Blood-Buffer-Lab PDF

Preview

Summary

Lab...

Description

APPENDIX: Graphs and Calculations Calculated Data Determining volume H2PO4- needed to make phosphate buffer (D1) i) Volume (mL) 0.50M H2PO42-(aq) required to make phosphate buffer = Volume = [(molarity H2PO4-)/(((moles of acid gained -

1 0−desired pH pKa

* (# moles lost from conjugate base)/

1 0−desired pH 1 0−initial pH ¿∗1 0−initial pH 1 0−desired pH + ((moles of acid gained * (# moles lost ¿ ¿ ¿ pKa pKa pKa pKa ¿ −desired pH 10 -r)))](1000 mL/ L) from conjugate base)/ pKa ¿ −7.25 10 1 0−7.25 1 0−7.45 ¿∗1 0−7.45 = [(0.50M)/(((0.002 * (-0.002))/ pKa ¿ ¿ ¿ + ((0.0026.31E-8 pKa pKa ¿ −7.25 −7.25 1 0 10 * # (-0.002))/ 6.31E-8 -r)))] (1000 mL/L) 6.31E-8 ¿ ii)

= 36.0 mL Volume (mL) 1.0M NaOH(aq) needed to make phosphate buffer = Volume = [(molarity NaOH)/ ((moles of acid gained -

10

−desired pH

pKa ¿

1 0−desired pH pKa

* (# moles lost from conjugate base)/

-r)))] (1000 mL/L)

1 0−7.25 = (1.0 M)/ ((0.0026.31E-8

1 0−7.25 * # (-0.002))/ 6.31E-8 -r))) ¿

= 11.5 mL Determining acid buffer capacity (D2) Acid buffer capacity = =

[ Molarity acid ][ volume acid used (L)] volume of buffer

[ 0.100 M ] [ 0.01655 L] 0.02500 L

= 6.62E-2 M HCl per L of buffer Determining base buffer capacity (D3) Base buffer capacity = =

[ Molarity base ][ volume base used (L)] volume of buffer

[ 0.100 M ] [ 0.00540 L] 0.02500 L

= 2.16E-2 M NaOH per L of buffer

Determining amount of Restoration Solution needed to restore pH from 7.07 to 7.45 in 25.00 mL sample (D5) - Was determined via reading volume of the Restoration Solution (on the x axis of the graph) used until pH of 7.45

2 was met: = 32.05 mL Determining amount of Restoration Solution needed to restore pH from 7.07 to 7.45 in 7.0 L of patient blood (D7) Volume Restoration Solution (mL) needed to restore pH in 7 L of blood = [volume needed for 25.00 mL of blood]*[280] = [32.05 mL]*280 = 8974 mL Determining volume per drop

mass of all drops * (density drops)-1 ¿ of drops 71.9904 −66.3423 = * (1 mL/g) 143

Volume per drop =

= 0.039497202 mL

LAB:pHi nBl ood

Ma y aKobl a ns ki#993 38188

ABSTRACT Thee ffe c tofbuffe r sa ndt hef a c t or st ha ti nflue n c et he i rc a p a c i t yt or e gu l a t ep Hwa sde t e r mi ne db yt hr e edi ffe r e nc et e s t s .Ap buffe rc on t a i ni n g1. 0M Na OHa nd0. 5M2H PO42-(aq)wa sma det omode lt hehu ma nbl oodb uffe rs y s t e m.Thepho s p ha t eb uffe rwa t i t r a t e dwi t ha p pr e c i a bl ea mount sofHCla ndNa OHt oe xp l a i nt her e s po ns eoft hehuma nbl oodbuffe r .The n,pa t i e n tbl oodwa s t i t r a t e da ga i ns t“ Re s t o r a t i onSol u t i on”t ode t e r mi neho wmuc hwoul dbene e de dt os a v epa t i e nt sofdi ffe r e ntbodyma s s e s& bl ood pH.ApHp r obewa su s e dt omoni t ort hep Hoft hes ol ut i on.

I NTRODUCTI ON The bicarbonate buffer system is an acid-base homeostatic mechanism involving the balance of carbonic acid (H2CO3), bicarbonate ion(HCO3-) and carbon dioxide (CO2) in order to maintain pH in the blood and duodenum, among other tissues, to support proper metabolic function.6 Catalyzed by carbonic anhydrase, carbon dioxide (CO2) reacts with water (H2O) to form carbonic acid (H2CO3), which in turn rapidly dissociates to form a bicarbonate ion (HCO-3) and a hydrogen ion (H+) as shown in the following reaction:8,9,10 CO2 + H2O ⇌ H2CO3 ⇌ HCO3- ⇌ H+ As calculated by the Henderson-Hasselbalch equation, in order to maintain a normal pH of 7.4 in the blood (whereby the pKa of carbonic acid is 6.1 at physiological temperature), a 20:1 bicarbonate to carbonic acid must constantly be maintained; this homeostasis is mainly mediated by pH sensors in the medulla oblongata of the brain and probably in the kidneys, linked via negative feedback loops to effectors in the respiratory and renal systems.3 As with any buffer system, the pH is balanced by the presence of both a weak acid (for example, H2CO3) and its conjugate base (for example, HCO3- ) so that any excess acid or base introduced to the system is neutralized. Failure of this system to function properly results in acid-base imbalance, such as acidosis (pH7.45) in the blood.9 There are two forms of acidosis that can occur during vigorous exercise: respiratory and metabolic acidosis. Respiratory acidosis occurs when the body metabolism increases in one who is active, and more food molecules (e.g glucose) are metabolized for energy, with the CO 2 product shifting the H2CO3/ HCO3- buffer towards bicarbonate, and therefore H3O.1 Acidosis can occur here if the lungs can’t eliminate the CO 2 fast enough. Metabolic acidosis can happen if exercise continues, and the breathing can’t supply the tissues with enough oxygen, leading to the lactic acid; a by-product of part-way metabolism.7 Nevertheless, using carbonic acid isn’t suitable to model this issue because it isn’t stable in aqueous solution. However, a phosphate buffer can be successfully used because it behaves similarly to the carbonic acid buffer, and doesn’t decompose easily in aqueous solution. The phosphate buffer can be used to explore the question of how the human blood buffer works against excess acid and base. The phosphate buffer will have a greater acid buffer capacity than a base buffer capacity; like the carbonic acid buffer, it must be have a high conjugate-base concentration that is suitable to neutralize metabolic by-products that are typically far more acidic than basic.

EXPERI MENTAL 36. 00mLof0. 50 M H2PO4f r o m agr a dua t e dc y l i nde rwa smi x e dwi t has uffic i e ntv ol umeof1. 0M Na OH f r om ac a l i br a t e ddr op c ou nt e r ,unt i lapHme t e ri ndi c a t e dt ha tt hephos pha t eb uffe rha dap Hof7. 55. Thi ss ol ut i onwa st he nt r a ns f e r r e di nt oa100. 00mL v ol ume t r i cfla s ka ndfil l e dwi t hboi l e dwa t e r .Thei ni t i a lpHt i t r a t i onc ur v ewa sobt a i ne dt h r ought het i t r a t i onof25. 0 0mLof phos pha t ebuffe rwi t h0. 100M HClunt i lt hepHdr o ppe db y1uni t .Thes e c ondpHt i t r a t i onc u r vewa sobt a i n e dwi t has i mi l a r t e c hni quebutwi t h0. 100M Na OHa dde dt o25. 00 mLo fbuffe rs ol ut i onunt i lt hepHi nc r e a s e db y1un i t .La s t l y ,at i t r a t i onc ur veus i n “ t e s tbl ood”wa sob t a i n e dupont hea d di t i onofr e s t or a t i ons o l ut i onunt i lapHof~7 . 45wa sr e a c he d.Thi swa st he nus e dt oc a l c ul a t e t hea mou ntofr e s t o r a t i ons ol ut i onne e de dt o“ s a vet hepa t i e nt ”a ndg e tapHofa r ound7. 4 5.

3

RESULTS Ta bl eI :Vol ume s( mL)0. 50M H2PO4-a nd1. 0M Na OHne e d e dt op r e pa r ephos pha t ebuffe r ,a nda c i da ndba s eb uffe rc a pa c i t i e s Vol ume( mL)0. 50M H2PO4-(aq)ne e de dt opr e pa r eph os pha t ebuffe r 36. 0mL Vol ume( mL)1. 0M Na OH(aq)ne e de dt opr e pa r ephos p ha t ebu ffe r 11 . 5mL Ac i dbuffe rc a pa c i t yofphos ph a t ebu ffe r( mol e sofHClpe rLofbuffe r ) 6 . 620E2 Ba s ebuffe rc a p a c i t yofphos pha t ebuffe r( Mo l e sofNa OHp e rLofb uffe r ) 2 . 160E2 Ta bl eI I : Vo l u mer e s t o r a t i ons ol u t i o nu s e dt or e s t or e2 5mLofp H7 . 0 7b l o odt op H7 . 4 5 Vo l u mer e s t o r a t i ons ol u t i o nn e e de dt or e s t or e7 . 0Lo fp H7 . 0 7bl oo dt op H7 . 4 5 p Hp a t i e ntbl oo d Vo l u mer e s t o r a t i ons ol u t i o nn e e de dt or e s t or e2 5mLpa t i e n tbl o odt o~pH7 . 4 5

3 2 . 0 5mL 8 . 9 74L 7 . 1 1 ~26 . 07mL

Ta b l eI I I :( D8)Vo l u me s( mL)ofRe s t o r a t i o nSol ut i onne e de dt os a v epa t i e n t swi t hv a r y i n gb od yma s s e s( k g )a n db l oo dpH pH 7 . 4 0

Vo l u me s( mL)OfRe s t or a t i o nSol ut i on 80 0 88 0 9 60

1 04 0

1 12 0

12 00

1 2 80

7 . 3 0 7 . 2 0

26 00 48 00

2 8 60 5 2 80

3 12 0 5 76 0

33 8 0 62 4 0

36 4 0 67 2 0

39 00 72 00

4 1 60 7 6 80

7 . 1 0 7 . 0 7 We i gh t

58 00 65 00 5 0kg

6 3 80 7 15 0 5 5kg

6 96 0 7 80 0 6 0kg

75 4 0 84 50 6 5kg

81 2 0 9 10 0 7 0kg

87 00 97 50 7 5kg

9 2 80 1040 0 8 0kg



Thea c i dc a pa c i t yo ft heb uffe r ,6. 620E2( mo l e so fHClp e rLofbuffe r )i sgr e a t e rt ha ni t sba s ec a pa c i t y ,2. 160E2( mol e Na OHp e rLofbuffe r ) .



r t i ono ft hebuffe r ,be c a us ea ddi n gi ti nc r e a s e dt hepHoft he Cl e a r l y ,t h er e s t or a t i ons o l ut i onmus tc ont a i nt h eHPO42-(aq)po bl ood.

DI SCUSSI ON 2 I na l l ,t heh ypot he s i swa ss uppor t e d;t hea c i dbuffe rc a pa c i t ywa sgr e a t e rt ha ni t sba s ebuffe rc a pa c i t y .3 6. 00mL2 o P f O H 4 a nd~11. 5 0mLNa OHwa sus e dt oma ket hepho s pha t eb uffe r( D1) .Thea c i dbuffe rc a pa c i t ywa s6. 620E2( mol e sHClpe rLof buffe r )( D2) ,wh i l et heb a s ebu ffe rc a p a c i t ywa s2. 1 60E2( mol e sNa OHp e rLofbuffe r )( D3) .Thea c i dbuffe rc a pa c i t ywa s t ha nt h eba s ebuffe rc a pa c i t y ,whi c hma ke ss e ns ebe c a us emos tme t a bol i cpr oc e s s e sr e s ul t sa r ea c i di c ;bl oodmus tha v eahi ghe ra c i d buffe rc a pa c i t yt oc ompe ns a t ef ort hel o we r i ngo fpH( D4) .3 2. 50mLofRe s t or a t i onSol ut i o nwa sne e de dt obr i ngt hep Hof25. 00 s tbe mLoft he7. 07pHpa t i e ntbl oo dba c kt o~7 . 45( D5) .Si nc et h ei ni t i a lpHo ft hebl oo dwa s7. 07 ,t hec on j u g a t eba s e ,HPO42-(aq)mu 2 + a dd e dt or a i s et hepHt o~7 . 45.Thi si sbe c a us ea sHPO4 (aq),i tr e a c t swi t hH3O,l o we r i n gt hea c i di cc o nt e n toft hes ol ut i on( D6) . I f 0. 0 2500Lofbl oodp H7. 07b l oodr e qui r e d32. 05mLofRe s t or a t i onSol ut i ont or e s t or et hepHt o~7. 45,t he nt he napa t i e ntwi t h7L oft hes a mebl oodwoul dr e qui r e8974mLo fRe s t or a t i onSo l ut i o n( 280t i me st hepr e vi ousa mount )t odot h es a me( D7) .About26. 07 mLofRe s t or a t i onSol ut i onwa sne e de dt os a v e( br i ngb l oodpHba c kt o~7. 45 )pa t i e ntwi t hpH7. 11bl ood.

Th ea ddi t i onofHClt ot hebuffe rwa sdon et ot r yt os i mul a t ea c i do s i s .Ba s i c a l l y ,be c a us et hepHr a n g ewa s n ’ twi t hi nt he r a n geof7. 357 . 45,a ne s t i ma t i onofv ol ume sofr e s t or a t i ons ol u t i onwa sne e de df ort hea c i d os i spa t i e nt st oobt a i napHba c ki nt he nor ma lr a n ge .I nt hec a s ewhe r et hepHi sdr o ppe dt oada ng e r ousl e v e l ,r e s t or a t i ons o l ut i on“ t r e a t e dt hepa t i e n t ”be c a us ei tc ont a i ne d awe a kba s es ol ut i on .Ther e s t o r a t i ons ol ut i onr a i s e st hepHs i nc ei tc ont a i nst hec on j ug a t eba s e ,HPO42-(aq),t h usr e du c i n gt he c on c e nt r a t i ono fH3O+(aq),r e s ul t i ngi napHwi t hi nt henor ma lr a ng e . Bu mpsone a c hoft hepHc ur v e si ndi c a t et ha tt he r ema yha v ebe e ns o mee r r o r .Thi se r r o rwa smos tl i ke l yduet ot he e xpe r i me nt e ra d j us t i ngt hema gne t i cs t i r r e rs pe e dt hr oughoute a c he xpe r i me nt ;i fa ta nypoi nti twa ss t i r r i n gt oos l o wl y ,t hepHi nt he s ol ut i ona r oundt hepHme t e rma yha v ei na c c ur a t e l yr e pr e s e nt e dt hepHoft hee nt i r es ol ut i on.Cons e que nt l y ,t hei nf ut ur e e xpe r i me nt soft het ype ,t hee xpe r i me nt e rs ho ul ds e tt hes t i r r e rt oac ons t a nt ,a nda de qua t e l yf a s ts pe e d.

4

CONCLUSI ON 

Thea c i dbu ffe rc a p a c i t y ,6. 66 2E2( mol e sHClpe rLo fbuffe r ) ,h a sbe e nde t e r mi ne dt obel a r ge rt ha nt heb a s ebuffe rc a pa c 2 . 16E2( mol e sNa OHpe rLofbuffe r ) .32. 50mLo fRe s t or a t i o ns ol ut i onwa sne e de dt ob r i ngt hep Hof25. 00mLof7 . 07pH p a t i e ntbl oodt o~pH7. 4 5,a nd8974mLwoul dbene e d e dt od ot hes a met o7. 0Lofpa t i e ntbl ood.Abo ut26 . 07mLof Re s t or a t i onSol ut i onwa sne e de dt os a v e( br i n gbl oodpHba c kt o~7. 45)pa t i e ntwi t hpH7. 11bl o od.

BI BLI OGRAPHY (1) (2) (3) (4) (5) (6) (7) (8) (9)

Crook, M. Case Presentations in Chemical Pathology 1993, 123–126. Function; MCGRAW HILL HIGHER EDUCATION: London, 2014. E. P.; Raff, H.; Strang, K. T. Vanders Human Physiology:13th Revised edition: The Mechanisms of Body Health/Lippincott Williams & Wilkins: Philadelphia, 2013. Johnson, L. R.; Byrne, J. H. Essential medical physiology; Elsevier Academic Press: Boston, 2003. Krieg, B. J.; Taghavi, S. M.; Amidon, G. L.; Amidon, G. E. J. Pharm. Sci. 2014, 103 (11), 3473–3490. Kovesdy, C. P. Metabolic Acidosis 2016, 131–143. Meldrum, N. U.; Roughton, F. J. W. J. Physiol. 1933, 80 (2), 113–142. Oxtoby, D. W.; Butler, L. J.; Gillis, H. P. Principles of modern chemistry; Cengage learning: Boston, MA, 2016.; pp. 611-753....