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Zwt_x-8hM54-000|Let's look at the titration curve of a weak base with a strong acid. |
Zwt_x-8hM54-002|The formation of water is favored. |
Zwt_x-8hM54-003|So this reaction lies far towards products. |
Zwt_x-8hM54-004|And basically, every mole of acid you have will consume a mole of your weak base. |
Zwt_x-8hM54-006|So let's look at the various points along this curve. |
Zwt_x-8hM54-013|When I get to point, the equivalence point here, now I've converted all my base to the conjugate acid. |
Zwt_x-8hM54-014|I've added a mole of strong acid for every mole of base that I originally had, converting it all to its conjugate base. |
Zwt_x-8hM54-015|So you have a solution here of a conjugate acid. |
Zwt_x-8hM54-016|You can calculate the pH of a solution of a weak acid. |
Zwt_x-8hM54-017|As you can continue out, the pH will start to be dominated by the fact that you're just adding strong acid. |
Zwt_x-8hM54-018|So these are the regions of the titration curve for a weak base by a strong acid. |
yoq9Vx6FvSE-000|Among the most common equilibria that we'll look, and the most important, are acid base equilibria. |
yoq9Vx6FvSE-001|Acid base equilibria involve transferring a proton from one species to another. |
yoq9Vx6FvSE-003|Now I'm saying proton, by that I mean when this hydrogen leaves the HA compound, it leaves its electron behind. |
yoq9Vx6FvSE-004|So it's only the nucleus of the hydrogen that leaves. |
yoq9Vx6FvSE-005|And the nucleus of a hydrogen atom is a proton. |
yoq9Vx6FvSE-006|So we call this proton transfer. |
yoq9Vx6FvSE-011|Because this is an equilibrium. |
yoq9Vx6FvSE-013|So that'll be our definition of acid and base. |
yoq9Vx6FvSE-017|So I've abbreviated significantly here. |
yoq9Vx6FvSE-018|And of those four hydrogens only one is donated as a proton. |
yoq9Vx6FvSE-019|One is the acidic proton on acetic acid. |
yoq9Vx6FvSE-020|H3O plus, that's water with a proton added. |
yoq9Vx6FvSE-021|So it has an extra proton, and it will very readily donate that one. |
yoq9Vx6FvSE-025|So water can act as an acid or a base. |
yoq9Vx6FvSE-027|So here would be water donating a proton, you'd have OH minus leftover. |
yoq9Vx6FvSE-031|NH3, it can accept the proton, act as a base, and OH minus, as we've already discussed, the strongest of these three bases. |
yoq9Vx6FvSE-032|The most able to take a proton. |
yoq9Vx6FvSE-035|So our definition, an acid gives up a proton, a base accepts a proton. |
AHUBz-4aAbE-000|Let's look at some more orbital shapes and orientation. |
AHUBz-4aAbE-008|We're talking about orbitals in terms of their shape, and the number of nodes, and their orientation. |
AHUBz-4aAbE-009|Orbitals are described by three quantum numbers, n, l, and m sub l. |
AHUBz-4aAbE-010|n gives you the overall size and energy, l gives you the shape, and m sub l an orientation. |
AHUBz-4aAbE-011|Now what we want to know for this case is where and how are the nodes oriented. |
AHUBz-4aAbE-012|Well, let's look at nodes. |
AHUBz-4aAbE-013|Remember, for angular nodes, they go as the value of l. |
AHUBz-4aAbE-014|So if l is 0, there are 0 angular nodes. |
AHUBz-4aAbE-015|An s orbital, l equals 0, has 0 angular nodes. |
AHUBz-4aAbE-018|If you understand that there are angular nodes slicing up space, that pretty much tells you about the shape and orientation of the orbital. |
AHUBz-4aAbE-020|Or 2, 3 minus 1, or 1. |
AHUBz-4aAbE-021|So there's three nodes here, two nodes here, one node here. |
AHUBz-4aAbE-025|When l is 2, that's a d, there are two angular nodes. |
AHUBz-4aAbE-026|When l is 3, that's an f orbital, there are three angular nodes. |
AHUBz-4aAbE-027|So now we can easily tell what's going on, remember there's three nodes in the 4f, but they're all angular. |
AHUBz-4aAbE-028|So there are no radial nodes in a 4f orbital. |
AHUBz-4aAbE-029|The 3d, there's two total nodes, n minus 1, but they're both angular. |
AHUBz-4aAbE-030|Again, no radial nodes in the 3d. |
AHUBz-4aAbE-031|The 2s, there's one node, s equals 0, so the value of l is 0. |
AHUBz-4aAbE-032|So there are zero angular nodes, so that one node must be a radial node. |
AHUBz-4aAbE-033|So no radial nodes, no radial nodes, 1 radial node. |
9JM7zZk1j_g-000|Let's look at the isothermal compression of a gas. |
9JM7zZk1j_g-001|I'll take a gas, compress it, and I'd like to know-- what is the heat flow? |
9JM7zZk1j_g-010|We're talking about the isothermal compression of an ideal gas. |
9JM7zZk1j_g-011|Now isothermal is the key word because if the temperature doesn't change for an ideal gas, its energy didn't change. |
9JM7zZk1j_g-012|So no energy change in this system. |
9JM7zZk1j_g-013|Now it was a compression, so work is being done on the system. |
9JM7zZk1j_g-014|Delta-V will be negative, but we have a negative sign minus the external delta-V-- positive work. |
9JM7zZk1j_g-019|Now let's think about this practically. |
9JM7zZk1j_g-020|I compress the gas, I'm doing work on the gas. |
9JM7zZk1j_g-021|I'm giving energy to the gas. |
9JM7zZk1j_g-022|And you say, wait, but if I give energy to the gas, its temperature has to change. |
9JM7zZk1j_g-023|How can I give it energy without its temperature changing? |
9JM7zZk1j_g-024|Well, that energy has to flow out at the same time I give it. |
9JM7zZk1j_g-025|So how can that energy flow out? |
9JM7zZk1j_g-026|Well, I'll have work done on the system, but every joule of work will flow out as a joule of heat. |
9JM7zZk1j_g-027|And we call that negative heat or an exothermic process when heat leaves the system. |
9JM7zZk1j_g-028|Practically, it means you probably do this process very slowly. |
9JM7zZk1j_g-029|You compress it a little bit, wait for some heat to flow out. |
9JM7zZk1j_g-030|You compress a little more, wait for heat to flow out. |
9JM7zZk1j_g-031|Or you compress it and wait a while until all the heat flows out. |
9JM7zZk1j_g-032|In general, you can do work quickly on a system, but heat flow always takes time. |
9JM7zZk1j_g-033|So when you see isothermal, you often think slow, because I have to wait for the system to come back to the same temperature or become isothermal. |
9JM7zZk1j_g-034|That's an isothermal compression of an ideal gas. |
jzHAfhGPQxg-003|Well, this chemical reaction releases energy, delta H is negative it's exothermic. |
jzHAfhGPQxg-004|Delta acid is positive and you might have guessed that because there's a gas released from a solid and a liquid. |
jzHAfhGPQxg-005|So that's going to have many more microstate. |
jzHAfhGPQxg-006|You're going to have a dispersion of energy when a gas is created. |
jzHAfhGPQxg-007|So delta H negative, delta S positive. |
jzHAfhGPQxg-008|This reaction will have a negative delta G for all temperatures. |
jzHAfhGPQxg-013|We've actually seen this chemical reaction, but in reverse. |
jzHAfhGPQxg-014|We saw peroxide going to water and oxygen. |
jzHAfhGPQxg-015|And we know energy is released in this direction. |
jzHAfhGPQxg-016|So it has to be absorbed in this direction. |
jzHAfhGPQxg-017|Delta S, in this case, is negative. |
jzHAfhGPQxg-018|Again you might guess that. |
jzHAfhGPQxg-019|Gases forming liquids. |
jzHAfhGPQxg-020|So this reaction in this direction is never favorable, at any temperature. |
jzHAfhGPQxg-024|You're forming a liquid from a solid, so you'd expect an increase in entropy. |
jzHAfhGPQxg-028|This one it's a little less clear, everything's a solid. |
jzHAfhGPQxg-029|So entropy wise it's hard to predict. |
jzHAfhGPQxg-030|Entropy wise this is an exothermic chemical reaction, and entropically there there's a decrease in entropy as you go. |
jzHAfhGPQxg-031|So this reaction is favored at low temperatures. |
jzHAfhGPQxg-032|So we need, since we have this negative delta S contribution, we need T to be small. |
jzHAfhGPQxg-033|So that the overall and felt that contribution will have a contribution to delta G that is negative. |
jzHAfhGPQxg-034|So this reaction favored at low temperature, unfavorable at high temperature. |
pxxhXa3TZAU-001|Now, the change in concentration over time is the slope of the concentration versus time curve. |
pxxhXa3TZAU-002|And we could start at any initial concentrations and let this reaction go. |
pxxhXa3TZAU-011|We're talking about writing an initial rate for a chemical reaction, and we can write initial rates in terms of changes in concentration over a change in time. |
pxxhXa3TZAU-013|That is, two hydrogen ions will be consumed when a zinc ion is produced. |
pxxhXa3TZAU-014|So two things will happen. |
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