Lecture 08. Acids and Bases. Pt. 5.
Arasasingham, Ramesh D.
 UCI Chem 1C is the third and final quarter of General Chemistry series and covers the following topics: equilibria, aqueous acid-base equilibria, solubility equilibria, oxidation reduction reactions, electrochemistry; kinetics; special topics. 
 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 we have the first midterm exam. But in preparation for that midterm exam, the chem department tutors will be holding a review on Friday from 5 to 7. And that information is on the class website. During the first hour of that review, I will be there to answer any questions. So this is a good opportunity if you have any questions. Please come prepared. 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. So please come prepared. It's on Friday from 5 to 7 p.m. and the information is on the class website. Now we're going to pick up where we stopped last time. And if you guys remember, last time we talked about acid-based properties of salts. 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? So if you take the soluble salts, we said 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 in water, it dissociates completely. So I want to show you a molecular level picture of what happens. So we have sodium chloride. Let me start that again. I have the wrong one. So I'm going to have to go find the right one for you. They have categorized it incorrectly. So just give me a moment to pull it up. There we are. That's the right one. Okay, so I wanted 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 green ones are the chloride anions. The gray ones are the sodium cations. Remember cations are always smaller than anions. 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, 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. All right? And we're at room temperature. So you can see that they're vibrating, but 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 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 chloride anion surrounded by water molecules. And then you can see the sodium ion coming up there, the sodium ion. Occasionally they'll come and bump against each other and then they'll go away. So what's happened is the water molecules are in such large excess 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. So if you wanted to write down in the form of an equation what we just saw, 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 plus Cl minus aqueous. In reality if you take Na plus 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. 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 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 when we, this structure is represented as Na plus aqueous. So within parenthesis when we say it's aqueous, it means that it's surrounded by water molecules. Likewise you have CO minus. Now CO minus is bigger size wise and it's negatively charged. So now what happens is that the delta plus end of the water molecule 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. Now we said salts can be insoluble as well. When you look at insoluble salts it turns out their attraction for each other is far greater than their attraction for water and therefore insoluble salts don't dissolve in water. They sink to the bottom and we call those precipitates. Right now we're interested in looking at soluble salts. 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. 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 we said if the cation turns out to be a group one metal or group two metal it's going to be neutral. 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? And so if you look at the table that I, you know, that list of acids and bases that I give you that is on the class website. You can see that anything that's in red, these conjugate bases, 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 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. All right? If they're counter ion, if the cation is sodium, potassium, or group one metal, all right? And 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 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 puridinium ion and this has a positive charge therefore if you place it in water 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. All right? 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 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 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 any one of these examples that we've looked at you should be able to calculate where the pH is, all right? So in this example we are asked to calculate the pH of a 0.1 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, 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 obviously the ones that they're giving you should be a soluble salt because 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 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 0.1 molar therefore the concentration is because when they dissociate it's a one-to-one stud geometry. Every unit of this will give you one unit of NH4 plus and one unit of chloride minus. So if you start with 0.1 molar of this you know that when it dissociates you're going to get 0.1 molar of each of those. Now you look at this and let's look at the basic form. You know that this 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. 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 plus, right? 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 plus and place it in water we know that this is the proton donor. This will be the proton acceptor. 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? 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 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 what the hydrodynamic 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 0.01 molar of ammonium. So this would be 0.01 molar. We're not concerned about the concentration of water. To begin with we have none of the products being formed so that would be 0. 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 to establish an equilibrium so we know it's going to proceed in the forward direction. Therefore some of the reactants are 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 0.01 minus x, 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 with it's a weak acid, strong 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 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 0.10 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? So that means that I can make the approximation and my approximation is going to be that 0.01 minus x is going to approximate to 0.01. In other words, that x is going to be so small that I can neglect it, okay? So therefore I can go back and I can say 5.6 times 10 to the negative 10 equals x squared over 0.1. And so if you solve for x, x comes out to be 7.5 times 10 to the negative 6 molar, all right? Now what do I do next? Yes, don't go ahead and quickly calculate the pH. Not yet. You have to make sure whether the assumption is correct. Remember that is an assumption that we made and we have to check to see whether that assumption is appropriate, okay? So if I check the validity of my approximation, my approximation was what? That x is really small compared to 0.1. So small I can neglect it. So that means 7.5 times 10 to the negative 6 molar divided by 0.1 molar times 100 and that comes out to be 7.5 times 10 to the negative 3 percent. So that's 0.007. That's well below our 5 percent threshold, okay? So we do know that this approximation is pretty good. And so now I can say 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 you take the negative log of that and 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, alright? 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. Alright? 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 during discussion, 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 according to relative acidic strength or basic strength. In other words, can you rank them starting from the strongest acid to the weakest base or can you rank them starting from the strongest base to, on the other side, the strongest acid, okay? And when you deal with salts, you've got to separate them out and 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, you look at to see whether it's a strong base or a weak base, alright? 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, I'm going to quiz you on something, alright? Remember, so let's say we have a strong acid, which is HCl, alright? And the initial concentration of HCl is 1 times 10 to the negative 2 molar. Second, if this, since this is a strong acid, what is the pH of the solution? PAH is 2, alright? Because you don't need a calculator, alright? You know that the pH is 2, alright? So the initial concentration of a strong acid like HCl is 10 to the negative 2, you know that the pH is 2, alright? Now, if my initial concentration of acidic acid is 1 times 10 to the negative 9 molar, what would the pH of the solution come out to be? 9? Now, tell me, if pH is 9, is that solution acidic, basic or neutral? Basic. Basic. So a strong acid and end up with a solution that is basic? No. No. So there's a problem here. And the reason is that you're in water, alright? Up to this point, every example that we looked at are concentrations of acidic acids were so large that we ignored water. 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? 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, alright? And so what we focus on is just the 10 to the negative 2 because the difference between 10 to the negative 2 and the 10 to the negative 7 is an enormous difference so large that you can ignore it, alright? 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? 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? 7. Alright, it's going to be a neutral, alright? Because now the 10 to the negative 7 is 100 times greater than 10 to the negative 9, alright? And therefore this solution would have a pH of 7, alright? Now if the difference between 10 to the negative 7 and the acid that you add is large, then you take whichever is large. If they're in the approximately similar range, and that is between something like 10 fold within, so if you have an acid that's 10 to the negative 7, pure water that's 10 to the negative 7, both that's 10 to the negative 7, pH is going to be 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 10 fold difference, 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, alright? And so that means now when you want to figure out what the pH of that solution is, it's going to be a little bit more complex, alright? And at this level of class, we're not going to look at calculations where you have 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, alright? So whenever you look at it, see a question of this type, make sure that you go back and ask yourself, is my pH consistent with whether that solution is acidic or basic, alright? And so this is examples where your acid concentration is so dilute that now water has to play an important role, okay? Up to this point, every example that we took, the concentrations were acids were of the order of 1 molar, 0.1 molar, or 0.01 molar. So if you're dealing at that concentration range, you can ignore the contribution that comes from hydronium ion, alright? Okay, so we have kind of completed looking at chapter 11 of your textbook where we looked at acid-base chemistry, 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, okay? So application of Le Chatelier's principle to acid-base equilibria, and we call this the common ion effect, alright? 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 to figure out how a system responds if you disturb a system that's an equilibrium. So if you take a system that's an equilibrium and disturb it in some way, we know that ultramarine ultimately, that system is going to go back to equilibrium. And so we can make predictions about how it will respond in order to go back to equilibrium, okay? 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, alright? So this is a weak acid. 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 plus and F minus, okay? So this is the equilibrium. So let's say 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, alright? 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 anacetion. And if this is a soluble salt, what will happen? So instead I add that salt to that solution, that equilibrium, what is going to happen? It's going to dissociate, alright? So this will dissociate to give me Na plus aqueous plus F minus aqueous. 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, it will react in the direction in which it'll counteract that stress. So what was the stress or disturbance? I added extra fluoride minus. So what is my stress? Too much fluoride. So the system will respond to counteract that stress, so that means it has to reduce the amount of fluoride, okay? 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? 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 will respond by reacting in the reverse direction until it goes back to equilibrium, all right? And so this shift in equilibrium position, so this shift in equilibrium that occurs because of the addition of an ion already involved in the equilibrium reaction is called the common ion 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 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 of a base. A weak base is ammonia. So if I take NH3 equals in water, by now you should know that NH3 is a base. So if I put NH3 in water, 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 plus and OH minus, okay? Great. So this would be NH4 plus equals and OH minus, okay? And because we're writing this in the form of a base, we're looking at KB and KB equals 1.8 times 10 to the negative 5. So once again we know that this is a weak base. 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 equals plus Cl minus equals. So we know that when you put a salt, it dissociates completely to produce cations and anions. And once again you can see that because I'm adding a common ion, NH4 plus to a system that's already established equilibrium, I'm going to disturb that equilibrium. And what is my stress? Too much NH4 plus. And if the NH4 plus has gone up, that means the system is going to respond by proceeding in the reverse direction. So the reaction will shift towards reactants. And the net result is, so if you look at this, the net result is that hydroxide ion concentration decreases and therefore if hydroxide ion concentration decreases, what will happen to the pH of that solution? 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, in the previous example if the reaction proceeds this way, what happens to hydronium ion concentration? Because the equilibrium shifts towards reactants, you can see that the hydronium ion concentration goes down. So in this example, the net result is that hydronium ion concentration decreases. And if hydronium ion decreases, we know pH increases. So in each instance, not only do you have the equilibrium disturbed and you have this common ion effect, 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. 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. So not only does this common ion effect disturb that equilibrium, but it also will affect the hydronium ion or hydroxide ion concentration. So now we want to calculate that. We want to calculate and figure out how big is this effect. 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. So to do that, remember all of these problems on the class website, so you don't need to copy it. 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 and 0.1 molar sodium fluoride. So you have HF containing one molar of HF and you're adding one molar concentration of sodium fluoride. So we're adding sodium fluoride, so all of you recognize that because we're 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. 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 plus F minus and we are told that Ka is 7.2 times 10 to the negative 4. All right, and this is the fluoride ion that's there. So if we wanted to look at initial concentrations 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 a fluoride ion concentration is different because we've added fluoride ions from this. So can you see that the source to begin with, there's fluoride ion concentration and therefore that would be one. So this is the only difference between the previous problems where we just look at HF in water 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 minus, so before anything happens, before equilibrium is established, we have one molar of the weak acid and we have one molar of the F minus ion. So now solving this problem is going to take the same approach that we've done before. 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. Therefore this is going to be negative X, this would be plus X, this would be plus X, therefore at equilibrium I would have one minus X, X, this would be one plus X. So can everybody see, the only difference here is now we have F minus to begin with and that concentration of, if the reaction is proceeding in the forward direction, then the concentration of F minus is going to increase. And so now we can solve this like any other equilibrium problem. We can say Ka equals 7.2 times 10 to negative 4 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, 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. 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. 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 percent or 0.072 percent. All right and therefore we know that our approximation is pretty good. It's well below our 5 percent threshold. So now the question was calculate the pH 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. So the pH would be the negative log of that and if you work out the pH, pH comes out to be 3.14. And then percent dissociation would be 7.2 times 10 to the negative 4 that's X divided by the initial concentration times 100 which is 7.2 times 10 to the negative 2 percent or as I said before 0.072 percent. All right so we calculated the pH of the solution if you have a common ion involved in that. Okay 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 compared to the absence of an ath. 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. This is sort of 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 and I know Ka is 7.2 times 10 to the negative 4. So now I don't have any common ion in there. So my initial concentration would be just the one molar of HF, not considered about water since I don't have any added salt. 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 one minus X, X, X. Alright, so if I were to calculate the value of X we can say Ka equals 7.2 times 10 to the negative 4 which is the hydronium ion concentration times the fluoride ion concentration divided by the concentration of HF which gives me X squared over one minus X. Once again I can safely make that approximation so I can say my approximation in this problem would be one minus X will approximate to one so I can go back and I can say 7.2 times 10 to the negative 4 equals X squared over one and therefore if you work that out it comes out to be 2.7 so X equals 2.7 times 10 to the negative 2 molar. And if you check validity 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 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. Alright and if I want to calculate what the percent dissociation is 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 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 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. So you can see that this has without the salt you have a higher hydronium ion concentration 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 0.072%. So adding the salt that is what is called a 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 so we're going to stop there for today and next class we'll start looking at buffers.
 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 we have the first midterm exam.  But in preparation for that midterm exam, the chem department tutors will be holding  a review on Friday from 5 to 7.  And that information is on the class website.  During the first hour of that review, I will be there to answer any questions.  So this is a good opportunity if you have any questions.  Please come prepared.  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.  So please come prepared.  It's on Friday from 5 to 7 p.m. and the information is on the class website.  Now we're going to pick up where we stopped last time.  And if you guys remember, last time we talked about acid-based properties of salts.  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?  So if you take the soluble salts, we said 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 in water, it dissociates completely.  So I want to show you a molecular level picture of what happens.  So we have sodium chloride.  Let me start that again.  I have the wrong one.  So I'm going to have to go find the right one for you.  They have categorized it incorrectly.  So just give me a moment to pull it up.  There we are.  That's the right one.  Okay, so I wanted 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 green ones are the chloride anions.  The gray ones are the sodium cations.  Remember cations are always smaller than anions.  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, 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.  All right?  And we're at room temperature.  So you can see that they're vibrating, but 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 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  chloride anion surrounded by water molecules.  And then you can see the sodium ion coming up there, the sodium ion.  Occasionally they'll come and bump against each other and then they'll go away.  So what's happened is the water molecules are in such large excess 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.  So if you wanted to write down in the form of an equation what we just saw, 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 plus Cl minus aqueous.  In reality if you take Na plus 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.  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  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 when we, this structure is represented as Na plus aqueous.  So within parenthesis when we say it's aqueous, it means that it's surrounded by water molecules.  Likewise you have CO minus.  Now CO minus is bigger size wise and it's negatively charged.  So now what happens is that the delta plus end of the water molecule 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.  Now we said salts can be insoluble as well.  When you look at insoluble salts it turns out their attraction for each other is far  greater than their attraction for water and therefore insoluble salts don't dissolve  in water.  They sink to the bottom and we call those precipitates.  Right now we're interested in looking at soluble salts.  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.  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 we said if the cation turns out to be a group one  metal or group two metal it's going to be neutral.  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?  And so if you look at the table that I, you know, that list of acids and bases that I  give you that is on the class website.  You can see that anything that's in red, these conjugate bases, 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 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.  All right?  If they're counter ion, if the cation is sodium, potassium, or group one metal, all right?  And 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 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 puridinium ion and this has a positive charge therefore if you place it  in water 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.  All right?  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 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 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 any one of these examples that we've looked at you  should be able to calculate where the pH is, all right?  So in this example we are asked to calculate the pH of a 0.1 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, 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 obviously the ones that they're giving you  should be a soluble salt because 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 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 0.1 molar therefore the concentration is because when  they dissociate it's a one-to-one stud geometry.  Every unit of this will give you one unit of NH4 plus and one unit of chloride minus.  So if you start with 0.1 molar of this you know that when it dissociates you're going  to get 0.1 molar of each of those.  Now you look at this and let's look at the basic form.  You know that this 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.  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 plus, right?  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 plus and place it in water we know that this is the proton donor.  This will be the proton acceptor.  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?  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 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 what the hydrodynamic 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 0.01 molar of ammonium.  So this would be 0.01 molar.  We're not concerned about the concentration of water.  To begin with we have none of the products being formed so that would be 0.  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 to establish an equilibrium so we know it's going to proceed  in the forward direction.  Therefore some of the reactants are 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 0.01 minus x, 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 with it's a weak acid, strong 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 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 0.10 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?  So that means that I can make the approximation and my approximation is going to be that 0.01  minus x is going to approximate to 0.01.  In other words, that x is going to be so small that I can neglect it, okay?  So therefore I can go back and I can say 5.6 times 10 to the negative 10 equals x squared  over 0.1.  And so if you solve for x, x comes out to be 7.5 times 10 to the negative 6 molar, all  right?  Now what do I do next?  Yes, don't go ahead and quickly calculate the pH.  Not yet.  You have to make sure whether the assumption is correct.  Remember that is an assumption that we made and we have to check to see whether that assumption  is appropriate, okay?  So if I check the validity of my approximation, my approximation was what?  That x is really small compared to 0.1.  So small I can neglect it.  So that means 7.5 times 10 to the negative 6 molar divided by 0.1 molar times 100 and  that comes out to be 7.5 times 10 to the negative 3 percent.  So that's 0.007.  That's well below our 5 percent threshold, okay?  So we do know that this approximation is pretty good.  And so now I can say 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 you take the negative log of that and 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, alright?  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.  Alright?  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 during discussion, 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 according to relative  acidic strength or basic strength.  In other words, can you rank them starting from the strongest acid to the weakest base  or can you rank them starting from the strongest base to, on the other side, the strongest  acid, okay?  And when you deal with salts, you've got to separate them out and 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, you look at to see whether it's a strong base or a  weak base, alright?  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, I'm going to quiz you on something, alright?  Remember, so let's say we have a strong acid, which is HCl, alright?  And the initial concentration of HCl is 1 times 10 to the negative 2 molar.  Second, if this, since this is a strong acid, what is the pH of the solution?  PAH is 2, alright?  Because you don't need a calculator, alright?  You know that the pH is 2, alright?  So the initial concentration of a strong acid like HCl is 10 to the negative 2, you know  that the pH is 2, alright?  Now, if my initial concentration of acidic acid is 1 times 10 to the negative 9 molar,  what would the pH of the solution come out to be?  9?  Now, tell me, if pH is 9, is that solution acidic, basic or neutral?  Basic.  Basic.  So a strong acid and end up with a solution that is basic?  No.  No.  So there's a problem here.  And the reason is that you're in water, alright?  Up to this point, every example that we looked at are concentrations of acidic acids were  so large that we ignored water.  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?  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, alright?  And so what we focus on is just the 10 to the negative 2 because the difference between  10 to the negative 2 and the 10 to the negative 7 is an enormous difference so large that  you can ignore it, alright?  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?  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?  7.  Alright, it's going to be a neutral, alright?  Because now the 10 to the negative 7 is 100 times greater than 10 to the negative 9, alright?  And therefore this solution would have a pH of 7, alright?  Now if the difference between 10 to the negative 7 and the acid that you add is large, then  you take whichever is large.  If they're in the approximately similar range, and that is between something like 10  fold within, so if you have an acid that's 10 to the negative 7, pure water that's 10  to the negative 7, both that's 10 to the negative 7, pH is going to be 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 10 fold difference, 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, alright?  And so that means now when you want to figure out what the pH of that solution is, it's  going to be a little bit more complex, alright?  And at this level of class, we're not going to look at calculations where you have 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, alright?  So whenever you look at it, see a question of this type, make sure that you go back and  ask yourself, is my pH consistent with whether that solution is acidic or basic, alright?  And so this is examples where your acid concentration is so dilute that now water has to play an  important role, okay?  Up to this point, every example that we took, the concentrations were acids were of the  order of 1 molar, 0.1 molar, or 0.01 molar.  So if you're dealing at that concentration range, you can ignore the contribution that  comes from hydronium ion, alright?  Okay, so we have kind of completed looking at chapter 11 of your textbook where we looked  at acid-base chemistry, 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, okay?  So application of Le Chatelier's principle to acid-base equilibria, and we call this  the common ion effect, alright?  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 to figure out how a system responds  if you disturb a system that's an equilibrium.  So if you take a system that's an equilibrium and disturb it in some way, we know that ultramarine  ultimately, that system is going to go back to equilibrium.  And so we can make predictions about how it will respond in order to go back to equilibrium,  okay?  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, alright?  So this is a weak acid.  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 plus and F minus, okay?  So this is the equilibrium.  So let's say 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, alright?  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 anacetion.  And if this is a soluble salt, what will happen?  So instead I add that salt to that solution, that equilibrium, what is going to happen?  It's going to dissociate, alright?  So this will dissociate to give me Na plus aqueous plus F minus aqueous.  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, it will react  in the direction in which it'll counteract that stress.  So what was the stress or disturbance?  I added extra fluoride minus.  So what is my stress?  Too much fluoride.  So the system will respond to counteract that stress, so that means it has to reduce the  amount of fluoride, okay?  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?  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 will respond by reacting in the reverse direction until it  goes back to equilibrium, all right?  And so this shift in equilibrium position, so this shift in equilibrium that occurs  because of the addition of an ion already involved in the equilibrium  reaction is called the common ion 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 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 of a base.  A weak base is ammonia.  So if I take NH3 equals in water, by now you should know that NH3 is a base.  So if I put NH3 in water, 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 plus and OH minus, okay?  Great.  So this would be NH4 plus equals and OH minus, okay?  And because we're writing this in the form of a base, we're looking at KB and KB equals  1.8 times 10 to the negative 5.  So once again we know that this is a weak base.  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 equals plus Cl minus equals.  So we know that when you put a salt, it dissociates completely to produce cations and anions.  And once again you can see that because I'm adding a common ion, NH4 plus to a system  that's already established equilibrium, I'm going to disturb that equilibrium.  And what is my stress?  Too much NH4 plus.  And if the NH4 plus has gone up, that means the system is going to respond by proceeding  in the reverse direction.  So the reaction will shift towards reactants.  And the net result is, so if you look at this, the net result is that hydroxide ion concentration  decreases and therefore if hydroxide ion concentration decreases, what will happen to the pH of that  solution?  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, in the previous example if the reaction proceeds  this way, what happens to hydronium ion concentration?  Because the equilibrium shifts towards reactants, you can see that the hydronium ion concentration  goes down.  So in this example, the net result is that hydronium ion concentration decreases.  And if hydronium ion decreases, we know pH increases.  So in each instance, not only do you have the equilibrium disturbed and you have this  common ion effect, 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.  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.  So not only does this common ion effect disturb that equilibrium, but it also will affect  the hydronium ion or hydroxide ion concentration.  So now we want to calculate that.  We want to calculate and figure out how big is this effect.  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.  So to do that, remember all of these problems on the class website, so you don't need to  copy it.  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 and 0.1 molar sodium fluoride.  So you have HF containing one molar of HF and you're adding one molar concentration  of sodium fluoride.  So we're adding sodium fluoride, so all of you recognize that because we're 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.  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 plus F minus and we are told  that Ka is 7.2 times 10 to the negative 4.  All right, and this is the fluoride ion that's there.  So if we wanted to look at initial concentrations 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 a fluoride  ion concentration is different because we've added fluoride ions from this.  So can you see that the source to begin with, there's fluoride ion concentration and therefore  that would be one.  So this is the only difference between the previous problems where we just look at HF  in water 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 minus, so before anything happens, before equilibrium is established,  we have one molar of the weak acid and we have one molar of the F minus ion.  So now solving this problem is going to take the same approach that we've done before.  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.  Therefore this is going to be negative X, this would be plus X, this would be plus X,  therefore at equilibrium I would have one minus X, X, this would be one plus X.  So can everybody see, the only difference here is now we have F minus to begin with and  that concentration of, if the reaction is proceeding in the forward direction, then  the concentration of F minus is going to increase.  And so now we can solve this like any other equilibrium problem.  We can say Ka equals 7.2 times 10 to negative 4 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, 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.  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.  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 percent or 0.072 percent.  All right and therefore we know that our approximation is pretty good.  It's well below our 5 percent threshold.  So now the question was calculate the pH 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.  So the pH would be the negative log of that and if you work out the pH, pH comes out to  be 3.14.  And then percent dissociation would be 7.2 times 10 to the negative 4 that's X divided  by the initial concentration times 100 which is 7.2 times 10 to the negative 2 percent  or as I said before 0.072 percent.  All right so we calculated the pH of the solution if you have a common ion involved in that.  Okay 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 compared to the absence of an ath.  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.  This is sort of 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 and I know Ka is 7.2 times 10 to the  negative 4.  So now I don't have any common ion in there.  So my initial concentration would be just the one molar of HF, not considered about  water since I don't have any added salt.  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 one minus X, X, X.  Alright, so if I were to calculate the value of X we can say Ka equals 7.2 times 10 to the  negative 4 which is the hydronium ion concentration times the fluoride ion concentration divided  by the concentration of HF which gives me X squared over one minus X.  Once again I can safely make that approximation so I can say my approximation in this problem  would be one minus X will approximate to one so I can go back and I can say 7.2 times 10  to the negative 4 equals X squared over one and therefore if you work that out it comes  out to be 2.7 so X equals 2.7 times 10 to the negative 2 molar.  And if you check validity 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 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.  Alright and if I want to calculate what the percent dissociation is 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 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 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.  So you can see that this has without the salt you have a higher hydronium ion concentration  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 0.072%.  So adding the salt that is what is called a 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 so we're going to stop there for today and next class we'll start looking at buffers.
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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 we have the first midterm exam. But in preparation for that midterm exam, the chem department tutors will be holding a review on Friday from 5 to 7. And that information is on the class website. During the first hour of that review, I will be there to answer any questions. So this is a good opportunity if you have any questions. Please come prepared. 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. So please come prepared. It's on Friday from 5 to 7 p.m. and the information is on the class website. Now we're going to pick up where we stopped last time. And if you guys remember, last time we talked about acid-based properties of salts. 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? So if you take the soluble salts, we said 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 in water, it dissociates completely. So I want to show you a molecular level picture of what happens. So we have sodium chloride. Let me start that again. I have the wrong one. So I'm going to have to go find the right one for you. They have categorized it incorrectly. So just give me a moment to pull it up. There we are. That's the right one. Okay, so I wanted 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 green ones are the chloride anions. The gray ones are the sodium cations. Remember cations are always smaller than anions. 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, 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. All right? And we're at room temperature. So you can see that they're vibrating, but 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 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 chloride anion surrounded by water molecules. And then you can see the sodium ion coming up there, the sodium ion. Occasionally they'll come and bump against each other and then they'll go away. So what's happened is the water molecules are in such large excess 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. So if you wanted to write down in the form of an equation what we just saw, 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 plus Cl minus aqueous. In reality if you take Na plus 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. 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 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 when we, this structure is represented as Na plus aqueous. So within parenthesis when we say it's aqueous, it means that it's surrounded by water molecules. Likewise you have CO minus. Now CO minus is bigger size wise and it's negatively charged. So now what happens is that the delta plus end of the water molecule 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. Now we said salts can be insoluble as well. When you look at insoluble salts it turns out their attraction for each other is far greater than their attraction for water and therefore insoluble salts don't dissolve in water. They sink to the bottom and we call those precipitates. Right now we're interested in looking at soluble salts. 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. 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 we said if the cation turns out to be a group one metal or group two metal it's going to be neutral. 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? And so if you look at the table that I, you know, that list of acids and bases that I give you that is on the class website. You can see that anything that's in red, these conjugate bases, 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 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. All right? If they're counter ion, if the cation is sodium, potassium, or group one metal, all right? And 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 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 puridinium ion and this has a positive charge therefore if you place it in water 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. All right? 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 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 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 any one of these examples that we've looked at you should be able to calculate where the pH is, all right? So in this example we are asked to calculate the pH of a 0.1 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, 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 obviously the ones that they're giving you should be a soluble salt because 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 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 0.1 molar therefore the concentration is because when they dissociate it's a one-to-one stud geometry. Every unit of this will give you one unit of NH4 plus and one unit of chloride minus. So if you start with 0.1 molar of this you know that when it dissociates you're going to get 0.1 molar of each of those. Now you look at this and let's look at the basic form. You know that this 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. 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 plus, right? 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 plus and place it in water we know that this is the proton donor. This will be the proton acceptor. 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? 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 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 what the hydrodynamic 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 0.01 molar of ammonium. So this would be 0.01 molar. We're not concerned about the concentration of water. To begin with we have none of the products being formed so that would be 0. 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 to establish an equilibrium so we know it's going to proceed in the forward direction. Therefore some of the reactants are 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 0.01 minus x, 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 with it's a weak acid, strong 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 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 0.10 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? So that means that I can make the approximation and my approximation is going to be that 0.01 minus x is going to approximate to 0.01. In other words, that x is going to be so small that I can neglect it, okay? So therefore I can go back and I can say 5.6 times 10 to the negative 10 equals x squared over 0.1. And so if you solve for x, x comes out to be 7.5 times 10 to the negative 6 molar, all right? Now what do I do next? Yes, don't go ahead and quickly calculate the pH. Not yet. You have to make sure whether the assumption is correct. Remember that is an assumption that we made and we have to check to see whether that assumption is appropriate, okay? So if I check the validity of my approximation, my approximation was what? That x is really small compared to 0.1. So small I can neglect it. So that means 7.5 times 10 to the negative 6 molar divided by 0.1 molar times 100 and that comes out to be 7.5 times 10 to the negative 3 percent. So that's 0.007. That's well below our 5 percent threshold, okay? So we do know that this approximation is pretty good. And so now I can say 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 you take the negative log of that and 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, alright? 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. Alright? 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 during discussion, 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 according to relative acidic strength or basic strength. In other words, can you rank them starting from the strongest acid to the weakest base or can you rank them starting from the strongest base to, on the other side, the strongest acid, okay? And when you deal with salts, you've got to separate them out and 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, you look at to see whether it's a strong base or a weak base, alright? 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, I'm going to quiz you on something, alright? Remember, so let's say we have a strong acid, which is HCl, alright? And the initial concentration of HCl is 1 times 10 to the negative 2 molar. Second, if this, since this is a strong acid, what is the pH of the solution? PAH is 2, alright? Because you don't need a calculator, alright? You know that the pH is 2, alright? So the initial concentration of a strong acid like HCl is 10 to the negative 2, you know that the pH is 2, alright? Now, if my initial concentration of acidic acid is 1 times 10 to the negative 9 molar, what would the pH of the solution come out to be? 9? Now, tell me, if pH is 9, is that solution acidic, basic or neutral? Basic. Basic. So a strong acid and end up with a solution that is basic? No. No. So there's a problem here. And the reason is that you're in water, alright? Up to this point, every example that we looked at are concentrations of acidic acids were so large that we ignored water. 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? 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, alright? And so what we focus on is just the 10 to the negative 2 because the difference between 10 to the negative 2 and the 10 to the negative 7 is an enormous difference so large that you can ignore it, alright? 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? 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? 7. Alright, it's going to be a neutral, alright? Because now the 10 to the negative 7 is 100 times greater than 10 to the negative 9, alright? And therefore this solution would have a pH of 7, alright? Now if the difference between 10 to the negative 7 and the acid that you add is large, then you take whichever is large. If they're in the approximately similar range, and that is between something like 10 fold within, so if you have an acid that's 10 to the negative 7, pure water that's 10 to the negative 7, both that's 10 to the negative 7, pH is going to be 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 10 fold difference, 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, alright? And so that means now when you want to figure out what the pH of that solution is, it's going to be a little bit more complex, alright? And at this level of class, we're not going to look at calculations where you have 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, alright? So whenever you look at it, see a question of this type, make sure that you go back and ask yourself, is my pH consistent with whether that solution is acidic or basic, alright? And so this is examples where your acid concentration is so dilute that now water has to play an important role, okay? Up to this point, every example that we took, the concentrations were acids were of the order of 1 molar, 0.1 molar, or 0.01 molar. So if you're dealing at that concentration range, you can ignore the contribution that comes from hydronium ion, alright? Okay, so we have kind of completed looking at chapter 11 of your textbook where we looked at acid-base chemistry, 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, okay? So application of Le Chatelier's principle to acid-base equilibria, and we call this the common ion effect, alright? 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 to figure out how a system responds if you disturb a system that's an equilibrium. So if you take a system that's an equilibrium and disturb it in some way, we know that ultramarine ultimately, that system is going to go back to equilibrium. And so we can make predictions about how it will respond in order to go back to equilibrium, okay? 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, alright? So this is a weak acid. 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 plus and F minus, okay? So this is the equilibrium. So let's say 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, alright? 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 anacetion. And if this is a soluble salt, what will happen? So instead I add that salt to that solution, that equilibrium, what is going to happen? It's going to dissociate, alright? So this will dissociate to give me Na plus aqueous plus F minus aqueous. 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, it will react in the direction in which it'll counteract that stress. So what was the stress or disturbance? I added extra fluoride minus. So what is my stress? Too much fluoride. So the system will respond to counteract that stress, so that means it has to reduce the amount of fluoride, okay? 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? 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 will respond by reacting in the reverse direction until it goes back to equilibrium, all right? And so this shift in equilibrium position, so this shift in equilibrium that occurs because of the addition of an ion already involved in the equilibrium reaction is called the common ion 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 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 of a base. A weak base is ammonia. So if I take NH3 equals in water, by now you should know that NH3 is a base. So if I put NH3 in water, 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 plus and OH minus, okay? Great. So this would be NH4 plus equals and OH minus, okay? And because we're writing this in the form of a base, we're looking at KB and KB equals 1.8 times 10 to the negative 5. So once again we know that this is a weak base. 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 equals plus Cl minus equals. So we know that when you put a salt, it dissociates completely to produce cations and anions. And once again you can see that because I'm adding a common ion, NH4 plus to a system that's already established equilibrium, I'm going to disturb that equilibrium. And what is my stress? Too much NH4 plus. And if the NH4 plus has gone up, that means the system is going to respond by proceeding in the reverse direction. So the reaction will shift towards reactants. And the net result is, so if you look at this, the net result is that hydroxide ion concentration decreases and therefore if hydroxide ion concentration decreases, what will happen to the pH of that solution? 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, in the previous example if the reaction proceeds this way, what happens to hydronium ion concentration? Because the equilibrium shifts towards reactants, you can see that the hydronium ion concentration goes down. So in this example, the net result is that hydronium ion concentration decreases. And if hydronium ion decreases, we know pH increases. So in each instance, not only do you have the equilibrium disturbed and you have this common ion effect, 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. 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. So not only does this common ion effect disturb that equilibrium, but it also will affect the hydronium ion or hydroxide ion concentration. So now we want to calculate that. We want to calculate and figure out how big is this effect. 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. So to do that, remember all of these problems on the class website, so you don't need to copy it. 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 and 0.1 molar sodium fluoride. So you have HF containing one molar of HF and you're adding one molar concentration of sodium fluoride. So we're adding sodium fluoride, so all of you recognize that because we're 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. 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 plus F minus and we are told that Ka is 7.2 times 10 to the negative 4. All right, and this is the fluoride ion that's there. So if we wanted to look at initial concentrations 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 a fluoride ion concentration is different because we've added fluoride ions from this. So can you see that the source to begin with, there's fluoride ion concentration and therefore that would be one. So this is the only difference between the previous problems where we just look at HF in water 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 minus, so before anything happens, before equilibrium is established, we have one molar of the weak acid and we have one molar of the F minus ion. So now solving this problem is going to take the same approach that we've done before. 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. Therefore this is going to be negative X, this would be plus X, this would be plus X, therefore at equilibrium I would have one minus X, X, this would be one plus X. So can everybody see, the only difference here is now we have F minus to begin with and that concentration of, if the reaction is proceeding in the forward direction, then the concentration of F minus is going to increase. And so now we can solve this like any other equilibrium problem. We can say Ka equals 7.2 times 10 to negative 4 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, 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. 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. 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 percent or 0.072 percent. All right and therefore we know that our approximation is pretty good. It's well below our 5 percent threshold. So now the question was calculate the pH 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. So the pH would be the negative log of that and if you work out the pH, pH comes out to be 3.14. And then percent dissociation would be 7.2 times 10 to the negative 4 that's X divided by the initial concentration times 100 which is 7.2 times 10 to the negative 2 percent or as I said before 0.072 percent. All right so we calculated the pH of the solution if you have a common ion involved in that. Okay 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 compared to the absence of an ath. 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. This is sort of 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 and I know Ka is 7.2 times 10 to the negative 4. So now I don't have any common ion in there. So my initial concentration would be just the one molar of HF, not considered about water since I don't have any added salt. 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 one minus X, X, X. Alright, so if I were to calculate the value of X we can say Ka equals 7.2 times 10 to the negative 4 which is the hydronium ion concentration times the fluoride ion concentration divided by the concentration of HF which gives me X squared over one minus X. Once again I can safely make that approximation so I can say my approximation in this problem would be one minus X will approximate to one so I can go back and I can say 7.2 times 10 to the negative 4 equals X squared over one and therefore if you work that out it comes out to be 2.7 so X equals 2.7 times 10 to the negative 2 molar. And if you check validity 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 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. Alright and if I want to calculate what the percent dissociation is 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 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 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. So you can see that this has without the salt you have a higher hydronium ion concentration 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 0.072%. So adding the salt that is what is called a 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 so we're going to stop there for today and next class we'll start looking at buffers.
wrap themselves around the anion and this is the aqueous plus NH3 aqueous and this form of the principle to acid-base equilibria, and we call quarter, we talked about Le Chatelier's