Advanced Geochemistry of Natural Waters
Advanced Geochemistry of Natural Waters GEOL 578
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Date Created: 10/23/15
Advanced Geochemistry of Natural Waters LESSON 6 OxidationReduction E ectromohve Force the Standard Hydrogen E ectrode and tne Nemst Start Audio Lecture cmtsznmnmes kxmmgh maur Mm STRENG39ITH D REDUCING AND OXIDIZING z Fe e Zn2 Fe thch Wav Wm tne reaction go 30 e163 kcano e Zn 5 a stronger reducmg agent tnan Fe Fe Cu e Fe2 Cu AG 734 51 kcano e Fe 5 a stronger reductant tnan Cut came Ma mekmukw Wm WM ELECTROM 0T1 V E SERIES WEkst oxidant Strongst reductant Ze39 Strongest oxidant GEDL578 Advanced Geochemistry of Natural Waters Volimeter Anode e Caihode Porous paiiltion 03 va 50139 Cu 50 Zn Znquot2239 Cu 2e a Cu Figure 16 1 from Faure Schematic diagram of a ZnCu electrochemical cell The two metal electrodes are immersed in solutions onnSO4 and CuSO4 which are prevented from mixing by a porous partition GEOL578 Advanced39Geochemi shy of Natural waters ELECTROM OTIVE FORCE Electromotive force EMF The electrical potential generated by the half reactions of an electrochemical cell Consider Zn Cu2 lt gt Zn2 Cu At equilibrium the electrochemical cell has to obey Zn2 u2 K 1037 24 GEOL578 Advanced Geochemistry of Natural Waters AG0 2 23025RTlogK 508 kcal r M33 2 nSEO SS Faraday Constant 96489 C mol391 2306 kcalV391 mol391 E0 AGrOZS 508 kcalZ mol2306 kcal V391 mol39l llV GEDL578 Advanced Geochemistry ofNatul39al Waters lSTANDARD HYDROGEN ELECTRODE SHE o In order to assign a ranking of halfcell reactions we arbitrarily set E0 000 V for the reaction H2g lt gt 2H 2e39 when pH2 1 bar and pH 0 This is equivalent to the convention AGf H AGf e39 000 kcalmole 0 We then connect this SHE to any other electrode representing a halfcell reaction and we can obtain E0 for all halfcell reactions This is called the standard electrode potential 7 Table 163 Eleclromotive Series Presemed as I 39 I w n Elecuode Halfreactions in Order of 39 h Decreasing Slrcnglhs as Reducing Agems39 g aquot mumu 0 Anna 5 quotA d elm ad 5quot 9quot quot quot I W 2 lm 3 I 3 g c c o a u or one Li a Li39 H 04 III I14quot or s u x a K 2 7 94 N nquot on a n n a 2 9 llr39 I am 9 a 5x SK 1 90 n o n 39 2 v M c Ca 2 u zcr a 1 JI NI o 1 9 7 7 A II l39 Q Rb 111539 n e m n a n39 41 96 v lt yquot 3 7 A0 I39 o I In J Mg Mgquot 1 7 35 L lt Laquot 3 36 39H lhIH c u V is 1 ilk a If we always wr1te halfcell Th 4 Th 4 Bl Al Ai 357 7 70 react1ons w1th the electrons on U U 4 43 Mquot a Mnquot ne 419 the r1ght hand s1de E0 tells us Nb a N 1 I0 z vzv g 7 457 the pos1t1on of the react1on 1n the n a n a 0 6 g grquot 35 4m electromotwe ser1es relat1ve to F lt Ftquot 20 4 Cd a Cd 917 0 Co c CD Zr r 28 NI Niquot 1 7 24 Me a Mo Je39 20 SH lt Sn 167 l6 8 Ph Pb 1 T IJ l GEDL578 Advanced Geochemistry of Natural Waters lTHE NIERNlST EQUATION o The value E0 refers to the EMF of a halfcell reaction when all reactants are in the standard state eg for Zn lt gt Zn2 2e39 E0 is the EMF when a2n2 10 What if this is not the case EE0 23023RT10gUAP nJ In thws partmmar Ease AP an and n 2 5 2 3025 RT E E 39 1og2n Ls At 25 C we have m 1111 particular case 00592 logZn2 2 E 0029510g2n Or m the must general case at 25 C 00592 71 EE EE IOEUAP m gum a my MFA Mm DEFINITION OF Eh We de ne Eh to be the EMF between a ha ceH reactwoh m ahv state and the SHE For examp e Eh for the zmc reactwoh above is de ned bv the overaH reactxoh Zn 2H lt gt an H2g for Wh ch 00592 anqui 5 HT Eh E lo 2 However for the SHE by de mtwoh 9H2 1 bar and w 1 mo Lquot 50 Eh Equot 0029610g2n2 M Advanced Geochemistry of Natural Waters LESSO N 2 ACIDBASE CH E MISTRV MODULE 35 Numerical Equlllbnum Calculations Start Audio Lecture guis imgmasmmmehwnngmm Mm NUMERICAL EQUILIBRIUM CALCULATIONS Monoprotic acid What are the pH and the concentrations of all aqueous species in a 5 x 10394 M solution of aqueous boric acid 50Ha7 Steps to solution 1 Write down all species likely to be pr ent in solution H OH39 BOH3 BOH439 mama Ma metriqu mum w 2 Write the reactions and find the equilibrium constanls relating concentrations of all species H20 lt gt H OH39 K HWIOH 104 0 H20 BOH30 H20 lt gt BOH439 H H1BOm1 mow ii 8OH21H201 3 Gawain axmuw3 miwg 3 Write down all mass balance relationships 10394 M EB 30H439 5OH3 iii 4 Write down a single charge alance electroneutrality expressions H 30H439 OH39iV 5 Solve n equations in n unknowns gem a my MFA Mm EXACT NUMERICAL SOLUTION Eiminate OH in i and iv OH39 Kw OH39 KWW H 30H4 KWH H 39 30H439 KWH K w H H 7 BOH 1 V swam a gunmanquot Wm WM Solve iii for B0H3 B0H3 EB 130H439 HB0HX K ZBsB0HX quot HB0H439 KAZB B0H439D vi Now solve v for 13mm 130H439 KvH 11 130HI439 Hf KylH W mmz r KAEB KwH KAKW an 7x10391 H23 6x10quot3gtH1lt7X1039 0 We can solve thls by trlai and error computer or graphlcai methods From trlai and error We obtalrl W 6l1x1or7 M or pH 6 21 gem a my MN Mm OH39 KWlH OH39 10441076 21 OH39 1039779 M Bme H KNH B0H439 61x10397 7 162x1oE B0H439 594x10397 M BOH3 EB BOH439 B0H3 5x104 594x10397 M 499x10394 M swam m memmulw Wm WM APPROXIMATE SOLUTION I Lookfor terms in additive equations that are negligibly small mul plicative terms even 39f very c no eg ect ed H i and then BOH3 EB H B0HZ Wu H i H Kquot 50H H 0 H i 257m w Kimmw H2 mm 7 KA B o Tnis i5 a quadratic equation of tne tonn a 2 bx c 7 KA gtlt and can be soived using the quadratic formuia gmusn is mmgtqmunum Mus In our case thiS becomes Oniv tne poSitive root nas ariv phvsicai meaning n1 5 92 x 1077 We couid nave made tnis probiem even simpiet Because boric acid i5 a quite Weak acid i e very KA vaiue verv iittie otitWiii be ionized tnus BOH3D gtgt BOH EB e BOH3 5 x10quot M ammu m minimum mm mm em quotlleUI H1 gelEmma 7x10 A H rfmo39m 5x10 112 35 x 1045 7x104quot E Q E 10 7 M It is wise to check your assumptions by back substituting into the original equations H L 39 n ii matiou is probably justi ed because KA values are o en at 7 CALULATE THE pH OF A STRONG ACID 2 Mass actwon aws a103 H20 Ham 3 Mass ba ance Hon ct 2 x 107 M 4 cnatge ba ance H ct OH39 10 K H39Cl atom a my Mum1 Mm Assumptwo Ha Hat 5 a verv strong and so n1 gtgt OH39 and ct gtgt Hon Now tne or v source at Hand state tne tssoctatton ofHC 50 H CH apparent from tne cnatge ba ance 09 2 x10quot 3 70 and ct 2 x 7 HMS 5 ago Thus pH 7 OH39 KwH 1039N2 x 10394 5 x 10391 M H39HCV z n 61042 4 K N10 4 Han HCL 103 4x10 MM Advanced Geochemistry of Natural Waters LESSON 4 ation and Dissolution MODULE 2 Geochemwca wadeS and the Commonrlon Effect Start Audio Lecture govsumnmmmawmww Mm GEOCHEMICAL DIVIDES In the ast prob em me raho Moof Pb2 changed from 3gtlt103972gtlt1039E 15 to 2 EO4X103973 SOXIO39m 779 If Wu femte cohtmued to precwpwtate from thws somuoh as a result ofevaporauon or examp e the rano Moof 1Pb2 womd tend towards H v mty TH s an examp e ofa geochemical divide Tm somtmh s ehhched h mo ybdate and dep eted H v ead because e H H a rauo Moof Pb2 gt 1 Thus precwpwtatwon can O y cause thws rano to ncrease It can never decrease so me raho Moo sz cross Ifwe started wnh Moof39Pb2 lt 1 precwpwtatwon new never resu t m a rauo gt 1 Ma kltetraqu mun w as m Geochemical divide involving gypsum a Ca504p2H10 JV50139 v u I waI 4xur39 nxxoI Gypsum Pmdpllaied moles De lilzr Fxgue m 1 ram Faun 1991 pnmpns andApphcahms ufGeachzmmxy Imam 15xm1ana50317xmm c1503 xsxmua ygtl and Gawain 1Q1m4 wxmwmg Natura SO UUOHS of even swmme satts are more comp ex than the examp es We have ooked at so for Consider simmtaneous Saturatwon ot a SO UUOU thh both Wu femte DbMoOJ thh pKSu 1o poweHwte Cab 00 thh pKSu 7 94 Moo 9 Db2 MoOf M e Ca2 MOO 2 Both react ons cohthbote MoOf OUS to the SO UUOU b e ha actmtv Co centraho ot Moog must be the same for both eq m ha Sombmtv product expresswons become game 4 mew MNtmrA mes pb2Mooz 1on Ca MoOf 107W and ehmmatmg Moog We obtam pb2Caz 10151077 so we so the Concentrahon ot Db2 Wm be 23 7x109 Umes ess hah the Concentraho of Ca2 To determme actua concentrat on We star W th the expresswon M0031 whxch 5 vahd f Wwfemte and bowemte are the CNN sources ot pb2 Ca2 and Moog Now pb2 1039 MoOf Ca2 1077 wMoog ammo m kllemmulry mm mm so mm is much argerthan pb h the absence otoowemte pb2 Moog o v2 Km 10715 n v 107 s Gawain m 4354 gnome So pb2 10 M or about 4 orders ot magmtude more than m the presence ot powehrte However m the absence ot Wmfemte 239 v so Ca2 MOO v2 1077 v 10497 80 Ca 1039397 M or the Same as m the The commonrion ettect occurs when the presence ot sa t depresses the so ubmtv ot a ess so ub e sa t contammg a common on can a catron or amen Thrs ettect can be used m envtronmenta remedratron m 3 Q m m g c E gmbs r 4 mew new Mes REPLACEMENT We have estabhshed thata SO UUOU m equmbnum thh pomewe and Wu femte at ZSDC W H have Db2Ca2 107m at 1 2 g 2 3 E c I Q 3 g m a m 3 quot139 m some ore deposrts torm T rs phenomenon a so b edratron i am Advanced Geochemistry of Natural Waters LESSON 3 The Carbonate System Start Audio Lecture goiazarmmacamwumw Ms CONSERVATIVE PROPERTIES Conservative properties are those that are independent of pressure and temperature e g ANC ENC and CT it expressed as moiesKg Exampies of properties that are not conservative pH and tne concentration ofany individuai Species e g 2003 These properties are aiso independent of certain enanges in enemicai composition For exampie additio ofHZCOJquot or cnange 0 Dee WiH decrease tne pH and increase Acyj and CT but WiH not attect Aik because Hzco is tne reterence by Wnrcn Aik is de ned mum Ma antitruqu mu w Alk HCOS39 2c03239 OH39 H Acy 2H2c03 HCOS39 H OH39 Acidity is unaffected by changes in concentration of C032 so addition of salts such as Na2C03s or CaC03s will not affect acidity Capacth dwagrams r p ots WM conservatwe propemes e g 0 AM or Acv as coordmates and contoured W th pH or t rbonate specwes H Addmon or remova of sods or bases can be represented bv vectors For anv gwen pH AM is a Hnear mncnon of C T Vertwca hnes on these dwagrams wech ac d met c or a kah met c Utrat on curves 4 capacity diagram 5 mg 4 a 39nm smmm iz Mnrgan Aumcyvs c capamcy magma Ma a ataxmwkgm39mg EXAMPLE 7 MIXING OF TWO WATERS Two waters A pH 6 1 Atk 1 0 meqL B pH 9 AM 2 meqL are rmxed m equat proportrons What is the pH ot the mixture if CO2 rs os Readmg from graph CA 2 8 mmotesL CB 19 mmotesL errhg m equat proportrons wetds C 22 35 mmo esL A k212 15 meqL From graph We rea pH amaze 4 mm mm was mm u when mm mm EXACT SOLUTION To PROBLEM Wk Crwa 2amp2 OH39 5 W Cr WM 5 OH39 H Du 292 ForA a o 387 62 3 07 x 1075 50 c 1040 387 2 6 rhM ForB 610939 62 92 o c 2 x 1040 939 2 x o 0592 1 9 rhM For the mtxture 6 2 6 192 225 rhM and AM 15 rheq ah u 262a aja AMcT 1 52 25 o 66 and the above occurs at pH 66 Where a o 666 and 62 67 x 10 The pH ot a surtace Water wrth AME 1 meqL an pH oSrsto erarsedto H83 Cacuate the cornbosrtionat changes required rt thrs rs accornbhshed bv 1 Addmon ot NaOH 7 From the graph We read 9817 m ertrca hne up to pH 8 3 gwes us 87 meqL NaOHr 2 Addmon ot NaZCO 7 Need to draw hne wrth s ope 2 from rnrtra pomt pH o 5 and AM 1 meqL to rntersectron wrth pH 8 3 contour About 14 meqL or o 7 rnM NaZCOJ ammo Ma mambo mu ways 2 contmued Thrs prob em cou d have been sotved rnore easrw usmg acrdrtv dragrarn A Water rth pH 655 d 17 rnM as Acv 2 35 rne L Because addmon ot NaZCOJ cannot change acrdrtv the ength a honzonta hne segrnent over to pH 8 3 we ds o 7 rnM 3 EV rernova otco 7 Use atkahnrtv dragrarn because atkahnrtv rs mdependent ofcoz The ength ot a honzonta hne segrnent at AM 1rneg L orn pH65topH corresponds to o 7rnM C02 Addmanawa o v gammamcamewmmw we PH OTOSV NTHESIS AND RESPIRATIUN A kahmtv cannot cnange due to cnanges m p002 a one However assocrated uptake or excrenon of wow ddnng tnese processes cnange c arge ba ance and nence a kahmtv e g N033 NH Hpog ecornposmon of orgamc pg mterstma ake Water 1000 AME 12 meqL pH 0 a 104 M Assume reacnon cmsHmomNmp 10602 14w 9 106 co2 16NH HDOf39 name Red e d cornposmon CDSH2630DNSD ammo Ma quarriqu Wm WM Estrrnate cnange m tworstep process k 9 5 Q m m m m n o a or M C02 9 szw 14 53 Wk qua 2 0H r M 9 do747d ltlt 1 ago 747 10m 710 s 9 c 104 M However the C02 added pv decomposmon of QC addsOSx 1039 Mto so 1x1039305x10393 211x10393M 5 tpHu Aw 6 66X 1075 AME 12x103936 6gtlt1075 12 66 gtlt 10 ea Wk CT U1 0 600 1 1 K H1 10 l 10W1 H H 1667107u3 1 m 25 x 1077 PH661 Advaxmed Geochelnish v of Natural Waters LESSON 4 Precipitation and Dissolution MODULE 7 The So ubmtv of Do ormte sumdes and Phosphates sun Audio nun Condmons of do ormte precwpwtatwon are poorw under t Drecwpwtatwon of do ormte does not contro co osmons of hamm Waters Dwsso utwon of do ormte mav exercwse some contro Dubhshed vames Km range from 107165 to 10m quot mam quot lam cmnn39zoanum Emmang 310ng NE Dung CaMgC032 lt gt ca2 Mg 2mg 09 Ksm me 2ca2 2c03239lt gt 2caco3 09 K5 am CaMgC032 Caga 2CaC03 Mg 09 K exzhange 2 K 7 Kmmm Mg 1 mm 2 2 5 mm Ca 1 For somuohs m equmbnum wuh bom came and do ormte the MgCa ramo shou d be constant at a gweh T and P Assummg Ksuydm 5x101 we obtam Ksummm e 2 0x10 m agreementwnh some aboratory measurements 0 Fig 513 39om Stumm amp Mo gan Ca1 Mgz COZHZO at 25 C and 1 bar K5 for dolomite Seawater 39 39 quot 39 quot quot 39 39 L h nan um Advanten mammary nr Natural waters SOL BILITY 0F Sill LFIDES Estimate the solubility of uCdS at pH 45 and pH25 1atm 25 C I 1 mol L l We have the following thermodynamic data log cquotKPSOCdS S8 log cKHH25 105 log CK1HZS 690 log cK2HS 140 For now we assume that no hydroxide sulfide or carbonate complexes are form Emma Ahmad Ezndlgmlstry nf Natural Waters The constant Kpso refers to the reaction uCdSs 2H lt gt Cd2 HZSg 2 wK Cd pH2S 710758 p50 H2 7 S8 log Cd2 log pH25g ZpH S8 log cd2 0 9 log Cd2 gt148 cd2 1585X10 15 mol I1 cd2 178x10 8 ugL ppb yuanomnmcs mame nan su LFA39DES Constructa predommance dtagram m the svstem CdZtHZSrCOgHZO atpc02 1045 atm We need the foHong addtttona thermodvnamtc data 09 vasumdcol 09 003 09 KM 3 og zmcoz 7151 Ourdtagram Wm ptottog peeMS vet pH CdCOaCds boundary CdCOJG H28g e Cd85 H200 3029 tog K tog quotK954Cdcooquotermdsgtgt 6 44 r is 8 12 24 tog pmpee edscab boundary 1185 2H 9 mm H28g AK Cdf39tvat 107m 09 pm e5 8 09 mm 09 pHZS 2pH Assume Cd2 104 mo L1 718109 pH252DH a muzewwmm4 my We However to ptot thtg on the p dtagram We must Subtract 09 p 02 both Stdes of the equatton to get 09 pcoZszs v5 c from 718 e tog pcoz tog me 09 pcoz ZPH but og p 2735 50 17 tog szspcoz 29H 09 peepeg 29H 1 7 CdcoaCdv boundary CdCOJG 2H 9 Cd 0029 H200 Cd FL71 10 H m 7 09 pm o 44 09 Cdz og pcoz 2pH Set cow 1039 mo Lquotand pcoz 1 7 atm 6447473 5 2pH 97 gt ql57nmnmadhiamm br l VN ur j Wm ugh 735 m m 5 cucogs 1 Cd 2 56m r 2 deSG a mug my Woes vnqsmnas neumssnmnmx or men mu manta To ca cu ate a predominance dxag am we must be Concerned about the pHrdependent d St buUOn of phosphate and carbonate We Wnte the reactwons m ter s of the predommantspec es at the pH ot mtereat choose 09 mwhere PTWS the tota phosphate Concentrahon ahd pH as axes for out diagrams Assume cT 10 mo Lquot and Ca s conserved h the sohd phases pH lt pxphnsphate lt pxcarbnnate 1ncac01s 6H1P04D ZHIO e c m K 39 2 5 1mna H1 01 39 IBM n17 5m won 7 5m P 13 25 na vy pkphnsphate lt pH lt pxcarbnnate 1UC3C015 6H1P04 ma K 5 pxcarbnnat 2C015 EHIPO an 9 JDC 2 521 mm rem DEFY Hr eltp D nEH1C01 r nqH1PO a Pr 6w 15 35 n lt pKZphnsphat ZHIO e cawwo 75 may HCO 7 mg H me P New 7 4 as 5 ZHIO 9 came 50H1s mcholn 1 spH 2 0H71s 1ano 4w 1P04 r 4DH mg P Wm Lquot Hydrmyapame came Advanced Geochemistry of Natural Waters LE N 2 ACIDBASE CH E MISTRV Fundamental Concepts and Definition Start Audio Lecture E AMPALES OF ACIES ND BASES PRESENT I 39 Most important base HCOa39 Other bass BOH4 PO43 NH3 AsO4339 504239 239 etc Most important acid C02aq or H2C03El Other acids H4Si04 NHf BOH3 H2504 CH3COOHEl acetic HZCZO El oxalic etc mama Ma guaranty mum w 0 Most acidbase ractions in aqueous solutions are very fast almost namic principles yield correct swers Acidbase reactions involve proton but a bare proton H oes not exist in aqueous solution it is hydrated eg hydronium ion H30 or more likely H904 R BRONSTED DEFINITION 0 Acid A substance that can donate a proton to any other substance 0 Base A substance that can accept a proton from any other substance cm an my Maw was r IDS AND BASES ARE AILWAVS PAIRED IN REACTIONS Hzcoan H20 lt gt H30 HCOS39 mg H20 lt gt H30 NH3 CH3COOH H20 lt gt H30 CH3COO H20 H20 lt gt H30 OH39 swam a mums Wm WM s om E DEFINITIONS 0 Amphoteric A substance that can act as either an acid or a base eg H20 3 o Polyprotic acid or base An acid or base that c onate or accept rspectively more than one protoneg H3PO4 H2C03 H4EDTA gaging meurim PLE ME L ION A E ALSO All metal ions are hydrated in aqueous solution The a ched waters can lose protons and are therefore acids The positive charge of metal ion determines the strength of the acid ZnH2052 H20 lt gt H30 ZnH205OH CuH2042 3H20 lt gt 3H30 Cumzoxom Fans 4 my mom was CON UGATE ACIDBASE PAIRSquot HCI CI39 Hzcoan HCOS39 HSO439 50427 CH3COOH CH3COO39 ZnHzo52 ZnHzos0H wequot a an M w LEWIS D EFINITION 0 Acid Any substance that can accept an electron pair 0 Base Any substance that can donate an electron pair STRENGTH OF AN ACID OR BASE strengtn The tendency to donate or accept a proton he how readrlv does the substance do a ce proton Weak acrd nas Weak protonrdonatmg tendencv a strong acrd has a strong protonr donatrng tendech Srrn larlv tor bases Canno ne str 9 hr abso te sense strengtn depends on botn tne acrd and base mvolved m an acldrbase reactlon strengtn measured relatwe to some reterence m our case the solvent wa er gets hummer arde Mes STRENGTH MEASURED QUANTITATIVELV BV IHE JONXZATIQN GONSTA39NI HA H20 lt gt H30 A39 or HAEl gt H A39 KA mgr HA 1 The larger KA the stronger the acid the smaller KA the Weaker the aci ammo Ma mekmulw Mm wnen DEFINITION OF pKA AND pH PKA 09 KA Thus the larger pKA the weaker the acid the smaller pKA the stronger the acid Similarly PH 39 3909 H pOH log OH39 pX log X Gawain axmuw3 ml39wg STRENGTH OF A BASE A39 H20 lt gt HAEl OH39 HA OH39 K5 A H20 pKE 10g KB The larger pKE the Weaker the base the smaller pKE the stronger the base t w my my MFA Mm SELF IDNIZATION OF WATER AND VNEUTRAL pH 2o lt gt H OH39 Neutrality is defined by the condition W OH39 KW HlllaH l 1044 A we and H20 lbar Kw log Kw 210gH log Kw 2 log H 14 H l7 PHmuual swam m memmulw Wm WM CON U GATE ACIDSgtBASES H A39lt gt HA 1KA H20 lt gt H OH39 KW A39 H20 lt gt HA OH39 KB a WKA The stmnger an acid the WEker the conjugate base and vice versa DISSOLVED CO2 AND ACID BASE EQUILIBRIA INVOLVING CARBONATE CO2 is the most prevalent source of acidity in natural waters unperturbed by man and causes minerals to dissolve in weathering processes eg CaCO3calcite C02g H20l lt gt Ca2 ZHCO339 NaAlSl308albite C02g 112H20l lt gt Na HCO339 2H48i04O l2AlZSi205OH4kaolinite Sources of C02 volcanic emissions respiration fossil fuel combustion respiration decomposition of organic matter precipitation of carbonates Sinks of C02photosynthesis mineral dissolution CLOSED SYSTEM Treat carbonic acid as a non volatile acid Species involved C02aq H2CO3O HCO339 C0323 Hi OH39 Four equilibrium relations 7 Hydration of COzaq COzaq HZO1 lt gt HZCO30 7 First and second dissociation reactions of HZCO30 HZCO3 lt gt H Hcoj HCO339 lt gt H co 7 Ionization of water HZO1 lt gt H OH39 Electroneutrality H OH39 HCO339 2CO3239 An appropriate mass balance 3 A SIMPLIFICATION A common simplification is to define a species H2CO3 such that H2CO3l E C02aQl H2CO30l H2CO3 lt gt HCO339 H K1 H2C030 lt gt HCO339 H K10 H2C030 lt gt C02aq H20l K m 650 K10 N K10 K10 1 1K N Y E Thus H2CO3O is actually a much stronger acid than H2CO3 by a factor of 650 at 25 C and 1 bar 4 Diagrams relevant to a closed system involvi CO in a Figure 41ad from Stumm and Morgan fr h g 2 110393M es Waters stem 7 Y CT7 10 3 M EKA 53 KAZ1025 l 134 man a Figure 41effrom quot Stumm amp Morgan g 10 C 1 mm CT 23 x 10 3 M E pK 5 1 A1 g pK 93 A2 A E l 0quot BT 41 x 1039 M I I pKboric acid 88 I 1 39 n l l 3 6 I2 on 6 ID 39 SEAW TEE 39 Figure 4 2 from Stumm 1 at 25 C and 1 atm OPEN SYSTEM Same species as closed system Same chargebalance expression Same ionization constants No mass balance instead we have a Henry s law expression c02g H200 lt gt HZCOZquot HzCOJI Kxxpco2 10quot pcoz Kap c401 all KM HC 7K K 0 an 11PM W 11PM CQZ Eknpm If KHPL Q Charge balance H HCOZ39 2cozz OH39 At a given temperature we can de ne the entire system by specifying pm For a system composed ofpure water and co2 with pCOZ 1035 atm the pH is given by point P in Figure 43 because H o HCOZ39 This point corresponds to a pH 5 Thus pure rainwater in equilibrium with co2 as the only acid ie natural rainwater unaffected by pollution will have an acidic pH39 o Flgure 4 3 from Stumm a H ofcarbonater an 9K 5 3 pKM10 25 For a system with excess strong acid or base the charge balance becomes CB Ht HCO3 2CO32 OH C A where CB concentration of strong base added concentration of counter cation and C A concentration of strong acid added concentration of counter anion Rearranging and substituting we get CB CA CT0 1 2062 OH39 Hll ANCl f 0 Alkl KHPCOZ 0 0 alk 04 20amp0H H In this case we define the system by specifying pCO2 and Alk 11 If other bases are present they too must be added to the chargebalance expression Groundwaters or soil waters tend to have higher pCO2 than surface waters primarily because of respiration and bacterial decomposition of organic matter Problem Estimate the pH of the following solutions at pCO2 1035 atm i 10393 M KOH39 ii 5 X 10394 M NaQCO339 iii 10393 M NaHCO339 iV 5 X 10394 M MgO Solution In each of the above cases Alk 10393 eqL so all have the same pH at the given pCOZ From Figure 43 we can read ofpr 83 ANOTHER EXAMPLE One liter of a solution of 2 X 10393 M NaHCO3 solution is brought into contact with 10 mL of gaseous N2 How much CO2 will be in the gas phase after equilibration at 25 C pKH 15 pK1 63 pK2 Start with Hemy s law constant and ideal gas law pcoz CO KH 2W0 C02g 3 RT pCQ C02 gt1 KHRT 075 1 1 R 0820 atm L K mol COZg And now a mass balance Note that VgasVSOFH 001 comm m 2mg 0 M 001C02g 2x10 04 13 If Vgas is very small relative to V then it can be sol n assumed that oco does not change much Recall that 1 So we need to calculate pH so we 04 K1 KIK2 can get oco and then get C02g H l H 2 from the massbalance expression To calculate the pH we start with charge and massbalances for the system Charge balance Na H HCO339 2CO3239 OH39 Mass balance Na CT H2CO3 HCO339 CO3239 Proton condition H2CO3 H CO3239 OH39 and from speciation diagram we see that H2005 0032 OH Speciation diagram for the carbonate system with CT 2 X 10393 M Approximate solu ion 9 H2C03 CTOLO39 CO3239 CTOLZ39 OH39 KWHt so the proton condition becomes CTocO z CToc2 KWHt And the solution to this equation leads to pH 823 and oco 115 X 10392 Note that if we had assumed the proton condition to reduce to H2CO3 m CO3239 and then pH 8275 Now C02aq m 2x103 mm 2x103 C02aq 23 x 105 M 04 115x 0 C02g 23 x 105075 307 x 105 molesL Figure 4 4 from Stumm amp Morgan Scncm atic titration of a carbonate solunon With a strong acid lam oH39Alk aw i lCQg AEy 1 l 1 ml A i L The endpoints of titration ofcarbonate solutions to HZCO and coil occur at pH 4 5 and 10 3 cspccnvcly ese pH limits also represent tnc approximate limits beyond which 1le cannot al normally proceed asusu ALKALINITY IN CARBONATE SYST M ACIDNEUTRALIZING CAPACITY ANC Caustic alkalinity f 2 OHAlk OH Hcogr 2H2C03 H pAlkalinity f 1 PAlk OH39 C031 HzCOJI H1 Alkalinity f o Alk Hcogr 2c03139 OH H ACIDITY IN CARBONATE SYSTEMS BASENEUTRALIZING CAPACITY BNC Mineral acidity f 0 HACY H HC0339 2C0321 OH39 COZAcidity f 1 C02ACY H2C03 H C0321 OH39 Acidity f 2 Acy 2H2CO3 HCO339 H OH39 RELATIONSHIPS AMONG ACIDITY AND ALKALINITY Alk HAcy 0 Acy OHAlk 0 pAlk COZAcy 0 Alk COZAcy CT Alk Acy 2CT Alk pAlk CT COZAcy HAcy CT OTHER CONTRIBUTIONS TO ALKALINITY Other bases that could be titrated as alkalinity silicate borate ammonia organic bases e g acetate sulfides and phosphates Usually concentration of these is small compared to carbonate Borate most important in seawater 1034 M and silicate in freshwater 10394 10393 M In most waters borate and silicate have only minor effects on alkalinity owing to high pKA values pKBOH3 89 pKH4SiO4 95 Alk CToc1 2amp2 OH39 H BTO BOH439 SiTaH3SiO439 039BOH4 1H3Si0439 Thus at pH lt 9 ocBOH4 and dH3SiO4 ltlt 1 so Slhca and borate are rarely s1gmflcant contributors to alkalinity TRAUUN Mbll lUJJ Figure 413 from smmm amp Morgan ALKALIN39ITY TITRATION GRAN METHOD Dif culties chat can be encountered in alkalinity Dxffmulty senng endpan ofweak and punues 1m 7 Changes a er sampling best to do ntxauons m the eld lt o a E a 5 a o E 3N n 3quot E 39r E lt o E p 0 measurements well past the endpoint g 2f BASIS OF GRAN METHOD Let VO original volume of sample V volume of strong acid added at each point in titration C A normality eqL of strong acid V2 volume of strong acid needed to get to endpoint g 2 VI volume of strong acid needed to get to endpoint g l Alk HCO339 2CO3239 OH39 H PAlk OH39 C0321 H2C03l Hll V0Alk V2CA VolP39Alkl V1CA THE GRAN FUNCTIONS There are four Gran functions to calculate In the region well past the endpoint g 2 we have F1 VO V1039 PH m V V2CA Note that F1 0 at V V2 In the region between g l and g 2 we have F2 V2 v10 PH vVIK1 we need to have V2 to get V1 F3 2 V 39 V110 pH V2 39 VK1 In the region g lt l we have F4 V2 2V1 V10 PH V1 VK2 TRAUUN Mbll lUJJ Egme 413 from Stumm amp Morgan DETAILED S TEPS TO GET Alk FROM GRAN TITRATION 1 Tmate a known volume of s 1e well past endpoint g 2 using staudaullzed strong acid 2 Caleulate F for data near and beyond g 2 Plot F vs v 3 Flt a straight line to due linear part of curve and extrapolate 39t o vaxis read v2 from inursection ofline with vaxis 4 Calculate Alk according to Alk VZCAvu 5 Caleulate F2 and plot it vs v using v2 obtained in step 3 5 Flt straight line to F2 and extrapolate to vaxis to get v 7 Calculate pAlk as pAlk v CANE 3 Plot F and F to check values of v and v2 obtained GRAN TITRATION OF A SAMPLE OF MINE DRAINAGE 00035 00030 00025 00020 F1 00015 00010 00005 00000 0 5 10 15 20 25 30 mL titrant added 29 CONSERVATIVE PROPERTIES Conservative properties are those that are independent of pressure and temperature eg ANC BNC and CT if expressed as molesKg Examples of properties that are not conservative pH and the concentration of any individual species e g H2C03 These properties are also independent of certain changes in chemical composition For example addition of H2CO3 or change of pCO2 will decrease the pH and increase Acy and CT but will not affect Alk because H2CO3 is the reference by which Alk is defined Alk HCO339 2CO3239 OH39 Ht Acy 2H2CO3 HCO339 Ht OH39 Acidity is unaffected by changes in concentration of C0323 so addition of salts such as NaZCO3s or CaCO3s will not affect acidity CAPACITY DIAGRAMS Capacity diagrams plots with conservative properties eg CT Alk or Acy as coordinates and contoured with pH or activities of carbonate spec1es Constructed from the equation Alkl CT0 1 2062 OH39 Hll Addition or removal of acids or bases can be represented by vectors For any given pH Alk is a linear function of CT Vertical lines on these diagrams yield acidimetric or alkalimetric titration curves 32 EXAMPLE MIXING OF WO ERS Two waters A pH 51 Alk 10 meqL B pH Alk 2 meqL are mixed in equal proportions What is due pH ofthe mixture ifno co2 is 10517 Reading from graph CTA 28 mmolesL cTaa mmolesL N xing in equal proportions yields cT 28 192 235 mmolesL Alk 2 12 15 meqL me graph we read pH 55 EXACT SOLUTION TO PROBLEM Alk Cm 2 OH1 Hn CT Alk OH Hltoc1 2 For A ocl 0387 a2 307 X 10395 so CT m103930387 26 mM For B ocl 0939 a2 00592 so CT m 2 X 103930939 2 X 00592 19 mM For the miXture CT 26 192 225 mM and Alk 15 meq and ocl 20c2 m oc1mAlkCT 15225 0667 and the above occurs at pH 66 where ocl 0666 and a2 167 X 104 37 EXAMPLE INCREASE IN pH BY ADDITION OF BASE OR REMOVAL OF C02 The pH of a surface water with AlkO 1 meqL and pHO 65 is to be raised to pH 83 Calculate the compositional changes required if this is accomplished by 1 Addition of NaOH From the graph we read CT m 17 mM Vertical line up to pH 83 gives us 07 meqL NaOH 2 Addition of Na2CO3 Need to draw line with slope 2 from initial point pH 65 and Alk 1 meqL to intersection with pH 83 contour About 14 meqL or 07 mM NaQCO3 required l 0 5 a o 0 k l 3N 2 continued This pioblein could have been solved indie easily using acidity diagram A water with pH 65 and cT 17 mM has Acy 235 meqL Because addition ofNa2C03 cannot change aciditythelengd1 a nonzontal lin segment over to pH 83 yields 07 mM 3 By removal of co2 Use alkalinity diagram because alkalinity is independent ofCOZ The engtn of a lionzontal line segment at Alk 1 meqL from pH 65 to pH 83 conesponds to PHOTOSYNTHESIS AND RESPIRATION Alkalinity cannot change due to changes in P002 alone owever associated uptake or excretion ofions during these processes change charge balance and hence a decomposition of organic matter 5 ttg 0c in 1 mL interstitial lake water 10 C Alkn 12 meqL pHn 590 an 1073 M Assume reaction Cm HmOmNmP 10502 14H lt gt 105 co2 15m HP04139 106H20 Red eld composition Cm HmOmNmP a Estimate change in twostep process 1 CO2 increase at constant Alk 2 Alk change at constant CT 6 mgL organic carbon gt 05 X 10393 M CO2 pcK1 643 pCK2 1039 pCKW 1453 Alk Cm 206 OH1 Ha at pHO 69 061 0747 062 ltlt1 Alk CT0747 10763 1069 CT 161 X 10393 M However the CO2 added by decomposition of OC adds 05 X 10393 M to CT so CT 161X10393 05 X 10393 211 X 10393 M 43 AA1k 5 X1039414106 66 gt4105 Alkf 12 X10393 66 X10395 1266 X10393eqL Alk m CTocl 061 0600 061 1 1 H 1 K2 11071039 K1 107643 H1 ZWil H25X10397 pH661 J Advanced Geochemistry of Natural Wailers LESSON 5 Aqueous Complexes Coordination Chemistry ns and General Concepts Start Audio Lecture GEaLsu mm Guachumlnry al Natural wmrs WHY 1539 CH EMI CAL SPECIRT ION SO I M PORTANT o The biological availability bioavailability of metals an t eir p ysiological and toxicological effects depend on the actual species present Example CuCOJ Cuen2 and Cu all affect the growth of algae differently Example Mediylmercury CHthy is leadin formed in bolog cal rocesses kinetically inert and readily passes through cell walls It is far more toxic than lnolganc orms o Solubility and mobility depend on speciation Ii doubling day 0151Advancm Gaacllamislry a mull nl39wamrs Figure 620 from Stumm amp Morgan 30 Effect offree Cu or growth ofaigae in j seamer zo 1 lo j 0 i n x I I l v 39 3 Gawain QKMEIWW DEFINITIONS I Coordination complex formation 7 anv combinatwon ot canons W th mo ecu es or among contammg free pans ot e ectrons Bondmg mav be e ectrostatwc Centra atom nucleus a the meta canon Ligand r amon or mo ecu e thh thch a canon forms comp exeS Mutrdentate Ngand r a hgand W th more tnan one posswb e bmdmg swte Che h to anon r c0mp ex Format on Nth mumdentate ands Mutn or poyrnuCear Complexes 7 comp exes W th more tnan one centra atom or nudeus 1FELSII MuknkakxhemglqManAwtgu MULTIDEN T ATE LIG AN 3 Oxalate bxdentate t gtm ECU NH Ethylendmmme b denim Ethylendxammetetxaacen and Dr EDTA hexadentate 5 swam m memmulw Wm WM Chelau39on NH2 H2N N NHZ HzN S S 239 Z f f b b H H H i s M 0 g 55255 ngwH Game39thQ1m4 xmmmm39 D EFIN ITIUN S Spemes 7 refers to the actua form n Wmch a m ecme or on 5 presentm 5 ton Coordination number 7 tota number of hgands surroundwng a meta won Ligation number 7 number of a specmc tvpe of hgand surroundwng a meta ion Colold r Suspenmon ofpamdes composed of severa umts whereas n true sohmon we have nvdratwon of a 5wng e mo ecu e atom or won mum a cam mww as FORMS OF UCEURRENCE 0F METAL SPECIES 1013 1001 1000 mm mm oxenm Men s may Metals m mum Mmeme mm mm m new new mm mum Mme M mm Cu mom MeSR Mamas eooH quotmom re mo MeOOCRM moim M430 hm Mem 4 unchys Phi Cncof Mn V FeOOHox We mm We N A Ag AP Cdcl Zni Cooll came Coo rdi natiu n n u m hers L Me L LVKL CN z amen 1 39 40l L L 3V1 If CN 4 tetrahedral CN 6 mamde Cnnrdmannn numbers 2 4 a 89 and 12 are must mmmnn fur cannns 9 TAB LITV C STANTS MEASURE THE T H 0 OMPLEX ATION stelese constants K MLW ML L L w H M n MLHllLl Cumulative constants p MLW lMllLlquot Bi KiKzKaquotKn gamut a male aim wasa For a protonated ligand We a e MLWHH stepwtse complexatton M HLe ML H MLHHHL MLWHH 1 MHHLF Cumulative complexation H n nH The larger the value ofthe stability constant the moie stable the complex and the greater the piopom39oh of the complex formed relative to the simple ion ammo so camith aim ma STABILITY CONSTANTS FOR POLYNQQLEARFQMP AXES mM nLe Man 6 MmLs m MIWILY mM nHL e Man nH 7 MWL H m MlmlHLr Ifm l the seeoml subseiipt on am is omitted and the expression simpli es to the previous expressions foi mononuclear complexes mm mm axmuw3 miwg METAL IQN TITRATIONS 0 Metal ions can be titrated by ligands in the same way t at acids and bases can be titrated 0 According to the Lewis definition metal ions are 39ds because they accept electrons ligands are bases because they donate electrons in Q mum a can 39 Figure 673 39nm mm smmm iz Mnrgan Tmum nfll and Cu with mama and tekamme m m Jl i z a a l ms or mm m ms x39owu in mi mm m m mm a so y a Advanced Geochemistry Irf Namral Waters LESSON 3 The Carbonate System Start Audln Lecture email We meme m Ntm Wk39rs Fxgue A 4 mm smmm amp Mmgnn Schemauc mum Ufa cubamte mum mm a mug and AM IMAM pH g Romy 7 I i 7 1 Why A 1 Am B 1 The enapmms afmxahan af cubamte saluhansta H200 and co mum at pH e A 5 md m z xespechvely These pH hmns 515a xepxesentthe 3 mum meme awmwwmumm ALKALIle IN CARBONATE SYSTEMS ACID NEUTRALIZING CAPACITY ANC Causnc a kahmty CMMk DH39 7 HCDa39 r 2H2C03 r H gt D A k OH C032 H2C03 W A kahmtv f o Mk HCDa39 2mer DH39 7 W ACIDITV IN CARBUNATE S STEMS ASE NEUTPALIZING CAPACITY ENC Mrnera actdttv f 0 new W 7 Hood 7 2mm 7 0H COZVActdttv f 1 COZVACVJ HZCOJ W C03 OH39 Acrdrtv r 2 Acv 2H2CO3 HCOJ39 W s OH39 meta Nmm we clubs Maura RELATIONSHIPS AMONG ACIDITV AND A k HnAcv O E n S Mk V brmk c COZVACV H Acv Cr ammo m kluhumulry Mm wnen 70TH E R CONTRIB UTIQN S TU ALKALI N UV otner bases tnat coutd be trtrated as atkahnrty smcate borate ammoma orgamc ases e g acetate somdes and onoso a es UsuaHv concentration ot tnese is srnaH compared to carbonate orate rnost rrnbortant m seawater 104 M and smcate m tresnwater 107 7104 M In most Waters borate and smcate nave onw mmor ettects on a kahmtv owmg to non pKA va ues pKBOH3 8 9 oKHSro 9 5 emuquot mo hummuhydruiumw im Alk cad 2042 OH39 H Bramonu39 Siro naso amowf 5OH439Br anamo H35i043915ir s at pH lt 9 amou and aH35 04 ltlt 1 so silica and borate are rarely significant contributors to alkalinity am quotmumu Mirnao rig4m 413 a 5 mama ALKALIN ITV TITRATIDN m mmumtmmwduaumw im Df cutes tnat can be encountered m a kahmtv ttratons Dw cmty 522mg ennpamr nfwzal and 5 anges arte samphng best to an ntrat ans m tne e d T get around tnese prob ems We do a Gran twtratwon or hneanze tne twtratwon curve In essence vou perform tne ttraton Just as before butvou make measurements WeH past tne endpomtg 2 f o Let VD omgma vo ume otsampte v vo ume ot Strong add added at each pomt h Utrat on lt g c to d 1 3 m 1 1 to o D A k OH39 C031 HZCOJ W quk VZCA vapAk vcA gammamcamewmmw was THE GRAN FUNCTIONS There are four Gran funchons to ca cu ate In the tegmh weH past the e dpomtg 2 we ave F VD V1039quotPH v e v2CA No thatF Oatv v2 In the regmh between 9 1 ahdg 2 we have F2 v2 7 wow v r vK we heed to have v2 to get v F v r v10 vH v2 7 K1 In the regwon g lt1We have v2 7 2v v10 vH v r vK2 Gm nmnan Mimao rum 4 5quot a Morgan smuzs39idum ddl mdiq MINMme 39 DETAILED STE STU GET Alk FROM GRAN THRAUDN T 1 trate a known vomme ofsamme WeH pastendpomtg 2 uSH vg standarmzed strong add 2 Ca cu ate E for data near and beyond g 2 Mot E1 vs v 3 Ema straTght We to the hnear part of curve and extrapo ate t to View read v2 from mtersecnon of hne W th vrast 4 Ca cu ate Alk accordmg to AW vchvu 5 Ca cu ate F2 and p ot t V v usmg v2 obtamed m step 3 Ft 5 e to E2 and extrapo ate to V EXWS to get v 7 Ca cu ate prATk as prNk CAVD a Not E3 and E to check vames of v and v2 obbamed mum A and 04de as GRAN TITRATIDN OF A SAMPLE OF MINE DMIN GE no u 15 2a m Want added COORDINATION CHEMISTRY COMPLEXATION WHY IS CHEMICAL SPECIATION SO IMPORTANT The biological availability bioavailability of metals and their physiological and toxicological effects depend on the actual species present 7 Example CuCO30 Cuen20 and Cu all affect the growth of algae differently 7 Example Methylmercury CH3Hg is readily formed in biological processes kinetically inert and readily passes through cell walls It is far more toxic than inorganic forms Solubility and mobility depend on speciation Figure 520 from smmm ampMorgan Effect offree Cu on glowtila ofalgae in seawater 39 1 I I I I II I 30 E 39 39l ozo s m 1 quot lo I II u DEFINITIONS I Coordination complexformation any combination of cations with molecules or anions containing free pairs of electrons Bonding may be electrostatic covalent or a mix Central atom nucleus the metal cation Ligand anion or molecule with which a cation forms complexes Multidentate ligand a ligand with more than one possible binding site Chelation complex formation with multidentate ligands Multi orpolynuclear complexes complexes with more than one central atom or nucleus MULTIDENTATE LIGANDS HO O O O M I Tim 0 O Oxalate bidentate N EV HO O 2 Ethylendiamine bidemate Ethylendiaminetetraacetic acid or EDTA hexadentate 5 Chelation NH2 IHZN N1 NH2 HZN Polynuclear complexes 3 239 OH OH 2 H H H g g OH OH s f s sbzs 39 Hg30H42 S DEFINITIONS II Species refers to the actual form in which a molecule or ion is present in solution Coordination number total number of ligands surrounding a metal ion Ligation number number of a specific type of ligand surrounding a metal ion Colloid suspension of particles composed of several units whereas in true solution we have hydration of a single molecule atom or ion FORMS OF OCCURRENCE OF METAL SPECIES 10 A4 100 A 4 1000 A Free metal Inorganic Organic Metals Eghly Metals 39 ion pairs complexes boundto dispersed sorbed on and chelates high mol colloids colloids complexes wt species Cu CuzOHZZ MeSR Melipids FeOOH MeXOHy Fe PbCOf39 MeOOCR Me FeOH3 MeCOK humic MeSetc acid on clays Pb CuCOJ39 MnIV FeOOH or oxides MnIV on ox39de Na AgSH Ang A1 CdC1 Zn CoOH Coordination numbers LL L Me L Me L L CN 2 linear CN 4 square planar L L Lm I L I39Me39 Me L I VL L xL L CN 4 tetrahedral CN 6 octahedral Coordination numbers 2 4 6 8 9 and 12 are most common for cations 9 STABILITY CONSTANTS MEASURE THE STRENGTH OF COMPLEXATION Stepwise constants K MLn 1VILnl L lt MLn n Cumulative constants MLnl MnL lt gtMLn quot MLquot Bn For a protonated ligand we have Stepwise complexation ML HL ML H K 2 MLnlH 111 9 n quot MLHHHL Cumulative complexation MLnHn MnHL lt gtMLnnH quot MHLquot The larger the value of the stability constant the more stable the complex and the greater the proportion of the complex formed relative to the simple ion STABILITY CONSTANTS FOR POLYNUCLEAR COlVIPLEXES mM nL lt gt Man M anl 5m T M l L mM nHL lt gt Man nH 73m M HL If m l the second subscript on rm is omitted and the expression simplifies to the previous expressions for mononuclear complexes METALION TITRATIONS 0 Metal ions can be titrated by ligands in the same way that acids and bases can be titrated 0 According to the Lewis de nition metal ions are acids because they accept electrons ligands are bases because they donate electrons Figure 63 from Stumm amp Morgan Titmtion of H and Cu with ammonia and tetramine Lrien pH o39 WHOM you 5 39 4 e C H quotI JIDH81 J 1 I 2 3 0 I Z MOLES DF AMMON A PER MOLE won Cu m MOLE 75mm PER MOLE Or Cu m 12 w S 30 mi PNH3 E DTRIEN S A 00 so loo 0 50 Leo E0 50100 14 WLE 391 MDLE A MOLE 39I HYDROLYSIS The waters surrounding a cation may function as acids The acidity is expected to increase with decreasing ionic radius and increasing ionic charge For example ZnHZO62 lt gt ZnHZO5OH H Hydrolysis products may range from cationic to anionic For example Zn gt ZnOH gt ZnOH 2 ZnOO gt ZnOH339 HZnOZ39 gt ZnOH4239 ZnOZZ39 May also get polynuclear species Kinetics of formation of mononuclear hydrolysis products is rather fast polynuclear formation may be slow GENERAL RULES OF HYDROLYSIS The tendency for a metal ion to hydrolyze will increase with dilution and increasing pH decreasing H The fraction of polynuclear products will decrease on dilution Compare Cu2 H20 lt gt CuOH H log K1 80 Mg2 H20 lt gt MgOH H log K1 114 K1 MOH H M 2 l a MOH MOH MOH M0H 1 MOHMOIIIH 1 l aMOH 1 1 K1 At infinite dilution pH m 7 so ocCuOH 11039710398391 111 0091 ocMgOH 1103971039114391125119 4XlO395 Only salts with pK1 lt 12pKW or p3n lt n2pKW will undergo significant hydrolysis upon dilution Progressive hydrolysis is the reason some salts precipitate upon dilution This is why it is necessary to add acid when diluting standards 17 POLYNUCLEAR SPECIES DECREASE IN IMPORTANCE WITH DILUTION Consider the dimerization of CuOH 2CuOH lt gt Cu2OH22 log Kzz 15 Assuming we have a system where CuT Cu2t CuOH 2Cu2OH22 we can write Cu20H Cu20H CuOHT Cur Cu 2Cu20H12 K22 So Cu2OH22 is clearly dependent on total Cu concentration 18 HYDROLYSIS OF IRONIII Example 1 Compute the equilibrium composition of a homogeneous solution to which 10394 102 M of ironHI has been added and the pH adjusted in the range 1 to 45 with acid or base The following equilibrium constants are available atl 3 M NaClO4 and 25 C Fe3 H20 lt gt FeOH2 H log K1 305 Fe3 ZHZO lt gt FeOH2 2H log 32 631 2Fe3 ZHZO lt gt Fe2OH24 2H log 322 291 FeT Fey FeOHZ FeOH2 2Fe2OH24 19 3 FeT Fe3l K1 322 2Fe 92322 H l H l H Now we define oco Fe3FeT ocl FeOH2FeT a2 FeOH2FeT and a22 2Fe2OH24FeT 10 1 Kl BZZ 71 H H H 2 9K O OZFeTZBzz a01 K1 B22 10 117 117 117 This last equation can be solved for oco at given values of FeT and pH The remaining 0c values are obtained from the following equations ad2F9TBzz 0 22 H2 a1 10 lt1 12 0 0 B22 H H2 These equations can then be employed to calculate the speciation diagrams on the next slide 100 Feb FeT 10 M 80 a 60 FeOH2 2E 40 FeOH 20 0 100 Feb FeT 102 M 80 Fe2OH 11 FeOH 393 60 o 4o 20 FeOH 0 1 2 3 4 pH 22 Example 2 Compute the composition of a FeIH solution in equilibrium with amorphous ferric hydroxide given the additional equilibrium constants FeOH3s 3Hlt gt Fe3 3H20 log Kso 396 FeOH3s H20 lt gt FeOH439 H log KS4 187 Fe3 log Fey log KSO 3pH Fe0H439 10g FeOH439 10g Ks4 pH 23 FeOH FeOH3s 3H lt gt Fe 3r120 log Kso 396 Fe H20 lt FCOH2 H4r log K1 305 FeOH3s 2H lt gt Fe0H2 2H20 log Ks1 091 log FeOH2 log K1 2pH FeOH2 FeOH3s 3H lt gt Fe3 3r120 log Kso 396 Fe3 2Hp lt gt FeOH2 2H log 32 631 FeOH3s H lt gt FeOH2 HZO log Ks2 235 10g FeOH2l 10g Ksz 39 pH 24 Fe20H24 2FeOH3s 6H lt gt 2Fe3 6H20 210g Kso 792 2Fe3 ZHZO lt gt Fe2OH24 2H log 322 291 2FeOH3s 4H lt gt Fe2OH24 4H20 log Km 501 log Fe2OH24 log Km 4pH These equations can be used to obtain the concentration of each of the FeHI species as a function of pH They can all be summed to give the total solubility 5 0 C FeOH3s 9 E E 8 5 C O 0 U 2 FeOH4 3910 39 Fe20H FeOH 2 Fe3 FeOH 15 I I 0 2 4 6 8 10 12 14 pH Figure 64a from Stumm and Morgan Predominant pH range for L of 39 39 for 39ous quot quot ta STA39E OXIDATION te Figure 64b from Stumm amp Morgan The linear dependence of the rst hydrolysis constant on the ratio of the charge to the M0 distance zd for four groups of cations at 25 C 09 Kx 0 Figure 676 from stumm amp Morgan Correlation between solubility product of solid oxidehydroxide and the rst hydrolysis constant PEARSON HARDSOFT ACIDBASE HSAB THEORY I Hard ions class A 7 small 7 highly charged 7 electron clouds not easily defmmed e prefaquot to fonn ionic bonds I So ions class B lar e 7 low charge 7 d elecmm con guration 7 electron clouds easily defmmed e prefaquot to fonn covalmt bonds Pearson s Principle In a competitive situation hara acids tend to form complexes with hard bases and soft acia s tend to form complexes with soft bases In other words metals that tend to bond covalently preferentially form complexes with ligands that tend to bond covalently and similarly metals that tend to bond electrostatically preferentially form complexes with ligands that tend to bond electrostatically la i catinn nf metal and Ii and in term nf Pear nn s 1963 HSAB Drincinle Borderline So Acids Acids Acids H Fe Mn Co Ni Augt Ag 01 Li gt Na gt K gt Rb gt cs 01 Zn Pb Sn Hg gt Cd Be gt Mg gt Ca gt SF gt Ba As sh Bi pt gt Pd A1 gt Ga other PGE S gt Y REE LIE gt La T1 gt rt 064 Sn Ti gt TR Zr meM Or gt Cr Mo gt Mo gt M04 W gt W Nb Tau Re gt Re gt Re V gt V gt V4 Mn4 Fey Coy ASS Sb 1114 m 39gt in 39 39 PGE gt PGE etc Ru Ir Os ses ases Bases F39 HO OH39 0739 NH3 NOZ39 Cl39 I39gtBr39 CN CO CO gt HCOZ39 so gt HSOA39 sir gt HS39 gt RS PO gt HIPOAZ39 gt HZPOA39 organic phosphines REP carboxylates ie acetate organic thiols RP oxalate etc polysul de SnSZ39 M004 W04 thiosulfate 203 sul te SOZZ39 HSe S HTe Te2 AsSz39 39 32 ION PAIRS VS COORDINATION COMPLEXES 39 ION PAIRS 39 COORDINATION 7 formed solely by COMPLEXES electrostatic attraction 7 1arge covalent component to bonding 7 ligand and metal joined 7 shortlived association directly 7 ions often separated by coordinated waters 7 no de nite geometry 7 longerlived species 7 also called outer 7 de nite geometry sphere complexes 7 also called inner sphere complexes STABILITY CONSTANTS OF ION PAIRS CAN BE ESTIMATED FROM ELECTROSTATIC MODELS For 11 pairs eg NaClO LiFO etc logKe01I0 For 22 pairs eg CaSO40 MgCO3O etc log Km 15 24 I 0 For 33 pairs eg LaPO4O AlPO40 etc log Km 28 40 I 0 Stability constants for covalently bound coordination complexes cannot be estimated as easily COMPLEX FORMATION AND SOLUBILITY Total solubility of a system is given by MelT free Solubilities of relatively insoluble phases such as Agzs szo 50 HgS szo 52 FeOOH szo 38 CuO pKSO 20 A1203 pKSO 34 are probably not determined by simple ions and solubility products alone but by complexes such as AgHSO HgSZZ39 or HgSzH39 FeOH CuCO3O and AlOH439 Calculate the concentration of Ag in a solution in equilibrium with AgZS with pH 13 and ST 01 M 20 C 1 atm I 01 MNaClO4 K50 10497 Ag2sz At pH 13 H280 ltlt HS39 because pK1 668 and pK2 140 so ST HS39 8239 01 M 27 HS H ll 1 S 10 2 H H821 10 14 S27 S2 11S2 8239 9110393 M Ag2 1049710204 104766 Ag 102385 141X103924 M Obviously in the absence of complexation the solubility of AgZS is exceedingly low under these conditions The concentration obtained corresponds to 1 Ag ion per liter What happens if we take 100 mL of such a solution Do we then have 110 of an Ag ion No the physical interpretation of concentration does not make sense here However an Ag ionselective electrode would read Agt 102385 nevertheless 37 Estimate the concentration of all species in a solution of ST 002 M and saturated with respect to AgZS as a function of pH in other words calculate a solubility diagram Aglr Ag AgHSO AgHS239 2Ag233H2239l Kso 14512132 bUt 3239 0 st 5 Kso Ag12 0 st K50 At g STOCZ Ag HS39 lt gt AgHSO log K1 133 AgHSO Hs lt gt AgHS239 log K2 387 AgZSs 2Hs lt gt Ag2s3H22 log KS3 432 38 K2 AgHSO 1 AgHS H87 2 ST K1 2 AgHSO KloaSAAgw 5 061 T I K AgHSO K1O 1ST f2 AgltHS KzlAgHSmHS AgHS KzKlsiaf K50 STaZ s3 A8251st27 HST Ag2S3H227 K33Sa12 AgT 1 KISTocl K2K155a12 2KS3ST20512 T 2 20 or no uoueuueouoo 60 3 153 u Tam memN swig m k uoueuueouoo 60 Region 1 AgHSO and H280 are predominant Ag23s H280 lt gt 2AgHSO log AgHSO 1210g H280 1210g K aha1gb Z 0 0pH HZS Region 2 AgHS239 and H280 are predominant AgZSs 3H280 lt gt 2AgHS239 2H log AgHS239 3210g H280 1210g K pH aha1gb 1 0pH H S 2 Region 3 AgHS239 and HS39 are predominant AgZSs 3HS39 H lt gt 2AgHS239 log AgHS239 3210g HS39 1210g K 12pH aha1gb 1 m apH HS Region 4 AgZS3H2239 and HS39 are predominant AgZSs ZHS39 lt gt AgZS3H2239 log AgZS3H2239 210g HS39 log K aha1gb 0 apH HS 44 22 Region 5 AgZS3H2239 and 8239 are predominant AgZSs 2s2 2H lt gt Ag2s3H22 log AgZS3H2239 Zlog 8239 log K 2pH alogtAgiT 2 apH SZ THE CHELATE EFFECT Multidentate ligands are much stronger complex formers than monodentate ligands Chelates remain stable even at very dilute concentrations whereas monodentate complexes tend to dissociate 23 WHAT IS THE CAUSE OF THE CHELATE EFFECT AGYO AHrO TA s0 For many ligands AHrO is about the same in multi and monodentate complexes but there is a larger entropy increase upon chelation CuHZO4Z 4NH3 lt gt CuNH342 4HZO CuHZO4Z N4 lt gt CuN4Z 4HZO The second reaction results in a greater increase in AS 0 Figure 611 39 quot39 39 39 39 4 C H N NO COMPLEX u 1 mm 2 4 N PCu 5 HZNCHZCHZNHz 390 HENCHECHZNHCHEIZ 39 39039 39 I I I I 0 I0quot Io39 I0quot 10quot I039 ID I I0quot mm mm mm v qu n requot n t tr I m Figure 612a from Stumm amp Morgan Complexing of FeIII The degree of complexation is expressed as ApFe for various ligands at a concentration of 1039Z 39EL M The complexng 5 effect is highly pH a 0quot dependent because of Li quotWquot a the competing effects of 1 0 3 H and OH39 at low and e f high pH respectively so OM g 395 1 w AnnLou ZmrlZ39n u 6an D 10 Em F COLlN YE mmuummcitm t I WW we 5 0D MKAHUNWW CITRM39E l G 7 A 9 u D METALION BUFFERS Analogous to pH buffers Consider Me L lt gt MeL K MeL M 1 1 L If we add MeL and L in approximately equal quantities Me will be maintained approximately constant unless a large amount of additional metal or ligand is added If MeL L then pMe pK Example Calculate Ca2 of a solution with the composition EDTA YT 195XlO392 M CaT 982X10393M pH 513 andl 01 M 20 C For EDTA pK1 20 pK2 267 pK3 616 and pK4 1026 27 CZaY 11F KcaY 210106 gaHY l KCQHY 21035 Ca Y Ca HY 139 Car Ca2CaY2 CaHY Ca21 KWW WKCHYK1HY4 Ca2 0671 Ca Ca2 ac CaT 52 26 ii Y H4Y er HZYZ HY3 Yquot CaYZ39 CaHY Y4 a 1 Ca 1W 1K6 Ca2HYquot K1KcHY 71 an Yr 1HHr H13 H14 4 4 2W W t K K4K3 K4K3K2 K4K3K2K1 10 4 Equations i and ii must be solved by trial and error We know pH so we can calculate 0c directly We can then assume that ZHiY439i m YT CaT This permits us to calculate Y439 and then solve i for Ca2 This approach leads to CaY239 966X10393 M CaHY39 109XlO394 M Ca2 412X10395M Y439 605X10399 M H3Y39 307xlO395 M H2Y239 8810393 M HY3 821XlO394 M H4Y0 226xlO398 M 53 MIXED COMPLEXES Examples ZnOH2C12239 HgOHHSO PdCl3Br239 etc Generalized complexation reaction MmAnB lt gtMAmBn m n IOgBJIAMBK log BMW log 3MB 10gS m n m 71 Log S is a statistical factor For example the probability of forming MAB relative to MA2 and MB2 is S 2 because there are two distinct ways of forming MAB ie AMB and BMA The probability of forming MAZB relative to MA3 and MB3 is S 3 54 27 In simple cases We can use the following S m n m 11 general mixed complexes usually only predomlnate under a Ver restricted set of conditions Figure 6715 from an H In N seawatexHgle ptedommates mom s COMPETITION FOR LIGANDS The ratio of inorganic to organic substances in most natural waters are usually very high Does a large excess of say Ca2 or Mgz decrease the potential of organic ligands to complex trace metals Example Fe3 Ca2 and EDTA Fe3 Y439 lt gt FeY39 Ca2 Y439 lt gt CaY239 log KFeY 251 log KCaY 107 These data suggest that Fe3 should be complexed by EDTA 5 But let us combine the two above expressions to get CaY239 Fe3 lt gt FeY39 Ca2 log Kexchgnge 144 Ca2 14 4 CaYZ Fey 39 FeY Thus the relative importance of the two EDTA complexes depends also on the ratio of calcium to iron in solution For an exact solution to this problem we also need to consider the species FeYOH and FeYOH2 29 F1 ure 6 173 f Stumm lt2 Morgan Compeume effect of Ca2 on complexauon ofFeOIIw1Lh EDTA FeOH3sprec1p1Lates aLpH gt 8 6 cmLEuvnvm Fxgure 6 17b from Stumm lt2 Morgan Compemwe effect ofCa2 on complexauon ofFeOIIw1Lh EDTA L hq un u u rv m39 o I Fxgure a 17 fmm smmm ampMurgan Cumpemxve effect an un cumplexa unufFeUH wah citrate ACIDBASE CHEMISTRY EXAIVIPLES OF ACIDS AND BASES PRESENT IN NATURAL WATERS Most important base HCO339 Other bases BOH439 PO43 NH30 AsO4339 SO42 C0323 etc Most important acid C02aq or H2C030 Other acids H 4Si040 NHf BOH30 HZSO40 CH3COOH0 acetic H2C2040 oxalic etc 0 Most acidbase reactions in aqueous solutions are very fast almost instantaneous thermodynamic equilibrium is attained and thermodynamic principles yield correct answers Acidbase reactions involve proton but a bare proton H does not exist in aqueous solution it is hydrated eg hydronium ion H3O or more likely H9043 BRONSTED DEFINITION Acid A substance that can donate a proton to any other substance Base A substance that can accept a proton from any other substance ACIDS AND BASES ARE ALWAYS PAIRED IN REACTIONS HCo30 H20 lt gt H3o HCO339 NH H20 lt gt H30 NH30 CH3COOH0 H20 lt gt H30 CH3COO39 H20 H20 lt gt H30 0H SOME DEFINITIONS Amphoteric A substance that can act as either an acid or a base eg H20 HCO339 Polyprotic acid or base An acid or base that can donate or accept respectively more than one proton eg H3PO 0 H2C030 H4EDTA SIMPLE METAL IONS ARE ALSO ACIDS All metal ions are hydrated in aqueous solution The attached waters can lose protons and are therefore acids The positive charge of metal ion determines the strength of the acid ZnHZO62 H20 lt gt H3O ZnH205OH CuHZO42 3HZO lt gt 3H3O CuHZOOH339 7 CONJUGATE ACIDBASE PAIRS HClO c1 HZCO30 HCO339 HSO439 sog CH3COOH0 CH3COO39 ZnH2062 ZnHZO5OH LEWIS DEFINITION Acid Any substance that can accept an electron pair Base Any substance that can donate an electron pair STRENGTH OF AN ACID OR BASE Strength The tendency to donate or accept a proton ie how readily does the substance donate or accept a proton Weak acid has weak protondonating tendency a strong acid has a strong protondonating tendency Similarly for bases Cannot define strength in absolute sense Strength depends on both the acid and base involved in an acidbase reaction Strength measured relative to some reference in our case the solvent water STRENGTH MEASURED QUANTITATIVELY BY THE IONIZATION CONSTANT HA0 H20 lt gt H3o A 01 HA0 lt gt H A39 H A A HA0 The larger K A the stronger the acid the smaller K A the weaker the acid 11 DEFINITION OF pKA AND pH pKA 39 10g KA Thus the larger pKA the weaker the acid the smaller pKA the stronger the acid Similarly pH log Ha pOH log OH pX log X STRENGTH OF A BASE A H20 lt gt HA0 0H HA0 OH 3 A HHZO pKB 39 10g KB The larger pKB the weaker the base the smaller pKB the stronger the base SELFIONIZATION OF WATER AND NEUTRAL pH H20 lt gt H OH39 Neutrality is de ned by the condition H OH39 HOH 210714 At 25 C and W H20 1 bar Kw H12 log KW 2 log H log KW 2 log H 14 2 pH pH 7 neutral 14 CONJU GATE ACIDSBASES Ht A39 lt gt HA0 lKA H20 lt gt H OH39 KW A39 H20 lt gt HA0 OH39 KB KB K xKA The stronger an acid the weaker the conjugate base and Vice versa ACTIVITY SCALES INFINITE DILUTION SCALE VA 2 aACA Based on the fact that solutions approach ideality as the total concentration of all ions in solutions approaches zero In other words yA gt 1 as cA Zci gt 0 ACTIVITY SCALES IONIC MEDIUM SCALE Applied to solutions with a dominant concentration of a relatively inert electrolyte to maintain a constant ionic medium y A gt 1 as cA gt 0 but 2ci is constant IchiH 10 cA then y AHl Seawater is a good example with approximately constant composition of dominantly NaCl pH CONVENTIONS In nite dilution scale pH log H 10g Ht 10g vH Ionic medium pH log H1 NBS NIST scale de nes pH relative to a series of standard buffers OPERATIONAL ACIDITY C ON S TANT S 1 Infinite dilution scale K A w HA w 2 Ionic medium scale C K A 0 concentration quotient HA or conditional constant 3 Mixed constant K A w 19 IONIC STRENGTH 0 A quantity required to calculate activity coef cients 0 Attempts to account for effects of both concentration and charge of ion on activity coef cients 1 101192 i1 ACTIVITY COEFFICIENT EXPRES S I CNS 1 DebyeHuckel Limiting Law 2 Full DebyeHuckel Equation Valid at I lt 0005 M Valid atI lt 01 M 2 logyiz AZf xI 10gy2AZi 1 BN7 3 Guntelberg Equation 4 Davies Equation Valid atIlt0l M Valid atIlt05 M Useful for mixed electrolytes 2 AZi 10g AZ J7 02 logyi 1xI 1 I A05 B033 at 25 C and 1 bar 21 NUMERICAL EQUILIBRIUM CALCULATIONS Monoprotic acid What are the pH and the concentrations of all aqueous species in a 5 X 10394 M solution of aqueous boric acid BOH3 Steps to solution 1 Write down all species likely to be present in solution HZ OH39 BOH30 BOH439 2 Write the reactions and nd the equilibrium constants relating concentrations of all species H20 lt gt H 0H KW 2 m 210714 i H 20 BOH30 H20 lt gt BOH439 Ht KA HBEOHZ 27x10710 ii 30H3H20 3 Write down all mass balance relationships 5 x 10394 M 2B BOH439 BOH3 iii 4 Write down a single chargebalance electroneutrality expressions H BOH439 OH39 1V 5 Solve n equations in n unknowns EXACT NIHVJERICAL SOLUTION Eliminate OH39 in i and iv HOH39 2 KW OH39 Kw H1 H BOH439 Kw H1 H1 BOH439 2 KW H1 K W H H B0H 1 V Solve iii for BOH30 B0H30 B BOH4 HBlt0H K EB Blt0H A HB0H439 KAZB BOH439 vi Now solve V for BOH439 BOH439 K H1 H1 BOH439 H KWH Substitute this into Vi 26 HltH KWHgt K4213 Ha KWH HT Kw KAEB mm KAKWHa H3 KWH KAEBH KAH2 KAKW H3 KAH2 KAEB KWH KAKW 0 H3 7X103910H2 36X103913H 7X103924 0 We can solve this by trial and error computer or graphical methods From trial and error we obtain H 61X10397 M or pH 621 27 OH39 KwHW OH39 101410621 OH39 10779M BOH439 H KWH BOH439 61X10397 162x108 BOH439 594x107 M BOH30 213 BOH439 BOH30 5x104 594x107 M 499x104M 28 APPROXIMATE SOLUTION Look for terms in additive equations that are negligibly small multiplicative terms even if very small cannot be neglected Because we are dealing with an acid we can assume that H gtgt OH39 so that the mass balance becomes Hll BOH439 and then BOH3O EB H A 2 H B 0HZ 27x10 ii 30H3 H20 2 KA L EB Hr H2 K AZB K AH H2 KAH KAZB 0 This is a quadratic equation of the form ax2 bx c 0 and can be solved using the quadratic equation x b ixlb2 4ac 20 30 In our case this becomes KArlKj 4KAZB H l 2 Only the positive root has any physical meaning H 592 X 10397 We could have made this problem even simpler Because boric as is a quite weak acid ie very K A value very little of it will be ionized thus BltOH30 gtgt Blt0Hgt4 ZB m BOH30 5 X 10394M HB0Hgt 27mm A BOH H20 2 A H 0 7x10 10 BOH3 2 KA H 7x10 10 5x104 H2 35 gt41015 Ht 592X10397M It is wise to check your assumptions by back substituting into original equations If the error is S 5 the approxi mation is probably justi ed because K A values are at least this uncertain 32 CALULATE THE pH OF A STRONG ACID Compute the pH and equilibrium concentrations of all species in a 2 X 10394 M solution ofHCl 1 Species Hi Cl39 HClO OH 2 Mass action laws W H0H 21014 KA 2 Humor H 20 HCl 3 Mass balance HClO Cl39 2 X 10394 M 4 Charge balance H Cl39 OH39 K 103 Assumptions HCl is a very strong acid so H gtgt OH39 and Cl39 gtgt HClO Now the only source of H and C139 are the dissociation of HCl so Hll Cl39 this is also apparent from the charge balance Thus pH log 2 X 104 370 and Cl39 2 X 10394M OH39 KWH 10442 X 10394 5 X 103911 M KA 7111151 4x10 HM 74 2 103 Ha z M10 103 CALCULATE THE pH OF A WEAK MONOPROTIC BASE Compute the pH and equilibrium concentrations of all species in a 1045 M solution of sodium acetate 1 Species Hi Nat Ac39 HAcO OH39 2 Mass action laws Hl14057107470 KWHlOH7l HAc H 20 3 Mass balances HAcO Ac39 1045 M C Na 1045 M C 4Electroneutrality Na H Ac39 OH39 Combine 3 and 4 to get proton condition HAcO Ht OH 35 K 714 A lO We cannot make any approximations relative to the concentrations of H and OH because acetate is a weak base and total acetate concentration is low However because base is weak Ac39 gtgt HAcO so Ac39 m 1045 M C Substitute for OH39 in proton condition HACO Hll K H HAcO KWH H Now substitute into ionization constant expression 7 KAZW2107470 fix Ilt Jr H40 1 V H 1 H 1H1C KAK1Hn K11H1 1H12C KAKW KAH12 1H1ZC 1H12KA KAKW 1H12ltC K1 KAKW W KAKWKC K1 1H1 KAKWC Kim5 H lo4710141045 10470 Ht 62 x108 M pH 72 pOH pKW pH 140 72 68 OH39 161 x 107 Rearranging the proton condition we get HAco OH39 Ht 16 X 10397 62 X 10398 98 X 10398 Check of assumption AC39 C HAco 10450 98 X 10398 so Ac m 1039450 and the assumption made is valid CALCULATE THE pH OF AN AMPHOLYTE Calculate the pH ofa 103937 M solution of sodium hydrogen phthalate NaHP 1 Species H2P0 HP P2 H OH39 Na r 2 Mass action expressions HP39 00H coom 2 K 10 1 54H1P HIP 10 Hr KWH 0H39 H20 3 Mass balance expressions PT 103937 M H2P0 HP39 13239 10quot4 PT 103937 M Na 4 Charge balance H Na OH39 HP39 2P239 Now substitute 3 into 4 to get proton condition H1 HzPU HP39 P21 OH HP39 2PZ39 H1 HzPU OH P21 Because both pK values are less than 7 assume OH39 ltlt H H HzPU 13239 H PT 2P2391 HP39 WWW H PT H H PT HPT 2sz 1HP HP EL 21g2 391 H HP HP 71Ksz H PT HlampHP KA1 11 PT H2P0HP P2 PT K1 A p mgj HPT KAI H1 219 1 2K P K 139 TPT H 1 Af HPT H KA1 H 2K P 2 PK 139 T PT H HKA72 PTH PT T f2 H M K H 3 2 ZKAVZPT H H2 KA2H PTKLZ KA39l KAJ 21 2KA2KA1PT H 3 KA1H 2 KA2KA1H fir H2 PTKA2KA1 H3 KA1PTH2 KA2KA1H KA2KA1PT O H 24 x 105 pH 462 CALCULATION OF THE pH OF A POLYPROTIC ACID Compute the pH and concentrations of all species in equilibrium in a 10393 M H3PO4 solution 1 Species H OH39 H3PO40 H2P0439 HPO439 PO4339 2 Mass action expressions KWH 0H 21014 KA11021H H2POO4 20 39 H3P04 KM 210770 2 HHPOZ39 KM 10422 HPOZ HZPOX HPOZ 44 22 3 Mass balance PT H3Po4ol H2PO4l HPO42l P043 4 Charge balance H OH39 H2P0439 2HPO4239 3PO4339 Assumptions Because phosphoric acid is an acid assume that H gtgt OH39 Also because K A2 and K A3 are quite small then HPO4239 and PO4339 are negligible compared to H3PO40 and H2PO439 The mass balance then becomes PT H3Po4ol H2PO439l And the charge balance expression becomes Hll H2PO439l which can be substituted into the expression for K A1 KA1 210 2391 2 H2 39 PT H KAIPT39KA1H Hll2 Hl2 KA1H 39 KAJPT 0 H KAJr4Kj 4KAPT 2 23 H 8986 x104M pH 305 pOH pKW pH 14 305 1095 OH39 1122 x1011M H3PO40 PT H2P0439 103 8986 x 104 H3PO40 1014 X10394M HHPOZ 112130 H2P0KA2 KA 2 210770M H KM 10 0 HPo 2111111113011 K 10712 2 13 HPOZ 10 3 05P027 KAY3 10712 2 1077 0 PO433910391615 M 7079 X 103917M A check of all the assumptions shows that they are all valid 24 CALCULATION OF pH OF A VOLATILE BASE Compute the pH and concentrations of all species of a solution exposed to an atmosphere of pNH3 10394 atm 1 Species NH30 NH4 OH39 H 2 Mass action expressions KW H0H 21014 KH 210175 NHh H20 PNH3 KB 210745NH1H0H NHglleOl 3 Charge balance NH4 H OH39 Assumptions NH3 is a moderate base so we can assume that OH39 gtgt H so the charge balance becomes NHI OH39 also NHBO pNH3KH 10400175 10225 M NHIHOH OH Y K 10 B NHg 10425 OH2 10675 OH 2 103375 M 422 X 10394 M 25 pH pKW pOH 14 3375 10625 so the assumption that OH39 gtgt H is valid The concentrations are then OH39 422 x 10394 M NH4 422 x 10394 M Ht 237 X103911M NH30 562 x10393 M GRAPHICAL APPROACH TO EQUILIBRIUM CALCULATIONS Consider the monoprotic acid HA KA 210755 HA HA0 CT 10393 HA0 A39 SO A39 CT HA0 KMHAO HA39 KMHAO HCT HAOD KAHA0 HCT HHA0 KMHAO HHA0 HCT CTH HA0 m 52 26 1At PH lt PKA H gtgt KA SO H KA H HA0 CTHH CT 10g HA0 log CT A39 CTKAH 10g A39 log CT pKA pH 2 pH pKA H KA so H KA 2H HA0 2 CTH2H 2 CT2 10g HA0 log CT 10g 2 log CT 0301 A39 2 CT H2H 2 CT2 10g A39 log CT 10g 2 log CT 0301 3 pH gt pKA H ltlt KA so KA H m KA HA0 CTHMltA 10 HA0 10 CT PKA 39 PH A39 CTKAKA CT log A1 CT 27 Speciation diagram for HA with pKA 55 and CT 10393 2 I HA0 A39 4 3 10 M HA OH g 6 103 M NaAc 2 X 8 H 10 0 2 4 6 8 10 12 14 pH To compute the composition of a 10393 M solution of HA we start with the charge balance Hll A39 OH39 Hll gtgt OH39 Hll z A39 To compute composition of 10393 M NaA solution start with proton condition HA0 Hll OH39 OH39 gtgt Hll HA0 OH39 28 SPECIATION DIAGRAM FOR A DIPROTIC SYSTEM Consider HZS with pK1 70 pK2 130 ST 10393 M H280 HS39 8239 0 S S H25 T HS T 1 K1 K1K22 H1 2 H H K1 H S 527 2 T H H 11 KIKZ K2 1 PH lt PK1lt PKz H gt K1gt K2 0 ST N 0 HZS 1 K1 K1K2 NST 10gHZS logsT H 1572 HS ST K 11 logHslogsTK1pH 11 i 1 K1 H K1 10gSz3910gSKK 52 MST HPHTH H H 1 H KIKZ K2 KIKZ 29 2 PH PK1 lt PKz H K1 gt K2 S S 130 z Tz T logH SO10gS 0301 2 1 K1 ltle 2 2 T H HT Hsi ST mi 10gHS3910g ST 0301 2 2 1 10g 8239 10g S K 2 ST z ST pH T 2 3 PK1lt PH lt PK2K1gt H gt K2 S S H50 T z T logHs0logs K 2 1 K1K1K2 K1 2 pH T1 H HT H H37 ST E ST 10gHS3910g ST 1 2 K1 H 1 Sz39l SK 52 ST zsr 1 21 2 2 H H1 H 4 PK1 lt PK2 PH K1 gt H K2 H2S0 z 2S 10g H280 11 STK12 H HT H H57 ST E i 10gHS3910g sT 0301 H1 i 2 1 527 HT STU 1 zSYT 10gSz3910g sT 0301 KIK2 K2 1 5 PK1 lt pK2lt PH K1 gt K2gt H S S H230 T E T 10 H230 10gSTK1K2 K KK KK 1 1 1 2 1 2 2pH H HT HT HS 1 ST K z 15 Iog1Hs1IogsTKzpH 1 2 2 K1 H H 2 T 10g 8239 10g ST S 7 2 z ST H 1 H 11 K1K2 K2 31 Speciation diagram for H28 with CT 10393 at 25 C 2 I I H280 Hs 4 S27 6 E U 2 8 10 OH H 12 I 0 2 4 6 8 10 12 14 pH IONIZATION FRACTIONS Monoprotic acid HB 0613 0615 BC KAKA H 06m 060 E HBVC HKA H ocl0c01 Diprotic acid H2A a0 E HZA C 1 32 a C H17I2 K H AZ L 1 a c H lH hl ltle K2 as l m m 4 5 5 N 11 mg 2Mmar7 Figure 3 10am from Stumm and Morgan TITRATION OF ACID OR BASE Titration curve plot of pH vs quantity of base added At any point on a titration curve electroneutrality must be maintained e g titration of HA with NaOH Nat Ht A39 OH39 but Na CB so CB A39 OH39 H By combining this equation and a log C vs pH diagram speciation diagram we can contruct a titration curve C Na Equivalent fraction f E B C C What about dilution during titration If we initially have vO mL of acid the concentration V 0 of acid at any point becomes C 2 CO V V0 This can be substituted into the previous expression Equivalence point the point where you have added just enough base to just neutralize the acid ie f CBC l or CB C Starting with A39 Cot1 and CB A39 OH39 H we obtain CB Cot1 OH39 H C OH 7 H l f B 051 w C C At the equivalence point we have H20 A39 lt gt OH39 HA0 and the only HA0 is that due to dissociation of A39 so OH39 e HA0 Now let s titrate the conjugate base of the weak acid HA ie KA with a strong acid eg HCl Again we must have electroneutrality so K H Cl39 C Ht A OH CA CA 2 C A39 Hll OH39 CA HA0 Hll OH39 CACoc0 H 69 Define the equivalent fraction of the titrant here as C H l OH 7 g E A 050 C C At the endpoint g 1 so CA C Note that g l f At this equivalence point H A39 because the only A39 is due to the reaction HAO lt gt H A39 Buffer HA and A39 present in near equal quantities acts as a buffer in that it slows down the change in pH as an acid or base is added HAO lt gt H A39 HA0 OH39 lt gt A39 H20 35 EXAMPLES OF BUFFERS 0 Any conjugate acidbase pair can work as a buffer eg HCO339CO3239 BOH3 BOH439 H2P0439HPO4239 Buffers are most effective at pH H pKA because then HA0 H A39 and acid and base are present in approximately equal quantities CALCULATION OF A TITRATION CURVE Titrate 01 M solution of a weak acid pKa 5 with 01 M NaOH 1 f 0 zero base added or start of titration HAO lt gt H A39 Assume HA0 m 01 M gtgt A39 H m A39 Kg WHO1 H12 2105 HA 01 H2106 pH 30 36 2f 01 10 titration A39HA0 1090 In the region between endpoints pH depends only on the A39HA0 ratio not on absolute concentrations Hm1 2105 2 H110 1 HA0 90 H 9 X 10395 pH 405 3f 05 50 titration H 0A 12105H 150H HA 50 pH 50 73 4f 08 80 titration KA 210 5 2 H 80 20 H 25 X 10396 pH 560 5 f 10 endpoint of titration Here all original HA has been converted to A39 Taking dilution into account A39 m 005 M The only HAO present is that due to the reaction A39 H20 lt gt HA0 OH39 also HA0 OH39 and pKB 140 pKA 140 50 90 74 37 7 0 7 2 KB 21090 2 0H HA 1 2 0H 1 A 005 OH 5x10 11 223x10 55 pOH 515 pH 140 515 885 6f 120 120 titration The pH is calculated from excess OH39 concentration alone OH39 is a stronger base than A39 Assume we started with 100 mL of01 M HA Atf 10 we also would have added 100 mL of 01 MNaOH Atf 12 we would have added an additional 20 mL of 01 M NaOH The total volume would then be 220 mL OH xm 00091M 220 pOH 204 pH 140 204 1196 75 CURVE FOR TITRATION OF WEAK ACID WITH STON G BASE 1 2 pKa 5 00 02 04 06 08 10 12 38 Buffer capacity number of moles of a strong base needed to raise the pH of 1 L of buffer by 1 pH unit 7 A H K 1 P P A gHAo DEMONSTRATION OF BUFFER CAPACITY EXAIVIPLE BUFFER CALCULATION In what ratio would ophthalic acid and sodium hydrogen phthalate have to be mixed to get a pH of 32 pKA1 292 H 1 1 HP 1 K 2107292 A H2130 HP H K 10 p p A1 gH2Po 32 292 log R 028 log R R 191 We can make this solution by miXing 01 moles of H2P and 0191 moles ofNaHP in 1 L of water 78 39 Titration of a strong acid with strong base at various concentrations of acid and base 1 3 12 1 1 10 D 8 7 pH 6 5 4 3 2 1 0 O 10 20 30 40 50 79 NaOH ml Titration of a strong base With strong acid 13 12 11 10 0 pH O Nwlgtmmlm 80 HCI m 40 Titration ofweak acids wich strong base Buffer regions in undoquot curves A me arul m r v mum My i mummy V runWW MM wvaud Um no 1 u g Vmamm a Titration of a weak acid with strong base in presence of indicators i 2 el 11 i I T m i i l i i I I I El i jPIquotHil il lhil l ll 39ll transition range 1 i I 2 i i L i39mE rhyi orange i 7 I 4 inai igi39ticln Hungary I N 7 H i 20 3 4393 51 il l39iM MESH ml Titration of a weak base with strong acid Methyl red transition range I Methyl orange transition range O 20 40 60 80 100 120 140 160 180 200 Titration 7o 84 42 What can titrations tell us Concentration of acid or base present analysis pK of acid or base present thennodynamics Titration ofmixtures ofacids orbases Ifthe pK s are su iciently di erent the stronger in titiation curve ie equivalence points iprK s 4 or more Same rule applies forpolyprotic acids Titration ofa mixture ofweak acids with a strong base t7 2quot IVAMUM m umrwnH quott new mun m BUFFER INTENSITY Buffer intensity A measure ofthe buffer intensity buffer capacity would be the inverse ofthe slope ofthe titration curve at any point E dCB 7 d CA de de For a monopro c system recalling that cE Cot OH39 1r B E dCB Cda1 dOH 7dH de de de de dH dH 23025dH 1 pH 123025d1nH dH H l 23025H dOH dOH d pH 123025d1nOH 230250H d0 1 CdH do 23025C KAUH C de de dH KA H2 0 1 Ki 1 H1 KAH KAH 0 7 C dog 23025050055 2 de 89 23025H0H C051050 HA0A 23025H0H71WJ The maximum buffer intensity occurs at the in ection point of the titration curve or where d2 051 Z 0 MPH2 This condition occurs where ocl CLO HA0 A39 or pH pKA A solution is well buffered at three points when H is dominant when OH39 is dominant and where pH PKA 9 45 395 m u mg CMma Fxgure 3 10 are from Stumm and Morgan as m m L 3 rm Figure 3 10d from Stumm and Morgan BUFFER INTENSITY FOR MIXTURES OF ACIDS 3 23025HOH CA0 H40 A CBaHBaB a 23025HOH ww io ml K HA lA l HB lB l J BUFFER INTENSITY FOR POLYPROTIC ACIDS B23025H0H HzTHHC J Hcii cz Hzc lIHC l HC lIC l The above is valid as long as KlK2 gt 100 GRAPHICAL DETERMINATION OF BUFFER INTENSITY 3 23025H OH HA0A At pH lt pKA HA0 A39 m HA0 so 3 m23HI OH39 A39 At pH gt pKA HA0 A39 m A39 so 3 m 23H OH39 HA0 Thus buffer intensity can be calculated from a speciation diagram by summing all concentrations represented by a line of slope i1 and multiplying by 23025 94 Base neutralizing capacity BNC the equivalent sum of all acids that can be titrated with a strong base to an equivalence point For a monoprotic acid HA BNC HA0 Hf OH39 BNC is the excess of protons above a reference ie pure NaA f l for which the proton condition is HA0 Ht OH39 Addition of NaA does not affect BNC Acid neutralizing capacity ANC the equivalent sumof all bases that can be titrated with a strong acid to an equivalence point For a monoprotic base ANC A39 OH39 Hll ANC is the deficiency of protons over a reference level ie pure HA f 0 for which A39 OH39 Ht Addition of HA does not affect ANC EXAlVIPLE PROBLEM Calculate ANC of the following solutions a NH4 NH30 5 X 10393 M pH 93 b NH4 NH30 10392 M pH 90 Which has the greater ANC pKA 93 ANC NH30 OH39 Hll ANC Coc1 OH39 H For a a KA 1079393 1 KA H10 9393 1043 ANC 055 X10393 M 1047 1093 ANC 25 X10393M 48 EXAMPLE PROBLEM CONTINUED For b KA 105quot3 a 033 1 KAH 10931090 ANC 03310392 M 1050 10399O ANC m 33 X 10393M Thus solution b has the higher acid neutralizing capacity even though it has the lower pH ie is more acidic than a ANC should not be confused with pH ANC AND BNC OF MULTIPROTIC ACIDBASE SYSTEMS For multiprotic acids we can define various reference levels f 012 Example Sulfidecontaining solution ANC with reference to the equivalence point f 0 g 2 ie a solution of pure H28 is ANCfO HS39 2SZ39 OH39 Ht ANCfo STltoc1 2 OH Hu Example Phosphoric acid system with reference to f 2 ie a solution of pure NaQHPO4 BNCf2 2H3PO40 H2P041 Hu P0431 OH1 BNCf2 PTaoco a1 a3 H OH 98 49
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