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Menu TOC Study Back Forward Main Guide TOC Textbook Website MHHE Website CHAPTER 16 Acid-Base Equilibria and Solubility Equilibria INTRODUCTION 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 HETEROGENEOUS SOLUTION 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 PRECIPITATION 16.8 THE COMMON ION EFFECT AND SOLUBILITY 16.9 pH AND SOLUBILITY 16.10 COMPLEX ION EQUILIBRIA AND SOLUBILITY 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 Chateliers 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 Chateliers 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 Ka[HA] [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 solution. It is important to remember that the Henderson-Hasselbalch equation is derived from the equilibrium constant expression. It is valid regardless of the source of the conjugate base (that is, whether it comes from the acid alone or is supplied by both the acid and its salt). In problems that involve the common ion effect, we are usually given the starting concentrations of a weak acid HA and its salt, such as NaA. 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 Equation (16.4). In the following example we calculate the pH of a solution containing a common ion. 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 were present? Register to View AnswerSodium 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] [CH3COO ] (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 Equation (16.3)]: 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: [H ] 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 most cases. 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 salt). 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 indicators 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 baseacid. Thus the sodium acetateacetic 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. Register to View AnswerH3PO4 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) Na2CO3/NaHCO3. 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] Ka [H ] 1.0 M and [CH3COO ] [H ][CH3COO ] [CH3COOH] 10 5 Ka[CH3COOH] [CH3COO ] (1.8 10 5)(1.0) (1.0) 1.8 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 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 CH3COO (aq) 0.10 mol H (aq) 88n CH3COOH(aq) 0.10 mol 0.10 mol the number of moles of acetic acid and the number of moles of acetate ions present are 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) 0.90 (1.8 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 2.2 1.8 10 10 5 5 M 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] 0 [acid] or 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 Equation (16.4). Answer H3PO4(aq) 34 H (aq) H2PO4 (aq) Ka1 7.5 10 3; pKa1 2.12 H2PO4 (aq) 34 H (aq) HPO2 (aq) 4 PO3 (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] [acid] 7.40 log pKa 7.21 log [HPO2 ] 4 [H2PO4 ] [HPO2 ] 4 [H2PO4 ] 0.19 Taking the antilog, we obtain [HPO2 ] 4 [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 1-L solution. 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 ACIDSTRONG BASE TITRATIONS The reaction between HCl, a strong acid, and NaOH, a strong base, can be represented by 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 acidstrong 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 13 12 11 10 9 8 pH 7 Equivalence point 6 5 4 3 2 1 0 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 ACIDSTRONG 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 p. 623). The following example deals with the titration of a weak acid with a strong base. FIGURE 16.4 pH profile of a weak acidstrong 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 13 12 11 10 9 Equivalence point 8 pH 7 6 5 4 3 2 1 0 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 the solution. (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 CH3COOH(aq) 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 [CH3COO ] (1.5 10 3)(1.8 1.0 10 2.7 pH 10 log (2.7 5 10 5) 3 M 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 CH3COOH(aq) 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 1L 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 1L 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 ACIDWEAK 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 acidweak 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 11 10 9 8 7 pH 6 Equivalence point 5 4 3 2 1 0 10 (a) 20 30 40 50 40 50 Volume of HCl added (mL) 10 9 8 7 6 pH Equivalence point 5 4 3 2 1 0 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 is 2.5 HCl(aq) 10 3 mol 2.5 NH3(aq) 88n NH4Cl(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) 0.050 0.00 x x Initial (M ): Change (M ): Equilibrium (M ): (0.050 x) x H (aq) 0.00 x x Step 2: From Table 15.4 we obtain the Ka for NH4 : [NH3][H ] [NH4 ] Ka x2 0.050 Applying the approximation 0.050 2 x 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: [HIn] 10 [In ] color of acid (HIn) predominates [In ] 10 [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. Register to View AnswerNear 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 Bromophenol blue Methyl orange Methyl red Chlorophenol blue Bromothymol blue Cresol red Phenolphthalein Red Yellow Orange Red Yellow Yellow Yellow Colorless Yellow Bluish purple Yellow Yellow Red Blue Red Reddish pink 1.22.8 3.04.6 3.14.4 4.26.3 4.86.4 6.0 7.6 7.28.8 8.310.0 *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 through it. 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 carbonic anhydrase: 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 anhydrase: 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 does HHb. 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 CA CO2 CO2 + H2O H2CO3 O2 + HHb HbO 2 + H+ HHbO2 H+ + HCO 3 CA CO2 CO2 + H2O H2CO3 H+ + HCO 3 HCO 3 HCO 3 Plasma Plasma (a) Oxygencarbon 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 25C. 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 complex: MgF2 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 4 Ag2CO3 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 25C 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 8.1 1.7 1.1 1.6 8.0 8.7 4.0 8.0 1.2 3.0 4.0 4.2 5.1 2.2 6.0 1.6 1.1 6.0 3.3 2.4 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 COMPOUND 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 6 10 72 28 9 11 6 26 29 21 8 12 20 37 14 36 19 14 4 2.0 4.1 1.4 3.4 4.0 1.2 3.0 3.5 4.0 1.4 7.7 8.1 1.6 8.3 1.4 6.0 1.6 3.8 1.0 1.8 3.0 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 14 8 8 28 5 11 14 18 54 24 13 12 10 17 5 51 9 7 26 14 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 Q [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 substances 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 solubility of compound Solubility of compound SOLUBILITY EQUILIBRIA 667 Concentrations of cations and anions Ksp of compound Molar solubility of compound Solubility of compound (a) Concentrations of cations 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 25C). 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 4 [Ca2 ] 4.9 10 3 M [SO2 ] 4 and 4.9 10 3 M Now we can calculate Ksp: Ksp [Ca2 ][SO2 ] 4 (4.9 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 compounds 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) 0.00 s Initial (M ): Change (M ): Equilibrium (M ): Br (aq) 0.00 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) 0.00 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 (s)(2s)2 2.2 10 4 20 s3 Solving for s, we get 2OH (aq) 0.00 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) EQUILIBRIUM CONCENTRATION (M ) COMPOUND Ksp EXPRESSION CATION ANION RELATION BETWEEN Ksp AND s AgCl BaSO4 [Ag ][Cl ] [Ba2 ][SO2 ] 4 s s s s Ksp 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 2 (Ksp) (Ksp) Ksp 4s3; s 4 Ksp 4s3; s 4 Ksp 27s4; s 27 Ksp 108s5; s 108 1 2 1 3 1 3 1 4 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 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 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 ), 4 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 patients 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 4 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 [SO2 ] 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 4 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 AgCl AgBr AgI Ksp 1.6 7.7 8.3 10 10 10 10 13 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 [Br ] 3.9 AgCl and (left) 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 ] Ksp [Cl ] 8.0 1.6 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 4 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 8.0 9.6 10 10 10 13 9 5 M The percent of Br remaining in solution (the unprecipitated Br ) at the critical concentration of Ag is % Br [Br ]unpptd [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: Q 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: AgCl(s) 34 Initial (M ): Change (M ): Equilibrium (M ): (6.5 Step 3: Ksp 1.6 Ag (aq) 6.5 10 3 s s) s [Ag ][Cl ] 10 10 3 10 Cl (aq) 0.0 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, we obtain 10 10 6.5 10 3 s s 1.6 2.5 10 8 M Step 4: At equilibrium [Ag ] [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 persons 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, [OH ] pOH 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 right: 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 SO2 4 PbSO4(s) Overall: Similar problem: 16.60. PbSO4(s) 34 Pb2 (aq) SO2 (aq) 4 (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 NH3(aq) 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. We write [Fe2 ][OH ]2 Ksp 1.6 Since FeCl2 is a strong electrolyte, [Fe2 ] [OH ]2 [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: NH3(aq) x 2.3 10 Initial (M ): Change (M ): Equilibrium (M ): (x 2.3 H2O(l ) 34 6 10 6) NH4 (aq) 0.00 2.3 10 2.3 10 6 6 OH (aq) 0.00 2.3 10 2.3 10 6 6 Substituting the equilibrium concentrations in the expression for the ionization constant, [NH4 ][OH ] [NH3] (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 x 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 6 ion. When we write Co(H2O)2 , 6 we mean the hydrated 2 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 6 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) 6 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): 4 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 4 Cu2 (aq) 4NH3(aq) 34 Cu(NH3)2 (aq) 4 for which the formation constant is Kf [Cu(NH3)2 ] 4 [Cu2 ][NH3]4 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 Cu(NH3)2 . 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 25C COMPLEX ION EQUILIBRIUM EXPRESSION Ag(NH3)2 Ag(CN)2 Cu(CN)2 4 Cu(NH3)2 4 Cd(CN)2 4 CdI2 4 HgCl2 4 HgI2 4 Hg(CN)2 4 Co(NH3)3 6 Zn(NH3)2 4 Ag Ag Cu2 Cu2 Cd2 Cd2 Hg2 Hg2 Hg2 Co3 Zn2 2NH3 2CN 4CN 4NH3 4CN 4I 4Cl 4I 4CN 6NH3 4NH3 FORMATION CONSTANT (Kf) 34 Ag(NH3)2 34 Ag(CN)2 34 Cu(CN)2 4 34 Cu(NH3)2 4 34 Cd(CN)2 4 34 CdI2 4 34 HgCl2 4 34 HgI2 4 34 Hg(CN)2 4 34 Co(NH3)3 6 34 Zn(NH3)2 4 1.5 1.0 1.0 5.0 7.1 2.0 1.7 2.0 2.5 5.0 2.9 107 1021 1025 1013 1016 106 1016 1030 1041 1031 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 4 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 . 4 Since Cu(NH3)2 does dissociate to a slight extent, we call the concentration of 4 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 ] 4 [Cu2 ][NH3]4 0.20 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? 4 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 ] [NH3]2 KspKf (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) 1.0 0.0 2s s Equilibrium (M ): (1.0 2s) s Cl (aq) 0.0 s 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) (1.0 2s)2 s2 3 2s)2 (1.0 Taking the square root of both sides, we obtain 0.049 s s 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 16.10). 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) 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 Hg2 2 Pb2 Bi3 Cd2 Cu2 Sn2 Al3 Co2 Cr3 Fe2 Mn2 Ni2 Zn2 Ba2 Ca2 Sr2 K Na NH4 PRECIPITATING REAGENTS INSOLUBLE COMPOUND HCl AgCl Hg2Cl2 PbCl2 Bi2S3 CdS CuS SnS Al(OH)3 CoS Cr(OH)3 FeS MnS NiS ZnS BaCO3 CaCO3 SrCO3 None None None A A g H2S in acidic solutions A g H2S in basic solutions A A A A A g Na2CO3 A A g No precipitating reagent Ksp 1.6 3.5 2.4 1.6 8.0 6.0 1.0 1.8 4.0 3.0 6.0 3.0 1.4 3.0 8.1 8.7 1.6 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 18 4 72 28 37 26 33 21 29 19 14 24 23 9 9 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 hydroxide: 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 +HCl Filtration Group 1 precipitates AgCl, Hg2Cl2, PbCl2 Solution containing ions of remaining groups +H2S Filtration Group 2 precipitates CuS, CdS, SnS, Bi2S3 Solution containing ions of remaining groups +NaOH Filtration Group 3 precipitates CoS, FeS, MnS, NiS ZnS, Al(OH)3, Cr(OH)3 Solution containing ions of remaining groups +Na2CO3 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 hens 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 hens 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 stops. The eggshell is largely composed of calcite, a crystalline form of calcium carbonate (CaCO3). Normally, the raw materials, Ca2 and CO 2 , are 3 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 ions: 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 chickens body than normal respiration does. According to Le Chateliers principle, panting will shift the CO2H2CO3 equilibrium shown above from right to left, thereby lowering the concentration of the CO2 3 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 chickens body fluids and shifts the CO2H2CO3 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 acidstrong base titrations, the pH at the equivalence point is 7; for weak acidstrong base titrations, the pH at the equivalence point is greater than 7; for strong acidweak 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 NH4Cl. BUFFER SOLUTIONS Review Questions 16.5 What is a buffer solution? What constitutes a buffer solution? 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 stronger acid? 16.8 Identify the buffer systems below: (a) KCl/HCl (b) NH3/NH4NO3 (c) Na2HPO4/NaH2PO4 (d) KNO2/HNO2 (e) KHSO4/H2SO4 (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 25C 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 buffer? 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 or Na2A/NaHA. 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? Why? ACID-BASE TITRATIONS 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 titrated. 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 Na . 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 NaOH. ACID-BASE INDICATORS 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 4.00? 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? SOLUBILITY EQUILIBRIA 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 compound AxBy. 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 the compound? 16.45 The solubility of an ionic compound M2X3 (molar mass 288 g) is 3.6 10 17 g/L. What is Ksp for the compound? 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.) FRACTIONAL PRECIPITATION 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 Na2CO3 solution. 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 AgNO3. (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 solution. 16.57 Calculate the molar solubility of AgCl in a solution made by dissolving 10.0 g of CaCl2 in 1.00 L of solution. 16.58 Calculate the molar solubility of BaSO4 (a) in water, (b) in a solution containing 1.0 M SO2 ions. 4 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? 4 16.68 Calculate the concentrations of Cd2 , Cd(CN)2 , 4 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 2.0 1033.] Al(OH)4 16.70 Calculate the molar solubility of AgI in a 1.0 M NH3 solution. 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 solution. QUALITATIVE ANALYSIS 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) from PbCl2(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 Na . 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 this? 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) 4 (b) 3Ca2 (aq) 2PO3 (aq) 34 Ca3(PO4)2(s) 4 (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 25C? 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 been added. 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 Pb(IO3)2? 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 patients body, what would be [Ba2 ]? For a good estimate, we may assume that the temperature is at 25C. 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 4 both compounds. 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 structure: O B COOH H H3C H3C G C D C S O NOC H AA C OCO NOCO CH2 B A O 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 25C 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? Why? 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. 4 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 hens 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 Albumen Yolk membrane Yolk Chalaza 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 ... View Full Document

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