The solubility of a substance is affected not only by temperature but also by the presence of other solutes. For example, the presence of an acid can have a major influence on the solubility of a substance. In the last section we considered the dissolving of ionic compounds in pure water. In this section we examine three factors that affect the solubility of ionic compounds: the presence of common ions, the pH of the solution, and the presence of complexing agents. We will also examine the phenomenon of amphoterism, which is related to the effects of both pH and complexing agents.
The presence of either Ca2+(aq) or F–(aq) in a solution reduces the solubility of CaF2, shifting the solubility equilibrium of CaF2 to the left:
This reduction in solubility is another application of the common-ion effect. In general, the solubility of a slightly soluble salt is decreased by the presence of a second solute that furnishes a common ion. Figure 17.15 shows how the solubility of CaF2 decreases as NaF is added to the solution. Sample Exercise 17.12 shows how the Ksp can be used to calculate the solubility of a slightly soluble salt in the presence of a common ion.
Figure 17.15 The effect of the concentration of NaF on the solubility of CaF2 demonstrates the common-ion effect. Notice that the solubility of CaF2 is on a logarithmic scale.
Calculate the molar solubility of CaF2 at 25°C in a solution that is (a) 0.010 M in Ca(NO3)2; (b) 0.010 M in NaF.
SOLUTION As in Sample Exercise 17.11, the solubility product at 25°C is
The value of Ksp is unchanged by the presence of additional solutes. Because of the common-ion effect, however, the solubility of the salt will decrease in the presence of common ions.
(a) We can again use our standard equilibrium techniques. In this instance, however, the initial concentration of Ca2+ is 0.010 M from the dissolved Ca(NO3)2:
Substituting into the solubility-product expression gives
This would be a messy problem to solve exactly, but fortunately it is possible to simplify matters greatly. Even without the common-ion effect, the solubility of CaF2 is very small. Assume that the 0.010 M concentration of Ca2+ from Ca(NO3)2 is very much greater than the small additional concentration resulting from the solubility of CaF2; that is, x is small compared to 0.010 M, and 0.010 + x 0.010. We then have
The very small value for x validates the simplifying assumption we have made. Our calculation indicates that 3.1 10–5 mol of solid CaF2 dissolves per liter of the 0.010 M Ca(NO3)2 solution.
(b) In this case the common ion is F–, and at equilibrium we have
Assuming that 2x is small compared to 0.010 M (that is, 0.010 + 2x 0.010), we have
Thus, 3.9 10–7 mol of solid CaF2 should dissolve per liter of 0.010 M NaF solution.
If you compare the results in parts (a) and (b), you will see that although both Ca2+ and F– reduce the solubility of CaF2, their effects are not the same. The effect of F– is more pronounced than that of Ca2+ because [F–] appears to the second power in the Ksp expression for CaF2, whereas [Ca2+] appears to the first power.
The value for Ksp for manganese(II) hydroxide, Mn(OH)2, is 1.6 10–13. Calculate the molar solubility of Mn(OH)2 in a solution that contains 0.020 M NaOH. Answer: 4.0 10–10 M
The solubility of any substance whose anion is basic will be affected to some extent by the pH of the solution. For example, consider Mg(OH)2, for which the solubility equilibrium is
A saturated solution of Mg(OH)2 has a calculated pH of 10.52 and contains [Mg2+] = 1.7 10–4 M. Now suppose that solid Mg(OH)2 is equilibrated with a solution buffered at a more acidic pH of 9.0. Then pOH is 5.0; that is, [OH–] = 1.0 10–5. Inserting this value for [OH–] into the solubility-product expression, we have
Thus, Mg(OH)2 dissolves in the solution until [Mg2+] = 0.18 M. It is apparent that Mg(OH)2 is quite soluble in this solution. If the concentration of OH– were reduced even further by making the solution more acidic, the Mg2+ concentration would have to increase to maintain the equilibrium condition. Thus, a sample of Mg(OH)2 will dissolve completely if sufficient acid is added (Figure 17.16).
The solubility of almost any ionic compound is affected if the solution is made sufficiently acidic or basic. The effects are very noticeable, however, only when one or both ions involved are at least moderately acidic or basic. The metal hydroxides, such as Mg(OH)2, are examples of compounds containing a strongly basic ion, the hydroxide ion. As we have seen, the solubility of Mg(OH)2 greatly increases as the acidity of the solution increases. As an additional example, the solubility of CaF2 increases as the solution becomes more acidic, because the F– ion is a weak base; it is the conjugate base of the weak acid HF. As a result, the solubility equilibrium of CaF2 is shifted to the right as the concentration of F– ions is reduced by protonation to form HF. Thus, the solution process can be understood in terms of two consecutive reactions:
The equation for the overall process is
Figure 17.17 shows how the solubility of CaF2 changes with pH.
Figure 17.17 The effect of pH on the solubility of CaF2. The pH scale is given with acidity increasing to the right. Notice that the vertical scale has been multiplied by 103.
Other salts that contain basic anions, such as CO32–, PO43–, CN–, or S2–, behave similarly. These examples illustrate a general rule: The solubility of slightly soluble salts containing basic anions increases as [H+] increases (as pH is lowered). The more basic the anion, the more the solubility is influenced by pH. Salts with anions of negligible basicity (the anions of strong acids) are unaffected by pH changes.
Which of the following substances will be more soluble in acidic solution than in basic solution: (a) Ni(OH)2(s); (b) CaCO3(s); (c) BaSO4(s); (d) AgCl(s)?
SOLUTION (a) We can conclude that Ni(OH)2(s) will be more soluble in acidic solution because of the basicity of OH–; the H+ ion reacts with the OH– ion, forming water:
(b) Similarly, CaCO3(s) dissolves in acid solutions because CO32– is a basic anion:
(c) The solubility of BaSO4 is largely unaffected by changes in pH because SO42– is a rather weak base and thus has little tendency to combine with a proton. However, BaSO4 is slightly more soluble in strongly acidic solutions.
(d) The solubility of AgCl is unaffected by changes in pH because Cl– is the anion of a strong acid and therefore has negligible basicity.
Write the net ionic equation for the reaction of the following copper(II) compounds with acid: (a) CuS; (b) Cu(N3)2. Answers: (a) CuS(s) + H+(aq) Cu2+(aq) + HS–(aq); (b) Cu(N3)2(s) + 2H+(aq) Cu2+(aq) + 2HN3(aq)
A characteristic property of metal ions is their ability to act as Lewis acids, or electron-pair acceptors, toward water molecules, which act as Lewis bases, or electron-pair donors. Lewis bases other than water can also interact with metal ions, particularly with transition-metal ions. Such interactions can have a dramatic effect on the solubility of a metal salt. For example, AgCl, whose Ksp = 1.8 10–10, will dissolve in the presence of aqueous ammonia because of the interaction between Ag+ and the Lewis base NH3, as shown in Figure 17.18. This process can be viewed as the sum of two reactions, the solubility equilibrium of AgCl and the Lewis acid-base interaction between Ag+ and NH3:
The presence of NH3 drives the top reaction, the solubility equilibrium of AgCl, to the right as Ag+(aq) is removed to form Ag(NH3)2+.
For a Lewis base such as NH3 to increase the solubility of a metal salt, it must be able to interact more strongly with the metal ion than water does. The NH3 must displace solvating H2O molecules (Sections 13.1 and 16.11) in order to form Ag(NH3)2+:
An assembly of a metal ion and the Lewis bases bonded to it, such as Ag(NH3)2+, is called a complex ion. The stability of a complex ion in aqueous solution can be judged by the size of the equilibrium constant for its formation from the hydrated metal ion. For example, the equilibrium constant for formation of Ag(NH3)2+ (Equation 17.24) is 1.7 107:
Such an equilibrium constant is called a formation constant, Kf. The formation constants for several complex ions are shown in Table 17.1.
Calculate the concentration of Ag+ present in solution at equilibrium when concentrated ammonia is added to a 0.010 M solution of AgNO3 to give an equilibrium concentration of [NH3] = 0.20 M. Neglect the small volume change that occurs on addition of NH3.
SOLUTION Because Kf is quite large, we begin with the assumption that essentially all of the Ag+ is converted to Ag(NH3)2+, in accordance with Equation 17.24. Thus, [Ag+] will be small at equilibrium. If [Ag+] is 0.010 M initially, then [Ag(NH3)2+] will be 0.010 M following addition of the NH3. Let the concentration of Ag+ at equilibrium be x. Then, at equilibrium
Because the concentration of Ag+ is very small, we can ignore x in comparison with 0.010. Thus, 0.010 - x 0.010 M. Substituting these values into the formation-constant expression, Equation 17.25, we obtain
Solving for x, we obtain x = 1.5 10–8 M = [Ag+]. It is evident that formation of the Ag(NH3)2+ complex drastically reduces the concentration of free Ag+ ion in solution.
Calculate [Cr3+] in equilibrium with Cr(OH)4– when 0.010 mol of Cr(NO3)3 is dissolved in a liter of solution buffered at pH 10.0. Answer: 1 10–16 M
The general rule is that the solubility of metal salts increases in the presence of suitable Lewis bases, such as NH3, CN–, or OH–, if the metal forms a complex with the base. The ability of metal ions to form complexes is an extremely important aspect of their chemistry. In Chapter 24 we will take a much closer look at complex ions. In that chapter and others we will see applications of complex ions to areas such as biochemistry, metallurgy, and photography.
Many metal hydroxides and oxides that are relatively insoluble in neutral water dissolve in strongly acidic and strongly basic solutions. These substances are soluble in strong acids and bases because they themselves are capable of behaving as either an acid or base; they are amphoteric. Examples of amphoteric substances include the hydroxides and oxides of Al3+, Cr3+, Zn2+, and Sn2+.
The dissolution of these species in acidic solutions should be anticipated based on the earlier discussions in this section. We have seen that acids promote the dissolving of compounds containing basic anions. What makes amphoteric oxides and hydroxides special is that they also dissolve in strongly basic solutions (Figure 17.19). This behavior results from the formation of complex anions containing several (typically four) hydroxides bound to the metal ion:
Amphoterism is often interpreted in terms of the behavior of the water molecules that surround the metal ion and that are bonded to it by Lewis acid-base interactions. For example, Al3+(aq) is more accurately represented as Al(H2O)63+(aq); six water molecules are bonded to the Al3+ in aqueous solution. As discussed in Section 16.11, this hydrated ion is a weak acid. As a strong base is added, Al(H2O)63+ loses protons in a stepwise fashion, eventually forming neutral and water-insoluble Al(H2O)3(OH)3. This substance then dissolves upon removal of an additional proton to form the anion Al(H2O)2(OH)4–. The reactions that occur are as follows:
Further proton removals are possible, but each successive reaction occurs less readily than the one before. As the charge on the ion becomes more negative, it becomes increasingly difficult to remove a positively charged proton. Addition of an acid reverses these reactions. The proton adds in a stepwise fashion to convert the OH– groups to H2O, eventually reforming Al(H2O)63+. The common practice is to simplify the equations for these reactions by excluding the bound H2O molecules. Thus, we usually write Al3+ instead of Al(H2O)63+, Al(OH)3 instead of Al(H2O)3(OH)3, Al(OH)4– instead of Al(H2O)2(OH)4–, and so forth.
The extent to which an insoluble metal hydroxide reacts with either acid or base varies with the particular metal ion involved. Many metal hydroxides—such as Ca(OH)2, Fe(OH)2, and Fe(OH)3—are capable of dissolving in acidic solution but do not react with excess base. These hydroxides are not amphoteric.
The purification of aluminum ore in the manufacture of aluminum metal provides an interesting application of the property of amphoterism. As we have seen, Al(OH)3 is amphoteric, whereas Fe(OH)3 is not. Aluminum occurs in large quantities as the ore bauxite, which is essentially Al2O3 with additional water molecules. The ore is contaminated with Fe2O3 as an impurity. When bauxite is added to a strongly basic solution, the Al2O3 dissolves because the aluminum forms complex ions, such as Al(OH)4–. The Fe2O3 impurity, however, is not amphoteric and remains as a solid. The solution is filtered, getting rid of the iron impurity. Aluminum hydroxide is then precipitated by addition of acid. The purified hydroxide receives further treatment and eventually yields aluminum metal.