil (A) and water (B)" mixture. Although different types of attractive forces operate in oil (van der Waals) and water (hydrogen bonds), oil and water both have relatively high boiling temperatures so both two liquids must have large positive vaporization enthalpies. The intermolecular forces in an oil-water mixture, on the other hand, are mainly van der Waals forces and are fairly weak (water molecules have small surfaces). This suggests that the vaporization enthalpy of oil-water should be smaller than that of either pure oil or pure water, and the overall enthalpy of mixing should be positive and unfavorable. In this case, the enthalpy change is consistent with the "like dissolves like" rule, but it does not mean "unlike" molecules repel. In this case, at least, oil and water simply don't attract each other as strongly as they attract molecules of their own kind.

Entropy of mixing. Entropy changes can affect mixing even more than enthalpy changes, but the connection between entropy, molecular structure, and mixing, can be hard to grasp.

Maybe the easiest place to begin is with a mixture of two gases. The molecules in an ideal gas do not interact at all so there are no intermolecular forces to worry about. ΔH_mixing = 0. Despite this, gases mix spontaneously because mixing raises the system's overall entropy. In mathematical terms, we would write ΔS_mixing > 0 and ΔG_mixing= –TΔS_mixing < 0.

A similar picture applies to the mixing of "like" liquids. As we saw above for the extreme case of two identical liquids, the enthalpy of mixing is negligible, yet mixing is strongly favored. The factor that favors mixing is the increase in entropy. "Like dissolves like" because of entropy. Mixing is not the result of any special attraction between "like" molecules.

Unfortunately, things get rather murky when we turn to the mixing of “unlike” substances, such as oil and water. We have already seen how mixing might be unfavorable from an enthalpy perspective, but why can't entropy overcome enthalpy as it does in "like" liquids and gases?

To understand the role of entropy, we must consider yet another factor that affects entropy: molecular orientation inside the liquid. The fewer orientations a molecule can adopt inside a liquid, the more "ordered" the molecule's structure will be. "More order" translates into lower entropy.

Molecules in liquid water adopt a relatively ordered (low entropy) structure to facilitate hydrogen bonding. By limiting themselves to particular orientations, neighboring molecules can achieve the linear O-H ---- O geometry optimal for hydrogen bonding. By limiting themselves further, many molecules can even make three or four simultaneous hydrogen bonds.

Surprisingly, placing a nonpolar molecule in water increases the structural order (lowers the entropy) of the liquid. Because a nonpolar molecule cannot hydrogen bond with its water neighbors, the latter are left with fewer effective hydrogen-bonding orientations to choose from. The water molecules that "coat" the nonpolar molecule become more ordered, and so do their water neighbors.

This effect of a nonpolar molecule on its surrounding waters is not unlike that of the President on his Secret Service bodyguards. If he's not around, they can mill about and talk to whomever they please, but the moment he appears, the guards closest to him turn their backs towards him so that they can better protect him from possible assailants. Their structure (not his) becomes more ordered.

I haven't figured out how to show the orientational choices available to a water molecule using a single molecular model, but I can show you how nonpolar molecules interfere with hydrogen bonding. The following model shows a combination of nonpolar pentane molecules (top) and water molecules (bottom). Hydrogen bonds between water molecules are many, but no hydrogen bonds cross the water-pentane interface. If you look carefully, you will also see that water molecules near the interface seem to make fewer hydrogen bonds. This factor should limit the orientational freedom of the interface molecules.

Molecular model of a pentane (C₅H₁₂)-water (H₂O) solution

The two-layer system shown above actually has higher entropy than a fully mixed system. Intermingling of pentane and water molecules is less effective at increasing entropy than simply preserving the orientational freedom of water molecules in the water layer. You may recall that ΔH_mixing was also unfavorable for mixing of "unlikes" so entropy and enthalpy work together to keep "unlikes" apart.

[Weird science note: Shaking water and oil vigorously to create millions of tiny interspersed bubbles might seem to increase the system's entropy, but this is not the case. The effect of bubble dispersal (increase entropy) is more than offset by the effect of increased contact area between water and oil (decrease entropy due to more water molecules at oil-water interfaces). Once you stop shaking, the two substances spontaneously separate (increase total entropy) to reduce their contact area.]

To recap, "like dissolves like" is an easy-to-remember rule for making solubility predictions, but it is easy to misinterpret. In particular, "like" molecules do not mix because of a special affinity for one another, and "unlike" molecules do not separate because they repel each other. To understand mixing, you must evaluate the factors that affect ΔH_mixing and ΔS_mixing, and you should remember that entropy is often in control.

Structural factors that affect mixing

In addition to measuring the solubility of different liquids in water, you will also build models of these molecules to help you interpret your results. You will not need to contemplate enthalpy and entropy changes. Instead, we will settle for the (far) simpler task of determining which molecules are more polar ("like water") and which are less so ("unlike water").

A common way to assess molecular polarity is to calculate a molecule’s dipole moment. The dipole moment reflects the degree of charge separation within a molecule, and one normally assumes that molecules with larger dipole moments are more polar.

Given the connection between van der Waals forces and molecular surface area, it also makes sense to calculate this area and divide it into polar and nonpolar regions. One might suppose that more polar molecules have larger polar surfaces, but it isn't immediately obvious whether it is more helpful to think about this in absolute terms (total polar surface area) or relative terms (fractional polar surface area).

You can identify polar and nonpolar surface regions using electrostatic potential maps. Our molecular modeling program (SPARTAN'08) can also tell you how much of the surface is polar and how much is nonpolar.

Instructions for using SPARTAN'08 to build molecular models, calculate dipole moments and surface areas, and display electrostatic potential maps, will be provided in class.