Lecture 08. Acids and Bases. Pt. 5.
This is a modal window.
The media could not be loaded, either because the server or network failed or because the format is not supported.
Formal Metadata
| Title |
| |
| Title of Series | ||
| Part Number | 8 | |
| Number of Parts | 26 | |
| Author | ||
| License | CC Attribution - ShareAlike 3.0 USA: You are free to use, adapt and copy, distribute and transmit the work or content in adapted or unchanged form for any legal purpose as long as the work is attributed to the author in the manner specified by the author or licensor and the work or content is shared also in adapted form only under the conditions of this | |
| Identifiers | 10.5446/18997 (DOI) | |
| Publisher | ||
| Release Date | ||
| Language |
Content Metadata
| Subject Area | ||
| Genre | ||
| Abstract |
|
00:00
SolubilitySaltChemical compoundChemical propertyWaterSodium chlorideChemistrySalt (chemistry)Sea levelMolecularityGroundwaterMaterials scienceLecture/Conference
01:53
VancomycinMan pageMagnetometerMaskierung <Chemie>Hydroxybuttersäure <gamma->Insulin shock therapyCigarDielectric spectroscopyBohriumPedosphärePeriodateMoleculeWaterfallSolventStress (mechanics)SodiumKatalaseChlorideMineralChemical structureElectronic cigaretteQuartzElektrolytische DissoziationBase (chemistry)PHSaltPrecipitation (chemistry)RapidConjugated systemAssetSeleniteMetalFunctional groupSolutionPotassiumWaterAcidWursthülleWeaknessCardiac arrestIce frontSodium chlorideSalt (chemistry)ThermoformingAqueous solutionRiver deltaUreaComputer animation
11:07
Sodium hydrideMonoamine oxidaseVancomycinMan pageTrihalomethaneMagnetometerMixtureWaterAcidMolar volumeConcentrateAmmoniumRiver sourceWursthülleSchmerzschwellePeriodateActive siteSaltAmmoniaConstitutive equationBase (chemistry)WeaknessGasSolutionChemical compoundThermoformingConjugated systemSalt (chemistry)Elektrolytische DissoziationAmmonium chlorideChlorideProtonationElectron donorWalkingWaterfallStoichiometryPHHydroxideSolubilityComputer animation
20:21
Man pageVancomycinCigarHydroxybuttersäure <gamma->AcidWaterConcentrateSystemic therapyTopicityChemistryAcetic acidSodium fluorideBase (chemistry)Shear strengthWeaknessSaltProtonationElectron donorFluorideStress (mechanics)ElektronenakzeptorAqueous solutionInitiationSolutionMolar volumeCalculus (medicine)PHSea levelMeat analoguePitch (resin)AssetMedical historyRiver sourcePeriodateCigaretteElectronic cigaretteStereoselectivityGasWursthüllePotenz <Homöopathie>Setzen <Verfahrenstechnik>Separator (milk)Controller (control theory)SteelSalt (chemistry)HydrogenSodium chlorideOctane ratingCast ironComputer animationLecture/Conference
29:35
Man pageHydroxybuttersäure <gamma->ExonMagmaLavaCigarBohriumVancomycinMixturePan (magazine)Monoamine oxidaseMashingSodium hydrideMagnetometerCommon landSodium fluorideChemical reactionCommon landConcentrateHydrogenSystemic therapyFluorideSolutionHydroxideWursthülleISO-Komplex-HeilweiseCigaretteIronElectronic cigaretteCell (biology)Active siteAreaTool steelRiver sourceMolar volumeElectron donorProtonationInitiationBase (chemistry)Joint (geology)SodiumElektrolytische DissoziationBiomolecular structureThermoformingChemical reactorAmmonium chlorideStress (mechanics)Medical historyWaterAcidAmmoniumChlorideHomöostaseSaltAqueous solutionAmmoniaElektronenakzeptorComputer animationLecture/Conference
38:48
PolyurethaneMan pageSodium hydrideMashingVancomycinLymphangioleiomyomatosisCycloalkaneConcentrateHydrogenElektrolytische DissoziationWursthülleAcidMolar volumeWaterSolutionSodium fluorideFireInitiationSaltAntigenCalculus (medicine)Soil conservationFluorideSchmerzschwelleFaserplatteChemical reactorChemical reactionCommon landWeaknessStress (mechanics)Aqueous solutionComputer animationLecture/Conference
48:02
FirnLecture/Conference
Transcript: English(auto-generated)
00:06
Okay. Can I have everyone's attention please? Let's go ahead and start I wanted to make one announcement and that is that you know that there's a midterm coming up next Wednesday. A week from today
00:22
we have the first midterm exam and in preparation for that midterm exam the chem department tutors will be holding a review on Friday from 5 to 7. All right? and that information is on the class website during the first hour of that review
00:42
I will be there to answer any questions. So so this is a good opportunity if you have any questions please come prepared. So you should have studied the materials, so when you have a review it is a review after you've studied the material rather than, you know, sort of coaching you. All right?
01:01
So please come prepared, it's on Friday from 5 to 7 p.m. and the information is on the class website. Okay? Now, we're going to pick up where we stopped last time, and if you guys remember last time we talked about acid-base properties of salts
01:21
and we said salts are ionic compounds, another word for salt is an ionic compound and we said if you take salts they can be either soluble in water or insoluble in water the majority of them are soluble. All right? and if you take the soluble salts, we said
01:40
the moment you put a soluble salt in water it dissociates completely. So if you take sodium chloride and you put sodium chloride in water the instant you mix it up with water it dissociates completely. So I wanted to show you a molecular level picture of what happens. Okay? So we have sodium chloride
02:05
and let me start that again
02:33
I have the wrong one so I'm going to have to go find the right one for you they have categorized it
02:41
incorrectly
03:18
so just give me a moment to pull it up
03:21
there we are. That's the right one Okay. So I wanted you to take a look at this picture, and what you see at the beginning is a sodium chloride as a crystal. So you can see that it's an ionic structure the yellow one, the green ones are the chloride anions the gray ones are the sodium cations, remember cations are always smaller than anions
03:44
in the ionic solid, they're packed together as a lattice, so you can see the lattice structure and then what happens is when you pour water on top of it you can see the whole structure disintegrating as water molecules wrap themselves around the cation and anion. So you can see in this corner here
04:01
you can see the ionic solid so here are the cations and anions the green ones are the chloride anions the gray ones are the sodium cations and so when you begin with, this is what the ionic solid looks like and we're at room temperature, so you can see that they're vibrating, but
04:21
their location is the same. So they vibrate about a fixed location. Now when you pour water over it, you can see now that the water molecules are attracted to the chloride anions and the sodium cations and so as a result of that the whole structure starts disintegrating and falling apart. So you can see
04:43
that the cations and anions are removed from the lattice structure and the whole structure disintegrates now once you form the sodium chloride solution this is what it looks like so here now after the sodium chloride has dissolved completely, you can see here's a
05:01
chloride anion, all right? surrounded by water molecules and then you can see the sodium ion coming up. There's the sodium ion occasionally they'll come and bump against each other and then they'll go away all right? so what's happened is the water molecules are in such large excess
05:20
that in the end every cation and anion will be surrounded by water molecules and they're separated by great distances because there's so much solvent molecules compared to the sodium chloride ions. Okay? So if you wanted to write down in the form of an equation what we just saw
05:44
so what we looked at was when you take sodium chloride this is a salt and in water we said it dissociates completely to give you Na+, plus Cl-, aqueous
06:01
in reality, if you take Na+, and if you want to represent that cation as a sphere remember they're in the same period so if you compare the size of the cation to the anion, we know that cations tend to be smaller so the smaller one would be the sodium cation and anions tend to be larger
06:21
and so if you take the cation, the cation is positively charged and we know water molecules are polar. So you have delta minus delta plus like that and so it turns out the reason they dissolve is because the attraction to the water molecules
06:40
is greater than their attraction for their counter ions so in the sodium chloride solid what holds it together is electrostatic attractions, but the moment you put it in water the attraction for the cation and anions to water is greater than the attraction for each other and therefore it dissolves and so
07:00
when we this structure is represented as Na+, aqueous. So within parenthesis when we say it's aqueous, it means that it's surrounded by water molecules likewise, you have Cl-, now Cl- is bigger, size-wise and it's negatively charged. So now what happens is that the delta plus end of the water molecule
07:23
will be attracted to the chloride anion and so these water molecules will wrap themselves around the anion and this is the reason why soluble salts dissolve.
07:41
All right? Now we said salts can be insoluble as well. When you look at insoluble salts, it turns out the attraction for each other is far greater than the attraction for water and therefore insoluble salts don't dissolve in water, they just sink to the bottom and we call those precipitates. All right? Right now we're interested in looking at soluble salts
08:00
and we said that soluble salts like sodium chloride can act as an acid base or a neutral substance in water and for you to figure out whether it's going to be an acid base or a neutral substance what you have to look at is the cation and anion that make up that substance, or salt
08:21
and we have rules for that. So if you take the anion we know that if the anion part of it is a conjugate base of a strong acid we said it's going to be neutral if the anion is the conjugate base of a weak acid it will be basic in water. All right? Those are the only two choices you have if you take the cation part of it
08:42
we said if the cation turns out to be a group one metal or a group two metal it's going to be neutral so if it's like sodium, potassium, and so on it's going to be neutral if the cation is the conjugate acid of a weak base then it will act as an acid. All right?
09:04
And so if you look at the table that I, you know, that list of acids and bases that I gave you that is on the class website you can see that anything that's in red, these conjugate bases
09:22
are these in red, these are the conjugate bases so these conjugate bases that are highlighted in red are the conjugate bases of strong acid. So if your anion of your salt is any one of those, you know it's going to be neutral in water. All right? if your anion happens to be
09:41
any one of these conjugate bases that have negative charges okay, so look for the ones that have negative charges. If you look at this urea is neutral. So if it's neutral, it can act as a cation or anion. All right? So leaving that out, if you look at everything else that's negatively charged all of these will act as a base in water.
10:01
All right? If they're counter ion, if the cation is sodium, potassium, or group one metal, all right? then these anions, the salts that these anions form will be basic in water. Okay? Now, if you're talking about conjugate acids of weak bases, if you look at these any one of those that are positively charged here, can you see this one is
10:23
positively charged that means it can act as a cation and therefore in water it will be acidic. So here's one example. And as you go down anything that has a positive charge here this one is the pyridinium ion and this has a positive charge therefore if you place it in water
10:41
this will be a weak acid. All right? Because this is the conjugate acid its base would be this and therefore this can act as a cation. Likewise, NH4 plus is the conjugate acid of the weak base NH3 therefore this can act as a cation.
11:00
So as long as you have this table in front of you, you can identify the ones that will act as conjugate acids in a salt. Because for a salt, you have to have a cation and an anion. The acid form, if it has a positive charge, it can act as a cation and then on the basic side if it has a negative charge
11:21
it can act as a weak base when it's a salt. Okay? So now all we have to do is work some examples of trying to figure out the pH of salts. Okay? So the next part is just calculating the amounts
11:40
of hydronium ion or hydroxide ion. Okay? So here's a question where, once again, remember, one of the kind of things that we've been developing is that you should be able to calculate the pH of any kind of solution. Whether you have a weak acid, strong acid. Whether you have a weak base, strong base. Whether you have a polyprotic acid. Regardless,
12:03
any one of these examples that we've looked at, you should be able to calculate what the pH is. All right? So in this example, we are asked to calculate the pH of a .1 molar ammonium chloride. So your starting point is you look at the substance that's given to you. You ask yourself overall, is that substance a weak acid, strong acid, weak base, strong base,
12:25
polyprotic, or is it a salt? All right? Because those are all the examples that we've looked at. So when you look at that compound, what would you, how would you categorize that? It's a salt. Okay? So that means if it's a salt, you know that
12:41
obviously the ones that they're giving you should be a soluble salt, because if it's if it's not soluble, you can't calculate the pH. Okay? So this is a soluble salt, and you know that if it's a soluble salt, the instant you put it in water, what will happen? It's going to dissociate. So we know that if I take ammonium chloride
13:01
I know that the instant this dissolves in water just like that picture that I showed you it's going to dissociate completely in water. All right? And we know that we started with .1 molar therefore the concentrations, because when they dissociate
13:22
it's a one-to-one stoichiometry every unit of this will give you one unit of NH4 plus and one unit of chloride minus. So if you start with .1 molar of this, you know that when it dissociates you're going to get .1 molar of each of those. Now you look at this, and let's look at the basic form. You know that this
13:40
is the conjugate base of a weak, of a strong acid. All right? Would the conjugate base of a strong acid be acidic, basic, or neutral? Neutral. So we know that this component is not going to affect the pH. Now let's look at the other part, which is the cation.
14:01
Look at this. This is the conjugate acid of a weak base. Is the conjugate acid of a weak base acidic, basic, or neutral? Acidic. All right? So that means that if we want to calculate the pH, we know the component that's going to affect the pH would be NH4+. Right?
14:23
So if it's NH4 that we're going to focus on, and we know that if you want to write the equilibrium, this is the acid form and therefore it's going to act as an acid in water, therefore it's going to be a proton donor. So if I take NH4+, and place it in water, we know that this is the proton donor, this will be the proton acceptor,
14:45
therefore we'll have H3O plus aqueous plus NH3 aqueous and this form of the equation gives us what? Ka. All right? But what do they give us?
15:03
They give us the Kb for NH3. Do you see that? They give us Kb, therefore we have to convert it to figure out what Ka is, and we know that Ka is what? Kw divided by Kb and therefore this would be 1 times 10 to the negative 14
15:23
divided by 1.8 times 10 to the negative 5 which gives me 5.6 times 10 to the negative 10 All right? So that's an extremely weak acid. All right? So the next step is now for us to figure out
15:42
what the hydronium ion concentration is in this equilibrium and that would give us pH. Okay? So how do we set about doing that? We start by saying the initial concentrations all right, in molarity and we know that we start with .01 molar of ammonium, so this would be .01 molar
16:03
we're not concerned about the concentration of water to begin with, we have none of the products being formed, so that would be zero then the change would be it's going to have to proceed in the forward direction because there are no products. We need to form some products
16:21
to establish an equilibrium so we know it's going to proceed in the forward direction therefore some of the reactant is going to be consumed and some product is going to be formed and the stoichiometry is 1 to 1 to 1 therefore at equilibrium I would have .01-x
16:41
x, x so you can see that there's a general pattern to solving all these problems almost every single problem that we've handled, whether it's a weak acid, strong acid, weak acid, weak base polyprotic acid any one of these examples is the same approach. All right? So now what I need to do is I know Ka
17:03
equals 5.6 times 10 to the negative 10 which equals the hydronium ion concentration times the concentration of ammonia divided by the ammonium cation all right? Which gives me x squared over .10
17:20
minus x. Now look at our K value. Our K value is 10 to the negative 10. That's a really really small Ka value so can I make an approximation here? Yes. So if Ka is that small I know that x is going to be really small because it's going to only react to a very small extent. All right?
17:41
So that means that I can make the approximation and my approximation is going to be that .01-x is going to approximate to .01. In other words, that x is going to be so small that I can neglect it. Okay?
18:02
So therefore I can go back and I can say 5.6 times 10 to the negative 10 equals x squared over .1 and so if you solve for x, x comes out to be 7.5 times 10 to the negative 6 molar. All right?
18:23
Now what do I do next? Yes. Don't go ahead and quickly calculate the pH. Not yet. You have to make sure whether your assumption is correct. Remember that that is an assumption that we made and we have to check to see whether that assumption is appropriate. Okay?
18:40
So if I check the validity of my approximation my approximation was what? That x is really small compared to .1, so small I can neglect it. So that means 7.5 times 10 to the negative 6 molar divided by .1 molar times 100
19:01
and that comes out to be 7.5 times 10 to the negative 3%. So that's .007 that's well below our 5% threshold. Okay? So we do know that this approximation is pretty good and so now I can say
19:20
my hydronium ion concentration equals x which is 7.5 times 10 to the negative 6 molar therefore if I take the pH of that that comes out to be 5.13. So I take the negative log of that and
19:41
my pH comes out to be 5.13 Now ask yourself remember our presumption at the beginning was that this is the conjugate acid of a weak base therefore it has to be acidic. All right? So now we've calculated the pH comes out to be around 5 when the solution has a pH of 5, is it acidic or basic? Acidic.
20:02
All right? So we know that it's in line with our prediction. Okay? So any questions up to this point? So we've looked at salts please, when you're studying at home and doing discussion,
20:21
I have given you lots of examples of problems of this type. So please make sure, you remember last time we went to if you're given salts and weak acids, strong acids whether you have the ability to rank them in according to relative acidic strength or basic strength. In other words, can you rank them
20:41
starting from the strongest acid to the weakest base or can you rank them starting from the strongest base to the on the other side the strongest acid. Okay? And when you deal with salts, you've got to separate them out, look at cation and anions if you're just dealing with acids alone, you just look at whether it's a strong acid or weak acid if you're looking at bases alone
21:02
you look at to see whether it's a strong base or a weak base. All right? And you should be able to rank this and this week's discussion worksheet has some examples and so during discussion you'll have an ample opportunity to practice that. Okay? So now I wanted to kind of ask you
21:22
I'm going to quiz you on something all right? So remember so let's say we have a strong acid which is HCl. All right? and the initial concentration of HCl is 1 times 10 to the negative 2. Molar.
21:42
So since this is a strong acid, what is the pH of the solution? pH is 2. All right? Because you don't need a calculator. All right? You know that the pH is 2. All right? So if the initial concentration of a strong acid like HCl
22:01
is 10 to the negative 2, you know that the pH is 2. All right? Now if my initial concentration of acetic acid is 1 times 10 to the negative 9 molar what would the pH of the solution come out to be? 9?
22:20
Now tell me if pH is 9, is that solution acidic, basic, or neutral? Basic. So can you add a strong acid and end up with a solution that is basic? No. So there's a problem here. And the reason is that you're in water. All right? Up to this point, every example that we looked at our concentrations of acid acids were so large that we ignored water.
22:45
So when you have the top one where it's 10 to the negative 2, we know that the hydronium ion that comes from HCl is 10 to the negative 2, but in the background you have water and what is the concentration of hydronium ion in pure water?
23:01
10 to the negative 7. So if you compare 10 to the negative 2, that comes from the strong acid, and the 10 to the negative 7 that's there in pure water, you can see that the 10 to the negative 7 is so small you can ignore it. All right? And so what we focus on is just the 10 to the negative 2, because the difference between 10 to the negative 2
23:20
and the 10 to the negative 7 is an enormous difference, so large that you can ignore it. Now when we go to a strong acid that has something like 10 to the negative 9, now what happens is HCl produces 10 to the negative 9 hydronium ion, but pure water produces what? 10 to the negative 7. Do you see that?
23:42
And so now that's a bigger number than this. So if you take a solution like this, then what do you think the pH of that solution should come out to be? Seven. All right? It's going to be a neutral. All right? Because now the 10 to the negative 7
24:02
is a hundred times greater than 10 to the negative 9. All right? And therefore this solution would have a pH of 7. All right? Now if the difference between 10 to the negative 7 and the acid that you add is large, then you take whichever is large.
24:21
If they're in the approximately similar range, and that is between something like tenfold within. So if you have an acid that's 10 to the negative 7, pure water that's 10 to the negative 7, both are 10 to the negative 7, pH is approximately 7. If you have the acid that you add is 10 to the negative 6 and water is 10 to the negative 7, so there's a tenfold difference,
24:42
then the problem gets to be more complicated because now they're in the same approximate ballpark and now what you have is you have two competing equilibria. All right? And so that means now if you want to figure out what the pH of that solution is, it's going to be a little bit more complex. All right? And at this level of class, we're not going to look at calculations where you have
25:03
two competing equilibria like that, but if you take a more advanced class like analytical chemistry, you will look at two competing equations, two competing equilibria, and figure out what the ultimate pH of that solution is. All right? So whenever you look at it and see a question of this type, make sure that you go
25:21
back and ask yourself is my pH consistent with whether that solution is acidic or basic. All right? And so this is examples where your acid concentration is so dilute that now water has to play an important role. Okay?
25:41
Up to this point, every example that we took, the concentrations of acids were of the order of 1 molar, .1 molar, or .01 molar. So if you're dealing at that concentration range, you can ignore the contribution that comes from hydronium ion. All right? Okay. So
26:01
we have kind of completed looking at chapter 11 of your textbook, where we looked at acid-base chemistries. Okay? Today we're going to move on to another topic that's related to aqueous equilibria and we're going to start looking at application of Le Chatelier's principle.
26:20
Okay? So application of Le Chatelier's principle to acid-base equilibria
26:42
and we call this the common ion effect. So if you recall when we looked at equilibria last quarter, we talked about Le Chatelier's principle and we said we use Le Chatelier's principle
27:02
to figure out how a system responds if you disturb a system, that's at equilibrium. So if you take a system that's at equilibrium and disturb it in some way, we know that ultimately that system is going to go back to equilibrium. And so we can make predictions about how it will respond
27:21
in order to go back to equilibrium. So we're looking at acid-base equilibria, and so let's take an example. So let's say we have a weak acid so we have HF in water all right? So this is a weak acid
27:40
so therefore HF will be the proton donor water will be the proton acceptor so can you tell me what the products of this equilibrium should be? H3O+, and? F-. Okay? So this is the equilibrium so let's say
28:01
we have a system, a weak acid that has established this equilibrium and we know it's a weak acid because Ka equals 7.2 times 10 to the negative 4 okay? So this acts as a weak acid and so let's say this system has established equilibrium. All right?
28:23
After equilibrium is established let's say we add the salt, sodium fluoride so all of you can recognize that what we're adding is a salt because it's got an anion and a cation
28:41
and if this is a soluble salt, what will happen? The instant I add that salt to that solution, that equilibrium, what is going to happen? It's going to dissociate. All right? So this will dissociate to give me Na plus aqueous plus F- aqueous
29:01
So can everybody see that when I add this salt to the system that's at equilibrium I'm actually increasing the fluoride ion concentration and so by increasing the fluoride ion concentration I'm disturbing that equilibrium, or I'm stressing that equilibrium and we know, according to Le Chatelier's principle, if you disturb an equilibrium
29:25
it will react in the direction in which it will counteract that stress so what was the stress on disturbance? I added extra fluoride minus so what is my stress? too much fluoride so the system will respond to counteract that stress
29:40
so that means it has to reduce the amount of fluoride so if you take the equilibrium, there are only two directions in which it can react. It can go in the forward direction, or it can go in the reverse direction so which direction does it have to proceed to overcome that stress? It's got to reduce the fluoride ion concentration it's got to go in the reverse. Because if it goes in the forward, it's going to produce even more fluoride ion. Okay?
30:02
So the system will respond by proceeding in the reverse direction until it goes back to equilibrium. All right? So we say that the system
30:22
will respond by reacting in the reverse direction until it goes back
30:41
to equilibrium and so this shift in equilibrium position so this shift in equilibrium
31:02
that occurs because of the addition of an ion
31:21
already involved in the equilibrium reaction is called the common ion
31:42
effect Okay? So we call this the common ion effect. So we said the shift in equilibrium position that occurs because of the addition of an ion already involved in the equilibrium is called the common ion effect. So if you're ever asked to explain what you mean by the
32:03
common ion effect it describes how the system will respond if you take a system that's at equilibrium and add a common ion to it that's involved in that equilibrium. Okay? So let's take another example so this is an example of a weak acid. So let's take an example
32:22
of a base. A weak base is ammonia so if I take NH3 aqueous in water by now you should know that NH3 is a base. So if I put NH3 in water,
32:40
can you tell me what are the products you're going to see? Now NH3 is the base. It's got to be the proton acceptor and water is going to be the acid, therefore water is the proton donor so can you tell me what the products of that reaction is going to be? NH4+, and OH-. Okay? Great. So
33:00
this would be NH4 plus aqueous and OH-. Okay? and because we're writing this in the form of a base we're looking at Kb and Kb equals 1.8x10-5. So once again we know that this is a weak base.
33:21
So let's say we start with this and we've established an equilibrium. Now, if I add a common ion, so let's say now into this I add a salt and the salt that I'm going to add is ammonium chloride. So now I end up with NH4 plus aqueous plus Cl- aqueous. So we know that when you put a salt
33:41
it dissociates completely to produce cations and anions. And once again you can see that because I'm adding a common ion NH4+, to a system that's already established equilibrium I'm going to disturb that equilibrium.
34:02
And what is my stress? Too much NH4+. And if the NH4 plus has gone up that means the system is going to respond by proceeding in the reverse direction. So that the reaction will shift towards reactants and the net result is
34:23
so if you look at this, the net result is that hydroxide ion concentration decreases decreases and therefore if hydroxide ion concentration decreases what will happen to the pH of that solution?
34:44
pH decreases. Now in the previous example the effect is, the net result is, so if we go back to the previous example that we looked at
35:03
in the previous example, if the reaction proceeds this way all right? What happens to hydronium ion concentration? Because the equilibrium shifts towards reactants, you can see that the hydronium ion concentration goes down. All right?
35:20
So in this example the net result is that hydronium ion concentration decreases and if hydronium ion decreases, we know pH increases. All right? So in each instance not only do you have the equilibrium disturbed and you have this common ion effect
35:41
but this common ion effect actually will affect the pH of that solution because in the first example when the reaction proceeds in the reverse direction you know that the hydronium ion concentration is going to go down if the hydronium ion concentration goes down we know that the pH is going to go up in the second example we're looking at hydroxide
36:01
and so in this case the net result is that it's proceeding this way, so hydroxide ion concentration decreases and therefore the pH would decrease. All right? So not only does this common ion effect disturb that equilibrium, but it also will affect the hydronium ion or hydroxide ion concentration.
36:20
So now we want to calculate that. We want to calculate and figure out how big is this effect? All right? So we're going to try to calculate what happens if you add a common ion and what happens to the pH of a solution. Okay? So to do that remember all of these problems on the class website, so you don't need to copy it
36:44
so here's an example of a problem involving the common ion effect so the question is calculate the pH and the percent dissociation of HF in a solution containing one molar
37:02
and .1 molar sodium fluoride. So you have HF containing one molar of HF and you're adding one molar concentration of sodium fluoride. Okay? So we're adding sodium fluoride, so all of you recognize that because we're
37:22
adding sodium fluoride we know sodium fluoride dissociates to give you the cation and anion and if we started with one molar of sodium fluoride, we know that we end up with one molar of sodium cations and fluoride anions. All right?
37:42
So the equilibrium that we're looking at is how this common ion affects the equilibrium that we're interested in looking at which is HF in water this is a weak acid so it's going to produce hydronium ion
38:01
plus F- and we are told that Ka is 7.2 times 10 to the negative 4 and this is the fluoride ion that's there. So if we wanted to look at initial concentrations
38:24
in molarity we know that we have one molar of HF to begin with we are not concerned with the concentration of water the acid hasn't reacted yet. So we're going to have zero hydronium ion, but
38:40
our fluoride ion concentration is different because we've added fluoride ions from this so can you see that the source that you begin with is fluoride ion concentration and therefore that would be one. So this is the only difference between the previous problems where we just looked at HF in water
39:00
and we figured out what the pH is now what we're looking at is not only do we have HF to begin with, but we're adding a salt that has F-, so before anything happens, before equilibrium is established, we have one molar of the weak acid and we have one molar of the F- ion. All right? So now solving this problem is going to take the same approach that we've done before
39:24
so what we say is now we're going to look at the change since we have no hydronium ion formed yet we know that the reaction has to proceed in the forward direction. All right? Therefore, this is going to be negative x this would be plus x
39:41
this would be plus x therefore at equilibrium I would have 1-x x this would be 1 plus x so can everybody see, the only difference here is now we have F-
40:01
to begin with and that concentration if the reaction is proceeding in the forward direction then the concentration of F- is going to increase. All right? and so now we can solve this like any other equilibrium problem we can say Ka equals 7.2 times 10 to the negative 4
40:21
equals the hydronium ion concentration times the fluoride ion concentration divided by HF and that gives me 1 plus x times x divided by 1 minus x now once again, if you look at Ka
40:42
it's 10 to the negative 4, so it's a small number so it's safe to make an approximation so now I can say my approximation is 1 plus x will approximate to 1 and 1 minus x will approximate to 1 as well
41:01
we're saying x is so small compared to 1 that we can actually neglect it so now I can go back to my equation, I can say 7.2 times 10 to the negative 4 equals 1 times x divided by 1 and therefore x is 7.2 times 10 to the negative 4 molar
41:26
I want to check my validity is my approximation valid? and I said x is really small compared to 1 and you can see that comes out to 7.2 times 10 to the negative 2
41:43
percent or .072 percent and therefore we know that our approximation is pretty good. It's well below our 5% threshold. So now the question was calculate the pH
42:03
and percent dissociation. So now if I calculate the pH I need the hydronium ion concentration and we've calculated that the hydronium ion concentration is x which gives me 7.2 times 10 to the negative 4 molar therefore the pH
42:21
would be the negative log of that and if you work out the pH, pH comes out to be 3.14. All right? and then percent dissociation would be 7.2 times 10 to the negative 4, that's x
42:43
divided by the initial concentration times 100, which is 7.2 times 10 to the negative 2 percent or, as I said before, .072 percent All right? So we've calculated the pH of the solution
43:00
if you have a common ion involved in that. Now what if you want to compare this to if you didn't have the common ion in there? How would you calculate that? So let's say compared to so let's compare this to
43:20
compared to the absence of NaF so let's compare this to the absence of sodium fluoride So if I just had the weak acid, then this would be the kind of problems that we worked before. Remember this is sort of
43:42
the classic problem where what is the pH of a weak acid? You're given one molar of a weak acid, calculate what the pH is you should be able to do this kind of in your sleep. So what do I do? Write the equation that gives me H3O plus aqueous plus F minus aqueous
44:02
and I know Ka is 7.2x10-4 So now I don't have any common ion in there so my initial concentration would be just the one molar of HF I'm not concerned about water since I don't have any added salt
44:21
my fluoride ion concentration to begin would be zero therefore the change would be minus X plus X plus X therefore at equilibrium I would have 1-x x x so if I were
44:42
to calculate the value of x, we can say Ka equals 7.2x10-4 which is the hydronium ion concentration times the fluoride ion concentration divided by the concentration of HF which gives me x squared over 1-x once again I can safely make that approximation so I can say my approximation
45:04
in this problem would be 1-x will approximate to 1 so I can go back and I can say 7.2x10-4 equals x squared over 1 and therefore
45:23
if you work that out it comes out to be 2.7 so x equals 2.7 times 10 to the negative 2 molar. All right? and if you check validity
45:41
you can see that it's 2.7 times 10 to the negative 2 divided by one molar times 100 which gives me 2.17% so once again we're okay that number is really small, we can ignore it and therefore now I can say
46:01
my hydronium ion concentration would be 2.7 times 10 to the negative 2 molar if that were the case pH would come out to be 1.57. All right? and if I want to calculate what the percent dissociation is
46:23
it's the amount that dissociated, which is 2.7 times 10 to the negative 2 molar divided by one molar times 100 which is 2.7% so I want you now to compare this with the previous problem. So in this case, if
46:41
I had no salt in there, my pH would be 1.57 if I had a salt, what does the pH become? 3.14 so you can see that that illustrates that added salt will shift the equilibrium in the reverse direction therefore the hydronium ion concentration goes down so you can see that the hydronium ion concentration to begin with
47:04
without any salt will be 2.7 times 10 to the negative 2 with the salt will be 7.2 times 10 to the negative 4. All right? so you can see that this has this has, without the salt, you have a higher hydronium ion concentration
47:21
with the salt, you have a lower if you take the pH the pH will be lower it's 1.57 you can see adding the salt raises the pH to 3.14 if you look at the percent dissociation without the salt, it's 2.7% dissociates if you add the salt, it goes to .072%
47:43
so adding the salt that is what is called the common ion effect it affects not only the pH of the solution because we said if you disturb an equilibrium by adding a common ion it's going to shift in the direction in which it counteracts that stress and so that net result is what you see here. Okay?
48:03
So we're going to stop there for today and next class we'll start looking at buffers. All right?