E.g.: cell membrane, kidneys, dialysis, fuel cells, liquid junction (beween

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1 Membranes and ons [cd www;mz donnan.html] 1/21 Sempermeable membrane; glass frt; daphragm concentratons of ons on both sdes dffer dfferent permeabltes of ons mechansms: = s proportonal on channels (n cell membrane) pores (wder, permeablty dffusvty) sorpton+dffuson (polymer membrane),... E.g.: cell membrane, kdneys, dalyss, fuel cells, lqud juncton (beween electrolytes) [cf. osmotc pressure] We are nterested n the membrane potental n equlbrum: one on permeates through the membrane zero dffuson (fast!) some ons permeate, other do not Donnan equlbrum We are nterested n the membrane pot. durng dffuson (rreversble!): thn membrane (e.g., cell): (bo)membrane potental (Goldman) electrolyss separated by a (thck) membrane: lqud juncton (dffuson) potental

2 Smple start: one on permeates [jkv -g nafon.png] 2/21 HCl, dfferent concentratons at both sdes of a membrane (glass, Nafon,... ) only catons H + can permeate Catons try to dffuse to places wth a lower concentraton. Snce the anons cannot follow them, a membrane potental arses. In equlbrum, the dfference n the chemcal potentals s compensated by the electrc potental, φ (also E, E): µ rght H + φ = φ rght φ left = RT µ left H + + zf φ = 0 zf ln a rght H + a left H + RT zf ln c rght H + H + < c rght φ = φ rght φ left Equvalently: the electrochemcal potentals µ = µ + z Fφ (z ncludes sgn) of ons H + left and rght are the same. Macroscopc concentratons of H + (HCl) are unchanged (electroneutralty), only concentratons close to surfaces (wthn double-layer) are affected.

3 Donnan equlbra 3/21 left : rght NaX : NaCl NaCl : anon X does not permeate The dfference of the electrochemcal potentals: µ rght Na + µ left Na + = RT ln crght Na + + F φ equlbrum = 0 Na + µ rght Cl µ left Cl = RT ln crght Cl Cl F φ equlbrum = 0 Sum of both equatons Generally for salt Kν Aν : ( Na + Cl = c rght Na + c rght Cl ) ν ( )ν = (c rght ) ν (c rght ) ν A A

4 Donnan equlbra membrane hydrolyss 4/21 In the left compartment, there s n = 0.01 mol of sodum p-toluensulfonate (NaTsO) n V left = 100 ml of water; n the rght compartment, there s V rght = 1 L of pure water. The membrane n mpermeable for TsO. Calculate ph n both compartments n equlbrum at 25 C. balance start equlbrum [mol] left : rght left : rght TsO n : n : Na + n : n x : x OH : 0 : x H + : x : 0 φ = RT F = RT F ln crght ln Na + Na + = V x/vrght (n x)/v left n x V left Na + OH = c rght K w x/v left = Na + c rght x V rght OH (or c rght x V rght Na + / Na + = c rght H + / H + ) Numercally (n mol, dm 3 ; more accurately by teratons) x = 3 K w (n x)(v rght ) 2 x n mol = mol ph left = 4.3, ph rght = 8.7

5 Dffuson potental at a thn membrane E.g., cell membrane (lpd double layer wth on channels) we know on concentratons and c rght statonary dffuson zero total current (after establshng voltage φ, whch s much faster) 5/21 φ = φ rght φ left sze of molecules neglected, membrane = delectrc contnuum electrc feld ntensty E = dx d φ across the membrane s homogeneous follows from the Posson equaton d 2 φ/dx 2 = ρ/ɛ : E(L) = E(0) + Lρ/ɛ E(0) pro L λ Nonzero dffuson flux = rreversble phenomenon Permeablty of the membrane for on : P = D K N /L u = D z F/RT λ = z Fu D = dffusvty n the membrane materal; D n a frt = D n. Nernst dstrbuton coeffcent K N here = sorpton coeffcent, dmensonless for sorpton from lqud J = P (c rght ) For unvalent ons: P D u λ. Ths s enough: we shall see that the voltage depends only on the rato of permeabltes.

6 Thn membrane: Goldman equaton 6/21 Flux of ons at x (left x = 0, rght x = L); [J ] = mol m 2 s 1 J = D grad c + λ c E z F dc = D dx + D c z FE RT = D RT c grad [µ + z Fφ] = D RT c grad µ J does not depend on x (statonary flux nothng accumalates). Equaton can be ntegrated (separaton of varables): L 0 dx = c rght D c z FED /RT J dc We calculate J from concentratons and E. After several steps: J RT = D z FE ɛz c rght ( ) FLE ɛ z, where ɛ = exp = exp 1 RT Zero total current: better K N c rght K N (K N cancel out) ( F φ ) RT NB sgns: φ = φ rght φ left = LE 0 = z J Addtonal smplfcaton: z = 1 (unvalent ons)

7 Thn membrane: Goldman equaton 7/21 Let s sum anons and catons separately, replacng D P lnear equaton for ɛ, after rearrangng: φ = RT F ln catons P c rght catons P Example. Relatve permeabltes of man ons n the mammalan plasmatc membrane are: P(K + ) = 1, P(Na + ) = 0.04, P(Cl ) = 0.45 Concentratons nsde the cell (n mmol dm 3 ): [K + ] rght = 400, [Na + ] rght = 50, [Cl ] rght = 50 Concentratons outsde the cell (n mmol dm 3 ): [K + ] left = 20, [Na + ] left = 500, [Cl ] left = 560 Calculate the restng potental of the membrane. φ = J mol 1 K K C mol 1 + anons P + anons P c rght credt: wkpeda ln = V The nsde of the cell ( rght ) s negatve

8 Cell membrane contnued 8/21 Numercal soluton for: L = 4 nm ɛ r = 4 K N = (membrane concentraton s 1000 smaller) φ/v c/mmol.dm [K + ] [Cl - ] [Na + ] φ model φ Goldman outsde nsde x/nm

9 Thck membrane (frt, lqud juncton) Irreversble process, complex problem (partal dfferental equaton). Smplfcaton: soluton of a un-unvalent salt at both sdes, conc. and c rght 9/21 the membrane s thck grad c = grad c = const It follows from the Posson equaton d 2 φ/dx 2 = z Fc /ε every layer s electroneutral, small dfference wll curve φ(x) as needed Formulas: µ = µ + z Fφ, j = c u µ, J = c D RT µ, u RT = D z F, z < 0 flux of catons: J = D d dx c D F RT c d dx φ flux of anons: J = D d dx c + D F RT c d dx φ x c rght Stady state: j = z J = J J = 0, c = c = c (D D ) dc dx = (D + D ) F RT cdφ, separaton of varables c, φ: dx D D D + D ln crght = F RT φ φ = (t t ) RT F ln crght t ± = D ± D + D

10 Thck membrane II 10/21 Applcatons: voltage loss at a lqud juncton (dffuson potental), e.g.: Ag AgCl HCl( ). HCl(c rght ) AgCl Ag For un-unvalent salt and t = t t holds φ = 0. Therefore n salt brdges there are solutons wth t t (e.g., KCl: t = 0.49, t = 0.51) Generalzaton for salt K z ν A z ν : φ = ( t z t ) RT z F ln crght Inaccurate dervaton (for un-unvalent salt): Let 1 mole of charge (1F) flows from left to rght. Ths s t anons mgratng left and t catons mgratng rght: +t anons ( ) :: t anons (c rght ) t catons ( ) :: +t catons (c rght ) G = t RT ln crght + t RT ln crght? = zf φ φ = (t t ) RT crght ln F But G work for an rreversble process fals for a thn membrane

11 Comparson of thn and thck membranes 11/21 1:1 electrolyte c rght : = 10 thn membrane (L λ) Controlled by electrc forces (they determne the local concentratons) The smplfed reasonng (prevous slde) fals! thck membrane (L λ) Controlled by dffuson (electrc forces screened off) φ / V For t = 1 (only catons permeate), both equatons gve t + φ = RT zf ln crght For t = 2 1, we have φ = 0 (symmetry catons:anons)

12 Osmoss 12/21 The membrane lets through the solvent (optonally wth small molecules). The solvents tres to permate to the place wth a lower chemcal potental osmotc pressure µ 1 (p A) µ 1 (p A, 1) 1 = solvent µ 1 (p A)! = µ 1B (p B, x 1B ) d. = µ 1 (p B) + RT ln x 1B = µ 1 (p B) + RT ln(1 x 2B ) V 1m Π = RT n 2 n V 1m =const, x 2 1 = µ 1 (p A) + V 1m (p } B {{ p A } ) RTx 2B Π Π = c 2RT J. H. van t Hoff, H. N. Morse n 2, c 2 are ncl. dssocaton

13 Osmoss 13/21 The osmotc pressure s a collgatve property t depends on the number of partcles (amount of substance) Osmolarty = amount of substance (not permeatng through the membrane) n unt volume Osmolalty = amount of substance (not permeatng through the membrane) par unt mass of the solvent Example. Calculate the osmolalty of 0.15 mol of NaCl n 1 kg of water. 0.3 mol of ons n 1 kg of water, osmolalty = 0.3 osmol kg 1 Approxmately ρ = 1 kg dm 3 osmolarty. = 0.3 osmol dm 3 Loosely 0.15 M NaCl = 0.3 Osm NaCl Osmotc pressure more accurately: Π = c 2 RT(1 + Bc 2 + δc 3/2 2 + Cc ) B: second osmotc vral coeffcent nteracton of a par of solutes for collod partcles determned manly by the excluded volume δ: onc nteractons (Debye-Hückel) C: trples of solutes

14 Osmoss 14/21 extra/ntracellular are sotonc solutons dalyss sea water desalnaton by reverse osmoss determnng molar masses credt: wkpeda Example. Calculate the mnmum pressure needed to desalnate sea water by reverse osmoss at 300 K, and the mnmum energy needed to produce 1 m 3 of fresh water. The total concentraton of ons n sea water s 1.12 mol dm bar; 2.8 MJ

15 Example 15/21 The osmotc pressure of a soluton of an enzyme n water (25 C) s c w g dm Π Pa Calculate the molar mass. c w = c 2 M After dvdng by M: From the plot: RT M Π/c w /(Pa dm 3 g -1 ) Π c w = RT M + RTB M 2 c w = 24 Pa g 1 dm 3 = 24 Pa kg 1 m 3 c w /(g dm -3 ) M = Pa m3 K K 24 Pa g 1 dm 3 = 103 kg mol kda

16 Van t Hoff factor 16/21 Eq. Π = c 2 RT s sometmes wrtten as Π = c 2 RT where c 2 s the analytc (formal) concentraton and s the van t Hoff factor,.e., the average number of molecules (not permeatng) the compound dssocates to, e.g.: (glucose) = 1 (NaCl) = 2 (CH 3 COOH) = 1 + α

17 Saturated vapor pressure of a soluton 17/21 Raoult + Dalton law p = py = x p s, p = p = x p s Apply to: 1 = solvent 2 = non-volatle solute (x 2 0 or n 2 n 1 ) f dsocated n, the fragments are counted n n 2 and x 2 p 1 = x 1 p s 1 = ps 1 x 2p s 1 or n 2 p = x 2 p s 1 = ps n 2 n 1 1 p s n 1 + n 1 2 n 2 n 1 The saturated pressure of a soluton s lower than that of a pure solvent at the same temperature.

18 Ebulloscopy 18/21 1 = solvent alternatve dervaton: 2 = solute (ncl. dssocaton) see next slde At constant p, p s compensated ( p) by ncreasng the bolng pont of the soluton by T. Usng the Clausus Clapeyron equaton: where T p p s 1 1 p RT 2 bol,1 p T vaph m RT 2 vap H 1,m = ps 1 n 2 n 1 p s 1 RT 2 bol,1 vap H 1,m = K E m 2 K E = RT 2 bol,1 M 1 vap H 1,m = ebulloscopc constant The bolng temperature of a soluton s hgher than that of a pure solvent at the same pressure. Or T = K E m 2, where m 2 s the chemcalformula-based molalty and s the van t Hoff s factor Example. Calculate the bolng pont of a soup (1% wt. NaCl) at normal pressure. K E (water) = K kg mol C

19 Cryoscopy 19/21 A compound dssolves n a lqud solvent, but there s no mxed crystal n the sold phase meltng (fuson) temperature decreases. Dervaton: 1 = solvent, 2 = solute x 1 = 1, T = T fus : µ s n 2 n 1 (T fus) = µ l 1 (T fus) 1 µ s x (T fus) = µ l 1 (T fus) x 1 < 1, T = T fus + T : µ s 1 (T fus + T) = µ l 1 (T fus + T) µ s 1 (T fus + T) = µ l 1 (T fus + T) + RT ln x 1 ( µ s) ( ) T 1 µ l = T 1 + RT ln(1 x T T 2 ) T (µ l T 1 µ s 1 ) = T( fus S m ) = T fush m T = RT ln(1 x 2 ) RTx 2 RTM 1 m 2 T = K K m 2 K K = M 1RT 2 fus fus H m = cryoscopc constant or K f = K K T = K f m non dssoc. 2 Example. Calculate the freezng temperature of beer (4.5 vol.% alcohol, densty of alcohol=0.8 g cm 3 ). K K,water = 1.85 K kg mol C

20 Vapor pressure osmometry 20/21 pure solvent evaporates vapor condenses on a soluton (lower vapor pressure) pure solvent cools down, soluton heats up T concentraton (collgatve property) credt:

21 Collgatve propertes summary 21/21... depend on the number of molecules (moles) dssolved. bolng temperature ncrease (ebulloscopy), T = K E m 2 = K E m 2 m 1 M 2 meltng temperature decrease (cryoscopy), T = K K m 2 T = K K m 2 m 1 M 2 e.g., camphor K K = 40 K kg mol 1, t fus = 176 C. osmotc pressure, Π = c 2 RT = m 2 VM 2 RT pressure of (deal) gas, p = nrt/v = crt = Usage: determnng molar masses m VM 2 RT Accuracy: ebulloscopy < cryoscopy < vapor pressure osmometry < membrane osmometry