PETER PAZMANY SEMMELWEIS UNIVERSITYCATHOLIC UNIVERSITY Development of Complex Curricula for Molecular Bionics and Infobionics Programs within a consortial* framework** Consortium leader PETER PAZMANY CATHOLIC UNIVERSITY Consortium members SEMMELWEIS UNIVERSITY, DIALOG CAMPUS PUBLISHER The Project has been realised with the support of the European Union and has been co-financed by the European Social Fund *** **Molekuláris bionika és Infobionika Szakok tananyagának komplex fejlesztése konzorciumi keretben * * * A p ro j e kt a z Eu ró p a i U n i ó t á m o g a t á sá v a l , a z E u r ó p a i S z o c i á l i s A l a p t á rsf i n a n sz í ro z á sá v a l v a l ó su l m e g . INTRODUCTION TO BIOPHYSICS (Bevezetés a biofizikába) THERMODYNAMICS OF ELECTROLYTES (Elektrolitok termodinamikája) GYÖRFFY DÁNIEL, ZÁVODSZKY PÉTER Introduction . While laws described in the previous chapter apply to uncharged particles, in the present chapter, we attempt to give a description ofsystems containing charged particles . Solutions containing charged particles, for example ions, are called electrolytes . Electrolytes are electrically conductive Thermodynamics of electorlytes . Let us suppose we have a NaCl solution with total free energy G . By the additivity rule G=.H 2 OnH2 O..NaCl nNaCl . We could imagine building up the solution byadding one Na+ ion at a time followed by one Cl- ion each time G=.H 2 OnH2 O..Na. nNa...Cl- nCl- . In our case nNaCl =nNa.=nCl- and .NaCl =.Na...Cl- . Now, we wonder how µNa+ relates to [Na+]? . First of all, let us generalize the problem byconsidering not Na+ and Cl- ions but a generalion with unit positive charge (+ sign insubscript) and another one with unit negativecharge (- sign in subscript) ..=.. 0 .RT ln .. c. and .-=.- 0 .RT ln .- c- where µ+ and µ- are the chemical potential, µ+0 and µ-0 are the standard chemical potential, .+ and .-are the activity coefficient, and c+ and c­are the molar concentrations of the positively and negatively charged particles, respectively . Summing the chemical potentials of thepositively and negatively charged ions we get .salt =....- =.. 0 ..- 0 .RT ln ...- c. c- . It turns out that we cannot really measure µ+ and µ experimentally - . But we can measure µ++µ-, that is µsalt . Let us define .. 0 ..- 0 0 . .± 2 which is called the mean ion standard chemical potential . Furthermore, we cannot measure .+ and .-but we can measure .++.­ . Let us define .± 2 ....- which is called the mean ion activity coefficient . Thus .salt =2 .± 0 .RT ln .± 2 c. c- . Now, let us suppose we put a protein with oneionizable group into chamber A of an osmometer . For simplicity, we will set VA =VB again, where VA and VB are the volumes of chamber A and B, respectively Osmometer with ions . - and + signs in the solution in the figuredenote the protein molecules with a singlenegative charge and the ions with a singlepositive charge, for example a Na+ ion, respectively . Since the molar concentration of the two typesof ions are the same [ P- ]=a and [ Na.]=a . Let us notice that if we do not take the positively charged ions into account then .=RT . ci i thus .=RT [ P-].RT [ Na.] and .=RT a. RT a=RT 2 a . Thus, the measured M2 will be off by a factor of 2 . For a protein with n ionizable groups .=RT na . To avoid this problem, let us add a lot ofadditional electrolyte, for example NaCl, toside B at a concentration b . Let us suppose that the amount of Na+ which has diffused over to A, at equilibrium is x . An equivalent amount x of Cl- will also diffuse over to maintain equivalent µ and electro­neutrality . Let us compare the initial concentrations of thedifferent ion types in side A and B with theseconcentrations at equilibrium Initial and equilibrium concentrations of the ions Initial concentrations of ions Concentrations of ions at equilibrium . To calculate the osmotic pressure, ., let us take the difference between the total concentrations in sides A and B and multiplyby RT . We can solve this problem for x in two ways First method . Intuitively, we can imagine that Na+ and Cl- can occasionally come together to form NaCl insolution Ka Na..Cl- . NaCl . The dissociation constant is 1 [ Na.]A[Cl- ]A [ Na.]B [Cl-]BKd == = Ka [ NaCl ]A [ NaCl ]B . Now, NaCl is uncharged and can freely diffusethrough the membrane giving the sameconcentration on both sides [ Na.]A[Cl- ]A=[ Na.]B[Cl-]B with symbolic letters .a. x. x=. b- x.2 and after rearrangement b2 x= a.2 b Second method . More rigorously, in a heterogeneous system .NaCl ,A=.NaCl ,B substituting the expressions describing thechemical potentials we obtain .2 .± 0 . p....RT ln .± 2 c. c- .A =.2 .± 0 . p ..RT ln .± 2 c. c- .B . After rearrangement ..± 2 c. c- .A0 00 2 .± ,A . p...-2 .± ,B . p.=V NaCl .=RT ln 2..± c. c- .B . Exponentiating both sides we get 2 0 / RT ..± c. c- .A -.V NaCl e = ..± 2 c. c- .B . Let us look at an example .=0.01 atm V NaCl 0 =0.015 T =298 K dm3·atm R=0.082 mol·K . Substituting these values into the expressionabove we get -.V NaCl / RT 0.01·0.015/ 0.082·298 6.14·10-6 e0 =e=e.1 . This says that the concentrations of salt onboth sides are about the same 22 .± ,A ..± ,B so .c. c- .A..c. c- .B and with symbolic letters .a- x. x=. b- x.2 . After rearrangement b2 x= a.2 b as with the first method . Let us return to the osmotic pressure (additivity rule) .=RT .[ P- ]A.[ Na.]A.[Cl- ]A-[ Na.]B-[Cl- ]B. with symbolic letters .=RT . a.. a.x.. x-.b-x.-. b- x.. .=RT . 2 a.4 x.2 b . . Since b2 x= a.2 b the expression above will be 2 a2.2 ab .=RT .. a.2 b . What if we do not add NaCl to side B? b=0 and .=RT 2 a . What if we add lots of NaCl to side B? b »a and .=RT a we can measure the molecular weight of theprotein . What if the protein releases n Na+'s? [ Na.]A =na. x and thus b2 x= na.2 b thus the osmotic pressure is .=RT . a.. na. x.. x-. b- x.-.b-x.. Donnan potential . The Donnan potential involves a closer examination of the forces keeping Na+ ions on side A even if no NaCl was added to side B . To be able to investigate the Donnan potential, we should evaluate the concept of electrical potential Electrical potential . The electrical potential, ., is the amount of electrical work that must be performed tomove one unit positive charge from a locationwhere .=0 to where the electrical potential is . . If ..0 we must do work . If ..0 the system can perform work, so we obtain work . The units of electrical potential are work/unitcharge . In physics, we use units of Joule /Coulomb=Volt . An important constant is the Faraday constant, F, which is a convenient conversion factor from a single charge to a mole of charges F = NA·e=96485 C mol-1 where NA=6.022·1023 is the Avogadro constant and e=1.602·10-19 is the elementary charge thatis the charge of a proton Michael Faraday (1791-1867) . The work to move one mole of charges is w =zN Ae.= zF . where z is the signed valence . Now we will discuss the mechanism that generates the Donnan potential . Connect two boxes (10 cm on a side) by asemipermeable membrane . The volume of each side will be one dm3 . Let us put a protein with one ionizable groupand Na+ ions into the side A Generation of the Donnan potential Initial concentration of ions . Does any Na+ goes across the membrane? – A little bit does diffuse across . At equilibrium [ Na.]B.10-11 M . This does not change [Na+]A significantly and the movement of Na+ ions creating a small charge separation is not significant for calculations . The Na+ which leaks across the membrane is found along the walls of chamber B . These positive charges repel each other andpush each other to the edge of the container (to get as far apart as possible) . There will be a small excess negative charge inchamber A . There will be an excess negative charge is also found along the walls of chamber A . Let us imagine that we take a unit positivecharge from a long distance (where .=0) and move it into B . We would perform .B work w =.B . Likewise, if we take a unit positive charge froma long distance and move it into A, we wouldperform .A work w =.A . Therefore to take a unit positive charge fromside A to side B requires .=.B-.A work . This work is called the Donnan potential . The Donnan potential in our case is ..=.B-.A.360 mV (1870-1956) Donnan potential . The (powerful) electrical field created at themembrane is what prevents Na+ from going across . .A and .B have opposite signs ..=.B-.A is the Donnan potential for a Na+ going from A to B . To calculate the Donnan potential, we use amodified version of the fundamental law . In a heterogeneous, closed system at equilibrium, the electrochemical potentialof any substance (charged or uncharged)is the same in all phases between whichit can freely pass . Let µ denote the electrochemical potential .G..= .. n..T ,p,n- is the change in free energy of a solution uponaddition of an infinitesimal amount of Na+ . Let us imagine that we can separate Na+ from its charge . Then we can put Na into the solution first andthen add the positive charge . We can then distinguish how much electricalwork is needed to add a positive charge ..=.. 0 .RT ln .. c.. zF . is the electrochemical potential . Now, ....A =....B . Let us subtract the expression for side A fromthat for side B .0 B.RT ln ... c..B. zF .B -.0 A.RT ln ... c..A. zF .A =0 . Since the standard chemical potentials dependonly on the properties of the substance andnot on the concentration of it, .B 0 =.0 A . Thus these terms disappear from the expression above ... c..B RT ln .zF ..B-.A .=0... c..A . After rearrangement we obtain ... c..B ..=.B-.A=-RT ln ... c..A . At 10-5 M concentration, ...1 so RT .c..B ..=- ln zF ..c..A . . Plugging in c+,A=10-5 M and c+,B=10-11 M, the Donnan potential is ...360 mV . This small potential keeps most of Na+ ions in the side A with the protein . Let us return to the osmometer problem and examine how dumping in salt suppresses the Donnan potential Donnan potential in an osmometer . After adding NaCl at an initial concentration b, x amount of NaCl diffuses to side A until equilibrium is reached . Now, we use either [Na+] or [Cl-] to calculate the Donnan potential, .., RT b- x ..=- ln zF a. x . Substituting b2 x= a.2b we get RT b ..=- ln zF a.b . What if we throw in lots of NaCl, that is b »a Then we can ignore a in the denominator and get RT ln b b =0 ..=- zF . We have suppressed the Donnan potential by throwing in a lot of salt b=0 Then 0 ln =-. a but .. does not go to infinity due to a little Na+ going across . We showed earlier that for this case ...360 mV Debye-Hückel theory . The Debye-Hückel theory allows us to calculatethe mean activity coefficient, .±, for electrolytes, taking nonideality into account ..=2 .± 0 . RT ln .± 2 c. c- or ..=2 .± 0 . RT ln .± 2 .RT ln c. c- . The second term on the right hand side is anenergy term related to the interactions between ions in the solution: / – Na+Na+ – Cl-/Cl­ – and Na+/Cl­ . .± is a measure of nonideality of electrolyte solutions . Let us recall that the ideal case assumes no interactions between particles in solution . There are strong ionic interactions between ions in solution leading to large deviations from ideality . The Debye-Hückel theory allows us to calculate .± for electrolytes and take nonideality into account . The result is log10 .±=-0.51|z. z-|I 1/ 2 where -0.51 is a constant for H2O as a solvent at 298 K, z+ and z-are the ionic charges and I is the total ionic strength Ionic strength . The ionic strength is I .. cizi 2 i where ci is the molar concentration of the i'th ion and zi is the charge of it Examples for ionic strengths . How much is .± for a non-electrolyte (sucrose)? . For an uncharged solute, . becomes significant, that is different from 1 by morethan 1% at a concentration at about 0.1 M . How much is . for electrolytes? . For electrolytes, . is already significant in the µM range . How good is our equation for predicting .±? . This was a simple model . A more sophisticated model that takes intoaccount ionic radii works up to a concentrationnear that where deviation from ideality occursfor non-electrolytes . The Debye-Hückel theory assumes that at lowconcentrations, deviations from ideality for electrolyte solutions are entirely due to coulombic forces . Let us imagine that we could take a snapshotof the Na+ and Cl- ions in a NaCl solution . On average, the density of Cl-ions in the vicinity of Na+ ions would be a little greater than the average density of Cl-ions . The same would be true of the density of Na+ ions around a Cl- ion Ionic density in the vicinity of other ions Ion atmosphere around a Na+ ion . Actually, the ions are undergoing violent Brownian motion, so a more realistic pictureinvolves the time averaged density . In the figure above, shading represents thetime-averaged excess negative charge densityaround a Na+ ion . The darker the shading the greater the density . The distribution of negative charge around theNa+ ion is spherically symmetrical . It is called the ion atmosphere . We can plot the ion atmosphere as a functionof distance, r, from the centre of the Na+ ion . It can be seen that at r=. c.=c- =c. where c- is the concentration of Cl- ion . The curves are symmetrical to the r axis The ion atmosphere as a function of r Peter Debye (1884-1966) (1896-1980) Coulomb's law . The force between two charges, F, can be calculated from Coulomb's law z1 z2F = Dr2 where D is the dielectric constant, z1 and z2 are ion charges and r is the distance between the ions . It is worth noting that D=1 in vacuum and D=80 in water . It means that the force between two charges is 80 times greater in vacuum than it is in water Charles Augustine de Coulomb (1736-1806) . The electrical field strength is the force per unit charge and is denoted by E . The electrical field around a charge, z, is determined by placing a positive test charge, z2, at some distance, r, from z, and measuring the force on z2 Fz1 E == z2 Dr2 . E is a vector quantity . It points away from z, if z1 is positive . The direction of E is now irrelevant for us; we only use its magnitude Electrical potential . The electrical potential is a quantity whosenegative derivative with respect to distance is the electric field, which represents the forceacting on a unit charge . The electrical potential is denoted by . .. -= E . r . And since F E = z2 the derivative of electric potential with respect todistance is .. F -= . rz2 . To obtain the electrical potential, we should integrate the expression above from . to r r rF .. .=-. . r .. z2 . We get rF..r.-....=-. . r . z2 . Since ....=0 rF ..r.=-. . r . z2 . As we know, force times distance is work . Therefore, .(r) is the work required to bring a unit positive test charge (z2) from a location where the potential is zero to where thepotential is .(r) . According to Coulomb's law F z1 = z2 D r2 so ..r.=- z1 D . . r . r r2 = z1 D r . We can plot . as a function of r, for a single Na+ ion in the absence of an ion atmosphere z1 =ez. where e is the elementary charge and z+ is the valence of the positive ion . So z. e ..r.= Dr Electrical potential vs. r . Now, we wonder whether there is a relationship between the electrical potential, ., of an ion and the ion atmosphere . We can calculate this relationship using the Maxwell-Boltzmann distribution and Gauss's law . A special consequence of Gauss' law is that fora spherically symmetrical charge distribution, the electric field at radius r will be equal to that caused by the sum of the charges inside asphere of radius r, acting as if they were at thecentre of the sphere (1777-1855) . We can relate the electric potential, ., to the ionic strength, I z -. r .= Dr e where 2 I 8. N Ae .2 = 1000 Dk BT where NA is the Avogadro constant and kB is the Boltzmann constant . If we plot . as a function of r both in the presence and absence of the ion atmospherewe see that .i.a...no i.a. where .i.a. are the electric potential and .no i.a.with and without ion atmosphere, since z -. r .i.a. = Dr e and if I=0 .no i.a. = z Dr Electric potential with and without an ion atmosphere . How much energy is required to put one positive charge on a neutral Na atom? . Let us recall that RT ln .. is the energy to create the ion atmosphere for 1 mole of Na+ ions, so kBT ln .. is the energy to create the ion atmosphere around one Na+ ion . Now, let us calculate the work to charge oneneutral Na atom z .e w =. . dz 0 . In absence of ion atmosphere z. ez. e z . ze.2 w1 =. . dz=. dz= 00 DrNa zDr Na . In presence of ion atmosphere z.ez.e z . ze.2 -. r Na dz-. r Na w2 =. . dz=. e= e 00 Dr Na zDrNa where rNa is the radius of Na . The difference between w1 and w2 is the work required to create the ion atmosphere . ze.2 -. r Na-1 . kBT ln ..= .e z DrNa . Since . rNa is very small, -. r Na.1-. rNa e . So kB T ln .. = . z e.2 z D rNa .1-. rNa-1 .=-. z e .2 z D . . Similarly, for negative charge kB T ln .-=-. z e.2 z D . .± 2 =...- the expression above can be written as 2 kBT ln .±=kBT ln ...kBT ln .- 2 2 e2 . z- e2 . 2 .' =- z.- =- e.z. 2 .z- 2 . I1 / 2 zD zD zD where . . '= I 1/ 2 . It can be shown in general that . z. 2 .z2 - .=|z. z-| . Now, for H2O at 298 K ln .±=-1.17|z. z-|I1 /2 or 1.17 log10 .±=- 2.303|z. z-|I1 / 2=-0.51|z. z-|I1 / 2 . This is an important equation because it allowsus to calculate a thermodynamic property, the mean ionic activity coefficient, .±, from molecular properties . Let us notice that ln .±.0 so .±.1 . Let us notice that the Debye-Hückel theoryalways predicts .±.1 . This occurs because the interaction between an ion and its ionic atmosphere is alwaysfavourable