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 =

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

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.

Page1 / 4

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 =

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online