C C = CA = M. = Na HPO

Transcription

1 CEE December 01 FINAL EXAM Clsed bk, three pages f ntes allwed. Answer all Three Questins. Please state any additinal assumptins yu made, and shw all wrk. If yu dn t have time t cmplete a sectin, please describe hw yu wuld slve the prblem (withut using a cmputer prgram such as MINEQL). Miscellaneus Infrmatin: R = cal/mle K = J/mle K Abslute zer = C 1 jule = 0.39 calries 1. Buffers (5%) Yu have prepared a 10 mm phsphate buffer designed t hld ph at 8.1 (5 C, assume I~0). The tw chemicals at yur dispsal are Na HPO 4 and KH PO 4. A. Hw much f each cmpund (in mm cncentratin) wuld yu add t create yur buffer? B. If yu were t elevate the slutin ph t 1.0, hw much NaOH wuld yu need t add (again in mm)? A. This is a gd use fr the empiricial Hendersn Hasselbach Equatin C ph = pka + lg C Where C A is the Na HPO 4 and C HA is the KH PO 4 CA ph pk a = = 10 = CHA And since C A + C HA = 10 mm, 1 A HA CA C = A CA = CA 8.943C A = CA = M = Na HPO C = M = KHPO HA 4 4 B. Nw determine the new mixture at ph 1 and determine amunt f prtns that need t be neutralized, but this time we re clser t the third K a, s C A is PO 4 3 and C HA is HPO 4

2 C C A HA ph pk a = = = CA C = CA = C 1.501C = C C A A HA A = M = PO A 3 4 = M = HPO 4 S the change in prtns is: H + in Phsphates = * = M r 4.66mM But yu als have significant hydrxide in the amunt f 10 M r 10 mm. Therefre the verall hydrxide additin must be equal t the sum r: NaOH dse = = M r 14.66mM. Redx (5%) Sulfide and sulfate are tw frms f sulfur that cmmnly exist in the aquatic envirnment. They ften are fund in grundwaters with bth reduced and xidized irn. A. Write a balanced equatin fr the xidatin f hydrgen sulfide (HS ) t sulfate by reactin with amrphus ferric hydrxide, liberating free ferrus irn. B. Determine the net cell ptential (E net) and the equilibrium cnstant fr this reactin and cmment n what thermdynamics tells yu abut whether this is a favrable reactin in a typical grundwater and surface water systems. C. At what ph des the eqilibrium rest halfway between the tw sulfur species? Assume a free ferrus irn cncentratin f 0.56 mg/l. Which species (sulfate r bisulfide) will predminate at lwer phs? S we use redx #1 and #14, but we need t reverse #14 and cut its stichimetry by a factr f s that the electrns balance. 14 Fe(OH) 3(am) + 3H + + e = Fe + + 3H O SO 4 + 9H + + 8e = HS + 4H O +0.5 This requires that we reverse the sign n the E, but since the ptentals are expressed n a perelectrn basis, the change in stichimetric factrs desn t result in any additinal change in E. Red 8Fe(OH) 3(am) + 4H + +8 e =8 Fe + + 4H O Ox HS + 4H O = SO H + + 8e 0.5 Net 8Fe(OH) 3(am) + 15H + + HS = 8 Fe + + SO 4 + 0H O +0.70

3 n lg K = Enet = (0.70) = [ Fe ][ SO4 ] 10 = + 15 [ H ] [ HS ] Nw substituing in fr 10 5 M Fe + (0.56 mg/l) we get: [10 ] [ SO4 ] 10 = + 15 [ H ] [ HS ] [ SO4 ] = 10 [ H ] [ HS ] S when the qutient n the left is unity, the ph must be 134.9/15 r 9.0. Belw this ph sulfate will predminate and abve it will be bisulfide. 3

4 3. Slubility (50%) Ferrus Sulfide (Trilite, a frm f Pyrrhtite) is a rather insluble mineral with a K s f abut One wuld expect this t frm in reduced grundwaters and sediments where ferrus irn and sulfide are majr frms f Fe and S. Fe(OH) (s) wuld, f curse frm under the right cnditins as well. # Reactin pk 1 1 FeS (s) = Fe + + S 18.1 Fe(OH) (s) + H + = Fe + + H O FeOH + + H + = Fe + + H O Fe(OH) + H + = Fe + +H O Fe(OH) 3 + 3H + = Fe + + 3H O 31.0 A. Prepare a cmbined Ferrus Hydrxide and Ferrus Sulfide slubility diagram assumng yu have a ttal sulfide cncentratin f 10 6 M. The xaxis shuld be ph and the yaxixs shuld be lg cncentratin. Shw lines fr the fur sluble irn species based n each precipitate. Identify the key precipitatin znes with Fe T lines. B. At what ph can bth precipitates cexist? C. If the sulfide cncentratin was t increase, wuld the ph where the precipitates cexist increase r decrease? D. Describe in qualitative terms hw the presence f 10 3 M ttal carbnate wuld affect the hydrxide and sulfide precipitatin znes. E. Describe in qualitative terms hw the presence f 10 6 M mercury wuld affect the hydrxide and sulfide precipitatin znes Develpment f hydrxide slubility diagram a. Using the equilibria fr ferric hydrxide frm abve: + [ Fe ] K = = 10 + [ H ] we get the fllwing fr the free aqu in: + lg[ Fe ] =+ 1.9 ph Frm this we use the hydrxide equilibria t get hydrxide species cncentratins: and fr the mnhydrxide: 1 Because f uncertainty ver K fr the sulfide system, this equilibrium are ften expressed in terms f HS rather than S. Hwever, t be cnsistant, I have cnverted it t the mre cnventinal Ks frm in a way that is internally cnsistent with ur value f fr the sulfide K. These pk values were taken frm Mrel & Hering,

5 + [ Fe ] K = = [ FeOH ][ H ] lg[ FeOH + ] = lg[ Fe + ] lg[ H + ] + + lg[ ] 9.5 lg[ ] FeOH = + Fe + ph + lg[ FeOH ] = (1.9 ph ) + ph + lg[ FeOH ] =+ 3.4 ph nw fr the dihydrxide: nw fr the trihydrxide: + [ Fe ] K = = 10 + [ Fe( OH ) ][ H ] + + lg[ Fe( OH ) ] = lg[ Fe ] lg[ H ] lg[ Fe( OH ) ] = (1.9 ph ) + ph lg[ Fe( OH ) ] = [ Fe ] [ Fe( OH ) 3 ][ H ] = + K = = lg[ Fe( OH ) ] 31.0 lg[ Fe ] 3lg[ H ] lg[ Fe( OH ) ] = (1.9 ph ) + 3pH 1 3 lg[ Fe( OH ) ] = ph 1 3 5

6 0 1 H + FeOH + Fe + 3 Fe Ttal 4 5 Lg C Fe(OH) (aq) OH Fe(OH) ph Develpment f Sulfidebased lines b. Using the slubility prduct cnstant fr ferrus sulfide frm abve: 18.1 K = [ Fe + ][ S ] = 10 b1. we get the fllwing fr the free aqu in: + lg[ Fe ] = 18.1 lg[ S ] And fr 10 6 M ttal sulfides: + lg[ ] = 18.1 lg[ T ] lg Fe S α + 6 lg[ Fe ] = 18.1 lg[10 ] lg + lg[ Fe ] 1.1 lg = 6 α α

7 Fr Sulfide Calculatins: 1 α = + [ H ] [ H + ] K1K K K1K Fr ph<7.0, α, r lgα ph [ H ] K Fr ph= , α, r lgα ph [ H ] Fr ph>13.9, α 1, r lgα 0 S fr ph<7.0, we have: + lg[ Fe ] = 1.1 ( ph ) Fe = + lg[ ] 8.8 And fr ph= , we have: + lg[ Fe ] = 1.1 ( ph ) + lg[ Fe ] 1.8 =+ And fr ph>13.9, we have: lg[ Fe + ] = 1.1 (0) lg[ Fe + ] = b. and fr the mnhydrxide: frm the Fe(OH) (s) based calculatins, we knw that: + + lg[ FeOH ] = lg[ Fe ] + ph And nw substituting in the FeS (s) based free irn equatin fr ph<7.0, we have: + lg[ FeOH ] = (8.8 ph ) + ph + lg[ FeOH ] = 0.68 And fr ph= , we have: + lg[ FeOH ] = (1.8 ph ) + ph lg[ FeOH + ] = And fr ph>13.9, we have: + lg[ FeOH ] = ( 1.1) + ph + lg[ FeOH ] = b3. nw fr the dihydrxide: 7

8 frm the Fe(OH) (s) based calculatins, we knw that: + + lg[ Fe( OH ) ] = lg[ Fe ] lg[ H ] And nw substituting in the FeS (s) based FeOH equatin fr ph<7.0, we have: lg[ Fe( OH ) ] = (8.8 ph ) + ph lg[ Fe( OH ) ] = And fr ph= , we have: lg[ Fe( OH ) ] = (1.8 ph ) + ph lg[ Fe( OH ) ] = And fr ph>13.9, we have: lg[ Fe( OH ) ] = ( 1.1) + ph lg[ Fe( OH ) ] = B4. nw fr the trihydrxide: frm the Fe(OH) (s) based calculatins, we knw that: 1 lg[ Fe( OH ) ] = lg[ Fe + ] 3lg[ H + ] 3 lg[ Fe( OH ) ] = lg[ Fe ] + 3pH And nw substituting in the FeS (s) based Fe(OH) equatin fr ph<7.0, we have: 1 lg[ Fe( OH ) ] = (8.8 ph ) + 3pH 3 lg[ Fe( OH ) ] =.18 + And fr ph= , we have: 1 lg[ Fe( OH ) ] = (1.8 ph ) + 3pH lg[ Fe( OH ) ] = 9. + And fr ph>13.9, we have: 1 lg[ Fe( OH ) ] = ( 1.1) + 3pH 3 lg[ Fe( OH ) ] =

9 Fe Ttal 0 FeOH + Fe + Fe Ttal 1 H Lg C 6 7 Fe(OH) FeOH OH Fe(OH) Fe(OH) 3 Fe(OH) 3 Fe ph Red lines are based n sulfide slubility and black are based n hydrxide 9

10 0 Fe H + FeOH + FeS Fe(OH) Fe(OH) 3 Lg C Fe(OH) OH Fe(OH) Fe(OH) 3 Fe + FeOH ph Nw cmbining, we find that the hydrxide cntrls at phs f and abve. Belw this level, the carbnate is less sluble. A. (30 pints) see abve B. (5 pints) The cexist at a ph f abut 11.1 C. (5 pints) ph increases; it frms ver a wider ph range D. (5 pints) it frms a 3 rd slid phase (FeCO 3 ) which wuld infringe n the FeS (s) zne mstly E. (5 pints) it wuld precipiate HgS quite extensively, depressing the sulfide cncentratins and thereby diminish the FeS (s) zne, allwing Fe(OH) (s) t expand int lwer phs. 10

11 4. Predminance (50%) Answer this r #3, yur chice Lead carbnate (PbCO 3(s) ) and red Lead Oxide (PbO (s)) are tw imprtant slid phases that may cntrl lead slubility in water. In the attached pages is a detailed slutin leading t a zinc hydrxide slubility diagram. Please use this t help slve the fllwing prblems. A. Prepare a slubility diagram (lg C vs ph) fr a water that is ptentially in equilibrium with lead hydrxide and lead carbnate. Assume the water has 10 3 M ttal carbnates (i.e., 1 mm C T ). Shw all sluble species alng with the Pb T line and indicate where precipitatin will ccur and the type f precipitate. T help yu, there is a full slubility diagram fr lead hydrxide nly belw. Feel free t wrk frm this t prepare the cmbined carbnatehydrxide diagram B. Belw is a partially cmpleted predminance diagram fr the lead xide lead carbnate system. This was prepared fr a ttal sluble lead cncentratin f 10 4 M (i.e., 0.1 mm Pb T ). Please cmplete this diagram shwing the line(s) separating the tw precipitates (C lines) and als determine and shw the last remaining B line that separates the PbCO3 precipitatin zne frm the Pb(OH) 3 zne (n precipitate). Shw yur wrk. Stumm & Mrgan Present the fllwing data in Table A6.1 OH CO 3 Pb + PbL 6.3 PbL 10.9 PbL s 13.1 PbL PbL s 15.3 They nte that the numerical values are Lg frmatin cntstants. Using these data yu can rearrange the equilibrium qutients t get the fllwing In equilibrium with PbO(s): Lg[Pb + ] = +1.7 ph Lg[PbOH + ] = +5.0 ph Lg[Pb(OH) ] = 4.4 Lg [Pb(OH) 3 ] = ph Sme mixed carbnatehydrxide species may, in fact, be the mst imprtant, but they will nt be cnsidered here fr the purpse f simplicity. 11

12 Pb Ttal 0 PbOH + Pb H + Pb(OH) Pb(OH) 3 Lg C OH ph In equilibrium with PbCO 3 (s): Lg[Pb + ] = 13.1 lg[co 3 ] Lg[PbOH + ] = 0.8 +ph lg[co 3 ] Lg[Pb(OH) ] = 30. +ph lg[co 3 ] Lg [Pb(OH) 3 ] = pH lg[co 3 ] 1

13 #B3a PbCO 3 (s) Lg CO 3T #B3b #B4b Pb(OH) 3 Pb + PbO (s) PbOH Pb T = 10 4 M ph #B1 #B #A1 Answer t A.1. 13

14 Pb Ttal Pb Ttal 0 PbOH + Pb H + Pb(OH) Pb(OH) 3 PbOH + Lg C OH Pb(OH) Pb(OH) 3 Pb ph 14

15 0 1 H + Pb + PbO Pb(OH) 3 3 PbOH + PbCO Pb(OH) Lg C OH PbOH + 10 Pb(OH) 11 1 Pb(OH) 3 Pb ph Answer t A.. Type A lines First, lets cnsider the bundary between the aqu in and the mnhydrxide (line A1) At this pint the tw are equal, s we can take bth equatins and set them equal: 15

16 Lg[Pb + ] = Lg[PbOH + ] +1.7 ph = +5.0 ph ph = ph = 7.7 fllwing the same apprach, we get the remaining A lines A: Lg[PbOH + ] = Lg[Pb(OH) ] +5.0 ph = 4.4 ph = ph = 9.4 A3: Lg[Pb(OH) ] = Lg [Pb(OH) 3 ] 4.4 = ph ph = ph =

17 H CO 3 HCO 3 CO 3 Lg C T PbOH + Pb(OH) Pb + Pb(OH) #L1 #A1 #A #A3 #L ph Type B lines First let s examine the hydrxide equilibria. By lking at the PbO slubility graph, its appears that nly free lead (Pb + ) and the trihydrxide (Pb(OH) 3 ) are the dminant sluble species in equilibrium with the hydrxide precipitate at a ttal sluble lead cncentratin f 10 4 M. B1: 17

18 S, first let s lk at the Pb + /PbOH + precipitate bundary. Frm the slubility prduct equatin: Lg[PbOH + ] = +5.0 ph And presuming that this sluble species is the dminant ne at the nset f precipitatin, we set the cncentratin equal t the Me T = 10 4 M: 4 = +5.0 ph ph = 9 this is in the range where the mnhydrxide predminates, s ur assumptin is crrect. B: Nw let s lk at the Pb(OH) 3 /PbO precipitate bundary. Frm the given slubility calculatins, we have: Lg [Pb(OH) 3 ] = ph And presuming that this sluble species is the dminant ne at the nset f precipitatin, we set the cncentratin equal t the Me T = 10 4 M: 4 = ph ph = 11.4 This falls beynd the A3 line, s we re in the Pb(OH) 3 1 zne, and the line is valid 18

19 Lg C T Pb + PbO (s) PbOH + Pb(OH) #B1 #B Pb(OH) Pb T = 10 4 M #A1 #A #A3 #L1 #L ph Inspectin f the carbnate/hydrxide slubility diagram leads ne t cnclude that free lead may be the nly sluble species that needs t be cnsidered fr the B lines defining carbnate slubility. B3 S, lets lk at the Pb + /PbCO 3 bundary: We may need t develp as many as three line segments due t carbnate s ph dependence.: Fr Carbnate Calculatins: 19

20 α + [ H ] K K 1 1 [ H + K = + ] + 1 K1K Fr ph<6.3, α, r lgα ph [ H ] K Fr ph= , α, r lgα ph [ H ] Fr ph>10.3, α 1, r lgα 0 Nw lking at the given equilibrium expressin: Lg[Pb + ] = 13.1 lg[co 3 ] And presuming that this sluble species is the dminant ne at the nset f precipitatin, we set the cncentratin equal t the Me T = 10 4 M: 4 = 13.1 lg[co 3 ] 4 = 13.1 lg[c T ] lgα Lg[C T ] = 9.1 lgα B3a: Substituting in fr the simplified lgα at ph < 6.3 Lg[C T ] = 9.1 ( ph) lg[c T ] = +7.5 B3b: And fr ph= , we have: 0

21 Lg[C T ] = 9.1 ( ph) lg[c T ] = +1. Nte that this is nly valid up t ph 7.7, the limit f free lead s predminance (A1 line) Nw we need t lk at the PbOH + /PbCO 3 bundary: Frm the slubility calculatins we had develped three line segments. We can cnveniently take the equatin fr each, retaining the ttal carbnate term in its general frm: B4 Nw lking at the next given equilibrium expressin: Lg[PbOH + ] = 0.8 +ph lg[co 3 ] And presuming that this sluble species is the dminant ne at the nset f precipitatin, we set the cncentratin equal t the Me T = 10 4 M: 4 = 0.8 +ph lg[co 3 ] 4 = 0.8 +ph lg[c T ] lgα Lg[C T ] = ph lgα B4b: And fr ph= , we have: Lg[C T ] = ph ( ph) 1

22 lg[c T ] = = Nte that this is nly valid up t ph 9.4, the limit f lead mnhydrxide s predminance (A line) At this pint, we re well int the PbO precipatin zne, s let s lk at the ther side where lead carbnate may reemerge at high phs. This is beynd the A3 line, s we must cnsider the trihydrxide. B5 Nw lking at the final given equilibrium expressin: Lg [Pb(OH) 3 ] = pH lg[co 3 ] And presuming that this sluble species is the dminant ne at the nset f precipitatin, we set the cncentratin equal t the Me T = 10 4 M: 4 = pH lg[co 3 ] 4 = pH lg[c T ] lgα Lg[C T ] = pH lgα B5c: And fr ph> 10.3, we have: Lg[C T ] = pH (0) lg[c T ] = > 11.0

23 Nte that this is nly valid beynd ph 11, the lwer limit f lead trihydrxide s predminance (A3 line) 0 1 #B3a PbCO 3 (s) #B5c 3 4 Lg CO 3T Pb + #B3b #B4b PbO (s) PbOH + Pb(OH) #B1 #B Pb(OH) Pb T = 10 4 M #A1 #A #A3 #L1 #L ph Type C lines 3

24 Finally, lets cnsider the bundary between the tw slid phases. This line will start n the lw ph side, were the B lines intersect (ph=9). This is well within the zne where bicarbnate is the dminant carbnate species. T determine this, we can pick frm any ne f the matched equatins. Each will give us the same answer, s let s just arbitrarily pick free lead. Nw we set them equal t each ther: Lg[Pb + ] = +1.7 ph = Lg[Pb + ] = 13.1 lg[co 3 ] +1.7 ph = 13.1 lg[co 3 ] lg[co 3 ] = ph r: lg[c T ] = ph lgα C1b: And fr ph= , we have: Lg[C T ] = ph ( ph) lg[c T ] = = C1c: And fr the carbnate zne; ph>10.3, we determined: Lg[C T ] = ph (0) Lg[C T ] = Nte that this is nly valid beynd ph 11.4, the upper limit f lead xide s inslubility (B line) 4

25 The cmplete lines are as fllws: 0 1 #B3a PbCO 3 (s) #B5c 3 4 #C1c Lg CO 3T Pb + #B3b #C1b #B4b PbO (s) PbOH + Pb(OH) #B1 #B Pb(OH) Pb T = 10 4 M #A1 #A #A3 #L1 #L ph And remving extra line segments and lead species that d nt exist at 0.1 mm cncentratin (e.g., thse under the precipitates, we get the final diagram: 5

26 0 1 #B3a PbCO 3 (s) #B5c 3 4 #C1c Lg CO 3T Pb + #B3b #C1b #B4b PbO (s) PbOH + Pb(OH) #B1 #B Pb(OH) Pb T = 10 4 M #A1 #A #A3 #L1 #L ph 6

27 0 1 #B3a PbCO 3 (s) #B5c 3 4 #C1c Lg CO 3T #B3b #C1b #B4b Pb(OH) 3 Pb + PbO (s) PbOH Pb T = 10 4 M ph #B1 #B #A1 7

28 0 1 PbCO 3 (s) Lg CO 3T Pb(OH) 3 Pb + PbO (s) PbOH Pb T = 10 4 M ph 8

29 Sme imprtant equilibrium cnstants: Equilibria Lg K Mg(OH) (s) = Mg + + OH 11.6 Mg + + H O = MgOH + + H MgCO 3 (s) = Mg + + CO CaCO 3(s) = Ca + + CO Ca(OH ) (s) = Ca + + OH 5.19 CaSO. 4 H O (s) = Ca + + SO 4 + H O 4.6 CaOH + = Ca + + OH 1.15 FeCO 3(s) = Fe + + CO AlOH + = Al +3 + OH 9.01 CdOH + = Cd + + OH 3.9 COH + = C + + OH 4.80 CuOH + = Cu + + OH 6.00 HgOH + = Hg + + OH NiOH + = Ni + + OH 4.14 PbOH + = Pb + + OH 6.9 ZnOH + = Zn + + OH 5.04 HgS (s) = Hg + + S 4.7 Sme imprtant halfcell reactins Equ# Half Cell Reactin E (Vlts) 1 O (aq) + 4H + + 4e = H O +1.3 Mn +3 + e = Mn Mn +4 + e = Mn MnO 4 + 8H + + 5e = Mn + + 4H O Fe +3 + e = Fe Cu + + e = Cu ½HOBr + ½H + + e = ½Br + ½H O O 3 (g) + H + + e = O (g) + H O Al e = Al (s) ½HOCl + ½H + + e = ½Cl + ½H O ½OCl + H + + e = ½Cl + ½H O BrO 3 +5H + +4e = HOBr + H O ½NH Cl + H + +e = ½Cl + + ½NH Fe(OH) 3(am) + 3H + + e = Fe + + 3H O SO 4 + 9H + + 8e = HS + 4H O S (s) + H + + e = H S (g) Zn + + e = Zn(s) Ni + + e = Ni(s) Pb + + e = Pb(s) ClO + e = ClO PbO (s) + 4H + + e = Pb + + H O

30 Prperties f Selected Elements Element Symbl Atmic # Atmic Wt. Valence Electrnegativity Aluminum Al Brmine Br ,3,5,7.74 Calcium Ca Carbn C ,4.50 Chlrine Cl ,3,5,7.83 Cpper Cu , 1.75 Hydrgen H Irn Fe ,, Magnesium Mg Manganese Mn ,3,4,6, Nitrgen N , Oxygen O Ptassium K Sdium Na Strntium Sr Sulfur S ,4,6.44 Zinc Zn Lead Pb ,4 Selected Acidity Cnstants (Aqueus Slutin, 5 C, I = 0) NAME FORMULA pka Perchlric acid HClO4 = H + + ClO4 7 Hydrchlric acid HCl = H + + Cl 3 Sulfuric acid HSO4= H + + HSO4 3 Nitric acid HNO3 = H + + NO3 0 Bisulfate in HSO4 = H + + SO4 Phsphric acid H3PO4 = H + + HPO4.15 Phthalic acid C6H4(COOH) = H + + C6H4(COOH)COO.89 phydrxybenzic acid C6H4(OH)COOH = H + + C6H4(OH)COO 4.48 Nitrus acid HNO = H + + NO 4.5 Acetic acid CH3COOH = H + + CH3COO 4.75 Aluminum in Al(HO)6 +3 = H + + Al(OH)(HO) Carbnic acid HCO3 = H + + HCO Hydrgen sulfide HS = H + + HS 7.0 Dihydrgen phsphate HPO4 = H + + HPO

31 Hypchlrus acid HOCl = H + + OCl 7.5 Hypbrmus acid HOBr = H + + OBr 8.71 Ammnium in NH4 + = H + + NH3 9.4 Bicarbnate in HCO3 = H + + CO Mnhydrgen phsphate HPO4 = H + + PO Sulfide HS = H + + S 13.9 Atmspheric Gases Gas Atmspheric Abundance Henry s Law Cnst 3 Nitrgen (N ) 780,840 ppmv (78.084%) 6.1 x 10 4 Oxygen (O ) 09,460 ppmv (0.946%) 1.3 x 10 3 Argn (Ar) 9,340 ppmv (0.9340%) 1.4 x 10 3 Carbn dixide (CO ) 387 ppmv (0.0387%) 3.4 x 10 Nen (Ne) ppmv ( %) 4.5 x 10 4 Helium (He) 5.4 ppmv ( %) 3.7 x 10 4 Methane (CH 4 ) 1.79 ppmv ( %) Kryptn (Kr) 1.14 ppmv ( %) Hydrgen (H ) 0.55 ppmv ( %) 7.8 x 10 4 Nitrus xide (N O) 0.3 ppmv ( %) Xenn (Xe) 0.09 ppmv (9x10 6 %) α fr a diprtic acid: α + [ H ] K K 1 1 [ H + K = + ] In: mles/l/atm 31

32 Lg C T ph 3