Chem271_F11_x1_combokey

Chem271_F11_x1_combokey - “IWLWBW ur ., e: rror. Jansen...

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Unformatted text preview: “IWLWBW ur ., e: rror. Jansen Kala: . Emmy of Maryland, College Park Your sin #; General Chemistry and Energetics ur ‘ #: Exam I (100 points total) October 12, 2011 You have 50 minuteS for this exam. N z Exams written in pencil or erasable ink will not be re-graded under any circumstances. (+0 Explanations should be concise and clear. I have given you more space an you should need. There is extra space on the last page if you need it. You will need a calculator for'this exam. No other study aids or materials are permitted. Partial credit will be given, i.e., if you don’t know, guess. Useful Equations: _ Kg '5 [If] [A_]/ [HA] PH = 4053*?) Kb = H+][HO']/‘[B] F=ma Jem+1=0 PV=nRT Kw = [11+][HO_] = 10—14 PH = PKa +10g([A_]/[HA]) PH (6-11) = (PKa1+ PKa2)/2 R = 0.08206 L-atm/mole K 0 °C = 273.15 K pKa = —log(Ka) K, = KART)“ P2/a3 = 47:2/MG ' x = m oor Bede: At the ed of I exam'mation time ; ease writ out thefoll W “I pledge on my honor that I have not given or received any unauthorized assistance on this examination.” +1 point extra, credit for filling in this box } a pHoflMHCl’L PH: “L9 ((MJ) ) Com-J - pOHofo.01MKoH (2 PM: “(35+ h; [m] * LUV / ' r pH of equal volumes of 0.1 M HOAc (pKu 4.75) + 0.1 M NaOAc LL} 3 pH of0.001 MNaOH ll Pow: +3 pK1i + pKi, = m 0" l H (for K1‘ and K, referring to a conjugate acid/base pair) [OH’] at pH 7 IO M . ’5 [If] at pH 6 lo M When Q > K, the reaction will proceed (circle one) forward. H H The Henderson—Hasselbach relationship is always +r “‘5 , sometimes 1.45 C & ( . K G “M The pH at the first equivalence point of a polyprotic acid titration is given by KC Score for the 0896 ‘ 27 60' 2x I 2 1 ts hort Answer 2 each i “X pHoflMHNO O P1424? W ' 1— P0“ 2 "9 [W] A pOH of 0.1 M NaOH pOH of equal volurnes of 0.1 M HOAc (pKa 4.75) + 0.1 M NaOAc q ‘ pk: Ll\}f+ [ca [VJ/[Mg] : («Wk/(9(1) :— ‘LQ-r/ imam/7741 pHof0.00lMNaOH H ’3 '-‘- ll pKa+pr= PM“) 0" I ‘ (forKaand Kb referring toaconjugate acid/base pair) 4 [H+]atpH7 lo M - -5 4% A- . a :[OH']atpH9 IO M (3- 1.0 Wl/tO 6‘ M " Jéw/Ewflj When Q < K, the reaction will proceed (circle one) 1 forward; backward. 2 The Henderson-Hasselbach relationship is always +V‘MC , sometimes Mgfié . . _ (“4 WM/ eWLm ./' I. eé) The pH at the first equivalence point of a polyprotic acid titration is given by L2 C 7“ E M4 2. 2 ' Score for the nage 2 an H ! .l_n E .n . Consider the pH obtained upon dissolving a weak monopmtic acid HA in water, as a function of its total concentration C0 and its K“. This is a problem you have done many times, here we are exploring a general formula. the equilibrium is of course HA :7 H" + A’ K, x [I-I*][A‘]/ [HA] (a; 12 pts) Initially, assume that “x” can be ignored in the denominator and show that pH =2 4/2 log(Ka x C0) 2 M W 4‘ 9:1 3f (xi:- Kata: Wt? @ ’X: W(hu1fl,¢0) @ _ pHéD—(W = «icy ‘3}. “0704(0) (b; 8 pts) Now, repeat the problem but do not assume that “x” is small, i.e. use the quadratic formula to. derive a general formula for the pH. ' ' @K. = fif—EI yzzKJa-w) :%G~K2’ Score for the name Chemistry 27] section 23xx Exam I 10/12 1 E (c; 4 pts) Show that the more complicated expression you just obtained reduces to the simpler expression from (a) if K“ << Co. (This gives us a more precise description of exactly when x is negligible in the denominator.) PM : ilog ('§<¢K<EWI4Q ’lél) @ 1022/- lza<<601 WW LlK‘iC-o >> KQZ M W46) >> KR. (H Pl42—lcy(é@w.a> 2 —/c§<‘/K¢GJJ3/—ZL/&7(I{q.aj (d; 6 pts) Physically, what happens to the % Dissociation of a weak acid as K, increases or C0 decreases? Give an explanation for the C0 effect based on LeChatelier’s principle or the dynamic balance of rates. A's Kai“ Col/2D WCWL® ® 1“: Ca flea/(AM F0 rle‘m I‘M A; Polk guide MM» - r (7.42 ‘0’ dissuaw xiv; [1,. W/ K“4Jfoc&./m l7 ‘4 eale Vii M 0“wa a." 3. 20 ts uffers and Titration Dilution Series of a Weak Acid/Strong Base Titration w v ‘ ’ i 0030:2186 5:12;: 2;; 56:: Iii/er - “l“"‘°”'“ 1"“ °' ‘l’m’ “lmm'm . . ._—co=o.oosu l / —' concentrations. of a weak ac1d .ouco = 0.0005 M ' - - ‘ ' I 0— C0 = 0.00005 M titrated With strong base. The x _ _Co = moms M axis is in terms of equivalents "" C° = “0000005 ” of base added relative to the I W _ _ “’2 MnIBnIMMiuI/zmm acid, so the actual concentration E of base added is also decreasing as CO decreases. As usual, we ignore dilution during each individual titration, so the total acid [ ] + J 18 consmnt' 0.0 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.0 1.1 1.2 Equivalents of Base Score for the 03 9e (a;6ts)WhatisthepKaoftheweakacidbeingused,andhowdoyouknow? @. «=5 — 44¢; a ,m J L Haw-,0ch ) . M firv‘v‘flm Cq/i/K/O—M ‘00 L 4.} 11* %'%"“/‘“ /°‘”"- @«e/M (b; 4 pts) Why does the pH at the equivalence point decrease as CO decreases? A qualitative answer is fine. a, A’l-KAqu/wa fattpl-MWL JoéLaUQWim/‘N wnsweag Pl «3 19ml. 45 a, 49er M£WWJWWM¢ Hagar-hi TWP/«WM— (c; 6 pts) The graph illustrates the critical features of buffers. How does it show us the utility of “10X” or “lOOX” reaction buffers in the lab? @“11; M19“ Cwm cm A M 0M M Mfi‘rflh’fiQ/zfig‘flcy“ A} Nil-W 4; ELM» film/«W Cay/[Majemug fl»- 4ww'M-4‘U. 114" % vertex/(amigc 6%.— a I @4554)? 00‘1"“ “‘7‘ fl‘ Wflfir Wffvok :Wuwz. (d; 4 pts) Explain why the pH at the end (12 equivalents, upper right area) of the C0 = 0.05 M titration is 12. M “PFC, = 0.05"» and M WW (tieyu’mA/L‘; 0.0 a {1.1. [barn]: 0064M I go 934i (ROUM/QMKJf—M _._——‘ (a; 6 ins) What is the pKa of the weak acid being used, and how do you lmow? [23w] ' (b; 4 pts) Why does the pH at the equivalence point decrease as CO decreases? A_ qualitative answer is fine. [23”] (c; 6 pts) The graph illustrates the critical features of buffers. How does it show us the utility of “10X” or “lOOX” reaction buffers in the lab? [23m] /‘ j .1 (d; 4 pts) Explain why the pH at the end (1.2 equivalents, top right corner) of the C0 = 05 M titration is 13. Score for the Dane ,H The proposed mechanism shown is the 0 .° 0 . 0 0 0 essence of catalysis by aspartyl 0 0 0 o proteases, a class of enzymes that e ‘— 0 includes HIV protease. They have a e e a classic bell-shaped pH rate profiles like those we have seen in class, with pK,’s typically at around 3 and 5. o + P" P. 31%“. “a; ,1 ,H @o 8 OH i o o I o," o o 0 G ..._.._.__. o o o G 0 O 0 9 9 o o e (a; 10 pts) Which residue is associaled with W mewdcmmic mm rampant Protease: the pKa of 3, and what is its function in the mechanism? In other words, why does the reaction fail when it is performed at a pH much below 3? Which residue has pKa = 5, and what is its function? Why does the reaction fail at basic pH? ‘W‘ [45’0218 i! 0590M) 5%) #1} /Ze f/v71, @ “M 5 la +w, «um/M (MI/a r” 4 =3- “ 3%- M £24 yin/an 17/ damn/c- WA“ Kmhmo/ @ «Ayala View « WW fiafl‘edz. xa Com; . Ma MD 7m as] may 2/; fr w WWW n» n M W? li‘” “yd/W’w‘ hemis 271 secti n2 xx Examl 10/ 2/11 7/ 5. {20 pts) Equilibrium Manipulations. I—IZSO4 HSO; _ Very weak No3 bases C1 ' We will confirm that any acid on the left column of the swag HNO3 , . . a“ 5 HCI table can protonalte any base that 18 below 1t1n the H O+ H 0 right column. For example, formic acid, HCOOH, H8103 I _ C1203- pKa = 375, should be able to protonate sulfite, 8032‘. Heloz CIOZ‘ The pKél of hydrogen sulfite (HSO3‘) is 6.97 . HSO; ' $042” (a; 3 pts) What is the net reaction for fomJic acid H2503 H503: protonating sulfite? H3113}: :ZPO“ 2' Q 09 9 HNO 7 —' NO - O 0 ‘9 2 ‘33:: 2 @ h C0 K +- 5 3 “C0 + “503 I—lCOOH HCOO‘ Weak CH3COOH 3 CH3COO' bases 'd ~3— — (b; 7 P13) Write down the base dissociation reaction for m 5 H220; " 3 sulfite and calculate 1ts pr and Kb. H50; 5032- 2 , _ e Hzpo; ripof- ) _ 0 y 0 ¢ 0 4.. 7 H00 C10 @ S 3 H” 3 ’10 NH4 NH3 HCO3’ c032- PO43’ @ pg :- («ma own = mg; 4 -: ? Very weak I 3 HS— 52— 1% ‘03 adds OH_ 02__ . ’1‘ Chemistry; The Science in Context 3Ie Figure 17.4 K —— Q 2012 WAN. Norton 8: Company, Inc. 5 i (c; 3 pts) Add the acid dissociation of formic acid and the base dissociation of sulfite to get a reaction including formic acid protonating sulfite. n‘e‘aou :2 1am Hcooe pr: («Mm/H) Join-147,0 :3 H303“? H0 .- l[b(§03 2‘) @é M cm” “031/120 6‘ H6006? magawn/vr Score for the 03 2e h——w__‘>m 4_4_ _ . . "——--_-. thmistry 271 Lsection 22x3 Exam l, 10/ 2/1 1 7 5. 12!! pts) Eguilibrium Manipulations. HZSO4 We will confirm that any acid on the left column of the S“??? HNO3 . . . a“ 5 HCl table can protonate any base that 18 below it in the H 0+ right column. For example, nitrous acid, HNOZ, pKa 2 H6103 3.62, should be able to protonate sulfite, 8032‘. The HCIOZ pKa of hydrogen sulfite (HSOS‘) is 6.97. H804“ (a; 3 pts) What is the net reaction for nitrous acid H2803 H3PO4 protonating sulfite? @ HMO; + soak—g: N0;+ H503». H Oi- VVeak CH3COOH acids HZCQ3 H28 5 H80; LA __ 1 fl H2130; Q 503 t H20 :2 [4503 + [40 H010; NH4+ ' (b; 7 pts) Write down the base dissociation reaction for sulfite and calculate its pr and Kb. HCO3‘ Q 2 [Lt-(ng 3’ W503”) HPO42‘ H20 9 m- M? = 7.03 acids OH- Z Chemistry: The Science in Context 319 Figure 17.4 @ 2 h ‘1 S ._ A g. O2012W.W.Norton &Company,lnc. a (c; 3 pm) Add the acid dissociation of nitrous acid and the base dissociation of sulfite to get a reaction including nitrous acid protonating sulfite. 'Hwo; —: Hum; M = Mam/ab) _ p 2— 503 2’ 4. [47/0 ~2— mog + H0 10> Mb (5’03 J ,- 2“ A + 142/0 2’ H62; MW; P/4’7Lf/4d” )9 2 [(4 [MW/2,) x [(5 (ft/L? 2] Score for the na 9e (d; p153); What ether equilibrium do we need to add to give us the net maqfiqfifiqm part, (a)? I .HkJelw’a H20 3/0”?" 1/’Ct./ j .2— 9. 9 “mo, 64 3 FM _. HCoerfa3 --»_ Hom- ##qu (e; 4 pts) Calculate the overall equilibde reaction constafit for the reaction of part (a). A (“Coo”) ' 2-) e ,w. K26, A P: .1042”:- 104'0’2' 10:" 2 [0'1" 2- 10 >100? "and h ' 27 ection 11/1 /8 \ (d; 3 pts) What other equilibrium do we need to add to give us the net reaction from part (a)? \ Ned f») 4414 H++M0F 2—3 WW [5; ’3 /16w "1 5‘ /0 (e; 4 pts) Calculate the overall equilibrium reaction constant for the reaction of part (a). Sum % gnu/LS -"—‘ HAW/Li 503 2-1:? ‘IUOL’HI- H505 —' @WW 5/ (67‘; 2 [(4 (HA/db). “Japflgz—J. Marx - - 463 I (144424.03) :— 10 3'u' l0 ' ,5"! :10 l . (HA 10.6.5") #235,. ‘2 IO 7— 10@ >(aoo 9d" [0 5 = 2210 Score for the Dage ...
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This note was uploaded on 01/29/2012 for the course CHEM 271 taught by Professor Staff during the Spring '08 term at Maryland.

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Chem271_F11_x1_combokey - “IWLWBW ur ., e: rror. Jansen...

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