cee115-Fall05-mt1-Sedlak-soln

cee115-Fall05-mt1-Sedlak-soln - Name 0C(fl—IOU Fall 2005...

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Unformatted text preview: Frequency 12 10 Name:, 0C (fl—IOU Fall 2005 CE 115: Water Chemistry Midterm Exam 1 This exam consists of three 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 y = 1.0, T = 25°C R ‘8.314J/molK Davies Equation (in water at 25°C): logy = —0.5z2[ 10.5 1+1"-5 where I = ionic strength (M) —— 0.21] If you are uncertain about the wording of a problem make a reasonable assumption. Write down any assumptions made to solve problems. H20 = 1‘]:+ + OH- Kw=10-l4 H2C03 = HCO3_ + H+ K] = 10-635 HCO3— = CO3} + H+ K2 = 10-1033 A table with alpha values for the carbonate system is provided on the last page of the exam. You may detach the last page if you wish to refer to it during the exam. Question Score Total _L_L l_-._L_.'. 10 20 3O 4O 50 60 7O 80 90 100 Score 1. Effect of disinfectants on drinking water pH 140 points) Chlorine is effective in the disinfection of drinking water because hypochlorous acid (HOCl) and hypochlorite (OCl') are toxic to most waterborne pathogens. The pKa of HOCl is 7.6. (Ca(OCl)2(S)). Calcium hypochlorite is relatively soluble and quickly and dissolves in water. Estimate the pH when 0.2 mM Ca(OCl)2(S) fully dissolves in water that does not contain dissolved carbon dioxide or any other added acids or bases. '76,": Ion)me Can 60, WW“; on" a 757“; IV flfl’b’t’M "f "ll/{0‘45 @ a. One way in which chlorine can be added to drinking water is as calcium hypochlorite Ref. rite/tr 0M7 f Lil/0va 2 mm m Wk! Ht 0 0 H“ a. o H‘ r w- ow MM 707 H96=fH’*J+[#0c1]~/'orf‘j "0r #W 2 0 [Margy zNickel“ flaf tum" mu" m rf-K s/Jt’ m1 761%] t ' , ' : -: _. z ' 7" -' / *[ZNI‘ C I f. L mam :— Tor 56M 0 5/ Key 1” ) fZ’M‘] ) [{ j (flaw?) 7 [4f] 0 'Ca/s J'IMm'f’CIcefx‘mJ': Adle [Jam] exfltcf‘ [I flange >‘) ] 4: h 7 - — 4+ 6/ ‘7 , “I Ur 0 "U: M7, hi) [78:] ' (KL? r [W £1.21?“ “:70 710 6 i ( am ,0- I' W H '/<’ Hfl” ,' H“) w . l </ T/ [ ) [It] 1*] : '~ ‘1'. ll) b. Set up an equation in terms of constants, known terms and [H+] that could be used to I L determine the equilibrium pH when 0.2 mM Ca(OCl)2(s) dissolves into water that initially has an alkalinity of 1.5 mM and a pH of 7.2. Assume that the water does not exchange gases with the atmosphere. / ("m/Mrm’r/od FKOM (I:- (0 CM; Alkfin‘lc : fllénuir = ['ft/O-J/L) + 7(a)qu : qx/O-{f/q Alma“! : 491/0"/4 -: [WW] v-[Hr‘] 4 [#(flf/fifc'flf ‘7 y l m ,- Cr ROM mil-rim (Md/17mm ,4! km -: [OH-j - [4+] +54”, CT f m1 C7. lav/H z. c [’/#'('/9,‘/>')[mf‘j II' fli/kM/tr A: of: CT . _ ' A? (:7, T. fl/IIM‘T‘ \ /,S—>(/O H I! V 0C.- 04-}; c. Chlorine also can be added to water by bubbling chlorine gas through the solution. The dissolution of chlorine gas occurs according to the following reaction: Clz(g)+ H20 9 + H+ + Cl- alkalinity of 1.5 mM and an initial pH of 7.2). After some time, the pH of the water is 6.8. Assuming that none of the chlorine has reacted with dissolved substances in the water, estimate the total chlorine concentration (i.e., [HOCl] + [OCl']). 4%.: —‘ M... -- m. = ('w-z'ww Maw 7(wa + My 7* L 54m Hate" 0F Held/18cm 0:50 If Ac’mmflm/ig/J fly / M,“- UF “"3 WWW Dfl‘dmrw’ AIR #7 //m9¢e, O Chlorine gas is bubbled into water that initially has the composition given in part (b) (i.e., |L /,j‘x10-!M LAW/0"}; "05/, = L/ «- [Ii/‘7 r 569 f 34% Cr loafwad. p14— «éfi .4 <- . .J. M..Y/O..JM_ €66} : ‘ v [0 +61. life/,pr r2;2,1"2¥1fl1/,H¥/0 +67l/JU/fll/r hardly-104% r @1654” nan/z)"; - fi/Wi’qi’ ’"hfiflZ/T 0691— = [WM/09M 2r IO'J'K 7 I, l : (__.. _ :} w /‘ )Iéela /0-;_&+ lamélf’ 4" ML: d. When drinking water contains bromide (Br') the following reaction readily occurs: HOCl + Br" 9 HOBr + Cl' . r \ CD The pKa of HOBr is 8.6. 1 mM NaBr is added to the water described in part (c), allowing the above reaction to occur. Will the pH increase, decrease or remain constant? Explain your reasoning. U0 #6!“ eFfiec 7 ad ,4c. k. Emu TWI‘ £64k I‘M/V. I ’4’th : /- wind/4 ~0A’rr 5 (17%?" (HUer 97* 5099' *K 0’54 I] mgr Ar H ‘. ~ WWW fl 6 a; [IAI/WOI’“ [Far/0&5: r : fl‘ urn," 7.0 I] 4 wem<g/L Ara/I3 73/MA/ #047. 2. Preservation of cyanide (35 points) Hydrocyanic acid (HCN) is a common waste product released during ore processing and other industrial processes. HCN (pKa = 9.24) is relatively volatile and the possible release of toxic HCN gas is a major concern for many industrial waste treatment systems. Volatilization of HCN also could result in underestimation of total cyanide concentrations in water samples if the samples cannot be analyzed immediately after collection. To minimize HCN volatilization and loss from water samples, the standard method for cyanide analysis recommends raising the pH of water samples to 12.0 immediately after collection. a. A wastewater sample from a mining operation is expected to contain 0.5 mM of HCN. Ignoring the possible contributions of other acids and bases in the sample, how much NaOH must be added to bring the sample pH to 12.0? TM: pal/cm ('tm is waived m 4 writ five/“(WI rep. slit-ms V ~I 0 +/' 'r/Lo / M" ace (4* HCN ml" #c‘U 7m" #3 [W] «(a/r) ~-f('A/j 61/ yr, (1/ For m s 14"» AM #661164 '70 667* 70 WNW : (""1" 1M)" r /T V q H: .- ,1, V b", 7'07“ Hm/ : R7434 "49' " .m' ' 7274/4) :HLUS—x/(J’iq = Mfr/0' ‘M M105, _‘\ b. Assuming that HCN is the only form of cyanide that can volatilize, and that cyanide volatilization rates are directly proportional to HCN concentrations (not total cyanide concentrations), how much slower will the volatilization of cyanide be in the pH—adjusted sample compared to the rate of loss prior to NaOH addition? Va” [cf/l #21190“ M (‘0) [RCA/JIM” «; {my M ('06,, : Ill/16%? demure 4cm «mi Milky t H. 0(‘0 #fYZqu/l’i 4 . I‘m : pay/rug : fl.0(0/77 a HAW}e AS FAJT 45’ (my C/UAG’IL (WIT/ML, cowflrf/w/I. 4 c. It is likely that water samples will be stored at 5°C. Recalculate your answer to part (b) at 5°C ignoring any effect of temperature on mass transport. The H g values for the species in this problem are listed below: Species Hot ng/mol) A _ u k_ _ Wt a, @ HCN 107.1 HKU '« wa‘ (/1 (a I: /0 CN‘ 150.6 A H4 » _ /m6 ’ c, H20 ~285.83 F MM o tsz H+ 0 OH' —230.0 . . -(W WW "'5 060 4T it 9 kMZ/fl/ A{Kn : 7350 i-_l) _. *meoffléd J.” i__ kn) ’l 7» 7: gait/awe W" 3‘1"“ ’9 :13? AH” : man—xver : L/xrtf/MI : ~ 4163’ ' ~I’) K” 1 0,131! if}, : 1,5; 7~WU C 10"“0 a“ ': ., ': fl-flfié/ a (7.5/72 0F IIUI7/I4L Mic /,éZ1»—</0“"’+ I0""" d. In your solution to part (a) you assumed ideal conditions. Show how you would more precisely estimate the amount of NaOH needed to reach pH 12.0 by including ionic strength effects in your calculation. It is acceptable for you to set up the solution-without actually making the calculations. TOT Hu‘ [HT/“[0117 ~[nuj we Ala-I) rd Arrow/r HA I ‘ ‘ _ (Ow/(t0 Fa/L , thv/fj-d‘ffzv—{Q- a‘ [#f')‘ biaf . _ , ~ _ - ,_ . _' , w [(7 Fm 16H} IQ, -dH,aM_ ~ am I'm/Mmp .. [oH) .‘MJWZ ~ ' w t' .. - ~ H" OH' oc+‘ 1:74, ; (It/r : C/(t:[’f’]c”'r t V CC I. C' k a (1? 9’9"“) (4: k 4 Fir-e- 3M“,— ~Iw 44.0 . _ l0 [0 _ C 7197' NW 5 '1—' "v f 7,. ‘ ._ __ A , h, :,' w l M la we 2;”? (x? 5,10 do [I] I", ‘ M 1‘ (+1- ;4 UJ-GAAVQ.‘ FCQIM~ ‘ 5 X :0? 3. Aerobic metabolism of organic compounds (25 points) Bacteria can metabolize benzene (C6H6) in the presence of oxygen according to the following reaction: C6H6 + 7.5 02 ‘9 6 C02 + 3 H20 a. Before benzene is added, the water had a dissolved oxygen concentration of 0.25 mM, an alkalinity of 0.65 mM and a pH of 8.0. At the start of the experiment, 0.1 mM of benzene is added. What is the total concentration of benzene remaining after all of the oxygen has disappeared? _ .. Y .u1, g fifth our!) PEA (6’45 0’- : 3.!!!70 JM (6746 War/3, @ [fix/0d“ (OF/c win/n — JJNIOTM : Aé'rr/U‘rM 69' “5 Karma/ma- 413nm 6L out-‘0. b. What is the pH "of the water after all of the oxygen has disappeared? @ Rmcrmu Hm N0 Hw/M'cr 0U flit/«turder flfaerrou :«Jcne'ruc’f C,— «97 [ 3,:‘x/0““’M - zua""’,q h—K ‘ 1r /,x,cr / LKL C,- AH‘u/q' = Afr/0%: : [my ~- [th + (mm-"l r 1/64, 1] 6.3‘K/o"”’p1 -~= (my a; c, “rm-«H “WW” ; aux/0"‘VM 67C. L _ _ ‘C {‘ACT 61.477 7 H33 5 rACr/ 2611‘ r 6/] 7' MAW! “ arr/film : (My ~ [Hf/7‘ [#65737] F 2170”: ‘7 L K’ “1 A, . I ‘ 5"“ ‘0 M — (WW) oq/g.mm.W--%/ a: mwflm _ y H- ,. . so, ‘ V ’ I f K6 71‘76’ r1 FILOM 7019c€ 0N fl-}/‘flf'("‘:é.,fi’ @ c. Bacteria also can metabolize 3,4,5-trichlorophenol (pKa=7.7) in the presence of oxygen according to the following reaction: C6H2C130H + 5.5 02 9 6 C02 + 3 C1" + 3 H+ Assuming the same experimental design as in parts a and b (initial conditions of 0.25 mM dissolved oxygen, an alkalinity of 0.65 mM and a pH of 8.0 and the addition of 0.1 mM of trichlorophenol) what is the pH after all of the oxygen has disappeared? Tm; fill/WNW ’5 Mom? (‘owfluflfi’ylx Hm any Mace" 0,1: 75/7 mad/zer [Mk t [H 3 Meter. ,4L50 Cerf/LC/Zao MTG—AI Mn. 4“ “MU/WU— &T(/‘f ACflL : ZIS‘UOW‘IH : 21:};XIO'WN {if . ‘ .' AW 1 ' n fill/(Flow? " “fl/um ’ M’k =14 W M ' few await] ‘ fi/J’éml '7 = [Mr/wart} F [Hwy/'1‘ M6173? “ [TC/’7 1’ [‘HCdjil)f (Tc/47 " CF f mic/l TC/Ft'r J’TIM K10 5‘. Hex/0 FM x V [‘1 “a.” Flu/Ac (IMO/1‘ (mu 5 0V.(CI~TACT) + 04TH, (TC/50%] a i. z - icy , x “ e ---v wax/0 "M a". (MM!) mama to] 1' 04’,“ (uni/0 H/ a, — WM WOW/M '\ "I" x . “ » s. Sl/M'V/o M - Kl (9.4%) M) tacmfl (sit/i340 M) I0“ TIL/m, +€AI1ch (/ IL" ’ULS mm 0 [hth 6/? Stand” (plows! 5‘ ‘1 5225110” 0 MN 6“ 7.6w10‘” (mm (1H: (‘35,, For the carbonate system 6.05 6.10 6.15 6.20 6.25 6.30 6.35 6.40 6.45 6.50 6.55 6.60 6.65 6.70 6.75 6.80 6.85 6.90 6.95 7.00 7.05 7.10 7.15 7.20 7.25 1.44E-05 1.75E-05 2. 1213-05 2.56E-05 3.07E—05 3.68E-05 4.40E-05 5.24E-05 6.21E-05 7.35E-05 8.66E-05 1.02E-04 1 . 1913—04 1.39E‘04 1.62E-04 1.88E-04 2.1 813—04 2.52E—04 2.90E—04 3.33E-04 3 .82E~04 4.37E-04 5 .00E-04 5.70E-04 6.49E-04 7.3 8E-04 6.91E-01 6.66E—01 6.40E-01 6.13E—01 5.85E—01 5.57E-01 5.29E—01 5.00E-01 4.71E-01 4.43E-01 4.14E-01 3.87E-01 3.60E-01 3.34E-01 3.09E-01 2.85E—01 2.62E—01 2.40E-01 2.20E-01 2.01E-01 1.83E-01 1.66E-01 1.51E—01 1.37E—01 1.24E-01 1.12E—01 7.35 7.40 7.45 7.50 7.55 7.60 7.65 7.70 7.75 7.80 7.85 7.90 7.95 8.00 8.05 8.10 8.15 8.20 8.25 8.30 8.35 8.40 8.45 8.3 8E-04 9.51E-04 1.08E-03 1.22E-03 1.38E-03 1.56E-03 1.76E-03 1.99E-03 2.24E-03 2.52E-03 2.84E-03 3.20E-03 3.60E-03 4.05E—03 4.55E-03 5 . 1 213—03 5.75E-03 6.46E—03 7.26E-03 8.15E-03 9.14E—03 1.03E-02 1.15E-02 1.29E-02 1.45E-02 1.01E-01 9.08E-02 8.17E—02 7.3 515-02 6.60E—02 5.93E-02 5 .3 1E-02 4.76E-02 4.27E-02 3 .82E-02 3.42E—02 3.06E-02 2.73E-02 2.44E-02 2.18E—02 1.95E-02 1.74E—02 1.55E-02 1.3 8E—02 1.23E-02 1.10E-02 9.80E-03 8.73E-03 7.78E-03 6.93E—03 ...
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This note was uploaded on 07/11/2009 for the course CEE cee taught by Professor Monteiro during the Fall '05 term at Berkeley.

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cee115-Fall05-mt1-Sedlak-soln - Name 0C(fl—IOU Fall 2005...

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