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Unformatted text preview: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website CHAPTER 16 Acid-Base Equilibria and
IN THIS CHAPTER WE WILL CONTINUE OUR STUDY OF ACID-BASE REAC- TIONS WITH A DISCUSSION OF BUFFER ACTION AND TITRATIONS. 16.1 HOMOGENEOUS VERSUS
EQUILIBRIA WE WILL ALSO LOOK AT ANOTHER TYPE OF AQUEOUS EQUILIBRIUM — THE ONE 16.2 THE COMMON ION EFFECT BETWEEN A SLIGHTLY SOLUBLE COMPOUND AND ITS IONS IN SOLUTION. 16.3 BUFFER SOLUTIONS
16.4 ACID-BASE TITRATIONS
16.5 ACID-BASE INDICATORS
16.6 SOLUBILITY EQUILIBRIA
16.7 SEPARATION OF IONS BY FRACTIONAL
16.8 THE COMMON ION EFFECT AND
16.9 pH AND SOLUBILITY
16.10 COMPLEX ION EQUILIBRIA AND
16.11 APPLICATION OF THE SOLUBILITY
PRODUCT PRINCIPLE TO QUALITATIVE
ANALYSIS 645 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 646 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 16.1 HOMOGENEOUS VERSUS HETEROGENEOUS SOLUTION EQUILIBRIA In Chapter 15 we saw that weak acids and weak bases never ionize completely in water. Thus, at equilibrium a weak acid solution, for example, contains nonionized acid
as well as H ions and the conjugate base. Nevertheless, all of these species are dissolved so the system is an example of homogeneous equilibrium (see Chapter 14).
Another type of equilibrium, which we will consider in the second half of this
chapter, involves the dissolution and precipitation of slightly soluble substances. These
processes are examples of heterogeneous equilibria — that is, they pertain to reactions
in which the components are in more than one phase. 16.2 THE COMMON ION EFFECT Our discussion of acid-base ionization and salt hydrolysis in Chapter 15 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 both sodium acetate and acetic acid are dissolved in the same solution, for example, they both dissociate and ionize to produce CH3COO ions:
H2O CH3COONa(s) 88n CH3COO (aq) Na (aq) CH3COOH(aq) 34 CH3COO (aq) H (aq) CH3COONa is a strong electrolyte, so it dissociates completely in solution, but
CH3COOH, a weak acid, ionizes only slightly. According to Le Chatelier’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 acid ionization 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 this chapter). 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 Chatelier’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
HA(aq) H2O(l ) 34 H3O (aq) A (aq) or simply
HA(aq) 34 H (aq) A (aq) The ionization constant Ka is given by
Ka Back Forward Main Menu TOC [H ][A ]
[HA] Study Guide TOC Textbook Website (16.1) MHHE Website 16.2 THE COMMON ION EFFECT 647 Rearranging Equation (16.1) gives
[A ] [H ] Taking the negative logarithm of both sides, we obtain
log [H ] log Ka log [HA]
[A ] log [H ] log Ka log [A ]
[HA] or So
pH pKa log [A ]
[HA] (16.2) where
pKa is related to Ka as pH is
related to [H ]. Remember that
the stronger the acid (that is,
the larger the Ka), the
smaller the pKa. pKa log Ka (16.3) Equation (16.3) is called the Henderson-Hasselbalch equation. A more general form
of this expression is
pH pKa log [conjugate base]
[acid] (16.4) 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
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. As long as 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 we can use
the starting concentrations as the equilibrium concentrations in Equation (16.1) or
In the following example we calculate the pH of a solution containing a common
EXAMPLE 16.1 (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
Answer (a) Sodium acetate is a strong electrolyte, so it dissociates completely in solution: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 648 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA H2O CH3COONa(s) 88n CH3COO (aq)
0.30 M Na (aq)
0.30 M As stated above, we can use the starting concentrations as the equilibrium concentrations; that is
[CH3COOH] 0.20 M and [CH3COO ] 0.30 M From Equation (16.1) we have
[H ][CH3COO ]
[CH3COOH] Ka or
[H ] Ka[CH3COOH]
(1.8 10 5)(0.20)
0.30 pH log [H ] 1.2 10 5 M Thus
10 5) log (1.2
4.92 Alternatively, we can calculate the pH of the solution by using the HendersonHasselbalch equation. In this case we need to calculate pKa of the acid first [see
pKa log Ka
10 5) log (1.8
4.74 We can calculate the pH of the solution by substituting the value of pKa and the
concentrations of the acid and its conjugate base in Equation (16.4):
pKa log [CH3COO ]
[CH3COOH] 4.74 log 0.30 M
0.20 M 4.74 pH 0.18 4.92 (b) Following the procedure in Example 15.8, we find that the pH of a 0.30 M
CH3COOH solution is:
pH 1.9 10 log (1.9 3 M
10 3) 2.72 Thus, without the common ion effect, the pH of a 0.20 M CH3COOH
solution is 2.72, considerably lower than 4.92, the pH in the presence of CH3COONa,
as calculated in (a). The presence of the common ion CH3COO clearly suppresses
the ionization of the acid CH3COOH.
Comment Similar problem: 16.3. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 16.3 BUFFER SOLUTIONS 649 PRACTICE EXERCISE What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK?
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
NH4 (aq) 34 NH3(aq) H (aq) [NH3][H ]
[NH4 ] Ka We can derive the Henderson-Hasselbalch equation for this system as follows.
Rearranging the above equation we obtain
Ka[N H4 ]
[N H3] [H ] Taking the negative logarithm of both sides gives
log [H ] log Ka log [NH4 ]
[NH3] log [H ] log Ka log [ N H 3]
[NH4 ] or
pH pKa log [ N H 3]
[NH4 ] 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. 16.3 BUFFER SOLUTIONS 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
A buffer solution must contain a relatively large concentration of acid to react
with any OH ions 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 simple buffer solution can be prepared by adding comparable amounts of acetic
acid (CH3COOH) and its salt sodium acetate (CH3COONa) to water. The equilibrium Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 650 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA FIGURE 16.1 The acid-base indicator bromophenol blue (added
to all solutions shown) is used to
illustrate buffer action. The indicator’s color is blue-purple above
pH 4.6 and yellow below pH
3.0. (a) A buffer solution made
up of 50 mL of 0.1 M
CH3COOH and 50 mL of 0.1 M
CH3COONa. The solution has a
pH of 4.7 and turns the indicator
blue-purple. (b) After the addition
of 40 mL of 0.1 M HCl solution
to the solution in (a), the color remains blue-purple. (c) A 100-mL
CH3COOH solution whose pH is
4.7. (d) After the addition of 6
drops (about 0.3 mL) of 0.1 M
HCl solution, the color turns yellow. Without buffer action, the
pH of the solution decreases
rapidly to less than 3.0 upon the
addition of 0.1 M HCl. (a) (b) (c) (d) concentrations of both the acid and the conjugate base (from CH3COONa) are assumed
to be the same as the starting concentrations (see p. 647). 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:
H2O CH3COONa(s) 88n CH3COO (aq) Na (aq) If an acid is added, the H ions will be consumed by the conjugate base in the buffer,
CH3COO , according to the equation
CH3COO (aq) H (aq) 88n CH3COOH(aq) If a base is added to the buffer system, the OH ions will be neutralized by the acid
in the buffer:
CH3COOH(aq) OH (aq) 88n CH3COO (aq) H2O(l ) As you can see, the two reactions that characterize this buffer system are identical to
those for the common ion effect described in Example 16.1. 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 above can be written as
CH3COONa/CH3COOH or simply CH3COO /CH3COOH. Figure 16.1 shows this
buffer system in action.
The following example distinguishes buffer systems from acid-salt combinations
that do not function as buffers.
EXAMPLE 16.2 Which of the following solutions are buffer systems? (a) KH2PO4/H3PO4, (b)
NaClO4/HClO4, (c) C5H5N/C5H5NHCl (C5H5N is pyridine; its Kb is given in Table
15.4). Explain your answer.
Answer (a) H3PO4 is a weak acid, and its conjugate base, H2PO4 , is a weak base
(see Table 15.5). 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 a H ion in solution to
form HClO4. Thus the system cannot act as a buffer system. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 16.3 Similar problem: 16.8. BUFFER SOLUTIONS 651 (c) As Table 15.4 shows, C5H5N is a weak base and its conjugate acid, C5H5N H
(the cation of the salt C5H5NHCl), is a weak acid. Therefore, this is a buffer system.
PRACTICE EXERCISE Which of the following are buffer systems? (a) KF/HF, (b) KBr/HBr, (c)
The effect of a buffer solution on pH is illustrated by the following example.
EXAMPLE 16.3 (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 L of the solution? Assume that the volume of the solution does
not change when the HCl is added.
(a) The pH of the buffer system before the addition of HCl can be calculated according to the procedure described in Example 16.1. Assuming that ionization of the acetic acid and hydrolysis of the acetate ions are negligible, at equilibrium we have Answer [CH3COOH]
[H ] 1.0 M and [CH3COO ] [H ][CH3COO ]
[CH3COOH] 10 5 Ka[CH3COOH]
(1.8 10 5)(1.0)
pH 1.8 1.0 M 10 log (1.8 5 M
10 5) 4.74 Thus when the concentrations of the acid and the conjugate base are the same, the
pH of the buffer is equal to the pKa of the acid.
(b) After the addition of HCl, complete ionization of HCl acid occurs:
HCl(aq) 88n H (aq)
0.10 mol Cl (aq)
0.10 mol Originally, 1.0 mol CH3COOH and 1.0 mol CH3COO were present in 1 L of the
solution. After neutralization of the HCl acid by CH3COO , which we write as
0.10 mol H (aq) 88n CH3COOH(aq)
0.10 mol the number of moles of acetic acid and the number of moles of acetate ions present
CH3COOH: (1.0 0.1) mol 1.1 mol CH3COO : (1.0 0.1) mol 0.90 mol Next we calculate the hydrogen ion concentration: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 652 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA Ka[CH3COOH]
[CH3COO ] [H ] 10 5)(1.1)
2.2 5 10 M The pH of the solution becomes
pH 10 5) log (2.2
4.66 Note that since the volume of the solution is the same for both species,
we replaced the ratio of their molar concentrations with the ratio of the number of
moles present; that is, (1.1 mol/L)/(0.90 mol/L) (1.1 mol/0.90 mol). Comment Similar problem: 16.15. PRACTICE EXERCISE Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH
after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution?
In the buffer solution examined in Example 16.3 there is a decrease in pH (the
solution becomes more acidic) as a result of added HCl. We can also compare the
changes in H ion concentration as follows
Before addition of HCl:
After addition of HCl: [H ] 1.8 10 5 M [H ] 2.2 10 5 M Thus the H ion concentration increases by a factor of
M 1.2 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 L of water, and compare the
increase in H ion concentration.
Before addition of HCl:
After addition of HCl: [H ] 1.0 [H ] 7 10 0.10 M M As a result of the addition of HCl, the H ion concentration increases by a factor of
1.0 0.10 M
10 7 M 1.0 106 amounting to a millionfold increase! This comparison shows that a properly chosen
buffer solution can maintain a fairly constant H ion concentration, or pH.
PREPARING A BUFFER SOLUTION WITH A SPECIFIC pH Now suppose we want to prepare a buffer solution with a specific pH. How do we go
about it? Equation (16.1) indicates that if the molar concentrations of the acid and its
conjugate base are approximately equal, that is, if [acid] ≈ [conjugate base], then
log [conjugate base]
pH ≈ pKa Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 16.4 ACID-BASE TITRATIONS 653 Thus, to prepare a buffer solution, we work backwards. First we choose a weak acid
whose pKa is close to the desired pH. Next, we substitute the pH and pKa values in
Equation (16.1) to obtain the ratio [conjugate base]/[acid]. This ratio can then be converted to molar quantities for the preparation of the buffer solution. The following example shows this approach.
EXAMPLE 16.4 Describe how you would prepare a “phosphate buffer” with a pH of about 7.40.
We write three stages of ionization of phosphoric acid as follows. The Ka
values are obtained from Table 15.5 and the pKa values are found by applying
Answer H3PO4(aq) 34 H (aq) H2PO4 (aq) Ka1 7.5 10 3; pKa1 2.12 H2PO4 (aq) 34 H (aq) HPO2 (aq)
4 Ka2 6.2 10 8; pKa2 7.21 4.8 13 HPO2
4 (aq) 34 H (aq) Ka3 10 ; pKa3 12.32 HPO 2
4 The most suitable of the three buffer systems is
/ H2PO4 , because the pKa
of the acid H2PO4 is closest to the desired pH. From the Henderson-Hasselbalch
equation we write
pH log [conjugate base]
7.21 log [HPO2 ]
[H2PO4 ] [HPO2 ]
[H2PO4 ] 0.19 Taking the antilog, we obtain
[H2PO4 ] Similar problems: 16.17, 16.18. 1.5 Thus, one way to prepare a phosphate buffer with a pH of 7.40 is to dissolve disodium hydrogen phosphate (Na2HPO4) 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
PRACTICE EXERCISE How would you prepare a liter of “carbonate buffer” at a pH of 10.10? You are provided with carbonic acid (H2CO3), sodium hydrogen carbonate (NaHCO3), and
sodium carbonate (Na2CO3).
The Chemistry in Action essay on p. 663 illustrates the importance of buffer systems in the human body. 16.4 ACID-BASE TITRATIONS Having discussed buffer solutions, we can now look in more detail at the quantitative
aspects of acid-base titrations, which we discussed briefly in Section 4.6. Recall that
titration is a procedure for determining the concentration of a solution using another Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 654 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA FIGURE 16.2 A pH meter is
used to monitor an acid-base
titration. solution of known concentration, called standard solution. 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. For this reason, they are
almost never carried out. Figure 16.2 shows the arrangement for monitoring the pH
during the course of a titration.
STRONG ACID–STRONG BASE TITRATIONS The reaction between HCl, a strong acid, and NaOH, a strong base, can be represented
NaOH(aq) HCl(aq) 88n NaCl(aq) H2O(l ) or, in terms of the net ionic equation,
H (aq ) OH (aq ) 88n H2O(l ) Consider the addition of a 0.10 M NaOH solution (from a buret) to an Erlenmeyer flask
containing 25 mL of 0.10 M HCl. Figure 16.3 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 by log (0.10), 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 a solution at every stage of titration. Here are
three sample calculations.
1. Back Forward Main Menu After the addition of 10 mL of 0.10 M NaOH to 25 mL of 0.10 M HCl.
The total volume of the solution is 35 mL. The number of moles of NaOH in
10 mL is TOC Study Guide TOC Textbook Website MHHE Website 16.4 FIGURE 16.3 pH profile of a
strong acid–strong base titration.
A 0.10 M NaOH solution is
added from a buret to 25 mL of
a 0.10 M HCl solution in an
Erlenmeyer flask (see Figure 4.21).
This curve is sometimes referred to
as a titration curve. ACID-BASE TITRATIONS 655 14
pH 7 Equivalence
10 20 30 40 50 Volume of NaOH added (mL) 10 mL 0.10 mol NaOH
1 L NaOH 1L
1000 mL 1.0 3 10 mol The number of moles of HCl originally present in 25 mL of solution is
25 mL 0.10 mol HCl
1 L HCl 1L
1000 mL 2.5 10 3 mol Thus, the amount of HCl left after partial neutralization is (2.5 10 3) (1.0
10 3), or 1.5 10 3 mol. Next, the concentration of H ions in 35 mL of solution is found as follows: Keep in mind that 1 mol
1 mol HCl.
NaOH 1.5 10 3 mol HCl
35 mL 1000 mL
1L 0.043 mol HCl/L
0.043 M HCl Thus [H ] 0.043 M, and the pH of the solution is
pH log 0.043 1.37 After the addition of 25 mL of 0.10 M NaOH to 25 mL of 0.10 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, [H ]
[OH ] and the pH of the solution is 7.00.
3. After the addition of 35 mL of 0.10 M NaOH to 25 mL of 0.10 M HCl.
The total volume of the solution is now 60 mL. The number of moles of NaOH
added is 2.
Neither Na nor Cl undergoes
hydrolysis. 35 mL 0.10 mol NaOH
1 L NaOH 1000 mL
1L The number of moles of HCl in 25 mL solution is 2.5
tralization of HCl, the amount of NaOH left is (3.5 Back Forward Main Menu TOC Study Guide TOC 3.5 10 3 mol 10 3. After complete neu10 3) (2.5 10 3), or Textbook Website MHHE Website 656 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 1.0 10 3 mol. The concentration of NaOH in 60 mL of solution is
1.0 10 3 mol NaOH
60 mL 1000 mL
1L 0.017 mol NaOH/L
0.017 M NaOH Thus [OH ]
lution is 0.017 M and pOH log 0.017 pH 14.00 pOH 14.00 1.77. Hence, the pH of the so- 1.77 12.23
WEAK ACID–STRONG BASE TITRATIONS Consider the neutralization reaction between acetic acid (a weak acid) and sodium hydroxide (a strong base):
CH3COOH(aq) NaOH(aq) 88n CH3COONa(aq) H2O(l ) This equation can be simplified to
CH3COOH(aq) OH (aq) 88n CH3COO (aq) H2O(l ) The acetate ion undergoes hydrolysis as follows:
CH3COO (aq) H2O(l ) 34 CH3COOH(aq) OH (aq) 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 16.4). Note
that this situation is analogous to the hydrolysis of sodium acetate (CH3COONa) (see
The following example deals with the titration of a weak acid with a strong base.
FIGURE 16.4 pH profile of a
weak acid–strong base titration.
A 0.10 M NaOH solution is
added from a buret to 25 mL of
a 0.10 M CH3COOH solution in
an Erlenmeyer flask. Due to the
hydrolysis of the salt formed, the
pH at the equivalence point is
greater than 7. 14
10 20 30 40 50 Volume of NaOH added (mL) Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 16.4 ACID-BASE TITRATIONS 657 EXAMPLE 16.5 Calculate the pH in the titration of 25 mL of 0.10 M acetic acid by sodium hydroxide after the addition to the acid solution of (a) 10 mL of 0.10 M NaOH,
(b) 25 mL of 0.10 M NaOH, (c) 35 mL of 0.10 M NaOH.
Answer The neutralization reaction is
CH3COOH(aq) NaOH(aq) 88n CH3COONa(aq) H2O(l ) For each of the three stages of the titration we first calculate the number of moles
of NaOH added to the acetic acid solution. Next, we calculate the number of moles
of the acid (or the base) left over after neutralization. Then we determine the pH of
(a) The number of moles of NaOH in 10 mL is
10 mL 1L
1000 mL 0.10 mol NaOH
1 L NaOH soln 1.0 3 10 mol The number of moles of CH3COOH originally present in 25 mL of solution is
0.10 mol CH3COOH
1 L CH3COOH soln 25 mL 1L
1000 mL 2.5 10 3 mol Thus, the amount of CH3COOH left after all added base has been neutralized is
(2.5 10 3 10 3) mol 1.0 The amount of CH3COONa formed is 1.0
1.0 10 3 mol 10 1.5
3 10 3 mol mole: NaOH(aq) 88n CH3COONa(aq)
1.0 10 3 mol
1.0 10 3 mol H2O(l ) At this stage we have a buffer system made up of CH3COONa and CH3COOH. To
calculate the pH of this solution we write
Since the volume of the solution
is the same for CH3COOH and
CH3COO , the ratio of the
number of moles present is
equal to the ratio of their
molar concentrations. [H ] [CH3COOH]Ka
(1.5 10 3)(1.8
2.7 pH 10 log (2.7 5 10 5)
10 5) 4.57 (b) These quantities (that is, 25 mL of 0.10 M NaOH reacting with 25 mL of
0.10 M CH3COOH) correspond to the equivalence point. The number of moles of
both NaOH and CH3COOH in 25 mL is 2.5 10 3 mol, so the number of moles
of the salt formed is
2.5 10 3 mol NaOH(aq) 88n CH3COONa(aq)
2.5 10 3 mol
2.5 10 3 mol H2O(l ) The total volume is 50 mL, so the concentration of the salt is
[CH3COONa] 2.5 10 3 mol
50 mL 0.050 mol/L Back Forward Main Menu TOC Study Guide TOC 1000 mL
0.050 M Textbook Website MHHE Website 658 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 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 15.13, we
find that the pH of the solution at the equivalence point is 8.72.
(c) After the addition of 35 mL of NaOH, the solution is well past the equivalence
point. At this stage we have two species that are responsible for making the solution basic: CH3COO and OH . However, since OH is a much stronger base than
CH3COO , we can safely neglect the CH3COO ions and calculate the pH of the
solution using only the concentration of the OH ions. Only 25 mL of NaOH are
needed for complete neutralization, so the number of moles of NaOH left after neutralization is
(35 25) mL 0.10 mol NaOH
1 L NaOH soln 1L
1000 mL 1.0 10 3 mol The total volume of the combined solutions is now 60 mL, so we calculate OH
concentration as follows:
1.0 [OH ] 10 3 mol
60 mL 0.017 mol/L
pOH 1000 mL
0.017 M log 0.017
1.77 pH 14.00 pOH 14.00 1.77 12.23 Similar problem: 16.92.
PRACTICE EXERCISE Exactly 100 mL of 0.10 M nitrous acid are titrated with a 0.10 M NaOH solution.
Calculate the pH for (a) the initial solution, (b) the point at which 80 mL of the base
has been added, (c) the equivalence point, (d) the point at which 105 mL of the base
has been added.
STRONG ACID–WEAK BASE TITRATIONS Consider the titration of HCl, a strong acid, with NH3, a weak base:
HCl(aq) NH3(aq) 88n NH4Cl(aq) or simply
H (aq) NH3(aq) 88n NH4 (aq) The pH at the equivalence point is less than 7 due to the hydrolysis of the NH4 ion:
NH4 (aq) H2O(l ) 34 NH3(aq) H3O (aq) or simply
NH4 (aq) 34 NH3(aq) H (aq) 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 16.5(a) shows the titration curve for this experiment and Figure 16.5(b) shows the titration curve for the case
in which a weak base is added from a buret to HCl. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 16.4 FIGURE 16.5 pH profiles of a
strong acid–weak base titration.
(a) A 0.10 M HCl solution is
added from a buret to 25 mL of
a 0.10 M NH3 solution in an
Erlenmeyer flask. (b) A 0.10 M
weak base solution is added
from a buret to 25 mL of a
0.10 M HCl solution in an
Erlenmeyer flask. As a result of
salt hydrolysis, the pH at the
equivalence point in both cases is
lower than 7. ACID-BASE TITRATIONS 659 12
7 pH 6 Equivalence
(a) 20 30 40 50 40 50 Volume of HCl added (mL)
6 pH Equivalence
10 20 30 Volume of base added (mL) (b) EXAMPLE 16.6 Calculate the pH at the equivalence point when 25 mL of 0.10 M NH3 is titrated by
a 0.10 M HCl solution.
The neutralization reaction is given above. The number of moles of NH3
in 25 mL of 0.10 M solution is
Answer 25 mL Back Forward Main Menu TOC 0.10 mol NH3
1 L NH3 Study Guide TOC 1L
1000 mL 2.5 10 3 mol Textbook Website MHHE Website 660 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA Since 1 mol NH3 1 mol HCl, at the equivalence point the number of moles of
HCl reacted is also 2.5 10 3 mol. The number of moles of the salt (NH4Cl) formed
10 3 mol 2.5 NH3(aq)
10 3 mol
2.5 10 3 mol The total volume is 50 mL, so the concentration of NH4Cl is
10 3 mol
50 mL 2.5 [NH4Cl] 0.050 mol/L 1000 mL
1L 0.050 M The pH of the solution at the equivalence point is determined by the hydrolysis of
NH4 ions. We follow the procedure on p. 608.
Step 1: We represent the hydrolysis of the cation NH4 , and let x be the equilibrium concentration of NH3 and H ions in mol/L:
NH4 (aq) 34 NH3(aq)
x Initial (M ):
Change (M ):
Equilibrium (M ): (0.050 x) x H (aq)
x Step 2: From Table 15.4 we obtain the Ka for NH4 :
[NH4 ] Ka x2
0.050 Applying the approximation 0.050
0.050 5.6 x 10 5.6 x 10
10 10 0.050, we get 2 x
0.050 5.6 10 10 x x 5.3 10 6 M Thus the pH is given by
pH log (5.3 10 6) 5.28 Similar problem: 16.27.
PRACTICE EXERCISE Calculate the pH at the equivalence point in the titration of 50 mL of 0.10 M methylamine (see Table 15.4) with a 0.20 M HCl solution. 16.5 ACID-BASE INDICATORS 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 to add 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 Chapter 4 that an indicator is usually a Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 16.5 ACID-BASE INDICATORS 661 weak organic acid or base that has distinctly different colors in its nonionized and ionized forms. These two forms are related to 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 below.
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:
HIn(aq) 34 H (aq) In (aq) If the indicator is in a sufficiently acidic medium, the equilibrium, according to Le
Chatelier ’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:
[In ] color of acid (HIn) predominates [In ]
[HIn] color of conjugate base (In ) predominates 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. 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 4.6 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 16.3, 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 is about 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.
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 16.6). Table 16.1 lists a number of indicators commonly used in acidbase titrations. The choice of a particular indicator depends on the strength of the acid
and base to be titrated. Example 16.7 illustrates this point.
EXAMPLE 16.7 Which indicator or indicators listed in Table 16.1 would you use for the acid-base
titrations shown in (a) Figure 16.3, (b) Figure 16.4, and (c) Figure 16.5(b)? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 662 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA FIGURE 16.6 Solutions containing extracts of red cabbage
(obtained by boiling the cabbage
in water) produce different colors
when treated with an acid and a
base. The pH of the solutions increases from left to right. Similar problem: 16.31. Answer (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.
PRACTICE EXERCISE Referring to Table 16.1, specify which indicator or indicators you would use for the
following titrations: (a) HBr versus CH3NH2, (b) HNO3 versus NaOH, (c) HNO2
versus KOH. TABLE 16.1 Some Common Acid-Base Indicators
COLOR INDICATOR IN ACID IN BASE pH RANGE* Thymol blue
Reddish pink 1.2–2.8
8.3–10.0 *The pH range is defined as the range over which the indicator changes from the acid color to the
base color. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 16.5 ACID-BASE INDICATORS 663 Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Back Maintaining the pH of Blood
All higher animals need a circulatory system to carry
fuel and oxygen for their life processes and to remove
wastes. In the human body this vital exchange takes
place in the versatile fluid known as blood, of which
there are about 5 L (10.6 pints) in an average adult.
Blood circulating deep in the tissues carries oxygen
and nutrients to keep cells alive, and removes carbon
dioxide and other waste materials. Using several
buffer systems, nature has provided an extremely efficient method for the delivery of oxygen and the removal of carbon dioxide.
Blood is an enormously complex system, but for
our purposes we need look at only two essential components: blood plasma and red blood cells, or erythrocytes. Blood plasma contains many compounds,
including proteins, metal ions, and inorganic phosphates. The erythrocytes contain hemoglobin molecules, as well as the enzyme carbonic anhydrase,
which catalyzes both the formation of carbonic acid
(H2CO3) and its decomposition:
CO2(aq) H2O(l ) 34 H2CO3(aq) The substances inside the erythrocyte are protected
from extracellular fluid (blood plasma) by a cell membrane that allows only certain molecules to diffuse
The pH of blood plasma is maintained at about
7.40 by several buffer systems, the most important of
which is the HCO3 /H2CO3 system. In the erythrocyte,
where the pH is 7.25, the principal buffer systems are
HCO3 /H2CO3 and hemoglobin. The hemoglobin
molecule is a complex protein molecule (molar mass
65,000 g) that contains a number of ionizable protons. As a very rough approximation, we can treat it
as a monoprotic acid of the form HHb:
HHb(aq) 34 H (aq) Hb (aq) where HHb represents the hemoglobin molecule and
Hb the conjugate base of HHb. Oxyhemoglobin
(HHbO2), formed by the combination of oxygen with
hemoglobin, is a stronger acid than HHb:
HHbO2(aq) 34 H (aq) HbO2 (aq) produced by metabolic processes diffuses into the erythrocyte, where it is rapidly converted to H2CO3 by
CO2(aq) H2O(l ) 34 H2CO3(aq) The ionization of the carbonic acid
H2CO3(aq) 34 H (aq) HCO3 (aq) has two important consequences. First, the bicarbonate ion diffuses out of the erythrocyte and is carried
by the blood plasma to the lungs. This is the major
mechanism for removing carbon dioxide. Second, the
H ions shift the equilibrium in favor of the nonionized oxyhemoglobin molecule:
H (aq) HbO2 (aq) 34 HHbO2(aq) Since HHbO2 releases oxygen more readily than does
its conjugate base (HbO2 ), the formation of the acid
promotes the following reaction from left to right:
HHbO2(aq) 34 HHb(aq) O2(aq) The O2 molecules diffuse out of the erythrocyte and
are taken up by other cells to carry out metabolism.
When the venous blood returns to the lungs, the
above processes are reversed. The bicarbonate ions
now diffuse into the erythrocyte, where they react with
hemoglobin to form carbonic acid:
HHb(aq) HCO3 (aq) 34 Hb (aq) H2CO3(aq) Most of the acid is then converted to CO2 by carbonic
H2CO3(aq) 34 H2O(l ) CO2(aq) The carbon dioxide diffuses to the lungs and is eventually exhaled. The formation of the Hb ions (due to
the reaction between HHb and HCO3 shown above)
also favors the uptake of oxygen at the lungs
Hb (aq) O2(aq) 34 HbO2 (aq) because Hb has a greater affinity for oxygen than
When the arterial blood flows back to the body
tissues, the entire cycle is repeated. As the accompanying figure shows, carbon dioxide Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA Tissues Lungs
Erythrocyte O2 Erythrocyte O2 + HHb O2
HbO 2 + H+ HHbO2
CO2 CO2 + H2O H2CO3 O2 + HHb
HbO 2 + H+ HHbO2 H+ + HCO –
CO2 CO2 + H2O H2CO3 H+ + HCO –
3 HCO –
3 HCO –
3 Plasma Plasma (a)
Oxygen–carbon dioxide transport and release by blood.
(a) The partial pressure of CO2 is higher in the metabolizing tissues than in the plasma. Thus, it diffuses into the
blood capillaries and then into erythrocytes, where it is
converted to carbonic acid by the enzyme carbonic anhydrase (CA). The protons provided by the carbonic
acid then combine with the HbO2 anions to form
HHbO2, which eventually dissociates into HHb and O2.
Because the partial pressure of O2 is higher in the erythrocytes than in the tissues, oxygen molecules diffuse
out of the erythrocytes and then into the tissues. The bi- 16.6 An aqueous suspension of BaSO4
is used to examine the digestive
tract. Back Capillary Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Capillary 664 Forward (b)
carbonate ions also diffuse out of the erythrocytes and
are carried in the blood plasma to the lungs. (b) In the
lungs, the processes are exactly reversed. Oxygen molecules diffuse from the lungs, where they have a higher
partial pressure, into the erythrocytes. There they combine
with HHb to form HHbO2. The protons provided by
HHbO2 combine with the bicarbonate ions diffused into
the erythrocytes from the plasma to form carbonic acid. In
the presence of carbonic anhydrase, carbonic acid is converted to H2O and CO2. The CO2 then diffuses out of the
erythrocytes and into the lungs, where it is exhaled. SOLUBILITY EQUILIBRIA Precipitation reactions are important in industry, medicine, and everyday life. For example, the preparation of many essential industrial chemicals such as sodium carbonate (Na2CO3) is based on precipitation reactions. The dissolving of tooth enamel, which
is mainly made of hydroxyapatite [Ca5(PO4)3OH], in an acidic medium leads to tooth
decay. Barium sulfate (BaSO4), an insoluble compound that is opaque to X rays, is
used to diagnose ailments of the digestive tract. Stalactites and stalagmites, which consist of calcium carbonate (CaCO3), are produced by a precipitation reaction, and so are
many foods, such as fudge.
The general rules for predicting the solubility of ionic compounds in water were
introduced in Section 4.2. Although useful, these solubility rules do not allow us to
make quantitative predictions about how much of a given ionic compound will dissolve in water. To develop a quantitative approach, we start with what we already know
about chemical equilibrium. Unless otherwise stated, in the following discussion the
solvent is water and the temperature is 25°C. Main Menu TOC Study Guide TOC Textbook Website MHHE Website 16.6 SOLUBILITY EQUILIBRIA 665 SOLUBILITY PRODUCT Consider a saturated solution of silver chloride that is in contact with solid silver chloride. The solubility equilibrium can be represented as
AgCl(s) 34 Ag (aq) Cl (aq) 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. We know
from Chapter 14 that for heterogeneous reactions the concentration of the solid is a
constant. Thus we can write the equilibrium constant for the dissolution of AgCl (see
Example 14.5) as
Ksp [Ag ][Cl ] where Ksp is called the solubility product constant or simply the solubility product. In
general, the solubility product of a compound is the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient
in the equilibrium equation.
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 more
MgF2(s) 34 Mg2 (aq) • Ksp [Mg2 ][F ]2 CO2 (aq)
3 Ksp [Ag ]2[CO2 ]
3 2PO3 (aq)
4 Ksp [Ca2 ]3[PO3 ]2
Ag2CO3(s) 34 2Ag (aq) • 2F (aq) Ca3(PO4)2
Ca3(PO4)2(s) 34 3Ca2 (aq) Table 16.2 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 we did not include Ka values for strong
acids in Table 15.3. 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.
A cautionary note: In Chapter 15 (p. 600) we assumed that dissolved substances
exhibit ideal behavior for our calculations involving solution concentrations, but this
assumption is not always valid. For example, a solution of barium fluoride (BaF2) may
contain both neutral and charged ion pairs, such as BaF2 and BaF , in addition to Ba2
and F ions. Furthermore, many anions in the ionic compounds listed in Table 16.2
are conjugate bases of weak acids. Consider copper sulfide (CuS). The S2 ion can
hydrolyze as follows
S2 (aq) H2O(l ) 34 HS (aq) OH (aq) HS (aq) H2O(l ) 34 H2S(aq) OH (aq) 3 And highly charged small metal ions such as Al and Bi3 will undergo hydrolysis
as discussed in Section 15.10. Both ion-pair formation and salt hydrolysis decrease the
concentrations of the ions that appear in the Ksp expression, but we need not be concerned with the deviations from ideal behavior here. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 666 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA TABLE 16.2 Solubility Products of Some Slightly Soluble Ionic Compounds at 25°C COMPOUND Ksp Aluminum hydroxide [Al(OH)3]
Barium carbonate (BaCO3)
Barium fluoride (BaF2)
Barium sulfate (BaSO4)
Bismuth sulfide (Bi2S3)
Cadmium sulfide (CdS)
Calcium carbonate (CaCO3)
Calcium fluoride (CaF2)
Calcium hydroxide [Ca(OH)2]
Calcium phosphate [Ca3(PO4)2]
Chromium(III) hydroxide [Cr(OH)3]
Cobalt(II) sulfide (CoS)
Copper(I) bromide (CuBr)
Copper(I) iodide (CuI)
Copper(II) hydroxide [Cu(OH)2]
Copper(II) sulfide (CuS)
Iron(II) hydroxide [Fe(OH)2]
Iron(III) hydroxide [Fe(OH)3]
Iron(II) sulfide (FeS)
Lead(II) carbonate (PbCO3)
Lead(II) chloride (PbCl2) 1.8
33 Ksp Lead(II) chromate (PbCrO4)
Lead(II) fluoride (PbF2)
Lead(II) iodide (PbI2)
Lead(II) sulfide (PbS)
Magnesium carbonate (MgCO3)
Magnesium hydroxide [Mg(OH)2]
Manganese(II) sulfide (MnS)
Mercury(I) chloride (Hg2Cl2)
Mercury(II) sulfide (HgS)
Nickel(II) sulfide (NiS)
Silver bromide (AgBr)
Silver carbonate (Ag2CO3)
Silver chloride (AgCl)
Silver iodide (AgI)
Silver sulfate (Ag2SO4)
Silver sulfide (Ag2S)
Strontium carbonate (SrCO3)
Strontium sulfate (SrSO4)
Tin(II) sulfide (SnS)
Zinc hydroxide [Zn(OH)2]
Zinc sulfide (ZnS) 9
23 For equilibrium reactions involving an ionic solid in aqueous solution, any one of
the following conditions may exist: (1) the solution is unsaturated, (2) the solution is
saturated, or (3) the solution is supersaturated. For concentrations of ions that do not
correspond to equilibrium conditions we use the reaction quotient (see Section 14.4),
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 is given by
Q [Ag ]0[Cl ]0 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
[Ag ]0[Cl ]0 Ksp
1.6 10 10 Unsaturated solution Q
[Ag ][Cl ] Ksp
1.6 10 10 Q
[Ag ]0[Cl ]0 Ksp
1.6 10 10 Saturated solution
Supersaturated solution; AgCl will precipitate
out until the product of the ionic concentrations
is equal to 1.6 10 10 MOLAR SOLUBILITY AND SOLUBILITY There are two other ways to express a substance’s solubility: molar solubility, which
is the number of moles of solute in one liter of a saturated solution (mol/L), and sol- Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 16.6 FIGURE 16.7 Sequence of
steps (a) for calculating Ksp from
solubility data and (b) for calculating solubility from Ksp data. Molar
compound Solubility of
compound SOLUBILITY EQUILIBRIA 667 Concentrations
and anions Ksp of
compound Solubility of
compound (a) Concentrations
and anions Ksp of
compound (b) ubility, which is the number of grams of solute in one liter of a saturated solution (g/L).
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 16.7(a).
Example 16.8 illustrates this procedure.
EXAMPLE 16.8 The solubility of calcium sulfate is found experimentally to be 0.67 g/L. Calculate
the value of Ksp for calcium sulfate.
Answer First we calculate the number of moles of CaSO4 dissolved in 1 L of so- lution:
0.67 g CaSO4
1 L soln 1 mol CaSO4
136.2 g CaSO4 4.9 10 3 mol/L The solubility equilibrium
Calcium sulfate is used as a drying agent and in the manufacture
of paints, ceramics, and paper. A
hydrated form of calcium sulfate,
called plaster of Paris, is used to
make casts for broken bones. SO2 (aq)
4 CaSO4(s) 34 Ca2 (aq) shows that for every mole of CaSO4 that dissolves, 1 mole of Ca2 and 1 mole of
SO2 are produced. Thus, at equilibrium
[Ca2 ] 4.9 10 3 M [SO2 ]
4 and 4.9 10 3 M Now we can calculate Ksp:
Ksp [Ca2 ][SO2 ]
2.4 Similar problem: 16.44. 10 3)(4.9
10 10 3) 5 PRACTICE EXERCISE The solubility of lead chromate (PbCrO4) is 4.5
ity product of this compound. 10 5 g/L. Calculate the solubil- Sometimes we are given the value of Ksp for a compound and asked to calculate
the compound’s molar solubility. For example, the Ksp of silver bromide (AgBr) is
7.7 10 13. We can calculate its molar solubility by the same procedure as outlined
on p. 608 for acid ionization constants. First we identify the species present at equilibrium. Here we have Ag and Br ions. Let s be the molar solubility (in mol/L) of Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 668 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA AgBr. Since one unit of AgBr yields one Ag and one Br ion, at equilibrium both
[Ag ] and [Br ] are equal to s. We summarize the changes in concentrations as follows:
AgBr(s) 34 Ag (aq)
s Initial (M ):
Change (M ):
Equilibrium (M ): Br (aq)
s s s From Table 16.2 we write
Ksp Silver bromide is used in photographic emulsions. 7.7 10 [Ag ][Br ] 13 (s)(s)
7.7 s 13 10 8.8 7 10 M Therefore, at equilibrium
[Ag ] 8.8 [Br ] 8.8 10 7 M 10 7 M Thus the molar solubility of AgBr also is 8.8 10 7 M.
The following example makes use of this approach.
EXAMPLE 16.9 Using the data in Table 16.2, calculate the solubility of copper(II) hydroxide,
Cu(OH)2, in g/L.
Answer Step 1: When Cu(OH)2 dissociates, the species in solution are Cu2 and OH ions. Step 2: Let s be the molar solubility of Cu(OH)2. Since one unit of Cu(OH)2 yields one Cu2 ion and two OH ions, at equilibrium [Cu2 ] is s and [OH ] is
2s. We summarize the changes in concentrations as follows:
Cu(OH)2(s) 34 Cu2 (aq)
s Initial (M ):
Change (M ): Copper(II) hydroxide is used as a
pesticide and to treat seeds. Equilibrium (M ):
Step 3: s
Ksp 2.2 10 20 2s [Cu2 ][OH ]2
4 20 s3 Solving for s, we get 2OH (aq)
2s s 1.8 10 7 5.5 10 21 M Knowing that the molar mass of Cu(OH)2 is 97.57 g/mol and knowing its
molar solubility, we can calculate the solubility in g/L as follows: Similar problem: 16.46. 1.8 10 7 mol Cu(OH)2
1 L soln 1.8 solubility of Cu(OH)2 10 5 97.57 g Cu(OH)2
1 mol Cu(OH)2 g/L PRACTICE EXERCISE Calculate the solubility of silver chloride (AgCl) in g/L. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 16.6 TABLE 16.3 669 SOLUBILITY EQUILIBRIA Relationship between Ksp and Molar Solubility (s)
CONCENTRATION (M ) COMPOUND Ksp EXPRESSION CATION ANION RELATION BETWEEN Ksp AND s AgCl
BaSO4 [Ag ][Cl ]
[Ba2 ][SO2 ]
Ksp Ag2CO3 [Ag ]2[CO2 ]
3 2s s Ksp PbF2 [Pb2 ][F ]2 s 2s Ksp Al(OH)3 [Al3 ][OH ]3 s 3s Ksp Ca3(PO4)2 [Ca2 ]3[PO3 ]2
4 3s 2s Ksp s2; s
s2; s 1
5 As the above examples show, solubility and solubility product are related. If we
know one, we can calculate the other, but each quantity provides different information.
Table 16.3 shows the relationship between molar solubility and solubility product for
a number of ionic compounds.
When carrying out solubility and/or solubility product calculations, keep in mind
the following important points:
The solubility is the quantity of a substance that dissolves in a certain quantity of
water. In solubility equilibria calculations, it is usually expressed as grams of solute
per liter of solution. Molar solubility is the number of moles of solute per liter of
• The solubility product is an equilibrium constant.
• Molar solubility, solubility, and solubility product all refer to a saturated solution.
• PREDICTING PRECIPITATION REACTIONS From a knowledge of the solubility rules (see Section 4.2) and the solubility products
listed in Table 16.2, 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 (C2O2 ),
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 16.10 illustrates the steps involved in precipitation reactions. A kidney stone. EXAMPLE 16.10 Exactly 200 mL of 0.0040 M BaCl2 are added to exactly 600 mL of 0.0080 M K2SO4.
Will a precipitate form? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 670 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA According to the solubility rules on p. 113, the only precipitate that might
form is BaSO4: Answer SO2 (aq) 88n BaSO4(s)
4 Ba2 (aq) The number of moles of Ba2 present in the original 200 mL of solution is
0.0040 mol Ba2
1 L soln 200 mL
We assume that the volumes
are additive. 1L
1000 mL 8.0 10 4 mol Ba2 The total volume after combining the two solutions is 800 mL. The concentration
of Ba2 in the 800 mL volume is
[Ba2 ] 8.0 10 4 mol
800 mL 1.0 The number of moles of SO2
4 M in the original 600 mL solution is 0.0080 mol SO2
1 L soln 600 mL 3 10 1000 mL
1 L soln 1L
1000 mL 4.8 10 3 mol SO2
4 The concentration of SO2 in the 800 mL of the combined solution is
4 4.8 10 3 mol
800 mL 6.0 10 3 1000 mL
1 L soln M Now we must compare Q and Ksp. From Table 16.2, the Ksp for BaSO4 is 1.1
10 10. As for Q,
Q [Ba2 ]0[SO2 ]0
6.0 10 (1.0 10 3)(6.0 10 3) 6 Therefore,
Q Ksp The solution is supersaturated because the value of Q indicates that the concentrations of the ions are too large. Thus some of the BaSO4 will precipitate out of solution until
[Ba2 ][SO2 ]
4 Similar problem: 16.49. 1.1 10 10 PRACTICE EXERCISE If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl2, will precipitation occur? 16.7 SEPARATION OF IONS BY FRACTIONAL PRECIPITATION 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. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 16.7 COMPOUND
17 SEPARATION OF IONS BY FRACTIONAL PRECIPITATION 671 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 the margin 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.
The following example 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 16.11 Silver nitrate is slowly added to a solution that is 0.020 M in Cl ions and 0.020
M in Br ions. Calculate the concentration of Ag ions (in mol/L) required to initiate (a) the precipitation of AgBr and (b) the precipitation of AgCl.
(a) From the Ksp values, we know that AgBr will precipitate before AgCl.
So for AgBr we write Answer Ksp [Ag ][Br ] Since [Br ] 0.020 M, the concentration of Ag that must be exceeded to initiate
the precipitation of AgBr is
[Ag ] Ksp
3.9 AgCl (left) and AgBr (right). Thus [Ag ] 3.9
(b) For AgCl 10 11 7.7
10 11 10
0.020 13 M M is required to start the precipitation of AgBr.
Ksp [Ag ] [Ag ][Cl ]
10 9 10
0.020 10 M 9 Similar problems: 16.51, 16.52. Therefore [Ag ] 8.0 10 M is needed to initiate the precipitation of AgCl.
To precipitate AgBr without precipitating Cl ions then, [Ag ] must be greater
than 3.9 10 11 M and lower than 8.0 10 9 M.
PRACTICE EXERCISE The solubility products of AgCl and Ag3PO4 are 1.6 10 10 and 1.8 10 18, respectively. If Ag is added (without changing the volume) to 1.00 L of a solution
containing 0.10 mol Cl and 0.10 mol PO3 , calculate the concentration of Ag
ions (in mol/L) required to initiate (a) the precipitation of AgCl and (b) the precipitation of Ag3PO4.
Example 16.11 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
[Br ] Back Forward Main Menu TOC Study Guide TOC Ksp
[Ag ] Textbook Website MHHE Website 672 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA [Br ] 7.7
5 M The percent of Br remaining in solution (the unprecipitated Br ) at the critical concentration of Ag is
% Br [Br ]unppt’d
[Br ]original 100% 9.6 10 5 M
0.020 M 100% 0.48% unprecipitated Thus, (100 0.48) percent, or 99.52 percent, of Br will have precipitated just before
AgCl begins to precipitate. By this procedure, the Br ions can be quantitatively separated from the Cl ions. 16.8 THE COMMON ION EFFECT AND SOLUBILITY In Section 16.2 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 following
process also contributes to the total concentration of the common silver ions in solution:
H2O AgNO3(s) 88n Ag (aq) NO3 (aq) If AgNO3 is added to a saturated AgCl solution, the increase in [Ag ] will make the
ion product greater than the solubility product:
At a given temperature, only the
solubility of a compound is
altered (decreased) by the
common ion effect. Its solubility
product, which is an equilibrium
constant, remains the same
whether or not other substances
are present in the solution. [Ag ]0[Cl ]0 Ksp To reestablish equilibrium, some AgCl will precipitate out of the solution, as Le
Chatelier ’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 ].
The following example shows the common ion effect on solubility.
EXAMPLE 16.12 Calculate the solubility of silver chloride (in g/L) in a 6.5
solution. Back Forward Main Menu TOC Study Guide TOC 10 3 M silver nitrate Textbook Website MHHE Website 16.9 pH AND SOLUBILITY 673 Answer Step 1: The relevant species in solution are Ag ions (from both AgCl and AgNO3)
and Cl ions. The NO3 ions are spectator ions.
Step 2: Since AgNO3 is a soluble strong electrolyte, it dissociates completely:
H2O AgNO3(s) 88n Ag (aq)
6.5 10 3 M NO3 (aq)
6.5 10 3 M Let s be the molar solubility of AgCl in AgNO3 solution. We summarize the
changes in concentrations as follows:
Initial (M ):
Change (M ):
Equilibrium (M ): (6.5 Step 3: Ksp
1.6 Ag (aq)
6.5 10 3
s) s [Ag ][Cl ] 10 10 3 10 Cl (aq)
s (6.5 10 3 s)(s) Since AgCl is quite insoluble and the presence of Ag ions from AgNO3 further lowers the solubility of AgCl, s must be very small compared with 6.5
10 3 . Therefore, applying the approximation 6.5 10 3 s 6.5 10 3,
10 10 6.5 10 3 s s 1.6 2.5 10 8 M Step 4: At equilibrium
[Cl ] (6.5
2.5 3 10
10 8 2.5 10 8) M 6.5 10 3 M M and so our approximation was justified in step 2. Since all the Cl ions must
come from AgCl, the amount of AgCl dissolved in AgNO3 solution also is
2.5 10 8 M. Then, knowing the molar mass of AgCl (143.4 g), we can
calculate the solubility of the AgCl as follows:
2.5 10 8 mol AgCl
1 L soln 3.6 solubility of AgCl in AgNO3 solution 10 6 143.4 g AgCl
1 mol AgCl g/L The solubility of AgCl in pure water is 1.9 10 3 g/L (see the Practice
Exercise in Example 16.9). Therefore, the answer is reasonable. Comment Similar problem: 16.55. PRACTICE EXERCISE Calculate the solubility in g/L of AgBr in (a) pure water and (b) 0.0010 M NaBr. 16.9 pH AND SOLUBILITY The solubilities of many substances also depend on the pH of the solution. Consider
the solubility equilibrium of magnesium hydroxide: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 674 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA Mg(OH)2(s) 34 Mg2 (aq) This is why milk of magnesia,
Mg(OH)2, dissolves in the acidic
gastric juice in a person’s stomach
(see p. 633). 2OH (aq) 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 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
Ksp [Mg2 ][OH ]2 1.2 11 10 Let s be the molar solubility of Mg(OH)2. Proceeding as in Example 16.9,
Ksp (s)(2s)2 4s3 4s3 1.2 10 11 3 3.0 10 12 s 1.4 10 4 s M At equilibrium, therefore,
pH 2 1.4 10 M 2.8 4 10 ) log (2.8
14.00 4 3.55 10 4 M 3.55 10.45 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 equilibrium condition, and more Mg(OH)2 dissolves. The dissolution
process and the effect of extra H ions can be summarized as follows: 2H (aq) Mg(OH)2(s) 34 Mg2 (aq)
2OH (aq) 34 2H2O(l ) Mg(OH)2(s) Overall: 2H (aq) 34 Mg2 (aq) 2OH (aq)
2H2O(l ) 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 BaF2 is
BaF2(s) 34 Ba2 (aq) 2F (aq) and
Ksp [Ba2 ][F ]2 In an acidic medium, the high [H ] will shift the following equilibrium from left to
H (aq) Recall that HF is a weak acid. F (aq) 34 HF(aq) 2 As [F ] decreases, [Ba ] 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: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 16.9 pH AND SOLUBILITY Overall: BaF2(s) 34 Ba2 (aq)
2F (aq) 34 2HF(aq) 2F (aq) 2H (aq)
BaF2(s) 2H (aq) 34 Ba2 (aq) 675 2HF(aq) The solubilities of salts containing anions that do not hydrolyze are unaffected by
pH. Examples of such anions are Cl , Br , and I .
The following examples deal with the effect of pH on solubility.
EXAMPLE 16.13 Which of the following compounds will be more soluble in acidic solution than in
water: (a) CuS, (b) AgCl, (c) PbSO4?
(a) CuS will be more soluble in an acidic solution because of the basicity of the S2 ion. The solubility and acid-base equilibria are summarized below: Answer S2 (aq) S (aq)
HS (aq) CuS(s) 34 Cu2 (aq)
H (aq) 34 HS (aq)
H (aq) 34 H2S(aq) CuS(s) 2H (aq) 34 Cu2 (aq) H2S(aq) 2 Overall: Since both HS and H2S are weak acids, the above equilibrium will lie to the right,
resulting in a greater amount of CuS dissolving in solution.
(b) The solubility equilibrium is
AgCl(s) 34 Ag (aq) Cl (aq) Since Cl is the conjugate base of a strong acid (HCl), the solubility of AgCl is not
affected by an acid solution.
(c) PbSO4 will be more soluble in an acidic solution because of the basicity of the
SO2 ion. The solubility and acid-base equilibria are summarized below:
4 PbSO4(s) Overall: Similar problem: 16.60. PbSO4(s) 34 Pb2 (aq) SO2 (aq)
(aq) H (aq) 34 HSO4 (aq)
H (aq) 34 Pb2 (aq) HSO4 (aq) However, because HSO4 has a fairly large ionization constant (see Table 15.5), the
above equilibrium is slightly shifted to the right. Consequently, the solubility of
PbSO4 increases only slightly in an acidic solution.
PRACTICE EXERCISE Are the following compounds more soluble in water or in an acidic solution:
(a) Ca(OH)2, (b) Mg3(PO4)2, (c) PbBr2?
EXAMPLE 16.14 Calculate the concentration of aqueous ammonia necessary to initiate the precipitation of iron(II) hydroxide from a 0.0030 M solution of FeCl2.
Answer The equilibria of interest are
2 Fe (aq) Back Forward Main Menu TOC H2O(l ) 34 NH4 (aq) OH (aq) 2OH (aq) 34 Fe(OH)2(s) Study Guide TOC Textbook Website MHHE Website 676 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA First we find the OH concentration above which Fe(OH)2 begins to precipitate.
[Fe2 ][OH ]2 Ksp 1.6 Since FeCl2 is a strong electrolyte, [Fe2 ]
[OH ] 14 10 0.0030 M and 1.6 10
0.0030 14 2.3 6 10 5.3 10 12 M Next, we calculate the concentration of NH3 that will supply 2.3 10 6 M OH
ions. Let x be the initial concentration of NH3 in mol/L. We summarize the changes
in concentrations resulting from the ionization of NH3 as follows:
2.3 10 Initial (M ):
Change (M ):
Equilibrium (M ): (x 2.3 H2O(l ) 34
6 10 6) NH4 (aq)
2.3 10 6
6 OH (aq)
2.3 10 6
6 Substituting the equilibrium concentrations in the expression for the ionization constant,
[NH4 ][OH ]
(2.3 10 6)(2.3 10 6)
(x 2.3 10 6) 1.8 10 5 1.8 Kb 10 5 Solving for x, we obtain
Similar problem: 16.64. 2.6 10 6 M Therefore the concentration of NH3 must be slightly greater than 2.6
initiate the precipitation of Fe(OH)2. 10 6 M to PRACTICE EXERCISE Calculate whether or not a precipitate will form if 2.0 mL of 0.60 M NH3 are added
to 1.0 L of 1.0 10 3 M FeSO4. 16.10 COMPLEX ION EQUILIBRIA AND SOLUBILITY Lewis acids and bases are discussed in Section 15.12. According to our definition,
Co(H2O)2 itself is a complex
ion. When we write Co(H2O)2 ,
we mean the hydrated
Co ion. Back Forward 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 as 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. In Chapter 22 we will discuss the chemistry of complex ions in more detail.
Transition metals have a particular tendency to form complex ions because they
have more than one oxidation state. This property allows them to act effectively as
Lewis acids in reactions with many molecules or ions that serve as electron donors, or
as Lewis bases. For example, a solution of cobalt(II) chloride is pink because of the
presence of the Co(H2O)2 ions (Figure 16.8). When HCl is added, the solution turns
blue as a result of the formation of the complex ion CoCl2 :
4 Main Menu TOC Study Guide TOC Textbook Website MHHE Website 16.10 COMPLEX ION EQUILIBRIA AND SOLUBILITY 677 FIGURE 16.8 (Left) An aqueous cobalt(II) chloride solution.
The pink color is due to the presence of Co(H2O)2 ions. (Right)
After the addition of HCl solution,
the solution turns blue because of
the formation of the complex
CoCl 2 ions.
4 4Cl (aq) 34 CoCl2 (aq)
4 Co2 (aq) 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:
Cu2 (aq) 2OH (aq) 88n Cu(OH)2(s) 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)2 (Figure 16.9):
Cu(OH)2(s) 4NH3(aq) 34 Cu(NH3)2 (aq)
4 2OH (aq) Cu(NH3)2
4 Thus the formation of the complex ion
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 Kf (also called the stability constant), which is the equilibrium constant for the complex ion formation. The larger Kf is, the more stable the complex ion is. Table 16.4 lists the formation constants of a number of complex ions.
The formation of the Cu(NH3)2 ion can be expressed as
Cu2 (aq) 4NH3(aq) 34 Cu(NH3)2 (aq)
4 for which the formation constant is
Kf [Cu(NH3)2 ]
5.0 1013 FIGURE 16.9 (Left) An aqueous solution of copper(II) sulfate.
(Center) After the addition of a
few drops of a concentrated
aqueous ammonia solution, a
light-blue precipitate of Cu(OH)2
is formed. (Right) When more
concentrated aqueous ammonia
solution is added, the Cu(OH)2
precipitate dissolves to form
the dark-blue complex ion
4 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 678 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA TABLE 16.4 Formation Constants of Selected Complex Ions in Water at 25°C COMPLEX ION EQUILIBRIUM EXPRESSION Ag(NH3)2
4NH3 FORMATION CONSTANT (Kf) 34 Ag(NH3)2
109 The very large value of Kf 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.
EXAMPLE 16.15 A 0.20-mole quantity of CuSO4 is added to a liter of 1.20 M NH3 solution. What is
the concentration of Cu2 ions at equilibrium?
Answer The addition of CuSO4 to the NH3 solution results in the reaction
Cu2 (aq) 4NH3(aq) 34 Cu(NH3)2 (aq)
4 Since Kf is very large (5.0 1013), the reaction lies mostly to the right. As a good
approximation, we can assume that essentially all the dissolved Cu2 ions end up
as Cu(NH3)2 ions. Thus the amount of NH3 consumed in forming the complex
ions is 4 0.20 mol, or 0.80 mol. (Note that 0.20 mol Cu2 is initially present in
solution and four NH3 molecules are needed to form a complex with one Cu2 ion.)
The concentration of NH3 at equilibrium therefore is (1.20 0.80) M, or 0.40 M,
and that of Cu(NH3)2 is 0.20 M, the same as the initial concentration of Cu2 .
Since Cu(NH3)2 does dissociate to a slight extent, we call the concentration of
Cu2 ions at equilibrium x and write
The fact that these species are in
the same solution allows us to use
the number of moles rather than
molarity in the formation
constant expression. Kf [Cu(NH3)2 ]
x(0.40)4 5.0 1013 5.0 1013 Solving for x, we obtain
x 1.6 10 13 M [Cu2 ] The small value of [Cu2 ] at equilibrium, compared with 0.20 M, certainly justifies our approximation. Comment Similar problem: 16.67. PRACTICE EXERCISE If 2.50 g of CuSO4 are dissolved in 9.0 102 mL of 0.30 M NH3, what are the concentrations of Cu2 , Cu(NH3)2 , and NH3 at equilibrium?
The effect of complex ion formation generally is to increase the solubility of a
substance, as the following example shows. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 16.10 679 COMPLEX ION EQUILIBRIA AND SOLUBILITY EXAMPLE 16.16 Calculate the molar solubility of AgCl in a 1.0 M NH3 solution.
Answer Step 1: Initially, the species in solution are Ag and Cl ions and NH3. The reaction between Ag and NH3 produces the complex ion Ag(NH3)2 .
Step 2: The equilibrium reactions are
AgCl(s) 34 Ag (aq)
Ag (aq) Cl (aq)
Ksp [Ag ][Cl ] AgCl(s) 10 10 2NH3(aq) 34 Ag(NH3)2 (aq)
Kf Overall: 1.6 [Ag(NH3)2 ]
[Ag ][NH3]2 2NH3(aq) 34 Ag(NH3)2 (aq) 1.5 107 Cl (aq) The equilibrium constant K for the overall reaction is the product of the equilibrium constants of the individual reactions (see Section 14.2):
K [Ag(NH3)2 ][Cl ]
(1.6 10 10 2.4 )(1.5 107) 3 10 Let s be the molar solubility of AgCl (mol/L). We summarize the changes in
concentrations that result from formation of the complex ion as follows:
Initial (M ):
Change (M ): AgCl(s) 2NH3(aq) 34 Ag(NH3)2 (aq)
s Equilibrium (M ): (1.0 2s) s Cl (aq)
s The formation constant for Ag(NH3)2 is quite large, so most of the silver
ions exist in the complexed form. In the absence of ammonia we have, at
equilibrium, [Ag ] [Cl ]. As a result of complex ion formation, however,
we can write [Ag(NH3)2 ] [Cl ].
Step 3: K
2.4 10 (s)(s)
s2 3 2s)2 (1.0 Taking the square root of both sides, we obtain
1.0 2s 0.045 M Step 4: At equilibrium, 0.045 mole of AgCl dissolves in 1 L of 1.0 M NH3 solution. The molar solubility of AgCl in pure water is 1.3 10 5 M. Thus, the
formation of the complex ion Ag(NH3)2 enhances the solubility of AgCl (Figure
Comment Similar problem: 16.70. PRACTICE EXERCISE Calculate the molar solubility of AgBr in a 1.0 M NH3 solution. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 680 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA FIGURE 16.10 (Left to right)
Formation of AgCl precipitate
when AgNO3 solution is added
to NaCl solution. With the addition of NH3 solution, the AgCl
precipitate dissolves as the soluble Ag(NH3)2 forms. All amphoteric hydroxides are
insoluble compounds. Finally we note that there is a class of hydroxides, called amphoteric hydroxides,
which 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:
Al(OH)3(s) 3H (aq) 88n Al3 (aq) 3H2O(l ) OH (aq) 34 Al(OH)4 (aq) 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. 16.11 APPLICATION OF THE SOLUBILITY PRODUCT PRINCIPLE TO QUALITATIVE ANALYSIS Do not confuse the groups in
Table 16.5, which are based on
solubility products, with those
in the periodic table, which
are based on the electron
configurations of the elements. In Section 4.6, we discussed the principle of gravimetric analysis, by which we measure the amount of an ion in an unknown sample. 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 twenty 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 16.5). Since an unknown solution may contain from
one to all twenty ions, any analysis must be carried out systematically from group 1
through group 5. Let us consider the general procedure for separating these twenty ions
by adding precipitating reagents to an unknown solution.
Group 1 cations. When dilute HCl is added to the unknown solution, only the Ag ,
Hg2 , and Pb2 ions precipitate as insoluble chlorides. The other ions, whose chlo2
rides 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
• M2 (aq) H2S(aq) 34 MS(s) 2H (aq) 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, and SnS (see Table 16.5).
• Group 3 cations. At this stage, sodium hydroxide is added to the solution to make Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 16.11 681 APPLICATION OF THE SOLUBILITY PRODUCT PRINCIPLE TO QUALITATIVE ANALYSIS TABLE 16.5 Separation of Cations into Groups According to Their Precipitation
Reactions with Various Reagents
GROUP 1 2 3 4 5 CATION Ag
NH4 PRECIPITATING REAGENTS INSOLUBLE COMPOUND HCl AgCl
g Na2CO3 A
g No precipitating
reagent Ksp 1.6
9 it basic. In a basic solution, the above equilibrium shifts to the right. Therefore, the
more soluble sulfides (CoS, FeS, MnS, NiS, 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
NaOH(aq) Because NaOH is added in group
3 and Na2CO3 is added in group
4, the flame test for Na ions
is carried out using the
original solution. NH4 (aq) 88n Na (aq) H2O(l ) NH3(g) 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 16.11).
Figure 16.12 summarizes this scheme for separating metal ions. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 682 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA FIGURE 16.11 (Left to right)
Flame colors of lithium, sodium,
potassium, and copper. FIGURE 16.12 A flow chart
for the separation of cations in
qualitative analysis. 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 form insoluble sulfides. Thus, if H2S 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 separation 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. Solution containing ions
of all cation groups
Filtration Group 1 precipitates
AgCl, Hg2Cl2, PbCl2 Solution containing ions
of remaining groups
Filtration Group 2 precipitates
CuS, CdS, SnS, Bi2S3 Solution containing ions
of remaining groups
Filtration Group 3 precipitates
CoS, FeS, MnS, NiS
ZnS, Al(OH)3, Cr(OH)3 Solution containing ions
of remaining groups
Filtration Group 4 precipitates
BaCO3, CaCO3, SrCO3 Solution contains
Na+, K+, NH + ions
4 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 16.11 APPLICATION OF THE SOLUBILITY PRODUCT PRINCIPLE TO QUALITATIVE ANALYSIS 683 Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry How an Eggshell Is Formed
The formation of the shell of a hen’s egg is a fascinating example of a natural precipitation process.
An average eggshell weighs about 5 grams and
is 40 percent calcium. Most of the calcium in an
eggshell is laid down within a 16 hour period. This
means that it is deposited at a rate of about 125 milligrams per hour. No hen can consume calcium fast
enough to meet this demand. Instead, it is supplied by
special bony masses in the hen’s long bones, which
accumulate large reserves of calcium for eggshell formation. [The inorganic calcium component of the bone
is calcium phosphate, Ca3(PO4)2, an insoluble compound.] If a hen is fed a low-calcium diet, her eggshells
become progressively thinner; she might have to mobilize 10 percent of the total amount of calcium in her
bones just to lay one egg! When the food supply is
consistently low in calcium, egg production eventually
The eggshell is largely composed of calcite, a
crystalline form of calcium carbonate (CaCO3).
Normally, the raw materials, Ca2 and CO 2 , are
carried by the blood to the shell gland. The calcification process is a precipitation reaction:
Ca2 (aq) CO2 (aq) 34 CaCO3(s)
3 In the blood, free Ca2 ions are in equilibrium with
calcium ions bound to proteins. As the free ions are
taken up by the shell gland, more are provided by the
dissociation of the protein-bound calcium.
The carbonate ions necessary for eggshell formation are a metabolic byproduct. Carbon dioxide
produced during metabolism is converted to carbonic
acid (H2CO3) by the enzyme carbonic anhydrase
(CA): KEY EQUATION SUMMARY OF FACTS
AND CONCEPTS Back Forward Main Menu • pH pKa log Chicken eggs.
X-ray micrograph of an eggshell,
showing columns of calcite. CO2(g) CA H2O(l ) 34 H2CO3(aq) Carbonic acid ionizes stepwise to produce carbonate
H2CO3(aq) 34 H (aq) HCO3 (aq) HCO3 (aq) 34 H (aq) CO2 (aq)
3 Chickens do not perspire and so must pant to cool
themselves. Panting expels more CO2 from the
chicken’s body than normal respiration does.
According to Le Chatelier’s principle, panting will shift
the CO2–H2CO3 equilibrium shown above from right
to left, thereby lowering the concentration of the CO2
ions in solution and resulting in thin eggshells. One
remedy for this problem is to give chickens carbonated water to drink in hot weather. The CO2 dissolved
in the water adds CO2 to the chicken’s body fluids
and shifts the CO2–H2CO3 equilibrium to the right. [conjugate base]
[acid] (16.4) Henderson-Hasselbalch equation. 1. The common ion effect tends to suppress the ionization of a weak acid or a weak base.
This action can be explained by Le Chatelier ’s principle.
2. A buffer solution is a combination of either a weak acid and its weak conjugate base (supplied by a salt) or a weak base and its weak conjugate acid (supplied by a salt); the solution
reacts with small amounts of added acid or base in such a way that the pH of the solution remains nearly constant. Buffer systems play a vital role in maintaining the pH of body fluids. TOC Study Guide TOC Textbook Website MHHE Website 684 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 3. The pH at the equivalence point of an acid-base titration depends on hydrolysis of the salt
formed in the neutralization reaction. For strong acid–strong base titrations, the pH at the
equivalence point is 7; for weak acid–strong base titrations, the pH at the equivalence point
is greater than 7; for strong acid–weak base titrations, the pH at the equivalence point is
less than 7.
4. Acid-base indicators are weak organic acids or bases that change color at the equivalence
point in an acid-base neutralization reaction.
5. The solubility product Ksp expresses the equilibrium between a solid and its ions in solution. Solubility can be found from Ksp and vice versa.
6. The presence of a common ion decreases the solubility of a salt.
7. The solubility of slightly soluble salts containing basic anions increases as the hydrogen
ion concentration increases. The solubility of salts with anions derived from strong acids is
unaffected by pH.
8. Complex ions are formed in solution by the combination of a metal cation with a Lewis
base. The formation constant Kf measures the tendency toward the formation of a specific
complex ion. Complex ion formation can increase the solubility of an insoluble substance.
9. Qualitative analysis is the identification of cations and anions in solution. KEY WORDS
Buffer solution, p. 649
Common ion effect, p. 646
Complex ion, p. 676 End point, p. 661
Molar solubility, p. 666
Formation constant (Kf), p. 677 Qualitative analysis, p. 680 Solubility, p. 666
Solubility product (Ksp), p. 665 QUESTIONS AND PROBLEMS†
THE COMMON ION EFFECT
Review Questions 16.1 Use Le Chatelier ’s principle to explain how the
common ion effect affects the pH of a solution.
16.2 Describe the effect on pH (increase, decrease, or no
change) that results from each of the following additions: (a) potassium acetate to an acetic acid solution; (b) ammonium nitrate to an ammonia solution; (c) sodium formate (HCOONa) to a formic acid
(HCOOH) solution; (d) potassium chloride to a hydrochloric acid solution; (e) barium iodide to a hydroiodic acid solution.
Problems 16.3 Determine the pH of (a) a 0.40 M CH3COOH solution, (b) a solution that is 0.40 M CH3COOH and
0.20 M CH3COONa.
16.4 Determine the pH of (a) a 0.20 M NH3 solution,
(b) a solution that is 0.20 M in NH3 and 0.30 M
Review Questions 16.5 What is a buffer solution? What constitutes a buffer
† Back 16.6 Define pKa for a weak acid. What is the relationship between the value of the pKa and the strength
of the acid? Do the same for a weak base.
16.7 The pKas of two monoprotic acids HA and HB are
5.9 and 8.1, respectively. Which of the two is the
16.8 Identify the buffer systems below:
(f ) HCOOK/HCOOH
Problems 16.9 Calculate the pH of the buffer system made up of
0.15 M NH3/0.35 M NH4Cl.
16.10 Calculate the pH of the following two buffer solutions: (a) 2.0 M CH3COONa/2.0 M CH3COOH,
(b) 0.20 M CH3COONa/0.20 M CH3COOH. Which
is the more effective buffer? Why?
16.11 The pH of a bicarbonate-carbonic acid buffer is 8.00.
Calculate the ratio of the concentration of carbonic
acid (H2CO3) to that of the bicarbonate ion (HCO3 ).
16.12 What is the pH of the buffer 0.10 M Na2HPO4/
0.15 M KH2PO4? The temperature is assumed to be 25°C for all the problems. Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS 16.13 The pH of a sodium acetate – acetic acid buffer is
4.50. Calculate the ratio [CH3COO ]/[CH3COOH].
16.14 The pH of blood plasma is 7.40. Assuming the principal buffer system is HCO3 /H2CO3, calculate the
ratio [HCO3 ]/[H2CO3]. Is this buffer more effective
against an added acid or an added base?
16.15 Calculate the pH of the 0.20 M NH3/0.20 M NH4Cl
buffer. What is the pH of the buffer after the addition of 10.0 mL of 0.10 M HCl to 65.0 mL of the
16.16 Calculate the pH of 1.00 L of the buffer 1.00 M
CH3COONa/1.00 M CH3COOH before and after the
addition of (a) 0.080 mol NaOH, (b) 0.12 mol HCl.
(Assume that there is no change in volume.)
16.17 A diprotic acid, H2A, has the following ionization
constants: Ka1 1.1 10 3 and Ka2 2.5 10 6.
In order to make up a buffer solution of pH 5.80,
which combination would you choose? NaHA/H2A
16.18 A student is asked to prepare a buffer solution at
pH 8.60, using one of the following weak acids:
HA (Ka 2.7 10 3), HB (Ka 4.4 10 6), HC
(Ka 2.6 10 9). Which acid should she choose?
Review Questions Problems 16.21 A 0.2688-g sample of a monoprotic acid neutralizes
16.4 mL of 0.08133 M KOH solution. Calculate the
molar mass of the acid.
16.22 A 5.00-g quantity of a diprotic acid was dissolved
in water and made up to exactly 250 mL. Calculate
the molar mass of the acid if 25.0 mL of this solution required 11.1 mL of 1.00 M KOH for neutralization. Assume that both protons of the acid were
16.23 In a titration experiment, 12.5 mL of 0.500 M H2SO4
neutralize 50.0 mL of NaOH. What is the concentration of the NaOH solution?
16.24 In a titration experiment, 20.4 mL of 0.883 M HCOOH
neutralize 19.3 mL of Ba(OH)2. What is the concentration of the Ba(OH)2 solution?
16.25 A 0.1276-g sample of an unknown monoprotic acid
was dissolved in 25.0 mL of water and titrated with Forward Main Menu 0.0633 M NaOH solution. The volume of base required to bring the solution to the equivalence point
was 18.4 mL. (a) Calculate the molar mass of the
acid. (b) After 10.0 mL of base had been added during the titration, the pH was determined to be 5.87.
What is the Ka of the unknown acid?
16.26 A solution is made by mixing exactly 500 mL of
0.167 M NaOH with exactly 500 mL of 0.100 M
CH3COOH. Calculate the equilibrium concentrations of H , CH3COOH, CH3COO , OH , and
16.27 Calculate the pH at the equivalence point for the following titration: 0.20 M HCl versus 0.20 M methylamine.
16.28 Calculate the pH at the equivalence point for the following titration: 0.10 M HCOOH versus 0.10 M
Review Questions 16.29 Explain how an acid-base indicator works in a titration. What are the criteria for choosing an indicator
for a particular acid-base titration?
16.30 The amount of indicator used in an acid-base titration must be small. Why?
Problems 16.19 Briefly describe what happens in an acid-base titration.
16.20 Sketch titration curves for the following acid-base
titrations: (a) HCl versus NaOH, (b) HCl versus
CH3NH2, (c) CH3COOH versus NaOH. In each
case, the base is added to the acid in an Erlenmeyer
flask. Your graphs should show pH on the y-axis and
volume of base added on the x-axis. Back 685 TOC 16.31 Referring to Table 16.1, specify which indicator or
indicators you would use for the following titrations:
(a) HCOOH versus NaOH, (b) HCl versus KOH, (c)
HNO3 versus CH3NH2.
16.32 A student carried out an acid-base titration by
adding NaOH solution from a buret to an
Erlenmeyer flask containing HCl solution and using phenolphthalein as indicator. At the equivalence
point, she observed a faint reddish-pink color.
However, after a few minutes, the solution gradually turned colorless. What do you suppose happened?
16.33 The ionization constant Ka of an indicator HIn is
1.0 10 6. The color of the nonionized form is red
and that of the ionized form is yellow. What is the
color of this indicator in a solution whose pH is
16.34 The Ka of a certain indicator is 2.0 10 6. The
color of HIn is green and that of In is red. A few
drops of the indicator are added to a HCl solution,
which is then titrated against a NaOH solution. At
what pH will the indicator change color?
Review Questions 16.35 Use BaSO4 to distinguish between solubility, molar Study Guide TOC Textbook Website MHHE Website 686 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA solubility, and solubility product.
16.36 Why do we usually not quote the Ksp values for soluble ionic compounds?
16.37 Write balanced equations and solubility product expressions for the solubility equilibria of the following compounds: (a) CuBr, (b) ZnC2O4, (c) Ag2CrO4,
(d) Hg2Cl2, (e) AuCl3, (f) Mn3(PO4)2.
16.38 Write the solubility product expression for the ionic
16.39 How can we predict whether a precipitate will form
when two solutions are mixed?
16.40 Silver chloride has a larger Ksp than silver carbonate (see Table 16.2). Does this mean that AgCl also
has a larger molar solubility than Ag2CO3?
Problems 16.41 Calculate the concentration of ions in the following
saturated solutions: (a) [I ] in AgI solution with
[Ag ] 9.1 10 9 M, (b) [Al3 ] in Al(OH)3 solution with [OH ] 2.9 10 9 M.
16.42 From the solubility data given, calculate the solubility products for the following compounds: (a)
SrF2, 7.3 10 2 g/L, (b) Ag3PO4, 6.7 10 3 g/L.
16.43 The molar solubility of MnCO3 is 4.2 10 6 M.
What is Ksp for this compound?
16.44 The solubility of an ionic compound MX (molar
mass 346 g) is 4.63 10 3 g/L. What is Ksp for
16.45 The solubility of an ionic compound M2X3 (molar
mass 288 g) is 3.6 10 17 g/L. What is Ksp for
16.46 Using data from Table 16.2, calculate the molar solubility of CaF2.
16.47 What is the pH of a saturated zinc hydroxide solution?
16.48 The pH of a saturated solution of a metal hydroxide MOH is 9.68. Calculate the Ksp for the compound.
16.49 If 20.0 mL of 0.10 M Ba(NO3)2 are added to 50.0
mL of 0.10 M Na2CO3, will BaCO3 precipitate?
16.50 A volume of 75 mL of 0.060 M NaF is mixed with
25 mL of 0.15 M Sr(NO3)2. Calculate the concentrations in the final solution of NO3 , Na , Sr2 , and
F . (Ksp for SrF2 2.0 10 10.)
Problems 16.51 Solid NaI is slowly added to a solution that is
0.010 M in Cu and 0.010 M in Ag . (a) Which
compound will begin to precipitate first? (b) Calculate [Ag ] when CuI just begins to precipitate. (c)
What percent of Ag remains in solution at this
point? Back Forward Main Menu TOC 16.52 Find the approximate pH range suitable for the
separation of Fe3 and Zn2 by precipitation of
Fe(OH)3 from a solution that is initially 0.010 M in
both Fe3 and Zn2 .
THE COMMON ION EFFECT AND SOLUBILITY
Review Questions 16.53 How does the common ion effect influence solubility equilibria? Use Le Chatelier ’s principle to explain the decrease in solubility of CaCO3 in a
16.54 The molar solubility of AgCl in 6.5 10 3 M
AgNO3 is 2.5 10 8 M. In deriving Ksp from these
data, which of the following assumptions are reasonable?
(a) Ksp is the same as solubility.
(b) Ksp of AgCl is the same in 6.5 10 3 M AgNO3
as in pure water.
(c) Solubility of AgCl is independent of the concentration of AgNO3.
(d) [Ag ] in solution does not change significantly
upon the addition of AgCl to 6.5 10 3 M
(e) [Ag ] in solution after the addition of AgCl to
6.5 10 3 M AgNO3 is the same as it would
be in pure water.
Problems 16.55 How many grams of CaCO3 will dissolve in 3.0
102 mL of 0.050 M Ca(NO3)2?
16.56 The solubility product of PbBr2 is 8.9 10 6.
Determine the molar solubility (a) in pure water,
(b) in 0.20 M KBr solution, (c) in 0.20 M Pb(NO3)2
16.57 Calculate the molar solubility of AgCl in a solution
made by dissolving 10.0 g of CaCl2 in 1.00 L of
16.58 Calculate the molar solubility of BaSO4 (a) in water, (b) in a solution containing 1.0 M SO2 ions.
pH AND SOLUBILITY
Problems 16.59 Which of the following ionic compounds will be
more soluble in acid solution than in water?
(a) BaSO4, (b) PbCl2, (c) Fe(OH)3, (d) CaCO3
16.60 Which of the following will be more soluble in acid
solution than in pure water? (a) CuI, (b) Ag2SO4,
(c) Zn(OH)2, (d) BaC2O4, (e) Ca3(PO4)2
16.61 Compare the molar solubility of Mg(OH)2 in water
and in a solution buffered at a pH of 9.0.
16.62 Calculate the molar solubility of Fe(OH)2 at (a) pH
8.00, (b) pH 10.00.
16.63 The solubility product of Mg(OH)2 is 1.2 10 11. Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS What minimum OH concentration must be attained
(for example, by adding NaOH) to decrease the
Mg2 concentration in a solution of Mg(NO3)2 to
less than 1.0 10 10 M?
16.64 Calculate whether or not a precipitate will form if
2.00 mL of 0.60 M NH3 are added to 1.0 L of 1.0
10 3 M FeSO4. 687 Pb2 remaining in solution.
16.77 Both KCl and NH4Cl are white solids. Suggest one
reagent that would allow you to distinguish between
these two compounds.
16.78 Describe a simple test that would allow you to distinguish between AgNO3(s) and Cu(NO3)2(s).
ADDITIONAL PROBLEMS COMPLEX ION EQUILIBRIA AND SOLUBILITY
Review Questions 16.65 Explain the formation of complexes in Table 16.3
in terms of Lewis acid-base theory.
16.66 Give an example to illustrate the general effect of
complex ion formation on solubility.
Problems 16.67 If 2.50 g of CuSO4 are dissolved in 9.0 102 mL
of 0.30 M NH3, what are the concentrations of Cu2 ,
Cu(NH3)2 , and NH3 at equilibrium?
16.68 Calculate the concentrations of Cd2 , Cd(CN)2 ,
and CN at equilibrium when 0.50 g of Cd(NO3)2
dissolves in 5.0 102 mL of 0.50 M NaCN.
16.69 If NaOH is added to 0.010 M Al3 , which will be
the predominant species at equilibrium: Al(OH)3 or
Al(OH)4 ? The pH of the solution is 14.00. [Kf for
16.70 Calculate the molar solubility of AgI in a 1.0 M NH3
16.71 Both Ag and Zn2 form complex ions with NH3.
Write balanced equations for the reactions.
However, Zn(OH)2 is soluble in 6 M NaOH, and
AgOH is not. Explain.
16.72 Explain, with balanced ionic equations, why (a)
CuI2 dissolves in ammonia solution, (b) AgBr dissolves in NaCN solution, (c) HgCl2 dissolves in KCl
Review Questions 16.73 Outline the general procedure of qualitative analysis.
16.74 Give two examples of metal ions in each group (1
through 5) in the qualitative analysis scheme.
Problems 16.75 In a group 1 analysis, a student obtained a precipitate containing both AgCl and PbCl2. Suggest one
reagent that would allow her to separate AgCl(s)
16.76 In a group 1 analysis, a student adds HCl acid to the
unknown solution to make [Cl ] 0.15 M. Some
PbCl2 precipitates. Calculate the concentration of Back Forward Main Menu TOC 16.79 The buffer range is defined by the equation pH
pKa 1. Calculate the range of the ratio [conjugate
base]/[acid] that corresponds to this equation.
16.80 The pKa of the indicator methyl orange is 3.46. Over
what pH range does this indicator change from 90
percent HIn to 90 percent In ?
16.81 Sketch the titration curve of a weak acid versus a
strong base like the one shown in Figure 16.4. On
your graph indicate the volume of base used at the
equivalence point and also at the half-equivalence
point, that is, the point at which half of the acid has
been neutralized. Show how you can measure the
pH of the solution at the half-equivalence point.
Using Equation (16.4), explain how you can determine the pKa of the acid by this procedure.
16.82 A 200-mL volume of NaOH solution was added to
400 mL of a 2.00 M HNO2 solution. The pH of the
mixed solution was 1.50 units greater than that of
the original acid solution. Calculate the molarity of
the NaOH solution.
16.83 The pKa of butyric acid (HBut) is 4.7. Calculate Kb
for the butyrate ion (But ).
16.84 A solution is made by mixing exactly 500 mL of
0.167 M NaOH with exactly 500 mL 0.100 M
CH3COOH. Calculate the equilibrium concentrations of H , CH3COOH, CH3COO , OH , and
16.85 Cd(OH)2 is an insoluble compound. It dissolves in
excess NaOH in solution. Write a balanced ionic
equation for this reaction. What type of reaction is
16.86 A student mixes 50.0 mL of 1.00 M Ba(OH)2 with
86.4 mL of 0.494 M H2SO4. Calculate the mass of
BaSO4 formed and the pH of the mixed solution.
16.87 For which of the following reactions is the equilibrium constant called a solubility product?
(a) Zn(OH)2(s) 2OH (aq) 34 Zn(OH)2 (aq)
(b) 3Ca2 (aq) 2PO3 (aq) 34 Ca3(PO4)2(s)
(c) CaCO3(s) 2H (aq) 34
Ca2 (aq) H2O(l ) CO2(g)
(d) PbI2(s) 34 Pb2 (aq) 2I (aq)
16.88 A 2.0-L kettle contains 116 g of boiler scale
(CaCO3). How many times would the kettle have to
be completely filled with distilled water to remove
all of the deposit at 25°C?
16.89 Equal volumes of 0.12 M AgNO3 and 0.14 M ZnCl2 Study Guide TOC Textbook Website MHHE Website 688 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 16.90
16.91 16.92 16.93 16.99 16.100 16.101 Mass of HgI2 formed 16.94 solution are mixed. Calculate the equilibrium concentrations of Ag , Cl , Zn2 , and NO3 .
Calculate the solubility (in g/L) of Ag2CO3.
Find the approximate pH range suitable for separating Fe3 and Zn2 by the precipitation of
Fe(OH)3 from a solution that is initially 0.010 M in
Fe3 and Zn2 .
A volume of 25.0 mL of 0.100 M HCl is titrated
against a 0.100 M CH3NH2 solution added to it from
a buret. Calculate the pH values of the solution (a)
after 10.0 mL of CH3NH2 solution have been added,
(b) after 25.0 mL of CH3NH2 solution have been
added, (c) after 35.0 mL of CH3NH2 solution have
The molar solubility of Pb(IO3)2 in a 0.10 M NaIO3
solution is 2.4 10 11 mol/L. What is Ksp for
When a KI solution was added to a solution of mercury(II) chloride, a precipitate [mercury(II) iodide]
formed. A student plotted the mass of the precipitate versus the volume of the KI solution added and
obtained the following graph. Explain the appearance of the graph. Volume of KI added 16.102
16.95 Barium is a toxic substance that can seriously impair heart function. For an X ray of the gastrointestinal tract, a patient drinks an aqueous suspension
of 20 g BaSO4. If this substance were to equilibrate
with the 5.0 L of the blood in the patient’s body,
what would be [Ba2 ]? For a good estimate, we may
assume that the temperature is at 25°C. Why is
Ba(NO3)2 not chosen for this procedure?
16.96 The pKa of phenolphthalein is 9.10. Over what pH
range does this indicator change from 95 percent
HIn to 95 percent In ?
16.97 Solid NaI is slowly added to a solution that is
0.010 M in Cu and 0.010 M in Ag . (a) Which compound will begin to precipitate first? (b) Calculate
[Ag ] when CuI just begins to precipitate. (c) What
percent of Ag remains in solution at this point?
16.98 Cacodylic acid is (CH3)2AsO2H. Its ionization constant is 6.4 10 7. (a) Calculate the pH of 50.0 mL
of a 0.10 M solution of the acid. (b) Calculate the
pH of 25.0 mL of 0.15 M (CH3)2AsO2Na. (c) Mix Back Forward Main Menu TOC 16.103 16.104 the solutions in part (a) and part (b). Calculate the
pH of the resulting solution.
Radiochemical techniques are useful in estimating
the solubility product of many compounds. In one
experiment, 50.0 mL of a 0.010 M AgNO3 solution
containing a silver isotope with a radioactivity of
74,025 counts per min per mL were mixed with
100 mL of a 0.030 M NaIO3 solution. The mixed
solution was diluted to 500 mL and filtered to remove all of the AgIO3 precipitate. The remaining
solution was found to have a radioactivity of 44.4
counts per min per mL. What is the Ksp of AgIO3?
The molar mass of a certain metal carbonate, MCO3,
can be determined by adding an excess of HCl acid
to react with all the carbonate and then “back-titrating” the remaining acid with NaOH. (a) Write an
equation for these reactions. (b) In a certain experiment, 20.00 mL of 0.0800 M HCl were added to a
0.1022-g sample of MCO3. The excess HCl required
5.64 mL of 0.1000 M NaOH for neutralization.
Calculate the molar mass of the carbonate and identify M.
Acid-base reactions usually go to completion.
Confirm this statement by calculating the equilibrium constant for each of the following cases: (a) A
strong acid reacting with a strong base. (b) A strong
acid reacting with a weak base (NH3). (c) A weak
acid (CH3COOH) reacting with a strong base. (d) A
weak acid (CH3COOH) reacting with a weak base
(NH3). (Hint: Strong acids exist as H ions and
strong bases exist as OH ions in solution. You need
to look up Ka, Kb, and Kw.)
Calculate x, the number of molecules of water in oxalic acid hydrate, H2C2O4 xH2O, from the following data: 5.00 g of the compound is made up to exactly 250 mL solution, and 25.0 mL of this solution
requires 15.9 mL of 0.500 M NaOH solution for neutralization.
Describe how you would prepare a 1-L 0.20 M
CH3COONa/0.20 M CH3COOH buffer system by
(a) mixing a solution of CH3COOH with a solution
of CH3COONa, (b) reacting a solution of
CH3COOH with a solution of NaOH, and (c) reacting a solution of CH3COONa with a solution of HCl.
Phenolphthalein is the common indicator for the
titration of a strong acid with a strong base. (a) If
the pKa of phenolphthalein is 9.10, what is the ratio of the nonionized form of the indicator (colorless) to the ionized form (reddish pink) at pH 8.00?
(b) If 2 drops of 0.060 M phenolphthalein are used
in a titration involving a 50.0-mL volume, what is
the concentration of the ionized form at pH 8.00?
(Assume that 1 drop 0.050 mL.) Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS 16.105 Oil paintings containing lead(II) compounds as constituents of their pigments darken over the years.
Suggest a chemical reason for the color change.
16.106 What reagents would you employ to separate the
following pairs of ions in solution? (a) Na and
Ba2 , (b) K and Pb2 , (c) Zn2 and Hg2 .
16.107 Look up the Ksp values for BaSO4 and SrSO4 in
Table 16.2. Calculate the concentrations of Ba2 ,
Sr2 , and SO2 in a solution that is saturated with
16.108 In principle, amphoteric oxides, such as Al2O3 and
BeO can be used to prepare buffer solutions because
they possess both acidic and basic properties (see
Section 15.11). Explain why these compounds are
of little practical use as buffer components.
16.109 CaSO4 (Ksp 2.4 10 5) has a larger Ksp value
than that of Ag2SO4 (Ksp 1.4 10 5). Does it
follow that CaSO4 also has greater solubility (g/L)?
16.110 When lemon juice is squirted into tea, the color becomes lighter. In part, the color change is due to dilution, but the main reason for the change is an acidbase reaction. What is the reaction? (Hint: Tea
contains “polyphenols” which are weak acids and
lemon juice contains citric acid.)
16.111 How many milliliters of 1.0 M NaOH must be added
to a 200 mL of 0.10 M NaH2PO4 to make a buffer
solution with a pH of 7.50?
16.112 The maximum allowable concentration of Pb2 ions
in drinking water is 0.05 ppm (that is, 0.05 g of Pb2
in one million grams of water). Is this guideline exceeded if an underground water supply is at equilibrium with the mineral anglesite, PbSO4 (Ksp
1.6 10 8)?
16.113 One of the most common antibiotics is penicillin G
(benzylpenicillinic acid), which has the following
C OCO NOCO CH2
HH It is a weak monoprotic acid:
P Ka 1.64 10 3 where HP denotes the parent acid and P the conjugate base. Penicillin G is produced by growing
molds in fermentation tanks at 25°C and a pH range
of 4.5 to 5.0. The crude form of this antibiotic is obtained by extracting the fermentation broth with an Back Forward Main Menu TOC organic solvent in which the acid is soluble. (a)
Identify the acidic hydrogen atom. (b) In one stage
of purification, the organic extract of the crude penicillin G is treated with a buffer solution at pH
6.50. What is the ratio of the conjugate base of penicillin G to the acid at this pH? Would you expect
the conjugate base to be more soluble in water than
the acid? (c) Penicillin G is not suitable for oral administration, but the sodium salt (NaP) is because it
is soluble. Calculate the pH of a 0.12 M NaP solution formed when a tablet containing the salt is dissolved in a glass of water.
16.114 Which of the following solutions has the highest
[H ]? (a) 0.10 M HF, (b) 0.10 M HF in 0.10 M NaF,
(c) 0.10 M HF in 0.10 M SbF5. (Hint: SbF5 reacts
with F to form the complex ion SbF6 .)
16.115 Distribution curves show how the fractions of nonionized acid and its conjugate base vary as a function of pH of the medium. Plot distribution curves
for CH3COOH and its conjugate base CH3COO in
solution. Your graph should show fraction as the y
axis and pH as the x axis. What are the fractions and
pH at the point where these two curves intersect?
16.116 Water containing Ca2 and Mg2 ions is called
hard water and is unsuitable for some household
and industrial use because these ions react with soap
to form insoluble salts, or curds. One way to remove the Ca2 ions from hard water is by adding
washing soda (Na2CO3 10H2O). (a) The molar
solubility of CaCO3 is 9.3 10 5 M. What is its
molar solubility in a 0.050 M Na2CO3 solution? (b)
Why are Mg2 ions not removed by this procedure?
(c) The Mg2 ions are removed as Mg(OH)2 by
adding slaked lime [Ca(OH)2] to the water to produce a saturated solution. Calculate the pH of a saturated Ca(OH)2 solution. (d) What is the concentration of Mg2 ions at this pH? (e) In general,
which ion (Ca2 or Mg2 ) would you remove first?
Answers to Practice Exercises: 16.1 4.01. 16.2 (a) and J HP 34 H 689 (c). 16.3 9.17; 9.20. 16.4 Weigh out Na2CO3 and NaHCO3 in
mole ratio of 0.60 to 1.0. Dissolve in enough water to make up
a 1-L solution. 16.5 (a) 2.19, (b) 3.95, (c) 8.02, (d) 11.39.
16.6 5.92. 16.7 (a) Bromophenol blue, methyl orange, methyl
red, and chlorophenol blue; (b) all except thymol blue, bromophenol blue, and methyl orange; (c) cresol red and phenolphthalein. 16.8 2.0 10 14. 16.9 1.9 10 3 g/L. 16.10 No.
16.11 (a) 1.6 10 9 M, (b) 2.6 10 6 M. 16.12 (a)
1.7 10 4 g/L, (b) 1.4 10 7 g/L. 16.13 (a) More soluble in
acid solution, (b) more soluble in acid solution, (c) about the
same. 16.14 A precipitate of Fe(OH)2 will form. 16.15 [Cu2 ]
1.2 10 13 M, [Cu(NH3)2 ] 0.017 M, [NH3] 0.23 M.
16.16 3.5 10 3 mol/L. Study Guide TOC Textbook Website MHHE Website C HEMICAL M YSTERY A Hard-boiled Snack M ost of us have eaten hard-boiled eggs. They are easy to cook and nutritious. But when was the last time you thought about the process of
boiling an egg or looked carefully at a hard-boiled egg? A lot of interesting chemical and physical changes occur while an egg cooks.
A hen’s egg is a complicated biochemical system, but here we
will focus on the three major parts that we see when we crack open an egg: the shell,
the egg white or albumen, and the yolk. The shell protects the inner components from
the outside environment, but it has many microscopic pores through which air can pass.
The albumen is about 88 percent water and 12 percent protein. The yolk contains 50
percent water, 34 percent fat, 16 percent protein, and a small amount of iron in the
form of Fe2 ions.
Proteins are polymers made up of amino acids. In solution, each long chain of a
protein molecule folds in such a way that the hydrophobic parts of the molecule are
buried inside and the hydrophilic parts are on the exterior, in contact with the solution.
This is the stable or native state of a protein which allows it to perform normal physiological functions. Heat causes protein molecules to unfold, or denature. Chemicals
such as acids and salt (NaCl) can also denature proteins. To avoid contact with water,
the hydrophobic parts of denatured proteins will clump together, or coagulate to form
a semirigid opaque white solid. Heating also decomposes some proteins so that the sulfur in them combines with hydrogen to form hydrogen sulfide (H2S), an unpleasant
smelling gas that can sometimes be detected when the shell of a boiled egg is cracked.
The accompanying photo of hard-boiled eggs shows an egg that has been boiled
for about 12 minutes and one that has been overcooked. Note that the outside of the
overcooked yolk is green.
What is the chemical basis for the changes brought about by boiling an egg? Shell Schematic diagram of an egg.
The chalazae are the cords that
anchor the yolk to the shell and
keep it centered. Membrane
Air space 690 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website CHEMICAL CLUES One frequently encountered problem with hard-boiled eggs is that their shells crack
in water. The recommended procedure for hard boiling eggs is to place the eggs in
cold water and then bring the water to a boil. What causes the shells to crack in
this case? How does pin holing, that is, piercing the shell with a needle, prevent
the shells from cracking? A less satisfactory way of hard boiling eggs is to place
room-temperature eggs or cold eggs from the refrigerator in boiling water. What
additional mechanism might cause the shells to crack?
2. When an eggshell cracks during cooking, some of the egg white leaks into the hot
water to form unsightly “streamers.” An experienced cook adds salt or vinegar to
the water prior to heating eggs to minimize the formation of streamers. Explain the
chemical basis for this action.
3. Identify the green substance on the outer layer of the yolk of an overcooked egg
and write an equation representing its formation. The unsightly “green yolk” can
be eliminated or minimized if the overcooked egg is rinsed with cold water immediately after it has been removed from the boiling water. How does this action
remove the green substance?
4. The way to distinguish a raw egg from a hard-boiled egg, without cracking the
shells, is to spin the eggs. How does this method work?
1. A 12-minute egg (left) and an
overcooked hard-boiled egg
(right). Iron(II) sulfide. 691 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website...
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- Spring '07