A. Equilibria of Sparingly Soluble Substances
When a sparingly soluble salt (the solubility of salts was discussed in Section 8.2D2) is added to water, a small amount dissolves until the solution is saturated at that temperature. An equilibrium is then present between the undissolved ions of the solid and those ions in solution. For example, when a saturated solution of calcium sulfate is prepared, the undissolved solute is in equilibrium with dissolved ions:

CaSO₄(s) Ca²⁺ + SO₄^2-

The equilibrium constant for this equation is

K_sp = [Ca²⁺][SO₄^2-] = 2.45 X 10^-5

Notice that solid calcium sulfate is not included in the equilibrium constant expression. The reason for this omission was given in Section 13.4C. To emphasize this omission, as well as the fact that this constant is for an ion product rather than a true equilibrium constant expression, the constant is known as the solubility product constant, K_sp. If the solubility of a substance is known, its solubility product constant can be calculated.

 

Example: The solubility of silver chloride is 6.56x10-4 g/L. Calculate the solubility product constant for silver chloride. Solution For silver chloride, Ksp = [Ag+][Cl-], where the brackets mean concentration in mol/L. For every mole of silver chloride dissolved, one mole of silver ion and one mole of chloride ion are found in solution. Therefore, the first step in solving the problem is to convert the solutility from g/L to mol/L. The formula weight of silver chloride is 107.8 + 35.4 = 143.3. The solubility of AgCl is 6.56x10-4g    1 L x   1 mol 143.3 g = 4.58 x 10-6 mol/L Therefore, [Ag+] = [Cl-] = 4.58 x 10-6 mol/L. Substituting this value into the solubility product expression gives: Ksp= [Ag+][Cl-] = (4.58x10-6)(4.58x10-6)       = 20.9x10-12 = 2.09x10-11

If the solubility product constant of a sparingly soluble salt is known, its solubility can be calculated. Table 13.3 lists the solubility product constants of several sparingly soluble salts.

Table 13.3 Solubility product constants

Sparingly soluble salt	Equilibrium	Ksp	pKsp
silver bromide	AgBr(s) Ag+ + Br-	5.2 X 10-13	12.3
calcium carbonate	CaCO3 Ca2+ + CO32-	4.7 X 10-9	8.3
lead(II) chloride	PbCl2 Pb2+ + Cl-	1.6 X 10-5	4.8
aluminum hydroxide	Al(OH)3 Al3+ + 3 OH-	5.0 X 10-33	32.3
calcium fluoride	CaF2 Ca2+ + 2F-	1.7 X 10-10	9.7
strontium sulfate	SrSO4 Sr2+ + SO42-	7.6 X 10-7	6.1
copper(II) sulfide	CuS Cu2+ + S2-	8.0 X 10-45	44.1

In the preceding exercises, the anion and cation of the salt were in a 1:1 ratio. If the ions of the salt are in some other ratio, such as the 1:2 ratio in lead(II) chloride, the concentrations of the ions are not equal. In a solution that contains only lead(II) chloride, the concentration of chloride ion is twice that of the lead ion. The equilibrium is

PbCl₂(s) Pb²⁺ + 2 Cl^-

and the solubility product constant expression is

K_sp = [Pb²⁺][Cl^-]²

The value of the solubility product constant of a salt depends on the solubility of the salt. Its usefulness is shown by the fact that any sparingly soluble salt will precipitate from a solution in which the product of the concentration of its ions, each raised to the power shown in the solubility product constant expression, is greater than the solubility product constant. For example, the solubility product constant of silver bromide is 5.2 X 10^-13. Whenever [Ag+][Br^-] is greater than 5.2 X 10^-13 in a solution, silver bromide will precipitate. The two concentrations need not be equal. Silver bromide will precipitate when

[Ag⁺] = 0.1 M, [Br^-] = 10^-11 M   because [Ag⁺][Br^-] = 10^-12

[Ag+] = [Br^-] = 7.4 X 10^-7 M    because [Ag⁺][Br^-] = 5.4 X 10^-13

[Ag⁺] = 10^-10 M, [Br^-] = 6 X 10^-2 M   because [Ag⁺][Br^-] = 6 X 10^-12

In each case, [Ag⁺][Br^-] is greater than 5.2 X 10^-13. In a saturated solution of silver bromide,

[Ag⁺][Br^-] = 5.2 X 10^-13

and

AgBr(s) Ag⁺Br^-

If more bromide ions are added to the solution, the equilibrium will shift to the left, forming more precipitate and decreasing the amount of silver ion in solution. Conversely, if bromide ions are removed from solution, the precipitate will redissolve until the product of [Ag⁺] and [Br^-] again equals 5.2 X 10^-13.

 

Example: A 1-L sample of solution contains 3.6x10-6 mol calcium ion. The equilibrium is CaC2O4(s) Ca2+ + C2O42-    Ksp = 1.8x10-9 Will calcium oxylate precipitae if 0.10 mol oxalate ion is added to the solution? Solution For this equilibrium Ksp = 1.8x10-9 In this solution, [Ca2+] = 3.6x10-6 M and [C2O42-] = 0.10 M. Therefore [Ca2+] [C2O42-] = (3.6x10-6)(0.10) =3.7x10-7 Because 3.7x10-7 > 1.8x10-9, calcium oxylate will precipitate.

Example: Solid lead(II) nitrate is added to 0.10 M sodium chloride. At what concentration of lead ion will lead chloride start to precipitate? Solution From Table 13.3, we know that the solubility product constant of lead chloride is Ksp = [Pb2+][Cl-]2 = 1.6x10-5 The [Cl-] is 0.10 M; the [Pb2+] is unknown. Substituting and rearranging, [Pb2+](0.10 M)2 = 1.6x10-5 [Pb2+] = 1.6x10-5 (0.10)2 = 1.6x10-3 M As soon as [Pb2+] = 1.6x10-3 M, lead(II) chloride will start to precipitate.

B. Equilibria of Weak Acids; Buffer Solutions
In Chapter 12 we discussed the equilibria between the molecules and ions of weak acids and showed how such equilibria respond to the stress of changing concentrations. Another interesting and important example of how such equilibria respond to concentration changes comes from the study of buffers. A buffer is a system that resists change. A chemical buffer system is one that resists change in hydrogen ion concentration (pH). Buffers are important because many chemical reactions, particularly those in biological systems, proceed best at a particular pH. If the reaction takes place in a solution that remains at that pH throughout the reaction, the most satisfactory results will be obtained.

A chemical buffer system that is designed to resist changes in hydrogen ion concentration will contain a weak acid and its conjugate base, both in such concentrations as will give the solution the desired pH. For example, if a buffer of pH 4.74 is needed, a solution that is 0.1 M in both acetic acid and acetate ion would be suitable (see Section 12.7A for the calculations that assign this pH to this solution). This buffer of pH 4.74 contains the following equilibrium:

HC2H3O2 0.1 M

H+ 1.8 X 10-5 M

+

C2H3O2- 0.1 M

Notice that the concentrations of both the acid and its conjugate base are much larger than that of the hydrogen ion. From our study of equilibria, we know that, if an acid (H⁺) is added to this solution, the equilibrium will shift to the left because the reverse reaction consumes the added hydrogen ions. If, on the other hand, hydroxide ion or some other strong base is added to the solution, that base will react with the hydrogen ions present; more acid will ionize to replenish the supply and keep the system in equilibrium. In the buffer we describe, if these added amounts are small, the concentrations of the acetic acid molecules and of acetate ion are sufficiently large that the pH of the solution will remain essentially constant. The calculations in the following example illustrate this point.

Example: Calculate the following: a. The pH of a buffer solution, 1 L of which contains 0.10 mol acetic acid and 0.10 mol acetate ion. The Ka of acetic acid is 1.8x10-5. b. The pH of 1.0 L of the buffer solution in part a after the addition of 0.010 mol hydrochloric acid to the solution. c. The pH of 1.0 L of the buffer solution in part a after 0.010 mol hydroxide ion as NaOH has been added. Solution The equilibrium involved is HC2H3O2 H+ + C2H3O2- The original concentrations are [HC2H3O2] = 0.10 M and [C2H3O2-] = 0.10 M a. We can calculate the pH of this solution using the Ka for acetic acid: Ka = [H+][C2H3O2-] [HC2H3O2] =1.8x10-5 Rearranging this equation gives [H+] = Ka[HC2H3O2] [C2H3O2-]   Solving for [H+] and substituting: [H+] = (1.8x10-5)(0.10 M)         0.10 M = 1.8x10-5 For this concentration of hydrogen ions, pH = 4.74 b. Before the addition of HCl, 1.0 L of the solution contains 0.10 mol each of acetate ion and acetic acid: [C2H3O2-] = 0.10 M   and   [HC2H3O2] = 0.10 M When 0.010 mol hydrochloric acid is added to the solution, 0.010 mol H+ is added to the solution. The added hydrogen ions combine with acetate ions to form more un-ionized molecules of acetic acid. The concentration of acetate ion decreases, and the concentration of acetic acid in the solution increases: [C2H3O2-] = 0.10 M - 0.010 M = 0.09 M [HC2H3O2] = 0.10 M + 0.010 M = 0.11 M To determine the hydrogen ion concentration in this solution, the same equality is used as was used for the original solution. Substituting these values into that equality: [H+] = (1.8x10-5)[HC2H3O2]       [C2H3O2-] [H+] = (1.8X10-5)(0.11 M) 0.09 M = 2.2x10-5 The addition to the buffer of 0.010 mol hydrogen ions as hydrochloric acid changed the pH to pH = -log[H+] = -log(2.2x10-5) = 4.66 a change of 0.08 pH units. c. Before addition of NaOH, 1 L of the solution contained 0.10 mol acetic acid and 0.10 mol acetate ion. The added NaOH (0.010 mol) reacts with the acetic acid to form 0.010 mol sodium acetate. This reaction increases the concentration of acetate by 0.010 mol and decreases the concentration of acetic acid by the same amount. [C2H3O2-] = 0.10 M + 0.010 M = 0.11 M [HC2H3O2] = 0.10 M - 0.010 M = 0.09 M For this solution [H+] = (1.8X10-5)(0.09 M) 0.11 M = 1.47x10-5 pH = -log[H+] = -log(1.47x10-5) = 4.83 a change of 0.09 pH units from the original solution.

FIGURE 13.10 pH versus concentration for a series of acetic acid-acetate ion solutions.

Figure 13.10 shows a plot of pH versus concentration of acetate ion for the buffer solution whose pH was calculated in Example 13.12. The graph shows a change of only two pH units as the acetate ion concentration changes from 0.02 M to 0.18 M. Within this range the solution acts as a buffer.

The addition of small amounts of either hydrogen or hydroxide ions to a buffer does not appreciably change the pH. Example 13.13 illustrates how different the results would be if, instead of using a buffer based on acetic acid, we used a solution of a strong acid with the same pH.

Example: A solution contains the same number of moles of HCl and NaCl and has a pH of 4.74. Calculate the final pH of 1.0 L of this solution after the addition of 0.01 mol of hydrogen ion.: Solution Because HCl is a strong acid, it is completely ionized. Cl- has no tendency to combine with H+ to form HCl molecules in water. At pH 4.74, the concentration of hydrogen ion is 1.8x10-6 M. Any addition of H+ will simply increase this concentration. The final [H+] of the solution will be [H+] = 0.01 M + 0.000018 M = =.010018 M and pH = 2 Thus, the addition of a small amount of H+ to a very dilute solution of a strong acid causes a large change in pH.

In choosing the buffer for a particular reaction, the following criteria should be considered.

1. pH. Buffering is most effective if the pH required for the reaction is close to the pK_a of the weak acid of the buffer. An effective buffer has a pH that equals the pK_a of the weak acid ±1 unit.

   For acetic acid, the pK_a is 4.74; therefore, a solution of acetic acid and sodium acetate will function as a buffer within a pH range of approximately 3.74-5.74.

2. Concentration. The concentration of a buffer refers to the total concentration of both the weak acid and the anion of the weak acid. A 0.1 M acetic acid-acetate ion buffer may be made up of 0.025 mol acetic acid and 0.075 mol sodium acetate in a liter of solution, or any other combination in which [HC₂H₃O₂] + [C₂H₃O₂^-] = 0.1 M. The buffer used in Figure 13.10 has a concentration of 0.2 M. At each point on the graph, the sum of the concentrations of acetic acid and acetate ion equals 0.2 M. The concentration of the buffer should be greater than the amount of H⁺ or OH^- that will be produced by the reaction being buffered.

3. Capacity. The capacity of a buffer, or its effectiveness, is the amount of hydrogen or hydroxide ion that the buffer can absorb without a significant change in pH. Capacity depends on concentration and pH. The most effective buffer is one with equal concentrations of a weak acid and its salt - that is, one in which the pH of the solution is equal to the pK_a of the weak acid. Although a solution of acetic acid and sodium acetate will buffer within the pH range 3.74-5.74, it is most effective as a buffer at, or very near, pH 4.74. Table 13.4 lists some common buffers and the pH at which each is most effective.