If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Boiling point elevation and freezing point depression

Video transcript

Let's think about what might happen to the boiling point or the freezing point of any solution if we start adding particles, or we start adding solute to it. For our visualization, let's just think about water again. It doesn't have to be water. It can be any solvent, but let's just think about water in its liquid state. The particles are reasonably disorganized because of their kinetic energy, but they still have that hydrogen bonds that wants to make them be near each other. So this is in the liquid state, and they have a reasonable amount of kinetic energy. You know, each of these particles is moving in some direction, rubbing against each other, bouncing off of each other. Now, to move it into the solid state, or to freeze it, what has to happen? The ice has to enter kind of a crystalline structure. It has to get pretty organized, so let's say it has to look something like this. The water molecules are going to have a regular structure where the hydrogen bonds dominate any kind of kinetic movement they want to do, and all the kinetic movement, they're just vibrating in place. So you have to get a little bit orderly right there, right? And then, obviously, this lattice structure goes on and on with a gazillion water molecules. But the interesting thing is that this somehow has to get organized. And what happens if we start introducing molecules into this water? Let's say the example of sodium-- actually, I won't do any example. Let's just say some arbitrary molecule, if I were to introduce it there, if I were to put something-- let me draw it again. So now I'll just use that same-- I'll introduce some molecules, and let's say they're pretty large, so they push all of these water molecules out of the way. So the water molecules are now on the outside of that, and let's have another one that's over here, some relatively large molecules of solute relative to water, and this is because a water molecule really isn't that big. Now, do you think it's going to be easier or harder to freeze this? Are you going to have to remove more or less energy to get to a frozen state? Well, because these molecules, they're not going to be part of this lattice structure because frankly, they wouldn't even fit into it. They're actually going to make it harder for these water molecules to get organized because to get organized, they have to get at the right distance for the hydrogen bonds to form. But in this case, even as you start removing heat from the system, maybe the ones that aren't near the solute particles, they'll start to organize with each other. But then when you introduce a solute particle, let's say a solute particle is sitting right here. It's going to be very hard for someone to organize with this guy, to get near enough for the hydrogen bond to start taking hold. This distance would make it very difficult. And so the way I think about it is that these solute particles make the structure irregular, or they add more disorder, and we'll eventually talk about entropy and all of that. But they make it more irregular, and it's making it harder to get into a regular form. And so the intuition is is that this should lower the boiling point or make it-- oh, sorry, lower the melting point. So solute particles make you have a lower boiling point. Let's say if we're talking about water at standard temperature and pressure or at one atmosphere then instead of going to 0 degrees, you might have to go to negative 1 or negative 2 degrees, and we're going to talk a little bit about what that is. Now, what's the intuition of what this will do when you want to go into a gaseous state, when you want to boil it? So my initial gut was, hey, I'm already in a disordered state, which is closer to what a gas is, so wouldn't that make it easier to boil? But it turns out it also makes it harder to boil, and this is how I think about it. Remember, everything with boiling deals with what's happening at the surface, and we talked about that in our vapor pressure. So at the surface, we said if I have a bunch of water molecules in the liquid state, we knew that although the average temperature might not be high enough for the water molecules to evaporate, that there's a distribution of kinetic energies. And some of these water molecules on the surface because the surface ones might be going fast enough to escape. And when they escape into vapor, then they create a vapor pressure above here. And if that vapor pressure is high enough, you can almost view them as linemen blocking the way for more molecules to kind of run behind them as they block all of the other ambient air pressure above them. So if there's enough of them and they have enough energy, they can start to push back or to push outward is the way I think about it, so that more guys can come in behind them. So I hope that lineman analogy doesn't completely lose you. Now, what happens if you were to introduce solute into it? Some of the solute particle might be down here. It probably doesn't have much of an effect down here, but some of it's going to be bouncing on the surface, so they're going to be taking up some of the surface area. And because, and this is at least how I think of it, since they're going to be taking up some of the surface area, you're going to have less surface area exposed to the solvent particle or to the solution or the stuff that'll actually vaporize. You're going to have a lower vapor pressure. And remember, your boiling point is when the vapor pressure, when you have enough particles with enough kinetic energy out here to start pushing against the atmospheric pressure, when the vapor pressure is equal to the atmospheric pressure, you start boiling. But because of these guys, I have a lower vapor pressure. So I'm going to have to add even more kinetic energy, more heat to the system in order to get enough vapor pressure up here to start pushing back the atmospheric pressure. So solute also raises the boiling point. So the way that you can think about it is solute, when you add something to a solution, it's going to make it want to be in the liquid state more. Whether you lower the temperature, it's going to want to stay in liquid as opposed to ice, and if you raise the temperature, it's going to want to stay in liquid as opposed to gas. I found this neat-- hopefully, it shows up well on this video. I have to give due credit, this is from chem.purdue.edu/ gchelp/solutions/eboil.html, but I thought it was a pretty neat graphic, or at least a visualization. This is just the surface of water molecules, and it gives you a sense of just how things vaporize as well. There's some things on the surface that just bounce off. And here's an example where they visualized sodium chloride at the surface. And because the sodium chloride is kind of bouncing around on the surface with the water molecules, fewer of those water molecules kind of have the room to escape, so the boiling point gets elevated. Now, the question is by how much does it get elevated? And this is one of the neat things in life is that the answer is actually quite simple. The change in boiling or freezing point, so the change in temperature of vaporization, is equal to some constant times the number of moles, or at least the mole concentration, the molality, times the molality of the solute that you're putting into your solution. So, for example, let's say I have 1 kilogram of-- so let's say my solvent is water. I'll switch colors. And I have 1 kilogram of water, and let's say we're just at atmospheric pressure. And let's say I have some sodium chloride, NaCl. And let's say I have 2 moles of NaCl. I'll have 2 moles. The question is how much will this raise the boiling point of this water? So first of all, you just have to figure out the molality, which is just equal to the number of moles of solute, this 2 moles, divided by the number of kilograms of solvent. So let's say we have 1 kilogram of solvent. This was, of course, moles. So our molality is 2 moles per kilogram. So we just have to figure out what this constant is, and then we'll know the temperature elevation. And actually, that same Purdue site, they gave a list of tables. I haven't run the experiments myself. They have some neat charts here. But they say, OK water, normal boiling point is 100 degrees Celsius at standard atmospheric pressure. And then they say that the constant is 0.512 Celsius degrees per mole. So let's just say 0.5. So it equals 0.5. So k is equal to 0.5. And I want to be very clear here because this is a very-- I won't say a subtle point, but it's an interesting point. So I said that there's 2-- the molality of-- I just realized I made a mistake. I said the molality of sodium chloride is 2. 2 moles per kilograms. But that would be if sodium chloride stayed in this molecular state, if it stayed together, right? But what happens is that the sodium chloride actually disassociates, and we learned all about it in that previous video. Each molecule or each sodium chloride pair disassociates into two molecules, into a sodium ion and a chlorine anion. And because of that, because this disassociates into two, the molality is actually going to be two times the number of moles of sodium chloride I have. So it's going to be two times this. So my molality will actually be 4. And this is an interesting point. If I was dealing with-- and I wrote it here. So this right here is glucose, and this is sodium chloride, or at least sodium chloride in its crystal form. One molecule, I guess you can view it, or one salt of it. I guess you could just view it as one of these little pairs right here. But the interesting thing is is you could have the same number of moles of sodium chloride when you view it as a compound and glucose. But glucose, when it goes into water, it just stays as one molecule of glucose. So a mole of glucose will disassociate into a mole of glucose in water. Well, I guess it won't disassociate. It'll just stay as one mole, while a mole of sodium chloride will turn into two moles because it disassociates. It turns into two separate particles. So in my example, when I start with a mole of this, I end up-- actually, once I dissolve it in water, I ended up with 4 moles per kilogram of molality, because this turns into two particles. So given that the molality is 4 moles. 2 moles of sodium, 2 moles of chloride per kilogram. So I just use that constant that I just got from Purdue. And I get the change in temperature is equal to that constant, 0.5, times 4, which is equal to 2 degrees. So my boiling point will be elevated by 2 degrees. Now, if I had the same number of moles, if I had 2 moles of glucose dissolved into my water, I'd only get half as much, half as much of an increase. Because the molality would be half as much. Because it doesn't turn into two particles. In some textbooks, you'll actually see it written like this. You'll actually see the same formula written like change in boiling temperature, or vapor temperature, or whatever you want to think, is equal to k times m times i, where they'll say this is the molality of the compound you're talking about. In this case, this number would be 2, and i is the number of molecules or the number of things that it disassociates into. So in this case, this would have been 2. And that's where we would have gotten 4 times k, which is 0.5, which is 2. In the case of water, this would be-- oh, sorry, in the case of the glucose, this would still be 2. But it only turns into one particle when it goes in the water, so that would be 1. So you would only have a 1 degree increase in the boiling point of water. Now, freezing point is the same thing. Change in freezing point is also proportional to the molality. And you can either say the molality of the original non-in-water compound times the number of compounds it disassociates into, although this k is going to be different for freezing than it is for boiling. Of course, this k changes at different pressures and for different elements. But the really big takeaway is just to realize that even if you have a mole of this and a mole of that, and they're going to be dissolved into the same amount of water, because this dissociates into two particles and this disassociates into only one for every-- or this disassociates into two moles for every mole of the crystal you have-- this doesn't disassociate; it just stays as one-- this'll have twice as large of an effect on the freezing point change or on the boiling point elevation than the glucose will.