Lecture 25: Solutions

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Settle down, settle down. This is not some other class, this is 3.091. So, I have an important announcement. There won't be a lecture next Wednesday. Instead, there will be another celebration of learning.

You're going to come, and you're going to show us what you've learned. It's going to be so festive. And, I want to make sure we are all celebrating the same thing. I meant to patch in. It didn't get it patched in.

I apologize. So, let me tell you what it is. We're going to look at is the lectures of starting with lecture 17, which was on x-rays. And, halfway through that lecture was the break point for the last test.

So, we will start with the material beginning with Bragg's law. We didn't talk about diffraction on the last test. The coverage talked about generation of x-rays, but not their use in diffraction.

So, starting with x-ray diffraction up through the end of last day's lecture. And, the reason I'm doing this is that tomorrow is a holiday. There are no academic exercises tomorrow. And so, I didn't think it made a good pedagogical sense to teach you something today and have no recitation on Thursday.

You don't meet again with your recitation instructor until Tuesday and then Wednesday's the test. So, being mindful of that, I'm going to start talking about solutions today. None of that will be on the test.

So, starting with x-ray diffraction and ending with diffusion, so as I've got up here, we saw Fick's laws, first and second laws, error function, all that stuff will be fair game. And of course, we won't have a test, a weekly quiz, on Tuesday.

We'll have plenty to work on for Wednesday. So, let's leave it that way. So, last day, we completed the discussion of diffusion, going through analytical presentation of Fick's laws, and how they're used in a quantitative measure of rate phenomena in diffusion.

Today what I want to do is introduce another topic from the general realm of physical chemistry. And, this is the topic of solutions. And, for that, I'm going to go all the way back to the very first lecture in 3.091 when we looked at this figure from the text.

And, what we've been doing up until now is largely working over in the lower left corner of this taxonomy. We've been looking at the properties of elements. We've looked at a few compounds, but we really haven't gotten up into here.

Everything has been pure pretty much. But, we have had to make the odd foray. What we talked about glasses, we got into the discussion of solutions because we were talking about how you can modify the properties and tailor the properties by changing composition of a glass.

So, that was really in this domain of mixtures. And we talked about doping of semiconductors. When we dope a semiconductor we actually put phosphorus or boron substitutionally for silicon. So, in essence, we are making a solution.

So that's all up in here. But, I'm a little bit uncomfortable with this broad definition of mixtures. So, I've taken the liberty of refining that somewhat. And so, what I would prefer you recognize is that there is two types of mixtures.

There's homogeneous mixtures and heterogeneous mixtures. So, homogeneous mixtures are what we would call solutions. And, that gives you mixing right down at the molecular level, whereas heterogeneous mixtures are not mixed down to the molecular level.

They are mixed down to the level of clusters of molecules. So, for example, if I look at something like brine, brine is salt dissolved in water. It's visually uniform. We cannot tell that there's more than one component.

And, it's this homogeneity uniformity that makes it a solution, whereas if you look at something like milk, milk is a mixture. It's an aggregate. Milk actually consists of two phases: there's a fatty phase which is insoluble, and the aqueous phase.

And, what's happened is we've got a fine dispersion of the second phase in the first phase. And that fine dispersion floats. The fat is clear and colorless, and the aqueous phase is clear and colorless.

But that dispersion scatters light. And that's where we get the white in milk. And, I don't like the notion that whether it's milky or clear and colorless, in other words, coarsely mixed or finely mixed; we simply lump them under.

So, I'm going to be talking today about solutions and making sure that we understand the underlying chemistry. So, you might say, well, we signed up for solid-state chemistry. Why is he talking about solutions? Well, I'm going to give you three reasons.

First of all, we have many systems that we're looking at are solutions. It's very rare to use pure substances in engineering applications. So, we need to understand solutions from that standpoint as we've learned how to tailor the properties of glasses, how to tailor the properties of semiconductors, how to tailor the properties of metal alloys, and so on.

Secondly, many materials are processed out of solution. The way we make fine powders is not to make little molds and cast liquid into tiny, tiny little molds. We make fine powders typically by precipitation out of solution.

So, we need to understand the laws that govern dissolution, and exsolution, precipitation. And thirdly, one of the major units that we're going to cover before the end of the semester involves biochemistry.

And, the environment for most of biochemicals is a wet environment. So, I think it's requisite that we know something about that wet environment. So, that's sort of the background preamble. And so, what I want to do today is introduce some definitions.

So, we'll breeze through some material on the handouts, and ultimately get to the miscibility rules. That's what we really want. We want to know that miscibility rules, the rules that govern what to substances or three or more substances are going to mix and form solutions.

So, that's what we are heading to: miscibility rules. And then, finally, we're going to do some quantitative calculations by the end of the class. So, let's get on to the handout. On the one side, we've got the types of solutions.

So I'll just quickly go through and get these definitions under our belts. A solution consists of at least two substances. The majority substance is termed the solvent. And then, the minority substance, and there can be more than one, is the solute.

So, you can have a plurality of solutes, but you have one solvent. So I just want to make you understand that solutions are not simply aqueous solutions. Solutions span all of chemistry. So, let's go down this chart that's been organized according to state of aggregation.

So, first case we've got a solid solute and a liquid solvent. Brine is a good example of that. You've got solid sodium chloride dissolving in liquid water. Here's liquid-liquid. This is wine, which can be broken down as ethyl alcohol dissolving in water.

All right, and we can dissolve a gas in a liquid. And, seltzer is an example where we have carbon dioxide dissolved in water. And, I mean, it's chemically dissolved in the water. It's under high pressure.

When you release the pressure, you see the bubbles exsolve. But when we say solution, we are talking about something that is mixed at the molecular level. So, this is solvated carbon dioxide. So, now, let's use a gas as the solvent.

There's only the one case there: gas dissolving in a gas. So, we could say that air is a variety of gases dissolved in nitrogen as the solvent. Nitrogen is the majority species, about 80% nitrogen.

We have oxygen. We have argon. We have carbon dioxide. If we're living next to an aluminum smelter, we probably have some tetrafluoromethane. If you've lived next to a power plant, burning coal, you've probably got some sulfur dioxide, and so on and so forth.

So, this is a multi-component solution. Now, let's look at solids. This is 3.091, so how about some solid solutions? We've seen evidence of those already. Solid metal alloy: carbon dissolved in iron.

The carbon goes in interstitially, mixing at the molecular level right down to the atomic level here. We've looked at semiconductors, boron going into silicon. Boron sits on a silicon site, and actually bonds into that framework.

We looked at stabilized ceramics where we put calcia into zirconia at the end of the last lecture to make a high mobility oxide ion conductor that would act as a sensor for automotive applications. And then, we looked at putting modifiers into network formers.

We form a solution. So, these are solid solutions. And then lastly, we can look at putting liquid into solid, which is the primary example, is mercury going into silver. It's an amalgam. This is now past history, but certainly in my lifetime for a while I was walking around with a number of these amalgams in the shape of fillings in my mouth.

These were very conveniently made right in the dentist's office. But, people have come to appreciate that it would be a good idea not to have mercury in your mouth because over time it does leach out a little bit.

And we know it's not conducive to good physical well-being. So, this is now thanks to advances in materials science, been replaced by some advanced fine ceramics that can function in place of this alloy.

And lastly, we can put gas in solid. This one is topical today. As people think about a hydrogen economy and how we're going to put hydrogen on automobiles, one way is to put compressed hydrogen tanks full of compressed hydrogen at about 5 to 10,000 psi.

That's not very appealing. Another approach is to interpolate hydrogen into a compound that can dissolve large quantities of hydrogen. And this is one such compound. So, this is a solution of hydrogen dissolved in lanthanum nickel five.

So, that's the range of possibilities of solutions. And then, I want to look at a little bit more detail here in the fine structure of mixtures. And, in the case of mixtures, we can look at the solution as being something that's mixed at the molecular level where, say, the diameter, if you will, the diameter of the solute is down less than several nanometers.

So, we are effectively looking at a homogeneous solution. I've mentioned brine. And, there are simple tests that we can put to this. First of all, it's visually uniform. And, in fact, it's transparent to visible light because of the band gap that it presents.

But the main thing is that it's visually uniform. It's impossible to filter. If we wanted to separate sodium chloride, if we wanted to desalinate, we couldn't do so by filtering because the filter would have to be at the subatomic level in order to trap something at the atomic level.

Well, that's absurd. So, you can't separate solution by filtration. And lastly, it will not settle out. If you make a solution of sodium chloride and water and let it sit in a gravity field, it's not going to stratify because it's mixed at the atomic level.

But, these two here, the colloid and the suspension, are the heterogeneous type. They are aggregates because they are mixed at the cluster level. And, depending on what the particle size is, we end up with two conditions, and I want you to be aware of these because there is some interesting physics involved.

And, what we're talking about here is dispersions. It's a dispersion because we don't have homogeneity. Dispersion is fine particles. And I'm using the term particle to mean this cluster, fine particles distributed throughout some matrix.

And, what's the principle behind this? Why doesn't everything just settle out? Well, here's what's going on. When we look at a foreign particle, this is this cluster that's on the order of 2 to beyond even 1,000 nm.

And, why is this suspended in solution in the first place? Because we know we've got two forces acting on it. First of all, we've got a gravity force, which is acting to cause this to settle. So, this is a settling force thanks to the gravity field.

And, working in the other direction, this interface between the insoluble and the solvent, along that interface we have binding forces as well. And these binding forces I'll simply call interfacial forces.

And, these interfacial forces want to bind the particle to the solvent at the resistance to the gravity field. So, now, why does particle size come in to play? You know the gravity force is proportional to mass.

And, the mass of the particle is proportional to the volume of the particle. And volume scales with the cube of the radius, whereas this interfacial force, as the term implies, interface, the interfacial force is proportional to the area of contact between the particle and the solvent.

And, area scales as the square of the radius. And you know from your math that R cubed dominates R squared when R is large. That comes as no surprise. But, at very, very small values of R, R squared dominates R cubed.

And, depending on what the relative forces are is going to govern what the relative particle size has to be before we get particles that will be suspended in the second phase. And so, we have here the example of the colloids.

One example of the colloid is milk, which I described to you already. And even though we have two phases, both transparent to visible light, that interface scatters the light. These particles are large enough that they can be separated by filtration.

But it's unlikely that they'll separate in a gravity field owing to this. They don't settle. They don't settle. Actually, what has happened over the past two generations is a shift in the particle size of the fatty phase of milk.

When you go to the store today, you buy this red cap milk called homogenized milk. Well, go back two generations ago. There was no homogenized milk. Milk was simply pasteurized and sent out. And the particle size was so great in the fat globules that after several hours the fat would coalesce and rise to the surface.

So, you'd have a creamy phase on top. And people would take that out and use it in their coffee. And then you'd have what's effectively skim milk on the bottom. So, somebody came up with the idea, obviously somebody who had studied the rudiments of physical chemistry, and reasoned that if the fatty phase could be broken down to a finer particle size, then the buoyancy forces would dominate the gravity forces and it wouldn't separate.

So, homogenized milk is simply milk in which the fatty phase has been reduced in cluster size to value so small that by the time the gravity field is able to dominate the buoyancy field, the product is spoiled anyway.

So, for the practical shelf life of the product, you don't have to deal with starting by taking the milk out of the fridge and shaking it in order to get good mixing. That's a good example. In the case of something like blood, the cluster size is far too large for that to happen, and so it's not homogeneous under illumination.

It's easy to filter, and it will separate in a gravity field. So, what we're seeing here is a demonstration of the competition between the gravitational forces and the interfacial forces. And, the interfacial forces are a function of the kinds of bonds that form.

So, it's all there. It's all there. And, lastly, this colloid, I want you to be aware of the definitions here. I'm not going to quiz you on this. I simply want you to have it so that at some point you might want to go back and refer to it.

I'm not going to test your memory. But this breaks down the entire taxonomy of colloids where we see that there's a difference between what happens in the liquid dispersive medium, a gas dispersive medium, and a solid dispersive medium.

And, you can see many of the common products that we know are in fact these dual phase mixtures that involve very, very fine particles on the order of some hundreds of nanometers that we use in common products.

So, so much for the taxonomy. But now I want to turn to the question of miscibility. And, in order to do that, I want you to consider this example, which I've actually taken from your reading. I want to look at dissolution of substances in a liquid bilayer.

So, we're going to put two vessels side-by-side, and each will contain the same liquid bilayer. And, the liquid bilayer is going to consist of the bottom layer in both vessels is carbon tetrachloride.

And the upper layer is water. And, first of all, let's ask why these don't mix. They're both liquid at room temperature. Why don't they mix? Well, let's take a look at the electronic structure and bonding.

Whenever I ask why, you know, you've got a test coming up a week from now. When I ask why, try to work something into the answer that speaks to electronic structure because that's the clue. So, what do we know about carbon tetrachloride? It's nonpolar, so why does one carbon tetrachloride bond to another carbon tetrachloride? And, please don't tell me something to do with the electronegativity difference between carbon and chlorine.

Let's forget about that. It's nonpolar, so all we've got is weak van der Waals forces. So, we have a van der Waals liquid on the bottom whose density is about 1.4. So, it's going to lie under water.

And water, on the other hand, in contrast to being nonpolar, it's polar. And it has hydrogen bonding capability. So, these are very different substances. And they don't have the capacity to link to one another.

And so, they don't. They choose forces of cohesion over forces of adhesion. And so, we have two separate layers. So now, we're going to ask, what happens if we introduce a candidate solute? So, we really have a bilayer of solvents.

And now, we're going to introduce a solute. In the one case, we're going to put in solid crystalline iodine. In the other case, we'll put in solid crystalline potassium permanganate. And why do we choose these? Well, first of all, they're both solids, solid going into liquid.

And, secondly, they're both purple, very important. They're both purple. What happens is if we drop iodine into this bilayer, stir vigorously, mix everything up, and then wait. The two liquids will phase separate in a gravity field, and then we'll find at the end of the experiment that we have a solution here of iodine carbon tetrachloride solution mixed at the molecular level.

On the other side, shake vigorously, wait for system to equilibrate, layers phase separate. And now, here we have a solution of potassium permanganate in water. We can see with the naked eye, the upper layer is purple.

The lower layer is clear and colorless. In this case, the upper layer is clear and colorless. The lower layer is purple. So, it's clear that iodine has strong affinity for carbon tetrachloride. Potassium permanganate has a strong affinity for water.

So, what's going on? Well, let's look at the structure of the candidate solute. What's iodine? Iodine is a diatomic molecule. And it's nonpolar symmetric. So, it's a van der Waals solid. And, the van der Waals solid dissolves in a van der Waals liquid because together they've got fluctuating dipole interactions that can allow them to bond to one another and form a solution, whereas with water it's less favored.

In this case, potassium permanganate is an ionic compound that forms potassium cations, and permanganate anions. And so, we have an ionic solid dissolving in a polar liquid. So, the polar liquid is able to attract the positive end of the dipole attracts the anion.

And the negative end of the dipole attracts the cation. If you read the typical chemistry textbooks, they don't go into such an elaborate discussion. They like to simplify it to a slogan, a tagline.

So, you will see this. Well, I hope you don't see that. I hope you see it's spelled correctly. You will see like dissolves like. I look at that and I said it's too simplistic. And I'm going to give you some examples of where that doesn't happen.

Besides, I mean, I look at that and I wonder if I'm reading something from California. The like dissolves a like. [LAUGHTER] Tenure means never having to say you're sorry, so just enjoy it. Enjoy it.

So now, what I want you to do is to look at this table. This table is going to show that this is not such a smart tag like to work with. Again, I will not expect you to commit this table to memory.

I would never ask you to tell me what the solubility of something is by heart. I would, instead, turn around and say, for example, calcium copper bromide forms aqueous solutions. And then, let's take it from there.

I tell you what the fact is. But what I want you to notice first of all is that the table is broken into two regimes, soluble ionic compounds and insoluble ionic compounds. So it's not so simple. There's something else going on here.

And, let's take a look. There's one they say, for example, the chlorides are generally soluble. And the oxides are generally not soluble. So, what do we have to think about? Well, it's a competition, really.

It's a competition. When we look at sodium chloride, it exists as a solid crystal. And now, we have to compare whether it wants to go into aqueous solution. And, what we see is that the answer here is, yes, whereas the answer for magnesium oxide, which is also an ionic crystal, it does not form aqueous solutions.

So, what's going on here? Well, what's the major difference here? Sodium chloride exists as sodium cations and chloride anions, whereas magnesium is divalent. So, we have magnesium cations and oxide anions.

And in the crystal lattice, we have a much, much higher crystal lattice energy for magnesium oxide. If you just go through the Madelung derivation, you will have double, double. So, it's, in essence four times.

And sure enough, the melting point of sodium chloride is about 800° melting point, whereas the melting point of magnesium oxide is about 2,800°. And, what happens when we try to make this dissolve in solution is something that looks like this.

Here's a cartoon showing water dissolving sodium chloride. So, sodium chloride, just to get oriented, the green spheres represent the chloride ions, and the blue spheres represent sodium ions. And, just to keep things current, you can see up here one, two, three, four, five, the face of an FCC Bravais lattice, but we've got two atoms per lattice point as our basis, etc., etc.

And, here's the water molecule, which looks sort of like Mickey Mouse here with the red showing the oxygen, and the white showing the hydrogen. So, the oxygen end is the negative end of the dipole. And, what the cartoon showing is that the negative end of the dipole is presenting itself to the cation.

And, if the attractive force between the solute and the solvent exceeds the binding energy, the crystal binding energy, we will have dissolution. And that's borne out here where you see that we have a difference in Coulombic force of attraction that makes it accessible for sodium chloride for water to dissolve it, but inaccessible for magnesium oxide because the forces are too strong.

And then, you see the complement here, that the positive end of the dipole is presenting itself to the chloride anion. I'm always amused when people draw ions colored because I know they're trying to color code them, but think about it.

If you could isolate a chloride ion, what color do you think it would be? There's an answer to that question. What's its electronic structure? See, it always goes back to electronic structure. What's its electronic structure? It's isoelectronic with argon.

And what you think the color of argon is? What do you think the color of any noble gas is? So, actually, these things should all be clear and colorless. I just wanted to point that out. I know they're color-coded for your amusement, but I was staring at this green ion, and asking myself, should ions be green? No, no, they shouldn't.

OK, so now let's talk about measures of solubility. I'm going to put some numbers on this. So, we can express solubility in the following units. There are a variety of units, but the only one that we really need to worry about is the molar quantity, which represents moles of solute divided by liters of solution.

And, remember, the solution is the sum of all solutes plus the solvent. All solutes plus the solvent constitute the solution. And so, moles per liter, when we say sodium chloride dissolves at so many moles per liter, we can abbreviate that by uppercase M, which we term molar.

So if I have a one molar solution of sodium chloride and water, one molar is equivalent to, say, one mole of NaCl, in one liter of solution. I was about to write water. It's not. It's 1 L of solution total.

And, here, I think, you can see that it's not an all or nothing situation. In other words, certain substances are insoluble for reasons we've seen. But, other substances are soluble up to a point.

You know from experience you can add sugar to water, and a dissolves very readily up to a certain value, beyond which you can add the sugar, stir it, shake it, do whatever you want, but the sugar will not dissolve beyond a certain limit which we call the solubility limit.

Or, in aqueous systems, they simply call it the solubility and, at this limit, we've reached a condition termed saturation. We say the solution is saturated with respect to the solute, and can take no more.

And, there are two measures for saturation solubility. One is just denoted C star, or some people use C sub S, which means the concentration of solute at saturation. And that's a simple way of getting it.

And, it works when all we are dealing with is one solute in one solvent. But, to make matters richer, so that we can treat conditions where we have more than one solute, the other approach is to use a term known as solubility product.

And, the solubility product is best shown by example. Let's look at a salt that dissolves in water but very sparingly: silver chloride. Silver chloride has a lattice energy that's low enough that we can get some to dissolve in water but only sparingly so.

And so, I'm going to put an equal sign if we move from left to right. We're dissolving, so let's call that reaction dissolution. And, when we move from right to left, that's called precipitation. And so, silver chloride dissociates.

It dissociates to form silver cations. And, to show that they are dissolved in water, I'm going to write aq, meaning it's in aqueous solution, and the chloride anions, again, aq. And, this notion that when an ionic solid dissolves in water, it dissolves not as a molecule of silver chloride, but rather as cation and anion was in fact what won Arrhenius his Nobel prize.

He got his Nobel prize not for, I think, his really good work on activation energies and so on that paved the way for our understanding of catalysis, and so on. He got his Nobel Prize for this, which was called the theory of electrolytic dissociation, meaning that when an ionic solid dissolves, it forms an electrolyte, that is to say, an ionic conductor.

So, that was his Nobel prize. And, what we do is we can express the solubility of silver chloride by the solubility product denoted Ksp as the product of the instant concentration of each of the ions.

And, I know at first blush you are going to say, well, that's just the same as talking about the concentration of the salt dissolved. And, that is the case in this trivial example. We'll move on to something a little more sophisticated in a moment.

So, I'm going to take the concentration using square brackets to denote concentration. So, it's the product of the concentration of the silver cation, and the concentration of the chloride anion. And, concentrations here are in units of moles per liter.

So, I'll plug in the value of the silver ion concentration and the chloride ion concentration. And that gives me the solubility product. And then, this goes up to some master value. And, that's the saturation.

So now, I want to show you that we have some evidence for this. So, here's some evidence. What I'm showing you is a plot of conductivity, the conductivity of the solution as a function of the number of moles of silver chloride that has been added.

And, what you see? First of all, here we have the conductivity of pure water. Pure water is a very, very poor conductor. We should really properly call it an insulator. And, in units of, this is reciprocal ohm, which is the Siemen.

In units of Siemens per centimeter, the conductivity of pure water is less than ten to the minus seven. It's some number times ten to the minus eight Siemens per centimeter. So, what's happening? When we dissolve silver chloride, we are adding, this is the same as doping.

We are adding charge carriers, only instead of adding holes to the valence band, or electrons to the conduction band, that's not going to help us because water isn't a metal. Water is capable of dissolving ions.

So, we're going to make it an ionic conductor by injecting carriers. The more silver chloride we dissolve, the greater the number of carriers. The greater the number of carriers, the greater the conductivity.

So, by measuring conductivity as a function of concentration, we can determine if we are dissolving the silver chloride, as opposed to we keep adding it and it just sinks to the bottom. So, let's look here.

What we find is as we add silver chloride, the conductivity of the solution rises up to a maximum value at around ten to the minus five moles of silver chloride per liter of solution. So, that would be a concentration of 10 to the minus five molar.

And then, what happens? We keep adding silver chloride, but the conductivity doesn't change. So, what do you think is happening if you could stare at the beaker? You could see the silver chloride just falling to the bottom.

So, the knee in the curve indicates that we've reached saturation solubility. So that's good. So, we have evidence to support. Now, we can then compare the solubility with the solubility product in the following manner.

We can say that the relationship between the saturation solubility is really equal to the concentration of silver chloride dissolved, which Arrhenius has told us is really the concentration of silver ion, which is equal to the concentration of chloride ion.

All of these are the same. And so, with a little bit of algebra, we can show them that the relationship between this and Ksp is - since the concentration of silver equals the concentration of chloride, then we can take Ksp to the one half will then equal the C star, the saturation solubility.

And, for silver chloride, it turns out that the product, the solubility product, is 1.8 times ten to the minus ten, which then means that the saturation solubility for silver chloride is the square root of that, which is about 1.3 times ten to the minus five.

And, sure enough, that's what you see on this plot. You see that somewhere just a little bit higher than ten to the minus five molar, the plot has flattened out. But now, let's look at the more interesting case and put the solubility product to work and ask ourselves, what happens if we add silver chloride not to pure water, but to a solution that contains already some other dissolved salt.

Is that going to have an impact on the amount of silver chloride we can dissolve? So, let's give a specific example, and a specific example is add silver chloride to a solution already containing sodium chloride.

I'm going to add silver chloride to a solution of 0.1 molar NaCl. So, I already have chloride ion present. See, this is interesting. If I said that, instead of adding silver chloride to pure water, I'm going to add silver chloride to water that already contains some silver chloride, you say, well, that's obvious.

If it contains some silver chloride, I can't put as much in as I otherwise could. But now I'm going to say, it's not silver chloride. It's sodium chloride. Does the presence of sodium chloride have any impact on the dissolution of silver chloride? So, again, we'll look Arrhenius' theory of electrolytic dissociation.

So, silver chloride will dissociate and go into solution as silver cation and chloride anion. And, sodium chloride is present as sodium ion and chloride ion. So, we've got now, note here multiple sources of chloride ion.

We're getting some chloride ion from silver chloride, and we get some chloride ion from sodium chloride. So, the consequence of that is that I know I can no longer assume that the concentration of silver ion equals the concentration of chloride ion.

That's no longer the case because I've got some chloride ion that came from sodium chloride. So, that's a problem. That's a problem. So, what do I do? I'm going to try to find the saturation solubility of silver chloride under these new circumstances.

So, I'm going to go back to my solubility product, Ksp. And, Ksp is going to equal the product of the silver ion concentration times the concentration of the chloride ion. And, I can take a stab at what this is.

I told that this is on the order of 0.1 molar. And, we saw that in the case of pure silver chloride, it's ten to the minus five molar. It's about four orders of magnitude lower even when we have full dissociation here.

So, I'm going to assume that ten to the minus one molar is so much greater than ten to the minus five molar, which is what we got in pure silver chloride that I'm going to neglect the contribution of chloride ion from silver chloride, and say that the concentration of chloride ion is simply equal to 0.1 because 0.1 is so much greater than what is left over from the previous calculation.

And now, I can still make the connection between the dissolution of silver chloride salt, and the concentration of silver ion. That still exists as one to one. So, this will then represent C of silver chloride times 0.1.

And, that's equal to this 1.8 times ten to the minus, what is it, ten to the minus ten. So, I can solve for the concentration of silver chloride, and that gives me 1.8 times ten to the minus nine molar.

So, the presence of the chloride from sodium has had a tremendous repressive effect on the dissolution of silver chloride. This is called the common ion effect because I have a common ion from both solutes, and one influences the solubility of the other: common ion effect.

And, we can use the common ion effect to our advantage. If I want to process something in order to cause precipitation of very, very fine particles, I can load a solution with one solute, introduce a second solute, and trigger the precipitation, and in doing so, control particle size, and the shape, and thereby prepare a material for subsequent processing.

In fact, you can use this. If you want to determine if you've got water that's been chlorinated, simple thing you can do: if you go back to that table 8.7, you'll know that silver nitrate has a very, very high solubility in water.

So, I introduce silver nitrate into water, and the question is, is there any chlorine present? And, if there's any chlorine present, we know that the saturation solubility of silver chloride is ten to the minus five molar.

What will happen is if you've got chlorinated water, and you add silver nitrate, you will see a shower of very, very fine, white powder, which is the silver chloride. So, this is a very good test for residual chlorination.

So, now how can we use some of this? Let's take a look. I've got two things I want to show you. First of all, one thing going back to the last lecture dealing with diffusion, and I want to show how diffusion is important in forming bonds if we want to make coatings.

So, here's a carefully drawn cartoon by your professor to pose the question, suppose we want to put a coating of A on a substrate, C, and they don't stick. They don't stick very well. So, one strategy to enable the binding of coating A on substrate C is to find a substance, B, such that A will bind to B, and B will bind to C.

This is called an interlayer. And, it's used commonly in the industry when the coating won't stick to the native substrate. So we use an interlayer. So, here's a product where this is used on a regular basis.

I don't own stock in Gillette. I just happen to know that there's a ton of materials science that went into the design of a product as mundane as a safety razor. So, I want to look at the blade. It turns out, they wanted to cut the blade with a substance called diamond like carbon.

Diamond like carbon is a mix of sp3 hybridized carbon and sp2 hybridized carbon. It's not a diamond film. I want you saying that Professor Sadoway said that the Mach 3 razor has a diamond film on it.

It doesn't. If it did, it would cost more than it does now. And it's exorbitantly expensive as it is. So, it's sp3 and sp2. It's a mix. And, this material is very, very hard. It's very hard. You want something that's hard to give you the resistance to keep that cutting edge.

But you don't want to meet the whole blade out of something like that. Otherwise, it will be brittle, and it will shatter. So, you want the composite material. So, it turns out that diamond like carbon, if you put it on steel, since the blade is made of steel, there is very poor adhesion.

And, Gillette was determined to get diamond like carbon on the blade. So, they asked, what could be an interlayer? And they tried many different substances, and finally settled upon niobium. They went to a fairly exotic part of the periodic table.

They used niobium. It turns out that niobium, niobium as you know is a body centered cubic metal. And so, it can interdiffuse with steel and form a bond here. And, the other thing is that niobium is an element that has a very high propensity to form carbides.

So, niobium forms an interdiffusion bond with the steel, and on the free surface has a carbide which then is capable of bonding to the diamond like carbon, and thereby giving us the bond that we need.

And the last thing I want to show you is something that I just saw this morning. I was getting ready to go. Tom, could we shift to the document camera? I want to show you how we can take diffusion to make products.

Perfect, OK, this is a ceramic knife. It's made of zirconia. It's made of zirconia. Now, how do you suppose they make this knife? How is this knife manufactured? You start with zirconia. This stuff melts at 2,800 degrees centigrade.

Can you imagine? It's an ionic compound. So, it has no ductility. And please don't tell me because of dislocations; it has no ductility because the bonds are so strong. The only way you could perform zirconia would be to push, push, push, and then it breaks.

So, you can't make a billet of zirconia the way they do how they make a steel knife. They make a billet of steel, and then they, by mechanical working, they reduce the thickness, and then eventually sharpen the edge.

You can't do this with this material. So, what do they do? They use solid-state diffusion starting with particles of zirconia, fine particles which are compressed, raised to elevated temperature on the order of about 1,000 degrees Celsius, and held there for several days.

And, during that time, diffusion takes place to reduce the void space, bringing this to near 100% density. So, the vast majority of fine ceramics that you see in the marketplace, I'm talking true ceramics now, not the glassy stuff.

Pyrex, yes, Pyrex is castable. But not this. This is all processed by solid-state diffusion, a process of agglomeration known as sintering. Sintering is consolidation by solid-state diffusion. And then, in the end, with the diamond wheel, the edge is cut.

And, this stuff has a mho hardness of 8.2, making it second only to diamond, which has a hardness of ten. High corrosion resistance, no metallic flavor, very sharp, and it holds its edge, made by solid state diffusion.

OK, we'll see you on Friday.

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