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Unformatted text preview: BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE WebsiteCHAPTER16Acid-Base Equilibria andSolubility EquilibriaINTRODUCTIONINTHIS CHAPTER WE WILL CONTINUE OUR STUDY OF ACID-BASE REAC-TIONS WITH A DISCUSSION OF BUFFER ACTION AND TITRATIONS.16.1 HOMOGENEOUS VERSUSHETEROGENEOUS SOLUTIONEQUILIBRIAWE WILLALSO LOOK AT ANOTHER TYPE OF AQUEOUS EQUILIBRIUM THE ONE16.2 THE COMMON ION EFFECTBETWEEN A SLIGHTLY SOLUBLE COMPOUND AND ITS IONS IN SOLUTION.16.3 BUFFER SOLUTIONS16.4 ACID-BASE TITRATIONS16.5 ACID-BASE INDICATORS16.6 SOLUBILITY EQUILIBRIA16.7 SEPARATION OF IONS BY FRACTIONALPRECIPITATION16.8 THE COMMON ION EFFECT ANDSOLUBILITY16.9 pH AND SOLUBILITY16.10 COMPLEX ION EQUILIBRIA ANDSOLUBILITY16.11 APPLICATION OF THE SOLUBILITYPRODUCT PRINCIPLE TO QUALITATIVEANALYSIS645BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website646ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA16.1HOMOGENEOUS VERSUS HETEROGENEOUS SOLUTION EQUILIBRIAIn 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 acidas 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 thischapter, involves the dissolution and precipitation of slightly soluble substances. Theseprocesses are examples of heterogeneous equilibria that is, they pertain to reactionsin which the components are in more than one phase.16.2THE COMMON ION EFFECTOur discussion of acid-base ionization and salt hydrolysis in Chapter 15 was limitedto solutions containing a single solute. In this section we will consider the acid-baseproperties of a solution with two dissolved solutes that contain the same ion (cation oranion), called the common ion.The presence of a common ion suppresses the ionization of a weak acid or a weakbase. If both sodium acetate and acetic acid are dissolved in the same solution, for example, they both dissociate and ionize to produce CH3COO ions:H2OCH3COONa(s) 88n CH3COO (aq)Na (aq)CH3COOH(aq) 34 CH3COO (aq)H (aq)CH3COONa is a strong electrolyte, so it dissociates completely in solution, butCH3COOH, 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 bothCH3COOH and CH3COONa will be less acidic than a solution containing onlyCH3COOH 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 effectplays an important role in determining the pH of a solution and the solubility of a slightlysoluble salt (to be discussed later in this chapter). Here we will study the common ioneffect 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 solublesalt of the weak acid, such as NaA. We start by writingHA(aq)H2O(l ) 34 H3O (aq)A (aq)or simplyHA(aq) 34 H (aq)A (aq)The ionization constant Ka is given byKaBackForwardMain MenuTOC[H ][A ][HA]Study Guide TOCTextbook Website(16.1)MHHE Website16.2THE COMMON ION EFFECT647Rearranging Equation (16.1) givesKa[HA][A ][H ]Taking the negative logarithm of both sides, we obtainlog [H ]log Kalog[HA][A ]log [H ]log Kalog[A ][HA]orSopHpKalog[A ][HA](16.2)wherepKa is related to Ka as pH isrelated to [H ]. Remember thatthe stronger the acid (that is,the larger the Ka), thesmaller the pKa.pKalog Ka(16.3)Equation (16.3) is called the Henderson-Hasselbalch equation. A more general formof this expression ispHpKalog[conjugate base][acid](16.4)In our example, HA is the acid and A is the conjugate base. Thus, if we know Ka andthe concentrations of the acid and the salt of the acid, we can calculate the pH of thesolution.It is important to remember that the Henderson-Hasselbalch equation is derivedfrom the equilibrium constant expression. It is valid regardless of the source of theconjugate base (that is, whether it comes from the acid alone or is supplied by boththe 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 becauseHA is a weak acid and the extent of the hydrolysis of the A ion is generally verysmall. Moreover, the presence of A (from NaA) further suppresses the ionization ofHA and the presence of HA further suppresses the hydrolysis of A . Thus we can usethe starting concentrations as the equilibrium concentrations in Equation (16.1) orEquation (16.4).In the following example we calculate the pH of a solution containing a commonion.EXAMPLE 16.1(a) Calculate the pH of a solution containing 0.20 M CH3COOH and 0.30 MCH3COONa. (b) What would the pH of a 0.20 M CH3COOH solution be if no saltwere present?Answer(a) Sodium acetate is a strong electrolyte, so it dissociates completely insolution:BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website648ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIAH2OCH3COONa(s) 88n CH3COO (aq)0.30 MNa (aq)0.30 MAs stated above, we can use the starting concentrations as the equilibrium concentrations; that is[CH3COOH]0.20 Mand[CH3COO ]0.30 MFrom Equation (16.1) we have[H ][CH3COO ][CH3COOH]Kaor[H ]Ka[CH3COOH][CH3COO ](1.810 5)(0.20)0.30pHlog [H ]1.2105MThus10 5)log (1.24.92Alternatively, 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 [seeEquation (16.3)]:pKalog Ka10 5)log (1.84.74We can calculate the pH of the solution by substituting the value of pKa and theconcentrations of the acid and its conjugate base in Equation (16.4):pKalog[CH3COO ][CH3COOH]4.74log0.30 M0.20 M4.74pH0.184.92(b) Following the procedure in Example 15.8, we find that the pH of a 0.30 MCH3COOH solution is:[H ]pH1.910log (1.93M10 3)2.72Thus, without the common ion effect, the pH of a 0.20 M CH3COOHsolution 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 suppressesthe ionization of the acid CH3COOH.CommentSimilar problem: 16.3.BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website16.3BUFFER SOLUTIONS649PRACTICE EXERCISEWhat 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, suchas NH3, and a salt of the base, say NH4Cl. At equilibriumNH4 (aq) 34 NH3(aq)H (aq)[NH3][H ][NH4 ]KaWe can derive the Henderson-Hasselbalch equation for this system as follows.Rearranging the above equation we obtainKa[N H4 ][N H3][H ]Taking the negative logarithm of both sides giveslog [H ]log Kalog[NH4 ][NH3]log [H ]log Kalog[ N H 3][NH4 ]orpHpKalog[ 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.16.3BUFFER SOLUTIONSA buffer solution is a solution of (1) a weak acid or a weak base and (2) its salt; bothcomponents must be present. The solution has the ability to resist changes in pH uponthe addition of small amounts of either acid or base. Buffers are very important tochemical and biological systems. The pH in the human body varies greatly from onefluid to another; for example, the pH of blood is about 7.4, whereas the gastric juicein our stomachs has a pH of about 1.5. These pH values, which are crucial for properenzyme function and the balance of osmotic pressure, are maintained by buffers inmost cases.A buffer solution must contain a relatively large concentration of acid to reactwith any OH ions that are added to it, and it must contain a similar concentration ofbase to react with any added H ions. Furthermore, the acid and the base componentsof 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 itsconjugate base (supplied by a salt) or a weak base and its conjugate acid (supplied bya salt).A simple buffer solution can be prepared by adding comparable amounts of aceticacid (CH3COOH) and its salt sodium acetate (CH3COONa) to water. The equilibriumBackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website650ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIAFIGURE 16.1 The acid-base indicator bromophenol blue (addedto all solutions shown) is used toillustrate buffer action. The indicators color is blue-purple abovepH 4.6 and yellow below pH3.0. (a) A buffer solution madeup of 50 mL of 0.1 MCH3COOH and 50 mL of 0.1 MCH3COONa. The solution has apH of 4.7 and turns the indicatorblue-purple. (b) After the additionof 40 mL of 0.1 M HCl solutionto the solution in (a), the color remains blue-purple. (c) A 100-mLCH3COOH solution whose pH is4.7. (d) After the addition of 6drops (about 0.3 mL) of 0.1 MHCl solution, the color turns yellow. Without buffer action, thepH of the solution decreasesrapidly to less than 3.0 upon theaddition of 0.1 M HCl.(a)(b)(c)(d)concentrations of both the acid and the conjugate base (from CH3COONa) are assumedto be the same as the starting concentrations (see p. 647). A solution containing thesetwo substances has the ability to neutralize either added acid or added base. Sodiumacetate, a strong electrolyte, dissociates completely in water:H2OCH3COONa(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 equationCH3COO (aq)H (aq) 88n CH3COOH(aq)If a base is added to the buffer system, the OH ions will be neutralized by the acidin 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 tothose 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 asCH3COONa/CH3COOH or simply CH3COO /CH3COOH. Figure 16.1 shows thisbuffer system in action.The following example distinguishes buffer systems from acid-salt combinationsthat do not function as buffers.EXAMPLE 16.2Which of the following solutions are buffer systems? (a) KH2PO4/H3PO4, (b)NaClO4/HClO4, (c) C5H5N/C5H5NHCl (C5H5N is pyridine; its Kb is given in Table15.4). Explain your answer.Answer (a) H3PO4 is a weak acid, and its conjugate base, H2PO4 , is a weak base(see Table 15.5). Therefore, this is a buffer system.(b) Because HClO4 is a strong acid, its conjugate base, ClO4 , is an extremely weakbase. This means that the ClO4 ion will not combine with a H ion in solution toform HClO4. Thus the system cannot act as a buffer system.BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website16.3Similar problem: 16.8.BUFFER SOLUTIONS651(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 EXERCISEWhich 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 MCH3COONa. (b) What is the pH of the buffer system after the addition of 0.10 moleof gaseous HCl to 1 L of the solution? Assume that the volume of the solution doesnot 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 haveAnswer[CH3COOH]Ka[H ]1.0 Mand[CH3COO ][H ][CH3COO ][CH3COOH]105Ka[CH3COOH][CH3COO ](1.810 5)(1.0)(1.0)1.8pH1.81.0 M10log (1.85M10 5)4.74Thus when the concentrations of the acid and the conjugate base are the same, thepH 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 mol0.10 molCl (aq)0.10 molOriginally, 1.0 mol CH3COOH and 1.0 mol CH3COO were present in 1 L of thesolution. After neutralization of the HCl acid by CH3COO , which we write asCH3COO (aq)0.10 molH (aq) 88n CH3COOH(aq)0.10 mol0.10 molthe number of moles of acetic acid and the number of moles of acetate ions presentareCH3COOH:(1.00.1) mol1.1 molCH3COO :(1.00.1) mol0.90 molNext we calculate the hydrogen ion concentration:BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website652ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIAKa[CH3COOH][CH3COO ][H ]10 5)(1.1)0.90(1.82.2510MThe pH of the solution becomespH10 5)log (2.24.66Note 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 ofmoles present; that is, (1.1 mol/L)/(0.90 mol/L) (1.1 mol/0.90 mol).CommentSimilar problem: 16.15.PRACTICE EXERCISECalculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pHafter 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 (thesolution becomes more acidic) as a result of added HCl. We can also compare thechanges in H ion concentration as followsBefore addition of HCl:After addition of HCl:[H ]1.8105M[H ]2.2105MThus the H ion concentration increases by a factor of2.21.8101055MM1.2To appreciate the effectiveness of the CH3COONa/CH3COOH buffer, let us findout what would happen if 0.10 mol HCl were added to 1 L of water, and compare theincrease in H ion concentration.Before addition of HCl:After addition of HCl:[H ]1.0[H ]7100.10 MMAs a result of the addition of HCl, the H ion concentration increases by a factor of1.00.10 M10 7 M1.0106amounting to a millionfold increase! This comparison shows that a properly chosenbuffer solution can maintain a fairly constant H ion concentration, or pH.PREPARING A BUFFER SOLUTION WITH A SPECIFIC pHNow suppose we want to prepare a buffer solution with a specific pH. How do we goabout it? Equation (16.1) indicates that if the molar concentrations of the acid and itsconjugate base are approximately equal, that is, if [acid] [conjugate base], thenlog[conjugate base]0[acid]orpH pKaBackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website16.4ACID-BASE TITRATIONS653Thus, to prepare a buffer solution, we work backwards. First we choose a weak acidwhose pKa is close to the desired pH. Next, we substitute the pH and pKa values inEquation (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 16.4Describe 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 Kavalues are obtained from Table 15.5 and the pKa values are found by applyingEquation (16.4).AnswerH3PO4(aq) 34 H (aq)H2PO4 (aq)Ka17.510 3; pKa12.12H2PO4 (aq) 34 H (aq)HPO2 (aq)4PO3 (aq)4Ka26.210 8; pKa27.214.813HPO24(aq) 34 H (aq)Ka310; pKa312.32HPO 24The most suitable of the three buffer systems is/ H2PO4 , because the pKaof the acid H2PO4 is closest to the desired pH. From the Henderson-Hasselbalchequation we writepHlog[conjugate base][acid]7.40logpKa7.21log[HPO2 ]4[H2PO4 ][HPO2 ]4[H2PO4 ]0.19Taking the antilog, we obtain[HPO2 ]4[H2PO4 ]Similar problems: 16.17, 16.18.1.5Thus, 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 dissolve1.5 moles of Na2HPO4 and 1.0 mole of NaH2PO4 in enough water to make up a1-L solution.PRACTICE EXERCISEHow 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), andsodium carbonate (Na2CO3).The Chemistry in Action essay on p. 663 illustrates the importance of buffer systems in the human body.16.4ACID-BASE TITRATIONSHaving discussed buffer solutions, we can now look in more detail at the quantitativeaspects of acid-base titrations, which we discussed briefly in Section 4.6. Recall thattitration is a procedure for determining the concentration of a solution using anotherBackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website654ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIAFIGURE 16.2 A pH meter isused to monitor an acid-basetitration.solution of known concentration, called standard solution. We will consider three typesof 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 aweak base. Titrations involving a weak acid and a weak base are complicated by thehydrolysis of both the cation and the anion of the salt formed. For this reason, they arealmost never carried out. Figure 16.2 shows the arrangement for monitoring the pHduring the course of a titration.STRONG ACIDSTRONG BASE TITRATIONSThe reaction between HCl, a strong acid, and NaOH, a strong base, can be representedbyNaOH(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 flaskcontaining 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 acidis 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 atthe equivalence point (that is, the point at which equimolar amounts of acid and basehave 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 dropof the base can cause a large increase in [OH ] and in the pH of the solution. Beyondthe 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 arethree sample calculations.1.BackForwardMain MenuAfter 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 in10 mL isTOCStudy Guide TOCTextbook WebsiteMHHE Website16.4FIGURE 16.3 pH profile of astrong acidstrong base titration.A 0.10 M NaOH solution isadded from a buret to 25 mL ofa 0.10 M HCl solution in anErlenmeyer flask (see Figure 4.21).This curve is sometimes referred toas a titration curve.ACID-BASE TITRATIONS655141312111098pH7Equivalencepoint65432101020304050Volume of NaOH added (mL)10 mL0.10 mol NaOH1 L NaOH1L1000 mL1.0310molThe number of moles of HCl originally present in 25 mL of solution is25 mL0.10 mol HCl1 L HCl1L1000 mL2.5103molThus, the amount of HCl left after partial neutralization is (2.5 10 3) (1.010 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 mol1 mol HCl.NaOH1.510 3 mol HCl35 mL1000 mL1L0.043 mol HCl/L0.043 M HClThus [H ]0.043 M, and the pH of the solution ispHlog 0.0431.37After 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 reactionand 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 NaOHadded is2.Neither Na nor Cl undergoeshydrolysis.35 mL0.10 mol NaOH1 L NaOH1000 mL1LThe number of moles of HCl in 25 mL solution is 2.5tralization of HCl, the amount of NaOH left is (3.5BackForwardMain MenuTOCStudy Guide TOC3.5103mol10 3. After complete neu10 3) (2.5 10 3), orTextbook WebsiteMHHE Website656ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA1.0103mol. The concentration of NaOH in 60 mL of solution is1.010 3 mol NaOH60 mL1000 mL1L0.017 mol NaOH/L0.017 M NaOHThus [OH ]lution is0.017 M and pOHlog 0.017pH14.00pOH14.001.77. Hence, the pH of the so-1.7712.23WEAK ACIDSTRONG BASE TITRATIONSConsider 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 toCH3COOH(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 pHwill be greater than 7 as a result of the excess OH ions formed (Figure 16.4). Notethat this situation is analogous to the hydrolysis of sodium acetate (CH3COONa) (seep. 623).The following example deals with the titration of a weak acid with a strong base.FIGURE 16.4 pH profile of aweak acidstrong base titration.A 0.10 M NaOH solution isadded from a buret to 25 mL ofa 0.10 M CH3COOH solution inan Erlenmeyer flask. Due to thehydrolysis of the salt formed, thepH at the equivalence point isgreater than 7.14131211109Equivalencepoint8pH765432101020304050Volume of NaOH added (mL)BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website16.4ACID-BASE TITRATIONS657EXAMPLE 16.5Calculate 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.AnswerThe neutralization reaction isCH3COOH(aq)NaOH(aq) 88n CH3COONa(aq)H2O(l )For each of the three stages of the titration we first calculate the number of molesof NaOH added to the acetic acid solution. Next, we calculate the number of molesof the acid (or the base) left over after neutralization. Then we determine the pH ofthe solution.(a) The number of moles of NaOH in 10 mL is10 mL1L1000 mL0.10 mol NaOH1 L NaOH soln1.0310molThe number of moles of CH3COOH originally present in 25 mL of solution is0.10 mol CH3COOH1 L CH3COOH soln25 mL1L1000 mL2.5103molThus, the amount of CH3COOH left after all added base has been neutralized is(2.510310 3) mol1.0The amount of CH3COONa formed is 1.0CH3COOH(aq)1.0 10 3 mol101.53103molmole:NaOH(aq) 88n CH3COONa(aq)1.0 10 3 mol1.0 10 3 molH2O(l )At this stage we have a buffer system made up of CH3COONa and CH3COOH. Tocalculate the pH of this solution we writeSince the volume of the solutionis the same for CH3COOH andCH3COO , the ratio of thenumber of moles present isequal to the ratio of theirmolar concentrations.[H ][CH3COOH]Ka[CH3COO ](1.5 10 3)(1.81.0 102.7pH10log (2.7510 5)3M10 5)4.57(b) These quantities (that is, 25 mL of 0.10 M NaOH reacting with 25 mL of0.10 M CH3COOH) correspond to the equivalence point. The number of moles ofboth NaOH and CH3COOH in 25 mL is 2.5 10 3 mol, so the number of molesof the salt formed isCH3COOH(aq)2.5 10 3 molNaOH(aq) 88n CH3COONa(aq)2.5 10 3 mol2.5 10 3 molH2O(l )The total volume is 50 mL, so the concentration of the salt is[CH3COONa]2.510 3 mol50 mL0.050 mol/LBackForwardMain MenuTOCStudy Guide TOC1000 mL1L0.050 MTextbook WebsiteMHHE Website658ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIAThe next step is to calculate the pH of the solution that results from the hydrolysisof the CH3COO ions. Following the procedure described in Example 15.13, wefind 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 equivalencepoint. 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 thanCH3COO , we can safely neglect the CH3COO ions and calculate the pH of thesolution using only the concentration of the OH ions. Only 25 mL of NaOH areneeded for complete neutralization, so the number of moles of NaOH left after neutralization is(3525) mL0.10 mol NaOH1 L NaOH soln1L1000 mL1.0103molThe total volume of the combined solutions is now 60 mL, so we calculate OHconcentration as follows:1.0[OH ]10 3 mol60 mL0.017 mol/LpOH1000 mL1L0.017 Mlog 0.0171.77pH14.00pOH14.001.7712.23Similar problem: 16.92.PRACTICE EXERCISEExactly 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 basehas been added, (c) the equivalence point, (d) the point at which 105 mL of the basehas been added.STRONG ACIDWEAK BASE TITRATIONSConsider the titration of HCl, a strong acid, with NH3, a weak base:HCl(aq)NH3(aq) 88n NH4Cl(aq)or simplyH (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 simplyNH4 (aq) 34 NH3(aq)H (aq)Because of the volatility of an aqueous ammonia solution, it is more convenient to addhydrochloric 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 casein which a weak base is added from a buret to HCl.BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website16.4FIGURE 16.5 pH profiles of astrong acidweak base titration.(a) A 0.10 M HCl solution isadded from a buret to 25 mL ofa 0.10 M NH3 solution in anErlenmeyer flask. (b) A 0.10 Mweak base solution is addedfrom a buret to 25 mL of a0.10 M HCl solution in anErlenmeyer flask. As a result ofsalt hydrolysis, the pH at theequivalence point in both cases islower than 7.ACID-BASE TITRATIONS659121110987pH6Equivalencepoint54321010(a)203040504050Volume of HCl added (mL)109876pHEquivalencepoint543210102030Volume of base added (mL)(b)EXAMPLE 16.6Calculate the pH at the equivalence point when 25 mL of 0.10 M NH3 is titrated bya 0.10 M HCl solution.The neutralization reaction is given above. The number of moles of NH3in 25 mL of 0.10 M solution isAnswer25 mLBackForwardMain MenuTOC0.10 mol NH31 L NH3Study Guide TOC1L1000 mL2.5103molTextbook WebsiteMHHE Website660ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIASince 1 mol NH3 1 mol HCl, at the equivalence point the number of moles ofHCl reacted is also 2.5 10 3 mol. The number of moles of the salt (NH4Cl) formedis2.5HCl(aq)10 3 mol2.5NH3(aq)88n NH4Cl(aq)10 3 mol2.5 10 3 molThe total volume is 50 mL, so the concentration of NH4Cl is10 3 mol50 mL2.5[NH4Cl]0.050 mol/L1000 mL1L0.050 MThe pH of the solution at the equivalence point is determined by the hydrolysis ofNH4 ions. We follow the procedure on p. 608.Step 1: We represent the hydrolysis of the cation NH4 , and let x be the equilibriumconcentration of NH3 and H ions in mol/L:NH4 (aq) 34 NH3(aq)0.0500.00xxInitial (M ):Change (M ):Equilibrium (M ):(0.050x)xH (aq)0.00xxStep 2: From Table 15.4 we obtain the Ka for NH4 :[NH3][H ][NH4 ]Kax20.050Applying the approximation 0.0502x0.0505.6x105.6x1010100.050, we get2x0.0505.61010xx5.3106MThus the pH is given bypHlog (5.310 6)5.28Similar problem: 16.27.PRACTICE EXERCISECalculate 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.16.5ACID-BASE INDICATORSThe equivalence point, as we have seen, is the point at which the number of moles ofOH ions added to a solution is equal to the number of moles of H ions originallypresent. To determine the equivalence point in a titration, then, we must know exactlyhow much volume of a base to add from a buret to an acid in a flask. One way toachieve this goal is to add a few drops of an acid-base indicator to the acid solution atthe start of the titration. You will recall from Chapter 4 that an indicator is usually aBackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website16.5ACID-BASE INDICATORS661weak 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 fora particular titration depends on the nature of the acid and base used in the titration (thatis, 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 effectiveindicator, HIn and its conjugate base, In , must have distinctly different colors. Insolution, 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 LeChatelier s principle, shifts to the left and the predominant color of the indicator is thatof the nonionized form (HIn). On the other hand, in a basic medium the equilibriumshifts 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 topredict the perceived color of the indicator:[HIn] 10[In ]color of acid (HIn) predominates[In ] 10[HIn]color of conjugate base (In ) predominatesIf [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 arange 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 atthe 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 indicatoris colorless but that it begins to turn reddish pink when the pH exceeds 8.3. As shownin Figure 16.3, the steepness of the pH curve near the equivalence point means thatthe 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. Whatis important, however, is the fact that the steep portion of the pH profile includes therange over which phenolphthalein changes from colorless to reddish pink. Wheneversuch a correspondence occurs, the indicator can be used to locate the equivalence pointof the titration.Many acid-base indicators are plant pigments. For example, by boiling choppedred 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 acidand base to be titrated. Example 16.7 illustrates this point.EXAMPLE 16.7Which indicator or indicators listed in Table 16.1 would you use for the acid-basetitrations shown in (a) Figure 16.3, (b) Figure 16.4, and (c) Figure 16.5(b)?BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website662ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIAFIGURE 16.6 Solutions containing extracts of red cabbage(obtained by boiling the cabbagein water) produce different colorswhen treated with an acid and abase. The pH of the solutions increases from left to right.Similar problem: 16.31.Answer (a) Near the equivalence point, the pH of the solution changes abruptlyfrom 4 to 10. Therefore all the indicators except thymol blue, bromophenol blue,and methyl orange are suitable for use in the titration.(b) Here the steep portion covers the pH range between 7 and 10; therefore, the 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 EXERCISEReferring to Table 16.1, specify which indicator or indicators you would use for thefollowing titrations: (a) HBr versus CH3NH2, (b) HNO3 versus NaOH, (c) HNO2versus KOH.TABLE 16.1Some Common Acid-Base IndicatorsCOLORINDICATORIN ACIDIN BASEpH RANGE*Thymol blueBromophenol blueMethyl orangeMethyl redChlorophenol blueBromothymol blueCresol redPhenolphthaleinRedYellowOrangeRedYellowYellowYellowColorlessYellowBluish purpleYellowYellowRedBlueRedReddish pink1.22.83.04.63.14.44.26.34.86.46.0 7.67.28.88.310.0*The pH range is defined as the range over which the indicator changes from the acid color to thebase color.BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website16.5ACID-BASE INDICATORS663Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in ActionChemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action ChemistryBackMaintaining the pH of BloodAll higher animals need a circulatory system to carryfuel and oxygen for their life processes and to removewastes. In the human body this vital exchange takesplace in the versatile fluid known as blood, of whichthere are about 5 L (10.6 pints) in an average adult.Blood circulating deep in the tissues carries oxygenand nutrients to keep cells alive, and removes carbondioxide and other waste materials. Using severalbuffer 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 forour 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 protectedfrom extracellular fluid (blood plasma) by a cell membrane that allows only certain molecules to diffusethrough it.The pH of blood plasma is maintained at about7.40 by several buffer systems, the most important ofwhich is the HCO3 /H2CO3 system. In the erythrocyte,where the pH is 7.25, the principal buffer systems areHCO3 /H2CO3 and hemoglobin. The hemoglobinmolecule is a complex protein molecule (molar mass65,000 g) that contains a number of ionizable protons. As a very rough approximation, we can treat itas a monoprotic acid of the form HHb:HHb(aq) 34 H (aq)Hb (aq)where HHb represents the hemoglobin molecule andHb the conjugate base of HHb. Oxyhemoglobin(HHbO2), formed by the combination of oxygen withhemoglobin, 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 bycarbonic anhydrase:CO2(aq)H2O(l ) 34 H2CO3(aq)The ionization of the carbonic acidH2CO3(aq) 34 H (aq)HCO3 (aq)has two important consequences. First, the bicarbonate ion diffuses out of the erythrocyte and is carriedby the blood plasma to the lungs. This is the majormechanism for removing carbon dioxide. Second, theH 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 doesits conjugate base (HbO2 ), the formation of the acidpromotes the following reaction from left to right:HHbO2(aq) 34 HHb(aq)O2(aq)The O2 molecules diffuse out of the erythrocyte andare taken up by other cells to carry out metabolism.When the venous blood returns to the lungs, theabove processes are reversed. The bicarbonate ionsnow diffuse into the erythrocyte, where they react withhemoglobin to form carbonic acid:HHb(aq)HCO3 (aq) 34 Hb (aq)H2CO3(aq)Most of the acid is then converted to CO2 by carbonicanhydrase: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 tothe reaction between HHb and HCO3 shown above)also favors the uptake of oxygen at the lungsHb (aq)O2(aq) 34 HbO2 (aq)because Hb has a greater affinity for oxygen thandoes HHb.When the arterial blood flows back to the bodytissues, the entire cycle is repeated.As the accompanying figure shows, carbon dioxideForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE WebsiteACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIATissuesLungsErythrocyteO2ErythrocyteO2 + HHbO2HbO 2 + H+HHbO2CACO2CO2 + H2OH2CO3O2 + HHbHbO 2 + H+HHbO2H+ + HCO 3CACO2CO2 + H2OH2CO3H+ + HCO 3HCO 3HCO 3PlasmaPlasma(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 theblood capillaries and then into erythrocytes, where it isconverted to carbonic acid by the enzyme carbonic anhydrase (CA). The protons provided by the carbonicacid then combine with the HbO2 anions to formHHbO2, which eventually dissociates into HHb and O2.Because the partial pressure of O2 is higher in the erythrocytes than in the tissues, oxygen molecules diffuseout of the erythrocytes and then into the tissues. The bi-16.6An aqueous suspension of BaSO4is used to examine the digestivetract.BackCapillaryChemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action ChemistryCapillary664Forward(b)carbonate ions also diffuse out of the erythrocytes andare carried in the blood plasma to the lungs. (b) In thelungs, the processes are exactly reversed. Oxygen molecules diffuse from the lungs, where they have a higherpartial pressure, into the erythrocytes. There they combinewith HHb to form HHbO2. The protons provided byHHbO2 combine with the bicarbonate ions diffused intothe erythrocytes from the plasma to form carbonic acid. Inthe presence of carbonic anhydrase, carbonic acid is converted to H2O and CO2. The CO2 then diffuses out of theerythrocytes and into the lungs, where it is exhaled.SOLUBILITY EQUILIBRIAPrecipitation 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, whichis mainly made of hydroxyapatite [Ca5(PO4)3OH], in an acidic medium leads to toothdecay. Barium sulfate (BaSO4), an insoluble compound that is opaque to X rays, isused to diagnose ailments of the digestive tract. Stalactites and stalagmites, which consist of calcium carbonate (CaCO3), are produced by a precipitation reaction, and so aremany foods, such as fudge.The general rules for predicting the solubility of ionic compounds in water wereintroduced in Section 4.2. Although useful, these solubility rules do not allow us tomake 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 knowabout chemical equilibrium. Unless otherwise stated, in the following discussion thesolvent is water and the temperature is 25C.Main MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website16.6SOLUBILITY EQUILIBRIA665SOLUBILITY PRODUCTConsider a saturated solution of silver chloride that is in contact with solid silver chloride. The solubility equilibrium can be represented asAgCl(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 knowfrom Chapter 14 that for heterogeneous reactions the concentration of the solid is aconstant. Thus we can write the equilibrium constant for the dissolution of AgCl (seeExample 14.5) asKsp[Ag ][Cl ]where Ksp is called the solubility product constant or simply the solubility product. Ingeneral, 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 coefficientin 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 morecomplex:MgF2MgF2(s) 34 Mg2 (aq)Ksp[Mg2 ][F ]2CO2 (aq)3Ksp[Ag ]2[CO2 ]32PO3 (aq)4Ksp[Ca2 ]3[PO3 ]24Ag2CO3Ag2CO3(s) 34 2Ag (aq)2F (aq)Ca3(PO4)2Ca3(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 listedin the table for essentially the same reason that we did not include Ka values for strongacids in Table 15.3. The value of Ksp indicates the solubility of an ionic compoundthe smaller the value, the less soluble the compound in water. However, in using Kspvalues 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 substancesexhibit ideal behavior for our calculations involving solution concentrations, but thisassumption is not always valid. For example, a solution of barium fluoride (BaF2) maycontain both neutral and charged ion pairs, such as BaF2 and BaF , in addition to Ba2and F ions. Furthermore, many anions in the ionic compounds listed in Table 16.2are conjugate bases of weak acids. Consider copper sulfide (CuS). The S2 ion canhydrolyze as followsS2 (aq)H2O(l ) 34 HS (aq)OH (aq)HS (aq)H2O(l ) 34 H2S(aq)OH (aq)3And highly charged small metal ions such as Al and Bi3 will undergo hydrolysisas discussed in Section 15.10. Both ion-pair formation and salt hydrolysis decrease theconcentrations of the ions that appear in the Ksp expression, but we need not be concerned with the deviations from ideal behavior here.BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website666ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIATABLE 16.2Solubility Products of Some Slightly Soluble Ionic Compounds at 25CCOMPOUNDKspAluminum 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)1.88.11.71.11.68.08.74.08.01.23.04.04.25.12.26.01.61.16.03.32.4101010101010101010101010101010101010101010COMPOUND33KspLead(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)96107228911626292181220371436191442.04.11.43.44.01.23.03.54.01.47.78.11.68.31.46.01.63.81.01.83.0101010101010101010101010101010101010101010148828511141854241312101755197261423For equilibrium reactions involving an ionic solid in aqueous solution, any one ofthe following conditions may exist: (1) the solution is unsaturated, (2) the solution issaturated, or (3) the solution is supersaturated. For concentrations of ions that do notcorrespond 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 willform. Note that Q has the same form as Ksp except that the concentrations of ions arenot equilibrium concentrations. For example, if we mix a solution containing Ag ionswith one containing Cl ions, then the ion product is given byQ[Ag ]0[Cl ]0The subscript 0 reminds us that these are initial concentrations and do not necessarilycorrespond to those at equilibrium. The possible relationships between Q and Ksp areQ[Ag ]0[Cl ]0Ksp1.61010Unsaturated solutionQ[Ag ][Cl ]Ksp1.61010Q[Ag ]0[Cl ]0Ksp1.61010Saturated solutionSupersaturated solution; AgCl will precipitateout until the product of the ionic concentrationsis equal to 1.6 10 10MOLAR SOLUBILITY AND SOLUBILITYThere are two other ways to express a substances solubility: molar solubility, whichis the number of moles of solute in one liter of a saturated solution (mol/L), and sol-BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website16.6FIGURE 16.7 Sequence ofsteps (a) for calculating Ksp fromsolubility data and (b) for calculating solubility from Ksp data.Molarsolubility ofcompoundSolubility ofcompoundSOLUBILITY EQUILIBRIA667Concentrationsof cationsand anionsKsp ofcompoundMolarsolubility ofcompoundSolubility ofcompound(a)Concentrationsof cationsand anionsKsp ofcompound(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 atsome given temperature (usually 25C).Both molar solubility and solubility are convenient to use in the laboratory. Wecan use them to determine Ksp by following the steps outlined in Figure 16.7(a).Example 16.8 illustrates this procedure.EXAMPLE 16.8The solubility of calcium sulfate is found experimentally to be 0.67 g/L. Calculatethe value of Ksp for calcium sulfate.AnswerFirst we calculate the number of moles of CaSO4 dissolved in 1 L of so-lution:0.67 g CaSO41 L soln1 mol CaSO4136.2 g CaSO44.9103mol/LThe solubility equilibriumCalcium sulfate is used as a drying agent and in the manufactureof paints, ceramics, and paper. Ahydrated form of calcium sulfate,called plaster of Paris, is used tomake casts for broken bones.SO2 (aq)4CaSO4(s) 34 Ca2 (aq)shows that for every mole of CaSO4 that dissolves, 1 mole of Ca2 and 1 mole ofSO2 are produced. Thus, at equilibrium4[Ca2 ]4.9103M[SO2 ]4and4.9103MNow we can calculate Ksp:Ksp[Ca2 ][SO2 ]4(4.92.4Similar problem: 16.44.10 3)(4.91010 3)5PRACTICE EXERCISEThe solubility of lead chromate (PbCrO4) is 4.5ity product of this compound.105g/L. Calculate the solubil-Sometimes we are given the value of Ksp for a compound and asked to calculatethe compounds molar solubility. For example, the Ksp of silver bromide (AgBr) is7.7 10 13. We can calculate its molar solubility by the same procedure as outlinedon 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) ofBackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website668ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIAAgBr. 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)0.00sInitial (M ):Change (M ):Equilibrium (M ):Br (aq)0.00sssFrom Table 16.2 we writeKspSilver bromide is used in photographic emulsions.7.710[Ag ][Br ]13(s)(s)7.7s13108.8710MTherefore, at equilibrium[Ag ]8.8[Br ]8.8107M107MThus the molar solubility of AgBr also is 8.8 10 7 M.The following example makes use of this approach.EXAMPLE 16.9Using the data in Table 16.2, calculate the solubility of copper(II) hydroxide,Cu(OH)2, in g/L.AnswerStep 1: When Cu(OH)2 dissociates, the species in solution are Cu2and OH ions.Step 2: Let s be the molar solubility of Cu(OH)2. Since one unit of Cu(OH)2 yieldsone Cu2 ion and two OH ions, at equilibrium [Cu2 ] is s and [OH ] is2s. We summarize the changes in concentrations as follows:Cu(OH)2(s) 34 Cu2 (aq)0.00sInitial (M ):Change (M ):Copper(II) hydroxide is used as apesticide and to treat seeds.Equilibrium (M ):Step 3:sKsp2.210202s[Cu2 ][OH ]2(s)(2s)22.210420s3Solving for s, we get2OH (aq)0.002ss1.81075.51021MKnowing that the molar mass of Cu(OH)2 is 97.57 g/mol and knowing itsmolar solubility, we can calculate the solubility in g/L as follows:Similar problem: 16.46.1.810 7 mol Cu(OH)21 L soln1.8solubility of Cu(OH)210597.57 g Cu(OH)21 mol Cu(OH)2g/LPRACTICE EXERCISECalculate the solubility of silver chloride (AgCl) in g/L.BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website16.6TABLE 16.3669SOLUBILITY EQUILIBRIARelationship between Ksp and Molar Solubility (s)EQUILIBRIUMCONCENTRATION (M )COMPOUNDKsp EXPRESSIONCATIONANIONRELATION BETWEEN Ksp AND sAgClBaSO4[Ag ][Cl ][Ba2 ][SO2 ]4ssssKspKspAg2CO3[Ag ]2[CO2 ]32ssKspPbF2[Pb2 ][F ]2s2sKspAl(OH)3[Al3 ][OH ]3s3sKspCa3(PO4)2[Ca2 ]3[PO3 ]243s2sKsps2; ss2; s12(Ksp)(Ksp)Ksp4s3; s4Ksp4s3; s4Ksp27s4; s27Ksp108s5; s1081213131415As the above examples show, solubility and solubility product are related. If weknow one, we can calculate the other, but each quantity provides different information.Table 16.3 shows the relationship between molar solubility and solubility product fora number of ionic compounds.When carrying out solubility and/or solubility product calculations, keep in mindthe following important points:The solubility is the quantity of a substance that dissolves in a certain quantity ofwater. In solubility equilibria calculations, it is usually expressed as grams of soluteper liter of solution. Molar solubility is the number of moles of solute per liter ofsolution. The solubility product is an equilibrium constant. Molar solubility, solubility, and solubility product all refer to a saturated solution.PREDICTING PRECIPITATION REACTIONSFrom a knowledge of the solubility rules (see Section 4.2) and the solubility productslisted in Table 16.2, we can predict whether a precipitate will form when we mix twosolutions or add a soluble compound to a solution. This ability often has 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 aprecipitate). The ability to predict precipitation reactions is also useful in medicine. Forexample, kidney stones, which can be extremely painful, consist largely of calcium oxalate, CaC2O4 (Ksp 2.3 10 9 ). The normal physiological concentration of calciumions in blood plasma is about 5 mM (1 mM 1 10 3 M ). Oxalate ions (C2O2 ),4derived 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 buildup in the kidneys. Proper adjustment of a patients diet can help to reduce precipitateformation. Example 16.10 illustrates the steps involved in precipitation reactions.A kidney stone.EXAMPLE 16.10Exactly 200 mL of 0.0040 M BaCl2 are added to exactly 600 mL of 0.0080 M K2SO4.Will a precipitate form?BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website670ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIAAccording to the solubility rules on p. 113, the only precipitate that mightform is BaSO4:AnswerSO2 (aq) 88n BaSO4(s)4Ba2 (aq)The number of moles of Ba2 present in the original 200 mL of solution is0.0040 mol Ba21 L soln200 mLWe assume that the volumesare additive.1L1000 mL8.0104mol Ba2The total volume after combining the two solutions is 800 mL. The concentrationof Ba2 in the 800 mL volume is[Ba2 ]8.010 4 mol800 mL1.0The number of moles ofSO24Min the original 600 mL solution is0.0080 mol SO241 L soln600 mL3101000 mL1 L soln1L1000 mL4.8103mol SO24The concentration of SO2 in the 800 mL of the combined solution is4[SO2 ]44.810 3 mol800 mL6.01031000 mL1 L solnMNow we must compare Q and Ksp. From Table 16.2, the Ksp for BaSO4 is 1.110 10. As for Q,Q[Ba2 ]0[SO2 ]046.010(1.010 3)(6.010 3)6Therefore,QKspThe 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 ]4Similar problem: 16.49.1.11010PRACTICE EXERCISEIf 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl2, will precipitation occur?16.7SEPARATION OF IONS BY FRACTIONAL PRECIPITATIONIn 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 ofsulfate ions to a solution containing both potassium and barium ions causes BaSO4 toprecipitate out, thereby removing most of the Ba2 ions from the solution. The otherproduct, K2SO4, is soluble and will remain in solution. The BaSO4 precipitate canbe separated from the solution by filtration.BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website16.7COMPOUNDAgClAgBrAgIKsp1.67.78.3101010101317SEPARATION OF IONS BY FRACTIONAL PRECIPITATION671Even 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 convertthem to insoluble silver halides. As the Ksp values in the margin show, the solubilityof the halides decreases from AgCl to AgI. Thus, when a soluble compound such assilver nitrate is slowly added to this solution, AgI begins to precipitate first, followedby 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 typesof ions if precipitates of differing solubility can be formed.EXAMPLE 16.11Silver nitrate is slowly added to a solution that is 0.020 M in Cl ions and 0.020M 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 writeAnswerKsp[Ag ][Br ]Since [Br ] 0.020 M, the concentration of Ag that must be exceeded to initiatethe precipitation of AgBr is[Ag ]Ksp[Br ]3.9AgCl (left) and AgBr (right).Thus [Ag ] 3.9(b) For AgCl10117.71011100.02013MM is required to start the precipitation of AgBr.Ksp[Ag ][Ag ][Cl ]Ksp[Cl ]8.01.6109100.02010M9Similar 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 greaterthan 3.9 10 11 M and lower than 8.0 10 9 M.PRACTICE EXERCISEThe 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 solutioncontaining 0.10 mol Cl and 0.10 mol PO3 , calculate the concentration of Ag4ions (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 welet [Ag ] 8.0 10 9 M. Then[Br ]BackForwardMain MenuTOCStudy Guide TOCKsp[Ag ]Textbook WebsiteMHHE Website672ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA[Br ]7.78.09.61010101395MThe percent of Br remaining in solution (the unprecipitated Br ) at the critical concentration of Ag is% Br[Br ]unpptd[Br ]original100%9.6 10 5 M0.020 M100%0.48% unprecipitatedThus, (100 0.48) percent, or 99.52 percent, of Br will have precipitated just beforeAgCl begins to precipitate. By this procedure, the Br ions can be quantitatively separated from the Cl ions.16.8THE COMMON ION EFFECT AND SOLUBILITYIn 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; precipitationof an ionic compound from solution occurs whenever the ion product exceeds Ksp forthat substance. In a saturated solution of AgCl, for example, the ion product [Ag ][Cl ]is, of course, equal to Ksp. Furthermore, simple stoichiometry tells us that [Ag ][Cl ]. But this equality does not hold in all situations.Suppose we study a solution containing two dissolved substances that share a common ion, say, AgCl and AgNO3. In addition to the dissociation of AgCl, the followingprocess also contributes to the total concentration of the common silver ions in solution:H2OAgNO3(s) 88n Ag (aq)NO3 (aq)If AgNO3 is added to a saturated AgCl solution, the increase in [Ag ] will make theion product greater than the solubility product:QAt a given temperature, only thesolubility of a compound isaltered (decreased) by thecommon ion effect. Its solubilityproduct, which is an equilibriumconstant, remains the samewhether or not other substancesare present in the solution.[Ag ]0[Cl ]0KspTo reestablish equilibrium, some AgCl will precipitate out of the solution, as LeChatelier 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 16.12Calculate the solubility of silver chloride (in g/L) in a 6.5solution.BackForwardMain MenuTOCStudy Guide TOC103M silver nitrateTextbook WebsiteMHHE Website16.9pH AND SOLUBILITY673AnswerStep 1: The relevant species in solution are Agions (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:H2OAgNO3(s) 88nAg (aq)6.5 10 3 MNO3 (aq)6.5 10 3 MLet s be the molar solubility of AgCl in AgNO3 solution. We summarize thechanges in concentrations as follows:AgCl(s) 34Initial (M ):Change (M ):Equilibrium (M ):(6.5Step 3:Ksp1.6Ag (aq)6.5 10 3ss)s[Ag ][Cl ]1010310Cl (aq)0.0s(6.5103s)(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 6.510 3 . Therefore, applying the approximation 6.5 10 3 s 6.5 10 3,we obtain10106.5103ss1.62.5108MStep 4: At equilibrium[Ag ][Cl ](6.52.53101082.510 8) M6.5103MMand so our approximation was justified in step 2. Since all the Cl ions mustcome from AgCl, the amount of AgCl dissolved in AgNO3 solution also is2.5 10 8 M. Then, knowing the molar mass of AgCl (143.4 g), we cancalculate the solubility of the AgCl as follows:2.510 8 mol AgCl1 L soln3.6solubility of AgCl in AgNO3 solution106143.4 g AgCl1 mol AgClg/LThe solubility of AgCl in pure water is 1.9 10 3 g/L (see the PracticeExercise in Example 16.9). Therefore, the answer is reasonable.CommentSimilar problem: 16.55.PRACTICE EXERCISECalculate the solubility in g/L of AgBr in (a) pure water and (b) 0.0010 M NaBr.16.9pH AND SOLUBILITYThe solubilities of many substances also depend on the pH of the solution. Considerthe solubility equilibrium of magnesium hydroxide:BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website674ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIAMg(OH)2(s) 34 Mg2 (aq)This is why milk of magnesia,Mg(OH)2, dissolves in the acidicgastric juice in a persons stomach(see p. 633).2OH (aq)Adding OH ions (increasing the pH) shifts the equilibrium from right to left, therebydecreasing 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 equilibriumfrom left to right, and the solubility of Mg(OH)2 increases. Thus, insoluble bases tendto 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 firstcalculate the pH of a saturated Mg(OH)2 solution. We writeKsp[Mg2 ][OH ]21.21110Let s be the molar solubility of Mg(OH)2. Proceeding as in Example 16.9,Ksp(s)(2s)24s34s31.2101133.01012s1.4104sMAt equilibrium, therefore,[OH ]pOHpH21.410M2.8410 )log (2.814.0043.55104M3.5510.45In 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 tomaintain the equilibrium condition, and more Mg(OH)2 dissolves. The dissolutionprocess 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 isBaF2(s) 34 Ba2 (aq)2F (aq)andKsp[Ba2 ][F ]2In an acidic medium, the high [H ] will shift the following equilibrium from left toright:H (aq)Recall that HF is a weak acid.F (aq) 34 HF(aq)2As [F ] decreases, [Ba ] must increase to maintain the equilibrium condition. Thusmore BaF2 dissolves. The dissolution process and the effect of pH on the solubility ofBaF2 can be summarized as follows:BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website16.9pH AND SOLUBILITYOverall:BaF2(s) 34 Ba2 (aq)2F (aq) 34 2HF(aq)2F (aq)2H (aq)BaF2(s)2H (aq) 34 Ba2 (aq)6752HF(aq)The solubilities of salts containing anions that do not hydrolyze are unaffected bypH. Examples of such anions are Cl , Br , and I .The following examples deal with the effect of pH on solubility.EXAMPLE 16.13Which of the following compounds will be more soluble in acidic solution than inwater: (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:AnswerS2 (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)2Overall: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 isAgCl(s) 34 Ag (aq)Cl (aq)Since Cl is the conjugate base of a strong acid (HCl), the solubility of AgCl is notaffected by an acid solution.(c) PbSO4 will be more soluble in an acidic solution because of the basicity of theSO2 ion. The solubility and acid-base equilibria are summarized below:4SO24PbSO4(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), theabove equilibrium is slightly shifted to the right. Consequently, the solubility ofPbSO4 increases only slightly in an acidic solution.PRACTICE EXERCISEAre the following compounds more soluble in water or in an acidic solution:(a) Ca(OH)2, (b) Mg3(PO4)2, (c) PbBr2?EXAMPLE 16.14Calculate the concentration of aqueous ammonia necessary to initiate the precipitation of iron(II) hydroxide from a 0.0030 M solution of FeCl2.AnswerThe equilibria of interest areNH3(aq)2Fe (aq)BackForwardMain MenuTOCH2O(l ) 34 NH4 (aq)OH (aq)2OH (aq) 34 Fe(OH)2(s)Study Guide TOCTextbook WebsiteMHHE Website676ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIAFirst we find the OH concentration above which Fe(OH)2 begins to precipitate.We write[Fe2 ][OH ]2Ksp1.6Since FeCl2 is a strong electrolyte, [Fe2 ][OH ]2[OH ]14100.0030 M and1.6 100.0030142.36105.31012MNext, we calculate the concentration of NH3 that will supply 2.3 10 6 M OHions. Let x be the initial concentration of NH3 in mol/L. We summarize the changesin concentrations resulting from the ionization of NH3 as follows:NH3(aq)x2.3 10Initial (M ):Change (M ):Equilibrium (M ):(x2.3H2O(l ) 34610 6)NH4 (aq)0.002.3 102.31066OH (aq)0.002.3 102.31066Substituting 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)1.81051.8Kb105Solving for x, we obtainxSimilar problem: 16.64.2.6106MTherefore the concentration of NH3 must be slightly greater than 2.6initiate the precipitation of Fe(OH)2.106M toPRACTICE EXERCISECalculate whether or not a precipitate will form if 2.0 mL of 0.60 M NH3 are addedto 1.0 L of 1.0 10 3 M FeSO4.16.10 COMPLEX ION EQUILIBRIA AND SOLUBILITYLewis acids and bases are discussed in Section 15.12.According to our definition,Co(H2O)2 itself is a complex6ion. When we write Co(H2O)2 ,6we mean the hydrated2Co ion.BackForwardLewis acid-base reactions in which a metal cation combines with a Lewis base resultin 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 ionsare 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 theyhave more than one oxidation state. This property allows them to act effectively asLewis acids in reactions with many molecules or ions that serve as electron donors, oras Lewis bases. For example, a solution of cobalt(II) chloride is pink because of thepresence of the Co(H2O)2 ions (Figure 16.8). When HCl is added, the solution turns6blue as a result of the formation of the complex ion CoCl2 :4Main MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website16.10COMPLEX ION EQUILIBRIA AND SOLUBILITY677FIGURE 16.8 (Left) An aqueous cobalt(II) chloride solution.The pink color is due to the presence of Co(H2O)2 ions. (Right)6After the addition of HCl solution,the solution turns blue because ofthe formation of the complexCoCl 2 ions.44Cl (aq) 34 CoCl2 (aq)4Co2 (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, forexample) are colorless. Adding a few drops of concentrated ammonia solution to aCuSO4 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 blueprecipitate redissolves to produce a beautiful dark-blue solution, this time due to theformation of the complex ion Cu(NH3)2 (Figure 16.9):4Cu(OH)2(s)4NH3(aq) 34 Cu(NH3)2 (aq)42OH (aq)Cu(NH3)24Thus the formation of the complex ionincreases the solubility of Cu(OH)2.A measure of the tendency of a metal ion to form a particular complex ion is givenby 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 as4Cu2 (aq)4NH3(aq) 34 Cu(NH3)2 (aq)4for which the formation constant isKf[Cu(NH3)2 ]4[Cu2 ][NH3]45.01013FIGURE 16.9 (Left) An aqueous solution of copper(II) sulfate.(Center) After the addition of afew drops of a concentratedaqueous ammonia solution, alight-blue precipitate of Cu(OH)2is formed. (Right) When moreconcentrated aqueous ammoniasolution is added, the Cu(OH)2precipitate dissolves to formthe dark-blue complex ionCu(NH3)2 .4BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website678ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIATABLE 16.4Formation Constants of Selected Complex Ions in Water at 25CCOMPLEX IONEQUILIBRIUM EXPRESSIONAg(NH3)2Ag(CN)2Cu(CN)24Cu(NH3)24Cd(CN)24CdI24HgCl24HgI24Hg(CN)24Co(NH3)36Zn(NH3)24AgAgCu2Cu2Cd2Cd2Hg2Hg2Hg2Co3Zn22NH32CN4CN4NH34CN4I4Cl4I4CN6NH34NH3FORMATION CONSTANT (Kf)34 Ag(NH3)234 Ag(CN)234 Cu(CN)2434 Cu(NH3)2434 Cd(CN)2434 CdI2434 HgCl2434 HgI2434 Hg(CN)2434 Co(NH3)3634 Zn(NH3)241.51.01.05.07.12.01.72.02.55.02.910710211025101310161061016103010411031109The very large value of Kf in this case indicates that the complex ion is quite stable insolution and accounts for the very low concentration of copper(II) ions at equilibrium.EXAMPLE 16.15A 0.20-mole quantity of CuSO4 is added to a liter of 1.20 M NH3 solution. What isthe concentration of Cu2 ions at equilibrium?AnswerThe addition of CuSO4 to the NH3 solution results in the reactionCu2 (aq)4NH3(aq) 34 Cu(NH3)2 (aq)4Since Kf is very large (5.0 1013), the reaction lies mostly to the right. As a goodapproximation, we can assume that essentially all the dissolved Cu2 ions end upas Cu(NH3)2 ions. Thus the amount of NH3 consumed in forming the complex4ions is 4 0.20 mol, or 0.80 mol. (Note that 0.20 mol Cu2 is initially present insolution 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 .4Since Cu(NH3)2 does dissociate to a slight extent, we call the concentration of4Cu2 ions at equilibrium x and writeThe fact that these species are inthe same solution allows us to usethe number of moles rather thanmolarity in the formationconstant expression.Kf[Cu(NH3)2 ]4[Cu2 ][NH3]40.20x(0.40)45.010135.01013Solving for x, we obtainx1.61013M[Cu2 ]The small value of [Cu2 ] at equilibrium, compared with 0.20 M, certainly justifies our approximation.CommentSimilar problem: 16.67.PRACTICE EXERCISEIf 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?4The effect of complex ion formation generally is to increase the solubility of asubstance, as the following example shows.BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website16.10679COMPLEX ION EQUILIBRIA AND SOLUBILITYEXAMPLE 16.16Calculate the molar solubility of AgCl in a 1.0 M NH3 solution.AnswerStep 1: Initially, the species in solution are Agand Cl ions and NH3. The reaction between Ag and NH3 produces the complex ion Ag(NH3)2 .Step 2: The equilibrium reactions areAgCl(s) 34 Ag (aq)Ag (aq)Cl (aq)Ksp [Ag ][Cl ]AgCl(s)10102NH3(aq) 34 Ag(NH3)2 (aq)KfOverall:1.6[Ag(NH3)2 ][Ag ][NH3]22NH3(aq) 34 Ag(NH3)2 (aq)1.5107Cl (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]2KspKf(1.610102.4)(1.5107)310Let s be the molar solubility of AgCl (mol/L). We summarize the changes inconcentrations that result from formation of the complex ion as follows:Initial (M ):Change (M ):AgCl(s)2NH3(aq) 34 Ag(NH3)2 (aq)1.00.02ssEquilibrium (M ):(1.02s)sCl (aq)0.0ssThe formation constant for Ag(NH3)2 is quite large, so most of the silverions exist in the complexed form. In the absence of ammonia we have, atequilibrium, [Ag ] [Cl ]. As a result of complex ion formation, however,we can write [Ag(NH3)2 ] [Cl ].Step 3:K2.410(s)(s)(1.0 2s)2s232s)2(1.0Taking the square root of both sides, we obtain0.049ss1.02s0.045 MStep 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, theformation of the complex ion Ag(NH3)2 enhances the solubility of AgCl (Figure16.10).CommentSimilar problem: 16.70.PRACTICE EXERCISECalculate the molar solubility of AgBr in a 1.0 M NH3 solution.BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website680ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIAFIGURE 16.10 (Left to right)Formation of AgCl precipitatewhen AgNO3 solution is addedto NaCl solution. With the addition of NH3 solution, the AgClprecipitate dissolves as the soluble Ag(NH3)2 forms.All amphoteric hydroxides areinsoluble 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 formationof the complex ion Al(OH)4 in which Al(OH)3 acts as the Lewis acid and OH actsas the Lewis base. Other amphoteric hydroxides behave in a similar manner.16.11 APPLICATION OF THE SOLUBILITY PRODUCT PRINCIPLETO QUALITATIVE ANALYSISDo not confuse the groups inTable 16.5, which are based onsolubility products, with thosein the periodic table, whichare based on the electronconfigurations 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 aqueoussolution. 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 fromone to all twenty ions, any analysis must be carried out systematically from group 1through group 5. Let us consider the general procedure for separating these twenty ionsby 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 chlo2rides 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 asM2 (aq)H2S(aq) 34 MS(s)2H (aq)Adding acid to the solution shifts this equilibrium to the left so that only the leastsoluble metal sulfides, that is, those with the smallest Ksp values, will precipitateout of solution. These are Bi2S3, CdS, CuS, and SnS (see Table 16.5). Group 3 cations. At this stage, sodium hydroxide is added to the solution to makeBackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website16.11681APPLICATION OF THE SOLUBILITY PRODUCT PRINCIPLE TO QUALITATIVE ANALYSISTABLE 16.5 Separation of Cations into Groups According to Their PrecipitationReactions with Various ReagentsGROUP12345CATIONAgHg22Pb2Bi3Cd2Cu2Sn2Al3Co2Cr3Fe2Mn2Ni2Zn2Ba2Ca2Sr2KNaNH4PRECIPITATING REAGENTSINSOLUBLE COMPOUNDHClAgClHg2Cl2PbCl2Bi2S3CdSCuSSnSAl(OH)3CoSCr(OH)3FeSMnSNiSZnSBaCO3CaCO3SrCO3NoneNoneNoneAAgH2Sin acidicsolutionsAgH2Sin basicsolutionsAAAAAgNa2CO3AAgNo precipitatingreagentKsp1.63.52.41.68.06.01.01.84.03.06.03.01.43.08.18.71.61010101010101010101010101010101010101847228372633212919142423999it basic. In a basic solution, the above equilibrium shifts to the right. Therefore, themore 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)3and 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 fromsolution, sodium carbonate is added to the basic solution to precipitate Ba2 , Ca2 ,and Sr2 ions as BaCO3, CaCO3, and SrCO3. These precipitates too are removedfrom solution by filtration. Group 5 cations. At this stage, the only cations possibly remaining in solution areNa , K , and NH4 . The presence of NH4 can be determined by adding sodiumhydroxide:NaOH(aq)Because NaOH is added in group3 and Na2CO3 is added in group4, the flame test for Na ionsis carried out using theoriginal 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 contactwith) the solution. To confirm the presence of Na and K ions, we usually use aflame 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. Eachtype 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 thatof Cu2 ions is green (Figure 16.11).Figure 16.12 summarizes this scheme for separating metal ions.BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website682ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIAFIGURE 16.11 (Left to right)Flame colors of lithium, sodium,potassium, and copper.FIGURE 16.12 A flow chartfor the separation of cations inqualitative analysis.Two points regarding qualitative analysis must be mentioned. First, the separationof the cations into groups is made as selective as possible; that is, the anions that areadded 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 thegroup 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 ionsof all cation groups+HClFiltrationGroup 1 precipitatesAgCl, Hg2Cl2, PbCl2Solution containing ionsof remaining groups+H2SFiltrationGroup 2 precipitatesCuS, CdS, SnS, Bi2S3Solution containing ionsof remaining groups+NaOHFiltrationGroup 3 precipitatesCoS, FeS, MnS, NiSZnS, Al(OH)3, Cr(OH)3Solution containing ionsof remaining groups+Na2CO3FiltrationGroup 4 precipitatesBaCO3, CaCO3, SrCO3Solution containsNa+, K+, NH + ions4BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website16.11APPLICATION OF THE SOLUBILITY PRODUCT PRINCIPLE TO QUALITATIVE ANALYSIS683Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action ChemistryChemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action ChemistryHow an Eggshell Is FormedThe formation of the shell of a hens egg is a fascinating example of a natural precipitation process.An average eggshell weighs about 5 grams andis 40 percent calcium. Most of the calcium in aneggshell is laid down within a 16 hour period. Thismeans that it is deposited at a rate of about 125 milligrams per hour. No hen can consume calcium fastenough to meet this demand. Instead, it is supplied byspecial bony masses in the hens long bones, whichaccumulate large reserves of calcium for eggshell formation. [The inorganic calcium component of the boneis calcium phosphate, Ca3(PO4)2, an insoluble compound.] If a hen is fed a low-calcium diet, her eggshellsbecome progressively thinner; she might have to mobilize 10 percent of the total amount of calcium in herbones just to lay one egg! When the food supply isconsistently low in calcium, egg production eventuallystops.The eggshell is largely composed of calcite, acrystalline form of calcium carbonate (CaCO3).Normally, the raw materials, Ca2 and CO 2 , are3carried by the blood to the shell gland. The calcification process is a precipitation reaction:Ca2 (aq)CO2 (aq) 34 CaCO3(s)3In the blood, free Ca2 ions are in equilibrium withcalcium ions bound to proteins. As the free ions aretaken up by the shell gland, more are provided by thedissociation of the protein-bound calcium.The carbonate ions necessary for eggshell formation are a metabolic byproduct. Carbon dioxideproduced during metabolism is converted to carbonicacid (H2CO3) by the enzyme carbonic anhydrase(CA):KEY EQUATIONSUMMARY OF FACTSAND CONCEPTSBackForwardMain Menu pHpKalogChicken eggs.X-ray micrograph of an eggshell,showing columns of calcite.CO2(g)CAH2O(l ) 34 H2CO3(aq)Carbonic acid ionizes stepwise to produce carbonateions:H2CO3(aq) 34 H (aq)HCO3 (aq)HCO3 (aq) 34 H (aq)CO2 (aq)3Chickens do not perspire and so must pant to coolthemselves. Panting expels more CO2 from thechickens body than normal respiration does.According to Le Chateliers principle, panting will shiftthe CO2H2CO3 equilibrium shown above from rightto left, thereby lowering the concentration of the CO23ions in solution and resulting in thin eggshells. Oneremedy for this problem is to give chickens carbonated water to drink in hot weather. The CO2 dissolvedin the water adds CO2 to the chickens body fluidsand 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 solutionreacts 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.TOCStudy Guide TOCTextbook WebsiteMHHE Website684ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA3. The pH at the equivalence point of an acid-base titration depends on hydrolysis of the saltformed in the neutralization reaction. For strong acidstrong base titrations, the pH at theequivalence point is 7; for weak acidstrong base titrations, the pH at the equivalence pointis greater than 7; for strong acidweak base titrations, the pH at the equivalence point isless than 7.4. Acid-base indicators are weak organic acids or bases that change color at the equivalencepoint 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 hydrogenion concentration increases. The solubility of salts with anions derived from strong acids isunaffected by pH.8. Complex ions are formed in solution by the combination of a metal cation with a Lewisbase. The formation constant Kf measures the tendency toward the formation of a specificcomplex 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 WORDSBuffer solution, p. 649Common ion effect, p. 646Complex ion, p. 676End point, p. 661Molar solubility, p. 666Formation constant (Kf), p. 677 Qualitative analysis, p. 680Solubility, p. 666Solubility product (Ksp), p. 665QUESTIONS AND PROBLEMSTHE COMMON ION EFFECTReview Questions16.1 Use Le Chatelier s principle to explain how thecommon ion effect affects the pH of a solution.16.2 Describe the effect on pH (increase, decrease, or nochange) 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.Problems16.3 Determine the pH of (a) a 0.40 M CH3COOH solution, (b) a solution that is 0.40 M CH3COOH and0.20 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 MNH4Cl.BUFFER SOLUTIONSReview Questions16.5 What is a buffer solution? What constitutes a buffersolution?Back16.6 Define pKa for a weak acid. What is the relationship between the value of the pKa and the strengthof the acid? Do the same for a weak base.16.7 The pKas of two monoprotic acids HA and HB are5.9 and 8.1, respectively. Which of the two is thestronger 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/HCOOHProblems16.9 Calculate the pH of the buffer system made up of0.15 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. Whichis 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 carbonicacid (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.ForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE WebsiteQUESTIONS AND PROBLEMS16.13 The pH of a sodium acetate acetic acid buffer is4.50. Calculate the ratio [CH3COO ]/[CH3COOH].16.14 The pH of blood plasma is 7.40. Assuming the principal buffer system is HCO3 /H2CO3, calculate theratio [HCO3 ]/[H2CO3]. Is this buffer more effectiveagainst an added acid or an added base?16.15 Calculate the pH of the 0.20 M NH3/0.20 M NH4Clbuffer. What is the pH of the buffer after the addition of 10.0 mL of 0.10 M HCl to 65.0 mL of thebuffer?16.16 Calculate the pH of 1.00 L of the buffer 1.00 MCH3COONa/1.00 M CH3COOH before and after theaddition of (a) 0.080 mol NaOH, (b) 0.12 mol HCl.(Assume that there is no change in volume.)16.17 A diprotic acid, H2A, has the following ionizationconstants: 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/H2Aor Na2A/NaHA.16.18 A student is asked to prepare a buffer solution atpH 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 TITRATIONSReview QuestionsProblems16.21 A 0.2688-g sample of a monoprotic acid neutralizes16.4 mL of 0.08133 M KOH solution. Calculate themolar mass of the acid.16.22 A 5.00-g quantity of a diprotic acid was dissolvedin water and made up to exactly 250 mL. Calculatethe 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 weretitrated.16.23 In a titration experiment, 12.5 mL of 0.500 M H2SO4neutralize 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 HCOOHneutralize 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 acidwas dissolved in 25.0 mL of water and titrated withForwardMain Menu0.0633 M NaOH solution. The volume of base required to bring the solution to the equivalence pointwas 18.4 mL. (a) Calculate the molar mass of theacid. (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 of0.167 M NaOH with exactly 500 mL of 0.100 MCH3COOH. Calculate the equilibrium concentrations of H , CH3COOH, CH3COO , OH , andNa .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 MNaOH.ACID-BASE INDICATORSReview Questions16.29 Explain how an acid-base indicator works in a titration. What are the criteria for choosing an indicatorfor a particular acid-base titration?16.30 The amount of indicator used in an acid-base titration must be small. Why?Problems16.19 Briefly describe what happens in an acid-base titration.16.20 Sketch titration curves for the following acid-basetitrations: (a) HCl versus NaOH, (b) HCl versusCH3NH2, (c) CH3COOH versus NaOH. In eachcase, the base is added to the acid in an Erlenmeyerflask. Your graphs should show pH on the y-axis andvolume of base added on the x-axis.Back685TOC16.31 Referring to Table 16.1, specify which indicator orindicators 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 byadding NaOH solution from a buret to anErlenmeyer flask containing HCl solution and using phenolphthalein as indicator. At the equivalencepoint, 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 is1.0 10 6. The color of the nonionized form is redand that of the ionized form is yellow. What is thecolor of this indicator in a solution whose pH is4.00?16.34 The Ka of a certain indicator is 2.0 10 6. Thecolor of HIn is green and that of In is red. A fewdrops of the indicator are added to a HCl solution,which is then titrated against a NaOH solution. Atwhat pH will the indicator change color?SOLUBILITY EQUILIBRIAReview Questions16.35 Use BaSO4 to distinguish between solubility, molarStudy Guide TOCTextbook WebsiteMHHE Website686ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIAsolubility, 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 ioniccompound AxBy.16.39 How can we predict whether a precipitate will formwhen two solutions are mixed?16.40 Silver chloride has a larger Ksp than silver carbonate (see Table 16.2). Does this mean that AgCl alsohas a larger molar solubility than Ag2CO3?Problems16.41 Calculate the concentration of ions in the followingsaturated 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 (molarmass 346 g) is 4.63 10 3 g/L. What is Ksp forthe compound?16.45 The solubility of an ionic compound M2X3 (molarmass 288 g) is 3.6 10 17 g/L. What is Ksp forthe 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 50.0mL of 0.10 M Na2CO3, will BaCO3 precipitate?16.50 A volume of 75 mL of 0.060 M NaF is mixed with25 mL of 0.15 M Sr(NO3)2. Calculate the concentrations in the final solution of NO3 , Na , Sr2 , andF . (Ksp for SrF2 2.0 10 10.)FRACTIONAL PRECIPITATIONProblems16.51 Solid NaI is slowly added to a solution that is0.010 M in Cu and 0.010 M in Ag . (a) Whichcompound will begin to precipitate first? (b) Calculate [Ag ] when CuI just begins to precipitate. (c)What percent of Ag remains in solution at thispoint?BackForwardMain MenuTOC16.52 Find the approximate pH range suitable for theseparation of Fe3 and Zn2 by precipitation ofFe(OH)3 from a solution that is initially 0.010 M inboth Fe3 and Zn2 .THE COMMON ION EFFECT AND SOLUBILITYReview Questions16.53 How does the common ion effect influence solubility equilibria? Use Le Chatelier s principle to explain the decrease in solubility of CaCO3 in aNa2CO3 solution.16.54 The molar solubility of AgCl in 6.5 10 3 MAgNO3 is 2.5 10 8 M. In deriving Ksp from thesedata, 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 AgNO3as in pure water.(c) Solubility of AgCl is independent of the concentration of AgNO3.(d) [Ag ] in solution does not change significantlyupon the addition of AgCl to 6.5 10 3 MAgNO3.(e) [Ag ] in solution after the addition of AgCl to6.5 10 3 M AgNO3 is the same as it wouldbe in pure water.Problems16.55 How many grams of CaCO3 will dissolve in 3.0102 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)2solution.16.57 Calculate the molar solubility of AgCl in a solutionmade by dissolving 10.0 g of CaCl2 in 1.00 L ofsolution.16.58 Calculate the molar solubility of BaSO4 (a) in water, (b) in a solution containing 1.0 M SO2 ions.4pH AND SOLUBILITYProblems16.59 Which of the following ionic compounds will bemore soluble in acid solution than in water?(a) BaSO4, (b) PbCl2, (c) Fe(OH)3, (d) CaCO316.60 Which of the following will be more soluble in acidsolution than in pure water? (a) CuI, (b) Ag2SO4,(c) Zn(OH)2, (d) BaC2O4, (e) Ca3(PO4)216.61 Compare the molar solubility of Mg(OH)2 in waterand in a solution buffered at a pH of 9.0.16.62 Calculate the molar solubility of Fe(OH)2 at (a) pH8.00, (b) pH 10.00.16.63 The solubility product of Mg(OH)2 is 1.2 10 11.Study Guide TOCTextbook WebsiteMHHE WebsiteQUESTIONS AND PROBLEMSWhat minimum OH concentration must be attained(for example, by adding NaOH) to decrease theMg2 concentration in a solution of Mg(NO3)2 toless than 1.0 10 10 M?16.64 Calculate whether or not a precipitate will form if2.00 mL of 0.60 M NH3 are added to 1.0 L of 1.010 3 M FeSO4.687Pb2 remaining in solution.16.77 Both KCl and NH4Cl are white solids. Suggest onereagent that would allow you to distinguish betweenthese two compounds.16.78 Describe a simple test that would allow you to distinguish between AgNO3(s) and Cu(NO3)2(s).ADDITIONAL PROBLEMSCOMPLEX ION EQUILIBRIA AND SOLUBILITYReview Questions16.65 Explain the formation of complexes in Table 16.3in terms of Lewis acid-base theory.16.66 Give an example to illustrate the general effect ofcomplex ion formation on solubility.Problems16.67 If 2.50 g of CuSO4 are dissolved in 9.0 102 mLof 0.30 M NH3, what are the concentrations of Cu2 ,Cu(NH3)2 , and NH3 at equilibrium?416.68 Calculate the concentrations of Cd2 , Cd(CN)2 ,4and CN at equilibrium when 0.50 g of Cd(NO3)2dissolves in 5.0 102 mL of 0.50 M NaCN.16.69 If NaOH is added to 0.010 M Al3 , which will bethe predominant species at equilibrium: Al(OH)3 orAl(OH)4 ? The pH of the solution is 14.00. [Kf for2.0 1033.]Al(OH)416.70 Calculate the molar solubility of AgI in a 1.0 M NH3solution.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, andAgOH 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 KClsolution.QUALITATIVE ANALYSISReview Questions16.73 Outline the general procedure of qualitative analysis.16.74 Give two examples of metal ions in each group (1through 5) in the qualitative analysis scheme.Problems16.75 In a group 1 analysis, a student obtained a precipitate containing both AgCl and PbCl2. Suggest onereagent that would allow her to separate AgCl(s)from PbCl2(s).16.76 In a group 1 analysis, a student adds HCl acid to theunknown solution to make [Cl ] 0.15 M. SomePbCl2 precipitates. Calculate the concentration ofBackForwardMain MenuTOC16.79 The buffer range is defined by the equation pHpKa 1. Calculate the range of the ratio [conjugatebase]/[acid] that corresponds to this equation.16.80 The pKa of the indicator methyl orange is 3.46. Overwhat pH range does this indicator change from 90percent HIn to 90 percent In ?16.81 Sketch the titration curve of a weak acid versus astrong base like the one shown in Figure 16.4. Onyour graph indicate the volume of base used at theequivalence point and also at the half-equivalencepoint, that is, the point at which half of the acid hasbeen neutralized. Show how you can measure thepH 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 to400 mL of a 2.00 M HNO2 solution. The pH of themixed solution was 1.50 units greater than that ofthe original acid solution. Calculate the molarity ofthe NaOH solution.16.83 The pKa of butyric acid (HBut) is 4.7. Calculate Kbfor the butyrate ion (But ).16.84 A solution is made by mixing exactly 500 mL of0.167 M NaOH with exactly 500 mL 0.100 MCH3COOH. Calculate the equilibrium concentrations of H , CH3COOH, CH3COO , OH , andNa .16.85 Cd(OH)2 is an insoluble compound. It dissolves inexcess NaOH in solution. Write a balanced ionicequation for this reaction. What type of reaction isthis?16.86 A student mixes 50.0 mL of 1.00 M Ba(OH)2 with86.4 mL of 0.494 M H2SO4. Calculate the mass ofBaSO4 formed and the pH of the mixed solution.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) 34Ca2 (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 tobe completely filled with distilled water to removeall of the deposit at 25C?16.89 Equal volumes of 0.12 M AgNO3 and 0.14 M ZnCl2Study Guide TOCTextbook WebsiteMHHE Website688ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA16.9016.9116.9216.9316.9916.10016.101Mass of HgI2 formed16.94solution 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 ofFe(OH)3 from a solution that is initially 0.010 M inFe3 and Zn2 .A volume of 25.0 mL of 0.100 M HCl is titratedagainst a 0.100 M CH3NH2 solution added to it froma 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 beenadded, (c) after 35.0 mL of CH3NH2 solution havebeen added.The molar solubility of Pb(IO3)2 in a 0.10 M NaIO3solution is 2.4 10 11 mol/L. What is Ksp forPb(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 andobtained the following graph. Explain the appearance of the graph.Volume of KI added16.10216.95 Barium is a toxic substance that can seriously impair heart function. For an X ray of the gastrointestinal tract, a patient drinks an aqueous suspensionof 20 g BaSO4. If this substance were to equilibratewith the 5.0 L of the blood in the patients body,what would be [Ba2 ]? For a good estimate, we mayassume that the temperature is at 25C. Why isBa(NO3)2 not chosen for this procedure?16.96 The pKa of phenolphthalein is 9.10. Over what pHrange does this indicator change from 95 percentHIn to 95 percent In ?16.97 Solid NaI is slowly added to a solution that is0.010 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) Whatpercent 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 mLof a 0.10 M solution of the acid. (b) Calculate thepH of 25.0 mL of 0.15 M (CH3)2AsO2Na. (c) MixBackForwardMain MenuTOC16.10316.104the solutions in part (a) and part (b). Calculate thepH of the resulting solution.Radiochemical techniques are useful in estimatingthe solubility product of many compounds. In oneexperiment, 50.0 mL of a 0.010 M AgNO3 solutioncontaining a silver isotope with a radioactivity of74,025 counts per min per mL were mixed with100 mL of a 0.030 M NaIO3 solution. The mixedsolution was diluted to 500 mL and filtered to remove all of the AgIO3 precipitate. The remainingsolution was found to have a radioactivity of 44.4counts 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 acidto react with all the carbonate and then back-titrating the remaining acid with NaOH. (a) Write anequation for these reactions. (b) In a certain experiment, 20.00 mL of 0.0800 M HCl were added to a0.1022-g sample of MCO3. The excess HCl required5.64 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) Astrong acid reacting with a strong base. (b) A strongacid reacting with a weak base (NH3). (c) A weakacid (CH3COOH) reacting with a strong base. (d) Aweak acid (CH3COOH) reacting with a weak base(NH3). (Hint: Strong acids exist as H ions andstrong bases exist as OH ions in solution. You needto 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 solutionrequires 15.9 mL of 0.500 M NaOH solution for neutralization.Describe how you would prepare a 1-L 0.20 MCH3COONa/0.20 M CH3COOH buffer system by(a) mixing a solution of CH3COOH with a solutionof CH3COONa, (b) reacting a solution ofCH3COOH with a solution of NaOH, and (c) reacting a solution of CH3COONa with a solution of HCl.Phenolphthalein is the common indicator for thetitration of a strong acid with a strong base. (a) Ifthe 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 usedin a titration involving a 50.0-mL volume, what isthe concentration of the ionized form at pH 8.00?(Assume that 1 drop 0.050 mL.)Study Guide TOCTextbook WebsiteMHHE WebsiteQUESTIONS AND PROBLEMS16.105 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 thefollowing pairs of ions in solution? (a) Na andBa2 , (b) K and Pb2 , (c) Zn2 and Hg2 .16.107 Look up the Ksp values for BaSO4 and SrSO4 inTable 16.2. Calculate the concentrations of Ba2 ,Sr2 , and SO2 in a solution that is saturated with4both compounds.16.108 In principle, amphoteric oxides, such as Al2O3 andBeO can be used to prepare buffer solutions becausethey possess both acidic and basic properties (seeSection 15.11). Explain why these compounds areof little practical use as buffer components.16.109 CaSO4 (Ksp 2.4 10 5) has a larger Ksp valuethan that of Ag2SO4 (Ksp 1.4 10 5). Does itfollow 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: Teacontains polyphenols which are weak acids andlemon juice contains citric acid.)16.111 How many milliliters of 1.0 M NaOH must be addedto a 200 mL of 0.10 M NaH2PO4 to make a buffersolution with a pH of 7.50?16.112 The maximum allowable concentration of Pb2 ionsin drinking water is 0.05 ppm (that is, 0.05 g of Pb2in one million grams of water). Is this guideline exceeded if an underground water supply is at equilibrium with the mineral anglesite, PbSO4 (Ksp1.6 10 8)?16.113 One of the most common antibiotics is penicillin G(benzylpenicillinic acid), which has the followingstructure:OBCOOHHH3CH3CGCDCSONOC HAAC OCO NOCO CH2BAOHHIt is a weak monoprotic acid:PKa1.64103where HP denotes the parent acid and P the conjugate base. Penicillin G is produced by growingmolds in fermentation tanks at 25C and a pH rangeof 4.5 to 5.0. The crude form of this antibiotic is obtained by extracting the fermentation broth with anBackForwardMain MenuTOCorganic solvent in which the acid is soluble. (a)Identify the acidic hydrogen atom. (b) In one stageof purification, the organic extract of the crude penicillin G is treated with a buffer solution at pH6.50. What is the ratio of the conjugate base of penicillin G to the acid at this pH? Would you expectthe conjugate base to be more soluble in water thanthe acid? (c) Penicillin G is not suitable for oral administration, but the sodium salt (NaP) is because itis 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 reactswith 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 curvesfor CH3COOH and its conjugate base CH3COO insolution. Your graph should show fraction as the yaxis and pH as the x axis. What are the fractions andpH at the point where these two curves intersect?16.116 Water containing Ca2 and Mg2 ions is calledhard water and is unsuitable for some householdand industrial use because these ions react with soapto form insoluble salts, or curds. One way to remove the Ca2 ions from hard water is by addingwashing soda (Na2CO3 10H2O). (a) The molarsolubility of CaCO3 is 9.3 10 5 M. What is itsmolar 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 byadding 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) andJHP 34 H689(c). 16.3 9.17; 9.20. 16.4 Weigh out Na2CO3 and NaHCO3 inmole ratio of 0.60 to 1.0. Dissolve in enough water to make upa 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, methylred, 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 inacid solution, (b) more soluble in acid solution, (c) about thesame. 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.416.16 3.5 10 3 mol/L.Study Guide TOCTextbook WebsiteMHHE WebsiteCHEMICALMYSTERYA Hard-boiled SnackMost 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 ofboiling 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 wewill 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 fromthe 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 50percent water, 34 percent fat, 16 percent protein, and a small amount of iron in theform of Fe2 ions.Proteins are polymers made up of amino acids. In solution, each long chain of aprotein molecule folds in such a way that the hydrophobic parts of the molecule areburied 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. Chemicalssuch 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 forma semirigid opaque white solid. Heating also decomposes some proteins so that the sulfur in them combines with hydrogen to form hydrogen sulfide (H2S), an unpleasantsmelling 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 boiledfor about 12 minutes and one that has been overcooked. Note that the outside of theovercooked yolk is green.What is the chemical basis for the changes brought about by boiling an egg?ShellSchematic diagram of an egg.The chalazae are the cords thatanchor the yolk to the shell andkeep it centered.MembraneAlbumenYolk membraneYolkChalazaAir space690BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE WebsiteCHEMICAL CLUESOne frequently encountered problem with hard-boiled eggs is that their shells crackin water. The recommended procedure for hard boiling eggs is to place the eggs incold water and then bring the water to a boil. What causes the shells to crack inthis case? How does pin holing, that is, piercing the shell with a needle, preventthe shells from cracking? A less satisfactory way of hard boiling eggs is to placeroom-temperature eggs or cold eggs from the refrigerator in boiling water. Whatadditional mechanism might cause the shells to crack?2. When an eggshell cracks during cooking, some of the egg white leaks into the hotwater to form unsightly streamers. An experienced cook adds salt or vinegar tothe water prior to heating eggs to minimize the formation of streamers. Explain thechemical basis for this action.3. Identify the green substance on the outer layer of the yolk of an overcooked eggand write an equation representing its formation. The unsightly green yolk canbe 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 actionremove the green substance?4. The way to distinguish a raw egg from a hard-boiled egg, without cracking theshells, is to spin the eggs. How does this method work?1.A 12-minute egg (left) and anovercooked hard-boiled egg(right).Iron(II) sulfide.691BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website...
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