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Problem Set 1 20100211

Problem Set 1 20100211 - 05:02p m:Lwa/eug My. 1 We 0 if...

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Unformatted text preview: Feb 11 10 05:02p m..:Lwa/eug My.» 1/ We 0 if; Z/H/QJQ/E} p.1 CH 339K Problem Set 1 1) Dihydroxyacetone phosphate (DHAP)and Glyceraldehyde—3-Phosphate (GSP) are both intermediates in the metabolic breakdown of glucose. They can be interconverted by the enzyme Triose Phosphate Isomemse. e a ,0 ox .9 P’ w I I \ so ‘0 0' 0e 1 F—I—I- 5 o '1 3 4"”0 0H DHAP G3P For the conversion of DHAP to GSP, AGE" = 7.5 lemol a) Calculate Keq and the equilibrium fraction of (BF at 37° C. b) What is AG if [GSP] = 11100 [DHAP]? 2) A Cl" ion is separated from a Na]r ion by a distance of 0.5 nm. What is the interaction energy (in J ouleslmole} in a. Water (3 = 78.54) b. n—Pentane (a = 1.84) 3) What is the pH of each of the following unbuffered solutions? a. 0.35 M HCl b. 0.35 M Acetic Acid c. 0.035 M Acetic Acid 4) What is the pH of the following buffer mixtures? a) 1 M Acetic Acid and 0.5 M Sodium Acetate? Feb ‘11 10 05:02p h} 0.3 M H3P04 and 0.8 M KH1P04? 5) You want a buffer of pH 7, made using KH2P04 and Na1HPO4: a) What is the conjugate acid in this system? What is the conjugate base? b) If the buffer is 0.1 M KH2P04, what must the concentration of NaIHPO4 be? c) You want the same buffer, but with a total phosphate concentration of 0.3 M. What concentrations of each species do you need? 6) You have a beaker containing 500 ml of 0.10 M potassium formate buffer pH 3.75. What is the 1111 if you add 5 ml of 1.0 M KOH? (Hint -— the change in volume can be ignored) 7) On my benchtop, I have a jar of the amino acid glycine, a solid powder with MW = 75.07 glmol. I also have a bottle of 1 M NaOH, all the distilled, deionized water I can use, and a 2 liter graduated cylinder. Tell me how to make 2.0 liters of 0.1 M glycine buffer, pH 9. Useful Physical Constants - r _. constant: Avagadro’s Number Boltzmann’s Constant Charge on electron Gas Constant R Faraday’s Constant k:1/4TEE __ . 7 . Value _ 8.99 X 10 Nm coulomb 2.303 log (1:) Dielectric constant of water Feb 11 10 05:03p p3 gLUwLJJ {(07 1Q 3 E/ [£8915 0 q OHM a?) g3? 56": 7,5 LS/ma/ FWD mm 4m xix-(3:1, Hatch»; “3170: -7 «Juno/era! : a. [Wag] : @F‘C‘Zz Vic? 3’ OI ‘ 3;. :3, 0554? (£393 +Cflrfirflj=f bis-58‘ 1 [my o.asqs“()-Eéépij LDSQS—CGESPB: ,ostif (as—3?]: 0.0’5‘aar $279 a a :5 AG“ 55’ I539}; To; {9,1549} 56:: AGO Jr 14 [filaé {3%.JQT33 écrsfé 5-3733 —. 1569+ flT/fl [55’53 ‘ g wrj ; flsoo 3/”! 1— $37,"; Bloom L4 (5L0 ficmof Feb 11 10 05:03p p4 q) W‘V‘Jfif ~ :1? J73”??- o:_ 1/... a}??? "t r 9 ‘5 //W602.xra"7C>(,f,®Zfi/a—z?c> a'DRE—ak ._ (at " Jig/9')“; M fl l J/Mol 5) {1924;(R’LC D: L“; 319.)“ r g 7.. :— 9 2} fig“ (F U GOU‘IO—Hq \1 (a CJZBKJQZ ( “CL “WK 521.59 / vest — “P4579 w Feb 11 10 05:04p p.5 Feb 11 10 05:04p p.6 9H5; “ :05 [Hi-E) '* " i=5) (0‘35) : (DJ-{E Q 0g (3' M Afifiré‘ti 4cm) ’Wefiivi‘fi [HM]: 0,33%- ij Feb 11 10 05:04p i [am @H: pin“? [‘95 (Gan-3'] 05M :. ‘I' 10; : (4,44 0‘3 M “Spa, OW) N: '9‘“ + Job E595; EQCEQj : 24x4 [195 056 0'5 0% m M H1903 Feb 11 10 05:05p p.8 5) mm“, 7,0 pH :2 MHVPDH; f Mayflécf ‘1. 54> Canduaeica.’ 62>???»— IS HPDV (@130 “j “with “114:5 {5 Mlpafif’ 6) 1% quCer :5 CHM LLHLPDL! 934140 NEW: 1:53: a P“: pt“ 1% 5:33 W = (936 + £95 u/ LHPaq‘Q] [K W0]; [9 “q; [H9553 .. 0-: H Ariditf: {Magi} : 0.1675 M M ; [Ease-:3 '10 Qt}; ~Cbfisa} ops 10H ‘ :0 L! l: Ea;st 5 C(vgje.) O 3) re”; [62.3.33 Cf +10 M!) -: it! [6:95.532 0.5 I :0 ‘E Feb 11 10 05:05p p.9 Feb 11 10 05:06p p.10 7) CICKB5‘1TfCIo’Li MW: 7.5a”? b/r/rwl lHMaOH dam/F 1Q O 3\\/C4/LQ_ {L‘Mgu-r/ 9-6 a 1%avmq :3? o Wig 75:07 Ts/my 23K : 6 M 5 GM . I _... Cline)”: " :E’iafiac : CM M (— DL‘E: pkg + [as L. 6:15.13 I C9119) 9,0 = Cm raj C {W523 {556633 I - '6 fl tease.) “a r {‘5an CWCIQ} ratlé; ééfistl 925’ = :C’bq‘s:3 {,Zy’) [GIG—Laj 2 OJ M CC; {131 z 03 M [E'iji =-— 01”“! => R99 '52“ Mac? F9 CGW‘J‘I/ ?L fittq} CPUI : (1 ML {9351. ‘ ' '1,bb{0Qer Home] MquL/ weird“ :‘7 @Ycrq: " Mei/7mg Cad—Jle Lgfig WKg/ ...
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