Lecture 05. Acids and Bases. Pt. 2.
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ZigarreMannoseVancomycinMaischeInsulinkomabehandlungMedroxyprogesteronChemische ForschungWasserMolekülWasserfallDeprotonierungSäureMeeresspiegelLactitolElektrolytische DissoziationThermisches KrackenEssigsäureBaseChlorwasserstoffSalzsäureChemische FormelWässrige LösungKonjugateThermoformenAstheniaScreeningAllmendeGasphaseBlauschimmelkäseAktivierung <Chemie>ISO-Komplex-HeilweiseNeutrale LösungStahlLösungWasserstoffVorlesung/KonferenzComputeranimation
08:49
MannoseAltbierDifferentielle elektrochemische MassenspektrometrieMagnetometerBohriumVancomycinMaischeNitrosamineDeprotonierungKonzentratKonjugateWasserSäureBaseSystemische Therapie <Pharmakologie>AstheniaNeutrale LösungElektronenakzeptorAmmoniakMethanolThermoformenLösungPotenz <Homöopathie>FleischersatzElektronische ZigaretteProteinkinase AHydroxideElektronendonatorZunderbeständigkeitWässrige LösungElektronegativitätElektrolytische DissoziationScherfestigkeitMolvolumenMineralAktivierung <Chemie>Fülle <Speise>Initiator <Chemie>BasenpaarungWasserstandClusterAlterungGasphaseWursthülleEisenDeformationsverhaltenComputeranimation
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MannoseBohriumMischenSauerstoffkonzentrationImpedanzspektroskopieDeprotonierungBaseChemische FormelAstheniaCHARGE-AssoziationSäureWasserstoffHydroxideNatriumhydroxidMolvolumenElektrolytische DissoziationKonjugateBasenpaarungWasserAtombindungElektronendonatorElektronenakzeptorMolekülChemische ReaktionChemische VerbindungenSterische HinderungKochsalzChemische StrukturWässrige LösungTetraederstrukturpk-WertKonzentratNatriumThermoformenAmmoniakSetzen <Verfahrenstechnik>KrankengeschichteAlterungBukett <Wein>IonenbindungAktivierung <Chemie>QuerprofilVerbrennungLösungWursthülleDissoziationskonstanteHybridisierung <Chemie>MineralComputeranimation
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MannoseImpedanzspektroskopiePolyurethaneMagnetometerVancomycinSäureElektronische ZigaretteHydroxideDeprotonierungWasserThermoformenChemische ReaktionAktivierung <Chemie>BaseFließgrenzeTiefseebeckenBrillenglasKonjugateAlterungGasphaseLösungMemory-EffektKrankengeschichteAcetateBodenschutzKonzentratPeriodateGenexpressionWildbachChemische FormelHämatitBasenpaarungEssigsäureHydroxyethylcellulosenElektrolytische DissoziationWässrige LösungAmmoniumverbindungenGleichgewichtskonstanteElektronenakzeptorComputeranimation
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MannoseMaischeHydroxybuttersäure <gamma->MagnetometerBohriumBaseSäureZunderbeständigkeitBegasungThermoformenLösungWasserWursthülleGasphaseChemische ForschungHämatitDeprotonierungTiefseebeckenQuerprofilKarsthöhleSalzsäureAstheniaAusgangsgesteinPhasengleichgewichtAktivierung <Chemie>Chemischer ProzessPeriodateReaktionsgleichungChemische VerbindungenIsotopenmarkierungHausmittelKonjugateFleischerSingulettzustandSenseEssigsäureAcetateHydroxideScherfestigkeitElektronenakzeptorComputeranimation
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DiethylstilbestrolMagmaEthylen-Vinylacetat-CopolymereZigarreMannoseSymptomatologieNatriumhydridMagnetometerMalzStickstofffixierungArgininHydroxybuttersäure <gamma->AdenosylmethioninVancomycinBaseEssigsäureThermoformenSäureProteinkinase AKonjugateLösungKonzentratChemischer ProzessFülle <Speise>ScherfestigkeitInitiator <Chemie>WasserAcetateDeprotonierungNeutrale LösungZunderbeständigkeitChemische VerbindungenChemische ReaktionAstheniaWursthülleZigaretteMineralAssimilatorische SulfatreduktionGletscherzungeBasenpaarungEisenWasserstoffAktivierung <Chemie>StoffgesetzOperonComputeranimation
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Konkrement <Innere Medizin>KonzentratVerrottungHydroxyethylcellulosenVerbundwerkstoffVorlesung/Konferenz
Transkript: Englisch(automatisch erzeugt)
00:07
Okay. Let's go ahead and start if I can have your attention. So I wanted to remind you that the chemistry department offers help
00:20
they offer free tutoring so if anybody needs assistance you know, the information is provided here there's also a link from the class website and you can, all this information is on the class website as well and so if you need help on a drop-in basis you can seek help from the chemistry department of tutoring services
00:43
or there's also lark, okay? and so please take advantage of these resources and make sure that you don't fall through the cracks. All right? Okay, great. So if you guys remember last time we talked about acids
01:03
and bases, and so we looked at the difference between strong acids and weak acids. And we said that the difference between a strong acid and a weak acid is that a strong acid dissociates completely. So it
01:20
dissociates 100% whereas a weak acid only dissociates partially and the difference is pretty dramatic so it's either 100% or the dissociation is something like 10% or less. Very often it's much smaller the weaker the acid the partial dissociation is going to be even
01:42
lesser. Okay? So remember it's important for you to have a mental picture of what it looks like at a molecular level. So we call this a microscopic view and we're looking at the dissociation, we're going to look at the difference between a strong acid and a weak acid
02:01
and we're going to look at it at a molecular level. So we'll start with a strong acid, which is hydrochloric acid and so if you place hydrochloric acid in water you know that every single molecule of hydrochloric acid that's placed in water will dissociate. So you see one molecule
02:21
all right, you see what happens to HCl and it comes in as HCl, but then it gets deprotonated the proton is removed. All right? And you saw that it forms a hydronium ion. Now remember, we're looking at it in aqueous solution, and so in this picture the number of water molecules will be millions upon millions. All right?
02:45
And the amount of HCl molecules will be very few compared to water, which is the solvent. So it's like millions upon millions of water molecules and perhaps one HCl molecule. Now the reason that in this picture window
03:02
we show only one molecule because if we wanted to show more than sorry, one HCl molecule because they're going to be far apart from each other. So if you're looking at an aqueous solution because the relative ratio of HCl molecules versus the solvent is such a big number
03:22
that if I wanted to show you another HCl molecule, I would need a giant screen. All right? The screen, the size of a football field. Then you can see another HCl molecule because they're so far apart from each other at a molecular level, we're just zooming in on one region and therefore we see only one HCl molecule. So once again,
03:43
every single- so we're just looking at one and so every single HCl molecule that's placed in water will lose that proton. So you can see losing that proton to form a hydronium ion. All right? And then you can see that the proton gets transferred from one water
04:02
molecule to another. Now if I take acetic acid, now what I what is shown here is an acetic acid molecule. So this is carbon, there are three hydrogens, two on this side, this is on the other side, you can see this whole molecule as a whole is a molecule
04:21
acetic acid molecule. So you have CH3COOH is hidden behind. Now because acetic acid is a weak acid and we know that only partially dissociates. In other words, if you have hundred molecules of acetic acid
04:41
maybe one will dissociate. If you have a thousand, or if you have one hundred acetic acid molecules only about one percent will dissociate. All right? And because of that, you can see that if I wanted to show this picture, and I'm only looking at one molecule, the chances are that that molecule is going to be one that does not dissociate. So we're just looking at one molecule of acetic acid
05:03
and therefore if you look at this picture you'll see that this acetic acid molecule just wanders around, bumping into water, all right? And that's the acetic acid molecule, but you can see that there's no proton lost from that. Okay? And that is because it's a weak acid
05:22
and if I wanted to show the weak acid, I would have to have maybe one hundred to thousand acetic acid molecules and among those only one will dissociate. Did you get that? So that's the difference between a strong acid and a weak acid. So when we show
05:41
this picture I want you to remember the reason we're just showing you one molecule of the acid, the strong acid or the weak acid is because the window is too small. All right? So to recap what we talked about last time
06:01
we said that so a summary of what we said last time was that so if you want to summarize the take-home message from last class that was that one strong acids
06:21
have Ka greater than one weak acids have Ka less than one. All right? Two smaller the value
06:41
of Ka the weaker the acid. All right? Three substances that have Ka less
07:01
than one times 10 to the negative 14 do not as an acid in water. All right? And that is because
07:20
water is a stronger acid so water is a stronger acid and therefore will act as the acid.
07:41
All right? So if your Ka value is less than 10 to the negative 14, that substance cannot act as an acid in water. Okay? So that's the most important take-home message and then finally I wanted you to remember
08:00
can I move this? So finally I wanted you to remember that you need to print out this table and keep it with you at all times because this is a list of common acids. All right? And as I said last time, when we list acids,
08:21
we always list them first as the acid form and remember, they come as pairs so when you have an acid, you know that you have a conjugate base associated with that as well. All right? And so the acid form is written over here, the formula of the acid form is written here the formula of the conjugate base is written here
08:42
and the corresponding Ka values for each one of those acids are listed here and we said if Ka is greater than one it's going to act as a strong acid and so these are the acids that would dissociate completely and I've highlighted that in red. Okay? Now a cutoff point is hydronium ion.
09:02
So hydronium ion has Ka greater it has Ka1 and that is a strong acid as well. Okay? Now, these are acids. So between hydronium ion and water you have the range of weak acids. And so starting from 10-1
09:23
to you can see up to 10-13, these are all examples of weak acids. So these acids, if you place them in water, will partially dissociate to give you hydronium ions. All right? Now at the bottom is water and the Ka of pure water
09:41
is 1 times 10-14. All right? And so anything below that like methanol or ammonia these will not act so these are substances that have Ka values less than the value of water and these substances will not act
10:01
as an acid in water. Does that make sense? So if they do not act as an acid in water remember, water is a stronger acid so in water they're going to act as a base. Do you understand that? Because water now is a stronger acid than those substances. All right? Now one of the things that you have to practice
10:22
and know how to do is to rank relative acid strengths. All right? And in order to rank relative acid strengths you have to use this table. Okay? You can't memorize this stuff, so you have to look at the way you rank acids, and you know, if you're asked, given a series of five acids
10:41
and said rank this in relative acid strengths, starting from the strongest to the weakest. All right? and that means that you're going to look at the Ka values or sometimes you're given the pKa values. Remember, pKa is the negative log. So if you look at the scale, you can say that the Ka values decrease as you
11:03
start from here and you go down but the pKa is reversed because you're taking the negative log. All right? So if you look at pKa values strong acids have negative values, or small numbers the weakest acid will have a bigger pKa. All right?
11:22
So can everybody see that the scale is reversed? All right? Strong acids will have a pKa that's a negative number. All right? Among the weak acids, anything that has a higher pKa is a weaker acid than anything that has a smaller pKa value. All right?
11:40
So that's worth remembering. Now we're going to do the same thing with bases. And so if you guys remember last class, at the end of class, we started looking at bases and we can draw the same kind of parallel between acids and bases. So if you take bases
12:01
we said that an aqueous an aqueous solution is basic if it contains an excess
12:20
of hydroxide ions over hydronium ions so in other words we say that the hydroxide ion concentration has to be greater than the hydronium ion concentration. That's why that solution in water will be basic.
12:41
All right? Two, we said that bases can be strong bases or weak bases. And if we take a strong base, so let's start by looking at strong bases
13:02
so if we take a strong base, all right, we use the same analogy as acids So if you take strong bases, one is that a strong base has Kb
13:20
greater than one. All right? And two we said a strong base dissociates or ionizes completely
13:41
and when we say completely we mean 100%. All right? So if I take an example of a strong base an example of a strong base would be NH2-. All right? So since this is a strong base, and this is going to act as a base in water,
14:07
I can write the equilibrium that would be established as NH2- in water so this is the base, so it has to be the proton acceptor so this is the base
14:22
proton acceptor therefore water is going to be the acid, which is the proton donor we're going to end up with this accepting the proton, so now it becomes NH3 aqueous remember this has a negative charge it's picking up NH+, so it becomes neutral, all right?
14:41
and because this has lost an H+, what you end up with is OH- and because you're producing OH-, now you have an excess of OH- over hydronium ion this would be a basic solution now this is a strong base and the reason it's a strong base is if you look at its Kb value
15:04
this is of the order of 10 to the power 20 so this is an enormously big number and you can see that the Ka value indicates that the equilibrium lies entirely on the product side so when equilibrium is established, you're going to have lots and lots of hydronium ion
15:21
and NH3 formed and very little of the reactants, all right? and so because this is a strong base so let's just go back. This will be the conjugate acid acid and this would be the conjugate base
15:41
and if you look at the pairs, the conjugate acid-base pairs, you can see that the acid form is NH3, the basic form is NH2- and H2O NOH-. So these are the conjugate acid-base pairs in that solution
16:03
and because we are looking at a strong base let's say I start with initial amounts my initial concentrations, let's say, of NH2- is something like .10 molar. Okay? So this is molar
16:22
and if the initial concentrations are .1 molar we're not concerned about the concentration of water we know that to begin with I'm going to have no products being formed now if I allow the system to go to equilibrium, I know it's going to proceed in the forward direction
16:41
so since it's proceeding in the forward direction I know that the change in concentration is going to be negative x plus x plus x but because this is a strong base, and we know that this dissociation is complete, we know that x
17:02
equals .1 molar in other words, all of it dissociates. All right? and so because of that, at equilibrium I know that this is going to be approximately zero and we're going to get all of this converted to .10 molar
17:24
and .10 molar. All right? So because this is a strong base and the dissociation is complete you can see that the hydroxide ion concentration would equal x
17:42
which is .1 molar and therefore pOH this is 10-1 so pOH should be 1 and therefore pH would be 13. So you can calculate because this is a strong base, it dissociates completely
18:05
you can figure out what the hydroxide ion concentration is if you know the hydroxide ion concentration you can figure out what the hydronium ion concentration and you can figure out pH and pOH. All right? now another example of is if you take
18:21
sodium hydroxide, now sodium hydroxide is a salt or an ionic compound and so if I place sodium hydroxide in water, too
18:42
it will dissociate completely to give me Na plus aqueous and OH minus aqueous as well so this is another example hydroxide sodium hydroxide is a strong base as well and unlike previously where we've looked at covalent compounds
19:02
all right where you have a covalent bond being formed broken and formed here, this is an ionic compound and so if you take sodium hydroxide and take this ionic structure and put it in water I said the giant lattice structure breaks apart. It disintegrates and gives you
19:20
Na plus and OH minus as well so once again if you start with an initial let's say I start with 0.1 molar NaOH I'm going to have zero of that and this is going to dissociate
19:41
it's going to proceed in the forward direction so at equilibrium, because the dissociation is complete now this will be zero and you'll end up with 0.1 molar of Na plus and 0.1 molar I'm just going to take it to two significant figures
20:00
so 0.1 molar of hydroxide and now you know in this solution, too the hydroxide ion concentration is going to be 0.1 molar and if you know the hydroxide ion concentration is 0.1 molar, you know that the pH is going to be 13
20:20
and pOH is going to be 1 so these are examples of strong acids sorry, strong bases. All right? now like acids we can also look at now weak bases. So if you take the second example, which are the weak bases
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and if you take weak bases weak bases have Kb less than one. All right? So in weak bases Kb is less than one sorry, the other way around
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all right? So Kb is less than one and two weak bases dissociate only partially
21:21
so let's take an example of a weak base so an example of a weak base would be NH3. All right? So we looked at NH2-, which is a strong base all right, but if you take NH3 NH3 is a weak base
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and if you place NH3 in water the equilibrium that you would establish would be the following. So if I take NH3 aqueous and place this in water it acts as a base. So this would be the base
22:03
and that would be the acid and because it's acting as a base this would be the proton acceptor this will be the proton donor so if NH3 picks up an H+, what would the conjugate acid of NH3 look like? Can you give me the formula of the conjugate acid?
22:25
NH? You have to add one more hydrogen. It would be NH4 you're taking a neutral compound and adding an H+, so what should the charge of that come out to be? Positive. So it's going to end up with NH4+. So this is the base, it's accepting a proton
22:41
and therefore its formula will come out to be NH4+, aqueous plus OH- aqueous. Okay? And this would be the conjugate acid and this would be the conjugate base.
23:01
So if you want to look at the conjugate acid base pairs it would be one, you have the acid form of this would be NH4+, and NH3 and two,
23:21
you would have H2O and OH-. So these are the conjugate acid base pairs. So if you want to see what's going on you start with NH3 which has this Lewis structure All right? So all of you know what the Lewis structure of NH3 looks like
23:41
it's neutral, so we're going to call that B. Now when this picks up a proton it goes to form this that's NH4+, and if you want to write this in the general form you end up with
24:00
BH+. All right? So by picking up a proton ammonia, when it picks up an H+, gives you that. All right? Which is NH4+, and if you look at the steric number it has a steric number of four, it has no lone pairs, so what do you think the shape of this molecule should look like? Tetrahedral. All right? So this would be a tetrahedral molecule.
24:23
Whereas this would be a trigonal pyramid. All right? Okay. So if I take this equilibrium which is NH3, so let's go back to this equilibrium there we are and
24:41
this is Kb because we know that we're writing this where it's the formation of hydroxide ion and ammonia behaves as a base in water and therefore this reaction has to be a base dissociation reaction. And since this is a base dissociation reaction,
25:01
I can say that Kb equals the concentration of NH4+, times hydroxide divided by NH3 gives me the base dissociation constant for this reaction. So this is
25:22
the base dissociation constant or the base ionization constant.
25:40
Now if you take this equation, and we know that this Kb, because this is a weak base, we know that this number has to be less than one. Okay? It's got to be a small number. And that dissociates only partially because the equilibrium lies entirely on the reactant side. You're going to have lots of NH3, but very little NH4 plus being formed. Got it?
26:02
But now we can take this equation that I've just written, and remember we know we have seen that Kw equals hydronium ion concentration times hydroxide ion concentration. All right? Therefore, we know that the hydroxide ion concentration
26:21
is Kw divided by the hydronium ion concentration. So if we want to rearrange that equation that we just wrote on top, but now we want to replace it with terms that represent hydronium. So if I take this term and I'm going to replace
26:40
hydroxide with that, now I have an equation that gives me the values in terms of hydronium rather than hydroxide. So if I go back to that equation, I can say that Kb equals
27:01
NH4 plus times, now I'm going to replace hydroxide with that. So it would be Kw if I replace this with that, I end up with Kw divided by NH3 times hydronium ion. So essentially what I'm doing is I'm replacing the hydroxide
27:23
term not in terms of the concentration of hydroxide, but because I know the relationship between hydroxide and hydronium, I'm going to replace it with a term that represents the hydronium ion concentration. All right? Now if you look at this term so let's leave Kw out, but if you look at this term
27:45
this is actually 1 over Ka. In other words, remember we have our conjugate acid-base pair is what? Our conjugate acid-base pair
28:00
in this solution is we have NH4 plus and NH3. All right? Now if I take the NH4 plus can all of you see that that's the acid part of it? That's the conjugate acid. All right? Now when I take an acid and place it in water
28:22
what happens? So now I'm looking at an acid and if I take the for this conjugate acid-base pair if I take the acid form of that pair and put it in water can somebody tell me what would I expect on the product side? Now this is acting as an acid and how do acids behave?
28:42
What do they produce? H3O+. All right? So can you tell me on the product side what would I have? I have H3O+, what else would I have? NH3. All right? So this is right, taking, remember, these all come in pairs
29:00
you always have a conjugate acid-base pair so if I take the acid form of this conjugate acid-base pair in water this would behave as this H3O+, NH3, aqueous. Now what does this equation describe? If I look at this equilibrium
29:21
how would I describe the equilibrium constant that describes this equilibrium? Ka. Can everybody see that? So this represents Ka. So if I take the acid form of the conjugate acid-base pair, and I take the acid and place the acid in water this equilibrium
29:43
describes Ka. All right? Now let's take, so this is this part this is the acid part now let me take the base part and if I take the base part and if I take a base and place it in water what is the equilibrium I would establish? Now we're looking at a base
30:03
and if you have a base in water can you tell me what are the products of that reaction? Now it's acting as a base. So if it's a base, it has to be a proton acceptor. So you'd end up with what? NH4+, and?
30:22
Hydroxide. Can you see that? So in this conjugate acid-base pair if you put the acid form only in water, you would establish that equilibrium if I take only the base form and place it in water it's going to give me NH4+, plus OH-, aqueous
30:41
and now how would I describe this equilibrium constant that describes this equilibrium? Kb. All right? So this would be Kb. All right? and so that describes this. So now can you see that if I take this equilibrium in the acid form
31:02
this describes H2O. So if I take Ka, I know Ka equals what here? Ka equals hydronium ion concentrations times NH3 divided by NH4+. All right? So if you take the acid form,
31:21
it is H3O+, times NH3 divided by NH4+. Can you see this term is actually this is NH3, H3O+, divided by NH4+. So can you see that if I go back to this equation, I can say Kb
31:41
equals Kw over Ka and therefore Kw equals Ka times Kb. Can you guys see that? All right? So if I write the equilibrium expression for Kb
32:02
and now replace the hydroxide term with the term that describes the hydronium ion concentration then what I can do is I can derive an expression where Ka times Kb will give me Kw. All right? So what you want to keep in mind is that when you take
32:21
for every conjugate acid-base pair you have an acid form which is this that would represent Ka you have a basic form and that would represent Kb. All right? So if I take if I take acetic acid, so let's take another conjugate acid-base pair
32:43
so if I take a conjugate acid-base pair and let's say now I'm looking at the acid form is acetic acid if this is the acid, can you tell me what would the corresponding conjugate
33:00
base turn out to be? what is the the conjugate base of acetic acid? Acetate. And what is the formula of acetate? CH3CO2-. Okay? So it's lost a proton, so it would be this. Now the acid form of
33:22
acetic acid is when it establishes an equilibrium, this would be acetic acid in water now this is the acid, this is the base, so you're going to end up with hydronium ion, all right? Plus acetate anion. So this is
33:40
the equilibrium that will be established if you take the acid part only and place it in water when you do that this would be Ka equals 1.76 times 10 to the negative 5 so that is Ka remember the table that I showed you? you can look up the Ka value for acetic acid
34:02
and it comes out to be 1.76 times 10 to the negative 5 now if I take its conjugate base so now I've taken care of the acid now if I take in this conjugate acid base pair, if I want to look at the base part of it and if I take the basic part of this
34:21
that would be the base in water now since this is acting as a base, this will be the proton acceptor so if the acetate anion accepts a proton what does it become? Acetic acid. Do you see that? Now it's accepted the proton from water
34:40
so the base part of it will accept a proton from water and then go to acetic acid and what else? now the water has lost its proton. So what does it become? OH-. Now since it's producing OH-, this would be the base. All right? and if it's a base
35:01
this is Kb Kb so if we know what Ka is, for the acid form, we can figure out what the Kb for the basic form is because we know that Ka times Kb equals Kw therefore Kb equals Kw over Ka
35:23
which is 1 times 10 to the negative 14 divided by 1.76 times 10 to the negative 5 all right? and so if you put that in your calculator, this number comes out to be 5.68
35:40
times 10 to the negative 10. All right? and so if you know what the Ka of the acid form of this conjugate acid-base pair is you can always figure out what the corresponding Kb is. All right? So this is important because what you will realize is that
36:05
this is the only table that you need. All right? and so if you go look at your textbook, if you go through like handbooks you will never find Kb values listed. All right? The only values that are listed are Ka values.
36:23
So you never find Kb values listed anywhere. and the reason is that if you know what Ka is for the conjugate base, you can always calculate what Kb is. All right? So if you look at this table, we said that these are all listed as conjugate acid-base pairs
36:43
so all the Ka values are given. All right? And so if you know the Ka value, so if you go to just pick an acid. So we just picked acetic acid. So if I go to acetic acid on this list, I will just go look for 10 to the negative 5
37:01
and here we are. This is acetic acid. You have the acid form and you have its conjugate base. Now what you will find in any table anywhere is only the Ka values. But if you need Kb, what do you need to do? Take the Ka value and Kw divided by Ka will give you Kb. All right?
37:23
And similarly for any base, so if I go to hydrochloric acid the Ka value is 10 to the 7. So if I want to figure out what its Kb value is, it'll be Kw divided by that Ka will give me
37:40
Kw. I'm sorry, Kb. Do you understand that? So you will never find Kb values in any table. Now if you're- just like we can rank compounds based on relative acid strengths, we can also rank compounds based on relative base strengths. All right?
38:01
And so when you're ranking on relative base strengths, you have to start with the Ka value and flip it and figure out what Kb is and rank them. So if I take the acids only if I take acids only can everybody see that you start at the top
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Ka values are large strong acids, as you go down, Ka values decrease until you hit here and that is the weakest acid that you can find. The Ka values keep decreasing, but these will have Ka values less than 10 to the negative 14 and therefore
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these won't act as acids in water. Now if I take Kb's, so now if I look at the conjugate bases now these represent bases and therefore they represent Kb values. All right? So can you see that the strongest soul
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now since it's Ka times Kb equals Kw, can everybody see that if Ka is large, what will Kb turn out to be? Small. All right? So in this scale, you can see now everything is reversed. All right? When you take conjugate bases now hydroxide
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will have a Kb of 1 so that yellow line is the cutoff between strong bases and weak bases. So the strong bases will be the ones down here. So these two are considered strong bases. All right? So these are the strong bases
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anything below Kb1 will be a weak base so starting from here as you go up the base gets weaker and weaker and weaker and weaker and weaker until you hit here. So if I take this base this has a Ka value of 10 to the negative 13
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so what does that tell me about the Kb value? It has to be 10 to the negative 1. Do you see that? Now if I take this, this is 10 to the negative 1, so its Kb value will be 10 to the negative 13. So the weakest base would be this and among all of the weaker bases, these are all starting from here to here, these are all weak bases but this will be the stronger of the weaker bases
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and this will be the weakest of the weaker bases now once you hit here now we're in water and so any of these will have Kb values less than 10 to the negative 14. What does that mean? Those substances will not act as a base in water
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because water now is a stronger base. Do you see that? And so water will act as the base in those solutions. Does that make sense to everybody? So it's the same parallel as acids, but now everything has been reversed. And so one of the things that you have to practice
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is that when I ask you to rank these in terms of relative basicities all right? Rather than acids are pretty straightforward because you can just look at that table and figure it out now if you're talking about ranking them in terms of relative bases, you have to be able to flip it around and know how to do that
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now just as we have looked at Kw, Ka, and Pkw and K Kw we can say that we now know that Ka times Kb
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gives us Kw. All right? now we said that these ranges are very broad, and very often in acid-base chemistry, we always like to take the logarithmic scale so that we can compress that scale so we said that if I take the negative log of Ka
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and remember, now I'm moving to the logarithmic scale, so when you have a multiplication in the logarithmic scale, it becomes a plus so now this would be negative log Kb so I'm taking the logarithmic log of this, negative log of that, that would on the other side
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I have to do the same operation, so it would be negative log Kw all right? So that means PKa plus PKb equals PKw which is 1 times 10 to the negative 14 so now, just to make life a little more complicated, not only should you be able
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to rank things based on Ka but now you have to be able to rank them in terms of PKa and remember everything gets flipped around. So let's go back to this table so
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now that we know this now we go to the scale and if we look at acids, we said that if you look at Ka values the numbers decrease as you go down if you look at PKs because you're taking the negative log of that, now this scale is reversed now if you're looking at Ka's
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you know how to figure out Kbs so if you look at Ka's as you go down, and as the Ka values get smaller and smaller and smaller, it gets to be a weaker acid all right? if you take PKs, it's flipped around all right? As the numbers get bigger and bigger and bigger, it means they're
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getting to be a weaker acid. All right? Now, if you go down, we said if Ka decreases that means Kb goes up all right? so as Kb goes up, so the order is reversed so in this table, weak
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strong bases will be at the bottom and weak bases will be at the top. All right? now if you take, so Kb values would decrease as you go down now when you take the negative log of that, that flips it around. So now if you're looking at PKb, PKb increases going down.
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Did you guys get that? So when you're at home, I want you to kind of think about this because we have actually four scales that we're looking at we're looking at Ka for acids and pKas that's one scale if you're looking at the conjugate base our next scale is Kb and pKb
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and you have to be able to see the relationship between everything you should be able to see the relationship between Ka and Kb and you should be able to see the relationship between pKa and pKb. All right? and you should be quickly able to figure out what pKas and pKbs are going to be. All right?
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so if you know that this number is 18 you know that the pKb of this base would be 14-0.18. Did you see that? if you know Ka of acetic acid is 4.75 then the Kb of the acetate anion would be 14-4.75. All right?
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but the most important take-home message is that you have to be able to rank compounds based on this relative scale. All right? So and in the homework in the discussion worksheet for next week
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I have placed some examples where I want you to convert from one to the other and rank these in terms of relative acid-base strengths or given five compounds you should pick out the one that is the strongest acid or pick the one that is the weakest base, and so on. All right? If you're given a list of compounds,
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you should be able to do that. Okay? All right. So now that we've looked at that and you have a feel for this and in fact I would suggest you print this out and pin it on the wall next to where you study and keep looking at this every time. When you're doing homework
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look at that. All right? Because there is a unit in Wiley Plus where they'll ask you to rank relative acids and bases and so on and so you don't memorize this stuff. You have to look at the Ka values. And so on an exam, too, I will attach a copy of this so that you have that to look up as you rank
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acids and bases. All right? Now we still have a few minutes so what I wanted to do is just take one example of some problems. So now we are in a position to work problems and this one is a problem that we worked before
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but I just kind of want to go through the process of how do you calculate the pH of a weak acid. So this problem is I don't want you to copy this down because I put all the examples that I'll be working in class is already on the class website so you can print out these problems
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so I don't want you to spend time copying the problem because you already have a copy of it but I want to show you the general strategy of solving these problems. Okay? So we're told that acetic acid in water donates a single hydrogen ion with a Ka of 1.76x10-5
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So remember these all come as conjugate acid-base pairs what are we starting with? The acid form or its conjugate base form? the acid form therefore it's going to act as an acid in water. So for any equilibrium problem the starting point is writing down the appropriate equilibrium equation
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so they tell us that we are taking acetic acid in water and they tell us that it donates a proton to water to give you hydronium ions and we know that the other product is going to be an acetate anion
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we know that Ka is 1.76x10-5 now, we are told that the initial concentration of acetic acid before anything happens is one mole in one liter. So what would the concentration come out to be? You have one mole in one liter
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so it's going to be one it's given to three significant figures and before anything happens this is what we have. All right? now we need to figure out the pH of this solution if we want to figure out pH, what does pH represent? what does pH represent?
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the hydronium ion concentration so we need to figure out, if we want to figure out pH, we need to figure out what the hydronium ion concentration and that we have to figure out from this so we know that the reaction proceeds in the forward direction and we know that there's we don't need to concern ourselves with the concentration of water and therefore now we follow the usual process
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we know it's going to proceed in the forward direction therefore if x is consumed you're going to form x and x on the product side therefore at equilibrium our concentrations are going to be 1-xx x
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now this number represents what? this number 1-x represents the amount of acetic acid that's left over. Do you see that? it's not the amount of acetic acid that reacted the amount of acetic acid that reacted is what? X. All right? so what we have is, after equilibrium is established
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it's the amount of acetic acid that's left over or is the unreacted part of it. All right? so now I can say Ka equals 1.76 times 10 to the negative 5 equals hydronium ion concentration
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times the acetate ion concentration divided by the concentration of acetic acid. All right? which gives me x squared over 1-x. All right? now can I make an approximation here? what is Ka? 10 to the negative 5.
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so can I ask you guys to figure out at home make the approximation can you calculate what the hydronium ion concentration is? if you know the hydronium ion concentration, can you calculate pH? Yes. and once you calculate that, you can also figure out percent decomposition so at the beginning of class next time we'll go through the answers. All right?