cee115-fall00-mt1-Sedlak-soln

cee115-fall00-mt1-Sedlak-soln - CE 115 Water Chemistry...

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Unformatted text preview: CE 115: Water Chemistry Midterm Exam 1 rimm— Fall 2000 Unless specified otherwise, assume Y = 1.0, T = 25°C, P = 1 atm, P002 = 10'” mm Ifyou are uncertain about the wording of a problem make a reasonable assumption. Write down Exam 1 “Nu—w- Answer all questions in the space provided. One sheet of notes is allowed. R= 8.314 Jlmol K any assumptions made to solve problems. H20 = H+ + 0H C023) + H20 7-}12C()3‘I H2CO3* = HCO3- + H" HC03' = C032‘ + I-F Question V R RULE '- 1 31-90 HEDMU I 2 b? He 2 H m A Z 2'?- (3) 3 4 ‘ Z 10 {5) 3" 1’. 6L (9| Total I 2. ST 5 L} 5 ‘- 3 if; ( L] Kw=10'” [(3 = 10"-“"r K] = 10-635 K2 = 10-1033 Score AH‘J (kJImole) 55.8 ~20.3 7.6 14.9 1. AcidlBase Chemisfl (40 mints! (:1) Calculate the pH of a 3 mM solution of NaOCl (pKa = 7.5). '45:“4'6- Ase“: maug aye! ' Wu! ‘ M = to” '1 (M), M = [m « [adj mm; m = 5;. (w) = 0a; : (Had) doggy we“ H0314; rum: (“05045471 {m} glad '[00/7 Riguitc on. IIHFCIFY” I “'96” MW} ‘ [0K7 (£461: 70"“? [m - 1. H“ Van My To me 6M6. amscnv 17 If [Tim‘yw’ " (m) :fit’} WAY IJzFF/mcr To rzyflchY, {lymrFICAnoUh - 499mm em; Farm? “WWW” Lofl‘) )9 [19") Mod12>LH7 - Emu. AH» Illa (kc: NW”) 1, _ _ .r “300"”? "r 49.9? __ %w' =Fii1£ ’0 I” W?) T (b) A solution ofNa2(tartarate) (p .1 = 3.0; pKag = 4.2) has a pH of 8.0. Assuming that no other acids or bases are present, calculate the total concentration of tartarate. 0 0H “LT? H" +Ifr- Kai o Inf-#44,?“ km. tartarate USMC» 4A6, [HT‘J + 1M?) + (H? = {NH oer," + wqu Hm) : to 45’ .. my 27'] ,_ agile " ’0 a”) (ff-5’ I - ~———-—-—-- '3'— 6; ’1 061' ’.fif,/0‘4+Linlnr¥/o‘qj “— Fnom Tafice'au um" flfifit" bums cut/.6, CUM} 1* (W) = (OH? 1‘ [#T‘} + LIT"! (RMIU-CHUW L/fE dmtcfl‘fj HM = up”. = 2M1pum+4117 10(3) + um; f luv; Mm) : (My +(Hr‘j + LU“) 11’th [Hr‘) + {at} .-.- [UZHj (Mr/t 70 Me'} (c) Tartaric acid is produced by addition of H1804 to sodium tartarate. Calculate the volume of l N H2504 required to acidify a 1 Liter 0.01 M Na2(tartarate) solution to pH 3. Ignore dink“. usngfig‘tbe‘ ‘ u (ll-uoae HIGH,” firmwa Jud) (yaw m1 = tau-1+ Mr} it UP] + ling, 1 Zfia’vl «.- [wlj+(f{" ‘(0H‘)-[Hf'} —L[Tj (w) 2 1 TNT Fm. um' ..~ -k - r “1M7” QUUL, l: lTwrw‘CW} 2-5:) “I. M’ Ll‘r "' - .rr " 0,0: 2/0.0I)+[10‘9&(’l%-_ij s-(oag/iqu (2400 x/o ){ } 2509“) 2 _ __ Lam erc’cw I. 5‘ 47.019917: 0.01971 {.7071} : 0.00HH4 (W : act/rm [m I 9 mm, M964 W357 (d) A groundwater sample isolated from the atmosphere has an alkalinity of 10'3 eq/L and a pH of 6.35. The solution contains only dissolved C02. When it is removed from the aquifer and placed in an open beaker will it gain or lose dissolved carbon dioxide as it equilibrates with the atmosphere? In other words, is the solution over- or under-saturated with respect to the atmosphere? Show your work. WHEU 7H:- Aorta-4 )1 (WOW?! TD ATWOIl/F’t‘flf' fl/fi Mu. M CWIUC-C. urmuvf _ (flat 2: (04‘1-[5'1‘14 [#6041 + lit-501‘} r". I” gru— mzll‘fct: {#1017 x; 4h we" DWI} >> [Mimi] ' AM i (Wm ; “.07 = ’U'I’M . t~¢r€pht -'- Cr ‘-'— Qd’a‘it‘i“ we.) IVS'IEM! A“ *- {diff/L. In.) ‘v » f _ a .I I. ' t‘ .l ‘ ’0 ‘1 3 fl:er {3-0: L f - _ 1—7 “(’{7‘ ,0 filgopfi’ofl ‘ s, , — [0.5 Cr :- Lima I ‘ _.___m 95L '_ l “lodefiit. ‘ 2 Jinn. «l .- f 'I - I _._. 2. Alkalini 3S ints : As discussed in class, diurnal pH fluctuations often are observed in lakes. (a) Assume that water collected near the surface of a lake has a temperature of 25'C and a pH of 8.0. Ifthe surface water is in equilibrium with the atmosphere, calculate the alkalinity of the lake water. Ignore contributions to alkalinity from borate and other weak bases. 41k at [on - (H’) + [HMS‘J + 1- H0: V) : Lu - [H'j+ Ejkquk L klkgtfih (Wu: Mfr—- f “ML car :09 w- 31' ._ - -9. .u; . .. , ~ .I. u ‘ 1%“ - zofl @1141? +(2iro I0 3:0 /0 I U (IosDL ': 7:? fi___——.____ (b) During the night, the temperature of water collected near the surface of the lake decreases to 15°C and the pH decreases to 7.92. Is the shift in pH attributable to temperature flucmations? -Asrmc 4ft. :5 Want-“(ch W 7MP. -At>wrr ['3' En Tam/'- : imYJ. -1.) g "Alla (I. -1. } llIn IL 7: T, 6'Jfl/J'...,J“'k*' lift/r 195nm FM TL= tfth’ 7-, 3 WIN!" Kr 4.9"” (When/0‘7) '- l ’H' 297.45" .. Jim (Mn/o“) _ I4 : mwé J _._ m 6394:. If . ’0 40-36 “(’VJl’WXIa'f’f/o") ‘ {0 40.42,: L — s k 2 -('7€M(I:Vuc'rj: ’0- ll N .. - 471;:00 Demo“ ‘ '3 £22:ch ) j: {0 N3? J‘llff _. Jam WV“ mr -gmz 44711 4M 4“ til/kn: {Zlmdomt to 10 10 142W luv 110 7w.” ’0 {0-1411 {IO‘JSV " ‘ 917—9 :{flgj'L—M—JH a. «(A H {MC/(4406M ‘ YET TeX-(fl. I! I4 Wt}th €XflM4/m " m—thflf __ _ ____#__ _/’ __._—__ - f..- _"——'--‘=--__ (c) You collect a sample from a different lake during daytime (i.e., under the conditions listed in part (a)). The sample is stored in a closed bottle with no head space. After all of the organic matter has been metabolized by aerobic respiration, the pH of the sample is 7.2. Ifthe initial oxygen concentration was 8 m g/L, predict the final oxygen concentration. Assume that aerobic respiration can be approximated as 0120 + 02 = C0: + H20. wad 50L rs Mower», pl! l; Aur Alfi=aum4m wnmLLV, Oflt‘d XVI ft‘HI pH :9 A”; :. fir/0‘ #17 IL 1' ‘ - 3 #1 -J. I ‘ - CT, w; r 2-. kl! I}. 1.. _ .-.: ' = y 731/04!” 0 a; - (Z.Iixia"j ‘ in I A t rmuflrweaf: q = (mm {If/0;! t 11W] m “WWW; m MM”! )(KleVT-lfikgga. CWUO'”7/0“") r /0'3 = 7.9/u'0‘1c1 “WM”, CWT!" JHFCH, Mkzttrm‘fivflb ,o/f: 4‘1. AM WWWWZ = {My ~[I/7 p {((fQ‘j + 2(601‘] ‘4 k {g} - (Hf/“i 6r: 6T {' LWLCT ._- - Sty--1“ ~. cm?” Ho ’ " _ __. r. mix/0'? 0.975} + Hid/4 via") A :; _ . C’ Cusp»; CIT/g“— l5.flx‘-/O M‘- 549nm” 2536 HUI/1 3. Smoke gets in your lung; (25 mints): Nicotine is an alkaloid found in tobacco that is responsible for addiction to cigarettes. The structure of nicotine is shown below. N /°H3 \ / New} Nicotine can be protonated at each of the nitrogen groups, resulting in charged (+1 and +2) $3212 a) in“? w + M L, pKfiznio-a Hut .12 p H} 42L KH=5 xlO5 Mfatm (note: we have defined K3 in the same way as we have for C02). "‘ “2a? “at”: k“ (3) Assume that the av of the gs can be modeled as a water-coated surface. Assuming that a smoker breathes in air with Pco2 = 10'” atm and Pnimfinfilff8 atm, derive an equation that can be used to predict the equilibrium pH of water on the surface of the aveoli. Express your equation in terms of constants and [H4], EJJJE. Maw-7+ (w; +019 - [0W 4 mag} + "2.06187 KHZ-‘45! " {U1:kHPU Pu K7,. -. mm [aw]: LEM ; Uth [HA/'1 Km. Kat. kg, = (am..lu.~“/ 4m : [Wee my”! 12, lg! km. {a “Fifi/7t- WT LIHULQP‘, 1L rig) + PM + li(kltfiff§_ a, to... (“at [m1 m7 my 755w:— F'E/L (Ky H/ P—_ F—jffihfi (b) One way of analyzing the concentration of nicotine in air is to bubble air through an aqueous solution and analyze the concentration bf nicotine dissolved in water. Assume that equilibrium is achieved between the air and the water during this process. What would you add to water to increase the concentration of nicotine in the aqueous phase? Explain the basis for your answer. 0. . [1.1/“1+ (M + [u] 2 W13)? [My 4 k’Hfl} _.._— h‘lk‘l £61, _F;_F M701 0’ 4 315754 I“); {bk 777T —CFJ/_. Egg-EJ740710 AC/fl. (My "CU \ Alpha values: Carbonate system Tartarate system Idpha 0 21pm 1 spin 2 pH alpha 0 alpha 1 alpha 2 3.0 9.996501 4.465504 2.00051 1 3.0 4.047501 4.047501 3.050502 3.2 9.993501 7.074504 524451 1 3.2 3.645501 5.777501 5.777502 3.4 9.909501 1.121503 1.317510 3.4 2.550501 6424501 1.010501 3.6 9.902501 1 .775503 3.305510 3.6 1.672501 6.656501 1.672501 3.0 9.972501 2.010503 0294510 3.0 1.010501 6424501 2.550501 4.0 9.956501 4.447503 2.000509 4.0 5.777502 5.777501 3.645501 42 9.930501 7.030503 521 1509 4.2 3.050502 4.047501 4.047501 4.4 9.009501 1.110502 1.304500 4.4 1.517502 3.010501 6.030501 4.6 9.025501 1.747502 3.253500 4.6 7.102503 2.027501 7.102501 4.0 9.726501 2.741502 0.090500 40 3.172503 2.001501 7.967501 5.0 9.572501 4.276502 2.000507 5.0 1.365503 1.366501 0.620501 5.2 9.339501 6.61 1502 4.901507 5.2 5.733504 9.006502 9.006501 5.4 0.991501 1.009501 1.105506 5.4 2.362504 5.934502 9.404501 5.6 0.490501 1.510501 2.011506 - 5.6 9.616505 3.020502 9.616501 50 7.001501 2.199501 6.409506 5.0 3.003505 2.450502 9.755501 5.0 6.91 ZE-OI 3.000501 1 .4445-05 6.0 ‘l 550E435 1 5605-02 9.844501 6.2 5.055501 4.145501 3.073505 6.2 6.247506 9.901503 9.901501 6.4 4.712501 5.207501 6.212505 6.4 2.49 6506 6.270503 9.937501 6.6 3.599501 6.400501 1.192504 6.6 9.960507 3.965503 9.960501 6.0 2.610501 7.379501 2170504 6.0 3.971507 2.506503 9.975501 7.0 1.029501 0.160501 3.020504 7.0 1.502507 1.502503 9.904501 7.2 1.237501 0.757501 6.491504 7.2 6.303500 9.990504 9.990501 7.4 0.174502 9.172501 1.070503 7.4 2.510500 6.306504 9.994501 7.6 5.315502 9.451501 1.760503 7.6 9.996509 3.979504 9.996501 7.0 3.41 7502 9.630501 2.042503 7.0 3.900509 2.51 1 504 9.997501 0.0 2.100502 9.736501 4.554503 0.0 1.505509 1.505504 9.99 6501 0.2 1.303502 9.709501 7.257503 0.2 6.309510 9.999505 _‘ 9.995501 0.4 0.732503 9.790501 1.151502 0.4 2.512510 6.309505 9.999501 0.6 5.490503 9.763501 1.010502 7 0.6 1 .000510 3.901505 1.000500 0.0 3.435503 9.660501 2.057502 0.0 3.90151 1 2.512505 1.000500 9.0 2.1 34E-O3 9.533601 4.459902 9.0 ‘I .53 55-1 1 1.585505 1 .OODE+OO ...
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