Yeah. Yeah. Yeah. Yeah. Yeah. Yeah. >> All right. >> Good afternoon. Today we will be working on chapter 15, but let me remind you that chapter 15 will not appear. This coming Friday is exact chapters for 1314 are fair game for your exam coming up this Friday. Also reminder that on Wednesday at 815 will be having a review session 08:15 PM in room 220 brown Lab, which is just on the opposite side of Brown. Laugh, but where we are right now. >> All right, title of chapter 15. >> As I mentioned at the end of last class, chemical equilibrium. What does that mean? >> Well, what it means is that some chemical reactions are reversible. >> Previously, when we were talking about energy of activation, we gave you the analogy of rolling a rock up a hill with the idea of trying to shove it over onto the other side. All right, so suppose you're the guy who's rolling the rocks up the hill and shoving them over all of the other side because you want to get those rocks off your property. Problem was a guy who lives at the bottom of the hill, the other side, and he's not too pleased about these rocks coming down onto his property, so he's going to roll them back up the hill and put them back where they came from. In short, it is possible for reaction to go in the forward direction. But as the concentration of, the concentration of the products piles up, it is sometimes possible for that same reaction to go in the reverse direction. Here's a couple of pictures that might help illustrate the basic point. Suppose you have a reaction where a is being converted into B. We talked earlier about the fact that there are two ways you can follow the rate of the reaction. You can either follow the rate of disappearance of a or follow the rate of appearance of B. But the point is, as time goes on and more and more a gets converted into B, is the concentration of B increases, it makes it more likely that the reverse reaction can occur and B can react to form a. And at some point when you have a forming B and B forming a at the same rate, you have achieved this state called dynamic chemical equilibrium Which may seem like a contradiction in terms at first glance. Because when you say something is dynamic, you mean that it's changing. But when you say something is at equilibrium, what you're saying is that it's a constant steady state. So which is it? Is it changing or is that a constant steady state? And the answer is, it depends on whether you're looking at from the point of view of the individual molecules or the system as a whole once you've achieved a state of equilibrium. But it is equally possible for a to become B or from B to become a. So the individual molecules are going back and forth between being a and b and b. But the system as a whole should have the same number of molecules in the a state or in the beach date at any given time once it has achieved a state of dynamic equilibrium. Another way of thinking about what this means, the rate of the forward reaction versus the rate of the reverse reaction become equal once we've achieved a state of dynamic equilibrium. Let me show you a slightly more complex version of this. If a and B react to form C and D. And if the balanced equation for this reaction has coefficients little a, little b, little c, little d. Well, from what we were talking about in the context of kinetics, you can write a rate equation for the forward reaction. And the forward reaction is gonna depend on the concentrations of a and B reacting to form the products. Or do you think of the reaction going in reverse? You can write a rate equation for the reverse reaction. And that's going to depend on the concentrations of C and D as the reaction goes in the opposite direction. By definition, according to this, or according to this, if the thing has achieved a state of dynamic equilibrium, the rate in the forward direction is going to be the same as the rate in the reverse direction. We can set these two things equal to each other. Which means we can set these two things equal to each other. And what I'm going to show you next is what happens when we set the two right sides equal to each other. If K forward times concentration of a to the a power times concentration of B to the b power is equal to K reverse times concentration of c to the c power times concentration of D to the D power. Then another way of writing that expression is to gather all the constant terms on one side and gather all the concentration terms on the other side. And this is typically written this way in the context of talking about dynamic chemical equilibrium. The ratio of the rate constants is just a number. And that number is generally given the symbol K EQ, referring to the equilibrium constant. The EQ represents equilibrium that's equal to the ratios of the concentration terms, which is commonly given the symbol Q. And Q is called the reaction quotient. And in general, the way Q is written is concentrations of products over concentrations of reactants, that is concentrations of things. On the right side of the equation, Over concentrations of things on the left side of the equation with everything raised to the exponent that corresponds to the coefficient in the balanced equation. Now it turns out that for 3D equations, you can't normally make the assumption that the coefficients of the balanced equation are going to be the same as the exponents in the rate equations. But for equilibrium expressions, you can make that assumption. If anybody wants to know more details about why that's true, we can have that conversation during my office hours sometime. But the main point is that if you have a balanced equation for an equilibrium reaction written here, you should be able to write a reaction quotient looking like this. Concentrations of products in the numerator, concentrations of reactants are the denominator, all concentrations raised to the exponent that matches the coefficient in the balanced equation. And at equilibrium, the reaction quotient Q will be equal to the equilibrium constant K EQ, which is just a number for any given reaction. So for the moment, does a terminology and do the symbols make sense? We'll look at some specific examples in a moment. >> Hopefully it'll make more sense there everybody. >> Okay, it may need more time with this line. >> All right, let's look at a specific example of a dynamic equilibrium and kind of situation. And the specific example that I'm going to use is a reaction known as the Haber process, but turns out to be a very important agricultural reaction. Because it turns out that what the Haber process is good for is converting nitrogen and hydrogen into ammonia, NH3. And here's a picture showing you what those molecules look like. If you were to draw pictures of the molecules. Here's the molecule nitrogen reacting with three molecules of hydrogen to form two molecules of ammonia. But in this drawing, it's not represented as an equilibrium. This is a normal reaction arrow drawn this way to show going from reactants to products if the reaction is reversible. Generally what's done is to draw arrows going in both directions. Forward reaction represented by this arrow, reverse reaction represented by this arrow. So when you see the two arrows going in opposite directions like that, that's intended to suggest that this is at equilibrium and if the reaction is reversible. Now, given that this is the balanced equation for the Haber process, what does the equilibrium expression? That is, what does Q, the reaction quotient, look like for this equilibrium? >> Well, >> He said before. >> Q, the reaction quotient is products over reactants right side, overlap side with each concentration term raised to the exponent that matches the coefficient in the balanced equation. >> So the only product we have here is ammonia. Therefore, the only concentration term that appears in the numerator here is the concentration of ammonia. And that's raised to the second power because the coefficient of ammonia in the balanced equation is two. That's the numerator for this equilibrium expression looks like. What would the denominator look like? >> Yes, the reactants still in the denominator, the reactant concentrations are nitrogen, N2, and hydrogen H2. >> Nitrogens coefficient is one. We don't generally bother writing in the one exponent there, although we could if we wanted to, the coefficient of hydrogen is three. Therefore, the concentration of hydrogen is going to be cubed. So the point is, given a balanced equation for equilibrium like this, you should be able to write down Q, the reaction quotient. Just remember, products on top react, it's on bottom product. So things on the right side, reactants or the things on the left side. And then just make sure you raise each concentration to the exponent that matches up with its coefficient in the balanced equation. Still making sense, at least for the moment. >> Ok, how can you determine an equilibrium constant? >> Well, here's one way you could do it. You can actually do the experiment. Notice that everything that's involved in the Haber process is a gas. Nitrogen is the gas that makes up roughly three-quarters of the atmosphere. Hydrogen is also a gas and ammonia. And most of the time when people think of ammonia, they think of the cleaning solution that we saw before was a weak electrolyte. But ammonia in this case is actually a gas. Ammonia, that is the cleaning solution is actually ammonia gas dissolved in water. Point is everything. And now everything involved in this reaction is a gas. And we know that gases tend to expand, to fill up whatever volume of whatever container they find themselves in. So suppose you took an empty one liter box and pumped into it nitrogen, hydrogen, ammonia. Let the reaction take place until it has reach this state of dynamic equilibrium where the concentrations are changing anymore. And then if you were to somehow measure the amount of each substance that you have in the box. Turns out the box contains 8.2 moles of ammonia. Point 1-0-0 mole nitrogen, and point 1-0-0 mole of Hydrogen. Given that information, how would you solve for the numerical value of K? Eq Bear in mind that at equilibrium Q and K, Q must be the same plate. So how would you figure out the value of K? >> Yes, that is correct. >> But let me just make sure everybody understood what you just said. There is one little refinement that I want to introduce based on some problems that you'll see in your textbook. These concentrations that are given here are not in moles. Concentration, by definition is in moles per liter. So technically what you have to do, for example, with ammonia, is to take the 8.2 moles of ammonia that you have and divide it by the volume of the box. Now in this case, that straightforward, because the volume of the box is one liter. So 8.2 moles divided by one liter is 8.2 moles per liter. That's the number that goes in MIT. And likewise, these other two are going to be point 1-0-0 moles per liter. And those are the numbers that go in there. Now it's convenient that we were using a one liter box because dividing by one is no big deal. But suppose, just for argument sake, that this had been a 2 liter box instead of a 1.00 liter box, what would you have to do before you inserted these numbers into the equation? >> Yeah. >> Not divide everything by two. >> Sure. >> 8.2 moles in a two liter box would be a concentration of 4.1 moles per liter. So the point is you would have to divide everything by the volume of the box. But here the volume of the box is one liter, so that's no big deal. So the rest is just arithmetic. The equilibrium expression looks like what we just wrote up here. At equilibrium, Q and K eq, you are going to be the same thing. And since the gasses expand to fill up the entire box, you can assume that the volume of each gas is one liter. And therefore you can insert the numbers as they are except change the units to m, which is moles per liter. And if you do all that And do the arithmetic and solve for the value of K EQ. In this case, you get 6.7 times ten to the fifth power. >> Now here's the good news about equilibrium constants. >> We saw in chapter 14 that rate constants sometimes have strange-looking units depending on whether the reaction is a first order reaction or a second-order reaction, or a third order reaction, whatever. When it came to equilibrium, constants, whoever was in charge of those units are a nuisance. So we're just not going to worry about the equilibrium constants are just numbers. So 6.7 times ten to the fifth with no units is the value of K Q for this reaction. Just as an aside, I'm going to hopefully play captain obvious here. This number, 6.7 times ten to the fifth large number or small number? >> Large number. >> If it had been, say, 6.7 times ten to the negative fifth, that'd be a small number. The significance of that is that equilibrium expressions have products in the numerator, reactants have been denominator. Therefore, if the equilibrium constant is a large number, what that tells you is that by the time this reaction is done, by the time this reaction comes to a state of equilibrium, you mostly have products, not very much reactants. On the other hand, if it was something like 6.7 times ten to the negative fifth, which would be a small number that tells you that when the reaction is done, you mostly have the reactants. Not very much of the products makes sense. We'll see cases later on where that becomes an important point. But for now, I'm hoping that the basic idea make sense. It'll be okay. I might need more time with the slide. >> All right, well, it's time to get out of this guy. >> I'm told that names that are spelled something like this. And Le Chatelier's principle is discussed in your textbook beginning on page 689. Names are spelled something like this, are very common in France. So I don't believe that all real Shockley a for Le Chatelier's principle is name is any relation. There's a zillion Smith who's a brilliant Smith's in this country. There's a billion chateau yeas and various forms of brains I do will ever, in my biased opinion, think Le Chatelier's principle is pretty important, and I hope to convince you of that before we're done here today. Here's what Le Chatelier's principle says. If you have a reaction that has achieved a state of dynamic equilibrium And then you do something to disturb that reaction so that it's not at a state of equilibrium anymore, then the system will adjust itself. However, it has to, to try to get back to a state of dynamic equilibrium. When I say adjust itself, what I mean is that either the forward reaction will go faster or the reverse reaction will go faster until once again, we're back to a state of dynamic equilibrium. A moment ago we said that if your reaction has achieved a state of equilibrium, then Q, the reaction quotient, the ratio of the concentration terms will be equal to k EQ, which is the equilibrium constant, which is just a number. If you disturb the equilibrium, this will no longer be true, which means one of two things must be true. Either q will be less than k q, or q will be greater than k EQ. If q is less than k EQ, that is, if the number is too small, what that means is you have too much of the denominator are not enough in the numerator. That is too much of the reactance part, not enough in the product sport. In that case, the forward reaction will go faster. Reactants will be converted into products. That's called a shift to the right will give you an analogy for something like this in a moment. But the point is that's what the system will have to do to get back to a state of equilibrium. Or if you disturb the equilibrium so as to make q greater than k EQ, that means you have too much in the numerator, which is where the products are not enough of the denominator, which is where the reactants are. How does the system compensate by making the reverse reaction go faster? That's called a shift to the left because you're converting products on the right side into reactants on the left side. This terminology will come up in the laboratory experiment that you do this week. And I'm going to show you an example in a few moments that I hope will make it a little bit more obvious what all this means. But for the moment we'll let people work on this slide. Everybody have it they need here. Alright, let's look at a few examples of Le Chatelier's principle in operation. And the examples are all going to sit around the reaction we showed you earlier, that is the Haber process. Let me put the Haber process equation back up here. >> Now I want to show you as the example. >> Alright, what this is a graph of is concentration of the three gases involved in the Haber process. Hydrogen, ammonia, nitrogen, as a function of time. And the point is, if the system is initially at a state of equilibrium, that means those concentrations don't change. But now suppose you do something to disturb the system. And disturbance in this case is going to be adding some hydrogen gas. What that does, if you just shoot some hydrogen gas into the vessel where all of this is taking place, then immediately increases the concentration of hydrogen. Now at that point, you have disturbed the equilibrium. So it's not a state of equilibrium anymore. You've added more hydrogen right here. >> What's that mean? >> Well, what that means is you have more stuff among the reactance at less stuff among the products that you're supposed to, how the system compensates for that. That extra hydrogen is going to react with some of the nitrogen that's there and for more ammonia. In other words, the immediate response is that the concentration of hydrogen is going to go down, so as the concentration of nitrogen, because those two are reacting with each other, but the concentration of ammonia is going to go up because that extra hydrogen that you added is reacting with some of the nitrogen to become ammonia. Until finally we get back to a point where the system has once again come to a state of equilibrium. And the way you can tell that these are the concentrations don't change. Did that make any sense? I'm not getting much of an enthusiastic response. Ok, so here comes my silly playground analogy. I'm gonna have to ask you folks this question because it's been a long time since I've been an element preschool. And they still let kids play on a seesaw? Or has it been decided it's too dangerous? I hope this Dalit gets play at the seesaw. Okay, here's a seesaw. >> Here's kids, first graders. >> We know they all weigh the same rate. So 3 first graders on each side of the seesaw, hopefully not standing up, but there's only so much my artwork will allow. Point is if he had three kids on each side and all the kids weigh the same. >> Seesaw is balanced except now to more first graders come along and jump on the right side of the seesaw. >> What's going to happen yet till kids on the other side go flying and they decide to shut down the playground. >> Those other two kids created a disturbance in what had been a balanced equilibrium kind of situation. >> If you don't want to have that kind of a disturbance, if you want to get back to a state of balance or equilibrium, what do you have to do but do more kids? >> On the other side? Just one problem. This is a very small elementary school that's the entire first rate right there, so we don't have anymore kids. What's the other way to solve that problem? >> Yep. Yeah. >> Move one kit instead of two kids jumping on the ones, add shift one, Get over here and now you're back to being balanced again. >> That's Le Chatelier's Principle. >> If you had equilibrium before and you add to one side, so you don't have equilibrium anymore, shift to the other side until you do that. Make sense? Okay, now let's go back to when we were looking at a moment ago. Here's the balanced equation. But if we unbalance the situation by adding hydrogen, which is what we're doing right here. Now, we got too many kids jumping on the left side of the seesaw. How do we fix it? Moves stuff over to the right side of the seesaw. And what that means is the extra hydrogen is going to react with nitrogen to form ammonia. So these two concentrations go down, and this concentration goes up. That's what they're trying to show you in the graph. Here's the concentration going up because the other two going down until you're back to a state of equilibrium make any more sense now? >> Yeah, question because what's happening to compensate for the fact that you added hydrogen here. >> That hydrogen is reacting with the nitrogen to form ammonia. So that means this concentration and this concentration are both going down because these two are reacting with each other to make this, these two concentrations going down at this concentration is going up. >> Makes sense. >> Let's try a few other examples using this same reaction. Suppose instead of adding hydrogen, we had added nitrogen. Well, that's going to disturb the equilibrium. So the equilibrium is going to have to shift one way or the other question is, is it going to shift to the right? Which means the forward reaction does faster, or shift to the left, which means the reverse reaction goes faster. If you add nitrogen, which way does it shift? >> Yes. Correct. >> You're adding nitrogen. You're adding to the left side. It's like adding kids to the left side of the seesaw. It, you add nitrogen. That added nitrogen is going to react with some of the hydrogen and make more ammonia. >> So net effect is the forward reaction does faster. >> That's called a shift to the right. These two concentrations go down, this concentration goes up. >> Suppose you had added ammonia shift to the right or shift to the left? Yes. Correct. >> You're adding ammonia, you're adding to the right side sort of like adding kids to the right side of the seesaw, which is what we did my example to fix it, you shift to the left side, some of the ammonia that you've added breaks down to form nitrogen and hydrogen. So these concentrations go up and this concentration goes down. >> You buying all this? >> This is Le Chatelier's principle question. >> We're not done yet. >> What happens if you increase the temperature? Now, before you answer this question, take a look at this value of delta H up here. I assume you know from Chem 101 what delta h means, right? Baby? That symbol look familiar? Okay, well, I know you've heard words like exothermic and endothermic before, right? Okay, take a look at the numerical value for delta H. Exothermic reactions are endothermic reaction. And how can you tell? >> Yes, how can you tell me? I remember Matt's right. >> If the value of delta H is a negative number, that's an exothermic reaction. That means the reaction gives off heat. If this had been a positive number and be an endothermic reaction, which means the reaction can CMC. In other words, as this reaction takes place, heat is being given off. If you like, you can think of heat as being a product here. So if you increase the temperature by adding heat, which way does a chip? Yes, if you're adding heat, you're adding to the right side, you add the right side, the shift goes in the opposite direction, shifts to the left side. But that's true in this case because it's an exothermic reaction and heat can be considered to be on the product side. What happens if you squish the box so that its volume is now decreased. >> Wow, there's a couple of ways to go here. >> You can either set up the equilibrium expression with the concentration terms in it and figure out what effect it's going to have on each of those concentration terms and react accordingly, or in a qualitative sense, you can think of it this way. Compressing the container forces the molecules closer together. That stresses the molecules out. So the way they fix that problem is to shift in whatever direction creates a smaller number of molecules. Take a look at the coefficients in the balanced equation. How many total moles of gas do we have on the left side? Yes. Or how many total moles of gas do we have on the right side to it, we compress the box, it's going to shift to the right. The idea being four moles of gas on this side, two moles of gas on this side, it shifts in whichever direction makes a smaller number of molecules of gas. Back to your question. Are there things we can add that have no affect? The answer is yes. And one of them would be something like a catalyst. We saw earlier what a catalyst does. It lowers the energy of activation. It's like digging a tunnel through the hill that you're trying to roll the Rocco. But the point is that tunnel goes in both directions. If you add a catalyst, it speeds up before reaction, but it also speeds up the reverse reaction. It makes it easier for both reactions to go in the opposite direction. Adding a catalyst does not pause an equilibrium to shift because it lowers the energy of activation in both directions. What a catalyst might do is speed up the rate at which your reaction gets to a state of equilibrium. If it's not already there. But once it's there, adding a catalyst has no impact, doesn't shift to the right. Does it shift to the left? It doesn't do anything. Are there questions about these examples? So when in doubt, go back to the kids on the seesaw and think about which way you have to shift a kid to make the seesaw come out level again. Does anyone need more time with this slide? >> Alright, why does any of this matter? >> Well, I can think of many reasons why this would matter, but I'll just hit you with a couple of white down. Well, let you ponder others. One way in which Le Chatelier's principle matters. Have you ever heard of the principle of homeostasis? Okay. What does it mean? I'll rephrase the question and I'll rephrase it sort of in the context of what's in the back of everybody's mind these days. >> This being March 2020. >> Happy march, everyone. Because everyone feeling well today as everyone been pretty much feeling well during this lecture. >> Hope so that's homeostasis. >> Homeostasis means if you feel well, you generally keep on feeling well until you get sick. And if he gets sick, then what happens is a lot of the chemical reactions in your body that are equilibria tends to shift to try to make it feel better. Sometimes when you get sick, you run a fever and people think, Oh God, I've got a fever, you gotta bring the temperature down well, yeah, to a safe level. But thought has always been, you know, running a fever. Is it your body trying to speed up certain chemical reactions that get back to a state of equilibrium for it. So that's really your body trying to help you get better. Anyway, homeostasis basically says you feel well most of the time and when you get sick. That's when Le Chatelier's principle kicks in and tries to help you get back to a state of equilibrium. There's a couple of other examples. Here is an equilibrium that actually has something to do with the solubility rules. And hopefully it's an equilibrium you take advantage of every day. I have a feeling that most people would not recognize the term hydroxyapatite, which is actually the name of this mineral. However, I trust you do recognize that this is the fluoride ion. What's one way you hopefully everyday use fluoride ions? >> Yeah, yeah. >> Hopefully when you brush your teeth, you brushed your teeth using fluoride toothpaste. You know why they put fluoride toothpaste. Hydroxyapatite is the form that calcium takes in your bones and your teeth. Your bones and your teeth are not pure calcium. Calcium is a metal, it is calcium phosphate hydroxide, which is hydroxyapatite. But if you brush your teeth with fluoride containing toothpaste, that makes the forward reaction proceed to gave me this new mineral called floral appetite. And it turns out that the fluoride binds more strongly with the calcium to make a stronger material which is more resistant to the formation of cavities. Now, how does this tie in with the solubility rules? Okay, hydroxyapatite is a hydroxide. Our hydroxides generally soluble or insoluble in water. Insoluble in water. >> That's a good thing. >> I would hope that drinking a glass of water, it wouldn't dissolve your teeth, right? But what our hydroxide soluble in acids. The problem is, there's a great many things that people eat and drink. They tend to be on the acidic side. So as you over time continue to expose your tooth enamel too acidic foodstuffs, little by little, this starts to dissolve. Many get holes in your teeth, which we call cavities. And then you'd have to go visit the dentist a lot, but maybe less. So if you've been taking advantage of Le Chatelier's principle by brushing your teeth with fluoride toothpaste and driving this equilibrium to the right to make the more cavity resistant material. So here's one thing. Le Chatelier's principle is good for brush your teeth and use fluoride toothpaste when you do. And hopefully that means fewer trips to the dentist. Here's one more application of Le Chatelier's principle. This comes up in poultry farming. Now, this is pretty straightforward. >> Water, carbon dioxide, carbonic acid, carbonate ions involved. >> Okay, that doesn't seem like a big b. Uh, why is this a big deal? >> Well, chickens, now I'm just going to put this in context. >> Everybody here has a blue hen. Re, I think you should know something about chickens before you get out of blue hen school with your degree, right? Chickens need carbonate ions in their bloodstream because egg shells are made of calcium carbonate. And so the chicken is trying to make more chickens. They do so by laying eggs with egg shells made of calcium carbonate. If all goes well, get baby chickens. But in the summer months when the weather gets hot, chicken is get rid of excess body heat by painting like dogs do. That's because chickens have no sweat glands. And the problem is when the chicken pants, that increases the rate at which CO2, which is what the chicken exhales, leaves the chickens body that shifts this equilibrium to the left. So in the summer months, it's hard to keep carbonate ion levels in the chickens bloodstream as to what they should be. Unless you take advantage of Le Chatelier's principle and put carbon dioxide and the chickens drinking water. Perrier, maybe a bit extreme, something like tonic water would probably be. Okay. But the point is what poultry farmers tend to do so as to make their chickens lay healthy eggs in the summertime, infuse the chickens drinking water with carbon dioxide shifts the equilibrium to the right, increase as the carbonate concentration. If they don't do that, there's not enough carbonate and the chickens bloodstream to make a proper egg shell hit. The eggs get laid at all. They get laid with egg shells that are too thin and the developing chick doesn't get to develop. But if the farmers are smart, they put that in. The drinking water shifts the equilibrium to the right, or radon levels or what they should be. Healthy eggshells, healthy chickens, healthy blue ends. Those are just a few examples. Those of you who are going to take other courses in chemistry beyond this one, are going to see a lot more examples of why Le Chatelier's principle is important for now, hopefully Office at that, they make sense that way. When you go to lab this week, and I have to apply Le Chatelier's Principle, you'll be able to do that. >> We'll see you on Wednesday, ruby to 20 to 21. >> But to show you where it is, I shall have to do is walk out the door to your left and to your left again, right there. 815. >> We got the first loud. Right. Okay. >> Just on Fridays. >> I know the only other legal lab session is Wednesday, Thursday, Thursday. I so like what's the fact? >> Because every time he did the first lap or well, okay point is if you missed the first lab, there's no way to make that goes that week is over. You're allowed to miss one lab without her having an impact. Your elaborate too much to just try and make it all the other labs. So that should hopefully be okay. >> And I think it okay. So I was kind of like that. >> Okay? >> For the moment we have only talked about two weak electrolytes. So those are the two we used UP demonstrations. That's ammonia and acetic acid. For right now, there's a weak electrolyte. >> You read them out. >> Later I will talk more about weak acids in general, weak bases in general, and all those are weak electrolytes. >> Do we have enough knowledge? >> I'm kind of like your lights are generally molecular compounds like sugar or things like that, which started making lightbulb, light up. Ionic compounds like thoughts generated electron. >> Now pretty **** well, and I get strong. It's generally a nonelectrolyte other than a week after, ought to be the ones that we already have a 3.2 molal solution. >> That way to predict where the benzene but information. >> So please immortal colligative property. >> All right, so the point is we're working with zeta, this problem of the salt boiling point, the value of K b would you use with also the freezing point of the value? >> We can use it just with acquainted freezing or providing a willing they told you what the allowance he told me that in order to allow for the blending, multiply KB times little allowance and then add that to this boiling point. That's what winning with other words, these are the normal boiling points, a, B, and C to E. >> But wait five minutes. >> But when you dissolve something in it, that's going to change. >> They told you the molality. Well, yeah, usable by the molality constancy. In the other they get, remembers for boiling 0.81, which was not the freezing point goes down. >> There were times that counter that by point, but it is, that is the concentration would want to lord. >> His name is, Oh, wow, that's moles per kilogram. For the wave number, is that number times this. Subtract from that the number of times this added boiling point. Try it on that basis.
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From Dana Chatellier March 02, 2020
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