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# Struc BIS 102

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This 89 page Class Notes was uploaded by Charity Daniel on Tuesday September 8, 2015. The Class Notes belongs to BIS 102 at University of California - Davis taught by Staff in Fall. Since its upload, it has received 10 views. For similar materials see /class/187638/bis-102-university-of-california-davis in Biological Sciences at University of California - Davis.

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E S mm Mudu e 2 omnsrvee EnEmv GlbbS Free Energy Part I Introduction Goa s R ewewthe eeneem e1 Gmbsnee enevgy AG R ewewthe ve auunshwp bemeen spemaneeus veammns andmew assuma ed AG va ues R ewewmemeamngmAG Wham a bmchemma veammn Wm eeeuv spemaneeusw s uepenuem un WEI enEvmethe emapr UHhe 123mmquot hea evmved and me emvupv UHhe 123mmquot the amuum evmsmuey pmuueeu The eeneem memes nee enevgy AG dwe uped byJ WHan wabs cumbmesthese Fm a gwen veacuun dwecuun sayA e a WAG 5 negawe the veammn Wm pmeeeu spumaneuus y 3 M be pmuueeu u s pusmve me veacuun Wm M pmeeeu spumaneuus y A W be pmuueeu HAG sEthe veammn s at ethbnum neuhev nu am he pmuueeu AG 5 uepenuem my me eeneennauen e1 pvudums and veactams m su uhun Tu expvessthws ve auunshwp u s cunvememm uevme a evm knuwn asme veammn queuenu a Censmeyme hvduwsws veadmm A H20 x 2v Undenhese cundmuns Wheve A x and v 312012 mu av cuncenhauuns M e thew vespemwe spemes m a 5mm physma chemws v sensE We shm d use acmues ms1ead see Swe pm 172 Huwevev assumme mm mm suhmuns ave dHu E WE can savew use mu av eeneenuauens ms ead NV My y W W V m m ssmnveva Thwswsacummun Anmhevwav ewumne W5 s u savmatwe ave assumme We aeuww uNatev s umw m We Wm use MS cunvemmn emuswew Seeevs bunk mseussesms and uthev cunvemmns an nu 155459 The unns e1 mew dependmn an 012 numbev e1 Spemes mvmveu mme veammn When a veactmn has veached Eumhbnum n5 0 va ue s Eumva emm a newtevm Keq knuwn asthe ethbnum cuns1am Knuwmg mm a and Keq AG can be ca cu a ed asmHqu AG e R T u j m 1 w emweae an Mudme z GmbsteeEnemvjnmvedxmcd E S mm Mudu e 2 omnsrvee EnEmv Wheve AG has unus u L R sthe gas eens1sm m malK T 5 absu me empevamve m K cundmmvs e a 5 3 um va ue eve th whmh m yeveyenee 01252 veadmns Ev veammn n cunvemwumthws sundan sae swheve 3 012 WWW anmams and pvuuuns ave at 1 M uneemvanuns Themmmatstandavd sme and AG R T 1 Keq Thwsva ue ems undev smeamsa e cundmuns 5 eeuwawemm a newtevm Aciwheve AG 4 T lrKeq mu Keq bemg ubsevved a tempevamve T May m Tu w T Ken m a wequot veacuun The AG vame Ma yeammn summerva descnbesme veadmn s enevgeucs such mm AG can be ca m a ed a uthey nunssnusmsae cundmuns R eanangmg uuy uemmen e1 AG mm abuve AGRTlrQrRTlrKeq and subsmmmg Aewm 7R T lrKeq y E ds AG mmvms mo and AG AGRTlrQAG WMhAG havmg been detevmmed attempevamve T Anme abum Euna s When eeusuens avE mesemed m The veHuwrbeme en vemuns as shave theywm u en use me Eun ean eeusws sen vamevman me ass gnmem upevamv The yessenvenms Tsmemm 1 m avmd geneysung a vedrcmuved enm e g AG 4 T mer duemtheuse e1 unuevmeu tevms nd 2n avm makmu an assmnmem ha vuu suequ may mean Euna s sen m uenevate eeusuensmsT an e s e u make an vuuv uwn m essencE WewwH use The EU nm have any ma hemauca meamngwnhm Memes Mudme z GmbsteeEnemvjnmvedxmcd BIS 102d Module 2 Gibbs Free Energy Part II Units and Temperature Goals Set up unit constants Practice using C and Ktemperature scales Before working some AG problems let39s first establish some unit constants that are not already builtin to Mathcad In the region below use the assignment operator to set M to be You can do this by 1ter typing Mmoliter 201 2 m01 I liter Based on this assignment of M create an assignment below for lel by setting it to be 10393M You can do this by typing lel10 3spacebarspacebarM 202 mM 10 3M Create similar assignments below for M to be 10395M and nM to be 10399M You can find the p symbol in the Greek Symbol Toolbar or you can type m followed by ctrlg 203 7 6 7 9 pM 10 M nM 10 M Create an assignment below for kJ to be 103J 204 k 103 The temperature of biochemical reactions is often given in terms of C or even F In contrast calculating Gibbs free energy requires use of absolute temperature K When calculating AG with pencil and paper you will likely convert temperature from C to K before proceding with your calculation In Mathcad this conversion step is not strictly necessary because Mathcad is happy to do the conversion for you However temperature is actually an odd kind of unit so let39s take some time practicing its usage Module02 3 GibbsFreeEnergyAnsweredxmcd BIS 102d Module 2 Gibbs Free Energy When one converts between say feet and meters you multiply by some conversion factor Mathcad does this easily Ask Mathcad below what 1ft is You can do this by typing 1ft 1ft 0305 In You should get that 1ft 0305m If you try the same thing with 1 C you get an error instead of an answer 1 C I The reason for this is that the conversion from C to K does not involve mulitplication by some conversion factor Instead you must add 27315 to convert from C to K Mathcad will do this but you have to be explicit that the conversion is not multiplicative In this sense C and F are quotspecialquot units in Mathcad and must be accessed in a particularway notjust typed as with other nits Ask Mathcad below what 37 C is To do this call up the quotCustom Charactersquot toolbar by clicking on quotViewquot in the top menu then quotToolbarsquot and then quotCustom Charactersquot it39s at the bottom ofthe list Now click in the white space below and type 37 Then click on the C button in the Custom Characters Toolbar Then type 206 37 C 31015K You should get that 37 C 31015K Ask Mathcad below what 986 F is You should get 31015K here as well 207 986 F 31015K Module02 4 GibbsFreeEnergyAnsweredxmcd E S mm Meme 2 ennsrvee Enevev Part III Keq and AG Goa s Ca cu a e Keq gwen ethbnum euneennanuns Ca cu a e w gwen Keq CunswdenhemHuwmu W mem um vuuHex 6amp6 W 741quot An enzymatic hydrnrys39s nHruclnserLP rueensew o H20 3 lrutlnse P was nnly 6 25 x m M Czlculmelhe eq um cnnslznllnrlhis marlinquot andlhe standardrslzle r22 energy change nl nydmys39s numemsew Wewm need m mm nsmad usethe Eun ean Emma s smn m cveate the uuuwmu exmessmn uctase n Ke q fucmsel nwanm be evama ed byMamcad Thwsexpvessmnsjus1sumemmgwewammwme exmesswuna a uumm euan n 3m fructase p Ke fructase unhat be uw Use the asswgnmem upevamv m seHvucu521P m be 5 25 x m39SM mz uctasel 5 25 x m39SM Mudme z GmbsteeEnemvjnwEvedxmcd E S mm Mama 2 omnsrvee EnEmv The mma cuncemvatmn uHmduseWWas a su gwen asEI 2M Let us de ne Mann be Makemat asswgnmen uv ucms pi be uw m uctaselPi n M ave 22ch m m y mm H mum at ethbnum Funhev we knuwmat mum and p must be equax Makethe VuHqun Mu dummv enuauuns be uwm ve ec hese ve atmnshms 1 huasupg hump hum uctase p 3m huasupg hump hum p hum Cnpy yum dummy equatmn m Keq mm abuve and paste n have be uw Usmg me abuve necessawm amve a 2 Km hump n 7 hump hump m5 hum p hum hum K K hump hump Kn hump hump hump hump hump hump n 7 hump2 Keq hump Mudme z GmbsteeEnemvjnmvedxmcd E S mm Mama 2 omnsrvee EnEmv swgnm an asswgnmem upevamv m armame equauun ma fructasel n 7 mummy Keq uctasel Ask Mama be uw Wm Keq 5 3m Keq6396xlU5m Vuu 5mm am an anwev m 5 295 x 1n5m WME cunem we d pmbamy be muve mtevested m 522mg me answev m tevms ummay cuncemvatmn M Let s u ma cunvevsmn R Vuu mm 522 a Vedangh av mack p acehu dev appeanu me mm mm answev m m p acehu dev emev M m Keq 529 SM Vuuv anmv 5mm chameWu be 539 EM whmh sthe answenu 012 ms pan ume pmmem veammn 77 AG Wham m begm ma ca cu auun 5 m pan a ma a uvpeysunax pyeveyem behave n 5 must ms1mclw2 m hamquot at uuv dehnmun u AG c R T 1 amp Keq What sthe va ue me when AG Aewunms yeacuw What mum sthws7 mm at ethbnum7 What 5 a at ethbnum mm yeacuunv any a 1M Tm sthe standavdrs am cunumunwheve pvudums and veac ams ave a 1 M Tm 5 nm a ethbnum a 539 EM 3 equmbnummnm veammn Mudme z GmbsteeEnemvjnmvedxmcd BlS102q Module 2 Gibbs Free Energy Knowing that Q 1M when AG AG use the above definition for AC and some dummy equations below to arrive at AGO iRT1nKeq You can generate the A symbol by clicking on it in the Greek Symbol Toolbar Alternatively you can type DCtrlg You can generate the symbol by holding down the alt key typing 0176 on the number pad and then releasing the alt key 310 AG R T Keq 1 AGquot RTl Keq AGO RT1nKeq 1 AGO iRTlnKeq Copy your last dummy equation from above and paste it below Change the Boolean equals sign to an assignment operator to quotactivatequot the equation 311 AGO iRT1nKeq Mathcad should give you an error Keq will be red That39s OK We39ll fix that in a bit But first let39s take the time to define R and T since it appears that we39ll need them forthe calculation Below use the assignment operator to define R as being 8315 1 Next to that region mo make another assignment for Tto be 25 C Remember that you39ll have to use the C button in the Custom Characters Toolbar to correctly assign the temperature units 312 I W 8315 Kimol a 25 C Module02 8 GibbsFreeEnergyAnsweredxmcd E S mm Meme 2 omnsrvee EnEmv s m tevms ewc Such a mwsma ch weum be measueus NW2 W212 kamg W5 pyemem Wm peneu and papev hmk Exam Hewevemn Mame ms speneeuyme As ung asyuu pmpev ydelme yum we Mathcad Wm hand e awe cunvevsmnswnhum any mervermun mm yuu Su hey saywe 2 n Jus dun t puHWs 51mm an exam ask Mame be uw WhatT s m T e 29215K Vuu shumd get an anwev e1 ZSEJEK Whmh e1 cuuvse stvue New M s 12mm m xmg mm enm m yuuvAG asswgnmem m Aswan E M m m m w m we Keq has unns e M And Mame Wm nuHake me namva mg n M because We vesuh s essenuauy meamngxess Tu get Mathcadm u W5 ca cu atmn yuu H have m dmde Keq by M m vemuve Ms um s bemve me namva ug upevatmn akes p ace m w e 7R T 1n E M m Mhe n m memvs ku muketc That sOK Justchamemeumtsbvvemacmutheb ackmacehumevunme quotgm uHhe ansNEHu be i 316 Mudme z GmbsteeEnemvjnwEvedxmcd BIS 102d Module 2 Gibbs Free Energy You should find that AG 716017 which is the rest ofthe answer to the original problem mo Under standardstate conditions is the hydrolysis of fructosei P spontaneous Why do you think so 317 Yes under standardstate conditions the hydrolysis of fructoseiP is spontaneous thinkthis because the AG forthis reaction is negative Modue02 10 GibbsFreeEnergyAnsweredxmcd BIS 102q Module 2 Gibbs Free Energy Part IV Q AG and Reaction Progression Goals Create a function to calculate Q given a fructose1 P concentration Make a graph of Q over a range of fructose1 P concentrations Create a function to calculate AG given a value of Q Make a graph of AG over a range of fructose1 P concentrations Let us explore the fructose1 P reaction described above in more general terms For example we were told that the initial concentration of fructose1P was 02M and that the reaction proceeded until fructose1 P was 625 X 10 5M In the problem we were asked to calculate Q at that equilibrium state But what happens to Q inbetween the initial condition and the equilibrium state How does AG change as the reaction progresses And what would happen if the reaction were somehow forced to go beyond equilibrium to concentrations of fructose1 P lessthan 625 X 10 5M Formally let us now answerthe following question What are the corresponding values of Q and AG over the range of fructose1P concentrations from 1pJV to 02M As before we will assume 1 hen fructose1 P 02M the concentrations of fructose and Pi are zero 2 The total concentration Of fructoselP fructose fructoselPio 3 The temperature is 25 C Begin by assigning fructose1P to be a range variable Its first value should be 1pJVI You can get the p symbol from the Greek Symbol Toolbar or alternatively by typing mCtrIg The second value should be 10M and the last value should be 02M fructoselP 1pIlI10pM 02M uw fructoselP 1pM10pM 02M vavwwvwww To calculate the Q values over this entire range of fructose1 P concentrations let39s create a function for Q that accepts a value of fructose1 P as an input Let us call this function Qf for Qcalculating function rather thatjust Q That way we can still use the letter Q to refer to specific values Module02 l l GibbsFreeEnergyAnsweredxmcd E S mm Mama 2 omnsrvee EnEmv mm m undwun Qi lruclnseH L be uw usmg We asswgnmem upevamv Use We same m 2 Mann nimmgm mm Q P uctasel 2 WWW 7 mama r Maw Q7 fructasel Iimeln dn sn Take Me me nuwm save yuukuvk an We mwmrmudme and came backm m pmm 4m 1 Cveate a new maph be uw b2 fructasel 4 Senhe mm SEHHE yawsm be M Qrltrnwmsew2 m M my 5 nm much m see an ms maph un ess Vuu 52 une m mm mm axestu be an a mu sca e Ammayw M s uuk at Hwyth xrawsse u a ug scab n r r r r nm m m m 3mm m 1 mm Mn Q fmmsel M Mn lxl n lxl 6 xl 3 I name M Mudme z 2 GmbsteeEnengAnwEvedxmcd BIS 102d Module 2 Gibbs Free Energy Describe what you see in the graph How does Q change when fructose1P decreases 405 As fructose1 P gets smaller Q increases Do you think Q would ever be negative Under what conditions 406 No Q being a ratio of positive quantities concentrations can never be negative Previously you found that when fructose1P was 625 X 10 5M Q Keq 6396 M Based on what you see in the graph would you be willing to believe that Explain your reasoning 407 Yes at fructose1P 100M Q is about 5000M and Q decreases as fructose1P gets bigger so I39d be willing to believe that Q 600M at fructose1 P 625 M Having created a way to easily calculate Q for any value of fructose1 P let us now focus on creating something similar for calculating AG Our strategy will be to create a AG function that accepts Q as an input parameter Create the function AGQ below Use the assignment operator to define this function to be RTl Keq I Q A 1 GQ RT Kqu Let us graph this function in two ways First we39ll set Q to be a range variable and plot AG over that range of Q values Then later we39ll use the Qf function from above and use the output of that function as the input to the AG function In this way we can graph AG over the range of fructose1 P values that we defined previously Use the assignment operator to define Q below as a range variable The first value should be 01M the second should be 1M and the last value should be 104M 409 Q 01M1M104M Module02 13 GibbsFreeEnergyAnsweredxmcd axsmzu Mame omnsrveegnemv Cveate a new maph be uw 2n 2 Senhe mm be 3 M man 3 Senhe yamm be may N u m 2 ma same mm be an a my sca e n1 1 m mnnnznn m man u m m izn 7m m m mu uheamams and pmducts7 Duesma make sense7 Exp am yum veasumng AG becames mm pusmve Wm mcveasmg va ues me As a women m mdmates me cuncemvatmn uvpmums s mmeasmg mauve m m cuncemvauun ahead ncveasmg a mm mm m up use mm mm muve pusmve means exac y ms ems Thus w vspumaneuus pmdummn uvpmums AG been Mudme z m GmbsteeEnemvjnmvedxmcd BIS 102d Module 2 Gibbs Free Energy Based on what you see in the graph what is AG at Q 1 How does this compare to the value of AG you calculated earlier Why do you thinkthat is 412 k k At Q 1 It looks like AG Is about 718 This Is pretty close to the 716 we mol mol calculated for AG earlier Indeed at Q 1 AG AG so this makes sense We will now use yourAG function togetherwith the Qf function you created earlier This will enable us to graph AG against the fructoselP concentration range you defined earlier 1 Create a new graph below 2 Set the xaxis to be W M 3 Set the yaxis to be kJ mol 4 Set the xaxis to be on a log scale 5 Make the graph wide enough so that you can clearly read the labels on the Xaxis 413 AGQ7ffmctoselP 2039 kJ 4o 60 7 80 I I 7 lUU 1x10 6 1x10 4 001 1 fructoselP M Module02 15 GibbsFreeEnergyAnsweredxmcd E S mm Meme 2 omnsrvee EnEmv spumaneuusmmm an mma cundnmn ummesewm 2M dues ms make sense7 m A eeumes mme negawe asmemsew cuncemvauun meyeases Tms makes sense because We knuwmamuemsew spumaneuus y kymmyses at a cuncemvauun u u 2M a veactmn ma euesm mach ethbnum unm mus1 unkememsew has been hydm yzed mm m rm k k k k n r n7 euesms make sense7 415 When AG mms pusmve the wave dwemmn nuke veammn s nu ungev spumaneuu Ekaeny Wham AG El the veammn s at ethbnum On me gyapmms cunespundsm a yuemsew va ue beMeen m 5M and mm m Pan m We W212 mm 013 25 x m39SM wasme ethbnum euneemvauun quumuseWP 5 W5 makes sense Mudme z GmbsteeEnemLAnmvedkmcd BIS 102d Module 2 Gibbs Free Energy Part V Physiological pHNeutral Standard State amp AG Goals Review the meaning of AG Calculate AG given AG Recall that AG is calculated under standardstate conditions in which all products and reactants are at 1M concentration In considering biochemical reactions that either consume or produce H ions this implies that AG is calculated at H 1M or pH 0 This is so acidic that it is hardly biologically relevant Most enzymes would be horribly denatured under such conditions Forthis reason biochemical reactions will often be presented in reference to a modified standard state conditioned where pH 7 We will call this the physiological or pHneutral standard state and AG underthese conditions will be designated as AG This is the symbol used in Segel39s book Your GampG text uses AG 39 instead There are a few different ways to generate an accent symbol The apostrophe is not going to work in equations because Mathcad interprets the apostrophe key as parentheses For simplicity let39s decide instead to use the accent mark in the upper left of the keyboard underneath the tilda sign In text regions this accent will look like it39s going the quotwrong wayquot but in equations it39ll look just fine In reactions that do not involve H ions or 0H ions AG AG so there is no difference between these terms In reactions that DO involve H ions AG 2 AG Instead we must calculate AG AG when all species are at 1M except for H ions and OH39 ions which are at 10397M pH 7 Considerthe following problem from your text GampG pg74 5 ATP hydrolysis at pH 70 is accompanied by release of a hydrogen ion to the medium ATP439 H20 3 ADP339 HP04239 W If the AG for this reaction is 7305k I1 at 25 C what is AG that is the free energy change mo for the same reaction with all components including H at a standard state of 1M Let us begin by making assignments for what is given in the problem k In the space below assign AG to be 7305 mol 501 AG 7305 mol Module02 17 GibbsFreeEnergyAnsweredxmcd BIS 102q Module 2 Gibbs Free Energy Check below that T is still assigned to be 25 C by asking Mathcad what T is The answerwill initially have units of Kelvin K Enter C in the units placeholder on the far right of region to get Mathcad to return the answer in C If for some reason T is not assigned to 25 C make an assignment to set it to this value here as you did in Excercise 312 T 25 C Recall from Part I that AG can be defined in terms of Q and AG AG RT1nQ AGO In the problem we are charged with finding AG We will take the strategy of rearranging the above equation and then substituting AG39 forAG From there we need only determine Q under pHneutral standard state conditions Use dummy equations below to arrive at an equation for AG in terms of AG and Q AG RT1nQ AGO AGO AG 7 RT1nQ Copy your dummy last equation from above and paste it here below We39re now going to assume that AG AG so modify your equation below to make that substitution 504 AGO AG 7 RTnQ Now Q forthis reaction is defined as Q ADPHPO4H ATP 80 what is the value of Q for this reaction under pHneutral standard state conditions Make a dummy equation and don39t forget the units 505 Modue02 18 GibbsFreeEnergyAnsweredxmcd axsmzu Meme omnsrveegnevev hmshed asswgnmem shuum uuk hke W5 Then change We Eun ean eeuaxsm an asswgnmem upevamvm acwa e me equauun Vuuv Memeer 7M2 2 e7 2 m M geman M2 Mme space be uw ask Mathcad what AG 5 wei m1 Vuu shumd 32 an anwev u 9 wirwmch sthe answenu We pvumem u a vpHmeu va s andavdrs ate cundnmns wmthwsveac mn pvuceed spumaneuus W Exp am yum veasumng sue nwm pvucede spumaneuus y because AG 5 negawe gwen m We pycmem as can ma spumaneuusm Exp am yuuvveasumng ma The veammn s m spumaneuus at standavdrsme cundmuns because AG 5 pusmve k We demvmmed m be 9 459 Mudme z GmbsteeEnemvjnwEvedxmcd BIS 102d Module 2 Gibbs Free Energy In words explain why pH affects whetherthis reaction procedes spontaneously In particular contrast the two situations you just considered ie pH 7 vs pH 0 H ions are a product in this reaction Decreasing pH means increasing the product concentration ofthis reaction which will tend to make the forward reaction less favorable Thus it is not suprising that at pH 7 the reaction procedes spontaneously but at pH 0 it does not Module02 20 GibbsFreeEnergyAnsweredxmcd E S mm Meme 2 omnsrvee EnEmv Part VI Coupled Processes Goa s Wenhe uvevaH Kee va ue ma paw Mcuup ed pvucesses Ca cu ate the wave AG enevgeucaHyVavuvame unes CunswdenhemHuwmg euesuen mm yuuHex 6amp6 pg 74 63 Fnrlhe pmeess A a 3 Ken AB is um 213 37 C Determine K 7 C Fnrlhe pmeess a a cam 13c1nnn 21 ee Keg AB and Ken 13c 1ACIhe equ hrium cnnslznl m me werle pmeess A a c mm y pvub em we HWanHu gwe hemaH umeuenames Lefsdemde un KiAB Ki C and KiAC The va uesmv KiAE and K750 ave warn and We ve chavged Wm ca cu a mg KiAC Make asswgnmems be uwmv KiAB m be um and mac m be1 KiAE e an ch e mun Mhe abuve veadmns me Vespecwe Ken s ave de ned asmHqu E K EC 2 K Ac 2 A E Tu sunk2012 pvub em Vuu must pm K Ac m tevms e K AB and K ac Theve ave a enume e1 Wavsm appvuach we Let stakeme stva euv Meemne m a m tevms MK AH and Anna then usmg 0m expvesswunmv a e subsmme m um expvessmn m K750 m tevms MA and KM KiAE E A a e A KiAE Mudme z 21 emueneegnevevAnweeWe BIS 102q Module 2 Gibbs Free Energy Use the expression you just made for B and the definition of K BC above create one or more dummy equation below that finally yields in terms of KAB and KBC K BC BAK AB B c K BC AKiAB c K BCK AB A Now use the definition of KAC to express KAC in terms of KAB and KBC below Change the Boolean operator in the final expression to an assignment operatorto quotactivatequot the equation KiAC KiBCKiAB Ask Mathoad below what KAC is KiAC 20 You should get an answer of 20 which is the answer to the problem Modue02 22 GibbsFreeEnergyAnsweredxmcd E S mm Mama 2 omnsrvee EnEmv Let us cunsmv anuthev Enumed veacnun pvub emhum vuunex 6amp6 nu 74w and ADP temperature glucnsdiP glucnsea Pi ATP ADP Pi U ma M H h m m m AG g ucuse ATP g ucuseEP ADP Aha a a m w m rm w swan Fm cummeteness We mmm be umun m vewme Me New veactmn m be g ucuse Pw ATP lt gmmseEP ADP Pw yu r m n m va uesmnhe Mu cuup ed veacuuns mm space be uw get Mamcad m ca cu a e Me New AG mm abuve veacuun Dun HuvgeHu mm mm ans k u arms ng m1 m1 ma ma The secund pan asks yuu m ca cu ate Me New Keq knuwmg AG Mudme z 23 GmbsteeEnemvjnmvedxmcd E S mm Mama 2 omnsrvss EnEmv Pecang ms dehnmunmv AG AG R T 1 i m ca cu ate Keq be uw Vuu Hwam m use yew dummy equauunsm su ve m Keq vs And dam c quotgm nuw assume m We mumgw am Nme L T31EIISK Keq 524 443 pyumsm Mudme z GmbsteeEnemLAnmvedxmcd E S mm Mudme 7 Ammu Acms Amino ACIds Part I Introduction Goa s R ewewme amdrbase pmpemes mammu amds Sm up um cun ams Set up cunvevsmn mm mnsmv pH H pm and Ka The amdrbase pvupemes u ammu amds ave smeaHu Muse u mhev pu ypvuu amds The New AH addmuna pvmun accemmgdunatmg gmup umhew 5mg chem and a 1 chavge my me ammu gmup mm m m m 2 2mmn2mva 51am ave knuwn as a zwmenuns Sumnuns Mzwmenun ammu amd ave um caHed sue ecm sumuuns le mm l mm mm um mum Au39mmr arm Fm mm chavge um WHHE m HA spemes Ma amne s a zwmenun m m chavge pmypmu amds Fm exampb we Wm 51m use We demmun mum amd mssumauun cuns am K m h h amdbase gmups Mudmem 1 AmmuAmusjem Anmveumm BIS 102q Module 7 Amino Acids H2AH HAH AH Kal Ka2 Ka3 H3A H2A HA where Ka1 Ka2 and Ka3 are the Ka values for the respective reactions all with units of M molar H is the molar concentration M of H Note that no effort is made to distinguish these Ka values based on where the groups are located on the amino acid or the type of acidbase group ie carboxyl amino phenol etc Only the relative values of each Ka is implied Namely that Ka1 gt Ka2 gt Ka3 Amino acid salts are formed by adding either strong acid or strong base to the zwitterion For example leucineHCI which we would represent as H2A Cl is known as leucine hydrochloride Similarly leucineNaOH which we would represent as A Na is known as sodium leucinate ln problems involving solutions of amino acid salts use ofthe HendersonHasselbalch HH equation is often appropriate Base pH pKa log Acid where Acid and Base are the molar concentrations M of weak acid and its conjugate base respectivel y pKa Is the log of Ka unitless Recall that the HH equation is most useful accurate near pH pKa where Acid Base Thus depending on the concentration ofthe acid species involved andorthe pH of the solution you must select the pKa value most appropriate for use in the HH equation From there your HH equation will only consider the two ions involved in that single reaction designating them as Acid and Base as appropriate This approximation using only a single HH equation for an amino acid solution is most accurate when the pKa values are sufficiently different that only two ion species exist at appreciable concentrations at the pH value in question Put another way we assume that the other reactions involved and ignored are all the way to the left or right and are unaffected by changes in pH or amino acid species concentrations This module makes use of Mathcad Find Solve Blocks If you haven39t already you should now complete the Find Solve Blocks MiniModule Welcome Back Module07 2 AminoAcidsSemiAnsweredx mcd BIS 102d Module 7 Amino Acids Before working some problems let39s first establish some unit constants that are not builtin to Mathcad In the region below use the assignment operator to set M to be You can do this by 1ter typing Mmoliter mol 397 liter Based on this assignment of M create an assignment below for lel by setting it to be 10393M You can do this by typing lel10 3spacebarspacebarM mM 10 3M As you39ve seen liter is a builtin Mathcad unit However ml is not Create an assignment for ml below to be 10393Iiter Make a similar assignment for mmol to be 10393mol 103 ml 107 31iter mmol 107 3mol Assign the equilibrium constant for water autoprotolysis Kw to be 103914M2 below You can do this by typing Kw10 14spacebarspacebarM 2 104 Kw 10 14M2 In this module it will be convenient for you to easily convert H into pH and vice versa 80 too with Ka and pKa Let39s anticipate this and create fourfunctions that will handle these conversions To avoid confusing these conversion functions with variables that you use in your problems the name of these functions will all be appended with an quotfquot 80 the conversion functions will be pHf Hf pKaf and Kaf Module07 3 AminoAcidsSemiAnsweredx mcd E S mm Mudme 7 Ammu Acms mm m mum mummy be uw Tm V NspH smmmn Wm take a va ue u H and 12mm 012 Huwmn m su ugamhm 5 Emma pm 14 pm 14 mm m quot rm mummy be uw Thwsmnmmn Wm take a pH va ue and vemmthe cunespundmg H cummanun Thetwu ave mated sucmha Huwmmm vesun uthsmm mn H pxn m39 PHM H pxx m pHM 2 caveM m pHM 5 mm same as m39PH M Be suve m mummy by M a enhe expunem s evama em nm bemve uw mam swm avmndmnsmv pmij and mij be uw MaMemaunHyJWse mums 10 and rum q n q he um s cum yuu H need m dmde M um um bemve akmg me Jug UVKa And 5 mu yuu H need m mummy by M a ev yuu ve taken m m m era puwev m7 pm m 4 am z m i K M Mudmem 6 AmmuAmusjem Anmveumm BIS 102q Module 7 Amino Acids Part II Titration Curve Goals Use a solve block to calculate pH and amino acid species concentrations during titration with strong base Graph a partial titration curve of an amino acid using a Find Solve Block Considerthe following problem from your text GampG pg101 5 Draw an appropriate titration curve for aspartic acid labeling the axes and indicating the equivalence points and the pKa values For aspartic acid pKa1 21 pKa2 39 and pKa3 98 This question is a little vague in terms of what an quotappropriatequot titration curve should look like For example should we start with aspartic acid hydrochloride and then titrate with strong base eg NaOH Alternatively perhaps we should start with isoelectric aspartate and then titrate in one direction with strong base and titrate in the other direction with strong acid In the absence of clear directions let us assume that we start with isoelectric aspartate This technique has the advantage that the acidic quotstarting pointquot of our titration curve is not fixed We can start and end at any pH we want Note that this setup differs from the solution presented in JampT Another parameter left to our discretion is the total molar concentration of aspartic acid in solution Titration curves often omit this opting instead for unitless base or acid quotequivalentsquot which are the ratio of the strong base or acid concentration to the total amino acid concentration While the generalized nature of equivalents is attractive Mathcad requires that the results of a Find Solve Block have consistent units because the result is a vector and all values within a vector must have the same units 80 if we express for example Na concentration in equivalents we must also express H concentration in equivalents which makes the calculation of pH somewhat awkward Forthis reason let us use molar concentrations exclusively Arbitrarily we39ll assume that the total aspartic acid concentration ie A HA H2A H3A is 1M As a final assumption we will assume that our solution has a constant volume 80 adding acid or base does not dilute the total amino acid concentration As implied above you will use a Find Solve Blockto construct the complete titration curve three solve blocks actually but we39ll get to that later The reason for using a solve block is that no single HendersonHasselbalch equation will allow use to compute pH for all amounts of added strong acid or base Let us begin by making assignments for what was given in the problem namely the pKa values Use the assignment operator below to assign pKa1 to be 21 pKa2 to be 39 and pKa3 to be 98 201 pKa1 21 pKa2 39 pKa3 98 Module07 5 AminoAcidsSemiAnsweredx mcd BIS 102q Module 7 Amino Acids In the solve block we will solve for H initially and then later convert that value to pH to answer the question This means that we39ll be using the definitions of K211 K212 and Ka3 in the solve block For that reason we39ll need to make assignments for those Ka values Use your Kaf function below to make appropriate assignments for Ka1 K212 and Ka3 below Below each assignment ask Mathcad for the value of that Ka Kal Ka7fpKa1 Ka2 Ka7fpKa2 Ka3 Ka7fpKa3 Kal 7943 X 10 3M Ka2 1259 X 10 4M Ka3 1585 X 10 10M You should find that Ka1 7943 X 10 3M Ka2 1259 X 10 4M and Ka3 1585 X 10 10M Before we construct the Find Solve Block we39ll need some guess values for the species we39ll be solving for We39ll solve for 4 aspartic acid species A HA H2A and H3A Of course we39ll also want to know H so we can calculate pH And finally we39ll also want to know 0H because at very basic pH OH ions will play an important role in our Conservation of Charge equation Our guess values need not conform to any of our equations In fact to get Mathcad to successfully find solutions it is often advantageous to offer guess values that do not conform to your equations For example we assume that total aspartic acid concentration is 1M However our subsequent Solve Block seems to work well at least under Mathcad 131 if all the aspartic acid species are set to 1M Of course this implies a total concentration of 4M not 1M but that39s OK These are just guess values The solve block will eventually find values that conform to our Conservation of Mass equation that requires the sum of all aspartic acid species be 1M Use the assignment operator below to assign a value of 1M to A HA H2A and H3A When you39re finished you should have 4 separate assignments 203 A1M HA1M H2A1M H3A1M Use the assignment operator below to assign the value of 10397M to both H and OH This arbitrarily sets the guess values for H and OH to that of a pHneutral solution 204 a 10 7M OH 10 7M We will initially be titrating aspartic acid with the strong base NaOH The amount of NaOH added will be reflected in the concentration of Na ions in solutions Na For now we39ll just assume that no base has been added Later Na will be a parameterthat you specify Use the assignment operator below to assign Na to be 0 zero Na 0 Module07 6 AminoAcidsSemiAnsweredx mcd E S mm Mudme 7 Ammu Acms Beam yum suNe muck nuw by ypmg thewuvd Given m m Wm space be uw m Gwzn sum b uck SD n duesn ma evwhmh m We s1anwnh m Let s heavywth K21 Kn L H HZA 22ch m mm M sumuun m m cunvevge equwabm mmn HZA Kn HZA H WMEH Mama seems much muve Wthg m wuvkwnh Use We Eun ean annex swan m mam memHuwmg nemmun m K21 be uw HZA Kn HZA H 2m HZA Kn HZA H mm svm av equatmns be uwmv K22 and K23 usmg me Eun ean equaxs swan HzAKnHAH HAKn 2m H2AKnHAH HAKnAH Eavhevwe assumed mnmax aspamc amd m mmquot qud be ma a 1M Let us nuw cvea e a Cunservauun u Mass equatmn thatwm EMDYEE 0m assumpuun H3AH2AHAAIM H3AH2AHAAIM Mudmem AmmuAcmsjem Anmveumm BIS 102q Module 7 Amino Acids The next equation to create is for Conservation of Charge That is all the positive charges in solution must equal all the negative ones We decided earlierthat this solve block will deal with the addition of strong base NaOH to aspartic acid Later we39ll create another solve blockthat handles titration with strong acid HCI The addition of NaOH is easily dealt with by simply tracking the concentration of Na ions in solution As a strong base every mole of NaOH added to the solution will produce 1 mole of Na The same cannot be said for OH39 because the autoprotolysis effect of water can combine added OH39 with existing H to make water In writing out the equation of conservation of charge one must carefully account for the net charge on the various amino acid species In the case of aspartic acid this is Species Charge A 2 HA 1 H2A 0 H3A 1 Use the Boolean equals sign to create the following Conservation of Charge equation HH3ANaHA2AOH HH3ANaHA 2A OH Note that H2A having no net charge is simply omitted from the conservation of charge equation Also note that we have included OH in the conservation of charge equation This means that we now have 6 unknowns but only 5 equations so far The sixth equation is simply the autoprotolysis of water Use the Boolean equals sign to create an equation for the autoprotolysis of water Kw HOH Kw HOH Having entered all the equations for the solve block we should now enter some constraints for our unknowns These constraints are that all the ion species except Na must be greaterthan zero The value of Na will be dictated by the amount of NaOH we add to the solution so we39ll soon explicitly specify Na Module07 8 AminoAcidsSemiAnsweredx mcd BlS102q Module 7 Amino Acids Create the following constraints below AgtU H2AgtD Hgt HAgtU H3AgtU OHDD 211 Agt0 H2Agt0 Hgt0 HAgt0 H3Agt0 OHgt0 Now that your solve block contains all six equations and six constraints we39ll end the solve block with a Find statement This will ask Mathcad to find values for our 6 unknowns Using the regular equals sign below ask Mathcad what FindAHAH2AH3AHOH is Set the displayed units to be M 212 AspiNaOHNa FindAHAH2AH3AHOH 161gtlt1078 0101 0799 In Mathcad 131 Find should return the vector M which contains from top to 9951 X 10 4 0 bottom the molar concentrations of A HA H2A H3A H and OH when no NaOH is added to the isoelectric aspartic acid solution If you39re using Mathcad 131 you39ll notice that our constraint of OH gt 0 wasn39t strictly met That39s OK OH is in the neighborhood of K 7 W 1005 X 10 11M which as far as Mathcad is concerned is the same as zero 9951 X 10 4M There are ways of forcing Mathcad to be more precise in its answers but we39re not going to bother Mathcad 14 should return the correct nonzero value for OH Module07 9 AminoAcidsSemiAnsweredx mcd E S mm Mudme 7 Ammu Acms Whatwe d hke m an nuw s paveme enze We sum Muck yuujus made Thwswm aHuw usm sum m nuns unknuwns a any avbmavy cuncsmvauun u NaOH n essence We Wm mm m smsmsm such mm m s asswgned m a newmnmmn m mm m H y m m m mns m suhmun 0H 1 H y y m mh mam m m hm N N2 2 0H Su m essence WW2 spsswy mum 0H Wm Maw mpm veaHy N2 aswejus mscussed mess veasuns M s caHWsmnctmn Mpi zo mzj Change We abuve Fwd smsmsm such mm m s asswgned m We mummy Mpi zo mzj AspiNaOIKNA FmdAHAH2AH3AIIOH m Ask Mama be uw Wm MILNZOHM 5 Change We mspxsysu unmsm M m AypiNaOIKEI s mm x m Vuu shumd gems samevecmvas yuu had pvewuus y ca cu ated abuve mm n h MILNZOHM m AypiNaOIKEDA 9951 x m39AM AspiNaOIKU 9951 x m39 AM Vuu shumd have sums up MW e vegmn ma uuks sumsmng hke W5 Mudmem m AmmuAmusjsm Anmvsumm BIS 102d Module 7 Amino Acids To make the titration curve we39ll need to calculate pH from H as returned by the AspNaOHNa function Just to make things easy on ourselves let39s create a new function based on AspNaOHNa that extractsjust the value for H and then uses our pHf function to convert that value to pH V thout thinking too much let39sjust call this new function pHNaOHNa Using the assignment operator below assign the function pH Na0HNa to be pH7fAsp7NaOHNa4 216 pHiNaOHNa pH7fAsp7NaOHNa4 Ask Mathcad below what pHNaOH0 is 217 pHiNaOH0 3002 You should get an answer of pH 3002 The fact that this is very close to w 3 is no accident We39ll explore this fact more carefully later in this module But for now let39s define a range of NaOH concentrations and create a graph of that titration curve Assign NaOH below to be a range variable The first value should be 0 the second value should be 20lel and the last value should be 3M 21a NaOH 020mM 3M 15 pHNa0HNaoH 10 5 Module07 l l AminoAcidsSemiAnsweredx mcd E15 1mm Mudme 7 Ammu Arms 1nme Mme Space above evea1e a new xrv PM Seme wawsm be 0 15 M Seme v39awsm be pH NaOHWaOH pH mommamu D enmnu an 012 Speed mum meme 1 5 mavtake Mama 3 enume mmmesm ea1eu1a1e 3 15m semuunsmv vuuvmaph M51 1 1 be pauem 1 2 111011 Sumethmg abuuHhe mam curve Mame pmduced m yuu 1s m eunee1 Thmk abum haw yuu 12 51111 s1hatMathcad has yeMmew Tu gem same msgm mm Wm 15 gumg awvy ask Mame bemw Wm MILNZOHEHML Mpi zo JMj andAslezOH JMj ave Change hemsp ayed unnsmbmhanwevsm M 221 n 994 3 1 6148 x m n 4789 x m7 n I mimosa HUM e n M mimosa 1M e n M 9 EDS x In 3 U n 1 6 148 x 1n393 1 8 687 x 1n399 n mimosa 2M e n M l 924 x In 5 12 Mudmem AmmuAmesjemAnweveemm BlS102q Module 7 Amino Acids I39m not entirely sure what39s going on here but apparently something very wrong happens to our solve block around NaOH 2M and H becomes 0 at least transiently My interpretation of this is that Mathcad 131 and Mathcad 14 have difficulty dealing with A and H when these variables are 1 each very important to the solution and 2 differ by more than 14 orders of magnitude Let this be a lesson that you should always be skeptical of the answers you get from a computer Always make sure you understand why the answer does or does not make sense Fortunately there is another way to compute solutions when NaOH is large we can assume H is zero and then compute pH indirectly by knowing that pH 14 pOH However this operation will require a new solve block 80 for now let39sjust not deal with NaOH values beyond 15M Copy your range variable assignment for NaOH and the graph of pHNaOHNaOH from above and paste them here below 222 NaOH 020mM 15M 1U 8 pHiNaOHNaOH 6 4 l l 0 05 1 1 5 NaOH M Change the range variable for NaOH above to end at 15M instead of 3M 223 Module07 13 AminoAcidsSemiAnsweredx mcd BIS 102q Module 7 Amino Acids Part III More of that Titration Curve Goals Create a solve block to calculate pH during titration beyond 2 equivalents of strong base Graph the titration curve over 0 to 3 equivalents of strong base As mentioned in the previous Part we can get around Mathcad39s unwillingness to accurately calculate pH for large values of NaOH by assuming that H is zero and calculating pOH instead In addition to further simplify the problem for Mathcad we can reasonably assume that H3A is zero as well in this range of basic pH One complication in assuming H is zero is that all our Ka definitions involve H There are a couple of ways to deal with that issue The most straightforward is to simply use the base dissociation constants Kb instead H3A OH H2A OH HA OH Kb1 Kb2 Kb3 H2A HA A where Km KW m KW Km Kal Ka2 Ka3 Before we create the solve block that will use these definitions let39s make sure we39re clear on the algebra behind these expressions Use the Boolean equals sign I below to create the following dummy equations HAH H2AOH Ka2 Kw HOH Kb2 H2A HA 301 HAH H2AOH Ka2 KW H OH Kb2 H2A HA Module07 14 AminoAcidsSemiAnsweredx mcd E S mm Mudme 7 Ammu Aems Km 7 K W K52 mz amp OH KW H2AK K52 HA H H m HA EL KbZZIIZALKW K52 HA H HA H 1032 Kw K52 Kw Km 7 m OED Use W5 yexanensm and me asswgnmem upevatuv be uwm make asswgnmemsluv Kid and Kid m evms emw K22 and K23 vespedwe y We Wem buthevwnh Kh1 becausewe H assume ma5 H A s zem and SD Wem meme a Kh1 equatmn m the Su ve week 1032 e Kl Kb Kl K52 K52 m 1032 e K W Kb K W K52 K52 HA H y u n m h because we un y be Wmka WM veaHy base pH va ues Use the asswgnmem upevatuv be uwm asan 0H m b21M and Na 5e b215M am we 1M gr 1 SM Gvand Nuw etsbm dthatsmveE uck Stamhe Sewe meek by ypmg me We Given be uw 3m Gwm Mudmem 15 AmmuAcmsjem Anweveemm BIS 102q Module 7 Amino Acids Use the Boolean equals sign I I to create definitions for Kb2 and Kb3 below As in Part II we39ll want to arrange these equations such that we39re not doing any division 80 you equations should look like this HAKb2 H2AOH AKb3 HAOH HAKb2 H2AOH AKb3 HAOH Construct your Conservation of Mass equation below Note that compared to the equation you used in Part II this one omits H3A H2AHAA1M H2AHAA1M Construct your Conservation of Charge equation below Note that compared to the equation you used in Part II this one omits H3A and H Na HA 2A OH Na HA 2A OH We have four unknowns A HA H2A and OH And we now have four equations 80 that39s all we need eg we don39t need to deal with water autoprotolysis Create the following constraints below AgtU HAgtU H2AgtU OHgtU Agt0 HAgt0 H2Agt0 OHgt0 Module07 16 AminoAcidsSemiAnsweredx mcd BIS 102d Module 7 Amino Acids Using the regular equals sign end the solve block below by asking Mathcad what FindAHAH2AOH is Set the displayed units to be M AspiNaOH2Na FindAHAH2AOH 05 05 Mathcad should return the vector 6 297 10 7 M which contains from top to bottom the molar gtlt 6308 X 10 5 concentrations of A HA H2A and 0H when 15M NaOH is added to the isoelectric aspartic acid solution Let39s take a timeout to run some checks on this new Solve Block Your previous solve block will return a pH value of pHiNaOH15M 98 for this point on the titration curve Using the present solve block the same point would have pH of 14 7 pHif 6308 X 107 5M 98 Hopefully those two numbers are the same if not there39s a problem with your solve blocks Note also that 15M NaOH is 15 base equivalents added to our 1M isoelectric aspartic acid solution so the pH at this point should be equal to pKa3 the fact that A HA here is another clue that we39re at pH pKa3 And indeed pKa3 98 that should say 98 if you assigned it correctly back in Part II Now that we39ve got that out of the way let39s parameterize your new solve block so we can use it to make a graph To do that we39ll assign your Find statement to a new function V thout thinking about it too much let39sjust call this new function AspNaOH2Na since itjust extends the range of ourAspNaOHNa function from Part IL Change the above Find statement such that it is assigned to the function AspNaOH2Na Don39t forget the 2 AspNaOH2Na FindAHA HZAOH Ask Mathcad below what AspNaOH215M is Change the displayed units to M 05 05 As NaOH215M M p L6297gtlt10 7 6308 x10 5J Module07 17 AminoAcidsSemiAnsweredx mcd BIS 102d Module 7 Amino Acids You should get the same vector as you had previously calculated above Similar to what we did in Part II let39s now make a function based on AspNa0H2Na that will extract OH and use that to calculate pH We39ll be a little tricky here and use the pHf function to calculate pOH first because the math is the same and then subtract that answer from 14 to get pH In keeping with Part II we39ll simply call this function pHNa0H2Na Using the assignment operator below assign the function pHNaOH2Na to be 14 7 pH7fAsp7NaOH2Na3 Again don39t forget the 239s pHiNaOH2Na 14 7 pH7fAsp7NaOH2Na3 Ask Mathcad below what pHNa0H215M is pHiNaOH215M 98 Again you should be getting pH 98 forthis point on the titration curve We39re now ready to make a graph with your new pHNa0H2Na function We know that this function works well at NaOH 15M so let39s start there and increase NaOH up to 3M 3 base equivalents Let39s not mess with the NaOH range that we defined in Part II though Instead we39ll assign this new range to Na0H2 Assign Na0H2 below to be a range variable The first value should be 15M the second value should be 152M and the last value should be 3M NaOH2 15M152M 3M Module07 18 AminoAcidsSemiAnsweredx mcd E S mm Mudme 7 Ammu Acme Cupvvuuvmaph cva Naonaonum Pan H and paste n have be uw Add an addnmna Hm venue u mom pHiNaOI39KNaOH 39 M Add an addnmna vrawsuace u PHJ39A DHW OHZ 5 pHiNaOHKNaOIIZ Depeneme an the speed mum machme m mawake Mamcad a enume mmmesm 2 ca cu ate 3 We sumnunsvm Vuuv maph NaOH mom m1 be panem 7 T payedImam U39 pHJaOHKMOIn 5 made m PanH7 Huwsu7 Mudmem AmmuAmesjem Anweveemm BIS 102q Module 7 Amino Acids Part IV The Last of that Titration Curve honesU Goals Create a solve block to calculate pH during titration with a strong acid HCI Graph the full titration curve Identify equivalence points and pKa values on the titration graph In the course of completing Part II and Part Hi you have created a partial titration curve for aspartic acid by adding NaOH to isoelectric aspartic acid In this Part you will complete that titration curve by titrating in the other direction adding a strong acid HCl to isoelectric aspartic acid This will require you to build a new solve block This new solve block like the solve block in Part II will use Ka definitions However to get to very acid pH values we will need to make some simplifying assumptions much like we did to get to very basic pH values in Part III Namely we will assume that OH and A are zero If you scroll backto Part II you should be able to convince yourself that these values are already close to zero even before we add HCl This new solve block requires minimal preparation We only need to beef up our guess value for H and assign CI to initially be 0 zero Similar to how we used Na to represent added NaOH in Parts ll amp lll here we will use the concentration of CI to quottrackquot the amount of added HCl Use the assignment operator below to set H to be 1M and set CI to be 0 g 1M C1 0 Now begin the solve block by typing the word Given below Given Construct equations for Ka1 and Ka2 using the Boolean equals sign below Because we39ll be assuming that A is zero there39s no need to make a Ka3 equation You can copy the Kai and Ka2 equations from Part II and paste them here if you like HBAKal H2AH HZAKaz HAH 403 H3AKa1 H2AH H2AKa2 HAH Construct your Conservation of Mass equation below Again we are omitting A HBA HZA HA 1M 404 H3AH2AHA1M Module07 20 AminoAcidsSemiAnsweredx mcd BIS 102d Module 7 Amino Acids Construct your Conservation of Charge equation below To represent the addition of HCI we now track the anion CI39 instead ofthe cation Na Note too that both OH and A are omitted because we assume their concentrations to be zero HH3AHAC1 HH3AHAC1 We have four unknowns H HA H2A and H3A And we now have four equations 80 that39s all we need eg we don39t need to deal with water autoprotolysis Create the following constraints below Hgt0 HAgt0 H2AgtD H3AgtU 406 Hgt0 HAgt0 H2Agt0 H3Agt0 Using the regular equals sign end the solve block below by asking Mathcad what FindHHAH2AH3A is Set the displayed units to be M 407 AspiHC1C1 FindHHAH2AH3A 9951 X 10 4 Mathcad should return the vector 039101 which contains from top to bottom the molar 0799 01 concentrations of H HA H2A and H3A when no HCI is added to the isoelectric aspartic acid solution You may recognize these values from Part II when we first did the quotfullquot solution with 6 unknowns and no added NaOH Let us parameterize this solve block now assigning it to the function AspHCICI Change the above Find statement such that it is assigned to the function AspHCC AspHCICI FindH HA HZA HBA Module07 21 AminoAcidsSemiAnsweredx mcd BIS 102d Module 7 Amino Acids Ask Mathcad below what AspHCI0 is Change the displayed units to M 9951 X 10 4W AspiHC10 0101 M 0799 01 You should get the same vector that you had previously calculated above Let us now make a make a function based on AspHCICI and pHf that will calculate pH for any arbitrary HCI concentration In keeping with past unimaginative names let39sjust call this function pHHCC Using the assignment operator below assign the function pHHCICI to be pH7fAsp7HC1C10 pHiHC1C1 pH7fAsp7HC1C1O Ask Mathcad below what pHHC0 is pHiHC10 3002 You should get an answer of pH 3002 You may recall that this is identical to the value of pHiNaOH0 3002 that you calculated in Part II This is an indication that our decision to ignore OH and A in this pH range is OK We39re now ready to define a range of HCI concentrations and make a titration graph with your new pHHCC function Use the assignment operator below to assign HCI to be a range variable The first value should be 0 the second value should be 20leI and the last value should be 2M 412 HC1 020mM2M Module07 22 AminoAcidsSemiAnsweredx mcd axsmzu Mudme 7 Ammu Arms Cveate a new gtltVV mm be uw 101 M W me Mamcad m mm mcveasmu How mm mmm M va henhan vmmm 2 The ma unmst became muve c eavwhen 2W5 HCHmaHun wnh Vuuv NaOH same mm be The negatwe swan vuu mvatmn 2 pHiHClUICD n same y awsm be pHiHCKHCD z 2 pH HCIHCI 1 Cnpy yum NaOH mvauun gvaph mm Pan m and paste n have be uw pHiNaOHOhOH m pHVmexxmom 5 pILHCKHCD Mudmem AmmuAmusjem Anmveumm E S mm Mudme 7 Ammu Arms Add vuuv How mvahuh m the ahuve maph Tu an thws 2 Add ah addmun Hm yahgevm 5 pH MOI1mm m an ah addmuna vrawsuace m P OH2 OH2 p How u H 110 am Chanqe aH menacesm be suhd hnEs L 4 uHhe same mm hm su vuuv maph duesn t uuk qune su mu ey 1n 2 4 2 NAOH mom ixxcx M M M 5 That sthe bastuHJmaHun curve Huwevehwewm a su asked m mdmate the muwa ence Th h Hh nK Hm m m y pKa s mavkevsWumwuvk Th hh h v m h N dummy vanab e man M s aHWs dummyvanab e x Use the ass gnmem upmmy wqu ass gn xm be 3 137132 VaHab E The mswamg shuum be 1 and the aswa ue shumd be 3 m x 72 3 Cnpy yuumhahuh gvaph hum abuve and paste n have be uw Add memHuwmg y awshacesmnhe muwa ence pumts Add memHuwmg ham puiucmwn N a h s smw vangesm get Evevylhmgm pH7N20H mm cunec y pH7N20H11M r1 pHleOHZ Mj n Add memHuwmg y awshacesmnhe pKa s 1 pm 2 mm x pKz3 that any Ema y awshaces shumd be mum agams he as Hm vangE Whmh h W5 case sx w Mudmem AmhuAmusjem Ahmveumm BIS 102d Module 7 Amino Acids pHiNaOHNaOH pHiNaOH2NaOH2 pHiHC1HC1 m pHiHC11M XXX pHiNaOH 0 pHiNaOH1M nun 5 pHiNaOH22M ooo pKal 72 71 0 1 2 3 NaOH NaOH2 7 HCl 71012x M M M To make you graph look appealing Make the graph wider once you make it wide enough Mathcad will change the default xaxis range to end at 3 and put a tick mark every 1 instead of every 2 Change the equivalence point traces to display a black symbol choose a different symbol for each trace Change the pKa traces to be dotted lines but different colors the should already be different colors 41a Module07 25 AminoAcidsSemiAnsweredx mcd E S mm Mudme 7 Ammu Acms pHiNaOI39K mom pHiNaOHZQM ltgtltgto pKal pm pm Vuuv gvaph 5mm uuk sman m une What mm eqwaxem pumts un yum gvapw Wham u they uncuy7 What many 5 an wuwa em pumW Mudmem AmmuAmusjem Anmveumm BIS 102q Module 7 Amino Acids In your graph above The third equivalence point at NaOH 1M is at a steep part ofthe titration curve yet the second equivalence point at NaOH 0 our starting point isn39t steep at all In your own words explain what is going on here 420 What39s going on at the outer edges of your graph Does the relative concentration of H3A change once the added concentration of HCI exceeds 1M Does the relative concentration of A change once the added concentration of NaOH exceeds 2M Why does pH keep changing once we get beyond these equivalence points 421 Module07 AminoAcidsSemiAnsweredx mcd BIS 102q Module 7 Amino Acids Part V pH of Isoelectric solutions Goal Understand how to combine two HendersonHasselbalch equations to calculate pH of an isoelectric amino acid solution In Part II you were asked to begin the titration assuming that we had an isoelectric amino acid solution That is a solution of the aspartate zwitterion As you later saw this quotstarting pHquot was actually at one of the equivalence points Based on what you39ve done in the past it39s not immediately clear how to calculate this pH value The zwitterion is both a weak acid and a weak base so in away the HendersonHasselbalch equation seems appropriate However with only one species in solution ie that zwitterion it39s unclearwhich pKa value to use and whether we should designate the zwitterion as HA or as A In truth no single HendersonHasselbalch equation will yield the correct answer We must use two However with afairly benign assumption we can avoid using a Solve Blockto arrive at a solution Considerthe following question from yourtext GampG pg101 9c Calculate the pH of a 03M solution of isoelectric leucine For leucine pKa1 24 pKa2 96 Let us begin by making assignments for what was given in the problem Use the assignment operator below to assign pKa1 to be 24 and pKa2 to be 96 501 24 96 How many protons does isoelectric leucine have associated with it That is should we specify this species as A HA H2A or H3A 502 The concentration of isoelectric leucine is given as 3M However in solution some ofthis HA species will loose a proton to become A and some of the HA will gain a proton to become H2A So the final values for A HA and H2A are unknown Forthis reason let us use the variable HA0 with a zero pronounced quotHAnaughtquot to represent the initial 3M of HA in solution Use the assignment operator below to assign HA0 to be 3M HA70 03M Module07 28 AminoAcidsSemiAnsweredx mcd E S mm Mudme 7 Ammu Acms HVH equauuns be uw usmg pH pKzL pKzZ A HA and HZA p p m p p HA ve atmnshm m suhmun H HZA A OH Hwe assume ma HZA 5 much gveatenhan H and halA 5 much gveatenhan OHJHEH We can veduce W5 equatmn m HZA hmmam HzAnu depend un HZA Agam 01252 shumd be made W m Eun ean squaxs swan su thanhey ave nm evamated by Mama 2 dummy equauuns p p A p p HA p p A p p HA Vuu shuum have made 01252 twu equatmns Mudmem AmmuAmusjsm Anmvsumm BIS 102q Module 7 Amino Acids Using as many dummy equations are you need below substitute one of these equations into the other to arrive at pH pKa1pKa2 2 506 H K1 1 H K2 1 A a0 a0 p p g A p p g HA pH pKal logHA 7 logA H39 K1 1 1 1 A a 70 70 p p g HA g H39 K1 1 1 1 A a 7 O O p p g HA g H39 K1 1 A 1 A 39 H K2 a 7 O O 7 a p p g HA g HA p p pH pKal 7 pH pKa2 2pH pKal pKa2 H pKal pKa2 p 2 Copy your last dummy equation from above and paste it here below Change the Boolean equals sign to an assignment operator to quotactivatequot the equation 507 H 7 pKal pKa2 P 2 Ask Mathoad what pH is pH6 Module07 30 AminoAcidsSemiAnsweredx mcd E S mm Mudme 7 Ammu Arms Vuu shumd nd mm pH E whmh sthe answenu he pvub em Huwevev we shumd nenee um um eeuanen m pH m W5 ease duesr amuaHv depend my me eeneenuanen uhsue ecmc eucme m suhmun u e HAi Tm s a eenseeuenee e um assumpuem at H2 and A ave much gveamnhan H and OH and Sn 5 nm smenyme Le us examme thwade uHha assummmn nuw A Nenhev HA quotEVA 5 mmedwate y knuwn Huwevev We du knuw HA7quot And mm Cunservauun e1 Masswe knuwthat HA A HA HzA and WEVE aheadv assumedthat HzA Whmh mphes HA 2A HA m HA HA 7 2A uHhe Ma s Heesm ma evwhmh equatmn yuu use A and HAquot A H K52la p p gHA 7 2A Mudmem AmmuAcmsjem Anweveemm BIS 102q Module 7 Amino Acids Use this new single HH equation to get Mathcad to return a value for A Use whatever technique you like If you do some algebra and then make an assignment make sure you show the intermediate steps with dummy equations If you solve symbolically you39ll may need to release Afirst by assigning it to itself If Mathcad returns more than one answerfor your symbolic solution try doing a little algebra first to eliminate the log Mathcad 131 has a known bug in its ability to symbolically evaluate logarithms If you have trouble changing the displayed units for your symbolic solution try assigning the solution to a variable like A 510 0000075318754513787126767 mol pH pKa2 log llter A solveA gt HA70 7 2A A H7 Ka2lo p p g HA7072Aj A 10pHepKaZ A solve A 0000075318754513787126768mol Mquot HA70 7 2A 11ter A 7532 x107 SM You should get an answer of A 7532 X 10 5 M Based on our assumption that H2A A this is also the value for H2A Module07 32 AminoAcidsSemiAnsweredx mcd BIS 102d Module 7 Amino Acids Get Mathcad to calculate the concentrations of H and OH below Use whatever techniques you prefer 511 Kw Hf H a 7 p H H1x10 6M OH1x10 8M For our assumption to be valid we can arbitrarily say that H2A and A must be 10X greater than H and OH Using this criteria was our assumption valid for this problem How might the validity of this assumption be affected by changes in concentration of isoelectric eucine ie HA0 A A 3 75319 7532 gtlt10 H OH Module07 33 AminoAcidsSemiAnsweredx mcd BIS 102q Module 7 Amino Acids Part VI More Problems Goal Use the tools you39ve created and the techniques you39ve learned to solve additional problems To solve the following problems feel free to use any of the functions or formulas you39ve created previously Rememberthat you can copy regions from earlier Parts and paste them here If you need to do any algebra show you intermediate steps with dummy equations using the Boolean equals sign To be clear about your answer white it out in words If you need additional space for your calculations you can create regions in the area to the right From GampG pg101 9a Calculate the pH of a 03M solution of leucine hydrochloride RKal 24 H2A70 03M W Ka7fpKa1 Ignoring water autoprotolysis HA H H Kal HZA H2A70 H2A HA H2A H2A70 7 HA H H H2A H2A70 7 H 0036606771128164557313 m01 H liter Hiq Kal solveH 7 H2A70 7 H 0032625699422629584805 m01 liter maxpH7fH7q 1486 Module07 34 AminoAcidsSemiAnsweredx mcd Module 7 Amino Acids B8102q From GampG pg101 9b Calculate the pH of a 03M solution of sodium leucinate 602 RKa2 96 A70 03M Ka2 Ka7fpKa2 K HAOH Kb2 W Kb2 m Ka2 A OH OH 00034758583877083425322m01 2 liter OH q Kb2 solveOH gt 7 A70 7 OH L 00034360476706529928071m01 J liter 14 7 pH7fmaxOH7q 11536 Module 07 Amino Acids Written by Eric Leaver 2009 Martin Wilson Last modified February 18 2009 1100am Module07 AminoAcidsSemiAnsweredx mcd BIS 102d Module 7 Amino Acids Help Improve This Module Completing this section is not required but will be helpful for future development ofthis module Your answers here will not affect your grade About how much time did you spend completing this module Were there any particular parts of this module that you found too difficult too confusing or too tedious How might these sections be improved Were there any particular parts that you found especially useful or helpful in understanding the material from BIS 102 Other suggestions kudos criticism Module07 36 AminoAcidsSemiAnsweredx mcd aTs Tum Mudme 5 HendevsuwHasseTba ch Eeuaueh The HendersonHasselbalch Equation Buffers Part I Introduction Goa s R ewewme bTthbTe cva bUWEvmg byweak acTus ahuTheTT saTTs R EVTEW HendevsuwHasse baTEH eeuaueh and NS apphcaueh Te caTcuTaTTeh e1 pH SeTubcbhyeTsTthuthTbhsTbTbH H bka ahuka e h SEI UHEIH a w ak acTu wTTHehu m uppuse unheT changes Th pH ThTs TsThe essehce cTT pH bUWEWvg ahu Ts ah ekampTe cube chaTeTTeT s pTTthpTe Tu ekaaTh ThTs TeT us bEgm wTTh buT uehhTTTcTh bnhe acTu uTsscTcTaTTcTh ED715137 K2 m ET HA wheTe HA Ts The cbhcehTTaTTcTh cTTweak acTu H Ts The cbhcehTTaTTcTh DVH TcThs A Ts The cbhcehTTaTTcTh bnhe weak acTu s ccThTugaTe base Because Ka Ts a ED715137 any chahee Th TH T musT TaT eeuTTTbTTum be meTwTTh sTmuTTahecTus chahees Th THAT andDY TAT The My scTuTce cTT ah mm uThe Em Tc HA s a depmmhahoh IoA hus any chahee Th H Hendstu be Uppused aT TeasT pamaHy by pTcTTcThaTTcTh cTTAcTT uebTbTbhaubh cTT HA 7 H Te chaTeTTeT Tabb me The same pTTthpTe apphesm cTTheT EYEHS EerDWYy acTusT such as HH WhTTe chavged NH Ts The eeuTyaTehT cTT HA Th buT thTaTTcTh because TT Ts a pTcTTcTh ubhbT NHJ Th TuTh wcTuTu be TepTesehTeu byA because TT Ts a mum acceprT The HendevsuwHasse ba ch HVH eeuaTTcTh Ts a useTuHcTcTT Th caTcuTaTThe The pH bTbuweTee Tweak acTu scTTuTTcThs AT TTs mus1 basTc evekthe HVH eeuaTTcTh Ts sTmpTy a cbhyehTehT TeaTTaheemehT m T e ka uehhTTTcTh such ThaT pH Ts ekaesseu Th Tehhs cTT pKz A and HA A a a 4 The HVH eeuaTTcTh can be aprTeuTcT a weak ach scTTuTTcTh ThaT has beeh pamaHy heuTTaTTzeu by a subhg base e g T HaoH ThTs snuaTTcTh Ts ahaTbgbusTb mTkThg a weak ach HA and TTs saTT e g NeA ahuTheh askmg abbunhe TesuTTahT scTTuTTcTh s pH The saTT Th ThTs case Ts hbThThg chTe Thah The bTbuucT cTTa weak ach heuTTaTTzeu by a subhg base The HVH eeuaTTcTh was ueyepreu Tc uescTTbe chaTbcThaTe bu eTThg Th chTcTu by LaWYEnEE u Hehuesth Th Tana ahu TaTeT pm Th bgaTTThmTc Tehhs by kaTTA HasseTbaTch u 1 kehueTsuhTeva 1942 MeeuTenajHjemTAhweTeekmee T BlS102q Module 5 HendersonHasselbalch quation As the HH equation is commonly used a couple of important assumptions are made In this module each assumption will be made explicitly as needed to simplify the mathematics However let us prepare ourselves by listing these assumptions up front 1 The contribution to H and OH by the autoprotolysis spontaneous dissociation of water is assumed to be small enough to ignore 2 The change in HA and A due to spontaneous dissociation as opposed to neutralization by the strong base is assumed to be small enough to ignore Conversely we will also ignore the changes in HA and A due to quotreassociationquot the simultaneous production of HA and OH What these assumptions should suggest to you is that the HH equation is only appropriate for calculating pH under certain circumstances The HH equation is most accurate where 1 which is where pH pKa Far away from this point the assumptions above become invalid and the HH equation breaks down in a way that is dependent on the pKa ofthe acid And in all cases the HH equation simply blows up if either HA or A is zero A note about equals When equations are presented in the yellowbeige text regions as above they will often use the Boolean equals sign I I ratherthan the assignment operator I I The reason for this is twofold 1 to avoid generating a redcolored error eg Ka if due to the use of undefined terms and 2 to avoid making an assignment that you should later make on your own In essence we will use the Boolean equals sign to generate equations that do not have any mathematical meaning within Mathcad Before working some problems let39s first establish some unit constants that are not builtin to Mathcad In the region below use the assignment operator to set M to be You can do this by 1ter typing Mmolliter Based on this assignment of M create an assignment below for lel by setting it to be 10393M You can do this by typing lel10 3spacebarspacebarM mM 10 3M Module05H HSemiAnsweredxmcd 2 Mudme 5 HendevsuwHasse ba ch Euuanun axsmzu cyeam svm av asswgnmems be uvauM m be 1n SM and quotM m be 1n 5M Van can nd m H symbm m We Gveek Symbu Tuu bav m yuu can type m uHuwed byCIr g seem my 5 a bum Mamcad um Huwevav rm 5 n 0mm an asswgnmen uv ml 3 I Asyane be uwtu be 1n Iiler Hexuwyumass gnmen askMamcauwhammby y m m m4 ml m Emu ml 1 xm39zy Mama Wm awe yuu me answev m tevms u L Whmh s an mdmatmn mm L mm mm m m Sum Wm K3 and pKa Let s antwcwpatethws and name ummncuunsmawm mm mesa cunvevsmns 1 su pHJ HJ pm and K2 mm m mum mummy be uw Thwsmnmmn Wm take a va ue u H and 12mm 012 V us pH Huwem m Su ugamhmwsexecmed H Hf 47 7 M M m5 pm 14 MuumenajHjem Anmveumm axsmzu Mudme 5 HendevsuwHasse ba ch Ewan mm m quot rm mummy be uw TMsmmmn Mu take a pH Va UE and vemmthe cunespundmg H cummanun Thetwu ave mated sucht a uwevemhe vesun unmmmmn H pxn m39 PHM H pxx m pHM 2 caveM m pHM s m evennhan m39PH M Eesuve m mummy by M a euhe expunem s evama em nm bemve Nuw mam swm avmndmnsmv pmij and mij be uw MaMemaunHyJWse mums 10 and HJ H q n q um s cum yuu H need m dmde M um n K bemve akmg mg Jug UVKa And 5 mu yuu H need m mummy by M May yuu ve taken mm m era puwev m7 pm 45 K5 W Smme WE H 25 these Vuncuuns shumy MuumenajHjem Anmveumm Module 5 HendersonHasselbalch BIS 102d quation Part II Titration of Weak Acid with Strong Base Goal Use the HendersonHasselbalch equation to calculate pH of weak acid solutions titrated with strong base Considerthe following problem from your text GampG pg49 4b The Ka for formic acid is 178 X 107 4M 150ml of 01M NaOH is added to 200ml of 01M formic acid and water is added to give a final volume of 1iter What is the pH of the final solution To use the HH equation to solve this problem you will need the pKa value of formic acid Assign Ka below to be 178 X 10 4M And then below that assign pKa appropriately using your pKa ffunction Adjacent to that confirm that your pKa ffunction is working correctly by asking what both pKa is and what 710g E is 201 7 4 Ka17810 M pKa pKaif Ka 1 E i 3 75 pKa375 Og M 39 In each case you should get an answer of 375 Assign a new variable NaOH to be lliter 202 NaOH 150m101M lliter The problem gives us neither HA nor A directly Instead we only know the original untitrated concentration of weak acid 20m 002M Let us introduce a new variable HA0 with a lliter zero to represent this quantity knowing that it will always be equal to the sum of A and HA such that HA70 HA A Module05H HSemiAnsweredxmcd 5 E S mm Mudme 5 HendevsuwHasse ba ch Euuanun Assng HAi be uwm be mt39 m HA U Z ml n 1M mm n can 235W ca cu ateme mhev A s We pvudud mm pvucesses 1 Neuuahzatmn u HAbv NaOH 2 Dwssumauun ENHAth H and A de evmmmu A Lltn mm ANHOHH m m u m y And Mequot usmg um dehmtmn uvHAi We mm HA HA 7 Any HA HA 7 NaOH 7 H Gwen 012 HH 2mm m thwsmvm A p p HA dummyequanunwm u nv N Mama NHOH H H Ka la me have We COULD yeananue mm and sum We vesuumu uuamauc euuanun Emwewum nsmad we Wm swmphwthe euuanun bv makmu an assummmn We Wm assume mum NHOH H NHOH Tm assummmn sma wequot a Mm mm hrm m equwaen y4haMNa0H Undevwhat cundmuns mam yuu expectthws assumpnun m be quesuunamev Huw mwgh he vahdny mum assumpnun be a emed bythe K3 va ue mum amd7 MuumenajHjem Anmveumm a BIS 102q Module 5 HendersonHasselbalch quation Copy your dummy equation from above and paste it here below Edit the equation using the new assumption of NaOH H NaOH to remove both instances of H from the right side of the equation Once you39ve made those edits change the Boolean equals sign to an assignment operator to quotactivatequot the equation 206 H K 1 NaOH a o p p g HAioi NaOH Ask Mathcad what pH is below 207 pH 4227 You should get an answer of 4227 which is the answer to the problem Now that you have an answer compare the values of H and NaOH below You can use your Hf function to convert your pH answer to a concentration 208 5 H7fpH 5933 X 10 M NaOH 0015M Was our assumption that NaOH NaOH Hvalid in this case Explain your answer Feel free to make any necessary math regions below to support your answer 209 NaOH H7fpH 0015M NaOH 0015M In terms of the relative values of A and H what must be true for this assumption to be valid Module05H HSemiAnsweredxmcd 7 E s mm Mudme 5 HendevsuwHasse ba ch Eduanun Lers mewe bEvundthws duesnun and ask mme denevauv n mhekuvds aswe mvate HA n U Wenme amd wdn u 1M NaOH keepmg VD UNE cuns am at lmenwha duesthe manen edwe uuk hke7 Tu cveate msmaph efsuse amnmmn Fv n m n vuuvvevsmn dune HVH dn n n n m n Cnpy yum asswgnmem en pH new abuve and paste n have be uw cnande mwsm be a NaOHrdependem mummy Nuw we aneady have ene pH Vuncuun pHJJha dues sumemmg cnande We 2 swde dune asswgnmem be uwhum pH m pHiHHtionj Ndoxx HHHNdo e Kdnd 7 p H p EKHAi rNaOII m Ndoa H HHNaO mud v 7 lt H v e W Nuw we need a yande e1 NaOH va ues Hewevenwe need m be eayem abundth smaH vahd ov BENJst mat m Mn 5 gumg m depend un HA and pKa buHm edwames e1 HA7 n U M and pm 75 mmassumpnen svahdduwnm abumEmMNaOH 6mM H7fpH7HH6mM OHM q n nn HVHEuuauuncan HeHusWhatha n Anna pennmne tevm Ndoa becames negawe 5 ds ugamhm s nu ungev 123 HA 7 men n n 1mM n Sm Mums ave kem vea VDU knuw me m USHWVM m keep d 123 212 We 6mM6 lmM HA 7 n SmM MuduxenajHjemnAnwevedmed E S mm Mudme 5 HendevsuwHasse ba ch Euuanun llynu haven t already ynu shnum lake Ihellme nnwln cnmplete the Graphing M mndule m 1 mm a new maph be uw 2 Senhe xrawsm be m mM 3 Sen 2 y awstu be 1quot HH1N20H 4 mm Vuuv answenu m umma pyumem pH HHltNaOH m addmn a vrawstvace a 15mm 1M lSEIml n 1M H HH 7 v7 J 7 mm xxx 5 Senmssecumyacemshuwsvmbms u s unEvuuth 5 Tu m m secund quotace m mm as a 1m mm mm add an addmuna 3st mm m lSEImJ n 1M NaOH T va ue u m mM 7 same xax sm begm am m m 15 1mm unu What 5 happemng amund pH pKa a 2 pH 3 my m MuumenajHjem Anmveuxmm a E S mm Mudme 5 HendevsuwHasse ba ch Euuanun Anm pumL yuu ve answeved um genem mmquot abum haw pH changesmv a vange u NaOH cuncemvauuns Huwevemms mm m swam spem Fm exammg n deaxs un ywnh mm What is the msulum 1quot ma range 11quoton cnncenlrminns and a range anz n MH v Hannvh uvev HA asweH Cnpy yum mum mummy mm abuve and paste n have be uw Muuwyme e m mum asswgnmemm add pKz as an mpm pavemetev The mm m shne m way u 5 mos H HH moxx Kn z mm 7 W P j P minimal mos mos K s mm w P j P minimal pKa va ues NaOH mm a new gvaph be uw SEHHE mm be M same y awsm be pHiHH leH l j same mm begm am 217 45 pH I INAOH35 A 25 3 m NAOH mM mm m y Because mum yum mm abuve 5mm uukvevy sman m m yuu made pvevmus y MuumenajHjem Anmveu mm m axsmzu Mudme 5 HendevsuwHasse ba ch Euuanun Cupyme abuve gvaph and paste M1212 be uw Addyrawshacesmv K puium a 5n pugImmoxLzs K pH HH1N20H55 FEMWOHM pH7HHNaOH5U pH HHNAOH65 5 pug1m NAOH 2 n m n NaOH m pH I INAOH35 8 pH I INAOH8U A 5 m 15 NaOH pKa mum amd mum pH mum vesuham sumnunv Why u yuu mum s w m MuumenajHjem Anmveumm BIS 102d Module 5 HendersonHasselbalch quation Earlier we were careful to not designate the range of NaOH as beginning too small This was because we didn39t want to run afoul of our assumption that NaOH is much greater than H For more positive values of pKa where less H is produced by acid dissociation this is less ofa problem For example we arbitrarily decided that if NagH gt 10 then our assumption was OK At pKa 35 6mM this is barely tenable 8132 but at pKa 80 our assumption is qUIte H7fpH7HH6mM35 6mM 5 sound 2571gtlt 10 H7fpH7HH6mM80 This suggests that we could extend the range of NaOH concentrations down to smaller values for the more positive pKa values However having an inconsistent NaOH range would make our calculations complex and the result would be somewhat awkward to graph Let us instead keep the NaOH range fixed with the lower end dictated by the results for pKa 35 Having plotted pH against NaOH for several different pKa values let39s now plot pH against pKa for several different NaOH values Define pKa below to be a range variable The first value should be 35 the second value should be 36 and the last value should be 80 220 m 353680 We already discussed how at low pKa values we run into trouble with low concentrations of NaOH That is low values ofA relative to H At very positive pKa values we also have a problem this time with high concentrations of NaOH At high concentrations of NaOH the concentration ofA becomes large compared to HA and so there is a tendency for A to quotreassociatequot That is forA to strip a proton from water producing HA and OH However we have assumed that dissociation and reassociation have no effect on the concentrations ofA and HA This is only valid if the concentration of HA is much greater than that of OH In essence this is a corollary of our criteria at low concentrations of NaOH that A be much greater than H We can calculate the concentration of OH OH by using the equilibrium constant of water Kw OH E H and calculate the concentration of HA as being HA HAio 7 NaOH based on our assumption that A NaOH From there we can inferthat our assumption is valid at 195mM NaOH up to about pKa 80 HA 0 7 195mM 12821 10714M2 H7fpH7HH195mM 80 Module05H HSemiAnsweredxmcd 12 BIS 102d Module 5 HendersonHasselbalch quation Create a new graph below Set the xaxis to be pKa Input the following traces on the yaxis pHHH5mMpKa pHHH10mMpKa pHHH15mMpKa pHHH199mMpKa 221 IU pHiHHSmM pKa pHiHH10mMpKa pHiHH15mMpKa 639 39 pHiHH195mMpKa 4 I I I I I 3 4 5 6 7 8 9 pKa Describe what you see in your graph above How do changes in pKa affect pH at the various concentrations of NaOH 222 As you can probably appreciate by now your pHiHH function is a simultaneous function of both NaOH and pKa Thus the results of pH HH is actually a twodimensional surface that can39t be fully represented by a 1dimensional line out approximation techniques in the graphs above notwithstanding To get an idea of what this surface looks like in 3D l have created a new function pHiHH3D that is based on your pH HH function M pHiHH3DNaOHpKa pHiHH NaOH m pKa mo 3 m In the space below I39ve made a surface plot ofthis function You may get an error here if you haven39t properly set up your pHHH function Module05H HSemiAnsweredxmcd 13 BIS 102q Module 5 HendersonHasselbalch Equation In this surface plot NaOH ranges from 6mM to HA 0 20 mM 195mM and pKa ranges from 35 to 80 7 These ranges are the same as in your previous graphs Note thatincreasing values ma be from left to right OR from right to left depending on the graph39s orientation The surface plot has been colored according to its contours Areas ofthe surface with the same color correspond to areas of similar pH This coloring gives an additional cue as to how pH changes as you move around on the surface 49 1 5 pKa NaOHmM pHiHH3D Rotate the plot around by clicking and dragging within the mesh Rotate the 3D graph around to get a good idea ofhow the pH surface looks Spin it all the way around Flip it up and look at it from the bottom Look down each ofthe NaOH and pKa axes head on Follow the grid lines which correspond to constant values ofeither NaOH or pKa and note how the lines move through different color contours indicating changes in pH 223 farwe have generalized our exploration of the HH equation to cover a range of base concentrations and a range opra values for the aci owever we continue to assume a particular acid concentration left over from the original problem That is HA 0 20 mM It would be nice ifwe could plot pH for a range ofHA 0 and range opra and a range ofNaOH all at once However that would be a 4dimensional graph something we can39t easily represent in Instead we can make a graph that combines the quantities of both NaOH and HA 0 into 0 e variable We39ll call this variable base eq for base equivalents and de ne it as the ratio ofNaOH to HA 0 NaOH HA70 Because the HH equation cannot deal with situations in which NaOH is greaterthan HA 0 base eq will range from 0 to 1 Being a ratio of concentrations base eq is unitless In the graphs 6 M that you have done so far you have implicitly explored the range of baseeq from m 20mM baseieq 03 to 195mM 0975 As you Will see using baseeq Instead of NaOH does not qualitatively change the pH surface that you see above Module05HHSemiAnsweredxmcd 14 E S mm Mudme 5 HendevsuwHasse ba ch Euuanun Eemrmmuwnh We H7H equauun aswe ve used m 5 an NaOH H7 Kala p p 4 in NaOH use dummy equatmns be uw usmg Eun ean mews and me dehmuun m haseieq mm abuve m amve at We uHuwmu mam base eq H7 Kala p p 1 7 butqu NAOH H7 Kala p p HA 7 NAOH H7 Kala p p g HA base eq H7 Kala p p 17 b5527qu JR and paste n have be uw Change We Eun ean equaxsm an asswgnmem upevamv Change me 2 5mm be me new mummy pH7HH1haseiqusz base 1 am 2 K 7 mm 7 W 7w gt P an an Wbaseiequa 7 pk la MuumenajHjem Anmveu mm 15 BIS 102q Here is what your new pHHH function looks like in 3D using baseeq instead ofjust NaOH You may be getting an error here if you haven39t set up your pHHHbaseeqpKa function properly In this surface plot baseeq ranges from 03 to However now that we are using base eq this graph is generalized over all concentrations of weak acid HA0 Module 5 HendersonHasselbalch Equation pKa baseeq Rotate the plot around by clicking and dragging within the mesh How does this graph look compared to the 3D graph you saw earlier the one that was for only 20 mM HA0 226 In broad terms what does this mean aboutthe relationship between pH HA0 and NaoH in such a titration curve Module05HHSemiAnsweredxmcd 16 227 E S mm Mudme 5 HendevsuwHasse ba ch Euuanun Part III Weak acids and Their Salts Goa Ca cu a e pH an a suhmun cumammg a mmuve evweak and and Ms san mamemaucaHy mamma w a suhmun cumammg a mmuve e1 aweak amd and Ms sah Asan m W5 cumex sthe pvuduct uMquahzauun Ma weak and m addmun m Wa ev Sn an m u e We mssumate m wa en genenaung We cumuga e base enne weak and m euesmn We have 5 an used We vaname Am vepvesem the eeneenuanen uths cumuga e base We saw m Pan L at pH pKa pH changes mus s uw ywnh changes m sunny has As e cuncemvatmnuvmuvegenevaHy base ee Asu utmnnthwspHvangessawdmbe bu eved Atp K amuums Tnn bu ev su mmm m snan Men Censmenneveuewng pvub emhum yuuHeM 6amp6 pg 494 nm1M aeeme hu erzlz pH Mia pyemem Assng pKz be uwm be was and IN a 55 pk e m m 5 A My Fan H yuu used We HVH equauun m su vemv pH Heweven m W5 pmb em pH 5 gwen su yuu H need a an senne a gebva Wme the HVH eeuanun be uw usme the Eun ean Emma s sun 3 dummv euuanun And then snumne m evmemate amps neeessame amve at A DPH PK HA A A a KW a he p p a p p a mpHspKaei HA MuumenajHjennnAnweveu mm 17 E S mm Mudme 5 HendevsunrHassemamh Euuanun We ave mm u make a u 1M sudmm aee a e Men mums eeneenuanen s neunenAnen HA nsmad u sthe sum MA and HA a uuanmvwe have tevmed HA wheve HA HA Ms 5 eenvemenL because u meansmatwe nuw enw need m su ve en eunen HA uvA Gwen n eas y detevmme We mnen Le us avbmamy demde m neus un su vmgmv HA une we ea Assng HAi be uwm be 1M an M U M Use a dummy equatmn be uwm expvess A Mevms DVHAi and HA am A HA 7 HA Cnpy yum ates anangemem enne HVH equatmn nem abuve the ene Wun un ene 5M2 and paste u nene be uw Use We abuve dehmtmn ev HA m shuw ha HA L mPH P 1 Cvea e as New dummv eeuauens as Vuu need m snemne mmvmedwate s ens m5 mpHspKa 1 HA HA mPH39PK HA 7 HA HA m 39 HA HA HA Hii39Fi a 1 HA HA 7 HA mPH PKH1 MuemenajHjem Anweveexmm BIS 102q Module 5 HendersonHasselbalch quation Copy your dummy equation for HA from above and paste it here below Change the Boolean equals sign to an assignment operator to quotactivatequot the equation Below the assignment ask Mathcad what HA is Change the displayed units to lel 306 HA 0 HA 10PH pKa 1 HA 18639mM You should get an answer of 18639mM Copy your dummy equation for A in terms of HA0 and HA from above and paste it here below Change the Boolean equals sign to an assignment operator to activate the equation Below the assignment ask Mathcad what A is Change the units of your answerto lel 307 AHA OiHA MN 7 A 81361mM Knowing the final concentrations of HA and A the initial concentrations of the stock solutions 01M and the final volume of the buffer 1iter you should be able finish this problem on your own below What volumes of the stock solutions need to be added together and does any water need to be added to the final buffer solution In addition to any math regions you create below also write your answer in words 01MxHA 01MyA 11 11 11 11 X HA y A 01M 01M x 186388m1 y 813612m1 117Xy0L Module05H HSemiAnsweredxmcd 19 BIS 102q Module 5 HendersonHasselbalch quation In using the HH equation to arrive at the above answer we made a couple implicit assumptions List them below and provide an explanation or evidence as to why these assumptions are OK for the given situation 309 In Part II you created atitration plot using the HH equation where you plotted pH against NaOH concentration for a given amount of acid Mathematically that situation is no different from mixing different amounts of weak acid and its salt Let39s briefly examine this by creating a graph that plots pH for a 01 M buffer while varying the concentration of weak acid39s salt A Because we39 assume that HA0 remains constant at 01 M weak acid HA will make up the remainder ofthe buffer Create a pHHHAHA function below and assign it to be pKa 104 A H HH AHA Ka 10 8 g p HA Now we need a range forA the final concentration ofthe salt We can safely begin the A range at SmM because 15143 and we39ll need to end the range at H7fpH7HH5mM01M 7 5mM 99mM to keep the HH equation real HA will be 105X greater than OH at A 99mM so we39re good on the top end as well Assign A below to be a range variable The first value should be SmM the second value should be M 6mM and the last value should be 99m 311 av SmM 6mM 99mM Module05H HSemiAnsweredxmcd 20 axsmzu Cveate a new maph be uw A Senhe mm be mM same vrawsm be pH HHtA 1MA HA HA 7 A SH yuu Camus1 M u Vuuv pH HH u e wan v uncuunvmuse as HA Add an addnmna v39awstvace ma 5 5W x 5 me mam pH mum mumem Vuu um mu 5 Add an addnmn y aws quotace ma 5 pKz pH7HHAEI mm 5 A pKa Mudme 5 HendevsuwHasse ba ch Euuanun pHJ KAII 1mm 54 5 abuve gvapw Exp am yum answev m yum gvaph wheve dues he pKa hne mmsm yuquace mum HVH equauunv Knuwmg mum A7 MuumenajHjem Anmveumm BIS 102g Module 5 HendersonHasselbalch quation Part IV More Buffers Volume Considerations Goal Use the HH equation to calculate pH in solutions of nonconstant volume Considerthe following problem from your text GampG pg 50 8 BICINE is a compound containing a tertiary amino group whose relevant pKa is 83 Given 1iter of 005M BICINE with its tertiary amino group in the unprotonated form how much 01N HCI must be added to have a BICINE buffer solution of pH 75 This problem reinforces the classification of a BrznstedLowry acid as any proton donor In this sense it does not matter if ourweak acid is charged or not in its protonated form The unprotonated form of BICINE while it may be uncharged is actually the conjugate base A of protonated BICINE HA Thus this problem begins with a solution where A 005M and we must add a strong acid to create a buffered solution For a review of Normality N vs Molarity M see Segel pg 23 Let us begin by making assignments for what is given in the problem Assign pKa below to be 83 and pH to be 75 m 83 W 75 As mentioned above the initial concentration of A is 005M but this is in 1iter of solution As soon as we start adding HCI the volume ofthe solution is going to change So it is not immediately clear what the volume of the solution will be once we reach pH 75 Forthis reason let us introduce two new variables VB the volume ofthe original BICINE solution VH the volume ofthe added HCI Note that the final volume ofthe solution will be VB VH Assign VB below to be lliter VB lliter From here we can generate expressions for HA and A the concentration of the protonated and unprotonated forms of BICINE respectively If we ignore H produced through water autoprotolysis then the number of moles of HA produced must be equal to the number of moles of HCI added If we further assume that any reassociation ofA to HA plus OH can be ignored then we can arrive at the following expression for HA 01MVH VB VH39 Module05H HSemiAnswereclxmcd 22 BIS 102d Module 5 HendersonHasselbalch ation If 01M VH moles of protonated BICINE are produced by addition of HCI then a similar number of moles of unprotonated BICINE must be eliminated This suggests the following expression for representing A A 005MVB 7 01MVH VB VH 39 Use the HH equation A H Ka 10 P P gHA and the above expressions for A and HA to write a dummy equation below expressing pH in terms VH of pKa VB and 403 005MVB701MVHW VBVH 01MVH VBVH pH pKa log Solve for VH Using dummy equations below show the intermediate steps necessary to arrive at 005MVB VH 01M10pmpKa 01M 005MVB701MVH 10pHipKa VBVH 01MVH VBVH 01MVH lopHipKa 005MVB 7 01MVH VB VH VB VH 01MVH10PH pKa 005MVB 7 01MVH 01MVH10pH pKa 01MVH 005MVB VH01M1opH pKa 01M 005MVB 005MVB VH 01M10pmpKa 01M Module05H HSemiAnsweredxmcd 23 BIS 102d Module 5 HendersonHasselbalch quation Copy your last dummy equation from above and paste it here below Change the Boolean equals sign to an assignment operator to quotactivatequot the equation Below that ask Mathcad what VH is 405 005MVB VH 01M10PHTPKa 01M VH 431597m1 You should get 431597ml which is the answer to the problem Let us now move beyond this particular answer and ask For a range of pH values what corresponding volumes of 01M HCI need to be added to 1iter of 005M BICINE You39ll make a graph for this answer shortly The pKa of BICINE is 83 Around pH 83 will adding 1ml HCI change the pH of a BICINE solution a little or a lot For example would the effect of 1ml HCI on pH be less ifthe BICINE solution was around pH 9 compared to around pH 83 Based on your answer you should be relatively confident that large amounts of HCI are going to be necessary to change the BICINE buffer39s pH in the region of pH 83 You might say that this is quotwhere the action isquot in terms of added HCI For that reason let39s create our pH range within a few units of 83 Assign pH to be a range variable below Set the first value to be 6 the second value to be 61 and the last value to be 10 This range won39t run us afoul of any of our HH assumptions 407 mi 661 10 Copy your assignment for VH from above and paste it below Mathcad will return an error because pH is now a range variable That39s OK Change the assignment to a function by changing the left side of the assignment from VH to VHpH 408 005MVB MPH 01M10pmpKa 01M Module05H HSemiAnsweredxmcd 24 BIS 102g Module 5 HendersonHasselbalch quation That should get rid ofthe error Now let39s make a graph ofthis function Create a new graph below Set the xaxis to be pH Set the yaxis to be VquH Plot your In answer to the above problem by adding a trace of to the yaxis To plot this as a single point add a second xaxis value of 75 Change this second trace to display a symbol choose one you like 409 UU I I VHpH 400 ml 300 431597m1 200 m1 x xx 100 I I I U 6 7 8 9 10 pH75 How is this graph different from other HH plots you39ve plotted in this module so far How is it similar 410 Explain in words what is happening in the middle part ofthe graph where your trace is steep ls pH well buffered in this range 411 Module05H HSemiAnsweredxmcd 25 BIS 102q Module 5 HendersonHasselbalch quation Part V More Problems Goal Use the tools you39ve created and techniques you39ve learned to solve additional problems To solve the following problems feel free to use any of the functions or formulas you have created in this module Rememberthat you can copy from earlier Parts and paste them here If you need to do any algebra show your intermediate steps with dummy equations using the Boolean equals sign To be clear about your answer write it out in words If you need additional space for your calculations you can create regions in the area to the right From Segel pg92 34 What is the pH of a solution containing 03M Trishydroxymethylaminomethane free base and 02M Tris hydrochloride pKa 81 501 A AHA Ka Ka 10 W P P HA pH7HHO3M02M81 8276 Module05H HSemiAnsweredxmcd 26 BIS 102q Module 5 HendersonHasselbalch ation From Segel pg92 32 What are the concentrations of NH3 and NH4CI in a 015M quotammoniaquot buffer pH 96 Kb 18 X 10 5M Recall that pr pKa 14 502 Kb 18107 5M W 015M pH 96 RKa 7 14 7 pKa Kb A HA70 7 HA H K 1 a0 P P gH H Ka 10 p p g HA70 7 A 7 A 10pH pKa HA70 7 A HA701opH pKa 7 A10pH pKa A HA701opH pKa A10pH pKa A HA701opH pKa A10PH pKa 1 HA o10PH pKa A 7 10PH pKa 1 A 0103M HA70 7 A 0047M Module05H HSemiAnswereclxmcd 27 BIS 102d Module 5 HendersonHasselbalch Equation From Segel pg92 42 When a sulfate ester is hydrolyzed an H ion is produced ROSO339 H20 gt ROH 804239 H The above reaction was carried out in 10ml of 002M Tris buffer pH 810 containing 001M ROSO339 and an enzyme called a sulfatase that catalyses the reaction At the end of 10 minutes the pH of the reaction mixture decreased to 797 How many moles of ROSO339 were hydrolyzed during the 10 minute incubation period The pKa of Tris is 81 503 a I RK 81 HA 0 002M le 81 pH2 7 97 r HAioIOPHipKa 1opH pKa 1 H7 K NH 1mm 10 6mm 10PHTPKa 1 ApH1 7 ApH21m1 1486pmol Module 05 HendersonHasselbalch Written by Eric Leaver 2009 Martin Wilson Last updated February 3 2009 130pm Module05H HSemiAnsweredxmcd 28 BIS 102d Module 5 HendersonHasselbalch quation Help Improve This Module Completing this section is not required but will be helpful for future development ofthis module Your answers here Will not affect your grade About how much time did you spend completing this module Were there any particular parts of this module that you found too difficult too confusing or too tedious How might these sections be improved Were there any particular parts that you found especially useful or helpful in understanding the material from BIS 102 Other suggestions kudos criticism Module05H HSemiAnsweredxmcd 29

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