Acid-Base Equilibria and Solubility

By Laird, B.B.

Edited by Paul Ducham


Our discussion of acid-base ionization and salt hydrolysis was limited to solutions containing a single solute. In this section, we will consider the acid-base properties of a solution with two dissolved solutes that contain the same ion (cation or anion), called the common ion.
The presence of a common ion suppresses the ionization of a weak acid or a weak base. If sodium acetate and acetic acid are dissolved in the same solution, for example, they both dissociate and ionize to produce CH3COO- ions:

CH3COONa is a strong electrolyte, so it dissociates completely in solution, but CH3COOH, a weak acid, ionizes only slightly. According to Le Châtelier’s principle, the addition of CH3COO- ions from CH3COONa to a solution of CH3COOH will suppress the ionization of CH3COOH (that is, shift the equilibrium from right to left), thereby decreasing the hydrogen ion concentration. Thus, a solution containing both CH3COOH and CH3COONa will be less acidic than a solution containing only CH3COOH at the same concentration. The shift in equilibrium of the acetic acidionization is caused by the acetate ions from the salt. CH3COO- is the common ion because it is supplied by both CH3COOH and CH3COONa.
The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. The common ion effect plays an important role in determining the pH of a solution and the solubility of a slightly soluble salt (to be discussed later in Section 12.7). Here we will study the common ion effect as it relates to the pH of a solution. Keep in mind that despite its distinctive name, the common ion effect is simply a special case of Le Châtelier’s principle.
Let us consider the pH of a solution containing a weak acid, HA, and a soluble salt of the weak acid, such as NaA. We start by writing
or simply
The ionization constant Ka was defined in Equation 11.10:

If the concentrations are sufficiently low, then the activities of solutes in Equation 11.10 can be replaced with molarities (without units) and the activity of pure water can be assumed to be unity, giving (Equation 11.11)

Rearranging this equation gives

Taking the negative logarithm (base 10) of both sides, we obtain

Using the definitions of pH (Equation 11.5) and pKa (Equation 11.15) gives (assuming dilute solution behavior)


Equation 12.1 is called the Henderson-Hasselbalch equation. A more general form of this expression is


In our example, HA is the acid and A- is the conjugate base. Thus, if we know Ka and the concentrations of the acid and the salt of the acid, we can calculate the pH of the solution.
It is important to remember that the Henderson-Hasselbalch equation is derived from the equilibrium constant expression. It is valid regardless of the source of the conjugate base (that is, whether it comes from the acid alone or is supplied by both the acid and its salt).
In problems that involve the common ion effect, we are usually given the starting concentrations of a weak acid HA and its salt, such as NaA. If the concentrations of these species are reasonably high (> 0.1 M), we can neglect the ionization of the acid and the hydrolysis of the salt. This is a valid approximation because HA is a weak acid and the extent of the hydrolysis of the A- ion is generally very small. Moreover, the presence of A- (from NaA) further suppresses the ionization of HA and the presence of HA further suppresses the hydrolysis of A-. Thus, in such cases, we can use the starting concentrations as the equilibrium concentrations in Equation 12.1.
In Example 12.1 we calculate the pH of a solution containing a common ion.
The common ion effect also operates in a solution containing a weak base, such as NH3, and a salt of the base, say NH4Cl. At equilibrium

We can derive the Henderson-Hasselbalch equation for this system as follows: Rear- ranging the preceding equation we obtain

Taking the negative logarithm of both sides gives


A solution containing both NH3 and its salt NH4Cl is less basic than a solution containing only NH3 at the same concentration. The common ion   NH4+ suppresses the ionization of NH3 in the solution containing both the base and the salt.

Example 12.1
(a) Calculate the pH of a solution containing 0.20 M CH3COOH and 0.30 M CH3COONa.
(b) What would the pH of a 0.20 M CH3COOH solution be if no salt were present?

Strategy (a) CH3COOH is a weak acid equation13 and CH3COONa is a soluble salt that is completely dissociated in solution  equation14The common ion here is the acetate ion (CH3COO-). At equilibrium, the major species in solution are CH3COOH, CH3COO-, Na+, H+, and H2O. The Na+ ion has no acid or base properties, and we ignore the ionization of water. Because Ka is an equilibrium constant, its value is the same whether we have just the acid or a mixture of the acid and its salt in solution. Therefore, we can calculate [H+] at equilibrium and hence pH if we know the equilibrium concentrations of both [CH3COOH] and [CH3COO]. For CH3COOH, Ka is 1.8 * 10-5 (see Table 11.3) giving a pKa of 4.74. (b) For a solution of CH3COOH, a weak acid, we calculate the [H+] at equilibrium, and hence the pH, using the initial concentration of CH3COOH. This problem can be solved by following the same procedure as in Example 11.11.

Solution (a) Sodium acetate is a strong electrolyte, so it dissociates completely in solution:
The initial concentrations, changes, and final concentrations of the species involved in the equilibrium are
Assuming that x is small, then 0.30 + x 
≈ 0.30 and 0.20 - x  0.20. The Henderson-Hasselbalch equation (Equation 12.1) then gives 
The value of x = [H+] corresponding to this pH is between 10-4 M and 10-5 M, which is significantly less than 5 percent of the initial concentrations of the acid or salt, so the approximation is valid.
(b) As in Example 11.11, x = [H
+] is determined from
In this case, x is less than 5 percent of the initial acid concentration, so the approximation is valid.

Check Comparing the results in parts (a) and (b), we see that when the common ion (CH3COO-) is present, according to Le Châtelier’s principle, the equilibrium shifts from right to left. This action decreases the extent of ionization of the weak acid. Consequently, fewer H+ ions are produced in (a) and the pH of the solution is higher than that in part (b). As always, you should check the validity of the assumptions.

Comment Note that, if the initial concentrations were small enough that the approximation used in part (a) were not valid, then the problem must be solved by substituting the concentrations [CH3COOH] = 0.20 -x, [CH3COO-] = 0.30 - x, and [H+] = x into the expression for Ka and solving the resulting quadratic equation for x.


A buffer solution is a solution of (1) a weak acid or a weak base and (2) its salt; both components must be present. The solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base. Buffers are very important to chemical and biological systems. The pH in the human body varies greatly from one fluid to another; for example, the pH of blood is about 7.4, whereas the gastric juice in our stomachs has a pH of about 1.5. These pH values, which are crucial for proper enzyme function and the balance of osmotic pressure, are maintained by buffers in most cases.
A buffer solution must contain a relatively large concentration of acid to react with any OHions that are added to it, and it must contain a similar concentration of base to react with any added H+ ions. Furthermore, the acid and the base components of the buffer must not consume each other in a neutralization reaction. These requirements are satisfied by an acid-base conjugate pair, for example, a weak acid and its conjugate base (supplied by a salt) or a weak base and its conjugate acid (supplied by a salt).
A simple buffer solution can be prepared by adding comparable molar amounts of acetic acid (CH3COOH) and its salt sodium acetate (CH3COONa) to water. The equilibrium concentrations of both the acid and the conjugate base (from CH3COONa) can generally be assumed to be the same as the starting concentrations because both the acid (CH3COOH) and its conjugate base (CH3COO-) are weak electrolytes. A solution containing these two substances has the ability to neutralize either added acid or added base. Sodium acetate, a strong electrolyte, dissociates completely in water:
If an acid is added, the H+ ions will be neutralized by the conjugate base in the buffer (CH3COO-) according to the equation
If a base is added to the buffer system, the OH- ions will be neutralized by the acid in the buffer:
As you can see, the two reactions that characterize this buffer system are identical to those for the common ion effect described in Example 12.1. The buffering capacity,  that is, the effectiveness of the buffer solution, depends on the amount of acid and conjugate base from which the buffer is made. The larger the amount, the greater the buffering capacity.
In general, a buffer system can be represented as salt-acid or conjugate base–acid. Thus, the sodium acetate–acetic acid buffer system discussed can be written as CH3COONa/CH3COOH or simply CH3COO-/CH3COOH. Figure 12.1 shows this buffer system in action.
Example 12.2 distinguishes buffer systems from acid-salt combinations that do not function as buffers.
The pH of a buffer solution is governed by the Henderson-Hasselbalch equation (Equation 12.2)
The form of this equation helps us understand the stability of a buffer solution to changes in pH from a mathematical perspective. In Equation 12.2, the logarithmic dependence of the pH on the ratio of conjugate base to acid means that in order for the pH to change by 1, the ratio of [conjugate base]/[acid] has to change by a factor of 10.
The effect of a buffer solution on pH is illustrated by Example 12.3.
In the buffer solution examined in Example 12.3, there is a decrease in pH (the solution becomes more acidic) as a result of the added HCl. We can also compare the changes in H+ ion concentration as follows
Thus, the H+ ion concentration increases by a factor of
To appreciate the effectiveness of the CH3COONa/CH3COOH buffer, let us find out what would happen if 0.10 mol HCl were added to 1.0 L of pure water (pH = 7), and compare the increase in H+ ion concentration.
As a result of the addition of HCl, the H+ ion concentration increases by a factor of 106, amounting to a million-fold increase! This comparison shows that a properly chosen buffer solution can maintain a fairly constant H+ ion concentration, or pH (Figure 12.2).

Example 12.2
Which of the following solutions can be classified as buffer systems: (a) KH2PO4/ H3PO4, (b) NaClO4/HClO4, and (c) C5H5N/C5H5NHCl (C5H5N is pyridine; its Kb is given in Table 11.4)? Explain your answer.

Strategy What constitutes a buffer system? Which of the preceding solutions contains a weak acid and its salt (containing the weak conjugate base)? Which of the preceding solutions contains a weak base and its salt (containing the weak conjugate acid)? Why is the conjugate base of a strong acid not able to neutralize an added acid?

Solution The criteria for a buffer system are that we must have a weak acid and its salt (containing the weak conjugate base) or a weak base and its salt (containing the weak conjugate acid).
(a) H3PO4 is a weak acid, and its conjugate base, H2PO4-, is a weak base (see Table 11.6). Therefore, this is a buffer system.
(b) Because HClO4 is a strong acid, its conjugate base, ClO4-, is an extremely weak base. This means that the ClO4- ion will not combine with an H+ ion in solution to form HClO4. Thus, the system cannot act as a buffer system.
(c) As Table 11.4 shows, C5H5N is a weak base and its conjugate acid, C5H5NH+ (the cation of the salt C5H5NHCl), is a weak acid. Therefore, this is a buffer system.

Example 12.3
(a) Calculate the pH of a buffer system containing 1.0 M CH3COOH and 1.0 M CH3COONa. (b) What is the pH of the buffer system after the addition of 0.10 mole of gaseous HCl to 1.0 L of the solution? Assume that the volume of the solution does not change when the HCl is added.

Strategy The pH of the buffer system before and after the addition of HCl can be calculated from the Henderson-Hasselbalch equation. The Ka of CH3COOH is 1.8 * 10-5 (see Table 11.3) which gives a pKa of 4.74.

Solution (a) Both the weak acid (CH3COOH) and its conjugate base (CH3COO- from CH3COONa) are weak electrolytes and will not dissociate appreciably [CH3COOH] ≈ [CH3COOH]0 = 1.0 M and [CH3COO-≈ [CH3COO-]0 = 1.0 M.(We must check this approximation at the end of the calculation.) The Henderson- Hasselbalch equation (Equation 12.2) gives
This corresponds to an H
+ ion concentration of between 10-4 M and 10-5 M, so the approximation is valid.
(b) When HCl (a strong acid) is added to the solution, it completely dissociates into H
+ ions and Cl- ions. The initial changes are
The H
+ ions are neutralized by the conjugate base (CH3COO-), and Cl- is a spectator ion in solution because it is the conjugate base of a strong acid and will have no tendency to neutralize the weak acid present. At this point, it is more convenient to work with moles rather than molarity. The reason is that in some cases the volume of the solution may change when a substance is added. A change in volume will change the molarity, but not the number of moles. The neutralization reaction is summarized next:
Finally, to calculate the pH of the buffer after neutralization of the acid, we convert back to molarity by dividing moles by 1.0 L of solution. Thus, the weak acid concentration, [CH
3COOH] after addition of HCl is 1.10 M and the conjugate base concentration [CH3COO-] = 0.90 M. Reapplying the Henderson-Hasselbalch equation gives




Now suppose we want to prepare a buffer solution with a specific pH. How do we go about it? Equation 12.2 indicates that if the molar concentrations of the acid and its conjugate base are approximately equal, that is, if [acid]  [conjugate base], then 

Thus, to prepare a buffer solution, we work backward. First we choose a weak acid whose pKa is close to the desired pH. Next, we substitute the pH and pKa values in Equation 12.2 to obtain the ratio [conjugate base]/[acid]. This ratio can then be converted to molar quantities for the preparation of the buffer solution. Example 12.4 shows this approach.

Example 12.4
Describe how you would prepare a “phosphate buffer” with a pH of about 7.40.

Strategy For a buffer to function effectively, the concentrations of the acid component must be roughly equal to the conjugate base component. According to Equation 12.2, when the desired pH is close to the pKa of the acid, pH  pKa,

Solution Because phosphoric acid is a triprotic acid, we write the three stages of ionization as follows. The Ka and pKa values are obtained from Table 11.6, and the pKa values are obtained by applying Equation 11.15
The most suitable of the three buffer systems is HPO
4-/H2PO4- because the pKa of the acid H2PO4- is closest to the desired pH. From the Henderson-Hasselbalch equation (Equation 12.2), we write
Thus, one way to prepare a phosphate buffer with a pH of 7.40 is to dissolve disodium hydrogen phosphate (Na
2HPO4) and sodium dihydrogen phosphate (NaH2PO4) in a mole ratio of 1.5:1.0 in water. For example, we could dissolve 1.5 moles of Na2HPO4 and 1.0 mole of NaH2PO4 in enough water to make up a 1-L solution.


Quantitative studies of acid-base neutralization reactions are most conveniently carried out using a technique known as titration. In a titration, a solution of accurately known concentration, called a standard solution, is added gradually to another solution of unknown concentration, until the chemical reaction between the two solutions is complete. If we know the volumes of the standard and unknown solutions used in the titration, along with the concentration of the standard solution, we can calculate the concentration of the unknown solution.
Sodium hydroxide is one of the bases commonly used in the laboratory. However, it is difficult to obtain solid sodium hydroxide in a pure form because it has a tendency to absorb water from air, and its solution reacts with carbon dioxide. For these reasons, a solution of sodium hydroxide must be standardized before it can be used in accurate analytical work. We can standardize the sodium hydroxide solution by titrating it against an acid solution of accurately known concentration. The acid often chosen for this task is a weak acid called potassium hydrogen phthalate (KHP), for which the molecular formula is KHC8H4O4. KHP is a white, soluble solid that is commercially available in highly pure form. The reaction between KHP and sodium hydroxide is
and the net ionic equation is
The apparatus for the titration is shown in Figure 12.3. First, a known amount of KHP is transferred to an Erlenmeyer flask, and some distilled water is added to make up a solution. Next, an NaOH solution is carefully added to the KHP solution from a buret until we reach the equivalence point, that is, the point at which the acid has completely reacted with or been neutralized by the base. The equivalence point is usually signaled by a sharp change in the color of an indicator in the acid solution. In acidbase titrations, indicators are substances that have distinctly different colors in acidic and basic media. One commonly used indicator is phenolphthalein, which is colorless in acidic and neutral solutions but reddish pink in basic solutions. Indicators will be discussed in more detail in Section 12.4. At the equivalence point, all the KHP present has been neutralized by the added NaOH and the solution is still colorless. However, if we add just one more drop of NaOH solution from the buret, the solution will immediately turn pink because the solution is now basic.
Having discussed buffer solutions, we can now look in more detail at the quantitative aspects of acid-base titrations. We will consider three types of reactions: (1) titrations involving a strong acid and a strong base, (2) titrations involving a weak acid and a strong base, and (3) titrations involving a strong acid and a weak base. Titrations involving a weak acid and a weak base are complicated by the hydrolysis of both the cation and the anion of the salt formed. These titrations will not be dealt with in this text. Figure 12.4 shows the arrangement for monitoring the pH during the course of a titration.




The reaction between a strong acid (say, HCl) and a strong base (say, NaOH) can be represented by
or in terms of the net ionic equation
Consider the addition of a 0.100 M NaOH solution (from a buret) to an Erlenmeyer flask containing 25.0 mL of 0.100 M HCl. Figure 12.5 shows the pH profile of the titration (also known as the titration curve). Before the addition of NaOH, the pH of the acid is given (approximately) by -log (0.100), or 1.00. When NaOH is added, the pH of the solution increases slowly at first. Near the equivalence point, the pH begins to rise steeply, and at the equivalence point (that is, the point at which equimolar amounts of acid and base have reacted) the curve rises almost vertically. In a strong acid–strong base titration, both the hydrogen ion and hydroxide ion concentrations are very small at the equivalence point (approximately 1 * 10-7 M); consequently, the addition of a single drop of the base can cause a large increase in [OH-] and in the pH of the solution. Beyond the equivalence point, the pH again increases slowly with the addition of NaOH.
It is possible to calculate the pH of the solution at every stage of titration. Here are three sample calculations.

1. After the addition of 10.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCl. The total volume of the solution is 35.0 mL. The number of moles of NaOH in 10.0 mL is
Thus, the amount of HCl left after partial neutralization is (2.50 * 10-3) - (1.00 3*10-3), or 1.50 * 10-3 mol. Next, the concentration of H+ ions in 35.0 mL of solution is found as follows:
Thus, [H+] = 0.0429 M, and the pH of the solution is
pH   -log10 0.0429 = 1.37

2. After the addition of 25.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCl. This is a simple calculation, because it involves a complete neutralization reaction and the salt (NaCl) does not undergo hydrolysis. At the equivalence point
and the pH of the solution is 7.00.

3. After the addition of 35.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCl. The total volume of the solution is now 60.0 mL. The number of moles of NaOH added is
The number of moles of HCl in 25.0 mL solution is 2.50 * 10-3 mol. After complete neutralization of HCl, the number of moles of NaOH left is (3.50 * 10-3) - (2.50 * 10-3), or 1.00 * 10-3 mol. The concentration of OH- in 60.0 mL of solution is
Thus, [OH-] = 0.0167 M and pOH = -log10 0.0167 = 1.78. Hence, the pH of the solution is
pH = 14.00 - pOH = 14.00 - 1.78 = 12.22



Consider the neutralization reaction between acetic acid (a weak acid) and sodium hydroxide (a strong base):
This equation can be simplified to
The acetate ion undergoes hydrolysis as follows:
Therefore, at the equivalence point, when we only have sodium acetate present, the pH will be greater than 7 as a result of the excess OH- ions formed (Figure 12.6). Note that this situation is analogous to the hydrolysis of sodium acetate (CH3COONa).
Example 12.5 deals with the titration of a weak acid with a strong base.

Example 12.5
Calculate the pH in the titration of 25.0 mL of 0.100 M acetic acid by sodium hydroxide after the addition to the acid solution of (a) 10.0 mL of 0.100 M NaOH, (b) 25.0 mL of 0.100 M NaOH, or (c) 35.0 mL of 0.100 M NaOH.

Strategy The reaction between CH3COOH and NaOH is
We see that 1 mol CH3COOH  1 mol NaOH. Therefore, at every stage of the titration, we can calculate the number of moles of base reacting with the acid, and the pH of the solution is determined by the excess acid or base left over. At the equivalence point, however, the neutralization is complete, and the pH of the solution will depend on the extent of the hydrolysis of the salt formed, which is CH3COONa.

Solution (a) The number of moles of NaOH in 10.0 mL is
The number of moles of CH3COOH originally present in 25.0 mL of solution is
We work with moles at this point because when two solutions are mixed, the solution volume increases. As the volume increases, molarity will change, but the number of moles will remain the same. The changes in the number of moles are summarized next:
At this stage, we have a buffer system made up of CH3COOH and CH3COO- (from the salt, CH3COONa). To calculate the pH of the solution, we can use the Henderson-Hasselbalch equation (Equation 12.2):

(b) These quantities (that is, 25.0 mL of 0.100 M NaOH reacting with 25.0 mL of 0.100 M CH3COOH) correspond to the equivalence point. The number of moles of NaOH in 25.0 mL of the solution is
The changes in number of moles are summarized next:
At the equivalence point, the concentrations of both the acid and the base are zero. The total volume is (25.0 + 25.0) mL or 50.0 mL, so the concentration of the salt is
The next step is to calculate the pH of the solution that results from the hydrolysis of the CH3COO- ions:
Following the procedure described in Example 11.17 and looking up the base ionization constant (Kb) for CH3COO- in Table 11.5, we write

(c) After the addition of 35.0 mL of NaOH, the solution is well past the equivalence point. The number of moles of NaOH originally present is 
The changes in number of moles are summarized next:
At this stage, we have two species in solution that are responsible for making the solution basic: OH- and CH3COO- (from CH3COONa). However, because OH- is a much stronger base than CH3COO-, we can safely neglect the hydrolysis of the CH3COO- ions and calculate the pH of the solution using only the concentration of the OH- ions. The total volume of the combined solutions is (25.0 + 35.0) mL or 60.0 mL, so we calculate OH- concentration as follows:



Consider the titration of HCl, a strong acid, with NH3, a weak base:
or simply
The pH at the equivalence point is less than 7 due to the hydrolysis of the NH4+ion:
or simply
Because of the volatility of an aqueous ammonia solution, it is more convenient to add hydrochloric acid from a buret to the ammonia solution. Figure 12.7 shows the titration curve for this experiment

Example 12.6
Calculate the pH at the equivalence point when 25.0 mL of 0.100 M NH3 is titrated by a 0.100 M HCl solution.

Strategy The reaction between NH3 and HCl is
We see that 1 mol NH3  1 mol HCl. At the equivalence point, the major species in solution are the salt NH4Cl (dissociated into NH4+ and Cl- ions) and H2O. First we determine the concentration of NH4Cl formed. Then we calculate the pH as a result of the NH4+ ion hydrolysis. The Cl-ion, being the conjugate base of a strong acid HCl, does not react with water. As usual, we ignore the ionization of water.

Solution The number of moles of NH3 in 25.0 mL of 0.100 M solution is
At the equivalence point, the number of moles of HCl added equals the number of moles of NH
3. The changes in the number of moles are summarized here:
At the equivalence point, the concentrations of both the acid and the base are zero. The total volume is (25.0 + 25.0) mL, or 50.0 mL, so the concentration of the salt is
The pH of the solution at the equivalence point is determined by the hydrolysis of NH
4+ ions
Check Note that the pH of the solution is acidic. This is what we would expect from the hydrolysis of the ammonium ion.



The equivalence point, as we have seen, is the point at which the number of moles of OH- ions added to a solution is equal to the number of moles of H+ ions originally present. To determine the equivalence point in a titration, then, we must know exactly how much volume of a base has been added from a buret to an acid in a flask. One way to achieve this goal is to add a few drops of an acid-base indicator to the acid solution at the start of the titration. You will recall from Section 12.3 that an indicator is used to determine the equivalence point of an acid-base titration. An indicator is a substance (usually a weak organic acid or base) that has distinctly different colors in its nonionized and ionized forms. Which of these two forms (ionized or nonionized) is dominant depends upon the pH of the solution in which the indicator is dissolved. The end point of a titration occurs when the indicator changes color. However, not all indicators change color at the same pH, so the choice of indicator for a particular titration depends on the nature of the acid and base used in the titration (that is, whether they are strong or weak). By choosing the proper indicator for a titration, we can use the end point to determine the equivalence point, as we will see in the following.
Let us consider a weak monoprotic acid that we will call HIn. To be an effective indicator, HIn and its conjugate base, In-, must have distinctly different colors. In solution, the acid ionizes to a small extent:
If the indicator is in a sufficiently acidic medium, the equilibrium, according to Le Châtelier’s principle, shifts to the left and the predominant color of the indicator is that of the nonionized form (HIn). On the other hand, in a basic medium, the equilibrium shifts to the right, and the color of the solution will be due mainly to that of the conjugate base (In-). Roughly speaking, we can use the following concentration ratios to predict the perceived color of the indicator:
If [HIn]  [In-], then the indicator color is a combination of the colors of HIn and In-.
The end point of an indicator does not occur at a specific pH; rather, there is a range of pH within which the end point will occur. In practice, we choose an indicator whose end point lies on the steep part of the titration curve, the position of which depends upon the strength of the acid or base being titrated. Because the equivalence point also lies on the steep part of the curve, this choice ensures that the pH at the equivalence point will fall within the range over which the indicator changes color. In Section 12.3, we mentioned that phenolphthalein is a suitable indicator for the titration of NaOH and HCl. Phenolphthalein is colorless in acidic and neutral solutions, but reddish pink in basic solutions. Measurements show that at pH = 8.3 the indicator is colorless but that it begins to turn reddish pink when the pH exceeds 8.3. As shown in Figure 12.5, the steepness of the pH curve near the equivalence point means that the addition of a very small quantity of NaOH (say, 0.05 mL, which isabout the volume of a drop from the buret) brings about a large rise in the pH of the solution. What is important, however, is the fact that the steep portion of the pH profile includes the range over which phenolphthalein changes from colorless to reddish pink. Whenever such a correspondence occurs, the indicator can be used to locate the equivalence point of the titration (Figure 12.8).
Many acid-base indicators are plant pigments. For example, by boiling chopped red cabbage in water, we can extract pigments that exhibit many different colors at various pHs (Figure 12.9).
Table 12.1 lists a number of indicators commonly used in acid-base titrations.
The choice of a particular indicator depends on the strength of the acid and base to be titrated. Example 12.7 illustrates this point.

Example 12.7
Which indicator or indicators listed in Table 12.1 would you use for the acid-base titrations shown in (a) Figure 12.5, (b) Figure 12.6, and (c) Figure 12.7?

Strategy The choice of an indicator for a particular titration is based on the fact that its pH range for color change must overlap the steep portion of the titration curve. Otherwise, we cannot use the color change to locate the equivalence point.

Solution (a) Near the equivalence point, the pH of the solution changes abruptly from 4 to 10. Therefore all the indicators except thymol blue, bromophenol blue, and methyl orange are suitable for use in the titration.
(b) Here the steep portion covers the pH range between 7 and 10; therefore, the suitable indicators are cresol red and phenolphthalein.
(c) Here the steep portion of the pH curve covers the pH range between 3 and 7; therefore, the suitable indicators are bromophenol blue, methyl orange, methyl red, and chlorophenol blue.





How can we predict whether a precipitate will form when a compound is added to a solution or when two solutions are mixed? It depends on the solubility of the solute, which was defined in Section 9.2 as the maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature and pressure. All ionic compounds are strong electrolytes, but they are not equally soluble. Table 12.2 classifies a number of common ionic compounds as soluble or insoluble. Keep in mind, however, that even insoluble compounds dissolve to a certain extent. Figure 12.11 shows several precipitates.
Example 12.8 applies the solubility rules in Table 12.2.

Example 12.8
Classify the following ionic compounds as soluble or insoluble: (a) silver sulfate (Ag2SO4), (b) calcium carbonate (CaCO3), and (c) sodium phosphate (Na3PO4).

Strategy Although it is not necessary to memorize the solubilities of compounds, you should keep in mind the following useful rules: all ionic compounds containing alkali metal cations; the ammonium ion; and the nitrate, bicarbonate, and chlorate ions are soluble. For other compounds, we need to refer to Table 12.2.

Solution (a) According to Table 12.2, Ag2SO4 is insoluble.
(b) This is a carbonate and Ca is a Group 2A metal. Therefore, CaCO3 is insoluble.
(c) Sodium is an alkali metal (Group 1A) so Na3PO4 is soluble.




Consider a saturated solution of silver chloride that is in contact with solid silver chloride. The solubility equilibrium can be represented as
Because salts such as AgCl are considered as strong electrolytes, all the AgCl that dissolves in water is assumed to dissociate completely into Ag+ and Cl- ions. Thus, we can write the equilibrium constant for the dissolution of AgCl in terms of the activities of the reactants and products (see Section 10.1):
where Ksp is called the solubility product constant or simply the solubility product. The activity of a pure solid in a reaction can be assumed to a very good approximation to be equal to 1. Also, if the concentrations of solutes are small enough that solute-solute interactions are negligible, then the solution can be considered ideal and we can approximate the activities of those solutes by their concentration relative to the standard concentration (1 mol L-1). With these approximations, Equation 12.3 becomes
(The concentrations in Equation 12.4 are dimensionless concentrations obtained by dividing the concentrations by 1 mol L-1.) In general, the solubility product of a compound can be approximated as the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient in the equilibrium equation. Keep in mind, however, that for solubility equilibria in which the concentration of ions is high (>0.001 M for singly charged ions, much less for multiply charged ions), the solubilities calculated using Equation 12.4 will differ somewhat from experimentally measured solubilities. It is possible to correct Equation 12.4 by taking into account the deviation of the activity from the molar concentration. 
Because each AgCl unit contains only one Ag+ ion and one Cl- ion, its solubility product expression is particularly simple to write. The following cases are slightly more complex:
Table 12.3 lists the solubility products for a number of salts of low solubility. 
Soluble salts such as NaCl and KNO3, which have very large Ksp values, are not listed in the table for essentially the same reason that Ka values for strong acids are not often reported. The value of Ksp indicates the solubility of an ionic compound—the smaller the value, the less soluble the compound in water. However, in using Ksp values to compare solubilities, you should choose compounds that have similar formulas, such as AgCl and ZnS, or CaF2 and Fe(OH)2.
Another factor, in addition to deviations from ideal solution behavior, that can affect the use of Ksp in determining solubility is the fact that many anions in the ionic compounds listed in Table 12.3 are conjugate bases of weak acids. Consider copper sulfide (CuS). The S2- ion can hydrolyze as follows
And highly charged small metal ions such as Al3+ and Bi3+ will undergo hydrolysis as discussed in Section 11.6. In such cases, it is necessary in accurate work to include the hydrolysis equilibrium expressions in addition to the solubility product to determine the equilibrium concentrations of all species in aqueous solution.
For concentrations of ions that do not correspond to equilibrium conditions, we use the reaction quotient (see Section 10.2), which in this case is called the ion product (Q), to predict whether a precipitate will form. Note that Q has the same form as Ksp except that the concentrations of ions are not equilibrium concentrations. For example, if we mix a solution containing Ag+ ions with one containing Cl- ions, then the ion product, assuming ideal solution behavior, is given by
The subscript 0 reminds us that these are initial concentrations and do not necessarily correspond to those at equilibrium. The possible relationships between Q and Ksp are



There are two other ways to express a substance’s solubility: molar solubility, which is the number of moles of solute in 1 L of a saturated solution (mol L-1), and solubility, which is the number of grams of solute in 1 L of a saturated solution (g L-1). Note that both these expressions refer to the concentration of saturated solutions at some given temperature (usually 25° C).
Both molar solubility and solubility are convenient to use in the laboratory. We can use them to determine Ksp by following the steps outlined in Figure 12.12(a).
Example 12.9 illustrates this procedure.
Sometimes we are given the value of K
sp for a compound and asked to calculate the compound’s molar solubility. For example, the Ksp for the dissolution of silver bromide
is 7.7 * 10
-13. We can calculate its molar solubility by the procedure outlined in Figure 12.12(b). First we identify the species present at equilibrium. Here we have Ag+ and Br- ions. Let s be the molar solubility (in mol L-1) of AgBr. Because one unit of AgBr yields one Ag1+and one Br- ion, at equilibrium, both [Ag+] and [Br-] are equal to s. Assuming ideal solution behavior, the solubility product expression for this reaction is given by
Thus, the molar solubility of AgBr also is 8.8 * 10
-7 M. This concentration is quite small, so the effect of nonideality will be minimal and the use of concentrations in the expression for Ksp is accurate. Example 12.10 makes use of this approach.
As Examples 12.9 and 12.10 show, solubility and solubility product are related. If we know one, we can calculate the other, but each quantity provides different information. Table 12.4 shows the relationship between molar solubility and solubility product for a number of ionic compounds, assuming ideal solution behavior.





From knowledge of the solubility rules (see Section 12.5) and the solubility products listed in Table 12.3, we can predict whether a precipitate will form when we mix two solutions or add a soluble compound to a solution. This ability often has practical value.
In industrial and laboratory preparations, we can adjust the concentrations of ions until the ion product exceeds Ksp in order to obtain a given compound (in the form of a precipitate). The ability to predict precipitation reactions is also useful in medicine. For example, kidney stones, which can be extremely painful, consist largely of calcium oxalate (CaC2O4, Ksp = 2.3 * 10-9). The normal physiological concentration of calcium ions in blood plasma is about 5 mM (1 mM = 1 * 10-3 M). Oxalate ions (C2O42-), derived from oxalic acid present in many vegetables such as rhubarb and spinach, react with the calcium ions to form insoluble calcium oxalate, which can gradually build up in the kidneys. Proper adjustment of a patient’s diet can help to reduce precipitate formation. Example 12.11 illustrates the steps involved in predicting precipitation reactions.




In chemical analysis, it is sometimes desirable to remove one type of ion from solution by precipitation while leaving other ions in solution. For instance, the addition of sulfate ions to a solution containing both potassium and barium ions causes BaSO4 to precipitate out, thereby removing most of the Ba2 ions from the solution. The other “product,” K2SO4, is soluble and will remain in solution. The BaSO4 precipitate can be separated from the solution by filtration.
Even when both  products are insoluble, we can still achieve some degree of separation by choosing the proper reagent to bring about precipitation. Consider a solution that contains Cl-, Br-, and I- ions. One way to separate these ions is to convert them to insoluble silver halides. As the Ksp values in Table 12.3 show, the solubility of the halides decreases from AgCl to AgI. Thus, when a soluble compound such as silver nitrate is slowly added to this solution, AgI begins to precipitate first, followed by AgBr and then AgCl.
Example 12.12 describes the separation of only two ions (Cl- and Br-), but the procedure can be applied to a solution containing more than two different types of ions if precipitates of differing solubility can be formed.
Example 12.12 raises the question: What is the concentration of Br- ions remaining in solution just before AgCl begins to precipitate? To answer this question, we let [Ag+] = 8.0 * 10-9 M. Then
Thus, (100 - 0.48) percent, or 99.52 percent, of Br- will have precipitated as AgBr just before AgCl begins to precipitate. By this procedure, the Br- ions can be quantitatively separated from the Cl- ions.



In Section 12.1, we discussed the effect of a common ion on acid and base ionizations. Here we will examine the relationship between the common ion effect and solubility. As we have noted, the solubility product is an equilibrium constant; precipitation of an ionic compound from solution occurs whenever the ion product exceeds Ksp for that substance. In a saturated solution of AgCl, for example, the ion product [Ag+][Cl-] is, of course, equal to Ksp. Furthermore, simple stoichiometry tells us that [Ag+] = [Cl-]. But this equality does not hold in all situations.
Suppose we study a solution containing two dissolved substances that share a common ion, say, AgCl and AgNO3. In addition to the dissociation of AgCl, the total concentration of the common silver ions in solution is also affected by the dissolution of silver nitrate:
To reestablish equilibrium, some AgCl will precipitate out of the solution, as Le Châtelier’s principle would predict, until the ion product is once again equal to Ksp. The effect of adding a common ion, then, is a decrease in the solubility of the salt (AgCl) in solution. Note that in this case [Ag+] is no longer equal to [Cl-] at equilibrium; rather, [Ag+[Cl-]. 
Example 12.13 shows the common ion effect on solubility.



The solubilities of many substances also depend on the pH of the solution. Consider the solubility equilibrium of magnesium hydroxide:
Adding OH- ions (increasing the pH) shifts the equilibrium from right to left, thereby decreasing the solubility of Mg(OH)2. (This is another example of the common ion effect.) On the other hand, adding H+ ions (decreasing the pH) shifts the equilibrium from left to right, and the solubility of Mg(OH)2 increases. Thus, insoluble bases tend to dissolve in acidic solutions. Similarly, insoluble acids tend to dissolve in basic solutions.
To explore the quantitative effect of pH on the solubility of Mg(OH)
2, let us first calculate the pH of a saturated Mg(OH)2 solution. We write
In a medium with a pH of less than 10.45, the solubility of Mg(OH)
2 would increase. This follows from the fact that a lower pH indicates a higher [H+] and thus a lower [OH-], as we would expect from Kw = [H+][OH-]. Consequently, [Mg2+] rises to maintain the a saturated solution, and more Mg(OH)2 dissolves. The dissolution process and the effect of extra H+ ions can be summarized as follows:
If the pH of the medium were higher than 10.45, [OH
-] would be higher and the solubility of Mg(OH)2 would decrease because of the common ion (OH-) effect.
The pH also influences the solubility of salts that contain a basic anion. For example, the solubility equilibrium for BaF
2 is
As [F
-] decreases, [Ba2+] must increase to maintain the equilibrium condition. Thus, more BaF2 dissolves. The dissolution process and the effect of pH on the solubility of BaF2 can be summarized as follows:
The solubilities of salts containing anions that do not hydrolyze are unaffected by pH. Examples of such anions are Cl
-, Br-, and I-.
Examples 12.14 and 12.15 deal with the effect of pH on solubility.




Lewis acid-base reactions in which a metal cation combines with a Lewis base result in the formation of complex ions. Thus, we can define a complex ion is an ion containing a central metal cation bonded to one or more molecules or ions. Complex ions are crucial to many chemical and biological processes. Here we will consider the effect of complex ion formation on solubility.
Transition metals have a particular tendency to form complex ions because they have more than one possible valence state. This property allows them to act effectively as Lewis acids in reactions with many molecules or ions that serve as electron donors (Lewis bases). For example, a solution of cobalt(II) chloride is pink because of the presence of the Co(H2O)62+ ions (Figure 12.13).
When HCl is added, the solution turns blue as a result of the formation of the complex ion CoCl42-:
Copper(II) sulfate (CuSO4) dissolves in water to produce a blue solution. The hydrated copper(II) ions are responsible for this color; many other sulfates (Na2SO4, for example) are colorless. Adding a few drops of concentrated ammonia solution to a CuSO4 solution causes the formation of a light-blue precipitate, copper(II) hydroxide:
The OH- ions are supplied by the ammonia solution. If more NH3 is added, the blue precipitate redissolves to produce a beautiful dark-blue solution, this time due to the formation of the complex ion Cu(NH3)42+ (Figure 12.14):
Thus, the formation of the complex ion Cu(NH3)42+ increases the solubility of Cu(OH)2.
A measure of the tendency of a metal ion to form a particular complex ion is given by the formation constant K
f  (also called the stability constant), which is the equilibrium constant for the complex ion formation. The larger the value of Kf, the more stable the complex ion. Table 12.5 lists the formation constants of a number of complex ions.
The very large value of K
f in this case indicates that the complex ion is quite stable in solution and accounts for the very low concentration of copper(II) ions at equilibrium.
The use of formation constants to determine aqueous equilibrium in solutions that exhibit complex ion formation is illustrated in Example 12.16.
The effect of complex ion formation generally is to increase the solubility of a substance, as Example 12.17 shows.
Finally, we note that there is a class of hydroxides, called amphoteric hydroxides, that can react with both acids and bases. Examples are Al(OH)
3, Pb(OH)2, Cr(OH)3, Zn(OH)2, and Cd(OH)2. Thus, Al(OH)3 reacts with acids and bases as follows:
The increase in solubility of Al(OH)
3 in a basic medium is the result of the formation of the complex ion Al(OH)4- in which Al(OH)3 acts as the Lewis acid and OH- acts as the Lewis base. Other amphoteric hydroxides behave in a similar manner.








Here we will briefly discuss qualitative analysis, the determination of the types of ions present in a solution. We will focus on the cations. There are some 20 common cations that can be analyzed readily in aqueous solution. These cations can be divided into five groups according to the solubility products of their insoluble salts (Table 12.6). Because an unknown solution may contain from 1 to all 20 ions, any analysis must be carried out systematically from group 1 through group 5. Let us consider the general procedure for separating these 20 ions by adding precipitating reagents to an unknown solution. 

  • Group 1 Cations. When dilute HCl is added to the unknown solution, only the Ag, Hg22+, and Pb2+ ions precipitate as insoluble chlorides. The other ions, whose chlorides are soluble, remain in solution. 
  • Group 2 Cations. After the chloride precipitates have been removed by filtration, hydrogen sulfide is reacted with the unknown acidic solution. Under this condition, the concentration of the S2- ion in solution is negligible. Therefore, the precipitation of metal sulfides is best represented as
    Adding acid to the solution shifts this equilibrium to the left so that only the least soluble metal sulfides, that is, those with the smallest Ksp values, will precipitate out of solution. These are Bi2S3, CdS, CuS, HgS, and SnS (see Table 12.6). 
  • Group 3 Cations. At this stage, sodium hydroxide is added to the solution to make it basic. In a basic solution, the equilibrium between M2+ and MS shifts to the right. Therefore, the more soluble sulfides (CoS, FeS, MnS, NiS, and ZnS) now precipitate out of solution. Note that the Al3+ and Cr3+ ions actually precipitate as the hydroxides Al(OH)3 and Cr(OH)3, rather than as the sulfides, because the hydroxides are less soluble. The solution is then filtered to remove the insoluble sulfides and hydroxides.
  • Group 4 Cations. After all the group 1, 2, and 3 cations have been removed from solution, sodium carbonate is added to the basic solution to precipitate Ba2+, Ca2+, and Sr2+ ions as BaCO3, CaCO3, and SrCO3. These precipitates too are removed from solution by filtration.
  • Group 5 Cations. At this stage, the only cations possibly remaining in solution are Na+, K+, and NH4+. The presence of NH4+ can be determined by adding sodium hydroxide:
    The ammonia gas is detected either by noting its characteristic odor or by observing a piece of wet red litmus paper turning blue when placed above (not in contact with) the solution. To confirm the presence of Na
    + and K+ ions, we usually use a flame test, as follows: A piece of platinum wire (chosen because platinum is inert) is moistened with the solution and is then held over a Bunsen burner flame. Each type of metal ion gives a characteristic color when heated in this manner. For example, the color emitted by Na+ ions is yellow, that of K+ ions is violet, and that of Cu2+ ions is green (Figure 12.16).
    Figure 12.17 summarizes this scheme for separating metal ions. Two points regarding qualitative analysis must be mentioned. First, the separation of the cations into groups is made as selective as possible, that is, the anions that are added as reagents must be such that they will precipitate the fewest types of cations. For example, all the cations in group 1 also form insoluble sulfides. Thus, if H
    2S were reacted with the solution at the start, as many as seven different sulfides might precipitate out of solution (group 1 and group 2 sulfides), an undesirable outcome. Second, the removal of cations at each step must be carried out as completely as possible. For example, if we do not add enough HCl to the unknown solution to remove all the group 1 cations, they will precipitate with the group 2 cations as insoluble sulfides, interfering with further chemical analysis and leading to erroneous conclusions.




Graphs of Exponential Functions

The graphs of for and 5 are shown in Figure 2. Note that all three have the same basic shape, and pass through the point (0, 1). Also, the x axis is a horizontal asymptote for each graph, but only as The main difference between the graphs is their steepness. Next, let’s look at the graphs of for and (Fig. 3). Again, all three have the same basic shape, pass through (0, 1), and have horizontal asymptote but we can see that for the asymptote is only as In general, for bases less than 1, the graph is a reflection through the y axis of the graphs for bases greater than 1. The graphs in Figures 2 and 3 suggest that the graphs of exponential functions have the properties listed in Theorem 1, which we state without proof. x S . b 6 1, y 0, 1 5 b 1 2 , 1 3 , y b x x S . b 2, 3, y b x These properties indicate that the graphs of exponential functions are distinct from the graphs we have already studied. (Actually, property 4 is enough to ensure that graphs of exponential functions are different from graphs of polynomials and rational functions.) Property 6 is important because it guarantees that exponential functions have inverses. Those inverses, called logarithmic functions, are the subject of Section 5-3.