cee115-fall01-mt2-Sedlak-soln

cee115-fall01-mt2-Sedlak-soln - CE 115 Water Chemistry Name...

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Unformatted text preview: CE 115: Water Chemistry Name: 1015 Midterm Exam 2 Fall 2001 This exam consists of two questions on a total of eight pages. Answer all questions in the space provided. Use the back of the page if you need additional space. One sheet of notes is allowed. Unless Specified otherwise, assume 7 = 1.0. T = 25°C, P = 1 atm, P002 = 10-15 atm Molecular weights (gm/mole): Ca=40.l C= 12.0 H=1.0 O=16.0 If you are uncertain about the wording of a problem make a reasonable assumption. Write down any assumptions made to solve problems. H20=H+ +OH' K“.=10'”M2 COzrg) + H20 = H2C03* KH = 10-1'47 M atm'l H2CO3* =HC03' + H+ K1=10'6'35 M Hcog‘ =c032' + F K2 =10‘“‘~33 M Question Score V2 ‘K A4 f LJ 2 n A (‘1) 1 2 3r 4- f L) 25? B H) 2 2 r0 8" ( q 2. ‘10 C + f ‘1 < 90 C (2-) Tm“ l. Solubility and Speciation in Sediments (55 points) As we discussed in class, reduced sulfur (S[-II]) and reduced iron (F e[II]) are important to the toxicity of metals in marine sediments. The following predominance diagram depicts the speciation of Fe(II) in the presence of S(-II) at a total Fe(II) concentration of 1 uM and an alkalinity ofS mEq/L. s1- refers to total sen) (i.e., [H25] + [HS'] + [52]). Fe2+ + OH' = FeOH’ log [5, = 4.5 Fe2+ + 2 OH' = Fe(om"2 log [32 = 7.4 Fe2+ + 3 OH' = Fe(on); log 3 = 11.0 Fe(OH)2(s) = Fe2+ + 2 0H log K50 : —15.1 FeCOMs) = Fe2+ + 0032' log Kso = -10.? res“) =Fe2+ + S" log Kso = -18.1 H28 = H“ + HS' log K.“ = -7.0 HS‘ =H+ + 32- log Ka2 = 43.9 6‘ r . é 4,. a2 -. .1. «I ~ 7% {carats} 2 3-. a a: pa a. What is the concentration of dissolved Fe(II) at: A (pH = 6, sT = 10"5 M) B (pH = 8.5, sT = 10"” M) C (pH = 8.5, ST =10's M) [This page may be used to complete your calculations from part a.] Pam: AIAccmuuG m Tiff Haunt, Uo IounJ AM“ ME‘QWZ THEAEFME' \_/ Fem = “m ‘ ’MH [(#1) poi”, 5! 4T B) Riff, Cflur’la‘ul- EMU flamrc: A: C E50,“, (aw/Law Jam/gm n/ EM; = [E (‘7 " (ROM? 4. [RP‘OLOZ'I- (RKWQI‘) ks. ‘ 10"“ (ems) {Fm : km 14, 54 EM" ' Eff“? 4Q L!- +5L- km E‘Haeékwi‘ ,9 Ha? 'éH‘I-ofif In; 553,417.02! Edgar-PH" = NM“): [54575 7M7 'I'IE0;7= £11 = 1341‘?! Czflk” VIM/‘74 Lowe: [mldfljflrf I” L (11%”? [IL/[”7 “Err [If] [EMF/7‘ : 3 = K J— m J + (JCS-7' 1'; a} 49%] m(/L” (T7, 0 re CH7 2'7 .. Ht! V‘s-v- 1' - EME— {/+L§Q’1’07021/0/0:) mzie . . v, in.“ : 71517074 FMUW War A kart/Ir WE' lap/0704119 06C;- [6”) 2 K __.___I( la El!!! art—2b.?” +5151 £4170 LfeT/fljj FMC, we'rlmuee Tmr arpH ff 4M2 (”(4/2 0461' - C, = 41k f‘x/o 3/1 ’95 I .. r _—"—"". -f #Tph’ YA" ML“ ~. l_0__ +10 +27) ’0 I .37 K‘ two—nr‘fH-IU may [0-126F+/0~:o.1: la“ 11' ”War 4.1 L [0—10.7- ‘L( 0M3, N {0714* u A“ Ce: 5.703x/a‘l1 1w Ag;S(s)=2Ag + s2 longo=-50.1 Ag +Sz=AgS logB1=l92 Ag + OH'= AgOH0 log pl = . Ag + 2 OH“ = Ag(0H)2' log [32 = 4.0 Ag + or = AgCl° log is, = . Ag+ + 2 C1' = AgClz' log 62 = 5.3 b. A water sample is collected from sediment under the conditions depicted by point B. In addition to sulfide, the water contains 1 mM Cl‘. Identify the predominant tom of dissolved Ag(I) at this point and estimate the fraction of dissolved Ag accounted for by each dissolved species. 4%,“ z (Aft) + 09.1% midi/”NM; (as?) 4— (1407 (1‘27 .c (4;, def} WW ( 1+ 5mm") =(Aj11/(H00Im’0 -sr IUHS'WJ‘] (“limo 0-”) (mt/if ’0 4+!) (Mm-103 “’01:; 9.17) “(fir/(1+, 10mm, 10“" “,4. 4 10"? f {0-4;} ”1/09,sz {ifs} H PRFJMWAUI' J'flé‘C/ff: WWW (9L WALL A - '9’”;- - £23, ”"7 _ m _ flip” - :0 3,;- 0§ ”jq é—z’hr}: :' I.” ’70 q A; f - la’rzfl- = $6140-: A/C/ ’ - [04],}. z - , g, ..___ JP I [gr-u}— '13:) W0- (6 [gflcfl- 3 A0310” ’0 c. Calculate the maximum possible concentration of dissolved Ag(I) at point B 141' l’omr 5‘ wt! 0431' Jame” NM" ,4 I“) {:XU'TI: L1°=Cflfivtaflj (@4231? ”II-(ll 0:39;)” :- [0-20.031 A’M’Ué' [15571-14 :{AJJ/Vflfitff 61"] [’0 MW Aura-fl «M71 rem ,F THE 1100 M'UN A] fine: Zfl‘fi‘J'J—lb [0'56” cat-my ,p 749' re‘ammrr 3 R”+I"~l’- r4") m'“ z i L ,2 ~12. 41mm law/gang haw—mac, law" a £4.11 ("U/=00 40 ) 4 i [R -/ 3/04'0'331 d. ChIoroform (CHC13) can undergo reduction to dichloromethane (CH2Clz): CHC13 + H+ + 2 e' = CH2C12 + C1- Chloroform is spilled in an anaerobic system with a low concentration of sulfide (point C in the figure). Write a balanced reaction for the reduction of chloroform by Fefill) under these conditions. Fe(III) is reduced to Fe(II). Under the conditions considered, the stable form of F6011) is F6(0H)3(5). Ehac [1‘ 65" [E ”h Raf CO!" é‘ £64m} W Rafi/)1“! f C-J—CUJL' (.1 560;“; 4, 304- 0fiu4ucf L Hm} Affl- ctr/war: CW: * ’1’” Le‘ a? cabaret-d“ “€60:ch 503’ €- Zfiwlf/f 4.1m!“ ~————\___£fl_x 2. law”, 4 W4 + Im- Mr 3 c 15/010,“) 2‘ cm; MM (4" gar IUD 0F 05:.- M0 a3 5M + W!" 2. Minerals in Open Systems (45 points) You are evaluating the possibility of using water from the Winnahoochamachongie River as a potable water supply. The concentration of Ca(II) in the water is 5 mM and the pH of the water is 7.4. You may assume that the river water is equilibrated with the atmosphere. Cacom = Ca2+ + co?- K50 =10“8 Ca2+ + OH: CaOI-F [31:10” a. To remove Ca(II) from the water, you decide to raise the pH by adding lime (112., Ca(OH)2(s)). If your target Ca(II) concentration is 0.1 mM, to what pH will you have to adjust the water to achieve your goal? Assume that the water remains equilibrated with the atmosphere after the addition of lime. Note: Lime is very soluble and will fully dissolve when added to water. : [(21053] ‘ a" 9- [Ca 13/ ' [[4019]? qud“ (Ca 3'] 4' {fact}? = I0 7” p' [to filo/1‘) - lawk‘iafl?‘ lfl‘l'fll . . t- . _ :‘r H<<l26 Mt! km ' [50"][503 )[G 17 ‘ E! M ”gamma (a [(7293 AKW'HUG. ([4 ‘/->> [Chat/r} [67' (Car In]: /’0 ‘1‘” ‘“ km km 5—.’ f L L [.Her MU;Z ’0 M‘fifl'slv )3 my?! -. , ’0 [giraf- (“W210 ,) / lg . ,4” PH 3-6 [Mom fiJJyMflTIOU- 'flHs'c [2%, )1? war: 141m" Ilsa/er 1F 74% Mac [105‘ (‘4 mi 1.. fat VET 67:04)) b What dose of lime (in mg lime/L water treated) will be required to reagh”EH7 thisC pH value? Onward“ 6M6“ 00" WM 41km” 49 flz’cgmm ’9 If ‘4L I d A g km“! (MM) atmmflpfliv) “kW/n, [M] {W} “”50: )4 1&0th "2: (H601) = khpwb )0‘51‘;0~IIV¥ Jr [#9 ‘ ¢____?__Io =lfl‘3'“ Mk”? : (”41'04‘h ”‘47 *lffflf‘] fiftims‘} 9. 10%}? my? 4w" )0 I” I" “75.59% = 10‘5”“ A’Qk a l0~ZJH H [0.3.]? .91 ”00/?6rMM1w/Mdéf “w ' T zwce qiktr x%c " [é'lmZ/L (hag/us; 0:00]? 3574 (Mada) c. Considering only Ca—containing minerals, estimate the mass of sludge produced per liter of water treated by this practice. / N QIO‘M '0 Q0!) IU Alum-e” a [All-Hf + Canada ‘- &£M41 = film‘s/t4 + Kilt/0‘11 - IJNO‘VH 1'- 3:7fix/0- 31"! GK stain/9‘3” my}, 100,100va 5450,“! 1 We. mag, (a ( [6110”.106- .407 #0:: WA 6’} d. The Winnahoochamachongie River water also contains 1 mM Mg”. Will you produce any Mg(II)—containing sludge when you add lime to the water? MgCO3(s) = Mg2+ + (1032' K =10”-45 Emwere' 47} Marina} 14/4 PX : . _ _,g -_ , LIAfl/iW/qulz ,0~a,.i‘£«£@ MW‘m’ m lg .km $1“? fuzz-:1. /0"' - XI. = l Pxflcnv 4r fist-rm 4710M _ - wue flawwr 4 VFGLMMLF HIM/yr 0F 1?.60} ' cal/T450404 fit/0616: e. Discuss the potential benefits of removing Ca(II) and Mg(II) from the Winnahoochamachongie River water. diemwuo 11; “U Q,” by,“ mm; .rmnr r0 Faw 9mm + Mu Aka/A mummy 0F mm mm. ~IT' Alan bum. Imam? fcnuw w m- omma fl/M-‘J. ...
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