Figure 17.8 708 We prepare a saturated solution by adding AgCl to water and stirring. Neither Agel nor a saturated solution of AgCl is purple. The color has been used in this illustration to clarify the process. In the resulting saturated solution, the concentrations of Ag + and Cl - are equal, and the product of their concentrations is equal to K sp [Ag+ ][Cl-] = 1.6 X 10- 10 Therefore, the concentrations are [Ag+] = l.3 X 10- 5 M and [Cl- ] = l.3 X 10- 5 M 2S.0C / After filtering off the solid AgCI, we dissolve enough NaCI to make the concentration of CI- = 1.0 M. Because the concentration of Cl - is now larger, the product of Ag+ and Cl- concentrations is no longer equal to K sp [Ag+][CI- ] = (1.3 X 10 - 5 M)(1.0 M) > 1.6 X 10 - 10 In any solution saturated with AgCl at 25 ° C, the product of [Ag+] and [Cl-] must equal the Ksp of AgCl. Therefore, AgCl will precipitate until the product of ion concentrations is again 1.6 X 10- 10 . Note that this causes nearly all the di ssolved AgCl - to precipitate. With a Cl- concentration of 1.0 M, the highest possible concentration of Ag+ is 1.6 X 10- 10 M. i [Ag+](1.0 M) = 1.6 X 10 - 10 therefore, [Ag+] = 1.6 X 10 - 10 M The amount of AgCI precipitated is exaggerated for emphasis. The actual amount of AgCI would be extremely small. What's the point? When two salts contain the same ion, the ion they both contain is called the "common ion." The solubility of a slightly soluble salt such as AgCl can be decreased by the addition of a soluble salt with a common ion. In this example, AgCI is precipitated by adding NaCl. AgCI could also be precipitated by adding a soluble salt containing the Ag + ion, such as AgN0 3 . 7 09 710 CHAPTER 17 Acid-Base Equilibria and Solubility Equilibria Think About It When a salt dissociates to give the conjugate ba se of a weak acid, H+ ions in an acidic solution consume a product (base) of the dissolution. This drives the equilibrium to the right (more solid dissolves) according to Le Chatelier's principle. Lewis acid-base reactions in which a metal cation combines with a Lewis base res ult in the format io n of comp l ex io ns. Sample Problem 17.11 demonstrates the effect of pH on solubility. . Sample Problem 17.11 Which of the following compounds will be more soluble in acidic solution than in water: (a) C uS , (b) AgC l, (c) PbS0 4 ? Strategy For each salt, write the dissociation equilibrium equation and determine whether it produces an anion that will react with H+ Only an anion that is the conju ga te ba se of a weak acid will react with H+. . Setup (a) CuS(s) , • Cu 2 +(aq) + S 2-( aq) S2 - is the conjugate ba se of the weak acid HS S2- reacts with H+ as follows: ( b)AgCl(s), • Ag +(aq) + C qaq) Cl- is the conjugate base of the strong acid HCI. Cl- does n ot rea ct with H+. (c) PbS0 4 (s) , • Pb 2+( aq) + SO ~ - ( aq ) SO ~- is the conjugate ba se of the weak acid HS0 4 . It reacts with H+ as follows: SO ~-( aq ) + H+(aq) -_. HS0 4 (aq) A salt that produces an anion that reacts with H+ will be more soluble in acid than in water. Solution CuS and PbS0 4 are more soluble in acid than in wate r. (AgCl is no more or less soluble in acid than in water.) Practice Problem A Determine if the following compounds are more soluble in acidic solution than in pure water: (a) Ca ( OH )2, (b) Mg 3 (P0 4 )2 , (c) PbBr 2' Practice Problem B Other than those in Sample Problem 17.11 and th ose in Practice Problem A, list three salts that are more soluble in acidic solution than in pure water. Complex Ion Formation A complex ion is an ion containing a central metal cation bonded to one or more molecules or • • • • • • • • • • • • • • ions. Complex ions are crucial to many chemi ca l and biological processes. Here we will consider the effect of complex ion formation on solubilit y. In Chapter 22 we will discuss the chemistry of complex ions in more detail. Transition metals have a particular tendency to form complex ions. For example, a solution of cobalt(II) chloride (CoCI 2 ) is pink because of the presence of the CO(H20)~ + ions (Figure 17.9 ). When HCI is added, the solution turns blue because the complex ion CoCl~ - forms: Copper(II) sulfate (CUS04) dissolves in water to produce a blue solutio n. The hydrated copper(II) ions are responsible for this color; many other sulfates (e.g., Na2S04) are colorless. Adding a few drops of concentrated ammonia solution to a CUS04 solution causes the formation of a light-blue precipitate, copper(II) hydroxide: Cu ?+ (aq) + 20H - (aq) • Cu(OHl2(s) The OH- ions are supplied by the ammonia solution. If more NH3 is added, the blue precipitate redissolves to produce a beautiful dark-blue solution, this time due to the formation of the com- plex ion CU(NH 3 )~ + (Figure 17.10): Thus, the formation of the complex ion Cu(NH3)~ + increases the solubility of Cu(OHh. A measure of the tendency of a metal ion to form a particular complex ion is given by the formation constant ( Kr ) (also called the stability constant), which is the equilibrium constant for the complex ion formation. The larger K f i s, the more stable the complex ion i s. Table 17.5 lists the formation constants of a number of complex ion s. Complex Ion Ag(NH 3 ); Ag(CNh Cu(CN) ~  Cu(NH 3)~ + Cd(CN)~  CdI~- HgCl ~ - HgI~  Hg(CN) ~  Co(NH 3)~+ Zn(NH 3)~+ Equilibrium Expression Ag + + 2NH 3 • • Ag(NH 3 ); Ag + + 2CN • Ag(CN)? Cu 2 + + 4CN - +.=~' Cu(CN) ~- Cu2+ + 4NH3 • • Cu(NH 3 )~ + Cd 2 + + 4CN- . · Cd(CN)~ - Cd 2 + + 41 • Cdl~- Hg2 + + 4Cl- . . HgCl ~ - Hg 2+ + 41 . HgI ~ - Hg 2+ + 4CN- . • Hg(CN)~ - Co 3+ + 6NH 3 +.=~' Co(NH3)~ + Zn 2+ + 4NH 3 • • Zn(NH 3)~ + , \ Formation Constant (K f ) 1.5 x 10 7 l.0 X 10 21 l.0 X 10 25 5.0 X 1013 7.1 X 10 16 2.0 X 10 6 l.7 X 10 16 2.0 X 10 30 2.5 X 10 41 5.0 X 10 31 2.9 X 10 9 Figure 17.9 (Left)Anaque ous cobalt(II) chloride solution. The pink color is due to the presence of CO(H 2 0) ~ + ions. (Right) After the addition of HCI solution, the solution turns blue because of the fonnation of the complex CoCl~ - ions. Fig u re 17. 10 (Left) An aqueous solution of copper(II) sulfate. (Center) After the addition of a few drops of concentrated aqueous ammonia solution, a light-blue precipitate of CU(OH)2 is fonned. (Right) When more concentrated aqueous ammonia solution is added, the Cu(OH)z is fonned. (Right) When more concentrated aqueous ammonia solution is added, the Cu(OHh precipitate dissolves to form the dar k- blue complex ion CU ( NH 3 )~+ ' • 7i ' 712 CHAPTER 17 Acid-Base Equilibria and Solubility Equilibria • Formati on of a complex ion consumes the metal ion produced by the di ssociation of a salt , increasing the salt's solubility si mpl y d ue to Le Chiitel ier 's principle [~ Section 15 . 5]. • - ! Multimedia Solutions Precipitation- fractional crystallization. The fOllnation of the CU(NH 3)~ + ion can be expressed as The corresponding fOlrnation constant is The large value of K f in this case indicates that the complex ion is very stable in solution and accounts for the very low concentration of copper(II) ions at equilibrium . Recall that K for the s um of two reactions is the product of the individual K values [ ~~ Section 15 .3] . The dissolution of silver chloride is represented by the equation The s um of this equation and the one representing the fOlmation of Ag(NH 3 )! is AgCl(s). . ~) + Cl- (aq) K sp = 1.6 X 10- 10 + aq) + 2NH 3( aq). • Ag(NH 3 )!( aq) K f = 1.5 X 10 7 AgCl(s) + 2NH 3 ( aq). • Ag(NH 3 )! (aq) + Cl - (aq) and the corresponding equilibrium constant is (1.6 X 10- 1 °)(1.5 X 10 7 ) = 2.4 X 10- 3 . This is significantly larger than the K sp value, indicating that much more AgCl will dissolve in the pres- . erlce ' oLiqiieo ui ammo ni a ' thin ' iii pure ' water. Although the actual calculation of solubility in these cases is somewhat complicated, the effect of complex ion formation generally is to increase the solubility of a substance. Finally, there is a class of hydroxide s, called amphoteric hydroxides, which can react with both acids and ba ses. Examples are Al(OH) 3, Pb(OHh, Cr(OH)3, Zn(OH)1, and Cd(OH)2' Al(OH)3 reacts with acids and bases as follows: Al ( OH Ms) + 3H+(aq) - -+. Al 3+ (aq) + 3H 2 0(l) Al(OHMs) + OH-(aq ) . • Al(OH)4 (aq) The increase in solubility of Al(OH)3 in a basic medium is the result of the formation of the com- plex ion Al(OH )4 in which Al(OH)3 acts as the Lewis acid and OH- acts as the Lewis base. Other amphoteric hydroxides react similarly with acids and bases. Checkpoint 17.5 Factors Affecting Soubility 17 .5 .1 Calculate the molar solubility of AgCl in 0.10 MCaCl 2 . a) 1.6 X 10 - 10 M b) 1.6 X 10- 9 M c) 8.0 X 10- 10 M d) 1.3 X 10- 5 M e) 2.6 X 10- 8 M 17 .5.2 Which of the following substances will be more soluble in acidic solution than in pure water? (Select all that apply.) a) PbC0 3 b) AgS c) AgJ d) Fe (OH)3 e) CaF2 Separation of Ions Using Differences in Solubility In chemical analysi s, it sometimes is necessary to remove one type of ion from solution by pre- cipitation while leaving other ions in solution. For instance, the addition of sulfate ions to a solu- tion containing both potassium and barium ions causes BaS04 to precipitate out, thereby removing most of the Ba 2+ ions from the solution. The other "product," K 2 S0 4 , is soluble and will remain in solution. The BaS0 4 precipitate can be separated from the solution by filtration . Fractional Precipitation Even when both products are insoluble, we can still achieve some degree of separation by choosing the proper reagent to bring about precipitation .• Consider a solution that contains Cl -, Br - , and 1- SECTION 17.6 Separation of Ions Using Differences in Solubili ty 713 ions. One way to separate these ions is to convert them to insoluble silver halides. As the K sp val- ues in the margin show, the solubility of the silver halides decreases from AgCl to AgI. Thus, when a soluble compound such as silver nitrate is slowly added to this solution, AgI begins to precipitate first, followed by AgBr, and then AgCl. This practice is known as fractional precipitation. Sample Problem l7.12 describes the separation of only two ions (Cl- and Br-), but the pro- cedure can be applied to a solution containing more than two different types of ions. ". ~ Sample Problem 17.12 ' , Silver nitrate is added slowly to a solution that is 0.020 M in Cl - ions and 0.020 Min Br - ions. Calculate the concentration of Ag + ions (in mol/L) required to initiate the precipitation of AgBr without precipitating AgCL Strategy Silver nitrate dissociates in solution to give Ag + and NO :; ions. Adding Ag + ions in sufficient amount will cause the slightly soluble ionic compounds AgCl and AgBr to precipitate from solution. Knowing the K sp values for AgCI and AgBr (and the concentrations of CI- and Br - already in solution), we can use the equilibrium expressions to calculate the maximum concentration of Ag + that can exist in solution without exceeding K sp for each compound. Setup The solubility equilibria, K sp values, and equilibrium expressions for AgCI and AgBr are AgCI(s) :;::. = ::t" Ag +(aq) + Cqaq ) AgBr(s) • " Ag +(aq) + Br-(aq) Ksp = 1.6 X 10- 10 = [Ag+][Cl- ] Ksp = 7.7 X 10- 13 = [Ag+ ][Br - ] Because the K sp for AgBr is smaller ( by a factor of more than 200), AgBr s hould prec ipitate first; that is, it will require a lower concentration of added Ag + to begin precipitation. Therefore, we first solve for [Ag +] using the equilibrium expression for AgBr to determine the minimum Ag + concentration necessary to initiate precipitation of AgBr. We then solve for [Ag +] again, using the equilibrium expression for AgCl to determine the maximum Ag + concentration that can exist in the solution without initiating the precipitation of AgC l. Solution Solving the AgBr equilibrium expres sion for Ag + concentration, we have + K sp [Ag ] = [Br ] and [ A +] = 7.7 X 10- 13 = 39 X 10- 11 M g 0.020 . For AgBr to precipitate from solution, the silver ion concentration mu st exceed 3.9 X 10- 11 M. Solving the AgCl equilibrium expression for the Ag + concentration, we have and A + _ K sp [ g ] - [CI ] [A +] = 1.6 X 10- 10 = 8.0 X 10- 9 M g 0. 020 For AgCI not to precipitate from solution, the silver ion concentration mu st stay below 8.0 X 10- 9 M. Therefore, in order to precipitate the Br - ions without precipitating the Cl - from this solution, the Ag + concentration must be greater than 3.9 X 10- 11 M and less than 8.0 X 10- 9 M. Practice Problem A Lead(II) nitrate is added slowly to a solution that is 0.020 M in CI- ions. Calculate the concentration of Pb z + ions (in mol/L ) required to initiate the precipitation of PbCl z . (K sp for PbCl z is 2.4 X 10- 4 .) Practice Problem B Calculate the concentration of Ag + (in mol/L) necessary to initiate the precipitation of (a) AgCI and (b) Ag 3 P0 4 from a solution in which [Cn and [PO ~ - ] are each 0.10 M. (K sp for Ag 3 P0 4 is 1.8 X 10- 18 .) Compound K,p Agel 1.6 x 10- 10 AgBr 7.7 x 10- 13 Agi 8.3 X 10- 17 Think About It If we continue adding AgN0 3 until the Ag + concentration is high eno u gh to begin the precipitation of AgCI, the concentration of Br - remaining in solution can also be determined using the K sp expression: 7.7 X 10- 13 8.0 X 10- 9 = 9.6 X 10- 5 M Thu s, by the time AgCI begins to precipitate, (9.6 X 10- 5 M) .;- (0.020 M) = 0.0048, so less than 0.5 percent of the original bromide ion remains in the solution. 714 CHAPTER 17 Acid-Base Equilibria and Solubility Equilibria Note that these group numbers do not correspond to groups in the periodic table. . • Multimedia • Chemical Analysis - flame tests of metals. Fig u re 17.11 Flame tests for sodium (ye llow flam e) and pota ss ium (violet flame ). Qualitative Analysis of Metal Ions in Solution The principle of selective precipitation can be used to identify the types of ions present in a solu- tion. This practice is called qualitative analysis. There are about 20 common cations that can be analyzed readily in aqueous solution. These cations can be divided into five groups according to the solubility products of their insoluble salts (Table 17.4). Because an unknown solution may cO · rita-iii from." r to · iil1 · 20 'lons; ' any an."<i1) ;sls ·iriiis die ·carried 'oiit systemiltlciiily fioill 'group 1 through group 5. The general procedure for separating the se 20 ions is as follows: • Group 1 cations. When dilute HCl is added to the unknown solution, only the Ag +, Hg ~+, and Pb 2 + ions precipitate as insoluble chlorides. The other ions, whose chlorides are soluble, remain in solution. • Group 2 cations. After the chloride precipitates have been removed by filtration, hydrogen sulfide is added to the unknown solution, which is acidic due to the addition of HCI. Metal ions from group 2 react to produce metal sulfides: In the pre se nce of H+, this equilibrium shifts to the left. Therefore, only the metal sulfides with the smallest K sp values precipitate under acidic conditions. These are Bi z S 3 , CdS, CuS, and SnS (see Table 17.4). The solution is then filtered to remove the insoluble sulfides. • Group 3 cations. At this stage, sodium hydroxide is added to the solution to make it basic. In a basic solution, the metal sulfide equilibrium shifts to the right and the more soluble sulfides (CoS , FeS, MnS, NiS, ZnS) now precipitate out of solution. The AI 3+ and Cr 3+ ions actually precipitate as the hydroxides Al(OH)3 and Cr(OH) 3, rather than as the sulfides, because the hydroxides are less soluble. The solution is filtered again to remove the insoluble sulfides and hydroxides. • Group 4 cations. After all the group 1, 2, and 3 cations have been removed from solu- tion, sodium carbonate is added to the basic solution to precipitate Ba 2 +, Ca 2 +, and Sr 2 + ions as BaC0 3 , CaC0 3 , and SrC0 3 . The se precipitate s, too, are removed from solution by filtration. • Group 5 cations. At this stage, the only cations possibly remaining in solution are Na +, K+, and NH t. The presence of NH t ions can be determined by adding sodium hydroxide: The ammonia gas is detected either by its characteristic odor or by observing a wet piece of red lit- mus paper turning blue when placed above (not in contact with) the solution. To confirm the pres- ence of Na + and K+ ions, a flame test is usually used in which a piece of platinum wire (chosen because platinum is inert) is dipped into the original solution and then held over a Bun sen burner flame. Na + ions emit a yellow flame when heated in this manner, whereas K+ ions emit a violet flame (Figure 17 .11). Figure 17.12 summarizes this scheme for separating metal ions. SECTION 17 .6 Separation of Ions Using Di ffe rences in So l ub ili ty 7 :> Solution containing ions of all cation groups + HCl Group 1 precipitates Filtration f AgCl, Hg1 C1 1 , PbCl 1 Solution containing ions of all remaining groups + H1S Group 2 precipitates Solution containing ions of all remaining groups T Na l Group 3 pr ecipitates f Filtration CoS, FeS, MnS, NiS, ZnS, Al(OH) 3' Cr( OH )3 Solution containing ions of all remaining groups IT .'!, 1- Group 4 precipitates f Filtration BaC0 3 , CaC0 3 , SrS0 3 Solution contains Na+ K+ NH + ions , , 4 Checkpoint 17.6 Separation of Ions Using Differences in Solubility 17.6.1 A solution is 0.10 M in Br - , CO ~ - , CC C and SO ~ - ions. Which compound will precipitate first as silver nitrate is added to the solution? a) AgBr b) Ag 1 C0 3 c) AgCl d) AgI e) Ag 2 S0 4 17.6.2 Barium nitrate is added slowly to a solution that is 0. 10 M in SO ~- ions and 0.10 M in F- ions. Calculate the concentration of Ba 2+ ions (in mol/L) required to initiate the precipitation of BaS0 4 without precipitating BaF 1 . a) 1.7 X 10- 6 M b) 1.1 X 10- 9 M c) 1.7 X 10 - 4 M d) 1.7 X 10 - 5 M e) 1.1 X 10 - 8 M • Figure 17.12 A flowchart for the separation of cations in qualitative analysis. 716 CHAPTER 17 Acid-Base E quilibria and Solubility Equilibria • Applying What You've Learned Most toothpastes contain fluoride, which helps to reduce tooth decay. The F- ions in toothpaste replace some of the OH - ions during the remineralization process: SCa 2 +(aq) + 3PO ~ - (aq) + F-(aq) +- • CaS(P04)3F(S) Because F- is a weaker base than OH - , the modified enamel, called fluoroapatite, is more resistant to the acid produced by bacteria . Problems: (a) Calculate the molar solubility of hydroxyapatite given that its K sp is 2 X 10 - 59 . [ ~~ Sample Problem 17. 7] (b) Calculate the K sp of fluoroapatite given that its molar solubility is 7 X 10- 8 M. [ ~~ Sample Problem 17.7] (c) Calculate the molar solubility of fluoroapatite in an aqueous solution in which the concentration of fluoride ion is 0.10 M. [ ~~ Sample Problem 17. 7] Cd) Calculate the molar solubility of hydroxyapatite in a buffered aqueous solution with pH = 4.0. [ ~~ Sample Problem 17.7] - CHAPTER SUMMARY =~~ Section 17.1 o The presence of a common ion s uppre sses the ioni zat ion of a weak acid or weak base. This is known as the common ion effect. o A common ion is added to a solution in the form of a sa lt. Section 17.2 o A solution that contains significant concentrations of both memb ers of a conjugate acid-base pair (weak acid -conjugate base or weak base - conjugate acid) is a buffer solution or simply a buffer. o Buffer solutions resist pH change upon addition of sma ll amounts of strong acid or s trong ba se . Buffers are imp o rtant to biological systems . o The pH of a buffer can be calculated us ing an equilibrium table or with the Henderson-Hasselbalch equation. o The pKa of a weak acid is - l og Ka. When the weak ac id and co njugate base concentrations in a buffer solution are equal, pH = pK a . o We can prepare a buffer with a specific pH by choosing a weak acid with a pK a clo se to the de sired pH. Section 17.3 o The titration curve of a strong acid - strong base titration ha s a long , steep region near the equivalence point. o Titration curves for weak acid-strong ba se or weak base-strong acid titrations have a sig nificantly s horter steep region. o The pH at the equivalence point of a strong acid-strong ba se titration is 7.00. o The pH at the equivalence point of a weak acid-strong ba se titration is above 7.00. o The pH at the equivalence point of a weak base - stro ng acid titration is below 7.00. o Acid-base indicators are usually weak organic acids that exhibit two different colors depending on the pH of the so lution. The end point of KEyWORDS Buffer, 683 Common ion effect, 683 Complex ion, 710 KEY EQUATIONS End point, 698 Formation constant (K r ), 710 Fractional pr ec ipitation, 713 KEY EQUATIONS 717 a titration is the point at which the color of the indicator changes. It is used to estimate the equivalence point of a titration. o The indicator used for a particular titration should exhibit a color change in the pH range corresponding to the steep region of the titration curve. Section 17.4 o The solubility product constant ( Ksp ) is the equilibrium constant that indic ates to what extent a s lightly soluble ionic compound dis soc iates m water. o Ksp can be u sed to determine molar solubility or solubility in gIL, and • VIce versa. o Ksp ca n also be used to predict whether or not a precipitate will form when two so lution s are mixed. Section 17.5 o Solubility is affected by common ion s, pH, and complex ion formation. Theformation constant (Kr) indicates to what extent complex ions form. o A sa lt that di ssoc iates to give a strong conjugate ba se s uch as fluoride ion will be more so luble in ac idic solution than in pure water. o A salt that dis sociates to give hydroxide ion will be more soluble at lower pH and le ss so luble at higher pH. o The solubility of an ionic compound increases when the formation of a complex ion consumes one of the products of dissociation. Section 17.6 o Ion s can be separated using fractional precipitation. o Fracti ona l precipit ation sc heme s can be designed ba sed on K,p values. o Group s of cations can be identified through the u se of selective precipitation. This is the ba sis of qualitative analysis. Henderson-Hasselbalch equation, 685 Molar so lubilit y, 701 Qualitative analysis, 714 Solubility, 701 Solubility product constant (K s p) , 70 1 17. I [A - ] P H = pK + 10" = , :- a b [HA] 17.2 pKa = -log Ka 17.3 [conjugate ba se ] pH = pK a + log [weak acid] [...]... ion in part (a) (c) The pH at which the dipolar ion predominates is called the isoelectric point, denoted by pl The isoelectric point is the average of the pKa values leading to and following the formation of the dipolar ion Calculate the pI of histidine (d) The histidine group plays an important role in buffering blood (the pH of blood is about 704) Which conjugate acid-base pair shown in part (a)... Figure 18.2 that lead to an increase in entropy of the system: Figure JB.2(a) In a solid, the particles (atoms, molecules, or ions) are confined to fixed positions [ ~~ Figure 18.2{a) Sliquid > > Melting: S solid ' Figure 18.2{b) svapor Section 12.3] and the number of microstates is small Upon melting, the particles can Sliquid Vaporization: SECTION 18.2 Figure 18.2(c) S aqueous > S pure Entropy 73... with solvent) and the dissociation of the compound into ions: NaCI(s) H 20 Na +(aq) + CI - (aq) A greater number of particles leads to a greater number of microstates However, we must also consider hydration, which actually causes water molecules to become more ordered around the ions This part of the process decreases entropy because it reduces the number of microstates of the solvent molecules (For... 2F~03(s) • 2H20(l) + 0 2(g) + 2H 20(l) • 2LiOH(aq) • Nig) + Hig) + 3H2 (g) The Second and Third Laws of Thermodynamics Recall that the universe is made up of two parts: the system and the surroundings [ ~~ Section 5.1] The system typically is the part of the universe we are investigating (e.g., the reactants and products in a chemical reaction) The surroundings are everything else Both the system and the... with an acid solution is by dilution (a) Consider a buffer solution made of 0.500 M CH3COOH and 0.500 M CH 3COONa Calculate its pH and the pH after it has been diluted lO-fold (b) Compare the result in part (a) with the pHs of a 0.500 M CH 3COOH solution before and after it has been diluted 10-fold 723 17.133 Histidine is one of the 20 amino acids found in proteins Shown here is a fully protonated histidine... Your graphs should show the pH on the y axis and the volume of base added on the x axis 17.25 Explain how an acid-base indicator works in a titration What are the criteria for choosing an indicator for a particular acid-base titration? 17.26 The amount of indicator used in an acid-base titration must be small Why? What is the pH of the buffer 0.10 M Na2HPOJ O.15 M KH ZP04 ? 17.15 (a) • 719 QUESTIONS AND... of food generates carbon dioxide gas and water vapor The?e gaseous products have far greater entropy than the solid foods that went into their production For us to cause a reduction in entropy in one part of the universe, the principles of entropy, free energy, and equilibrium require that we generate an even greater increase in entropy somewhere else • , In This Chapter, You Will Learn about the three... called a spontaneous process One that does not occur under a specific set of conditions is called nonspontaneous Table 18.1 lists examples of familiar spontaneous processes and their nonspontaneous counterparts These examples illustrate what we know intuitively: Under a given set of conditions, a process that occurs spontaneously in one direction does not also occur spontaneously in the opposite direction... the absolute entropy of a substance at 1 atm (Tables of standard entropy values typically are the values at 25°C because so many processes are carried out at room temperature-although temperature is not part of the standard state definition and therefore must be specified.) Table 18.2 lists standard entropies of a few elements and compounds Appendix 2 provides a more extensive listing The units of entropy... (CH3)2A sOzH Its ionizati on constant is 6.4 X 10- 7 (a) Calculate the pH of 50.0 mL of a 0.10 M solution of the acid (b) Calculate the pH of 25.0 mL of 0.15 M (CH 3lzAs0 2 a (c) M ix the solutions in parts (a) and (b) Calculate the pH of the resulting solution 17.107 Radiochemical techniques are useful in estimating the solubility product of many compounds In one experiment, 50.0 mL of a 0.010 M AgN0 . formation on solubilit y. In Chapter 22 we will discuss the chemistry of complex ions in more detail. Transition metals have a particular tendency to form complex ions. For example, a solution. increases the solubility of Cu(OHh. A measure of the tendency of a metal ion to form a particular complex ion is given by the formation constant ( Kr ) (also called the stability constant),. used to estimate the equivalence point of a titration. o The indicator used for a particular titration should exhibit a color change in the pH range corresponding to