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oqN1aemQnu0-012|Every point on this line, the solid and the liquid are in equilibrium. |
oqN1aemQnu0-013|Now, for carbon dioxide, it turns out that common equilibrium that we observe is the solid gas, or sublimation equilibrium. |
oqN1aemQnu0-016|And if I bring it here to the desktop, that's solid carbon dioxide. |
oqN1aemQnu0-017|And that vibrating that you hear is the gas subliming off the carbon dioxide. |
oqN1aemQnu0-018|That gas phase provides a cushion of gas. |
oqN1aemQnu0-020|That's the sublimation of carbon dioxide. |
oqN1aemQnu0-021|Now, let's talk about some other regions on the phase diagram. |
oqN1aemQnu0-022|Interestingly, there's a point where all three of the solid, liquid, and gas phases intersect. |
oqN1aemQnu0-023|This is called the triple point. |
oqN1aemQnu0-027|A particle can move from the solid phase to the liquid phase, liquid to solid, or solid to gas independently. |
oqN1aemQnu0-028|All three are equally likely at the triple point. |
oqN1aemQnu0-029|We also have the critical point. |
oqN1aemQnu0-035|Now, water is one of the only substances where the slope of the solid-liquid slopes backwards. |
oqN1aemQnu0-036|Now, that's interesting. |
oqN1aemQnu0-037|That says if you compress solid water, you go up in pressure-- so here's a solid water point. |
oqN1aemQnu0-038|I go up in pressure. |
oqN1aemQnu0-039|It should turn into the liquid. |
oqN1aemQnu0-040|And that's counterintuitive. |
oqN1aemQnu0-041|Most things, when you increase the pressure, you should continue to favor the solid, the more tightly compact phase of matter. |
oqN1aemQnu0-042|Turns out for water, the solid is not the most tightly compact phase. |
oqN1aemQnu0-043|The solid is actually slightly less dense than the liquid. |
oqN1aemQnu0-044|And we know that. |
oqN1aemQnu0-045|We've seen solid ice float on top of liquid ice. |
oqN1aemQnu0-046|In fact, it's the most common solid-liquid phase transition we know about. |
oqN1aemQnu0-047|So when we think about it, you'd think, oh, maybe all solids float on their liquids. |
oqN1aemQnu0-048|It's actually the reverse. |
oqN1aemQnu0-049|Just because that's a common substance, water, and we see the solid floating on the liquid, we get the impression that that's common. |
oqN1aemQnu0-050|It's actually the least common interaction. |
oqN1aemQnu0-051|Most solids sink in their own liquid. |
oqN1aemQnu0-055|So very near 0 Celsius, but a fraction of an atmosphere. |
oqN1aemQnu0-056|There are 760 torr in one atmosphere. |
OlehBTSyPDA-000|Let's do a calculation that's a titration calculation. |
OlehBTSyPDA-002|The question is, what is the pH after those two are mixed? |
OlehBTSyPDA-003|So I can think about this as a titration. |
OlehBTSyPDA-005|The question is, how far along on the titration curve does this addition of strong base take me? |
OlehBTSyPDA-006|Does it take me to the center of the buffer region? |
OlehBTSyPDA-007|Is it enough to take me to the equivalence point? |
OlehBTSyPDA-009|We have HAc plus water, forming acetic acid and H3O plus. |
OlehBTSyPDA-010|That's one of the equilibria we have to deal with for this problem. |
OlehBTSyPDA-011|But we want to find out, well, what are the initial conditions? |
OlehBTSyPDA-012|How much HAc, how much Ac minus, and how much H3O plus are there at a point? |
OlehBTSyPDA-013|And then we'll allow the equilibria to expand. |
OlehBTSyPDA-014|And this is how you do equilibrium calculations. |
OlehBTSyPDA-015|You find a point that you can nail down. |
OlehBTSyPDA-016|Say, OK, these will be my starting conditions. |
OlehBTSyPDA-017|And then you apply the equilibrium to see how it shifts from that one point. |
OlehBTSyPDA-018|And that's the great thing about equilibria, they approach the same equilibria no matter where you start from. |
OlehBTSyPDA-019|So as long as you can nail down one point that you know and then apply equilibrium from there, you're going to get to the correct equilibrium condition. |
OlehBTSyPDA-023|Well, molarity times volume is the number of moles. |
OlehBTSyPDA-024|So 0.01 moles of acetic acid. |
OlehBTSyPDA-025|Now this is assuming no dissociation. |
OlehBTSyPDA-026|That's fine. |
OlehBTSyPDA-027|I can choose any point along my equilibrium to be my starting point. |
OlehBTSyPDA-033|And I choose that because the weak acid reacts completely with the strong base. |
OlehBTSyPDA-034|That's a very strong reaction to favor the products. |
OlehBTSyPDA-038|So that's not really an equilibrium, that one just goes all the way towards products. |
OlehBTSyPDA-039|So if we start there and say, what happens after that occurs? |
OlehBTSyPDA-040|Are there an equilibrium in effect, that will calculate the pH of the solution. |
OlehBTSyPDA-041|So let's do that. |
OlehBTSyPDA-042|We'll say, these two will react completely. |
OlehBTSyPDA-043|So every mole of base will react with a mole of acid. |
OlehBTSyPDA-044|And I'll have-- this is essentially half of this. |
OlehBTSyPDA-045|So half of my base-- excuse me. |
OlehBTSyPDA-046|Half of my acid will react with this base, and I'll form 0.05 moles of the conjugate base. |
OlehBTSyPDA-047|So the acid reacts with the base to form the conjugate base. |
OlehBTSyPDA-049|Here, the acid reacting with the base to form the conjugate base. |
OlehBTSyPDA-050|In this case, I've added half the number of moles of HA that I originally had. |
OlehBTSyPDA-052|Half of these react, so I have equal concentrations essentially of this and this at this point in the titration. |
OlehBTSyPDA-053|So equal concentrations of these two. |
OlehBTSyPDA-054|Half of my acid has been converted into its conjugate base. |
OlehBTSyPDA-055|My OH minus concentration, it's gone to zero. |
OlehBTSyPDA-056|I've used up one for one. |
OlehBTSyPDA-057|One of those react with one of those, producing one of these. |
OlehBTSyPDA-058|So let's treat these now as the initial concentrations in an equilibrium calculation and see where that gets us. |
OlehBTSyPDA-063|But I know they're both in about 100 millimeters. |
OlehBTSyPDA-064|So I can convert these easily into concentrations, moles per liter. |
OlehBTSyPDA-070|Now this initial point is quite a ways into our calculation, but it's just the first point where I'm going to allow the equilibria to occur. |
OlehBTSyPDA-071|In all the previous steps, I've said they go to completion. |
OlehBTSyPDA-072|So now I'm at a point where everything has gone 100%, and I'm going to let the equilibria start to take over. |
OlehBTSyPDA-073|And I do that at this point. |
OlehBTSyPDA-074|And I can do that for equilibria at any point. |
OlehBTSyPDA-075|Choose a point, nail it down, say, this is where I'm going to start, and then let your equilibria table take over. |
OlehBTSyPDA-076|So the change to get to equilibrium is a little of this. |
OlehBTSyPDA-077|We'll dissociate. |
OlehBTSyPDA-078|And actually at this point, I'm not even sure. |
OlehBTSyPDA-079|I'm not sure if the equilibrium will go this way or this way. |
OlehBTSyPDA-080|I'm just going to assume a little of this goes away. |
OlehBTSyPDA-081|And if it does, it'll produce a little of this. |
OlehBTSyPDA-082|Since there's equal amounts, who knows? |
OlehBTSyPDA-083|Maybe it'll shift back this way. |
OlehBTSyPDA-084|I suspect it'll go this way though, because I think HAc is a strong enough acid to dissociate a little bit more. |
OlehBTSyPDA-085|That will produce a little H3O plus. |
OlehBTSyPDA-086|And I'm saying initially there was none of this. |
OlehBTSyPDA-093|So that's 10 to the minus 4.75. |
OlehBTSyPDA-094|4.75 is the Ka for acetic acid. |
OlehBTSyPDA-102|That is the H3O plus concentration. |
OlehBTSyPDA-103|And notice that I left the k like this, because I knew after I get the H3O plus concentration, I'm going to take minus log of it anyway. |
OlehBTSyPDA-104|So I'm going to go right back to the exponent. |
OlehBTSyPDA-105|So if I calculate the pH, that's minus log of 10 to the minus 4.75, which is 4.75. |
OlehBTSyPDA-106|So what I found is that the pH is numerically equal to the pKa. |
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