CoreChem:10.5 Solutions

From ChemEd Collaborative

10.5 SOLUTIONS

Up to this point we have discussed only the properties of pure solids and liquids. Of more importance to a chemist, though, are the properties of solutions. Very few chemical reactions involve only pure substances―almost all involve a solution of some sort.

In Sec. 3.5 we defined a solution as a homogeneous mixture of two more substances, that is, a mixture which appears to be uniform throughout. Under this definition we would refer to sugar or salt dissolved in water as solutions, but we would not apply the term to muddy water or to milk. A close inspection of muddy water reveals that it is not uniform in appearance but consists of small solid particles dispersed in water. We refer to such a mixture as a suspension. Under the microscope, milk can also be seen to be nonuniform, It consists of small drops of milk fat dispersed throughout an aqueous phase.

Our definition of a solution in terms of the homogeneity of a mixture is somewhat unsatisfactory since it does not tell us where to draw the line. Field-emission microscopes and electron microscopes have now been developed which can just about ”see” a single atom. With such a microscope virtually all matter looks nonuniform and hence not homogeneous. If our definition extends to such microscopes, then true solutions do not exist. In practice we draw the line somewhere around the 5 nanometer (nm) mark, even though some molecules (see Fig. 1.5) are larger than this.

On the molecular level a solution corresponds to the random arrangement of one kind of molecule or ion around another. In Fig. 10.14 the illustration on the left corresponds to a solution since each black molecule is randomly surrounded by black and red molecules, and vice versa. The illustration on the right is a suspension. The distribution is not random, and most red molecules have red neighbors, while most black molecules have black neighbors.

Strictly speaking, the term solution applies to any homogeneous mixture, but we will concentrate our discussion on those solutions which involve liquids since these are the most common. It should be realized, though, that other types of solutions also exist. Air is a solution of a large number of gases (oxygen is the most concentrated) in another gas (nitrogen). A 5-cent coin is made from an alloy in which one solid (nickel) is dissolved in another (copper). Solutions of hydrogen gas in solid palladium and some other metals are also possible.

As mentioned in the brief discussion of solutions in Chap. 3, it is sometimes difficult to decide which component of a solution is the solute and which is the solvent. Usually the amount of solvent is much larger than that of the solute. If the pure components were initially in separate phases (a gas and a solid, for example), the phase corresponding to the state of the solution is taken to be the solvent. In the case of H₂(g) and Pd(s) mentioned above, for example, Pd would be the solvent because the solution is a solid phase.

Saturated and Supersaturated Solutions

We often find that there is a limit to the quantity of solute which will dissolve in a given quantity of solvent. This is especially true when solids dissolve in liquids. For example, if 36 g KCl crystals is shaken with 100 g H₂O at 25°C only 35.5 g of the solid dissolves. If we raise the temperature somewhat, all the KC1 will dissolve, but on cooling to 25°C again, the extra 0.5 g KC1 will precipitate, leaving exactly 35.5 g of the salt dissolved. We describe this phenomenon by saying that at 25°C the solubility of KCl in H₂O is 35.5 g KC1 per 100 g H₂O. A solution of this composition is also described as a saturated solution since it can accommodate no more KCl.

Under some circumstances it is possible to prepare a solution which behaves anomalously and contains more solute than a saturated solution. Such a solution is said to be supersaturated. A good example of supersaturation is provided by Na₂S₂O₃, sodium thiosulfate, whose solubility at 25°C is 50 g Na₂S₂O₃ per 100 g H₂O. If 70 g Na₂S₂O₃ crystals is dissolved in 100 g hot H₂O and the solution cooled to room temperature, the extra 20 g Na₂S₂O₃ usually does not precipitate. The resulting solution is supersaturated; consequently it is also unstable. It can be “seeded” by adding a crystal of Na₂S₂O₃, whereupon the excess salt suddenly crystallizes and heat is given off. After the crystals have settled and the temperature has returned to 25°C, the solution above the crystals is a saturated solution—it contains 50 g Na₂S₂O₃.

Miscibility

When a solid dissolves in a liquid, we very seldom find that the liquid has any tendency to dissolve in the solid. In a saturated solution of potassium chloride, for example, essentially no water dissolves in the potassium chloride crystals. With liquids the situation is usually different. If equal quantities of 1-butanol and water are shaken together, the mixture slowly separates into two layers. The bottom layer is a saturated solution of 1-butanol in water—it contains about 8% 1-butanol by weight. The top layer is not pure 1-butanol but a saturated solution of water in 1-butanol. It contains about 32% water by weight. A pair of liquids, like 1-butanol and water, which separates into two layers is said to be partially miscible.

By contrast with the solubilities of solids in liquids, a great many liquid pairs are completely miscible. That is, regardless of the proportions in which the two liquids are mixed, each will dissolve completely in the other. There will be no phase boundary as in the case of partially miscible liquids like 1-butanlo and water. Ethanol and water provide a good example of two liquids which are completely miscible. If you have a source of pure ethanol, it is possible to mix a drink in any proportions you even up to 200 proof—without forming two separate liquid phases.

Measuring the Composition of a Solution

When speaking of solubility or miscibility, or when doing quantitative experiments involving solutions, it is necessary to know the exact composition of a solution. This is invariably given in terms of a ratio telling us how much solute is dissolved in a unit quantity of solvent or solution. The ratio can be a ratio of masses, of amounts of substances, or of volumes, or it can be some combination of these. For example, concentration was defined in Sec. 3.5 as the amount of solute per unit volume of solution:

The two simplest measures of the composition are the mass fraction w, which is the ratio of the mass of solute to the total mass of solution, and the mole fraction x, which is the ratio of the amount of solute to the total amount of substance in the solution. If we indicate the solute by A and the solvent by B, the mass fraction and the mole fraction are defined by

---

EXAMPLE 10.4 A solution is prepared by dissolving 18.65 g naphthalene, C₁₀H₈ in 89.32 g benzene, C₆H₆. Find (a) the mass fraction and (b) the mole fraction of the naphthalene.

---

Solution

a) The mass fraction is easily calculated from the masses:

It is sometimes useful to distinguish mass of solute and mass of solution for purposes of calculation. In such a case we can write

w_C10H8 = 0.1727 g C₁₀H₈ per g solution

b) In order to calculate the mole fraction, we must first calculate the amount of each substance. Since

we find    

                

Thus             or      0.1128 mol C₁₀H₈ per mol solution

The mass fraction is useful because it does not require that we know the exact chemical nature of both solute and solvent. Thus if we dissolve 10 g crude oil in 10 g gasoline, we can calculate w_{crude oil}= 0.5 even though the solute and solvent are both mixtures of alkanes and have no definite molar mass. By contrast, the mole fraction is useful when we want to know the nature of the solution on the microscopic level. In the above example, for instance, we know that for every 100 mol of solution, 11.28 mol is naphthalene. On the molecular level this means that out of each 100 molecules in the solution, 11.28 will, on the average, be naphthalene molecules.

The mass fraction of a solution is often encountered in other disguises. The weight percentage (strictly speaking, the mass percentage) of a solution is often defined by the formula

This definition is really the same as that of the mass fraction because the percent sign means “divided by 100.” Thus 100% is merely a synonym for 100/100, that is, the number 1, and we can write

w_C10H8 = 0.1727 × 1 = 0.1727 × 100% = 17.27%

When the mass fraction is very small, it is often expressed in parts per million (ppm) or parts per billion (ppb). These symbols can be handled in much the same way as a percentage if you remember how they are related to unity: 1 = 100% = 10⁶ ppm = 10⁹ ppb

In other words

---

EXAMPLE 10.5 A 1-kg sample of water from Lake Powell, Utah, is found to contain 10 ng mercury. Walleyed pike caught in the lake contain 0.427 ppm mercury. (a) What is the mass fraction of mercury (in ppb) in water from Lake Powell? (b) If you ate 2 lb of walleyed pike caught from the lake, what mass of mercury would you ingest?

---

Solution

a) The mass fraction of mercury is by definition

                   

Therefore      

In ppb           

b) Assuming the mercury to be uniformly distributed throughout the walleyed pike, we have

w_Hg = 0.427 ppm = 0.427 × 10^–7

By definition and

and      

                      

---

From Example 10.5 you can see that from nearly the same mass of fish, almost 40 000 times as much mercury would be obtained as from the water. Indeed, if you ate 2 lb of walleyed pike every day, you would exceed the minimum dosage (300 μg for a 70-kg human) at which symptoms of mercury poisoning can appear. Fortunately, most of us do not eat fish every day, nor are we as gluttonous as the example suggests. Nevertheless, the much higher mass fraction of mercury in fish than in water shows that very small quantities of mercury in the environment can be magnified many times in living systems. This process of bioamplification will be discussed more thoroughly in 20.3.

Solubility and Molecular Structure

Chemical theory has not reached the point where it can predict exactly how much of one substance will dissolve in another. The best we can do is to indicate in general terms the relationships between solubility and the microscopic structures of solute and solvent.

To begin with, moving particles of any kind tend to become more randomly distributed as time passes. If you put a layer of red marbles in the bottom of a can and cover it with a second layer of white marbles, shaking the can for a short time will produce a nearly random distribution. The same principle applies on the microscopic level. Moving molecules tend to become randomly distributed among one another, unless something holds them back. Thus gases, whose molecules are far apart and exert negligible forces on one another, are all completely miscible with other gases.

In liquid solutions, the molecules are much closer together and the characteristics of different types of molecules are much more important. In particular, if the solute molecules exert large intermolecular forces on each other but do not attract solvent molecules strongly, the solute molecules will tend to group together. This forms a separate phase and leaves the solvent as a second phase. Conversely, if the solvent molecules attract each other strongly but have little affinity for solute molecules, solvent molecules will segregate, and two phases will form.

A classic example of the second situation described in the previous paragraph is the well-known fact that oil and water do not mix—or if they do, they do not stay mixed for long. The reason is that oil consists of alkanes and other nonpolar molecules, while water molecules are polar and can form strong hydrogen bonds with each other. Suppose that alkane or other nonpolar molecules are randomly dispersed among water molecules, as shown in Fig. 10.15a.The constant jostling of both kinds of molecules will soon bring two water molecules together. Dipole forces and hydrogen bonding will tend to hold the water molecules together, but there are only weak London forces between water and nonpolar molecules. Before long, clusters of water molecules like those in Fig. 10.15b will have formed.

These clusters will be stable at room temperature because the energy of interaction between the water molecules will be larger than the average energy of molecular motion. Only an occasional molecular collision will be energetic enough to bump two water molecules apart, especially if they are hydrogen bonded. Given enough time, this process of aggregation will continue until the polar molecules are all collected together. If the nonpolar substance is a liquid, this process corresponds on the macroscopic level to the liquids separating from each other and forming two layers.

If instead of mixing substances like oil and water, in which there are quite different kinds of intermolecular attractions, we mix two polar substances or two nonpolar substances, there will be a much smaller tendency for one type of molecule to segregate from the other. Thus two alkanes like n-heptane, C₇H₁₆, and n-hexane, C₆H₁₄, are completely miscible in all proportions. The C₇H₁₆ and C₆H₁₄ molecules are so similar (see Table 8.3 for projection formulas) that there are only negligible differences in intermolecular forces. Thus the molecules remain randomly mixed as they jostle among one another.

For a similar reason, methanol, CH₃OH, is completely miscible with water. In this case both molecules are polar and can form hydrogen bonds among themselves, and so there are strong intermolecular attractions within each liquid. However, CH₃OH dipoles can align with H₂O dipoles, and CH₃OH molecules can hydrogen bond to H₂O molecules, and so the at tractions among unlike molecules in the solution are similar to those among like molecules in each pure liquid. Again there is little tendency for one type of molecule to become segregated from the other.

All the cases just discussed are examples of the general rule that like dissolves like. Two substances whose molecules have very similar structures and consequently similar intermolecular forces will usually be soluble in each other. Two substances whose molecules are quite different will not mix randomly on the microscopic level. In general, polar substances will dissolve other polar substances, while nonpolar materials will dissolve other nonpolar materials. The greater the difference in molecular structure (and hence in intermolecular attractions), the lower the mutual solubility.

---

EXAMPLE 10.6 Predict which of the following compounds will be most soluble in water:

a)       b)

---

Solution Since ethanol contains an OH group, it can hydrogen bond to water. Although the same is true of hexanol, the OH group is found only at one end of a fairly large molecule. The rest of the molecule can be expected to behave much as though it were a nonpolar alkane. This substance should thus be much less soluble than the first. Experimentally we find that ethanol is completely miscible with water, while only 0.6 g hexanol dissolves in 100 g water.

---

Our discussion of solubility in terms of microscopic structure concludes with one more point. The solubilities of other substances in solids are usually small. The constituent particles in a solid crystal lattice are packed tightly together in a very specific geometric arrangement. For one particle to replace another in such a structure is very difficult, unless the particles are almost identical. The most common solid solutions are alloys, in which one essentially spherical metal atom replaces another. Thus alloys are easily made by melting two metals and cooling the liquid solution. In many other cases, however, completely miscible liquids separate when a solid phase forms. A good example of this is benzene and naphthalene:

A naphthalene molecule is almost twice as big as a benzene molecule and cannot fit in the benzene lattice.

Ideal Solutions: Raoult’s Law

When two substances whose molecules are very similar form a liquid solution, the vapor pressure of the mixture is very simply related to the vapor pressures of the pure substances. Suppose, for example, we mix 1 mol benzene with 1 mol toluene

as shown in Fig. 10.16. The mole fraction of benzene, x_b, and the mole fraction of toluene, x_t, are both equal to 0.5. At 79.6°C the measured vapor pressure of this mixture is 516 mmHg, slightly less than 517 mmHg, the average of the vapor pressures of pure benzene (744 mmHg) and of pure toluene (290 mmHg) at the same temperature.

It is easy to explain this behavior if we assume that because benzene and toluene molecules are so nearly alike, they behave the same way in solution as they do in the pure liquids. Since there are only half as many benzene molecules in the mixture as in pure benzene, the rate at which benzene molecules escape from the surface of the solution will be half the rate at which they would escape from the pure liquid. In consequence the partial vapor pressure of benzene above the mixture will be one-half the vapor pressure of pure benzene. By a similar argument the partial vapor pressure of the toluene above the solution is also one-half that of pure toluene. Accordingly, we can write

p_b = ½ P_b^*      and      p_t = ½ P_t^*

where p_b and p_t are the partial pressures of benzene and toluene vapors, respectively, and P_b^* and P_t^* are the vapor pressures of the pure liquids. The total vapor pressure of the solution is

The vapor pressure of the mixture is equal to the mean of the vapor pressures of the two pure liquids.

We can generalize the above argument to apply to a liquid solution of any composition involving any two substances A and B whose molecules are very similar. The partial vapor pressure of A above the liquid mixture, p_A, will then be the vapor pressure of pure A, P_A^*, multiplied by the fraction of the molecules in the liquid which are of type A, that is, the mole fraction of A, x_A. In equation form

p_A = x_AP_A^*      (10.2a)

Similarly for component B

p_B = x_BP_B^*      (10.2b)

Adding these two partial pressures, we obtain the total vapor pressure

P = p_A + p_B = x_AP_A^* + x_BP_B^*      (10.3)

Liquid solutions which conform to Eqs. (10.2) and (10.3) are said to obey Raoult’s law and to be ideal mixtures or ideal solutions.

In addition to its use in predicting the vapor pressure of a solution, Raoult’s law may be applied to the solubility of a gas in a liquid. Dividing both sides of Eq. (10.2a) by P_A^* gives

Since the vapor pressure of any substance has a specific value at a given temperature, Eq. (10.4) tells us that the mole fraction x_A of a gaseous solute is proportional to the partial pressure p_A of that gas above the solution.

For an ideal solution the proportionality constant k_A is the reciprocal of the vapor pressure of the pure solute at the temperature in question. Since vapor pressure increases as temperature increases, k_A, which is 1/P_A^*, must decrease. Thus we expect the solubility of a gas in a liquid to increase as the partial pressure of gas above the solution increases, but to decrease as temperature increases. Equation (10.4) is known as Henry’s law. It also applies to gaseous solutes which do not form ideal solutions, but in such cases the Henry’s-law constant k_A does not equal the reciprocal of the vapor pressure.

In actual fact very few liquid mixtures obey Raoult’s law exactly. Even for molecules as similar as benzene and toluene, we noted a deviation of 517 mmHg – 516 mmHg, or 1 mmHg at 79.6°C. Much larger deviations occur if the molecules are not very similar. These deviations are of two kinds. As can be seen from Fig. 10.17, a plot of the vapor pressure against the mole fraction of one component yields a straight line for an ideal solution. For nonideal mixtures the actual vapor pressure can be larger than the ideal value (positive deviation from Raoult’s law) or smaller (negative deviation). Negative deviations correspond to cases where attractions between unlike molecules are greater than those between like molecules. In the case illustrated, acetone (CH₃COCH₃) and chloroform ([[Image:]]) can form a weak hydrogen bond:

Because of this extra intermolecular attraction, molecules have more difficulty escaping the solution and the vapor pressure is lower. The opposite is true of a mixture of benzene and methanol. When C₆H₆ molecules are randomly distributed among CH₃OH molecules, the latter cannot hydrogen bond effectively. Molecules can escape more readily from the solution, and the vapor pressure is higher than Raoult’s law would predict.