 Alright, in this lecture third of module 3, we look at electrically charged monolayers and the Gouy theory for explaining the potential in such charged monolayer systems. That is also known in literature as diffuse double layer theory, but before we go into that, let us quickly recap the components of delta v due to a monolayer. The first point that we noted was the absolute potential for pure water v is in the range minus 0.5 to plus 0.4 volts and many researchers agreed on even narrow range of between minus 0.1 and minus 0.2 volts. We do not deal with absolute potentials anyway and as far as the influence of a monolayer is concerned, whether it is a spread or adsorbed monolayer on a clean water surface, the dipole moments would contribute in basically three categories. At this point maybe I should explain that the words have been chosen carefully here, spread or adsorbed. By spread monolayer, we would mean that we have a very thin film of the material in the form of monolayer in spread form. While as in calling it adsorbed, it would mean that there is an affinity for the interface by virtue of amphipelic nature of a compound hydrophobic hydrophilic combination and what it means is that spread may mean insoluble monolayer whereas adsorbed means soluble monolayer. There are components of these molecules which could actually dissolve but will prefer to be at the interface. The three dipole moments categories which will contribute to the overall dipole moment arise from first the reoriented water molecules. Water dipoles get reoriented in presence of a dipole which is anchored into the surface because of the adsorbed or spread monolayer and therefore mu 1 rising out of orientation of water molecules whose its existence to mu 2 which is the orienting dipole moment. In case of a long chain amine for example, it would be the cNH2 group which is anchored in water which would carry a dipole moment of mu 2 in turn giving rise to mu 1 from water. The third component of delta V comes from the upper limit of the monolayer that is the cNH at the top. This is for electrically neutral films so mu 1, mu 2 and mu 3 if they all contribute in perfectly vertical orientation to the overall dipole moment. We could use Helmholtz formula with n dipoles per centimeter square for the array of the monolayer molecules. Delta V would be summed up as 4 pi n mu 1 plus 4 pi n mu 2 plus 4 pi n mu 3. Water the head group anchored in water and the cNH dipole moment at the top. Since mu 1 cannot be measured but is direct result of mu 2, we may choose to combine it with mu 2 and mu 3 is practically constant because in most surface active molecules we have paraffinic chains. So, the dipole moment at the top end mu 3 is constant. So, we could add up all these 3 dipole moments mu 1, mu 2 and mu 3 into an overall dipole moment mu d and when we have this electrically neutral films only dipole moments contribute to delta V. Now, depending on the nature of the molecule or the head group whether it is symmetric or unsymmetrical we might expect to see effect of surface pressure. If the head group the dipolar head is unsymmetrical then upon compression of the film so that more surface molecules get crowded together we might actually cause a difference in the orientation and therefore that may be reflected in the measured or observed dipole moment mu d. For ethyl permeate the head group has a moment of 535 milli device whereas, the calculated or predicted value is 525 milli device that is provided the area available per molecule is greater than this number 72 angstrom square for ethyl permeate. This background will help you understand what might happen when we come to electrically charged monolayers that should be one simple extension of the physical idea behind calculation of delta V. So with the electrical charge now coming into picture we expect delta V to differ. We will take an example like for example this octa-dacil quaternary ammonium salt. If we take octa-dacil quaternary ammonium chloride for example in water then like we discussed earlier these ionogenic molecules should get dissociated upon dissolution in water. The octa-dacil ammonium cations will tend to adsorb that at the interface. The chloride ions on the other hand will remain dissolved in dissolved form in water and they may be distributed over a certain length away from the interface. Maybe we can have a quick look at the diagram or the figure here before we return to this discussion. So, let this line indicate the interface between either air and water or between a paraffin oil and water. The paraffin oil could be decaying and the octa-dacil quaternary ammonium chloride molecules when they are added to water they would dissociate to have this cation over here and the chloride in dissolved form. So when you dissolve lots of these molecules we will have number of these chloride groups in dissolved form but clustered around the interface. At this point you could think of some simple minded pictures of what might happen. If one way to take the oversimplified picture one may say that chloride ions are all exactly in one plane at a very short distance from the interface. That will mean your cations and anions are in very close proximity in two parallel planes. But more realistic picture could be that the cations are anchored and in place here at the interface but the chloride will float around. They will be close to the interface but not in a very tightly packed fashion. They would be rather present in a diffuse layer over certain distance here and one might actually go on to estimate the mean distance of the anions from the interface from the Debye Huckel length that we had seen earlier. So one may take that view as a better representation of reality. So what is the difference between the potential that you may measure now compared to for the electric, the one for electrically neutral monolayers. We should have the ordinary dipole moments contributing like earlier those would still be present but in addition we will have an electrostatic term. This would be coming from unequal distribution of ions around the adsorbed monolayer. So there should be a simple generalization of the potential for charge monolayer compared to an electrically neutral monolayer. Now strictly speaking this potential that we expect from the unequal distribution of charges by definition it should be the psi potential. However in light of what I also just described the anions are not probably going to be in a very tightly packed form in a single plane at a fixed distance from the interface. It is likely that they would rather be present in a diffuse layer and if it is a paraffin oil in contact with water there is always that potential possibility of the anions distributing to an extent. So strictly speaking it is a psi potential but we take a different view here. We say that by very nature it is a distribution potential psi potential but it arises directly from the presence of monolayer more in line of what Bohr had thought. So here is a situation which is somewhere in between the situations visualize as extremes by Butner and Bohr. We definitely have adsorption here. There is a adsorbed monolayer. However the anions associated with this adsorbed monolayer are to an extent unequally distributed. So they would contribute some cations also could be present inside water. So they would cause a psi potential. But since this psi potential is linked intimately to adsorbed monolayer will rather include it in delta v. So the discussion that we had so far of what components delta v will have from the electrically neutral films will have only an additional term which is corresponding to the electrostatic potential. And we will call that additional contribution psi 0 that will be in nature of distribution potential but is being clubbed into delta v. Now why this subscript 0? By subscript 0 we want to indicate that this electrostatic potential which is originally distribution potential but is being attributed to the interface. So that means it is an electrostatic potential in the interface relative to the aqueous phase at a 0 distance from it. We just shift the potential all to the interface now. So that means that our equations which contribute individually 4 pi n mu 1 from the water dipoles, 4 pi n mu 2 from the cations but from the dipole moment part of it and 4 pi n mu 3 the dipole corresponding to the top end will have on top of it this psi 0. So we see delta v is getting a contribution from distribution potential but it is referred to a 0 distance from the interface. And like earlier we could replace the first 3 terms by 4 pi n mu d and we could add psi 0 to this or one may say that entire thing on the right hand side could be expressed as 4 pi n mu overall. So now we are choosing to express the entire contribution on the right hand side in terms of another overall dipole moment like term. We already looked at this diagram. The mean depth of counter ions or the mean thickness of the diffuse ionic double layer is 1 by chi. We then address the question of how to obtain psi 0. Calculation of psi at the surface in such situation has been handled by go in his 1910 theory which was based on some basic assumptions. Still even if we have air the counter ions they may not be all present at a fixed distance and they may not all be tightly packed. They will be distributed. They will be in a diffuse layer and they would still cause a potential which would not be operating from a single depth. So that all can be clubbed into the contact potential. So this one we will see here as you look at the Goethe theory it will become clear. First assumption in the Goethe theory is that the charge surface is impenetrable. Second the charge is uniformly distributed at this value sigma per centimeter square. Third the counter ions behave as point charges. They are able to write approach the surface right up to the plane of charges. So it is not necessary chloride ions remain only away from interface. They can come right up to the surface. But these are assumptions. Let us look at the pictorial representation. We have air or oil here the cations anchored in the interface and by these dots we indicate the chloride ions. In the bulk of water there is zero potential ok. So strictly speaking if it is air then they cannot be any chloride ions over here right. But they could be they could be some of these cations not just at the interface but inside this right. So you may say as an extreme explanation that you have zero or minimal concentration in air and a certain miniscule concentration here. So we may still attribute the distribution nature to such potential although we know that this is practically the contact potential. In that case we are safer because if it is oil then there will be a greater degree of nature of distribution potential in this extreme it is closer to delta v. So we might be even more justified in writing whatever distribution nature attributable now to a single contribution in the contact manner right. So with this go solve the equations of Boltzmann for distribution of cations and anions in terms of a potential psi near the positive surface relative to the bulk of liquid. Maybe we should look at some of these equations. Like earlier Boltzmann equations will be given by S c plus equal to c into exponential minus epsilon psi by k t. Likewise for anions it is S c minus equal to c into exponential plus epsilon psi by k t. And by solving these equations in terms of potential psi near the positive surface go a proposed that the psi 0 will be approximated by this psi g which is given as 2 k t upon epsilon sin h inverse sigma by c i to the power 1 by 2 500 pi by d r t to the power half where c is the ionic concentration. This subscript S represents the surface sigma is the uniform charge per area c i is the total univalent electrolyte concentration that is in moles per liter epsilon is the electronic charge that is your absolute temperature. And d is the dielectric constant for water it is 80 at 20 degree centigrade decreases somewhat at 25 degree centigrade centigrade to 78 r is the usual mass constant. Now this equation which is the solution of the Boltzmann equations could be simplified as we see in the next slide. At 20 degree centigrade we find that psi g becomes 2 k t by epsilon or that reduces to 50.4 sin hyperbolic inverse 134 by a c i to the power 1 by 2 where psi g is expressed in millivolts and area available for each ionogenic long chain is this capital A expressed as angstrom square per charge group. N is the number of charge groups per centimeter square 1 centimeter is 10 to the power 8 angstroms. So, we could express A in terms of N as 10 to the power 16 divided by N provided all ionis ionizable groups are dissociated here. At 50 degree centigrade our original equation reduces to this equation 38 55.3 sin h inverse 139 by a c i to the power 1 by 2. Now there is one special case which arises at high potentials psi g which is 2 k t by epsilon into psi h inverse 139 by a c i to the power 1 by 2 reduces to a different functional form. This is when the potentials are greater than about 100 millivolts high meaning greater than 100 millivolts. I would like you to spend a few minutes and see how you would get this equation 39 from 38 under these conditions. This is more to stimulate the idea that you should be also thinking mathematically in certain parts over here. The important thing is we are moving from sin hyperbolic inverse to a ln function the logarithmic function and the condition given to you is the potentials are high greater than 100 millivolts. So, think about how possibly this could arise. Exactly expand the sin hyperbolic inverse and then use this condition the potential is large and you would be able to see that it can be now expressed as a logarithmic function. Those of you who are not able to see it I recommend that you refer to the handbooks of mathematics work it out for yourself jot down this result the equation 39 should follow from 38. This is not difficult if we accept that result then we could also say that from 39 dou psi g by dou log c will be equal to minus 2.303 kT by epsilon. This equation could be alternatively obtained from Stern equations or from thermodynamics and therefore is not as selective as the basic equation 35 that go ahead provided. How good is the validity of this equation? That may be the next question or how far can we identify this go potential psi g with psi 0? Monolayer strictly do not obey our equation 35 that is the first equation, but the objections that charges are discrete as you can see from here the cations and anions are discrete charges. So, one might think that the electrical lines of force are not parallel as presumed in the theory that could be one possible objection to Gois's theory, but that is not very important in the sense that these lines of force actually become parallel very close to the interface. Although the charges are discreetly distributed the electrical lines of force are nearly equidistant and parallel to each other. All are perpendicular to the surface at a very short distance below the surface or interface as visualized in the Gois theory. We could look at this sketch to understand what I am saying there. We have these ionic groups over here. These are the associated chains and arising from this set of cations the lines of force may be represented roughly as shown here. At a quite short distance these lines of force become nearly parallel and the hydrophobic chains of course are oriented upward in air. The positive charge groups, these ionic groups are immersed to a depth of about 3 angstroms. In any case direct measurement of size 0 and size g cannot be made unless of course we know mu d. We will see later that the problem related to measurement of mu d could be handled in a different manner especially when we look at the reactions in the surface, but presuming that mu 1 the dipolar contribution from water dipoles is unaffected by electrical field mu d should be constant at constant area. And in neutral films mu d should be practically independent of the ionic strength of the electrolyte and if size 0 and size g are equal then we can write delta v minus 2 kT by epsilon sinh inverse 134 by Aci to the power 1 by 2 equal to 4 pi n mu d. The right hand side would be a constant at a given value of A or n, n being the number of adsorbed ionogenic groups. And if we make a plot of the left hand side versus log C, extended data reveal that there is a near constancy at a level of 400, at level of 200 millivolts. So, if you make a plot experimental plot of this versus log C, we get a value about 200 millivolts. What is surprising is the equation actually seems to be giving correct value for size 0 even at very high ionic strengths. What do you expect at very high ionic strengths? You expect the mean thickness of this ionic double layer to be quite small. At high ionic strengths it can be as low as 2 angstroms and yet size g seems to be a good approximation to size 0 as revealed from experiments. What is striking about this is that this distance of 2 angstroms is less than even the radius of hydrated chloride ions. So, the justification that the electrical lines of force are nearly parallel at a very short distance from the interface leading to size 0 being well approximated by size g works out to be fine. The electrical potential of a film of C16 quaternary compound like hexadecil ammonium ions at water oil interface decreases from about 300 millivolts to 200 millivolts when the area per long chain ion is increased from about 100 to 400 square angstroms and the size 0 and size g calculated agree well for a constant mu d of 450 millidibis independent of area A. Same value is found at air water interface for various long chain quaternary ions and this is possibly a consequence of symmetry of the structure of the quaternary ammonium ions. Contrast this against the asymmetrical n groups like sulphate long chain sulphates we do not find such constancy. We may also remember that the application of our equation 41 is restricted to a given orientation that is we are not altering the orientation of the adsorb molecules as a result of compression. For very large areas per group charge group each charge will act as a separate entity on its own counter ion and the Debye Huckel length could be calculated from this expression 1 by chi is 3 by C i to the power 1 by 2. We can look at some numbers arising out of this. For dilute solutions it will mean a significant ionic separation. If you take a sodium chloride solution 10th of a milli normal or 110,000th of a normal NaCl the Debye Huckel length is about 300 angstroms. If the concentration of NaCl is increased to about 2 normal then that shrinks to just about 2.1 angstroms. So, we should be able to estimate Debye Huckel length from an expression of this kind. One last point I want to make today is the effect of applied potential on the property that we have been dealing with again and again that is surface tension. The equation which allows you to account for variation of surface tension because of applied potential is as simple as this dou gamma by dou E is equal to minus sigma. E is the applied potential and the decrease in the surface tension with a unit increase in the electrical potential applied is simply equal to negative of the charge density. So, this will help you understand probably the behavior of interfaces or surfaces when an electrical potential is applied. This might also help you understand what might be happening in that extent where we visualized an interface between mercury and water and saw that upon application of electrical field we might be able to get spontaneous emulsification, one phase getting dispersed in another. That is not unexpected because when you apply this electrical potential according to this equation the surface tension is going to diminish. When you decrease it to a value momentary less than 0 we will have the opposite of the contraction. Now, the interface will continue to expand leading to buckling of the interface. So, the spontaneous emulsification becomes better understandable in terms of Liebman equation. All right, we will stop here for today and resume with the monolayers next time.