H-1 Calculations Using the Solubility Product Constant, K_sp. We wish now to consider the quantitative aspects of phase equilibria of slightly soluble salts. We will do this via several illustrative examples. We can deal with these problems readily using the general approach developed in Chapter 12 for treating problems in chemical equilibrium.

Simple Solubility Calculations. The simplest problem that we encounter is to calculate the molar solubility of a salt, given the value of its K_sp. Example H-1 shows the approach.

Example H-1. Calculate the solubility in moles/L of BaF₂. K_sp for BaF₂ = 1.7 * 10^-6 M³.

Solution. First, we write the chemical equation for the process being considered; second, we indicate the initial amounts of all substances involved; third, we write changes in these concentrations which occur when reaction takes place; and fourth, we write the equilibrium concentration of each species by adding the change to the initial concentration. This is done below.

Because the number of moles of BaF₂(s) that dissolves per liter of solution is unknown, it is represented as a variable. We know that for each mole of BaF₂(s) that dissolves, one mole of Ba²⁺ ions and 2 moles of F- ions are produced. This follows directly from the stoichiometry of the equation. Thus if y moles of solid dissolves per liter of solution, y moles of Ba²⁺ ions and 2y moles of F- ions will result. A few comments about the initial and equilibrium entries for solid BaF₂ are in order. When we seek the molar solubility of BaF₂(s) in water, we are interested in the number of moles of BaF₂ that dissolve in 1 L of water when the solution is saturated. As seen above, the only way to insure a saturated solution is to use excess (xs) of the solid salt. Thus we have indicated in the "initial" line that we begin with an xs of solid BaF₂. This means that we begin with enough of the salt to produce a saturated solution and to have undissolved salt left over to maintain equilibrium with the dissolved ions. The "change in concentration" line then indicates that an unknown number of moles of BaF₂ dissolves per liter of solution, where it is understood that this number is less than the amount of solid initially present. At equilibrium, then, we still have excess undissolved solid present, hence the xs designation here. Note that since the concentration of BaF₂(s) does not appear in the expression for K_sp, IT DOES NOT MATTER WHAT ITS AMOUNT AT EQUILIBRIUM IS, AS LONG AS SOME OF IT IS PRESENT.

The next step in the systematic approach is to substitute the equilibrium concentrations of Ba²⁺ and F^- into the expression for K_sp. Equation H-1-2 results.

Solving, we obtain y = 7.5 * 10^-3 M, which is the molar solubility of BaF₂(s).

At equilibrium, the concentrations of Ba²⁺ ion and F^- ion will be 7.5 * 10^-3 M and 15 * 10^-3 M, respectively, and there will be excess BaF₂(s) in equilibrium with the ions in solution.

Figure H-1a shows graphically how the concentrations of Ba²⁺ and F- increase with time and eventually level off at the equilibrium values. Notice that the fluoride concentration is twice the barium ion concentration at all times on the graph. Figure H-1b shows the relationship between [Ba²⁺] and [F-], based on the K_sp expression in H-1-2. At any point on the curve, [Ba²⁺] = K_sp/[F-]². The arrow from the origin to the curve shows how the two concentrations change from the initial situation, in which there is no dissolved BaF₂, to the final equilibrium situation, represented by the curve. The slope of the arrow is 1/2, because the [F-]changes twice as fast as [Ba²⁺].

Solubility and the Common Ion Effect. We will now show that in the presence of a substance containing an ion in common with the slightly soluble salt, the solubility of the salt decreases in a manner consistent with Le Chatelier’s Principle. The decrease in solubility caused by the presence of the common ion is called the Common Ion Effect. We again examine a quantitative example.

Example H-2. What is the solubility of BaF₂(s) in a 1.0 M solution of barium nitrate, Ba(NO₃)₂? K_sp for BaF₂(s) = 1.7 * 10^-6.

Solution. We recognize two facts. First, barium nitate has an ion in common with barium fluoride. Second, barium nitrate is soluble according to solubility rule 3 (Chapter 10). Since it is a salt, it is a strong electrolyte, so a 1.0 M solution of it will actually contain 1.0 M Ba²⁺ and 2 M NO₃^-. Having recognized these facts, we are in a position to solve the problem. We use the standard approach.

Here y represents the molar solubility of BaF₂. We substitute the equilibrium concentrations into the expression for K_sp.

If we expand the left side of this equation, a complex equation involving both y³ and y² will result. Although such equations can certainly be solved, it is not within our scope to do so. If possible, we must make a simplifying assumption. Because K_sp is very small, dissolution of the salt will take place to only a minor extent; i.e., the value of y will be small, probably much smaller than 1.0. If we assume that 1.0 >> y, then y can be neglected with respect to 1.0 in the first term on the left side of the equation. The equation simplifies to

This is now a simple quadratic equation which is readily solved to give y = 6.5 * 10^-4 M. Before accepting this result, we must make sure that the approximation made above is indeed valid. To do this, we compare the value of y with the number with respect to which we neglected it. If y is less than 5% of this number, the approximation is acceptable.

Since this is much less than 5%, the answer is acceptable. At equilibrium, when the solution is saturated with barium fluoride, the concentrations of barium and fluoride ions will be

The molar solubility of barium fluoride, given directly by y, is 6.5 * 10^-4 M.

Figure H-2 is a reproduction of Figure H-1b, and shows the relationship between possible equilibrium concentrations of Ba²⁺ and F- in the BaF₂ dissolution problem. The arrow shows the progress of dissolution from the initial conditions, [Ba²⁺] = 1 M and [F-] = 0 M, to the equilibrium concentrations calculated above.

In Example H-1, we saw that the solubility of barium fluoride in pure water is 7.5 * 10^-3 M. In Example H-2, we found that in water containing 1 mole of barium ions per liter of solution, the solubility of barium fluoride is 6.5 * 10^-4 M, over 10 times smaller than in pure water. This is the common ion effect. These two examples provide a quantitative illustration of the same phenomenon that we would predict qualitatively using Le Chatelier’s principle: addition of barium ions (in the form of a soluble barium salt) to a saturated solution of barium fluoride in water constitutes a stress on equilibrium H-1-3 that results in precipitation of barium fluoride as H-1-3 shifts left in response to the stress.

Selective Precipitation. Suppose we have a solution containing comparable concentrations of two anions that we want to separate from one another. How might we carry out the separation? The simplest way is to add to the solution a cation which forms an insoluble salt with only one of the anions. For example, addition of barium ion (in the form of a soluble barium salt) to a solution containing Cl^- and SO₄^2- anions would result in the formation of a precipitate of barium sulfate, BaSO₄, which is insoluble. Filtration of the mixture would give a solid containing almost all of the sulfate ion, and a solution containing a very small amount of sulfate ion and all of the chloride ion. By carrying out this simple procedure, we can separate two anions which were initially intimately mixed in solution. Such separations are very useful, and the technique described finds many applications in synthetic and analytical chemical processes. Now suppose that both anions in the original solution form insoluble salts with some cation. Can we still separate the anions by addition of the cation to a solution of both anions? Interestingly, we can, provided that the molar solubilities of the two salts differ sufficiently. We look at an example.

Example H-3. A solution contains 0.1 M Cl- and 0.1 M Br-. AgNO₃ is slowly added to the solution. What will the concentration of Br- be in the solution when AgCl just starts to precipitate? Assume that the volume of solution does not change when AgNO₃ is added. K_sp(AgCl) = 1.7 * 10^-10. K_sp(AgBr) = 5 * 10^-13.

Solution. Before doing calculations, we consider in qualitative terms what happens as silver ion is added. Both anions form slightly soluble salts with Ag⁺. However, since K_sp for AgBr is 300 times smaller than that for AgCl, AgBr is less soluble. As Ag⁺ is added to the solution, AgBr begins to precipitate first. What the problem asks, then, is how much of the original Br- ion is left in solution at the point where AgCl just starts to precipitate? This last phrase is the key to the problem. At the point where precipitation of AgCl just begins, two things must be true. First, the concentration of Cl- ion in solution must be 0.1 M, because precipitation of AgCl has not yet removed any chloride from solution. Second, the solution is just exactly saturated with AgCl. This means that K_sp for AgCl must be satisfied. Knowing [Cl-] and K_sp for AgCl allows us to calculate [Ag⁺] at the point where AgCl just begins to come down (i.e., to precipitate). Once we know [Ag⁺], we can use K_sp for AgBr (which must also be satisfied, since AgBr has been precipitating for quite a while) to compute [Br-] in solution. This is the concentration we need.

Calculation of [Ag⁺] at the point where AgCl precipitation starts:

Calculation of [Br-] consistent with this [Ag⁺]:

According to this calculation, we can reduce the concentration of Br- from the initial value of 0.1 M to 2.94 * 10^-4 M before AgCl starts to precipitate. The fraction of the original Br- left in solution is

Multiplying by 100, we find that only 0.29% of the original bromide remains behind. In other words, we have removed 99.7% of the bromide ion from solution without removing any chloride ion. Filtration gives us a solid containing almost all of the bromide, and a solution containing all of the chloride. The two ions have been effectively separated.

H-2 Methods of Influencing the Solubility of Slightly Soluble Salts. The Common Ion Effect. Using the methods illustrated in Example H-1, the solubility of BaF₂(s) in water according to H-2-1 at 25 ^oC is estimated to be 0.018 M.

Suppose we have a mixture of solid BaF₂ in equilibrium with a solution containing Ba²⁺ and F- ions. What will happen if some solid NaF is added to the solution? NaF is a soluble salt, so will dissolve to produce F- and Na⁺ ions. The additional F- ions place a stress on equilibrium H-2-1, which can be relieved by a shift to the left (Le Chatelier). This will cause the precipitation of some BaF₂(s), and when equilibrium is reestablished , there will be less Ba²⁺ and more F- than there were before addition of NaF. Thus the effect of the added NaF is to decrease the molar solubility of BaF₂(s). This effect is often called the Common Ion Effect, because it results from adding to the solution a substance having an ion in common with the substance in equilibrium. The common ion effect decreases the solubility of BaF₂, by Le Chateliers Principle. We now ask whether the solubility of a slightly soluble salt can be increased by another substance that may be dissolved in the same solution. The answer is yes, if the second substance somehow removes from solution one of the ions produced by dissolution of the salt. The combination of an unfavorable process and a favorable one in tandem fashion is called coupling of reactions. Coupling enables an otherwise unfavorable process to proceed. The coupling principle is used to advantage in biological processes, where favorable reactions are frequently used to drive unfavorable ones via coupling. The essence of coupling is that a product of the unfavorable process is a reactant of the favorable one. Coupling may be used in several ways to enhance solubility.

Reaction of the Anion of the Salt with H₃O⁺. If the anion of a slightly soluble salt is also the conjugate base of a weak acid, the solubility of the salt can be enhanced by adding acid to its solutions. In fact, this will work for barium fluoride, because the fluoride anion is the conjugate base of the weak acid HF. Dissolution of barium fluoride has been represented in equation H-2-1. Reaction of F- with hydronium ion resulting from addition of a strong acid is shown in reaction H-2-2. The dissolution of barium fluoride and the protonation of fluoride anion are shown in tandem below:

The very favorable protonation process, H-2-2, lowers the concentration of fluoride ion resulting from the dissolving of barium fluoride. By Le Chatelier’s Principle, there is a consequent shift of reaction H-2-1 toward products; in other words, more barium fluoride dissolves. Its solubility is enhanced.

Combination of H-2-1 and H-2-2 gives the net overall reaction shown in H-2-3. H-2-2 is taken twice to allow for the formation of two F- ions per formula unit of BaF₂ that dissolves.

The overall equilibrium constant for H-2-3 is K_sp/(K_a)², or 4.5. Although this is not large enough that we may assume that H-2-3 runs to completion, it is larger than 1, meaning that H-2-3 is slightly favorable. Thus the unfavorable process H-2-1 has been coupled to the favorable process H-2-2 to give the favorable process, H-2-3.

Reaction of an Ion of the Salt with an Oxidizing or Reducing Agent. If one of the ions produced by dissolution of a slightly soluble salt is readily reduced or oxidized, its concentration in solution can be effectively lowered by reaction with an appropriate reducing or oxidizing agent. The effect of depletion of the concentration of the ion will be further dissolution of the salt, hence an increased solubility. For example, if solid CuS(s) is placed in contact with water, a small amount of it dissolves to establish the phase equilibrium in H-2-4.

Ordinarily, the solubility of CuS is very low (K_sp = 9 * 10^-36), so in pure water the molar solubility is 3 * 10^-18 M. However, addition of nitric acid to the solution results in reaction H-2-5, in which nitrate ion, NO₃^-, oxidizes S^2- to elemental sulfur.

Reaction H-2-5 decreases [S^2-] in solution, causing a shift of H-2-4 to the right by Le Chatelier's’ principle. Thus the solubility is enhanced. The net of reactions H-2-4 and H-2-5 is H-2-6. Knowledge of the equilibrium constants for H-2-4 and H-2-5 enables us to obtain K_eq for H-2-6, from which the solubility of CuS ([Cu²⁺]) can be calculated.

This, too, is reaction coupling.

Complex Ion Formation. Many cations, particularly of the transition metals, combine with anions or molecules in solution to form discrete species called complex ions. Formation of such complex ions, or complexes, reduces the concentration of free cation in solution and thereby enhances the solubility of a slightly soluble salt. For example, silver ion, Ag⁺, binds strongly to ammonia, NH₃, to form a complex of formula Ag(NH₃)₂²⁺. Adding ammonia to a mixture of solid AgCl in equilibrium with Ag⁺ and Cl- results in the sequence of reactions H-2-7 and H-2-8.

Depletion of the silver ion concentration in solution via reaction H-2-8 shifts reaction H-2-7 to the right, enhancing the solubility of silver chloride.

H-3 Fractional Crystallization. The different temperature profiles of the solubilities of two salts can be used to separate a mixture of them, as Example H-4 shows.

Example H-4. A mixture of potassium dichromate, K₂Cr₂O₇ (a beautiful orange salt) and sodium chloride, NaCl, a white crystalline salt, contains 15 g of each. Using the solubility curves in Figure H-3, devise a procedure to separate the two salts to give a reasonable amount of each in pure form.

Solution. The solubility curves for K₂Cr₂O₇ and NaCl are reproduced in expanded form in Figure H-4. We will use this more detailed figure as the basis for designing a separation procedure. Let’s get an overview first. Figure H-4 shows that the temperature dependence of the solubility is much flatter for NaCl than for K₂Cr₂O₇ . Further, the solubility curves cross at 57^oC. Thus NaCl is less soluble than K₂Cr₂O₇ above 57^oC, and more soluble below 57^oC. By adding an amount of water to the mixture sufficient to dissolve all of the sodium chloride even at 0^oC (this is 42 g of water, since the solubility of NaCl at 0^oC is 35.7 g per 100 g water, and we have 15 g NaCl), we should be able to isolate pure potassium dichromate by cooling the mixture to 0^oC and filtering. This will give a solid (pure K₂Cr₂O₇ ) and a solution containing all of the NaCl and a small amount of K₂Cr₂O₇ . To recover the NaCl, we could then heat the solution to 100^oC and boil off the water. When the amount of water is reduced to the point where the solution is saturated in NaCl, white NaCl will begin to crystallize from the boiling solution. We can remove water until the solution is almost saturated in K₂Cr₂O₇ . At this point we will have crystallized a reasonable amount of NaCl without also crystallizing K₂Cr₂O₇ . Filtration of the hot solution will give pure NaCl. The solution resulting from the filtration will be saturated in NaCl and unsaturated in K₂Cr₂O₇ , at 100^oC.

A detailed step-by-step procedure can now be developed.

1. Add 42 g (about 42 mL) of water to the 30 g of solid mixture, and stir. Assuming the termperature of the water to be 25^oC, all of the NaCl will dissolve, and some of the K₂Cr₂O₇ will dissolve (4.6 g, according to the solubility curve). We are at point A on the diagram with respect to K₂Cr₂O₇, and point A’ with respect to NaCl.

2. Before cooling the mixture to 0^oC, warm it first to a temperature sufficient to dissolve all of the dichromate, and filter it. This will remove any small amount of insoluble material that may contaminate the mixture, and will allow us to "start clean." An initial complete dissolution of the materials to be separated is almost always a good policy, unless one of the materials is completely insoluble. Thus the second step is to warm the mixture to a temperature higher than 53^oC (which is the temperature that will just dissolve all of the dichromate in 42 g of H₂O), say 70^oC, with stirring. In doing this, we follow path ABC for dichromate, and path A’C for NaCl. (In Figure H-4, solid lines are solubility curves, dashed lines are paths followed by solutions when these do not coincide with the solubility curves.) When all or most of the solid has dissolved, filter the warm mixture to remove insoluble contaminants. The filtrate should be bright orange in color, and clear.

3. Place the filtrate (the filtered solution) in an ice/water bath to cool to 0^oC. When the temperature reaches 53^oC, orange crystals of K₂Cr₂O₇ begin to form. Precipitation of K₂Cr₂O₇ continues as temperature drops toward 0^oC, until at 0^oC a total of 13 g will have precipitated. (The solubility of the dichromate is 5 g per 100 g of H₂O at 0^oC. In 42 g of H₂O, 2 g of dichromate will remain dissolved, and 15 - 2 = 13 g will precipitate.) In this step, we follow path CBE for dichromate, path CE’ for NaCl.

4. Filter the cold solution to obtain pure potassium dichromate as an orange crystalline solid. The filtrate will contain 15 g of NaCl and 2 g of K₂Cr₂O₇, so its color will still be orange, but a much less intense orange than was observed at 70^oC in step 2.

5. Heat the filtrate to 100^oC (the boiling point of water--actually, the boiling point of the solution will be a bit higher than this due to the boiling point elevation colligative property). The solution follows path EF for dichromate, path E’F’ for NaCl. Boil the solution to remove water. When 4.5 g of water have been removed, white NaCl will begin to precipitate. Continue boiling until 30 g of water have been removed. Sufficient water now remains to keep the 2 g of dichromate in solution as long as T stays above 30^oC. During boiling, we have followed F’G’ for NaCl, FG for dichromate.

6. Quickly filter the solution to isolate white NaCl as a pure solid. Ideally, we should obtain 10.2 g of solid. The filtrate should contain 2 g of dichromate and 4.8 g of NaCl. These concentrations correspond to G and G on the diagram.

7. At this point we have 13 g of pure dichromate (a yield of 13/15 x 100 = 87%) and 10.2 g of pure NaCl (a 68% yield). This is not bad for only one series of operations. If desired, we could work further on the filtrate from step 6 to obtain additional small amounts of pure salts, but in this case the effort is probably not justified by the small returns. There are cases in which the extra effort is justified.

The general procedure illustrated in this example is called fractional crystallization, because it allows separation--or fractionation--of the components of a mixture of solids, using the technique of crystallization. Fractional crystallization is used extensively by synthetic chemists (chemists whose interest is in making new compounds) to purify their creations.

H-4 Factors Determining Solubility

Molecular Structure.

The Effect of T Changes on Systems in Equilibrium. As we learned in earlier chapters, all natural processes are governed by two driving forces:

In particular, we saw that a balance of these two driving forces is responsible for the equilibria among phases of a pure substance. By again considering phase transformations for a moment, we can make an observation about the second driving force that is applicable to any process. Conversion of solid to gas via liquid is accompanied by an increase in enthalpy, H, and an increase in entropy, S, as summarized in H-4-1.

Thus the drive to minimum enthalpy favors the solid phase; the drive to maximum entropy favors the gas phase.

Our experience with phase changes allows us to add a further bit of information to the scheme; specifically, no matter what the pure substance, the solid phase is favored at low T, and the gas phase at high T. Lowering temperature shifts H-4-1 toward the solid phase. Lowering T enough causes the substance to exist exclusively in the solid phase. Raising T shifts H-4-1 toward the gas phase. Raising temperature enough causes the substance to exist exclusively in the gas phase. The information in H-4-1 suggests that as T is lowered from some arbitrary point, processes shift in the direction that decreases enthalpy; and as T is raised from some arbitrary point, processes shift in the direction that increases entropy. This is certainly true for phase changes, and a large body of experimental data indicates that it is true for all processes, physical and chemical, that occur in the natural universe. To emphasize the importance of this generalization, we restate it and present it in a more official way.

The theoretical foundation for this guideline was laid in Chapter 11. It is a summary of experimental observation. We now apply it to enhance our understanding of solubility.

Application of the Disorder Guideline to Solubility. Although it is possible to have seven types of solutions (gas, liquid, or solid solute dissolved in liquid or solid solvent; gas solute and solvent), only three types occur commonly in the laboratory and we will consider two of these. They are solutions of a gas in a liquid, and of a solid in a liquid. We will apply the disorder guideline to these two types of solutions to see how solute solubility should vary with temperature.

Gas in Liquid. An example of such a solution is aqueous ammonia, consisting of ammonia gas dissolved in water. We represent the formation of the solution in H-4-2.

Water is not explicitly written, but its presence is indicated by the symbol "aq". Consider that we have ammonia in contact with water in a closed container at a fixed temperature. Under these conditions, gaseous NH₃ exists in phase equilibrium with a saturated solution of NH₃ in water. A certain amount of NH₃ is dissolved in the water, and this amount is the solubility of NH₃ in water at the given T. How do we expect the solubility of NH₃ to be affected by an increase in T? Clearly, ammonia molecules have more freedom in the gas phase than when confined in the aqueous phase. Thus the left side of equilibrium H-4-2 has higher entropy than the right side. Applying the disorder guideline, we conclude that an increase in temperature shifts equilibrium H-4-2 to the left, toward the higher S side. Thus the solubility of NH₃, and of any gas, decreases with an increase in T. This is the major reason why thermal pollution of rivers by nuclear power plants is of such concern. The solubility of O₂ in water warmed by cooling tower effluent is decreased to an extent that is injurious to fish.

Solid in Liquid. In this text we have seen numerous examples of solutions of solids in liquids. Any soluble or slightly soluble salt produces such a solution when contacted with water. For example, the equilibrium between solid sodium chloride and its saturated water solution is represented in H-4-3.

In solid sodium chloride, the Na⁺ and Cl^- ions are tightly confined in a highly organized lattice. This is a situation of high order, hence low entropy. In contrast, in the solution, the Na⁺ and Cl^- ions are separated and free to roam over the entire volume of the liquid phase. In this situation, we expect the disorder to be high. The right side of equilibrium H-4-3 thus has the higher entropy. According to the disorder guideline, an increase in temperature should shift H-4-3 to the right. The solubility of NaCl(s) should increase with increasing temperature.

From an experimental standpoint, it is found that the solubilities of many common salts, including NaCl, increase with increasing temperature. This agrees with our simple analysis of the entropy change accompanying dissolution. However, the rate at which solubility increases with temperature varies markedly from salt to salt. This can be seen in Figure H-5, where the solubilities of several common salts in water are plotted against T. Some salts, such as CaCl₂ and KNO₃, have steep solubility-temperature profiles; i.e., solubility goes up fairly quickly with temperature. Other salts, such as NaCl, have flat profiles. Only a bit more NaCl dissolves at 80^oC than at 25^oC.

Although in many cases, the entropy of ions in a crystal is lower than their entropy in solution, there are cases in which the ions in solution have lower entropy than they do in the solid. For example, for the dissolution of BaSO₄(s) (equation H-4-4) the change in entropy is negative, in contrast to what we would predict.

The highly charged ions cause a very marked ordering of water molecules by exerting intermolecular forces of the ion-dipole type. This solvent ordering is so pronounced that it more than accounts for the loss of order when the crystal lattice breaks up. Although negative DS values for salt dissolutions are uncommon, they do occur, particularly in the case of highly charged ions. For these exceptions, our simple order-disorder considerations fail. The disorder guideline still works; but the entropies of solid and solution are counterintuitive.

Le Chatelier’s principle is useful for predicting the temperature dependence of solubility, just as it was useful in discussing the temperature dependence of phase equilibrium and chemical equilibrium. If dissolution of a salt is endothermic, as in H-4-5, we expect an increase in temperature to increase the salt solubility.

Similarly, a salt that dissolves exothermically should show decreased solubility at higher temperature. It is found experimentally that most salts dissolve endothermically, consistent with an increase in their solubilities with increasing temperature. For example, when 4.4 moles of water are added to 1 mole of ammonium nitrate, NH₄NO₃, enough heat is absorbed to lower the temperature of the solution from 25 to -11 ^oC.

The Effect of Pressure on Solubility. In the last section of this chapter, we consider the effect of pressure on the solubilities of gases and solids in a liquid solvent.

As we might expect from our work in Chapter 10, a change in pressure has little effect on the equilibrium established between a pure solid and a saturated liquid solution of the solid (H-4-6).

Both the solid and solution (liquid) phases are condensed, hence incompressible--their volumes are essentially unaffected by pressure. Unless there is a change in volume with pressure, there can be no change in concentration (the number of molecules per unit volume); and unless there is a concentration change in a phase, there can be no change in the rate at which molecules or ions leave the phase as long as T is constant (recall that rate processes are related directly to concentration). All phase equilibria are dynamic equilibria involving a balance of opposing rates. Unless one of the rates is affected by pressure, a change in P will have no effect on the position of equilibrium. Since neither the forward nor the reverse rate of an equilibrium involving two condensed phases is affected by pressure, this type of equilibrium is unperturbed when P is varied. The equilibrium between a solid and its saturated solution is of this type, where in this case the two balanced rates are the rate of dissolution and the rate of crystallization. The equilibrium between s and l phases of a pure substance is very similar. The temperature at which such an equilibrium exists is unaffected by even huge changes in P. Thus the solubility of a solid in a liquid should be independent of pressure.

In contrast, the solubility of a gas in a liquid is markedly affected by the pressure at which the gas is maintained over the liquid. This is a consequence of the effect of pressure on the volume, hence the concentration, of a gas. Consider the experiment shown in Figure H-6. We begin with a quantity of water at 25 ^oC in a cylinder fitted with a piston and a gas inlet (Figure H-6a). Initially, the space above the water is free of all gas except water vapor, which exerts the characteristic vapor pressure at 25 ^oC. Suppose that we now fill the space above the water with a gas, B(g), to some pressure P_B1 by opening the gas inlet valve, while holding the piston fixed in place, until the desired pressure is reached (Figure H-6b). As soon as the gas is admitted to the cylinder, gas molecules begin to strike the walls of the cylinder and the surface of the water with a frequency that depends on the concentration, hence pressure, of gas molecules in the space above the liquid. Some of the molecules that strike the water surface enter the liquid--i.e., they dissolve. The rate of dissolution is directly proportional to the pressure of the gas and to the surface area of the water, as indicated in H-4-7.

Here SA is the surface area of the water, and k₁ is a proportionality constant. Gas molecules that dissolve move randomly in the liquid phase, and at any given time some of them will be moving toward the surface. Those with sufficient energy will escape the liquid and reenter the gas phase. The rate of escape is directly proportional to the surface area of the liquid, and to the number of molecules travelling toward the surface, which in turn is directly related to the concentration of dissolved gas. Thus

where k₂ is a proportionality constant. Immediately after the gas is introduced to the cylinder, the rate of escape is 0 because there is no dissolved gas, but the rate of dissolution is greater than 0 because P > 0. The initial imbalance of rates leads to a buildup of dissolved molecules, an increase in [B], and a corresponding increase in Rate_escape, until Rate_dissolution = Rate_escape. At this point, dynamic equilibrium between the gas and solution phases has been established according to H-4-9.

where the double arrow shows the opposing equal rates.

Equating the right sides of H-4-7 and H-4-8 and rearranging gives

Thus equilibrium is characterized by a particular ratio of the concentration of dissolved B to the pressure of B above the solution.

Now consider quickly moving the piston down (Figure H-6c) so that the volume above the liquid is halved. This doubles the pressure of the gas above the solution, and immediately doubles R_dissolution (H-4-7) because the number of collisions of gas molecules with the surface is now twice as great. However, R_escape is initially unaffected. The immediate effect of the pressure increase is thus to make R_dissolution > R_escape. This imbalance results in a net flow of molecules from the gas to the liquid phase, causing a gradual buildup in [B], until the two opposing rates again become equal. The new equilibrium is established with higher values of both P and [B], but their ratio, given by k₁/k₂, is the same as before the pressure increase. Thus the concentration of dissolved gas--its solubility--is directly proportional to the pressure of gas above the solution. We express this in mathematical terms by writing H-4-10 somewhat differently.

Equation H-4-11, and the verbal statement that it summarizes, is known as Henry's Law. K is called the Henry's Law constant, but we recognize it as a typical equilibrium constant. The values of K for dissolution of several gases in water are given in Table H-1. For a gas pressure of 1 atm, these constants are numerically the same as the molar solubilities of the gases.

H-1. Use your favorite spreadsheet to calculate the solubility of CaF₂ in 1 * 10^-4 M HCl. Chapter 12 presents a discussion of the use of spreadsheets in equilibrium calculations.