Well say we were dealing with a weak acid K a = 1x10, and had a formal concentration of.1m. What is the % dissociation of the acid?

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1 Chpter 9 Buffers Problems 2, 5, 7, 8, 9, 12, 15, 17,19 A Buffer is solution tht resists chnges in ph when cids or bses re dded or when the solution is diluted. Buffers re importnt in Biochemistry becuse mny of the enzymes tht mke your body run re designed to work t one prticulr ph, if the solution doesn t hve the right ph things go south. 91 Wht you mix is wht you get When you mke buffer you usully mix n cid nd its conjugte bse (or vice vers) While there is wek cid equilibrium going on, it is usully so wek tht it cn be ignored, Thus most of the time you ssume tht the wek cid stys in its wek cid form (HA) nd the conjugte bse stys in its bse form (A ) Why? 4 Well sy we were deling with wek cid K = 1x10, nd hd forml concentrtion of.1m. Wht is the % dissocition of the cid? HA H A 4 2 1x10 = X /(.1X) X =.003 % dissocited =.003/.1 = 3% so very little A Now wht hppens when we dd the conjugte bse (A ) tht mkes this buffer? First, wht is it s effect on the bove cid dissocition? (Common ion effect by le Chteliers even less A, so you cn ignore the cid 5 dissocition s mking contribution to A (ctul number 3.2x10 dissocition) Second does it mke much OH by bse ssocition with wter? No It ws wek cid to begin with, so it is n even weker bse!

2 2 92 The HendersonHsselblch Eqution Since we cn ignore the dissocition of the cid, we cn rerrnge the K eqution into more convenient form K = [H ][A ]/[HA] log K = log[h ] log( [A ]/[HA]) pk = ph log( [A ]/[HA]) ph = pk log( [A ]/[HA]) (for n cid) The equivlent HH eqution for bse is: ph = pk log ([B]/[BH ]) When A = HA, ph = pk Where pk refers to dissocition of the BH conjugte cid NOT the pk of the bse B If you study the HH equtions you will see tht when [HA]=[A ] you hve log 1/1 =log 1 =0 nd ph=pk. So when this occurs you hve n esy ph to clculte. When [A ]/[HA] chnges by fctor fo 10, ph chnges by 1 unit Sme ide sy [HA] = 10X [A] log [A ]/[HA] = x/10x log [A ]/[HA] = 1/10 = 1 nd Log (10/1) = 1 Also note: when [A ] > [HA] ph > pk When [A ]<[HA] ph< pk A HH type problem Simple forwrd: I m going to mix 50 mls of.5m Sodium cette with 25 ml of.1m cetic Acid,

3 3 wht is the ph of the resulting buffer solution? First, wht re the concentrtions of A nd HA in the mixed solution? I m mixing 50 mls of one solution with 25 of nother, so my totl volume is 75 ml Using M1V 1 = M2V2 Sodium cette will ionize completely so I hve.5m N nd.5m Acette nd cette = A Tking into ccount the complete dilution then [A ]:.5(50) = X (75) X =.5(50)/75) =.333M Acetic Acid is HA [HA] :.1(25) = X(75) X =.1(25/75) = ph = pk log [A ]/[HA]; pk cetic cid = ph = log (.333/.0333) ph = (note ph > pk nd this checks with [A ]>[HA] Now let s try hrder problem going in the other direction, strt with ph nd come up with [HA] nd [A ] Let s sy the totl (forml concentrtion of [HA] nd [A ] should be 1M, nd I wnt the ph to be 5, nd gin I will use cetic cid with pk of ie [HA] [A ] = 1M ph=5 pk = 4.757

4 4 Using the HH we hve 5 = log[a ]/[HA] Let s sy X= [A ], then [HA]=1X nd 5 = log (X/(1X)) =log(X/(1X)) 0.243=log (X/(1x)) = X/1X 1.75=X/1X X=X 1.75=2.75X X=1.75/2.75=.636 = [A ] so [HA] = = A buffer in ction The whole ide of buffer is tht it is solution tht resists chnges in ph. Let s test this out Sy I hve 1 l of pure wter, wht is its ph? (7) Now lets strt with 1 l of pure wter nd dd 5 ml of 10M HCl, wht hppens to the ph? Totl volume is now L [H ] using dilution eqution: M1V 1 = M2V 2;.005(10) = 1.005(X); X =.005(10)/1.005 = ph = 1.3 So ph dropped from 7 to 1.3, swing of 5.7 ph units Wht would hve hppened if we were using the buffer we just clculted? Initil ph = 5 How do we figure out wht the cid does??? Chemiclly the cid will rect with wht? The wek cid or its bse? (Its bse) So we hd just determined tht [HA] =.364, nd [A ] =.636

5 5 nd if we hve 1 l then Moles [HA] =.364, nd mole A =.636 But now the cid is going to rect with the bse A H HA Moles H =.005(10) =.05 moles So the rection tble A H HA Buffer Acid. 05 RXM so [A ] = ( )1.005 =.583 [HA] = ( )/1.005 =.412 ph = log.583/.412 = = so my ph hs dropped from 5 to 4.908, ll of ~.1 ph units. Yes this resisted chnge in ph!! There re two things I wnt you to notice 1.) one ws in the lst clcultion. Notice how I went from moles to molrity nd then took [A ]/[HA]. Did I hve to convert from moles to molrity? NO to covert to molrity divide by volume, but in next step volume/volume cncels out. So this is extr useless mth! 2.) How do buffers work? The cid rects with bse, nd the bse with cid, s long s I hve both cid nd bse left in the solution it will continue to buffer. Wht if I hd tried 5 ml of 10M NOH? ( tht would just bout hve blown the buffer wy) 94 Prepring Buffers While clcultion like the bove re common on qunt tests, wht hppened if you mke this solution up? You ctully never get the ph right. This is becuse rel ions usully ct little differently thn their concentrtions. There is correction for this clled n ctivity coefficient tht we will get to in chpter 11. For now just tke my word for it. Besides tht, you usully mke smll errors when you weigh thing nd the K s re temperture dependent s well, nd ll these combine to throw your finl ph off. When you re in the lb wht you usully do is to mix the chemicl you wnt to form buffer together, mke up the solution to bout 80% of its finl volume, check the

6 6 ph with ph meter, then dd little cid or bse to djust to exctly wht you wnt, then dd the remining liquid to mke the totl volume. If you wnt to be very exct, you lwys need to ctully check the ph of the buffer before you use it. Some buffers re sensitive to chnges in ionic strength, so their ph will chnge s they re diluted, others re sensitive to temperture, so their ph will shift s the temperture of the experiment chnges. Bottom line, double check ny criticl experiment! 95 Buffer Cpcity We won t spend too much time on buffer cpcity, but I do wnt you to remember tht when choosing n cid or bse for n experiment you need to chose one whose pk is s close s possible the ph you re trying to buffer for. Also the buffer works best in the region where you re ± 1 ph unit from the cids pk. IF you think bck to the cetic cid buffer exmple we hve been using you will see tht this is ctully good exmple. The pk. of Acetic cid is 4.757, so it work best s buffer between nd In our exmple problem we hve been buffering t ph of 5, nd this is inside the optimum rnge for this cid. Your tsk is to look t the Books tble of commonly used buffer (Tble ) nd come up with better buffer (There relly doesn t seem to be one!) 96 How Indictors Work Wht is nd indictor? Its just nother cid or bse, the only difference is tht its cid form hs different color thn the conjugte bse form. Let s choose one, Bromocresol green tht we will be using in couple of our lbs. It hs pk of Its structure is shown below: It is yellow in the HA form nd blue in the A form. If this is n cid, where is its buffer region? Around ph 4.7 Where so we hve the best buffer?

7 7 From 1 to 1 ph unit round the pk or Do you know why? (Yesterdy s lecture) At 3.7 A/HA=1:10 nd 5.7 =10:1 Lets use this to think bout the concentrtions of HA nd A ph HA A Color Y Y Y/G G G/B B B B So you see two distinct colors <ph4 yellow > ph 6 Blue nd one region of muddled color 45 vrying yellow/green 56 vrying green/blue Now wht if we re doing titrtion tht hs nd endpoint t ph 5? Well, if the titrtion is set up right, the ph should swing from 4 to 6 within drop of titrnt t the equivlence point. Thus before the equivlence point you will be yellow. And then, within one drop of the equivlence point the ph swings to 6, nd the color chnges directly to Blue, nd you never hve chnce to see the intermedite green color! This is idel if you cn t see intermedite color, you cn t get confused where the endpoint is Thus The pk of your indictor should mtch the ph t the equivlence point s closely s possible, t lest to within 1 ph unit. Now think little more... if your indictor is n cid or bse, do you wnt to hve tons of it in solution with your unknown? No you use up some mount of your bse titrting you indictor, ny you endpoint will be off by tht much. Alwys try to use the minimum mount of indictor! Actully not tht bd, colors re usully very intense. Did you ll red Box 92 the Secret of Crbonless Copy Pper? Acid/bse indictor is inside polymeric microcpsules in top sheet of pper the pressure of your pen brek the microcpsules nd releses the indictor the indictor relesed onto the lower sheet of pper the lower sheet contins Bentonite (mined just west of Belle) Bentonite is n cidic cly. This cid mkes the indictor chnge color nd tht is the color you see on the lower pge.

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