cee115-fall02-mt1-Sedlak-soln

cee115-fall02-mt1-Sedlak-soln - Count CE 115: Water...

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Unformatted text preview: Count CE 115: Water Chemistry Name: 50; Mflgflj Midterm Exam 1 Fall 2002 This exam consists of three questions on a total of six 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, Pco2 = 103'44 atm Constants needed for this exam are provided on page 6. You may tear this page out and use it for reference. If you are uncertain about the wording of a problem make a reasonable assumption. Write down any assumptions made to solve problems. Question Score 7% =4 0‘) 7 f? =3“ C‘U .:.B 50 60 TD 60 90 100 Score 1 1. Water Treatment and pH (35 points] A drinking water treatment plant uses water from the hypolimnion of a stratified lake. Initially, the water has a pH of 7.0 and an alkalinity of 0.5 mEq/L. Prior to treatment, the plant operators decide to equilibrate the water with the atmosphere by spraying it up in the air. a. Predict the pH of the water after it has been Sprayed into the air. talk 2 411?” (w Cir-mae- ru Albtuwry Add’ 70 ml, Fwtutuwuj Mitt filing a fi/o‘VM = [M‘J- £4") + [55037 ’ “2047‘? [H(@‘)>>[(4"} [W‘s/312;] [wiry] Antwan“ feature To [#6057 " J‘x/O‘qfi ‘5 (350;?) 3 [Ck/46:7; (m) [W]: kk‘f -65r‘ -/ : um. ‘ [0 3-1/0 '7’?. J “W7 JIM flmwfl' " WM an: l0 ’1 /M=¥-‘i5i b. Estimate the amount of carbon dioxide that was lost or gained by the water during this process. (Hoof) rfllkww 3" Ajkfin‘f : “1C1” I‘. : Cr _.___.KI WW: CT = MW ~. aux/0% 7, 0.2m- Elm-"v.5 fuwtu - . «V - tn, 7' Cr *.__._._4"‘/0 M 4.17% #1 j 0'7?2. T4416. . . I 1'6 . . . . . c. Llst two assumptions (not Simpllficatrons) that you have made 1n your answer to thIS problem. — 3’”? I (Mm—a. wvnrr/amj . Afl’vuc’ EQWCIAIUVM IJ' AC,“ng ‘ “02.11001”, OH; #7 Am: 7a: om Mex/es AFFc‘tr/dfiA/k. 7 2. Drinking Water Treatment (35 points) A different drinking water treatment plant wants to add sodium hypochlorite (N aOCl) as a disinfectant to their water. a. The NaOCl stock solution is produced by adding NaOCI to pure (deionized) water to give a 500 mM solution. The solution does not contain any dissolved carbon dioxide. Estimate the pH of the stock solution. [wry/J [M : [ox-1+ [oer] €41.61 Na]. = pa, = [Hog] 9 [od'] mac: [Hod] L + [44*] : [0!4‘) «19W [Nady (4+) = [M‘] M om. .. m .. a hfiffl“) fig fiJJUMé'I [71”) (<- [04’7 (r4an05: my cm}ij 014415- 74) Iago} (0.36., ,2 K4 >> “’0 (M’Me‘ mm Al. am I j k «mne- «rum; ‘41 0dr : a} { —-— —.9 ’ l k... [#1) W) - log)" [0"7/046 [0-10 64* 0a "1 10! T [0 {lg : ,0. b. Estimate the volume of 10 N HCl would you need to add to l L of the stock solution to achieve a pH of 8.0. Assume that the stock solution remains isolated from the atmosphere. Ignore the effect of dilution of the stock solution with acid. 5 0100!] 4 {14”} : [M7 {[07 [Ft/6' «2/ Cf“ term + awn/Pita Wow 45 49,7er [0(7: [Had] tap)- [04‘] : If)— Mr 4- HI“) "Q 567’”) 01+} 3 5.26:9?“(0571 4-10-7—[0‘6 :. 0'I‘IZM am a c l/é‘é‘fi 0.1% M 14* Mix/x x; =0.Iw_iq= Xena!“ L - may“, c. The drinking water treatment plant operators add a very small volume of the stock solution from part a (no acid added) to treated drinking water. The treated drinking water is equilibrated with the atmosphere and has a pH of 8.3 before hypochlorite addition. After the hypochlorite is added, the water remains in equilibrium with the atmOSphere and the pH of the water increases to 8.5. What is the total hypochlorite concentration in the water? WHEN me“ Moot Is A6660 THE" 41 k 1‘ H 196/7. Mk; *2 46km” + 06/1. MW : [011‘] ~[#*)+[ch;] + my") 1: [Hw~ = ~Mr _ J) I “K” 40L : m 0 lryf/afoy .2!“ 1H" [per-'13 : 41%,: : {WWW/M} 5/1/6017“ lag?" may 5: [#097 + [00"] A: “It/’60.. am c 4'75 [Hr] (- : 6/7 _ - . . , “MN? MW ’0 “akin/UWMW’T 0d7- " ‘ [0"{376 __/0_¢_7‘ [H’Ji-k a r. '('7‘_ ‘7-9‘ ~ “T €753- = “W ’m - f—-——'—'—-—-- 3. Dissolved gases (30 points! Hydrogen sulfide (H28) is a gas that is produced under anaerobic conditions. Measurements of H28 in soil gas (i.e., the gas in the pores of soil particles) can be used to predict the concentration of sulfide in groundwater. a. Use the data below to estimate the Henry’s Law constant for H28 at 10°C. S ecies G /mole H /mole H2S(3) -33.56 -2063 H28 :1 -27.87 -3975 EM Heuny‘: Mu): Hbicfl 3 Mia“ (H: [A45] PH; “a” 3"" Aéc=(56:;fim)‘(£§fimnr-J : '2 7.91%}:116) 56' :- {Kr-57 fir/mag! I“ k: exp z 47' (6’13 17' J/Mafifiqfimtfl L1,, 243cc). 10: 14 /¢ ha M" H CaItAEcr To lfl’c -' b. You measure a partial pressure of H28 of 0.001 atm in the soil gas. Assuming that the groundwater pH is 7.0, T = 10°C and the groundwater is equilibrated with the soil gas, estimate the total concentration of dissolved S(-II) (i.e., [H28] + [H8] + [82]). Note: The pK values given on the last page have been adjusted for temperature effects. p IL!) ~ 1‘ = (at) KL: =k Ir: [dcf]+f/{y) J-[f/ (Fucziflfig_£ll r: r?” a FHPNLYJ’ J- kflk’hfulj— __ L r [m] by»; L 1‘ {LP} ‘- [HLZJJKI 1H1. i r 5; .:.- (mm/xi 10~3+M)[ I + 12:??? [JIM/0-14.77 2-. 3' gym r1 /0' {0"9 Unless specified otherwise, assume 7 = 1.0, T = 25°C, P = 1 atm, Pco2 = 10'3‘4‘1 atm R = 8.314 J/mol K Equilibrium Constants: (Under standard conditions) H20 = H+ + OH” I<...=10'l4 com, + H20 = H2C03* KH = 10"-47 M atm" H2C03* = H+ + Hco; Kl = 106-35 HC03‘ = H+ + cof- K2 = 10""-33 HOCl = H*+ 001' K3 = 10”-60 (At 10°C) M = H++ HS' K4 = 10'6-73 Alpha values for H2CO3 H al haO alhal al ha2 Magi)“; 6. 6.91E-01 3.09E-01 1.44E-05 6.1 6.40E-01 3.60E-01 2.12E—05 6.2 5.85E-01 4.14E-01 3.07E-05 6.3 5.29E-01 4.71E-01 4.40E-05 6.4 4.71E-01 5.29E-01 6.21E-05 6.5 4.14E-01 5.85E-01 8.66E-05 6.6 3.60E-01 6.40E-01 1.19E-04 6.7 3.09E-01 6.91E-01 1.62E-04 6.8 2.62E-01 7.38E-01 2.18E-04 6.9 2.20E—01 7.80E—01 2.90E-04 7.0 1.83E-01 8.17E-01 3.82E-04 7.1 1.51E-01 8.49E-01 5.00E-04 7.2 1.24E-01 8.76E-01 6.49E-04 7.3 1.01E-01 8.98E-01 8.38E-04 7.4 8.17E-02 9.17E-01 LOSE-03 7.5 6.60E-02 9.33E-01 1.38E-03 7.6 5.31E-02 9.45E-01 1.76E-03 7.7 4.27E-02 9.55E-01 2.24E-03 7.8 3.42E-02 9.63E-01 2.84E-03 7.9 2.73E-02 9.69E-01 3.60E-03 8.0 2.18E-02 9.74E-01 4.55E-03 8.1 1.74E-02 9.77E-01 5.75E-03 8.2 1.38E—02 9.79E~01 7.26E-03 8.3 1.10E-02 9.80E-01 9.14E-03 8.4 8.73E-03 9.80E-01 1.15E-02 8.5 6.93E-03 9.79E-01 1.45E-02 8.6 5.49E-03 9.76E-01 1.82E-02 8.7 4.35E-03 9.73E-01 2.23E-02 8.8 3.43E-03 9.68E-01 2.86E-02 8.9 2.71E-03 9.62E-01 3.57E-02 9.0 2.13E-03 9.53E-01 4.46E-02 ———-—___.__________ 6 ...
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cee115-fall02-mt1-Sedlak-soln - Count CE 115: Water...

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