Chapter 8 Solutions

Chapter 8 Solutions - CHAPTER EIGHT ' APPLICATIONS OF...

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Unformatted text preview: CHAPTER EIGHT ' APPLICATIONS OF AQUEOUS EQUILIBRM Buffers i5. 16. t7. ' 2.8. t9. A bafiered sciatica must contain both a weak acid and a weak base. Most buffered solutions are ‘ prepared nsing a weak acid plus the conjugate base of the weak acid (which is a weak base). Buttered soiutions are useftii for controiiing the pH of a soiution since they resist pH change. When strong acid or strong base is added to a sodinni dihydrogen phosphatefsodiom hydrogen phosphate buffer mixture, the strong acidr’base. is neatralized. The reaction goes to compietion resetting in the strong acidz'baSe being repiaced with a weak acidfhase. This results in a new heifer sciatica. The reactions are: WWI) ‘1“ H904”(3Q) ‘"“‘* HzP04'(aq); 0316(1) + H2P0¢'(a<}) *’ Himftaci) "1" H200) The capacity of a buffer is a measure of how much strong acid or strong base the buffer can neutralize. All the buffers tinted have the same pH. (2 33KB m 4.74) since they aii have a 1:2 concentration ratio between the weak acid and the conjugate base.‘ The 1.0 Mhnfi‘er has the greatest capacity; the 0.01 Mbnft'er the ieast capacity. In genera}, the larger the concentrations of weak acid and conjugate base, the greater the buffer capacity, i.e., the greater the ahiiity to neutraiize added strong acid or strong base. {Mtg} {on “] NH3 + H20 w NH: K], w '_""""""""“W ; Taking the «tog of the Kb expression: . {N53} “3 K” m “3g {0}” tog [NM] I he [OH-3 m — tog 14., + iog [W453 {Nttsj’ {m3} _ [We] £Acid} 0H: K, “5“ I0 OHm: + to mum. P P b g {NHEJ or p pKfi g [Base] a. This isa weak acid prohiein. Let Hcgflsoz m HOFr and C3H502‘ w OPr‘. HOPdaq) x—“fi H*(aq) + 0Pr“(aq) Ka = 1.3 X 10'5 initial 0.100 M ~o a I - x tnth HCfik dissociates to reach equiiihriuin Change ~x , ' — a» +5.“ +3: Bqttii. 0.100 - x x ' x 2} 5 CHAPTER 8 APPMCATIONS OF AQUEOUS E UILEBRIA OH Wiil react eompieteiy with the best acid present, HOPr. HOPr + OH' ------ --> O'Pr‘ + 1-1.0 Before 0.20030 0.020M 0 Change 0020 0020 we +0020 Reacts centereme After 0.080 0 0.020 ' A buffer solution rescits after the reaction. Using the Hendersomflasseibaich equation: w. [Base] .. . (0.020) m H m + 10 - 4.89 + to m p 9K“ g- ate-id] g {0.080) . We have a week base and a strong base present at the same time. The amount of OH“ added by tits weak base wilt be negligibie. To prove it, lets consider the weak base equiiibrium: 4.29 01%“ + 1—120 % HOPr + OH' K.m7.?><10-*°- Initiai 0.100 M _ 0 0.020 M x mohr'L 01’1‘“ reacts with H20 to reach equiiibriem Change «x ------ "3» +x +x Equii. 0.100 ~ 3: x 0.020 + x [03"] m 0.020 +x e 0.020 M; 13013 x 1.?0; pH 3 £2.30 Assumption good. Note: The OH“ contribution from the weak base 0391“ was aegiigible (x "e 3.9 X 10’” M as compared to 0.020 MOH" from the strong base). The pH can be determined by oniy considering the amount of strong base present. ' This is a strong base in water. {OH} m 0.020 M; pOH m 1.70; pH = 12.30 . OH’ veil! react Completer virith HOPr. the best acid present. How + OH‘ ‘ m+ OPr + H20 Before 0.100 M 0.020 M 0.100 M Change 0020 41.020 —> +0020 Reacts compieteiy After 0.030 0 0.120 Using the Henderson—Hasselbaich equation to solve for the pH of the resetting heifer satiation: [Base] (0.120) _, pH = 13K.a + iog w - 5.0? m 4.39 + tog [Acid] (0.080) 222 ‘ CMPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 27. _ 24 mmoi 0C1 ” 250m " WWW. + . :14 log zémszmeCi “3’46 (4)035) 3 250 mL pH m 146 + 1 mol 82.03 g d. 26.4 g NaCIHSO2 X m 0.322 11102 NaCzfiso2 1 L x 6.00 moi HQ} 000 mL L 58.0 mL HG! >< 2 “‘ 0.300 mol HG} I? reacts oompietcly with best base present, C2H302‘. (Strong acids are always assumed to react to completion.) W + 021:1302‘ m») HCQHSOQI Before 0.300 moi 0.322 moi ' _ 0 _ Change «0.300 41.300 W} - +0380 Reacts compieteiy 0.922 mol 0.300 moi ' After 0 After reaction, a buffer sciatica resnits‘ Using the Henderson-Hasseibaich equation: _ c H 0“ , . pH=pKa+log $4.74fi-Iog WW3“), $24.74 ~ 1.13 mam {HC2H302j (0.300111013050423 L) K 44 (35st *W’r CsHsN Ks f m% m 5.9 x 10*; glam ~tog (5.9 x 104) = 5.23 b X _ We wilZ use the Henderson-Hasscibaich equation to caicuiaie the conmtration ratio necessary for each buffer. ‘ - ’- ' ' ' C H N pHm pKa+ log {233$} , pH W 523 +10g “Wm-mi 5 5 {amt} - {CSHSNH -; CHN CHN a. 4.50 w 5.23 + 10g Mimi—$3M»- b. 5.90 = 5.23 + log liming“ E(:SE-ISNVI‘I I9} Jr] CHN _ CHN 10g ME— 5 5 3+ m 43.73 . _ 10g mini—1L m .923 [CsfisNi-I ] {CSHSNH 1 CHN CHN ME 5 5 i = 180732019 W? 5 5 '1 m 20423 moss; {CSHSNH "3 .[CfifiSNH "’1 CHAPTER 8 APPLICATIONS OF AQUEOUS 135 gmmm ' 223 [CSHSNE . ' [QHSN] c. 5.23 =s.23 Hog W a. 5.50ms,23+zog W ‘ . {CSHSNH +1 {cg-gm 7 MCSHSN} 3 10°ma 2.0 wcsfism m 10mm 1.9 {<35};an +1 {CSHSNH *] 23. When 02-1- is added, it converts 124.150.1130. into (22:14:); 12022130. + 01; w» c.2330.- + H20 From this reaction, the inoies of CZH3Og produced eguais the moles of OR“ added. A250, the total _ concentration of acetic acid pius acetate ion must equal 2.0 M (assuming no vohnne change an addition ofNaOI—I). Summarizing for each soiution: {(321%qu + [110.1130] m 2.9 Mend {c.3305} produced m {03*} added CH0_ CH1)— {2 3 21‘ForpflmpKwiog { 2 3 21w!) a. p§IWpKn+logm~Tm, m I {HLZHSQZ} [141023302] C Hf} _ Therefore, “’ 2 3 {chngozl *2 Le and {(31305} = {HC2H302} Since [(31.05 + [HCzflaOzj m 2.0 M, then [0219:3023 m mango.) m 1.0 Mm {033 added To produce a LO M (321130; soiution we need to add 1.0 moi ofNaOH to 1.0 L of the 2.0 M HCZHJ)2 solution. The resultant solution wiii have pH W pKa = 4.74. {cg-230;] {(221492} . e _ M 1041’” W [Hcggogj {Iago}; h. 4.80 = 4.?4 + log [Czfisoz']; 0-13 [HCflst] 01' [P1023302] 2.5-6 {CZHIiOZ-j; Since [Cszofl “é” [chfiaozl = 2.0wM, en: ' ' ' ' {02330;} + 5.5 {61:19.1 = 2.0 M, {0.3.0.1 m 36%: 9.30 M: {on-1 added We need to add 0.30 11103 of NaOH to 1.0 L of 2.0 M HCQIQO2 sciatica to produce 0.39 M C1H302‘. The resultant soiution WiIl have pH = 4.80. [Cg-1302‘} {0530;} , — 109-26 - 1.3 [11021430.] {neg-1302; c. 5.00 = 4.74 + 10g 3-8 [Eczflzofl 2 [Czflaozj or #3023302} m 0'56 [Czfisoz']; Since [HCzH302] + [(233023 W 2.0 M, then: . 1.56 {02330;} :-~ 2.0 M. {(223305} .-~ 1.3 Mm {OH‘} added We need to add 1.3 mo! ofNaOH to 1.0 L of 2.0 M HQJIHJD2 to produce a soiution with pH = 5.00. CHAPTER 8 APPLICATIONSOF AQUEOUS EQUIme 225 by 4.9 gNaOH x 2 molNaOH x 1' moi-OH ' m 0.10 mo} OH; {0H,} 3 0.202102 m 0.16M 40.00 g me} NaOH 1.0 L CEHSNH; + OH' —> C6};st + I H20 Before 1.2 M_ 8.10M 0.50M Change «0.10 «0.10 —> +0.10 Afler 1.1 0 0.60 . . m 0.60 ;__ A bufi'er sohmon cxmts. pH m 4.58 + tog mini-w - 4.32 N .. wag] . 3.55wlcg(4.0>< IO“)+30g { 02} {meg} _ {HNOQ {N020 {N05} [meg -’ {ENC}; 32. pH m 13Ka + Iog 3.55=3.40+20g m IO‘Wm 1.4 Let 3: = volume (1..) HMO2 solution needed, then 1.00 w x :“r volume ofNaNOz soiution needed to _ _ form this buffer solution. (1 00 ) 0.50 moi NeNO2 .. _ x x “WNW-..” {N02} m1 4 m L _ 0.50 4150.5 {Hch} ' {3.501.101 RN02 0.503.: x x - m L . 0.70 x m 0.50 — 0.50 x , 1.20 x m 0.50, x = 0.42 L We acct} 0.42 L of 0.50 Junie“)2 and 1.00 - 0.42 e 0.58 L of 0.59 iii/{1031002 to form a pH = 3.55 1 buffer satiation. _ . '. ' HCO‘ HCO' 33. a. pH=pK,+iog {We}. 7.4omzogms X10‘?)+iog { 3] =6.37+2og { 3E {9.ng - - [32003; {212C031 Hcc“ ' HCO' ' CO [ flew-emu; { 3 3L i 23 = 1 macaw [1-15.03] [Hcog'é [HCOQ'J H ' Hm?" . H1302" h. 7.15=~20g(6.2x10~8)+zog [ 4w}.7.15=7.21+10g I 4w} ' {H2P04 [Hngd mo?” '13 P0" [ ‘mewmwoy, [2 ‘th 1 emu—1 {11220.3 {emf} 0‘9 c. A best buffer has approximately equal concentrations of weak acid and conjugate base so that pH #5 9K“ for a best buffer. 2328 13K; value for a H3PO4/H2PO; buffer is Jog (7.5 X 106) = 2.12. A pH of 7.1 is too high for a HsPOJHZPO; buffer to be effective. At this high of pH, there ' wcuid be so littie H31904 present that we could hardly consider it a buffer; this solution wouid 2101236 effective in resisfing pH changes, especiaity when strong base is added. CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 229 {HONH;]{0H‘J¥_ (2.0x IO“5 +x)x N {LOXEOWSM [Home] .i.{}>'<1{}"‘ ~x - it»: 10"4 Kh=l.i x10" — _ (Assnmingx<< 1.0 X 10".) x w {011‘} m 1.1 x IOJM; Assumption theta: << 1.0 X 10“ is good (x is I.i% of 1.0 X 10's). In the reguiar procednre to soive the buffer prohiem, the problem reduced down to the expression: {HONHflbKfi-i '] {HGNHZ} a This expression holds if x is negligibie as compared to [HONHJL and {HONHQO as it was in this _ problem. Now we want to know if we need to worry about the contribution of OH‘ from water. 23mm the equation for the exact treatment of buffers, ifQOR]: n K.) I {013‘} is much less than [HONiiflo and {HONHQm then the exact equation reduces to: ' K§m w {on "'1 {Homage Kb . inosnzgo This is the same expression we ended up with to soive the prehient using the regniar procedures. I I Checking the neglected term using the [OE-1'] caicuiated above: ' {0151‘}2 ~ Kw _ (1.1 x 10"?)2 ~ 1.0 x 2.0"“ {0H} 1.1 x to"? =13 X it)”8 This is indeed much smailer than [KONHJL and {HONHgfi (1.9 X 10'3 is 0.29% of 1.0 X 10"). . So for this prohiein we would calculate the "same {Off} using the exact equation as we calculated - i. using the regular procedures. In generai, we only need to use the exact eqnation when the buffering materiais have a concentration of 10”“ M or iess, I 41. Using reguiar procedures, pH 2 pig m wiog(i.6 X 10”) = 6.88 since [A1, e {HA}, in this buffer sotution. Kowcver, the pi! is very close to' that of neutral water so maybe we need to consider the B." contribution from water. Another problem with this answer is that x (= EH?) is not email as _i compared to {HA}a and [A1, , which was assumed when soiving asing the reguiar procedures. Since the concentrations of the buffer components are iess than 10‘6 M, then let us use the expression for _ the exact treatment of heifers to soive. {Am} i WW? H Kw] iH+E[5.0x 10“? + WM 32 ‘ m" 304% ° [H *1; _ _ - [He + 2 A . + 2 m ‘14 in 1 KW 5‘“ 20-7 q {H 3 1.9x10 [H *g ' {H *E - Soiving exactly requires sotving a cubicecination. Instead, we wilt use the method of successive approximations where our initiai guess for [H1 e' i .6 X' 10" M (the value obtained using the regular procedures). [H +] Ka £1.6X10'7“ {Rein ~ CHAFI‘ER 8 APPLICATIONS 0? Ag QUEOUS - 237 :1. - I " 32m. + H" —» 32ml; Before 10.0 mol 8.08 mmoi 0 After 2.0 moi 0 8.00 moi A buffer solution results. [0353} m + . 8.00 mmolNT - pH w pK. + log . . m 3.48 + (—0.60) m 7.88 i7 {301d} e. mmoi I? adéeé 4"» 50.0 >< 8.200 Mm 10.0 mmoi H” 3£«1E32\TI.~2‘£-I2 ~¥~ if HQNNHJ Before I00 moi 10.0 moi 8 After ' 0 0 10.8 mmoi As is aiways the case in a weak basefstrong acid titration, the pfi at the equivalence point is acidic because 0er a weak acid (HZNNHQ is present. Solving the weak acid equiiibrium probiem: HQNNH; w H" + KENNEL: Initial 10.0 111111001501) mL 0 0 Equil. {3.0667 ~ 3: x x x 2 x 2 . Ka=3.3><1{}“9— x—[IF]#1.SX10‘5M 6.0667 -x x 0.0667’ pi! m 4.82; Assumptions good. f. tomol H+ added 7—“ i000 mL x 0200 Me 20.0 mmol I?“ Home. , ‘ + IF. a HZNNH; Before 10.0 mo} 20.0 11311101 I 0 After 0 10.8 mo} 10.0 mo} ’Z‘wo acids are present past the eqnivalenoe point, but the excess H“ wili determine the pH of the soiution since HzNNHJ is a weak acid. 10.0 mmoi _ [m 100.0 mL + 290.0 mL *3 49. We wiil do sample caicuiations for the various parts of the titration. All resuits are snmmarized iii Tobie 8. I. at the end of Exercise 8.52. ' At the beginning of the titratioii, oniy the oieak acid Ii’CflI-ISO3 is present. . :i i . i i | | 238 " CHAPTER 8 APPLICATIONS 0? AS EUEOUS EQUILle files W I? + Lac” Ka=1{}”3‘“mi.4><10" HLac=HC3HSO3 Lac“ = €3H503‘ Initial ' {3.106 M ~0 O x molfL HLac dissociates to reach equilibrium Change -x ------> +1: +3: Equil. 0.190 « x I x x 2 2 ' 1.4 x 10“i - x 3‘ x -~ {14mw 3.7 x 103 M; 13H m 2.43 Assumptions good. 0.100 "x W 0.1007 Up to the stoichiometric point, we calculate the pH using the Hendersomflasseibaich equation. Thi is the buffer region. 2:01“ exazapie, at 4.0 m}; ofNaOH eddeé: ' 0. 100 mine} éhitial 2mm} HLac presth m 25.0 mL >< m =3 2.50 mmolHLac mo: OH” adcied m 4.0 mL x M m 0.4:} mmei 011' mL Note: The units mmoi are usuafly easier numbers to work with. The units for molarity are molest but are also equal} to mmoiesme. ' 3316 0.40 more! added OH“ converis 0.48 mmoies {flee to 9.40 moles Lae‘ according to the equation: . £11.39 + OH' -* Lae‘ + 2120 Reacts completer mmoi Hiae remaining a 2.50 ~ 0.48 m 2.10 mmoi; moi Lac “ produced =_0.48 name] We have a buffer soiution. Using the fiendersonéiesselbaich equation where :lea w 3.86: {fora} voiume cahceis, so we can use use the ratio of moies or moies.) $133.36 — 0.72 £3.14 Other points in the buffer region are calculated in a simiiar fashion. E’erform a stoiehiometry 'probiem first, foiiowed by a buffer problem. The hufi‘er region inehzdes a1} points up to 24.9 mi. OH‘added. At the stoiehiometrie point (25.0 ml. 0H" raided), we have added enough OH" to convert a}! ofthe Him: (2.50 1111110?) into its conjugate base, Lac“. All that is preseht is a weak base. "f0 determine the pH, we perform a weak base caieulation. [Lao]o ~—r m = {19590 M 25.0 ml. + 25.0 ml. 3 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 239 “Mum—mme 50. mm ‘ Lac‘ + H20 W mac + OH" K, x 3% m m x 10‘“ .. 3.4 x 10“1 Initiai 0.0508 M 0 0 x InoifL Lac“ reacts with H20 to reach equiiibrinnt Change or _ - ---- --> +x +x Equii. 0.8500 - x x x 2 2 Kb: “22.35%... “mi.” m7.2 x 39H 9.05f30 —x {3.0500 x = {OH} 3 1.9 X 10‘6 M; pOH w 5.72; pH at 8.28 Assumptions good. 9215f the stoiehiometrie point, we have added more than 2.50 moi of NaOH. The priII be determined by the excess OH" ion present. An example of this eatenlation foilows. At 25.: mL: on added = 25.1 mL x 533940;:me m 2.51 mot on“; 2.50 mmol on neuttaiizes a}! the week acid present. The remainder is excess OH". Excess OH" W 2.51 ~ 2.58 m 0.91 named OH" 0.01 mine} (25.0 + 25.1)2111. A1} results are listed in fable 8.1 at the end of the solution to Exercise 8.52. [on-1mm mzx 104M; p0§-I*3.7; prnm Resuits for ad points are summarized in Table 8.1 at the end ofthe soiution to Exercise 8.52. At the beginning of the titration, we have a weak acid probiem: HOPr @ if 4" Din“ H0131 m Hcsnsog ' 0?!“ m CSHSOZ' ' Initial 0.180 M ~{} 0 x inolfL HOE acid dissociates to reach equilibrium Change or +x - fix ' fiquii. 8.100 — x ’ x x K“: {$3.3{0P?}=1.3x10.5:m x2 2; x2 [HOPfl 0.200 w x 8.100 x m [11"] w 1.1 X I93 M; pH $2.96 Assnmptions good. The bnffer region is from 4.0 - 24.9 mL of OH” added. We wilt do a satnpie caicukatien at 24.8 mL OH” added. initial mine-i HOPI present m 25.0 1111.. X 0’ E'Ogmd m 2.50 mine! HOP)“ mine} OH" added m 24.0 2111. X 0100 mm] mm m 2.40 mmoi OH" mL . The added strong base ‘conven-s HOE into 01’1". ’ _____________ CHAPTER 8 APPLICATIONS-OF AQUEOUS BQUILIBRM 245 1111.. OH' added m 25. mmol 01-9 x M m 1.9 x :92 mi. OH“ 0.25 mmel H‘ + OK“ Wu} .2120 Eefofe 50. mo! _ 25 met . After 25 11111101 0 [H1 - 25 mm"; m =0.I3M; pH==0.89 m (109.9 +1.0x102)mL At the equivalence poifit of a strong ecidr’streng base titration, oniy eeutral species are present (Nata; H20) so the pH 3 2.00. 54. At equivalence point: 16.00 mi >< 0.125 moi/ml. : 2.00 mmoi Ofi' added; There must be 2.00 mmoi HX present initiafly. - 2.00 mi. NeOI-i added m 2.00 2112.. >< 0.125 mmoUmL m 0.250 mmoi OH ‘; 0.250 moi of OH‘ added wiii convert 0.250 mmoi HX into 0.250 mmei X'. RemainingHX = 2.00 ~ 0.250 = 1.25 mol RX; This is a buffer soiuzien where {17?} m Z063” m 1.22 x 107‘ M. Since totai voiume eaneeis: K3,“ {1—9; [X'"} 4 1.22 x 29'"? (9.259) . W 1.74 >< 20'“ [HX] 2.175 Note: We couid aiso soive for K“ using the Hendersonwfiasselbaleh equation. 55. - a. 1.00 L X 0.100 mom, m 0.100 mol HG! added to reach. stoiehiometrie point. 10.09 g I 0.100 met _ b. 300.0 mL of EC} addedrepresents the halfway point toeqeivaEenee. Se, pH m pig m 5.00 and Ky 1.0 X 30". At the eqeiveience point, engugh I"? has been added to convert 312 the A‘ present initiatiy into fiA. The concentration of at the equivalence-point is: The £0.00 g sampie must have contamed 0.100 mol ofNaA. m 100. g/mol {M10 = (“00 “531 = 9.9999 M ' 1.19 1. HA w_ H” + A“ Kawiflx 195 Znitiai ' 9.9999 M 9 o Bquii. 0.0989 - x x x 2 2 n21.9x104: m" a x 0.0909 x 0.0909 x = 9.5 X 10“ M= [W]; pfi m 3.02 Assumptions good. 56. pH 5 5 for bromeresoi green to be tine. 'pH < 8 for thyme! biue to he yeilew. The pH is between 5 sad 8. ’ 57. - a. yeilow ' b. green (Both yellow and blue forms are present.) . ....... ............ ’ 250 r ‘ . CHAPTER 8 APPLICATIQNS OF A UBOUS E UZLIBRIA 69. a. N3" is presem in aii soiutions. The added H“ from HCi reacts compieteiy with (1033‘ to convert it into HCOB'. After aii C032" is reacted (after point C, the first equivaience point), then if reacts completer with the next best base present, HC03‘. Point E represents the second equivalence point. The major species present at the various points after I? reacts compieteiy foiiow. A. (3033;1320 B. (3032-, HCOa", 1120,13? C. HCO3', I120, Ci‘ D. HCO3', C02 (HZCOQ, H20, C!” B. CO; (H4303), H20, Ci‘ F. 2-?" (excess), C03 (HQCOQ, 320, C1" b, Point A (initiaiiy): 2- Q ,, - 2» M w _ EON” .4 C03 +H20 «m 3003 +03 Kb(cog)ww~W-2axw ._ - Kaz 4.8X2011 Initiai 0.100.347 . 0 ~43 Eqnii. 0.108 ~ x x x IHCO M OHW' '3 3 wa2.1><10“*w [ 31% j m. x W : mil-w {€033} {3.100 e x 9.100 x = 4.6 X 19'3 Me {Off}; pH 2 11.66 Assumptions good. Riot B: 2316 first halfway point where {6032‘} m {HCOg} pH = pKazm wiog (4.8 X 20‘“) m 10.32 ' Assumptions gooé. Point 'First equivaience point (25 .00 mi. of0.100 M H01 added). The mphoteric ECO; is the major aeidibase species present. ' 3.3. pH m , pKa] w ~10g (4.3 >< 10-?) 56.37 pHm m E’oint Z}: ’i‘he secood haifway point Where [HC03‘] w {H1603}. pH = pKaf 6.3? Assumptions good. Point E: This is the second equivaience point where at! of the C032“ present initiain has been converted into HZCG3 by the added strong acid, 5&8 In}; H01 added. [1143033 m 2.50 mmoif’YSfl ml, 3 0.0333 M 32003 e»: H‘" + HCO; $4.3 >< 20er Initial 0.033334 " o " 0 . Equii. 0.8333~x x x CHAPTER 8 APPLICATIONS 0}: AQUEOUS EQUILIBRIA 251 x3 x2 K X106!“ W ::: 3* 0.9333 « x 0.0333 x = [If] * 1.2 X 10“ M; pH = 3.92 Assumptions good. 79. a. HA‘ vb H" + A2” K, = 3 X 10‘“; When m [A21 pH =pKa’2 * 8.00. 2‘11: titration reaction is A2“ + H’" 4 HA‘ (goes to campiction). Begin with 100.0 mL x 0.200 mofimL = 20.0 moi A2“. We need to convert 10.0 moi A?“ into HA‘ by adding 10.0 mo} E“. This wiii produce a solution where {HA'} 2 [A33 and- pH : pK32= 8.00. 10.0 moi = 1.00 mmoUmL >< V, V m 10.0 m}. HCI b. At the 2:16 stoichiomctric point, aii A} is converted into HZA. This requires 40.0 mmol HCZ which is 40.0 m}; of 1.00 M HG}. mat, m m = 9.143 M; Since K3} >> K32, 32A is the major source ofH“. 240.0131. HQA w 1—? + HA” Iaitiai 0.016 M 0 0 Equii. 0.143 ~x x x X2 X2 . K“, m w LO x 106 x , x m 0.012 M Check assumptions: = 0.143 w :1: 0.143 0.012 0 M3 X 100 m 8.4%; Can": neglcct x. Using successive approximations: -x m 0.01IS (cmying extra Sig. figs) so {If} = 0.0115 Maud pH = 1.94 Soiubiiity Equikibria 71. in our setmps, s W soiubiiity in moHL. Since Isoiids do not appear in the K3, expression, we do not need to worry about their initial or equilibrium amounts. 3. Agfi’OAs} W 3 AgYaq) + E’OfTaq) Znitiai 0 O .9 may}; of AggPOJs) éissoives to reach eqzziiibrinm Change «5 my +3.: "5? Equii. ' 33 5 Km 3 1.8 X 30'“; = {Agfis [P03] 3 (383(3) = 2754 2?; = 1.8 X I0”, s F (6.7x 302°)“ ='1'.6 X 20" molfL = moiar soiuhility 1.6x10'5moiA PO 418."? A PC) g3 ‘x g g3 4 =6.7x10'3gfL L moiAgsPfi)4 ...
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This note was uploaded on 04/16/2009 for the course CHEM 2150 taught by Professor Chirik,p during the Fall '08 term at Cornell.

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Chapter 8 Solutions - CHAPTER EIGHT ' APPLICATIONS OF...

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