Hwk 4 Solution EnvE 321 Sp09

Hwk 4 Solution EnvE - 2.25 3275 ppm[C02 = KH Pg 2 0.033363 mol/L-atm x 275x106 atm = 9.17 x 10-6m61/L using(2.46[H 12 = K[C02 mm 10-14[H]2 =

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Unformatted text preview: 2.25 @ 3275 ppm: [C02] = KH Pg 2 0.033363 mol/L-atm x 275x106 atm = 9.17 x 10-6m61/L using; (2.46): [H+12 = K] [C02 mm + 10-14 [H+]2 = 4.47 )(10‘7 x 9.17 x 10-6 +10-14 : 4.11 x 10-12 [W] = 2.0 x 10-6 pH = -log [H+] =. log (2.0x 10-6):5.69 @ 600 ppm: [C02] = 0.033363 mol/L—atm x 600x106 atm = 2.0 x10‘5 mol/L using (2.46): [H+ )2 = 4.47 x 10-7 x 2.0 x 10-5 + 10-14 = 8.95 x 10-12 [W] = 2.99 x 10-6 pH = — log [H+] = — 16g (2.99 x 10-6) = 5.52 2.26 Begin by writing the full set of equations that must be satisfied: [H*][HCO_; -7 “(CO 1 — = K, = 4.47x10 mol/L (2.37) H+ CO * LTfi—ELO 3]] = K2 2 4.68x10‘” mol/L (2.38) [ca2*][cof‘] = KSP = 4.57x10’9 mol/L (2.39) [H‘] + 2[Ca“] = [HCOS‘] + 2[CO32‘] + [OH‘] z [HCO3'] + [OH'] charge balance [C02(aq)] 2: KH Pg : 0.033363 mol/L—atm x 3602(10'6 atm = 1.2 x 10‘5m01/L Solve for the concentration of Ca2+ using (2.37), (2.38), and (2.39): K K [W] K [H+]‘ from the Charge balance: [H.“] = [HCO3‘] + [OH‘] — 2 [Ca2+] Pg 29 [H.]__K.[c0.]+1o-” 2 Kiwi: "‘ [if] [H+]_K,K2[COZ] 2 KSP[H*]3 I + 3 _ —14______________ ..() [H ] —KI[COZ]+10 KIKZICOZ] 2x4.57x10‘9[H+]3 [HT = 4.47x10'7x12x10‘5 +10“ ——-——_,——-—-—:fi—-——_—5 4.47x10 x4.68x10 x1.2x10 3 [11*]2 =5.37x10‘” -3.64x10“3[H*] or [H+]3 + 2.75 x 10-14 [H+]2 = 1.47 x 10-25 lmagine pH > 7 for example: [H+] < 10-7 then 2.75 x 10-14 [H+]2 < 2.75x10-28 which is negligible compared to 1.47 X 10‘25 so we will ignore the [H+] term... lhen [H+]3 z 1.47 x 10-25 so [W] = 5.27 x 10-9 pH = — log (5.27 x 10-9) = 8.3 2.30 a. dichloromethane b. trichloromethane c. 1,1—dichloroethylene Cl Cl C' H I I I I H—C-CI H—C-CI C=C I I I I H CI CI H d. trichlorofluoromethane e. 1,1,2,2—tetrachloroethane f. 0—dichlorobenzene CI CI CI CI I I I | FflC-CI H-C—C-H /C| I | I CI CI CI g. tetrachloroethene(PCE) h. dichlorofluoromethane CI CI CI I I l C == CI) F - C - H I I C! Cl CI Pg 2.11 2.31 Z33x ——> :a + 2253 a:262, b:86 fix ——> [3 + 3,:Y a:32,b=l6 2.32 64g——>32g—>16g—>8g——>4g—92g—>1g elapsed days: 60 120 180 240 300 360 days ...
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This note was uploaded on 05/26/2009 for the course ENVE 321 taught by Professor Dolan during the Spring '09 term at Oregon State.

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Hwk 4 Solution EnvE - 2.25 3275 ppm[C02 = KH Pg 2 0.033363 mol/L-atm x 275x106 atm = 9.17 x 10-6m61/L using(2.46[H 12 = K[C02 mm 10-14[H]2 =

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