acid-base equilibria and solubility equilibria€¦ · 576 chapter 17 acid-base equilibria and...

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Acid-Base Equilibria and Solubility Equilibria

ESSENTIAL CONCEPTSBuffer Solutions A buffer solution contains a weak acid and asalt derived from the acid. To maintain a relatively constant pH,the acid and base components of the buffer solution react withadded acid or base. Buffer solutions play an important role inmany chemical and biological processes.

Acid-Base Titrations The characteristics of an acid-base titra-tion depend on the strength of the acid and base involved.Different indicators are used to determine the end point of atitration.

Solubility Equilibria Another application of the equilibriumconcept is the solubility equilibria of sparingly soluble salts,which are expressed as the solubility product. The solubility ofsuch a substance can be affected by the presence of a commoncation or anion, or the pH. Complex-ion formation, an exampleof the Lewis acid-base type reaction, increases the solubility ofan insoluble salt.

C H A P T E R

Activity Summary

1. Animation: Buffer Solutions (17.2)2. Animation: Acid-Base Titrations (17.3)

3. Interactivity: Neutralization Reaction I & II (17.3)

CHAPTER OUTLINE

17.1 Homogeneous Versus Heterogeneous SolutionEquilibria 575

17.2 Buffer Solutions 575Preparing a Buffer Solution with a Specific pH

17.3 A Closer Look at Acid-Base Titrations 580Strong Acid–Strong Base Titrations •Weak Acid–Strong Base Titrations •Strong Acid–Weak Base Titrations

17.4 Acid-Base Indicators 58617.5 Solubility Equilibria 589

Solubility Product • Molar Solubility and Solubility •Predicting Precipitation Reactions

17.6 The Common Ion Effect and Solubility 59617.7 Complex Ion Equilibria and Solubility 59717.8 Application of the Solubility Product Principle

to Qualitative Analysis 600

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17.2 Buffer Solutions 575

17.1 Homogeneous Versus HeterogeneousSolution Equilibria

In Chapter 16 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 aswell as H� ions and the conjugate base. Nevertheless, all of these species are dissolved,so that the system is an example of homogeneous equilibrium (see Chapter 15).

Another important type of equilibrium, which we will study in the second halfof the chapter, involves the dissolution and precipitation of slightly soluble substances.These processes are examples of heterogeneous equilibria; that is, they pertain to reac-tions in which the components are in more than one phase. But first we will concludeour discussion of acid-base equilibria by considering buffer solutions and taking acloser look at acid-base titrations.

17.2 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 pHupon the addition of small amounts of either acid or base. Buffers are very impor-tant to chemical and biological systems. The pH in the human body varies greatlyfrom one fluid to another; for example, the pH of blood is about 7.4, whereas the gas-tric juice in our stomachs has a pH of about 1.5. These pH values, which are crucialfor the proper functioning of enzymes and the balance of osmotic pressure, are main-tained by buffers in most cases.

A buffer solution must contain a relatively large concentration of acid to reactwith any OH� ions that may be added to it and must contain a similar concentrationof base to react with any added H� ions. Furthermore, the acid and the base compo-nents of the buffer must not consume each other in a neutralization reaction. Theserequirements are satisfied by an acid-base conjugate pair (a weak acid and its conju-gate base or a weak base and its conjugate acid).

A simple buffer solution can be prepared by adding comparable amounts of aceticacid (CH3COOH) and sodium acetate (CH3COONa) to water. The equilibrium con-centrations of both the acid and the conjugate base (from CH3COONa) are assumedto be the same as the starting concentrations. This is so because (1) CH3COOH isa weak acid and the extent of hydrolysis of the CH3COO

� ion is very small and(2) the presence of CH3COO

� ions suppresses the ionization of CH3COOH, and thepresence of CH3COOH suppresses the hydrolysis of the CH3COO

� ions.A solution containing these two substances has the ability to neutralize either

added acid or added base. Sodium acetate, a strong electrolyte, dissociates completelyin water:

If an acid is added, the H� ions will be consumed by the conjugate base in the buffer,CH3COO

�, according to the equation

If a base is added to the buffer system, the OH� ions will be neutralized by the acidin the buffer:

CH3COOH(aq) � OH�(aq) ¡ CH3COO�(aq) � H2O(l)

CH3COO�(aq) � H�(aq) ¡ CH3COOH(aq)

CH3COONa(s) ¡H2O CH3COO�(aq) � Na�(aq)

Fluids for intravenous injectionmust include buffer systems tomaintain the proper blood pH.

Animation:Buffer solutionsARIS, Animations


576 CHAPTER 17 Acid-Base Equilibria and Solubility Equilibria

The buffering capacity, that is, the effectiveness of the buffer solution, depends on theamounts of acid and conjugate base from which the buffer is made. The larger theamount, 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 can be written as CH3COONa/CH3COOHor CH3COO

�/CH3COOH. Figure 17.1 shows this buffer system in action.

Figure 17.1The acid-base indicator bromophenol blue (added to all solutions shown) is used to illustratebuffer 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 MCH3COONa. The solution has a pH of 4.7 and turns the indicator blue-purple. (b) After theaddition 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 pHof the solution decreases rapidly to less than 3.0 upon the addition of 0.1 M HCl.

Which of the following solutions can be classified as buffer systems? (a) KH2PO4/H3PO4,(b) NaClO4/HClO4, (c) C5H5N/C5H5NHCl (C5H5N is pyridine; its Kb is given inTable 16.5). Explain your answer.

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

Solution The criteria for a buffer system is that we must have a weak acid and its salt(containing the weak conjugate base) or a weak base and its salt (containing the weakconjugate acid).

(a) H3PO4 is a weak acid, and its conjugate base, H2PO4�, is a weak base (see

Table 16.4). 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.

(c) As Table 16.5 shows, C5H5N is a weak base and its conjugate acid, (thecation 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) Na2CO3/NaHCO3.

C5H5N�

H

Example 17.1

Similar problems: 17.5, 17.6.

(a) (b) (c) (d)



(a) Calculate the pH of a buffer system containing 1.0 M CH3COOH and 1.0 MCH3COONa. (b) What is the pH of the buffer system after the addition of 0.10 mole ofgaseous HCl to 1.0 L of the solution? Assume that the volume of the solution does notchange when the HCl is added.

Strategy (a) The pH of the buffer solution before the addition of HCl can be calculatedfrom the ionization of CH3COOH. Note that because both the acid and the sodium salt ofthe acid are present, the initial concentrations of CH3COOH and CH3COO

� (fromCH3COONa) are both 1.0 M. The Ka of CH3COOH is 1.8 � 10

�5 (see Table 16.3). (b) Itis helpful to make a sketch of the changes that occur in this case.

Example 17.2

Recall that the presence of CH3COOHsuppresses the hydrolysis of CH3COO

�

and the presence of CH3COO�

suppresses the ionization of CH3COOH.

Solution (a) We summarize the concentrations of the species at equilibrium as follows:

CH3COOH(aq) 34 H�(aq) � CH3COO

�(aq)Initial (M): 1.0 0 1.0Change (M): �x �x �x

Equilibrium (M): 1.0 � x x 1.0 � x

Assuming 1.0 � x � 1.0 and 1.0 � x � 1.0, we obtain

or

Thus,

(b) When HCl is added to the solution, the initial changes are

HCl(aq) 88n H�(aq) � Cl�(aq)Initial (mol): 0.10 0 0Change (mol): �0.10 �0.10 �0.10

Final (mol): 0 0.10 0.10

The Cl� ion is a spectator ion in solution because it is the conjugate base of a strong acid.

The H� ions provided by the strong acid HCl react completely with the conjugate base of the buffer, which is CH3COO

�. At this point it is more convenient to work with moles rather than molarity. The reason is that in some cases the volume of the solution may change when a substance is added. A change

(Continued )

pH � �log (1.8 � 10�5) � 4.74

x � [H�] � 1.8 � 10�5 M

1.8 � 10�5 �(x)(1.0 � x)

(1.0 � x)�

x(1.0)

1.0

1.8 � 10�5 �(x)(1.0 � x)

(1.0 � x)

Ka �[H�][CH3COO

�]

[CH3COOH]



In the buffer solution examined in Example 17.2, there is a decrease in pH (thesolution becomes more acidic) as a result of added HCl. We can also compare thechanges in H� ion concentration as follows

Before addition of HCl:After addition of HCl:

Thus, the H� ion concentration increases by a factor of

To appreciate the effectiveness of the CH3COONa/CH3COOH buffer, let us findout what would happen if 0.10 mol HCl were added to 1 L of water, and comparethe increase in H� ion concentration.

Before addition of HCl:After addition of HCl:

As a result of the addition of HCl, the H� ion concentration increases by a factor of

0.10 M

1.0 � 10�7 M� 1.0 � 106

[H�] � 0.10 M[H�] � 1.0 � 10�7 M

2.2 � 10�5 M

1.8 � 10�5 M� 1.2

[H�] � 2.2 � 10�5 M[H�] � 1.8 � 10�5 M

Similar problem: 17.14

in volume will change the molarity, but not the number of moles. The neutralizationreaction is summarized next:

CH3COO�(aq) � H�(aq) 88n CH3COOH(aq)

Initial (mol): 1.0 0.10 1.0Change (mol): �0.10 �0.10 �0.10

Final (mol): 0.90 0 1.1

Finally, to calculate the pH of the buffer after neutralization of the acid, we con-vert back to molarity by dividing moles by 1.0 L of solution.

CH3COOH(aq) 34 H�(aq) � CH3COO

�(aq)Initial (M): 1.1 0 0.90Change (M): �x �x �x

Equilibrium (M): 1.1 � x x 0.90 � x

Assuming 0.90 � x � 0.90 and 1.1 � x � 1.1, we obtain

or

Thus,

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 thebuffer solution?

pH � �log (2.2 � 10�5) � 4.66

x � [H�] � 2.2 � 10�5 M

1.8 � 10�5 �(x)(0.90 � x)

1.1 � x�

x(0.90)

1.1

1.8 � 10�5 �(x)(0.90 � x)

1.1 � x

Ka �[H�][CH3COO

�]

[CH3COOH]



amounting to a millionfold increase! This comparison shows that a properly chosenbuffer solution can maintain a fairly constant H� ion concentration, or pH (Figure 17.2).

Preparing a Buffer Solution with a Specific pH

Now suppose we want to prepare a buffer solution with a specific pH. How do wego about it? Referring to the acetic acid–sodium acetate buffer system, we can writethe equilibrium constant as

Note that this expression holds whether we have only acetic acid or a mixture of aceticacid and sodium acetate in solution. Rearranging the equation gives

Taking the negative logarithm of both sides, we obtain

or

So

(17.1)

in which

(17.2)

Equation (17.1) is called the Henderson-Hasselbalch equation. In a more generalform, it can be expressed as

(17.3)

If the molar concentrations of the acid and its conjugate base are approximatelyequal, that is, [acid] � [conjugate base], then

or

Thus, to prepare a buffer solution, we choose a weak acid whose pKa is close to thedesired pH. This choice not only gives the correct pH value of the buffer system, butalso ensures that we have comparable amounts of the acid and its conjugate base pres-ent; both are prerequisites for the buffer system to function effectively.

pH � pKa

log [conjugate base]

[acid]� 0

pH � pKa � log [conjugate base]

[acid]

pKa � �log Ka

pH � pKa � log [CH3COO

�]

[CH3COOH]

�log [H�] � �log Ka � log [CH3COO

�]

[CH3COOH]

�log [H�] � �log Ka � log [CH3COOH]

[CH3COO�]

[H�] �Ka[CH3COOH]

[CH3COO�]

Ka �[CH3COO

�][H�]

[CH3COOH] Figure 17.2A comparison of the change inpH when 0.10 mol HCl isadded to pure water and to anacetate buffer solution, asdescribed in Example 17.2.

0

7654321

pH

0 0.02 0.04

Water

0.06Mole of HCl added

0.08 0.10

Buffer solution

pKa is related to Ka as pH is related to[H�]. Remember that the stronger theacid (that is, the larger the Ka), thesmaller the pKa.

Keep in mind that pKa is a constant, butthe ratio of the two concentration termsin Equation (17.3) depends on a particularsolution.



17.3 A Closer Look at Acid-Base Titrations

Having discussed buffer solutions, we can now look in more detail at the quantitativeaspects of acid-base titrations (see Section 4.6). We will consider three types of reactions:(1) titrations involving a strong acid and a strong base, (2) titrations involving a weak acidand a strong base, and (3) titrations involving a strong acid and a weak base. Titrations

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

Strategy For a buffer to function effectively, the concentrations of the acid componentmust be roughly equal to the conjugate base component. According to Equation (17.3),when the desired pH is close to the pKa of the acid, that is, when pH � pKa,

or

Solution Because phosphoric acid is a triprotic acid, we write the three stages ofionization as follows. The Ka values are obtained from Table 16.4 and the pKa valuesare found by applying Equation (17.2).

The most suitable of the three buffer systems is HPO42�/H2PO

�4, because the pKa of the

acid H2PO4� is closest to the desired pH. From the Henderson-Hasselbalch equation we

write

Taking the antilog, we obtain

Thus, one way to prepare a phosphate buffer with a pH of 7.40 is to dissolve disodiumhydrogen phosphate (Na2HPO4) and sodium dihydrogen phosphate (NaH2PO4) in amole ratio of 1.5:1.0 in water. For example, we could dissolve 1.5 moles of Na2HPO4and 1.0 mole of NaH2PO4 in enough water to make up a 1-L solution.

Practice Exercise How would you prepare a liter of “carbonate buffer” at a pH of10.10? You are provided with carbonic acid (H2CO3), sodium hydrogen carbonate(NaHCO3), and sodium carbonate (Na2CO3). See Table 16.4 for Ka values.

[HPO2�4 ]

[H2PO�4 ]

� 100.19 � 1.5

log [HPO2�4 ]

[H2PO�4 ]

� 0.19

7.40 � 7.21 � log [HPO2�4 ]

[H2PO�4 ]

pH � pKa � log [conjugate base]

[acid]

Ka3 � 4.8 � 10�13; pKa3 � 12.32HPO

2�4 (aq) Δ H�(aq) � PO3�4 (aq)

Ka2 � 6.2 � 10�8; pKa2 � 7.21H2PO

�4 (aq) Δ H�(aq) � HPO2�4 (aq)

Ka1 � 7.5 � 10�3; pKa1 � 2.12H3PO4(aq) Δ H

�(aq) � H2PO�4 (aq)

[conjugate base]

[acid]� 1

log [conjugate base]

[acid]� 0

Example 17.3

Animation:Acid-Base TitrationsARIS, Animations


17.3 A Closer Look at Acid-Base Titrations 581

involving a weak acid and a weak base are complicated by the hydrolysis of both thecation and the anion of the salt formed. These titrations will not be dealt with here.Figure 17.3 shows the arrangement for monitoring the pH during the course of a titration.

Strong Acid–Strong Base Titrations

The reaction between a strong acid (say, HCl) and a strong base (say, NaOH) can berepresented by

or in terms of the net ionic equation

Consider the addition of a 0.100 M NaOH solution (from a buret) to an Erlenmeyerflask containing 25.0 mL of 0.100 M HCl. For convenience, we will use only threesignificant figures for volume and concentration and two significant figures for pH.Figure 17.4 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 (0.100), or 1.00.When NaOH is added, the pH of the solution increases slowly at first. Near the equiv-alence point the pH begins to rise steeply, and at the equivalence point (that is, thepoint at which equimolar amounts of acid and base have reacted) the curve risesalmost vertically. In a strong acid–strong base titration both the hydrogen ion andhydroxide ion concentrations are very small at the equivalence point (approximately1 � 10�7 M); consequently, the addition of a single drop of the base can cause alarge increase in [OH�] and in the pH of the solution. Beyond the equivalence point,the pH again increases slowly with the addition of NaOH.

It is possible to calculate the pH of the solution at every stage of titration. Hereare three sample calculations:

1. After the addition of 10.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCl.The total volume of the solution is 35.0 mL. The number of moles of NaOH in10.0 mL is

10.0 mL �0.100 mol NaOH

1 L NaOH�

1 L

1000 mL� 1.00 � 10�3 mol

H�(aq) � OH�(aq) ¡ H2O(l)

NaOH(aq) � HCl(aq) ¡ NaCl(aq) � H2O(l)

Figure 17.3A pH meter is used to monitoran acid-base titration.

Interactivity:Neutralization Reaction I and IIARIS, Interactives



The number of moles of HCl originally present in 25.0 mL of solution is

Thus, the amount of HCl left after partial neutralization is (2.50 � 10�3) �(1.00 � 10�3), or 1.50 � 10�3 mol. Next, the concentration of H� ions in35.0 mL of solution is found as follows:

Thus, [H�] � 0.0429 M, and the pH of the solution is

2. After the addition of 25.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCl.This is a simple calculation, because it involves a complete neutralization reac-tion and the salt (NaCl) does not undergo hydrolysis. At the equivalence point,[H�] � [OH�] � 1.00 � 10�7 M and the pH of the solution is 7.00.

3. After the addition of 35.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCl. Thetotal volume of the solution is now 60.0 mL. The number of moles of NaOHadded is

The number of moles of HCl in 25.0 mL solution is 2.50 � 10�3 mol. Aftercomplete neutralization of HCl, the number of moles of NaOH left is

35.0 mL �0.100 mol NaOH

1 L NaOH�

1 L

1000 mL� 3.50 � 10�3 mol

pH � �log 0.0429 � 1.37

� 0.0429 M HCl

1.50 � 10�3 mol HCl

35.0 mL�

1000 mL

1 L� 0.0429 mol HCl/L

25.0 mL �0.100 mol HCl

1 L HCl�

1 L

1000 mL� 2.50 � 10�3 mol

Volume of NaOH added (mL)

pH

5010

Equivalencepoint

403020

14

0

13

11

12

10

9

8

7

6

5

4

3

2

1

Volume NaOHadded (mL)

0.05.0

10.015.020.022.024.025.026.028.030.035.040.045.050.0

1.001.181.371.601.952.202.697.00

11.2911.7511.9612.2212.3612.4612.52

pH

Figure 17.4pH profile of a strong acid–strong base titration. A 0.100 M NaOH solution is added from a buret to 25.0 mL of a 0.100 M HCl solution in an Erlenmeyer flask (see Figure 4.21). This curve is sometimes referred to as a titration curve.

Keep in mind that 1 mol NaOH � 1 mol HCl.

Neither Na� nor Cl� undergoeshydrolysis.



(3.50 � 10�3) � (2.50 � 10�3), or 1.00 � 10�3 mol. The concentration of NaOHin 60.0 mL of solution is

Thus, [OH�] � 0.0167 M and pOH � �log 0.0167 � 1.78. Hence, the pH ofthe solution is

Weak Acid–Strong Base Titrations

Consider the neutralization reaction between acetic acid (a weak acid) and sodiumhydroxide (a strong base):

This equation can be simplified to

The acetate ion undergoes hydrolysis as follows:

Therefore, at the equivalence point, when we only have sodium acetate present, the pHwill be greater than 7 as a result of the excess OH� ions formed (Figure 17.5). Note thatthis situation is analogous to the hydrolysis of sodium acetate (CH3COONa) (see p. 558).

CH3COO�(aq) � H2O(l) Δ CH3COOH(aq) � OH�(aq)

CH3COOH(aq) � OH�(aq) ¡ CH3COO�(aq) � H2O(l)

CH3COOH(aq) � NaOH(aq) ¡ CH3COONa(aq) � H2O(l)

� 12.22 � 14.00 � 1.78

pH � 14.00 � pOH

� 0.0167 M NaOH

1.00 � 10�3 mol NaOH

60.0 mL�

1000 mL

1 L� 0.0167 mol NaOH/ L


pH

5010 403020

14

0

13

11

12

10

9

8

7

6

5

4

3

2

1

Equivalencepoint

Volume NaOHadded (mL)

0.05.0

10.015.020.022.024.025.026.028.030.035.040.045.050.0

2.874.144.574.925.355.616.128.72

10.2911.7511.9612.2212.3612.4612.52

pH

Figure 17.5pH profile of a weak acid–strong base titration. A 0.100 M NaOH solution is added from a buret to 25.0 mL of a 0.100 M CH3COOHsolution in an Erlenmeyer flask. Due to the hydrolysis of the salt formed, the pH at the equivalence point is greater than 7.



Because the volume of the solution is thesame for CH3COOH and CH3COO

�, theratio of the number of moles present isequal to the ratio of their molarconcentrations.

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

Strategy The reaction between CH3COOH and NaOH is

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

Solution (a) The number of moles of NaOH in 10.0 mL is

The number of moles of CH3COOH originally present in 25.0 mL of solution is

We work with moles at this point because when two solutions are mixed, thesolution volume increases. As the volume increases, molarity will change but thenumber of moles will remain the same. The changes in number of moles aresummarized next:

Initial (mol): 2.50 � 10�3 1.00 � 10�3 0Change (mol): �1.00 � 10�3 �1.00 � 10�3 �1.00 � 10�3

Final (mol): 1.50 � 10�3 0 1.00 � 10�3

At this stage we have a buffer system made up of CH3COOH and CH3COO� (from

the salt, CH3COONa). To calculate the pH of the solution, we write

Therefore, pH � �log (2.7 � 10�5) � 4.57

(b) These quantities (that is, 25.0 mL of 0.100 M NaOH reacting with 25.0 mL of0.100 M CH3COOH) correspond to the equivalence point. The number of moles ofNaOH in 25.0 mL of the solution is

(Continued )

25.0 mL �0.100 mol NaOH

1 L NaOH soln�

1 L

1000 mL� 2.50 � 10�3 mol

�(1.50 � 10�3)(1.8 � 10�5)

1.00 � 10�3� 2.7 � 10�5 M

[H�] �[CH3COOH]Ka

[CH3COO�]

Ka �[H�][CH3COO

�]

[CH3COOH]


25.0 mL �0.100 mol CH3COOH

1 L CH3COOH soln�

1 L

1000 mL� 2.50 � 10�3 mol

10.0 mL �0.100 mol NaOH

1 L NaOH soln�

1 L

1000 mL� 1.00 � 10�3 mol


Example 17.4



Strong Acid–Weak Base Titrations

Consider the titration of HCl, a strong acid, with NH3, a weak base:

HCl(aq) � NH3(aq) ¡ NH4Cl(aq)

Similar problem: 17.19(b).

The changes in number of moles are summarized next:


Final (mol): 0 0 2.50 � 10�3

At the equivalence point, the concentrations of both the acid and the base are zero.The total volume is (25.0 � 25.0) mL or 50.0 mL, so the concentration of the salt is

The next step is to calculate the pH of the solution that results from the hydrolysisof the CH3COO

� ions. Following the procedure described in Example 16.13 andlooking up the base ionization constant (Kb) for CH3COO

� in Table 16.3, we write

(c) After the addition of 35.0 mL of NaOH, the solution is well past the equivalencepoint. The number of moles of NaOH originally present is

The changes in number of moles are summarized next:


Final (mol): 0 1.00 � 10�3 2.50 � 10�3

At this stage we have two species in solution that are responsible for making thesolution basic: OH� and CH3COO

� (from CH3COONa). However, because OH� is

a much stronger base than CH3COO�, we can safely neglect the hydrolysis of the

CH3COO� ions and calculate the pH of the solution using only the concentration of

the OH� ions. The total volume of the combined solutions is (25.0 � 35.0) mL or60.0 mL, so we calculate OH� concentration as follows:

Practice Exercise Exactly 100 mL of 0.10 M nitrous acid (HNO2) are titrated with a0.10 M NaOH solution. Calculate the pH for (a) the initial solution, (b) the point atwhich 80 mL of the base has been added, (c) the equivalence point, (d) the point atwhich 105 mL of the base has been added.

pH � 14.00 � 1.78 � 12.22 pOH � �log [OH�] � �log 0.0167 � 1.78

� 0.0167 mol/L � 0.0167 M

[OH�] �1.00 � 10�3 mol

60.0 mL�

1000 mL

1 L


35.0 mL �0.100 mol NaOH

1 L NaOH soln�

1 L

1000 mL� 3.50 � 10�3 mol

x � [OH�] � 5.3 � 10�6 M, pH � 8.72

Kb � 5.6 � 10�10 �

[CH3COOH][OH�]

[CH3COO�]

�x2

0.0500 � x

� 0.0500 mol�L � 0.0500 M

[CH3COONa] �2.50 � 10�3 mol

50.0 mL�

1000 mL

1 L




or simply

The pH at the equivalence point is less than 7 due to the hydrolysis of the NH�4 ion:

or simply

Because of the volatility of an aqueous ammonia solution, it is more convenient toadd hydrochloric acid from a buret to the ammonia solution. Figure 17.6 shows thetitration curve for this experiment.

17.4 Acid-Base Indicators

The equivalence point, as we have seen, is the point at which the number of molesof OH� ions added to a solution is equal to the number of moles of H� ions orig-inally present. To determine the equivalence point in a titration, then, we mustknow exactly how much volume of a base to add from a buret to an acid in aflask. One way to achieve this goal is to add a few drops of an acid-base indica-tor to the acid solution at the start of the titration. You will recall from Chapter4 that an indicator has distinctly different colors in its nonionized and ionizedforms. These two forms are related to the pH of the solution in which the indi-cator is dissolved. The end point of a titration occurs when the indicator changescolor. However, not all indicators change color at the same pH, so the choice of

NH�4 (aq) Δ NH3(aq) � H�(aq)

NH�4 (aq) � H2O(l) Δ NH3(aq) � H3O�(aq)

H�(aq) � NH3(aq) ¡ NH�4 (aq)

Volume of HCl added (mL)

pH

5010 4030200

Equivalencepoint

11

12

10

9

8

7

6

5

4

3

2

1

Volume HCladded (mL)

0.05.0

10.015.020.022.024.025.026.028.030.035.040.045.050.0

11.139.869.449.088.668.397.885.282.702.222.001.701.521.401.30

pH

Figure 17.6pH profiles of a strong acid–weak base titration. A 0.100 M HCl solution is added from a buret to 25.0 mL of a 0.100 M NH3solution in an Erlenmeyer flask. As a result of salt hydrolysis, the pH at the equivalence point is lower than 7.

An indicator is usually a weak organicacid or organic base. Only a smallamount (a drop or two) should be used ina titration experiment.


17.4 Acid-Base Indicators 587

indicator for a particular titration depends on the nature of the acid and base usedin the titration (that is, whether they are strong or weak). By choosing the properindicator for a titration, we can use the end point to determine the equivalencepoint, as we will see next.

Let us consider a weak monoprotic acid that we will call HIn. To be an effectiveindicator, HIn and its conjugate base, In�, must have distinctly different colors. Insolution, the acid ionizes to a small extent:

If the indicator is in a sufficiently acidic medium, the equilibrium, according to LeChâtelier’s principle, shifts to the left and the predominant color of the indicator isthat of the nonionized form (HIn). On the other hand, in a basic medium the equi-librium shifts to the right and the color of the solution will be due mainly to that ofthe conjugate base (In�). Roughly speaking, we can use the following concentrationratios to predict the perceived color of the indicator:

If [HIn] � [In�], then the indicator color is a combination of the colors of HIn and In�.The end point of an indicator does not occur at a specific pH; rather, there is

a range of pH values within which the end point will occur. In practice, we choosean indicator whose end point range lies on the steep part of the titration curve.Because the equivalence point also lies on the steep part of the curve, this choiceensures that the pH at the equivalence point will fall within the range over whichthe indicator changes color. In Section 4.6 we mentioned that phenolphthalein isa suitable indicator for the titration of NaOH and HCl. Phenolphthalein is color-less in acidic and neutral solutions, but reddish pink in basic solutions. Measure-ments show that at pH � 8.3 the indicator is colorless but that it begins to turnreddish pink when the pH exceeds 8.3. As shown in Figure 17.4, the steepness ofthe pH curve near the equivalence point means that the addition of a very smallquantity of NaOH (say, 0.05 mL, which is about the volume of a drop from theburet) brings about a large rise in the pH of the solution. What is important, how-ever, is the fact that the steep portion of the pH profile includes the range overwhich phenolphthalein changes from colorless to reddish pink. Whenever such acorrespondence occurs, the indicator can be used to locate the equivalence pointof the titration (Figure 17.7).

Many acid-base indicators are plant pigments. For example, by boiling choppedred cabbage in water we can extract pigments that exhibit many different colors atvarious pHs (Figure 17.8). Table 17.1 lists a number of indicators commonly used inacid-base titrations. The choice of a particular indicator depends on the strength ofthe acid and base to be titrated.

[HIn]

[In�]� 0.1 color of conjugate base (In�) predominates

[HIn]

[In�] 10 color of acid (HIn) predominates

HIn(aq) Δ H�(aq) � In�(aq)

Which indicator or indicators listed in Table 17.1 would you use for the acid-base titra-tions shown in (a) Figure 17.4, (b) Figure 17.5, and (c) Figure 17.6.

(Continued )

Example 17.5



Similar problem: 17.25.

Figure 17.8Solutions containing extracts ofred cabbage (obtained by boil-ing the cabbage in water) pro-duce different colors whentreated with an acid and abase. The pH of the solutionsincreases from left to right.


pH

5010 403020

14

0

13

11

12

10

9

8

7

6

5

4

3

2

1

Phenolphthalein

Methyl red

Thymol blue

Figure 17.7The titration curve of a strongacid with a strong base.Because the regions over whichthe indicators methyl red andphenolphthalein change coloralong the steep portion of thecurve, they can be used tomonitor the equivalence pointof the titration. Thymol blue, onthe other hand, cannot be usedfor the same purpose (see Table17.1).

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

Solution (a) Near the equivalence point, the pH of the solution changes abruptlyfrom 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 suitableindicators are cresol red and phenolphthalein.

(c) Here the steep portion of the pH curve covers the pH range between 3 and 7; there-fore, the suitable indicators are bromophenol blue, methyl orange, methyl red, andchlorophenol blue.

(Continued )


17.5 Solubility Equilibria 589

17.5 Solubility Equilibria

Precipitation reactions are important in industry, medicine, and everyday life. Forexample, the preparation of many essential industrial chemicals such as sodium car-bonate (Na2CO3) makes use of precipitation reactions. The dissolving of tooth enamel,which is mainly made of hydroxyapatite [Ca5(PO4)3OH], in an acidic medium leadsto tooth decay. Barium sulfate (BaSO4), an insoluble compound that is opaque toX rays, is used to diagnose ailments of the digestive tract. Stalactites and stalagmites,which consist of calcium carbonate (CaCO3), are produced by a precipitation reac-tion, and so are many foods, such as fudge.

The general rules for predicting the solubility of ionic compounds in water wereintroduced in Section 4.2. Although useful, these solubility rules do not enable us tomake quantitative predictions about how much of a given ionic compound will dis-solve in water. To develop a quantitative approach, we start with what we alreadyknow about chemical equilibrium.

Solubility Product

Consider a saturated solution of silver chloride that is in contact with solid silver chlo-ride. The solubility equilibrium can be represented as

Because salts such as AgCl are treated as strong electrolytes, all the AgCl that dis-solves in water is assumed to dissociate completely into Ag� and Cl� ions. We knowfrom Chapter 15 that for heterogeneous reactions the concentration of the solid is aconstant. Thus, we can write the equilibrium constant for the dissociation of AgCl as

]Ksp � [Ag�][Cl�

AgCl(s) Δ Ag�(aq) � Cl�(aq)

BaSO4 imaging of the humanlarge intestine.

Downward-growing stalactitesand upward-growing stalagmites.

Color

Indicator In Acid In Base pH Range*

Thymol blue Red Yellow 1.2–2.8

Bromophenol blue Yellow Bluish purple 3.0–4.6

Methyl orange Orange Yellow 3.1–4.4

Methyl red Red Yellow 4.2–6.3

Chlorophenol blue Yellow Red 4.8–6.4

Bromothymol blue Yellow Blue 6.0–7.6

Cresol red Yellow Red 7.2–8.8

Phenolphthalein Colorless Reddish pink 8.3–10.0

TABLE 17.1 Some Common Acid-Base Indicators

*The pH range is defined as the range over which the indicator changes from the acid color to the base color.

Practice Exercise Referring to Table 17.1, specify which indicator or indicators youwould use for the following titrations: (a) HBr versus CH3NH2, (b) HNO3 versus NaOH,(c) HNO2 versus KOH.



in which Ksp is called the solubility product constant or simply the solubility prod-uct. In general, the solubility product of a compound is the product of the molar con-centrations of the constituent ions, each raised to the power of its stoichiometriccoefficient in the equilibrium equation.

Because each AgCl unit contains only one Ag� ion and one Cl� ion, its solu-bility product expression is particularly simple to write. The following cases are morecomplex.

• MgF2

• Ag2CO3

• Ca3(PO4)2

Table 17.2 lists the Ksp values for a number of salts of low solubility. Solublesalts such as NaCl and KNO3, which have very large Ksp values, are not listed in thetable.

Ca3(PO4)2(s) Δ 3Ca2�(aq) � 2PO3�4 (aq) Ksp � [Ca2�]3[PO3�4 ]2

Ag2CO3(s) Δ 2Ag�(aq) � CO2�3 (aq) Ksp � [Ag�]2[CO2�3 ]

MgF2(s) Δ Mg2�(aq) � 2F�(aq) Ksp � [Mg2�][F�]2We ignore both ion pair formation andsalt hydrolysis (see p. 533).

Compound Ksp Compound Ksp

Aluminum hydroxide [Al(OH)3] 1.8 � 10�33 Lead(II) chromate (PbCrO4) 2.0 � 10

�14

Barium carbonate (BaCO3) 8.1 � 10�9 Lead(II) fluoride (PbF2) 4.1 � 10

�8

Barium fluoride (BaF2) 1.7 � 10�6 Lead(II) iodide (PbI2) 1.4 � 10

�8

Barium sulfate (BaSO4) 1.1 � 10�10 Lead(II) sulfide (PbS) 3.4 � 10�28

Bismuth sulfide (Bi2S3) 1.6 � 10�72 Magnesium carbonate (MgCO3) 4.0 � 10

�5

Cadmium sulfide (CdS) 8.0 � 10�28 Magnesium hydroxide [Mg(OH)2] 1.2 � 10�11

Calcium carbonate (CaCO3) 8.7 � 10�9 Manganese(II) sulfide (MnS) 3.0 � 10�14

Calcium fluoride (CaF2) 4.0 � 10�11 Mercury(I) chloride (Hg2Cl2) 3.5 � 10

�18

Calcium hydroxide [Ca(OH)2] 8.0 � 10�6 Mercury(II) sulfide (HgS) 4.0 � 10�54

Calcium phosphate [Ca3(PO4)2] 1.2 � 10�26 Nickel(II) sulfide (NiS) 1.4 � 10�24

Chromium(III) hydroxide [Cr(OH)3] 3.0 � 10�29 Silver bromide (AgBr) 7.7 � 10�13

Cobalt(II) sulfide (CoS) 4.0 � 10�21 Silver carbonate (Ag2CO3) 8.1 � 10�12

Copper(I) bromide (CuBr) 4.2 � 10�8 Silver chloride (AgCl) 1.6 � 10�10

Copper(I) iodide (CuI) 5.1 � 10�12 Silver iodide (AgI) 8.3 � 10�17

Copper(II) hydroxide [Cu(OH)2] 2.2 � 10�20 Silver sulfate (Ag2SO4) 1.4 � 10

�5

Copper(II) sulfide (CuS) 6.0 � 10�37 Silver sulfide (Ag2S) 6.0 � 10�51

Iron(II) hydroxide [Fe(OH)2] 1.6 � 10�14 Strontium carbonate (SrCO3) 1.6 � 10

�9

Iron(III) hydroxide [Fe(OH)3] 1.1 � 10�36 Strontium sulfate (SrSO4) 3.8 � 10

�7

Iron(II) sulfide (FeS) 6.0 � 10�19 Tin(II) sulfide (SnS) 1.0 � 10�26

Lead(II) carbonate (PbCO3) 3.3 � 10�14 Zinc hydroxide [Zn(OH)2] 1.8 � 10

�14

Lead(II) chloride (PbCl2) 2.4 � 10�4 Zinc sulfide (ZnS) 3.0 � 10�23

TABLE 17.2 Solubility Products of Some Slightly Soluble Ionic Compounds at 25�C



For the dissolution of an ionic solid in aqueous solution, any one of the follow-ing conditions may exist: (1) The solution is unsaturated, (2) the solution is saturated,or (3) the solution is supersaturated. Following the procedure in Section 15.3, we useQ, called the ion product, to represent the product of the molar concentrations of theions raised to the power of their stoichiometric coefficients. Thus, for an aqueous solu-tion containing Ag� and Cl� ions at 25C,

The subscript 0 reminds us that these are initial concentrations and do not necessarilycorrespond to those at equilibrium. The possible relationships between Q and Ksp are

Q � Ksp Unsaturated solution[Ag�]0[Cl

�]0 � 1.6 � 10�10

Q � Ksp Saturated solution[Ag�][Cl�] � 1.6 � 10�10

Q � Ksp Supersaturated solution; AgCl will precipitate[Ag�]0[Cl

�]0 � 1.6 � 10�10 out until the product of the ionic

concentrations is equal to 1.6 � 10�10

Molar Solubility and Solubility

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 sol-ubilities, you should choose compounds that have similar formulas, such as AgCl andZnS, or CaF2 and Fe(OH)2. There are two other quantities that express a substance’ssolubility: molar solubility, which is the number of moles of solute in 1 L of a satu-rated solution (moles per liter), and solubility, which is the number of grams of solutein 1 L of a saturated solution (grams per liter). Note that all these expressions referto the concentration of saturated solutions at some given temperature (usually 25C).Figure 17.9 shows the relationships among solubility, molar solubility, and Ksp.

Both molar solubility and solubility are convenient to use in the laboratory. Wecan use them to determine Ksp by following the steps outlined in Figure 17.9(a).

Q � [Ag�]0[Cl�]0

Look up the Ksp value for AgCl in Table17.2.

Solubility ofcompound

(a)

Concentrationsof cationsand anions

(b)

Ksp ofcompound

Molarsolubility ofcompound

Solubility ofcompound

Molarsolubility ofcompound

Ksp ofcompound

Concentrationsof cationsand anions

Figure 17.9Sequence of steps (a) for calcu-lating Ksp from solubility dataand (b) for calculating solubil-ity from Ksp data.

The solubility of calcium sulfate (CaSO4) is found to be 0.67 g/L. Calculate the valueof Ksp for calcium sulfate.

(Continued )

Example 17.6Calcium sulfate is used as adrying agent and in themanufacture of paints, ceramics,and paper. A hydrated form ofcalcium sulfate, called plaster ofParis, is used to make casts forbroken bones.



Sometimes we are given the value of Ksp for a compound and asked to calculatethe compound’s molar solubility. For example, the Ksp of silver bromide (AgBr) is7.7 � 10�13. We can calculate its molar solubility by the same procedure as outlinedfor acid ionization constants. First, we identify the species present at equilibrium. Herewe have Ag� and Br� ions. Let s be the molar solubility (in mol/L) of AgBr. Becauseone 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) � Br�(aq)Initial (M): 0.00 0.00Change (M): �s �s �s

Equilibrium (M): s s

From Table 17.2 we write

s � 27.7 � 10�13 � 8.8 � 10�7 M

7.7 � 10�3 � (s)(s)

Ksp � [Ag�][Br�]


Silver bromide is used inphotographic emulsions.

Strategy We are given the solubility of CaSO4 and asked to calculate its Ksp. Thesequence of conversion steps, according to Figure 17.9(a), is

solubility of 88n molar solubility 88n and Ksp ofCaSO4 in g/L of CaSO4 CaSO4

Solution Consider the dissociation of CaSO4 in water. Let s be the molar solubility (inmol/L) of CaSO4.

Initial (M): 0 0Change (M):

Equilibrium (M): s s

The solubility product for CaSO4 is

First we calculate the number of moles of CaSO4 dissolved in 1 L of solution

From the solubility equilibrium we see that for every mole of CaSO4 that dissolves,1 mole of Ca2� and 1 mole of SO4

2� are produced. Thus, at equilibrium

and

Now we can calculate Ksp:

Practice Exercise The solubility of lead chromate (PbCrO4) is 4.5 � 10�5 g/L.Calculate the solubility product of this compound.

� 2.4 � 10�5 � (4.9 � 10�3)(4.9 � 10�3)

Ksp � [Ca2�][SO2�4 ]

[SO2�4 ] � 4.9 � 10�3 M[Ca2�] � 4.9 � 10�3 M

0.67 g CaSO41 L soln

�1 mol CaSO4

136.2 g CaSO4� 4.9 � 10�3 mol/L � s

Ksp � [Ca2�][SO2�4 ] � s

2

�s�s�s

CaSO4(s) Δ Ca2�(aq) � SO2�4 (aq)

[SO2�4 ]¡[Ca2�]



Therefore, at equilibrium

Thus, the molar solubility of AgBr also is 8.8 � 10�7 M. Knowing the molar solu-bility will enable us to calculate the solubility in g/L, as shown in Example 17.7.

[Br�] � 8.8 � 10�7 M [Ag�] � 8.8 � 10�7 M


As Examples 17.6 and 17.7 show, solubility and solubility product are related. Ifwe know one, we can calculate the other, but each quantity provides different infor-mation. Table 17.3 shows the relationship between molar solubility and solubilityproduct for a number of ionic compounds.

When carrying out solubility and/or solubility product calculations, keep in mindthese important points:

Copper(II) hydroxide is used asa pesticide and to treat seeds.

Using the data in Table 17.2, calculate the solubility of copper(II) hydroxide, Cu(OH)2,in g/L.

Strategy We are given the Ksp of Cu(OH)2 and asked to calculate its solubility in g/L.The sequence of conversion steps, according to Figure 17.9(b), is

Ksp of 88n [Cu2�] and 88n molar solubility 88n solubility of

Cu(OH)2 [OH�] of Cu(OH)2 Cu(OH)2 in g/L

Solution Consider the dissociation of Cu(OH)2 in water:

Initial (M): 0 0Change (M):

Equilibrium (M): s 2s

Note that the molar concentration of OH� is twice that of Cu2�. The solubility productof Cu(OH)2 is

From the Ksp value in Table 17.2, we solve for the molar solubility of Cu(OH)2 as follows:

Hence

Finally, from the molar mass of Cu(OH)2 and its molar solubility, we calculate thesolubility in g/L:

Practice Exercise Calculate the solubility of silver chloride (AgCl) in g/L.

� 1.8 � 10�5 g/L

solubility of Cu(OH)2 �1.8 � 10�7 mol Cu(OH)2

1 L soln�

97.57 g Cu(OH)21 mol Cu(OH)2

s � 1.8 � 10�7 M

s3 �2.2 � 10�20

4� 5.5 � 10�21

2.2 � 10�20 � 4s3

� (s)(2s)2 � 4s3 Ksp � [Cu

2�][OH�]2

�2s�s�s

Cu(OH)2(s) Δ Cu2�(aq) � 2OH�(aq)

Example 17.7



• The solubility is the quantity of a substance that dissolves in a certain quantityof water. In solubility equilibria calculations, it is usually expressed as grams ofsolute per liter of solution. Molar solubility is the number of moles of solute perliter of solution.

• 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 productslisted in Table 17.2, we can predict whether a precipitate will form when we mix twosolutions or add a soluble compound to a solution. This ability often has practicalvalue. In industrial and laboratory preparations, we can adjust the concentrations ofions until the ion product exceeds Ksp in order to obtain a given compound (in theform of a precipitate). The ability to predict precipitation reactions is also useful inmedicine. For example, kidney stones, which can be extremely painful, consist largelyof calcium oxalate, CaC2O4 (Ksp � 2.3 � 10

�9). The normal physiological concen-tration of calcium ions in blood plasma is about 5 mM (1 mM � 1 � 10�3 M ).Oxalate ions (C2O4

2�), derived from oxalic acid present in many vegetables such asrhubarb 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 canhelp to reduce precipitate formation.

Compound Ksp Expression Cation Anion Relation Between Ksp and s

AgCl ] s s

BaSO4 ] s s

Ag2CO3 ] 2s s

PbF2 s 2s

Al(OH)3 s 3s

Ca3(PO4)2 3s 2s Ksp � 108s5; s � aKsp

108b

15

[Ca2�]3[PO3�4 ]2

Ksp � 27s4; s � aKsp

27b

14

[Al3�][OH�]3


4b

13

[Pb2�][F�]2


4b

13

[Ag�]2[CO2�3

Ksp � s2; s � (Ksp)

12[Ba2�][SO2�4

Ksp � s2; s � (Ksp)

12[Ag�][Cl�

TABLE 17.3 Relationship Between Ksp and Molar Solubility (s)

A kidney stone.

Exactly 200 mL of 0.0040 M BaCl2 are added to exactly 600 mL of 0.0080 M K2SO4.Will a precipitate form?

Strategy Under what condition will an ionic compound precipitate from solution? Theions in solution are Ba2�, Cl�, K�, and SO4

2�. According to the solubility rules listedin Table 4.2 (p. 98), the only precipitate that can form is BaSO4. From the informationgiven, we can calculate [Ba2�] and [SO4

2�] because we know the number of moles ofthe ions in the original solutions and the volume of the combined solution. Next wecalculate the reaction quotient Q (Q � [Ba2�]0[SO4

2�]0) and compare the value of Q(Continued )

Example 17.8



with Ksp of BaSO4 to see if a precipitate will form, that is, if the solution issupersaturated. It is helpful to make a sketch of the situation.

Solution The number of moles of Ba2� present in the original 200 mL of solution is

The total volume after combining the two solutions is 800 mL. The concentration ofBa2� in the 800 mL volume is

The number of moles of SO42� in the original 600 mL solution is

The concentration of SO42� in the 800 mL of the combined solution is

Now we must compare Q with Ksp. From Table 17.2,

As for Q,

Therefore,

The solution is supersaturated because the value of Q indicates that the concentrationsof the ions are too large. Thus, some of the BaSO4 will precipitate out of solution until

Practice Exercise If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 MCaCl2, will precipitation occur?

[Ba2�][SO2�4 ] � 1.1 � 10�10

Q 7 Ksp

� 6.0 � 10�6Q � [Ba2�]0[SO

2�4 ]0 � (1.0 � 10

�3)(6.0 � 10�3)

Ksp � 1.1 � 10�10BaSO4(s) Δ Ba2�(aq) � SO2�4 (aq)

� 6.0 � 10�3 M

[SO2�4 ] �4.8 � 10�3 mol

800 mL�

1000 mL

1 L soln

600 mL �0.0080 mol SO2�4

1 L soln�

1 L

1000 mL� 4.8 � 10�3 mol SO2�4

� 1.0 � 10�3 M

[Ba2�] �8.0 � 10�4 mol

800 mL�

1000 mL

1 L soln

200 mL �0.0040 mol Ba2�

1 L soln�

1 L

1000 mL� 8.0 � 10�4 mol Ba2�

We assume that the volumes are additive.




17.6 The Common Ion Effect and Solubility

As we have noted, the solubility product is an equilibrium constant; precipitation ofan ionic compound from solution occurs whenever the ion product exceeds Ksp forthat 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 acommon ion, say, AgCl and AgNO3. In addition to the dissociation of AgCl, the fol-lowing process also contributes to the total concentration of the common silver ionsin solution:

If AgNO3 is added to a saturated AgCl solution, the increase in [Ag�] will make the

ion product greater than the solubility product:

To reestablish equilibrium, some AgCl will precipitate out of the solution, as LeChâtelier’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 equi-librium; rather, [Ag�] � [Cl�].

Q � [Ag�]0[Cl�]0 7 Ksp

AgNO3(s) ¡H2O Ag�(aq) � NO3

�(aq)

At a given temperature, only the solubilityof a compound is altered (decreased) bythe common ion effect. Its solubilityproduct, which is an equilibrium constant,remains the same whether or not othersubstances are present in the solution.

Calculate the solubility of silver chloride (in g/L) in a 6.5 � 10�3 M silver nitratesolution.

Strategy This is a common-ion problem. The common ion here is Ag�, which issupplied by both AgCl and AgNO3. Remember that the presence of the common ionwill affect only the solubility of AgCl (in g/L), but not the Ksp value because it is anequilibrium constant.

Solution Step 1: The relevant species in solution are Ag� ions (from both AgCl andAgNO3) and Cl

� ions. The NO3� ions are spectator ions.

Step 2: Because AgNO3 is a soluble strong electrolyte, it dissociates completely:

Let s be the molar solubility of AgCl in AgNO3 solution. We summarize thechanges in concentrations as follows:

AgCl(s) 34 Ag�(aq) � Cl�(aq)Initial (M): 6.5 � 10�3 0.00Change (M): �s �s �s

Equilibrium (M): (6.5 � 10�3 � s) s

Step 3:

(Continued )1.6 � 10�10 � (6.5 � 10�3 � s)(s)

Ksp � [Ag�][Cl�]

6.5 � 10�3 M6.5 � 10�3 MAgNO3(s) ¡H2O Ag�(aq) � NO�3 (aq)

Example 17.9


17.7 Complex Ion Equilibria and Solubility 597

17.7 Complex Ion Equilibria and Solubility

Lewis acid-base reactions in which a metal cation (electron-pair acceptor) combineswith a Lewis base (electron-pair donor) result in the formation of complex ions:

acid base

Thus, we can define a complex ion as an ion containing a central metal cation bondedto one or more molecules or ions. Complex ions are crucial to many chemical and bio-logical processes. Here we will consider the effect of complex ion formation on sol-ubility. In Chapter 20 we will discuss the chemistry of complex ions in more detail.

Transition metals have a particular tendency to form complex ions. For example,a solution of cobalt(II) chloride is pink because of the presence of the Co(H2O)6

2�

ions (Figure 17.10). When HCl is added, the solution turns blue as a result of the for-mation of the complex ion CoCl4

2�:

Copper(II) sulfate (CuSO4) dissolves in water to produce a blue solution. Thehydrated copper(II) ions are responsible for this color; many other sulfates (Na2SO4, forexample) are colorless. Adding a few drops of concentrated ammonia solution to a CuSO4solution causes a light-blue precipitate, copper(II) hydroxide, to form (Figure 17.11):

Cu2�(aq) � 2OH�(aq) ¡ Cu(OH)2(s)

Co2�(aq) � 4Cl�(aq) Δ CoCl2�4 (aq)

Ag�(aq) � 2NH3(aq) Δ Ag(NH3)�2 (aq)


Lewis acids and bases are discussed inSection 16.11.

Because AgCl is quite insoluble and the presence of Ag� ions from AgNO3 furtherlowers the solubility of AgCl, s must be very small compared with 6.5 � 10�3.Therefore, applying the approximation 6.5 � 10�3, we obtain

Step 4: At equilibrium,

and so our approximation was justified in Step 3. Because all the Cl� ionsmust come from AgCl, the amount of AgCl dissolved in AgNO3 solution alsois 2.5 � 10�8 M. Then, knowing the molar mass of AgCl (143.4 g), we cancalculate the solubility of AgCl as follows:

Check The solubility of AgCl in pure water is (see the Practice Exer-cise in Example 17.7). Therefore, the lower solubility (3.6 � 10�6 g/L) in the presenceof AgNO3 is reasonable. You should also be able to predict the lower solubility usingLe Châtelier’s principle. Adding Ag� ions shifts the solubility equilibrium to the left,thus decreasing the solubility of AgCl.

Practice Exercise Calculate the solubility in g/L of AgBr in (a) pure water and in(b) 0.0010 M NaBr.

1.9 � 10�3 g/L

� 3.6 � 10�6 g/L

solubility of AgCl in AgNO3 solution �2.5 � 10�8 mol AgCl

1 L soln�

143.4 g AgCl

1 mol AgCl

[Cl�] � 2.5 � 10�8 M [Ag�] � (6.5 � 10�3 � 2.5 � 10�8) M � 6.5 � 10�3 M

s � 2.5 � 10�8 M 1.6 � 10�10 � (6.5 � 10�3)s

6.5 � 10�3 � s �

According to our definition, Co(H2O)62�

itself is a complex ion. When we writeCo(H2O)6

2�, we mean the hydrated Co2�

ion.



in which the OH� ions are supplied by the ammonia solution. If an excess of NH3 isthen added, the blue precipitate redissolves to produce a beautiful dark-blue solution,this time as a result of the formation of the complex ion Cu(NH3)4

2�, also shown inFigure 17.11:

Thus, the formation of the complex ion Cu(NH3)42� increases the solubility of Cu(OH)2.

A measure of the tendency of a metal ion to form a particular complex ion isgiven by the formation constant Kf (also called the stability constant), which is theequilibrium constant for complex ion formation. The larger Kf is, the more stable thecomplex ion is. Table 17.4 lists the formation constants of a number of complex ions.

The formation of the Cu(NH3)42� ion can be expressed as

for which the formation constant is

The very large value of Kf in this case indicates the great stability of the complexion in solution and accounts for the very low concentration of copper(II) ions atequilibrium.

� 5.0 � 1013

Kf �[Cu(NH3)

2�4 ]

[Cu2�][NH3]4

Cu2�(aq) � 4NH3(aq) Δ Cu(NH3)2�4 (aq)

Cu(OH)2(s) � 4NH3(aq) Δ Cu(NH3)2�4 (aq) � 2OH�(aq)

Figure 17.11Left: An aqueous solution ofcopper(II) sulfate. Center: Afterthe addition of a few drops of aconcentrated aqueous ammoniasolution, a light-blue precipitateof Cu(OH)2 is formed. Right:When more concentratedaqueous ammonia solution isadded, the Cu(OH)2 precipitatedissolves to form the dark-bluecomplex ion Cu(NH3)4

2�.

Figure 17.10(Left) An aqueous cobalt(II)chloride solution. The pinkcolor is due to the presence ofCo(H2O)6

2� ions. (Right) Afterthe addition of HCl solution,the solution turns blue becauseof the formation of the complexCoCl4

2� ions.


17.7 Complex Ion Equilibria and Solubility 599

Complex Ion Equilibrium Expression Formation Constant (Kf)

Ag(NH3)2� Ag� � 2NH3 34 Ag(NH3)2

� 1.5 � 107

Ag(CN)�2 Ag� � 2CN� 34 Ag(CN)2

� 1.0 � 1021

Cu(CN)42� Cu2� � 4CN� 34 Cu(CN)4

2� 1.0 � 1025

Cu(NH3)42� Cu2� � 4NH3 34 Cu(NH3)4

2� 5.0 � 1013

Cd(CN)42� Cd2� � 4CN� 34 Cd(CN)4

2� 7.1 � 1016

CdI42� Cd2� � 4I� 34 CdI4

2� 2.0 � 106

HgCl42� Hg2� � 4Cl� 34 HgCl4

2� 1.7 � 1016

HgI42� Hg2� � 4I� 34 HgI4

2� 2.0 � 1030

Hg(CN)42� Hg2� � 4CN� 34 Hg(CN)4

2� 2.5 � 1041

Co(NH3)63� Co3� � 6NH3 34 Co(NH3)6

3� 5.0 � 1031

Zn(NH3)42� Zn2� � 4NH3 34 Zn(NH3)4

2� 2.9 � 109

TABLE 17.4 Formation Constants of Selected Complex Ions in Water at 25�C

A 0.20-mole quantity of CuSO4 is added to a liter of 1.20 M NH3 solution. What is theconcentration of Cu2� ions at equilibrium?

Strategy The addition of CuSO4 to the NH3 solution results in complex ion formation

From Table 17.4 we see that the formation constant (Kf) for this reaction is very large;therefore, the reaction lies mostly to the right. At equilibrium, the concentration of Cu2�

will be very small. As a good approximation, we can assume that essentially all thedissolved Cu2� ions end up as Cu(NH3)4

2� ions. How many moles of NH3 will reactwith 0.20 mole of Cu2�? How many moles of Cu(NH3)4

2� will be produced? A verysmall amount of Cu2� will be present at equilibrium. Set up the Kf expression for thepreceding equilibrium to solve for [Cu2�].

Solution The amount of NH3 consumed in forming the complex ion is 4 � 0.20 mol,or 0.80 mol. (Note that 0.20 mol Cu2� is initially present in solution and four NH3molecules are needed to form a complex ion with one Cu2� ion.) The concentration ofNH3 at equilibrium is therefore (1.20 � 0.80) mol/L soln or 0.40 M, and that ofCu(NH3)4

2� is 0.20 mol/L soln or 0.20 M, the same as the initial concentration of Cu2�.[There is a 1:1 mole ratio between Cu2� and Cu(NH3)4

2�.] Because Cu(NH3)42� does

dissociate to a slight extent, we call the concentration of Cu2� at equilibrium x and write

Solving for x and keeping in mind that the volume of the solution is 1 L, we obtain

Practice Exercise If 2.50 g of CuSO4 are dissolved in 9.0 � 102 mL of 0.30 MNH3, what are the concentrations of Cu

2�, Cu(NH3)42�, and NH3 at equilibrium?

x � [Cu2�] � 1.6 � 10�13 M

5.0 � 1013 �0.20

x(0.40)4

Kf �[Cu(NH3)4

2�]

[Cu2�][NH3]4

Cu2�(aq) � 4NH3(aq) Δ Cu(NH3)2�4 (aq)

Example 17.10


The very small value of [Cu2�] justifiesour approximation.



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. For example, aluminum hydroxide reacts with acids andbases as:

The increase in solubility of Al(OH)3 in a basic medium is the result of the forma-tion of the complex ion [Al(OH)�4 ] in which Al(OH)3 acts as the Lewis acid andOH� acts as the Lewis base (Figure 17.12). Other amphoteric hydroxides behave ina similar manner.

17.8 Application of the Solubility ProductPrinciple to Qualitative Analysis

In Section 4.6, we discussed the principle of gravimetric analysis, by which we mea-sure the amount of an ion in an unknown sample. Here we will briefly discussqualitative analysis, the determination of the types of ions present in a solution. Wewill focus on the cations.

Twenty common cations can be analyzed readily in aqueous solution. Thesecations can be divided into five groups according to the solubility products of theirinsoluble salts (Table 17.5). Because an unknown solution may contain any one or upto all 20 ions, analysis must be carried out systematically from group 1 through group 5.Let us consider the general procedure for separating these ions by adding precipitat-ing reagents to an unknown solution.

• Group 1 cations. When dilute HCl is added to the unknown solution, only theAg�, Hg2

2�, and Pb2� ions precipitate as insoluble chlorides. The other ions,whose chlorides are soluble, remain in solution.

• Group 2 cations. After the chloride precipitates have been removed by filtration,hydrogen sulfide is reacted with the unknown acidic solution. Under this condi-tion, the concentration of the S2� ion in solution is negligible. Therefore, the pre-cipitation of metal sulfides is best represented as

M2�(aq) � H2S(aq) Δ MS(s) � 2H�(aq)

Al(OH)3(s) � OH�(aq) Δ Al(OH)�4 (aq)

Al(OH)3(s) � 3H�(aq) ¡ Al3�(aq) � 3H2O(l)

All amphoteric hydroxides are insolublecompounds.

Figure 17.12(Left to right) Formation ofAl(OH)3 precipitate when NaOHsolution is added to an Al(NO3)3solution. With further addition ofNaOH solution, the precipitatedissolves due to the formationof the complex ion Al(OH)�4 .

Do not confuse the groups in Table 17.5,which are based on solubility products,with those in the periodic table, whichare based on the electron configurationsof the elements.


17.8 Application of the Solubility Product Principle to Qualitative Analysis 601

Adding acid to the solution shifts this equilibrium to the left so that only the leastsoluble metal sulfides, that is, those with the smallest Ksp values, will precipitateout of solution. These are Bi2S3, CdS, CuS, HgS, and SnS.

• Group 3 cations. At this stage, sodium hydroxide is added to the solution to makeit basic. In a basic solution, the preceding equilibrium shifts to the right. Therefore,the more soluble sulfides (CoS, FeS, MnS, NiS, and ZnS) now precipitate out ofsolution. Note that the Al3� and Cr3� ions actually precipitate as the hydroxidesAl(OH)3 and Cr(OH)3, rather than as the sulfides, because the hydroxides are lesssoluble. 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 fromsolution, sodium carbonate is added to the basic solution to precipitate Ba2�,Ca2�, and Sr2� ions as BaCO3, CaCO3, and SrCO3. These precipitates too areremoved from solution by filtration.

• Group 5 cations. At this stage, the only cations possibly remaining in solutionare Na�, K�, and NH�4. The presence of NH

�4 can be determined by adding

sodium hydroxide:

The ammonia gas is detected either by noting its characteristic odor or by observ-ing a piece of wet red litmus paper turning blue when placed above (not in contact

NaOH(aq) � NH�4 (aq) ¡ Na�(aq) � H2O(l ) � NH3(g)

Group Cation Precipitating Reagents Insoluble Compound Ksp

1 HCl AgCl

Hg2Cl2PbCl2

2 H2S Bi2S3in acidic CdSsolutions CuS

HgS

SnS

3 H2S Al(OH)3in basic CoS

solutions Cr(OH)3FeS

MnS

NiS

ZnS

4 Na2CO3 BaCO3CaCO3SrCO3

5 No precipitating Nonereagent None

NoneNH�4

Na�K�

1.6 � 10�9Sr2�8.7 � 10�9Ca2�8.1 � 10�9Ba2�3.0 � 10�23Zn2�1.4 � 10�24Ni2�3.0 � 10�14Mn2�6.0 � 10�19Fe2�3.0 � 10�29Cr3�4.0 � 10�21Co2�1.8 � 10�33Al3�1.0 � 10�26Sn2�4.0 � 10�54Hg2�6.0 � 10�37Cu2�8.0 � 10�28Cd2�1.6 � 10�72Bi3�2.4 � 10�4Pb2�3.5 � 10�18Hg

2�2

1.6 � 10�10Ag�

TABLE 17.5 Separation of Cations into Groups According to Their Precipitation Reactions with Various Reagents

8888n8888n

88888888888n8888n



with) the solution. To confirm the presence of Na� and K� ions, we usually usea flame test, as follows: A piece of platinum wire (chosen because platinum isinert) 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 17.13).

Figure 17.14 summarizes this scheme for separating metal ions.

Because NaOH is added in group 3 andNa2CO3 is added in group 4, the flametest for Na� ions is carried out using theoriginal solution.

Figure 17.13Left to right: Flame colors of lithium, sodium, potassium, and copper.

Solution containing ionsof all cation groups

Group 1 precipitatesAgCl, Hg2Cl2, PbCl2

Solution containing ionsof remaining groups



Solution containsNa+, K+, NH 4

+ ions

Group 2 precipitatesCuS, CdS, HgS, SnS, Bi2S3

Group 3 precipitatesCoS, FeS, MnS, NiS

ZnS, Al(OH)3, Cr(OH)3

Group 4 precipitatesBaCO3, CaCO3, SrCO3

Filtration

Filtration

Filtration

Filtration

+HCl

+H2S

+NaOH

+Na2CO3

Figure 17.14A flow chart for the separationof cations in qualitative analysis.


Key Words 603

Two points regarding qualitative analysis must be mentioned. First, the separa-tion of the cations into groups is made as selective as possible; that is, the anions thatare added as reagents must be such that they will precipitate the fewest types ofcations. For example, all the cations in group 1 also form insoluble sulfides. Thus, ifH2S were reacted with the solution at the start, as many as seven different sulfidesmight precipitate out of solution (group 1 and group 2 sulfides), an undesirable out-come. Second, the removal of cations at each step must be carried out as completelyas possible. For example, if we do not add enough HCl to the unknown solution toremove all the group 1 cations, they will precipitate with the group 2 cations as insol-uble sulfides; this too would interfere with further chemical analysis and lead us todraw erroneous conclusions.

KEY EQUATIONS

pKa � �log Ka (17.2) Definition of pKa.

(17.3) Henderson-Hasselbalch equation.pH � pKa � log [conjugate base]

[acid]

1. Equilibria involving weak acids or weak bases in aque-ous solution are homogeneous. Solubility equilibria areexamples of heterogeneous equilibria.

2. A buffer solution is a combination of a weak acid andits weak conjugate base; the solution reacts withsmall amounts of added acid or base in such a waythat the pH of the solution remains nearly constant.Buffer systems play a vital role in maintaining the pHof body fluids.

3. The pH at the equivalence point of an acid-base titrationdepends on the hydrolysis of the salt formed in the neu-tralization reaction. For strong acid–strong base titra-tions, the pH at the equivalence point is 7; for weakacid–strong base titrations, the pH at the equivalencepoint is greater than 7; for strong acid–weak base titra-tions, the pH at the equivalence point is less than 7.

Acid-base indicators are weak organic acids or basesthat change color at the end point in an acid-base neu-tralization reaction.

4. The solubility product Ksp expresses the equilibrium be-tween a solid and its ions in solution. Solubility can befound from Ksp and vice versa. The presence of a com-mon ion decreases the solubility of a salt.

5. Complex ions are formed in solution by the combina-tion of a metal cation with a Lewis base. The forma-tion constant Kf measures the tendency toward theformation of a specific complex ion. Complex ionformation can increase the solubility of an insolublesubstance.

6. Qualitative analysis is the identification of cations andanions in solution. It is based largely on the principlesof solubility equilibria.

SUMMARY OF FACTS AND CONCEPTS

Buffer solution, p. 575Complex ion, p. 597

Solubility, p. 591Solubility product (Ksp), p. 590

End point, p. 586Formation constant (Kf), p. 598

Molar solubility, p. 591Qualitative analysis, p. 600

KEY WORDS



QUESTIONS AND PROBLEMS

Buffer Solutions

Review Questions

17.1 Define buffer solution.

17.2 Define pKa for a weak acid and explain the relation-ship between the value of the pKa and the strength ofthe acid. Do the same for pKb and a weak base.

17.3 The pKas of two monoprotic acids HA and HB are5.9 and 8.1, respectively. Which of the two is thestronger acid?

17.4 The pKbs for the bases X�, Y�, and Z� are 2.72,

8.66, and 4.57, respectively. Arrange the followingacids in order of increasing strength: HX, HY, HZ.

Problems

17.5 Specify which of these systems can be classified asa buffer system: (a) KCl/HCl, (b) NH3/NH4NO3,(c) Na2HPO4/NaH2PO4.

17.6 Specify which of these systems can be classified as abuffer system: (a) KNO2/HNO2, (b) KHSO4/H2SO4,(c) HCOOK/HCOOH.

17.7 The pH of a bicarbonate–carbonic acid buffer is8.00. Calculate the ratio of the concentration of car-bonic acid to that of the bicarbonate ion.

17.8 Calculate the pH of these two buffer solutions:(a) 2.0 M CH3COONa/2.0 M CH3COOH, (b) 0.20 MCH3COONa/0.20 M CH3COOH. Which is the moreeffective buffer? Why?

17.9 Calculate the pH of the buffer system 0.15 MNH3/0.35 M NH4Cl.

17.10 What is the pH of the buffer 0.10 M Na2HPO4/0.15 MKH2PO4?

17.11 The pH of a sodium acetate–acetic acid buffer is4.50. Calculate the ratio [CH3COO

�]/[CH3COOH].

17.12 The pH of blood plasma is 7.40. Assumingthe principal buffer system is HCO�3/H2CO3, cal-culate the ratio [HCO�3 ]/[H2CO3]. Is this buffermore effective against an added acid or an addedbase?

17.13 Calculate the pH of 1.00 L of the buffer 0.80 MCH3NH2/1.00 M CH3NH3Cl before and after the ad-dition of (a) 0.070 mol NaOH and (b) 0.11 mol HCl.(See Table 16.5 for Ka value.)

17.14 Calculate the pH of 1.00 L of the buffer 1.00 MCH3COONa/1.00 M CH3COOH before and after theaddition of (a) 0.080 mol NaOH and (b) 0.12 molHCl. (Assume that there is no change in volume.)

17.15 A diprotic acid, H2A, has the following ionization con-stants: � 1.1 � 10�3 and � 2.5 � 10�6. ToKa2Ka1

make up a buffer solution of pH 5.80, which combina-tion would you choose: NaHA/H2A or Na2A/NaHA?

17.16 A student wishes to prepare a buffer solution at pH �8.60. Which of these weak acids should she chooseand why: HA (Ka � 2.7 � 10

�3), HB (Ka � 4.4 �10�6), or HC (Ka � 2.6 � 10

�9)?

Acid-Base Titrations

Problems

17.17 A 0.2688-g sample of a monoprotic acid neutralizes16.4 mL of 0.08133 M KOH solution. Calculate themolar mass of the acid.

17.18 A 5.00-g quantity of a diprotic acid is dissolved inwater and made up to exactly 250 mL. Calculate themolar mass of the acid if 25.0 mL of this solution re-quired 11.1 mL of 1.00 M KOH for neutralization.Assume that both protons of the acid are titrated.

17.19 Calculate the pH at the equivalence point for thesetitrations: (a) 0.10 M HCl versus 0.10 M NH3,(b) 0.10 M CH3COOH versus 0.10 M NaOH.

17.20 A sample of 0.1276 g of an unknown monoproticacid was dissolved in 25.0 mL of water and titratedwith 0.0633 M NaOH solution. The volume of baserequired to reach the equivalence point was 18.4 mL.(a) Calculate the molar mass of the acid. (b) After10.0 mL of base had been added to the titration, thepH was determined to be 5.87. What is the Ka of theunknown acid?

Acid-Base Indicators

Review Questions

17.21 Explain how an acid-base indicator works in a titration.

17.22 What are the criteria for choosing an indicator for aparticular acid-base titration?

Problems

17.23 The amount of indicator used in an acid-base titra-tion must be small. Why?

17.24 A student carried out an acid-base titration byadding NaOH solution from a buret to an Erlen-meyer flask containing HCl solution and using phe-nolphthalein as indicator. At the equivalence point,he observed a faint reddish-pink color. However, af-ter a few minutes, the solution gradually turned col-orless. What do you suppose happened?

17.25 Referring to Table 17.1, specify which indicator orindicators you would use for the following titrations:(a) HCOOH versus NaOH, (b) HCl versus KOH,(c) HNO3 versus NH3.


Questions and Problems 605

17.26 The ionization constant Ka of an indicator HIn is 1.0 �10�6. The color of the nonionized form is red andthat of the ionized form is yellow. What is the colorof this indicator in a solution whose pH is 4.00?(Hint: The color of an indicator can be estimated byconsidering the ratio [HIn]/[In�]. If the ratio isequal to or greater than 10, the color will be that ofthe nonionized form. If the ratio is equal to orsmaller than 0.1, the color will be that of the ionizedform.)

Solubility and Solubility Product

Review Questions

17.27 Define solubility, molar solubility, and solubilityproduct. Explain the difference between solubilityand the solubility product of a slightly soluble sub-stance such as BaSO4.

17.28 Why do we usually not quote the Ksp values for sol-uble ionic compounds?

17.29 Write balanced equations and solubility productexpressions for the solubility equilibria of thesecompounds: (a) CuBr, (b) ZnC2O4, (c) Ag2CrO4,(d) Hg2Cl2, (e) AuCl3, (f) Mn3(PO4)2.

17.30 Write the solubility product expression for the ioniccompound AxBy.

17.31 How can we predict whether a precipitate will formwhen two solutions are mixed?

17.32 Silver chloride has a larger Ksp than silver carbonate(see Table 17.2). Does this mean that the former alsohas a larger molar solubility than the latter?

Problems

17.33 Calculate the concentration of ions in these saturatedsolutions:

(a) [I�] in AgI solution with [Ag�] � 9.1 � 10�9 M(b) [Al3�] in Al(OH)3 with [OH

�] � 2.9 � 10�9 M

17.34 From the solubility data given, calculate the solubil-ity products for these compounds:

(a) SrF2, 7.3 � 10�2 g/L

(b) Ag3PO4, 6.7 � 10�3 g/L

17.35 The molar solubility of MnCO3 is 4.2 � 10�6 M.

What is Ksp for this compound?

17.36 Using data from Table 17.2, calculate the molar sol-ubility of calcium phosphate, which is a componentof bones.

17.37 The solubility of an ionic compound M2X3 (molarmass � 288 g) is 3.6 � 10�17 g/L. What is Ksp forthe compound?

17.38 Using data from Table 17.2, calculate the solubilityof CaF2 in g/L.

17.39 What is the pH of a saturated zinc hydroxide solution?

17.40 The pH of a saturated solution of a metal hydroxideMOH is 9.68. Calculate the Ksp for the compound.

17.41 A sample of 20.0 mL of 0.10 M Ba(NO3)2 is added to50.0 mL of 0.10 M Na2CO3. Will BaCO3 precipitate?

17.42 A volume of 75 mL of 0.060 M NaF is mixed with25 mL of 0.15 M Sr(NO3)2. Calculate the concentra-tions in the final solution of NO�3, Na

�, Sr2�, andF�. (Ksp for SrF2 � 2.0 � 10

�10.)

The Common Ion Effect

Review Questions

17.43 How does a common ion affect solubility? UseLe Châtelier’s principle to explain the decrease insolubility of CaCO3 in a Na2CO3 solution.

17.44 The molar solubility of AgCl in 6.5 � 10�3 MAgNO3 is 2.5 � 10

�8 M. In deriving Ksp from thesedata, which of these assumptions are reasonable?

(a) Ksp is the same as solubility.(b) Ksp of AgCl is the same in 6.5 � 10

�3 M AgNO3as in pure water.

(c) Solubility of AgCl is independent of the concen-tration of AgNO3.

(d) [Ag�] in solution does not change significantlyon the addition of AgCl to 6.5 � 10�3 M AgNO3.

(e) [Ag�] in solution after the addition of AgCl to6.5 � 10�3 M AgNO3 is the same as it wouldbe in pure water.

Problems

17.45 How many grams of CaCO3 will dissolve in 3.0 �102 mL of 0.050 M Ca(NO3)2?

17.46 The solubility product of PbBr2 is 8.9 � 10�6. Deter-

mine the molar solubility (a) in pure water, (b) in0.20 M KBr solution, (c) in 0.20 M Pb(NO3)2 solution.

17.47 Calculate the molar solubility of AgCl in a solu-tion made by dissolving 10.0 g of CaCl2 in 1.00 Lof solution.

17.48 Calculate the molar solubility of BaSO4 (a) in waterand (b) in a solution containing 1.0 M SO4

2� ions.

Complex Ions

Review Questions

17.49 Explain the formation of complexes in Table 17.4 interms of Lewis acid-base theory.

17.50 Give an example to illustrate the general effect ofcomplex ion formation on solubility.

Problems

17.51 Write the formation constant expressions for thesecomplex ions: (a) Zn(OH)4

2�, (b) Co(NH3)63�,

(c) HgI42�.



17.52 Explain, with balanced ionic equations, why (a) CuI2dissolves in ammonia solution, (b) AgBr dissolvesin NaCN solution, (c) Hg2Cl2 dissolves in KClsolution.

17.53 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)42�, and NH3 at equilibrium?

17.54 Calculate the concentrations of Cd2�, Cd(CN)42�,

and CN� at equilibrium when 0.50 g of Cd(NO3)2dissolves in 5.0 � 102 mL of 0.50 M NaCN.

17.55 If NaOH is added to 0.010 M Al3�, which will be thepredominant species at equilibrium: Al(OH)3 orAl(OH)4

�? The pH of the solution is 14.00. [Kf forAl(OH)4

� � 2.0 � 1033.]

17.56 Calculate the molar solubility of AgI in a 1.0 M NH3solution. (Hint: You need to consider two differenttypes of equilibria.)

Qualitative Analysis

Review Questions

17.57 Outline the general principle of qualitative analysis.

17.58 Give two examples of metal ions in each group(1 through 5) in the qualitative analysis scheme.

Problems

17.59 In a group 1 analysis, a student obtained a precipitatecontaining both AgCl and PbCl2. Suggest onereagent that would allow her to separate AgCl(s)from PbCl2(s).

17.60 In a group 1 analysis, a student adds hydrochloricacid to the unknown solution to make [Cl�] � 0.15 M.Some PbCl2 precipitates. Calculate the concentra-tion of Pb2� remaining in solution.

17.61 Both KCl and NH4Cl are white solids. Suggest onereagent that would enable you to distinguish be-tween these two compounds.

17.62 Describe a simple test that would enable you to dis-tinguish between AgNO3(s) and Cu(NO3)2(s).

Additional Problems

17.63 A quantity of 0.560 g of KOH is added to 25.0 mL of1.00 M HCl. Excess Na2CO3 is then added to the so-lution. What mass (in grams) of CO2 is formed?

17.64 A volume of 25.0 mL of 0.100 M HCl is titratedagainst a 0.100 M NH3 solution added to it from aburet. Calculate the pH values of the solution (a) af-ter 10.0 mL of NH3 solution have been added, (b) af-ter 25.0 mL of NH3 solution have been added,(c) after 35.0 mL of NH3 solution have been added.

17.65 The buffer range is defined by the equation pH �pKa � 1. Calculate the range of the ratio [conjugatebase]/[acid] that corresponds to this equation.

17.66 The pKa of the indicator methyl orange is 3.46. Overwhat pH range does this indicator change from 90%HIn to 90% In�?

17.67 Sketch the titration curve of a weak acid versus astrong base such as that shown in Figure 17.5. Onyour graph indicate the volume of base used at theequivalence point and also at the half-equivalencepoint, that is, the point at which half of the base hasbeen added. Show how you can measure the pH ofthe solution at the half-equivalence point. UsingEquation (17.3), explain how you can determine thepKa of the acid by this procedure.

17.68 A 200-mL volume of NaOH solution was added to400 mL of a 2.00 M HNO2 solution. The pH of themixed solution was 1.50 units greater than that of theoriginal acid solution. Calculate the molarity of theNaOH solution.

17.69 The pKa of butyric acid (HBut) is 4.7. Calculate Kbfor the butyrate ion (But�).

17.70 A solution is made by mixing exactly 500 mL of0.167 M NaOH with exactly 500 mL 0.100 MCH3COOH. Calculate the equilibrium concentrationsof H�, CH3COOH, CH3COO

�, OH�, and Na�.

17.71 Cd(OH)2 is an insoluble compound. It dissolves in aNaOH solution. Write a balanced ionic equation forthis reaction. What type of reaction is this?

17.72 Calculate the pH of the 0.20 M NH3/0.20 M NH4Clbuffer. What is the pH of the buffer after the addi-tion of 10.0 mL of 0.10 M HCl to 65.0 mL of thebuffer?

17.73 For which of these reactions is the equilibrium con-stant called a solubility product?

(a) Zn(OH)2(s) � 2OH�(aq) 34 Zn(OH)4

2�(aq)(b) 3Ca2�(aq) � 2PO4

3�(aq) 34 Ca3(PO4)2(s)(c) CaCO3(s) � 2H

�(aq) 34Ca2�(aq) � H2O(l) � CO2(g)

(d) PbI2(s) 34 Pb2�(aq) � 2I�(aq)

17.74 A student mixes 50.0 mL of 1.00 M Ba(OH)2 with86.4 mL of 0.494 M H2SO4. Calculate the mass ofBaSO4 formed and the pH of the mixed solution.

17.75 A 2.0-L kettle contains 116 g of calcium carbonateas boiler scale. How many times would the kettlehave to be completely filled with distilled water toremove all of the deposit at 25C?

acid-base equilibria and solubility equilibria€¦ · 576 chapter 17 acid-base equilibria and...

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