Problem Set 5 Solutions

Problem Set 5 - C HAPTER 1 0 ACID-BASE T ITRATIONS l0-1 P rior t o t he e quivalence oint,t he r eactioni s O H H 2O a nd t he p tl i s p H

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Unformatted text preview: C HAPTER 1 0 ACID-BASE T ITRATIONS l0-1. P rior t o t he e quivalence oint,t he r eactioni s O H- * H + - --+ 2O a nd t he p tl i s p H computedf rom t he e xcess H-. A t t he e quivalence oint, t he s olutionc ontains p O just n eutrals alt i n w ater a nd t he p H i s 7 . A fter t he e quivalence oint, t herei s p excess + p resent nd t he p H i s d etermined y t he c oncentration f e xcess +. H a b o H l0-2. T he i nitial s olutionc ontainsa k nown c oncentration f t he w eak a cid a nd t he p H i s o determined rom t he a cid d issociation eaction: H A - --> + + A -. B etweent he f r H initial p oint a nd t he e quivalence oint, e achi ncrement f O H- n eutralizes p o a correspondingncrement f H A, l eavinga k nown m ixture o f H A a nd A -. A ha! A i o buffer! T he p H i s g iven b y p H : p Ka+ l og([A-]/[HA]). A t t he e quivalence oint p the s olutionc ontains he w eak b ase,A -, a t a d iluted c oncentrationrom t he o riginal t f HA. T he p H i s d etermined y t he b aseh ydrolysisr eaction:A - + H 2O--- l -lA + b OH-. B eyondt he e quivalence oint t herei s e xcess H- t itrant i n t he s olution. T he p O pH i s d etermined rom t he c oncentration f e xcess H-. f o O 10-3. T he i nitial s olutionc ontainsa k nown c oncentration f t he w eak b asea nd t he p H i s o determined rom t he h ydrolysisr eaction: A - + H 2O - -' H A + O H-. B etweent he f initial p ointa nd t he e quivalenceoint,e achi ncrement f H + n eutralizes p o a correspondingncrement f A -. l eavinga k nown m ixture o f H A a nd A -. A ha! A i o bffir! T he p H i s g iven b y p H : p Ka+ l og([A-]/[HA]). A t t he e quivalence oint p the s olutionc ontains he w eak a cid. H A. a t a d iluted c oncentrationrom t he t f original A -. T he p H i s d etermined y t he a cid d issociation eaction: b r HA - - H + + A -. B eyondt he e quivalence oint t here i s e xcess + t itrant i n t he p H solution. T he p H i s d etermined rom t he c oncentration f e xcess +. f o H 10-4. l mmediatelyp rior t o t he e quivalence oint, t he p H i s c ontrolledb y t he b uffer p actiono f t he w eak b ase,B , a nd i ts c oniugate cid. B H+. A s w e g et v ery c loset o a the e quivalenceoint,a lmosta llthe b asei s g onea nd t he c oncentrationf B H+ i s p o . e ssentially onstant.T he l og t erm i n t he H enderson-Hasselbalch c e quation, v Ull/[BH+], c hanges ery r apidly w ith e acht iny i ncremento f a ddedH +. A t t he equivalenceoint,t he c oncentrationf H + i s - 10-6.5M . E achs ubsequentddition p o a of s tronga cid f rom t he b uret i ncreasesH*]by a r elativelyl argea mounta nd t he [ pH c ontinues o d rop r apidly u ntil t here i s a s ubstantial oncentration f e xcess t c o stronsa cid. 6l 64 5.10m L: 6.00m L: Chapterl 0 loH-l : - (0.l0 m L X 0 . 5 0 M ) : 9 . 0 7 x l0 -4 M 0 = p H: 1 0 . 9 6 5 tffi,nL ( 1 .00 L )(0 M) m : - s6.0,r.,L . 5 0 0 : g . 9 3 ' l 0 -3M = p H: 1 1 . 9 5 lotl-l lz 11 a ? o a' 10 q B o- 6 4 ? 34 V6( mL) r0-9. M m (I/.(mL))(0.0500 ) : ( 25.0 LX0.0100) ) M mmolO Hmmol A Ve:5.00 mL. V 6=0mL: H A + 0 0.010- x -2 0 0.010 - , V6= 2 '50 m L'. :. .:- K a : 2 . 2 6 x r Q -rr pKa= 1 0.645 = Ka ) HA v - - 4 .76x l 0-7 = P H : - log-r : 6 .32 + O H- - + 0 .125 0.125 A+ H zO lnitialm mol:0 .250 m Final mo l: 0 . 1 2 5 + + pH: p Ku l ogffi : r o.64sl oghi# : 1 0'64 o f A -: c m Vb: V ":5.00 L: F ormaloncentration F ' : f fii# (25.0 l=)(0 . 0 1 0 0) : 8 . 3 3 x M m l 0 -3M 30.0m L A- + H 2O i : F ' -.r 12 : H A + O Hx x : K 6: K e,lKu 4 '42" l 0-4 Kw r Zu= 1 0.00 nl-: E xcessOH-l : t (5.0m LlQ.0I! ! M ) = 7 .14x l 0-3M ' : S.On,'t_ A cid-Base itrations T 65 tz aa o+ a I ve The s mooth a urve f or t he g raph was c alculated xactly u sing a e spreadsheets d escribedl ater i n a this c hapter. T here i s n o b reak a t the e quivalence oint b ecause p pKa f or t he a cid B H+ i s t oo h igh w hich i s a notherw ay o f 1n o 8 7 6 0' l 3456 Vo( m L) 10 sayingt hat t he a cid i s j ust t oo weak. ( $-*"1 10-10. r irration eacrion: r \-,/ - + oH ; ( -)NH2 \-/ B + H 20 v ,: t 00 L m BH+( pKa: 4 .601) 0 m L: B H+ i l 0.100-x B + H+ x x x: K u : 2 .51 x l 0-5 p H : 2 .80 + H rO x2 : Ka ) 0.100_; 1 .57x l 0-3M = BH++OH----+B V 0.100 e: 1 0.0 t.: m Initialm mol: 1 0.00 1 .00 Final mol m 9 .00 1 .00 : : pFI p Ku(forBH+)+ l og-'tr = 4 .601*rogffi 3 .65 0.500 e:50.0m L: p H : P Ka:4'60 V qoo V 0.900 e:90.0m L: p H = 4 .601+ l ogffi : 5 .56 Z ": 1 0 0 .0 L : F': [ B ] : m B +H zO F ' -x *2 Ft --r 1 0 . 0 mo l : m o 0.050 M 2 0 0, L uo:* = 3 . e 8x l Q -lo + B H+ + o Hx :K6 = x: 4 .46x 1 0-6M p H:8.65 K,,, lH-l : ;:2.24, lO-e = A cid-Base itrations T 67 HA l0 -13. oHx Ax + H zO Initial m mol: 5 .857 Finalm mol: 5 .857-x pH : e .l3 : p Kunrog f fi - - s .3s+logt8#; = j r : 2 .077 mol m : rym| toH-l : o .oer8M 10-r4. ( a) Titration eaction: A + O H- - -- A - + H 2O r H mmolH A = m molO H- r equired r each e: ( 27 63m L)(0.093 l M ) : to V . 8 2.592mmoL c oncentration T he o fH A'rffi## : 0.0259 2M. (b) At t hee quivalence p oint, hei nitialv olume f 1 00.0 L o f 0 .025 2 M H A t o m 9 hasb een iluted p t o 1 27.63 L a ndc onverted A -. T he f ormal d u m to concentration- i s( 0.025zN ,uffi e ofA : 0 .020l M . 3 (c) Becausehe p H i s 1 0.99, OH-] : 9 .77 x l 0-4 a nd w e c an w rite t I A -+HrO x (0 .0 2 0 3 1 -9 .7 7 l 0 -4) = HA + O H- \ 9.77 r c-4 9 .77\ r c-4 (9.77 " l 0-4)2 -- 4 .94 x l 0-5 f, _ ,.\b K ^: t HAlfoH-l tA-l K* f r, 0 . 0 2 0 3 1 -(9 . 7 7 ,l,0 -a ) PKa: 9.69 : 2 .0 3x l Q -1 0 ) (d) F orthe1 9.47-nL oint,wehave H A + p Initialmmol: Finalm mol: O H- - ' t .826 A - + F I2O t.826 2.592 0.766 tA-t t .826: 1 0 . 0 7 + pH : p ,(u l o g f f i : 9 . 6 e+ lo gf f i 10-15. T itrationeaction: + H r- - --) B H+ r B V a: 0 I . 00 5.00 9 .00 9.00 8 .05 K u : 1 0-5'00 9.90 7. 00 V":10.0 mL 1 0 . 0 0 1 0 . 1 0 12.00 5 .02 3 .04 1.75 pH: I I . 00 9 .9s 68 Representative c alculations: 0mL: B + H 2O=B H+ O H0 .1 0 0 x x x t2 Chapter0 l 0.16fi : 1 g- 5' oo = x =9.95x l 0- 4 M M K,,, lH*l:;=pH:11.00 1.00 L: m B + H+ Initialmm o l: 0 . 0 1 Finalm mol: 9 .0 ---) BH+ 1.00 1.00 : 9 .00 + l og 9.0 = 9 .95 1.0 pH : p KaH++ l osJ * ." t, [BH+] 10.00 L: B H+ i = B + H + m (100 LX0.l00-x) m llO m l x x Kw orto-t; = K6 =.tr:9.53 = p H:5.02 a x 1 0-6 10.10 L: e xcess m [ H+] 't2 10 B a (0 . 1m L X l. 0 0M ) : 9 . 0 8 x l 0 -4 M = p H: 3 . 0 4 110.1 L m a ao 4 2 46 %( m L ) a a 10-16. T itration eaction: H3CH2CO)+ H + - --+C H3CH2CO2H r C FindV s:( I/eX0.083M ) = ( 100.0 l)(0.0a00 ) = V ":47.79 m L 7 m M mmolH CI mmolN aA 0 m L: A -+H 2 o = H A + OH0.0400--r x x * o: E : 7.46x l Q- r o x2 0.0400 _ , = , (b > .x : 5 .47 x I 0 -6M p H= - t o g f = 8 . 7 4 72 Chapter| 0 10 o O. a a a n 024681012 % ( mL) l0-20. No. w hen a w eaka cidi s t itrated ith w a s trong ase, b 70-21. Yellow, g reen,b lue t hes orution ontains - a t c A thee quivalence p oint. A s olution f A - n tust avea p H o h a bove . T 10-22. Therei s n ot a n a bruptc hange i n p H a t v ". T hecoror f t he i ndicator o w ourd change eryg radualty. v 10-23. (a) c olorless- -+ p ink (b) I f y ou d o t hee xperiment s rowry, t oo t he p inkk eeps adinga ndy ou k eep f adding oreN aoH t o t urnt he s orution ink m p a gain.T herei s a s ystematic errori n w hichi t a ppears m oreN aOH t hat i s r equiredhans hould e. t b 10-24. Two c andidates:resor ed ( orange* c r - r ed)a ndp henorphtharein ( cororress r ed). - -, 10-2s. q) 30000 20000 = Endp oint 10.727 L m .= 10 000 fD o o 0 -10000 -20000 -30000 1 0 .6 9 1 0 .7 0 1 07 1 10.72 1 0.73 1 0.74 10.75 1 0.76 Volume ( mL) E C c! ...
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This note was uploaded on 04/06/2011 for the course CHEM 2011 taught by Professor Buhlman during the Spring '10 term at Minnesota.

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