ceaglex/VisualTTS_Chem · Datasets at Hugging Face

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fysCY3d9zAU-034\|And of course K would be equal to 1, because you would put in 1 for every pressure or every concentration, and you'd get a K equal to 1.
fysCY3d9zAU-036\|They stay at 1.
fysCY3d9zAU-037\|Standard free energy 0, K equal 1.
fysCY3d9zAU-040\|K is less than 1, delta G greater than 0.
fysCY3d9zAU-041\|K is greater than 1, delta G less than 0.
fysCY3d9zAU-042\|So that behaves like a natural log function.
fysCY3d9zAU-043\|Here, if you take the natural log of K, and you go to K is greater than one, that's positive.
fysCY3d9zAU-044\|Put a negative sign in front of that, that would give you negative delta G's.
fysCY3d9zAU-045\|So K's greater than 1 give you negative delta G's for this natural log function.
fysCY3d9zAU-046\|And when you go to a fraction less than 1, the natural log function becomes negative.
fysCY3d9zAU-047\|Multiply that times a negative, and you get a positive sign.
fysCY3d9zAU-048\|So that accounts for this relationship.
fysCY3d9zAU-049\|A K less than 1, a fraction, gives you a negative natural log and a positive delta G.
fysCY3d9zAU-053\|So these two relationships allow us to switch between free energies and equilibrium constants analytically.
fysCY3d9zAU-055\|So you can see the power of the thermodynamics as they relate to each other.
fysCY3d9zAU-056\|Tables of standard state free energies give you equilibrium constants and tell you how a reaction is favored at equilibrium, products or reactants.
foJavVRkARU-000\|Strong acids and strong base is completely dissociated in water.
foJavVRkARU-001\|So if I take a mole of strong acid, HCl, hydrochloric acid, and I put it in water, I get a mole of H30 plus and Cl minus.
foJavVRkARU-002\|That complete dissociation is what characterizes it as a strong acid.
foJavVRkARU-003\|Some acids and bases don't completely dissociate, and we call those weak acids or weak bases.
foJavVRkARU-007\|The acetate ion is actually H3CCOO minus.
foJavVRkARU-008\|Just Ac minus here for convenience.
foJavVRkARU-014\|So there's another difference between the strong acid and the weak acid.
foJavVRkARU-015\|At equilibrium, the strong acid favors the products-- complete dissociation.
foJavVRkARU-016\|The weak acid favors the reactants.
foJavVRkARU-017\|You can look at it in terms of free energy.
foJavVRkARU-018\|Since this K is less than one, the standard state free energy difference is positive.
foJavVRkARU-019\|So this is a non-spontaneous reaction.
foJavVRkARU-020\|Again, non-spontaneous does not mean it doesn't go at all.
foJavVRkARU-021\|It just means at equilibrium the reactants are favored.
foJavVRkARU-022\|For a strong acid, by contrast, the standard state free energy difference is negative.
foJavVRkARU-023\|That's a spontaneous reaction.
foJavVRkARU-026\|So this is a pKa of 4.7.
foJavVRkARU-027\|And the smaller the pKa, the larger the Ka.
foJavVRkARU-028\|So a small pKa means a relatively stronger acid.
foJavVRkARU-029\|The smaller the pKa, the stronger the acid.
foJavVRkARU-030\|Let's look at the conjugate base.
foJavVRkARU-034\|The reverse of the acid reaction would be AC minus reacting with H3O plus.
foJavVRkARU-035\|Here's AC minus reacting with H2O.
foJavVRkARU-036\|So when I write the equilibrium constant for this, I'm writing a reaction with H2O and Ac minus.
foJavVRkARU-039\|This is a weak base.
foJavVRkARU-040\|This favors the reactants.
foJavVRkARU-041\|You should also notice that Kas and Kbs, even though they're not reverse-- that is, Ka isn't one over Kb-- they are correlated.
foJavVRkARU-042\|The larger the acid dissociation constant, the smaller the base dissociation constant.
foJavVRkARU-043\|And that should be obvious because the stronger the acid, the more this equilibrium lies towards Ac minus in solution.
foJavVRkARU-044\|So Ac minus should be favored in solution.
foJavVRkARU-045\|That means this equilibrium constant must be small.
01ZxF6TcY-A-000\|Let's talk about the photoelectric effect in terms of ChemQuiz.
01ZxF6TcY-A-001\|If we shine a beam of light on a certain metal, it has no effect.
01ZxF6TcY-A-002\|The question I have for you is, what change in that beam should I make in my best hope to eject electrons?
01ZxF6TcY-A-011\|If you have a photon that won't eject an electron, what do you need?
01ZxF6TcY-A-012\|You need photons with more energy.
01ZxF6TcY-A-013\|You have to increase the energy of the photon.
01ZxF6TcY-A-014\|Increasing the number of photons won't do.
01ZxF6TcY-A-015\|You'll just strike a lot of electrons with a little bit of energy.
01ZxF6TcY-A-023\|So here you could have answered B or C, and you would have got an electron ejected from that metal.
2hikObHZrXg-000\|Let's try to sketch the titration curve for a polyprotic weak acid.
2hikObHZrXg-001\|So I've chosen here histidine.
2hikObHZrXg-005\|There are three of them, with pKa1, 1.81, pKa2, 6, and pKa3, 9.85.
2hikObHZrXg-006\|The associated sites are as follows.
2hikObHZrXg-010\|PKa2 is associated with this site here.
2hikObHZrXg-013\|The final pKa of 9.85 is associated with this amino group here.
2hikObHZrXg-018\|So here I've drawn the molecule again.
2hikObHZrXg-019\|And I want to start by just talking about the first proton to be removed from solution.
2hikObHZrXg-021\|So these two molecules, you noticed they're identical except for this proton has been removed.
2hikObHZrXg-022\|And these two forms will be equal in concentration at pHs equal to the pKa of 1.81.
2hikObHZrXg-025\|That's something I just defined.
2hikObHZrXg-026\|I said, let's start at a low pH and go to a high pH.
2hikObHZrXg-027\|So the solution is prepared at low pH and the initial pH will be less than 1.8.
2hikObHZrXg-028\|And what I'm going to do is start adding base, raising the pH.
2hikObHZrXg-029\|And as I get to pHs around the first pKa, I'll have an equal mixture of these two forms.
2hikObHZrXg-034\|Now I can continue to add base.
2hikObHZrXg-035\|And when I do, I'll go over to the equivalence point of the first proton.
2hikObHZrXg-036\|So I'll completely remove the first proton.
2hikObHZrXg-037\|Remember at half equivalent, I have equal concentrations of these.
2hikObHZrXg-038\|As I keep adding base, I get more and more of the base form until all of the acid form is used up.
2hikObHZrXg-039\|So at this point, the first equivalence point, all of the acid form, is converted into the base form.
2hikObHZrXg-040\|But notice, it's only for this first titration, this first proton.
2hikObHZrXg-041\|These other two protons are still attached.
2hikObHZrXg-046\|So this second plateau occurs at pH 6.
2hikObHZrXg-047\|So let's look at that.
2hikObHZrXg-050\|The first proton stays gone, it's completely gone.
2hikObHZrXg-051\|And now the second proton is being titrated.
2hikObHZrXg-055\|I'll get to the second equivalence point.
2hikObHZrXg-056\|At that second equivalence point, I've been converted to all this form.
2hikObHZrXg-057\|All these protons are gone.
2hikObHZrXg-058\|And that's the second endpoint, or the second equivalence point.
2hikObHZrXg-059\|It's funny to say endpoint in a multiple titration-- equivalence point is certainly the better term.
2hikObHZrXg-060\|So the second equivalence point, it's going to occur at less than pH 9, but more than pH 6.
2hikObHZrXg-061\|So somewhere between 9 and 6, the two pKas, I'll have the second equivalence point.
2hikObHZrXg-062\|At that second equivalence point, the form in solution is this.
2hikObHZrXg-063\|That is, the second proton has been completely removed.
2hikObHZrXg-064\|So let's continue.
2hikObHZrXg-068\|Notice this proton is now being titrated, or this proton is being removed.
2hikObHZrXg-069\|And when the pH is exactly equal to the pKa, there'll be an equal mixture of these two species in solution.
2hikObHZrXg-070\|Now I can continue to the third endpoint.
2hikObHZrXg-071\|And the third endpoint is where I've removed all of these protons, and the form in solution is exclusively completely deprotonated now.
2hikObHZrXg-072\|And that pH will be above 9.8.
2hikObHZrXg-075\|We've sketched out the titration curve.
2hikObHZrXg-076\|And what we'd like to say is, at pH 8 which species are present in solution.

fysCY3d9zAU-034|And of course K would be equal to 1, because you would put in 1 for every pressure or every concentration, and you'd get a K equal to 1.

fysCY3d9zAU-036|They stay at 1.

fysCY3d9zAU-037|Standard free energy 0, K equal 1.

fysCY3d9zAU-040|K is less than 1, delta G greater than 0.

fysCY3d9zAU-041|K is greater than 1, delta G less than 0.

fysCY3d9zAU-042|So that behaves like a natural log function.

fysCY3d9zAU-043|Here, if you take the natural log of K, and you go to K is greater than one, that's positive.

fysCY3d9zAU-044|Put a negative sign in front of that, that would give you negative delta G's.

fysCY3d9zAU-045|So K's greater than 1 give you negative delta G's for this natural log function.

fysCY3d9zAU-046|And when you go to a fraction less than 1, the natural log function becomes negative.

fysCY3d9zAU-047|Multiply that times a negative, and you get a positive sign.

fysCY3d9zAU-048|So that accounts for this relationship.

fysCY3d9zAU-049|A K less than 1, a fraction, gives you a negative natural log and a positive delta G.

fysCY3d9zAU-053|So these two relationships allow us to switch between free energies and equilibrium constants analytically.

fysCY3d9zAU-055|So you can see the power of the thermodynamics as they relate to each other.

fysCY3d9zAU-056|Tables of standard state free energies give you equilibrium constants and tell you how a reaction is favored at equilibrium, products or reactants.

foJavVRkARU-000|Strong acids and strong base is completely dissociated in water.

foJavVRkARU-001|So if I take a mole of strong acid, HCl, hydrochloric acid, and I put it in water, I get a mole of H30 plus and Cl minus.

foJavVRkARU-002|That complete dissociation is what characterizes it as a strong acid.

foJavVRkARU-003|Some acids and bases don't completely dissociate, and we call those weak acids or weak bases.

foJavVRkARU-007|The acetate ion is actually H3CCOO minus.

foJavVRkARU-008|Just Ac minus here for convenience.

foJavVRkARU-014|So there's another difference between the strong acid and the weak acid.

foJavVRkARU-015|At equilibrium, the strong acid favors the products-- complete dissociation.

foJavVRkARU-016|The weak acid favors the reactants.

foJavVRkARU-017|You can look at it in terms of free energy.

foJavVRkARU-018|Since this K is less than one, the standard state free energy difference is positive.

foJavVRkARU-019|So this is a non-spontaneous reaction.

foJavVRkARU-020|Again, non-spontaneous does not mean it doesn't go at all.

foJavVRkARU-021|It just means at equilibrium the reactants are favored.

foJavVRkARU-022|For a strong acid, by contrast, the standard state free energy difference is negative.

foJavVRkARU-023|That's a spontaneous reaction.

foJavVRkARU-026|So this is a pKa of 4.7.

foJavVRkARU-027|And the smaller the pKa, the larger the Ka.

foJavVRkARU-028|So a small pKa means a relatively stronger acid.

foJavVRkARU-029|The smaller the pKa, the stronger the acid.

foJavVRkARU-030|Let's look at the conjugate base.

foJavVRkARU-034|The reverse of the acid reaction would be AC minus reacting with H3O plus.

foJavVRkARU-035|Here's AC minus reacting with H2O.

foJavVRkARU-036|So when I write the equilibrium constant for this, I'm writing a reaction with H2O and Ac minus.

foJavVRkARU-039|This is a weak base.

foJavVRkARU-040|This favors the reactants.

foJavVRkARU-041|You should also notice that Kas and Kbs, even though they're not reverse-- that is, Ka isn't one over Kb-- they are correlated.

foJavVRkARU-042|The larger the acid dissociation constant, the smaller the base dissociation constant.

foJavVRkARU-043|And that should be obvious because the stronger the acid, the more this equilibrium lies towards Ac minus in solution.

foJavVRkARU-044|So Ac minus should be favored in solution.

foJavVRkARU-045|That means this equilibrium constant must be small.

01ZxF6TcY-A-000|Let's talk about the photoelectric effect in terms of ChemQuiz.

01ZxF6TcY-A-001|If we shine a beam of light on a certain metal, it has no effect.

01ZxF6TcY-A-002|The question I have for you is, what change in that beam should I make in my best hope to eject electrons?

01ZxF6TcY-A-011|If you have a photon that won't eject an electron, what do you need?

01ZxF6TcY-A-012|You need photons with more energy.

01ZxF6TcY-A-013|You have to increase the energy of the photon.

01ZxF6TcY-A-014|Increasing the number of photons won't do.

01ZxF6TcY-A-015|You'll just strike a lot of electrons with a little bit of energy.

01ZxF6TcY-A-023|So here you could have answered B or C, and you would have got an electron ejected from that metal.

2hikObHZrXg-000|Let's try to sketch the titration curve for a polyprotic weak acid.

2hikObHZrXg-001|So I've chosen here histidine.

2hikObHZrXg-005|There are three of them, with pKa1, 1.81, pKa2, 6, and pKa3, 9.85.

2hikObHZrXg-006|The associated sites are as follows.

2hikObHZrXg-010|PKa2 is associated with this site here.

2hikObHZrXg-013|The final pKa of 9.85 is associated with this amino group here.

2hikObHZrXg-018|So here I've drawn the molecule again.

2hikObHZrXg-019|And I want to start by just talking about the first proton to be removed from solution.

2hikObHZrXg-021|So these two molecules, you noticed they're identical except for this proton has been removed.

2hikObHZrXg-022|And these two forms will be equal in concentration at pHs equal to the pKa of 1.81.

2hikObHZrXg-025|That's something I just defined.

2hikObHZrXg-026|I said, let's start at a low pH and go to a high pH.

2hikObHZrXg-027|So the solution is prepared at low pH and the initial pH will be less than 1.8.

2hikObHZrXg-028|And what I'm going to do is start adding base, raising the pH.

2hikObHZrXg-029|And as I get to pHs around the first pKa, I'll have an equal mixture of these two forms.

2hikObHZrXg-034|Now I can continue to add base.

2hikObHZrXg-035|And when I do, I'll go over to the equivalence point of the first proton.

2hikObHZrXg-036|So I'll completely remove the first proton.

2hikObHZrXg-037|Remember at half equivalent, I have equal concentrations of these.

2hikObHZrXg-038|As I keep adding base, I get more and more of the base form until all of the acid form is used up.

2hikObHZrXg-039|So at this point, the first equivalence point, all of the acid form, is converted into the base form.

2hikObHZrXg-040|But notice, it's only for this first titration, this first proton.

2hikObHZrXg-041|These other two protons are still attached.

2hikObHZrXg-046|So this second plateau occurs at pH 6.

2hikObHZrXg-047|So let's look at that.

2hikObHZrXg-050|The first proton stays gone, it's completely gone.

2hikObHZrXg-051|And now the second proton is being titrated.

2hikObHZrXg-055|I'll get to the second equivalence point.

2hikObHZrXg-056|At that second equivalence point, I've been converted to all this form.

2hikObHZrXg-057|All these protons are gone.

2hikObHZrXg-058|And that's the second endpoint, or the second equivalence point.

2hikObHZrXg-059|It's funny to say endpoint in a multiple titration-- equivalence point is certainly the better term.

2hikObHZrXg-060|So the second equivalence point, it's going to occur at less than pH 9, but more than pH 6.

2hikObHZrXg-061|So somewhere between 9 and 6, the two pKas, I'll have the second equivalence point.

2hikObHZrXg-062|At that second equivalence point, the form in solution is this.

2hikObHZrXg-063|That is, the second proton has been completely removed.

2hikObHZrXg-064|So let's continue.

2hikObHZrXg-068|Notice this proton is now being titrated, or this proton is being removed.

2hikObHZrXg-069|And when the pH is exactly equal to the pKa, there'll be an equal mixture of these two species in solution.

2hikObHZrXg-070|Now I can continue to the third endpoint.

2hikObHZrXg-071|And the third endpoint is where I've removed all of these protons, and the form in solution is exclusively completely deprotonated now.

2hikObHZrXg-072|And that pH will be above 9.8.

2hikObHZrXg-075|We've sketched out the titration curve.

2hikObHZrXg-076|And what we'd like to say is, at pH 8 which species are present in solution.